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AQA
A-level
Physics
Nick England
Carol Davenport
Jeremy Pollard
Nicky Thomas
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© Nick England, Carol Davenport, Jeremy Pollard, Nicky Thomas 2019
First published in 2019 by
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www.hoddereducation.co.uk
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Cover photo © pixbox77 - stock.adobe.com
Illustrations by Aptara
Typeset in 11/13 pt ITC Berkeley Oldstyle by Aptara, Inc.
Printed by Replika Press Pvt. Ltd. in Haryana, India
A catalogue record for this title is available from the British Library.
ISBN: 978 1 5104 6988 4
469884_FM_AQA_A-Level_Physics_i-vi.indd 2
17/04/19 8:54 AM
469884_FM_AQA_A-Level_Physics_i-vi.indd 3
Get the most from this book
v
1
Particles and nuclides
1
2
Fundamental particles
18
3
Electrons and energy levels
42
4
Particles of light
57
5
Waves
70
6
Combining waves
95
7
Introduction to mechanics
116
8
Motion and its measurement
133
9
Newton’s laws of motion
151
10 Work, energy and power
168
11 Momentum
183
12 Properties of materials
200
13 Current electricity
221
14 Electrical circuits
244
15 Circular motion
274
16 Simple harmonic motion
285
17 Gravitation
308
18 Thermal physics
329
19 Electric fields
358
20 Capacitance
381
21 Magnetic fields
403
22 Magnetic flux
420
23 Alternating currents and transformers
435
24 The evidence for the nucleus
449
25 Radioactive decay
468
26 Nuclear energy
485
Contents
Contents
17/04/19 8:54 AM
27 Optional topic: Astrophysics
501
28 Maths in physics
546
29 Developing practical skills in physics
571
30 Preparing for written assessments
584
Index
593
Acknowledgements
600

Go online to www.hoddereducation.co.uk/AQAPhysics
for answers.
iv
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Welcome to the AQA A-level Physics Student’s Book. This book covers all
content for the AQA A-level Physics specification.
The following features have been included to help you get the most from
this book.
Prior knowledge
7
TEST YOURSELF ON PRIOR KNOWLEDGE
Introduction to
mechanics
Scalar and vector quantities
This is a short list of topics that
you should be familiar with before
starting a chapter. The questions
will help to test your understanding.
1 Explain why speed is a scalar quantity and velocity is a vector quantity.
2 Make an estimate of the size of the following forces:
a) The weight of an apple.
b) The push on a pedal when you cycle.
c) The drag on a car moving at 20 m s–1.
3 A parachutist falls at a constant speed of 5 m s–1. Which of the
following statements is true?
a) The pull of gravity on the parachutist is greater than the drag.
b) The pull of gravity on the parachutist is less than the drag.
c) The pull of gravity on the parachutist is the same size as the drag.
4 a) Explain why it is easier to undo a nut using a spanner with a long
handle.
b) Calculate the turning moment when a force of 30 N is applied at
right angles to the handle of a spanner with a length of 0.25 m.
PRIOR KNOWLEDGE
A scalar quantity has only size, for example a mass of 2 kg.
A vector quantity has both size and direction, for example a 5N force
acting downwards.
A force is a push or a pull. Forces are measured in newtons, N.
The size of a turning effect is called a turning moment or torque.
Turning moment = force applied × perpendicular distance from
the pivot.
● Newton’s first law of motion states that when no force acts, or
balanced forces act on an object, that object will remain at rest or
continue to move in a straight line at a constant speed.
●
●
●
●
●
Get the most from this book
Get the most from this book
●● Scalar and vector quantities
A scalar quantity is one which has size only.
A vector quantity is one which has size and
direction.
A resultant vector is the vector that results
from adding two or more vectors in a vector
sum (e.g. resultant force or resultant
displacement).
A scalar quantity is one that only has size. Scalar quantities include:
mass, temperature, energy, distance, speed and time. We say that ‘room
temperature is 19°C’; it makes no sense to say ‘19°C westwards’.
A vector quantity is one that has magnitude (size) and direction. Vector
quantities include: force, acceleration, velocity and displacement. Velocity
and speed seem like similar quantities, as they both can be measured in
m s–1. But, if you write that a car is travelling at 30 m s–1, you are talking
about a speed; when you write that a car is travelling at 30 m s–1 due
east, you have used a vector quantity, which is called velocity. In a similar
way, the word displacement is used to describe a distance travelled in a
particular direction.
You have learnt to add or subtract simple vectors. In Figure 7.3 a two forces
acting in the same direction add up to make a larger resultant force of
200 N, and in Figure 7.3 b two forces acting in opposite directions make a
smaller resultant force of 300 N.
Figure 7.1 Formula 1 racing cars
moving at speed around a corner.
Figure 7.2 A modern bridge with central
pillar chords.
100 N
800 N
100 N
Engineers who build bridges, or who design Formula 1 cars, need a
profound understanding of mechanics. Engineers use the laws of physics,
and their knowledge of the strength and flexibility of materials, to calculate
the size and shape of material required at each stage of design and
construction. In this chapter you will meet ideas about the vector nature of
forces, and also the concept of turning moments. Both these ideas are used
by design and construction engineers.
116
500 N
117
800 N
500 N
100 N
100 N
200 N
300 N
Figure 7.3 (a) two forces acting in the same direction and (b) two forces acting in
opposite directions.
26/03/19 8:59 PM 469884_07_AQA_A-Level_Physics_116-132.indd 117
469884_07_AQA_A-Level_Physics_116-132.indd 116
26/03/19 8:59 PM
These practical-based activities
will help consolidate your learning
and test your practical skills. AQA’s
required practicals are clearly
highlighted.
Mica
window
e
Cathod
Atom
500 V
detector
inside the charged
inert gas
e of
-pressure ducing a cascad odes. The small
). The low
d pro
the
d electr
radiation on) is ionise
ly charge is detected by e a
site
arg
(gamma
po
or
to op
neon
rs hav
charge
attracted by the moving
r counte in that they
(helium,
Ionising
that are
t’. Geige
n
radiatio
particles
produced registers a ‘coun cloud chambers
current
pulse of counter, which rk counters and
ic
spa
r
on
ove
on.
electr
advantage types of radiati
Electron
ee
distinct
ect all thr
Ion
can det
R
Counter
Anode
Figure
1.12 A Gei
ger-Mulle
counte
r (GM)
r.
0
muth-21 of beta par ticles
m of bis
.
energy
spectru
wn below
kinetic
ACTIVITY rticle emission nt measuring the rements are sho
asu
ta pa
experime th-210. Her me
The be
d out an
pe bismu
t carrie
A studen the radioisoto
by
emitted
176 186
144 160
112 128
96
80
64
48
0.2 0.0
32
of
16
1.5 0.6
0
energy
2.9
etic
Kin
ticles,
6.0 4.4
beta par
10.0 8.0
−15 J)
E (× 10
12.0 11.5
(x-axis).
a
rgy
11.5
bet
0
10.
ty of
ticle ene
th-210
Intensi (arbitr ary
t beta par second bismu
le
a
agains
possib
particles
d with
ine the
(y-axis)
units)
ng emitte
intensity
to determ
Table 1.2
e
−15 J).
graph
ticles bei
par ticl
rce.
(186 × 10
of beta
beta par units. Use the second sou
is 186 fJ
a graph
ry
asures
the
th-210
1 Plot
arbitra
dent me
es from
m bismu ity.
ity of 11
ond stu
par ticl
itted fro
ens
2 A sec with an intens emitted beta
ticles em d at this int
the
par
a
of
source
itte
bet
of
energies
inos em
energy
kinetic
kinetic
antineutr
ximum
es of the
3 The ma e the energi
Calculat
e
that tak
reactions
nuclear
cribe the
d to des
servation
can be use ay.
ple con
A X notation
two sim
dec
e
ay,
e
ctiv
dec
Z
Th
ioa
ctive
ring rad
le (α)
ma radioa
place du
ha partic –
a and gam
4
He – alp
(β )
alpha, bet
2
particle
0
During e into play:
0 – beta
served
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rictly 0γ)
ay
ays con
–1
rules com
a ray (st 0
the dec
ved.
A, is alw
γ – gamm
A before
number, is always conser
rictly 0ν) 0
on
(st
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value for (and, similarly,
ino
● nu
er, Z,
dition)
ν – neutr
rictly 0ν)
decay
at the
ton numb
total (ad
utrino (st
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that the ue of A after the decay). Look
for
ν – antine
s means
AX notation
r the
) val
thi
afte
e
ion
of
dit
ctic
Z
and
ore
al (ad
1.18 The (Note: values
In pra
ks,
Figure
same bef
on
to the tot
t the
nite roc is
g radiati
withou
is equal for Z remain the
ionisin
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written
ium-22 radon-222. Th
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the totals two examples:
ay of rad
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ha decay e and the rad
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producin ay can be wr
4 = 226
stable and
r 222 +
dec
ber afte
t are un
4 e
TIP
nuclear
ission
clei tha
leon num
222 Rn + 2H
Total nuc
Some nu gamma ray em after
226 Ra → 86
ns
uatio
●● Nuclear eq
w
URSELF
right sho
TEST YO
es to the
es,
’ added
decay via
letter ‘m
meaning
have the on number,
mple
cle
their nu
good exa , a
ble’. A
99m
‘metasta
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is techn ray emitter
s
of this
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ES AND
88
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ber bef
leon num
Total nuc
226 Ra
88
Figure
ore = 88
ber bef
ton num
Total pro
cay
1.19 De
9
+
4 He
222 Rn + 2
86
2 = 88
r 86 +
+
ber afte
ton num
Total pro
of radium
.
v
y, counts
intensit
These short questions, found
throughout each chapter, are useful
for checking your understanding as
you progress through a topic.
NUCLIDE
S
e pictur alpha par ticl
10 Th
inside
hs of
X-r ays
the pat
ticles and
beta par
er:
chamb
mber
a cloud
ud
cloud cha
1.14 Clo
a) Which e(s) show the
Figure Picture B.
er
ud
pictur
chamb
1.13 Clo
?
Figure Picture A.
following par ticle tracks
er
chamb
i) alpha
ay track
ck
tra
ii) an X-r
e
a par ticl
er
iii) a bet y cloud chamb of
types
lain wh
b) Exp e C show two
pictur
ture D.
r
mber Pic
es?
ud cha
par ticl
rk counte es
1.16 Clo
why a spaalpha par ticl
Figure
lain
a
ture C.
11 Exp
ily detect
ect bet
mber Pic
can eas y cannot det
ud cha
s.
all
1.15 Clo
Figure
but usu or gamma ray
es
used to
par ticl
nter is alpha
iger cou
of
5000
12 A Ge the intensity d from
itte
detect
being em
ing
par ticles source contain es,
4000
top
a radium emitting iso
6.
-22
haradium
two alp
3000
-223 and ty (counts per
radium
of intensi ge in air
ran
A graph
t
in
) agains rce is shown
2000
minute
this sou
(cm) for .
1.17
Figure
7
1000
its alpha
6
rgy
223 em
5
Radium- with higher ene graph
4
3
the
es
e
ticl
)
Us
2
0
par
6.
1
in air (cm
ium-22
of alpha
8
range
in air.
0
than rad ine the range each
range
by
ty ver sus
to determ in air emitted
.
of intensi
ing
ph
son
Gra
1.17
your rea
par ticles
Figure
Explain
isotope.
ICL
1 PART
Test yourself questions
ns
equatio
Nuclear
Activities and Required
practicals
26/03/19
d 8
1-017.ind
hysics_00
-Level_P
1_AQA_A
469884_0
7:22 PM
d 9
1-017.ind
hysics_00
-Level_P
1_AQA_A
26/03/19
7:22 PM
469884_0
Tips
These highlight important facts,
common misconceptions and signpost
you towards other relevant topics.
469884_FM_AQA_A-Level_Physics_i-vi.indd 5
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Practice questions
Practice qu
estions
L PARTICLE
2 FUNDAM
ENTA
atomic parti
Σ +, sigma +
cle Quar
k struc ture
Baryon OR
ud
Meson
meson
udd
Relative charg
e
Baryon numb
0
a) Copy and
comp
er Stran
gene
lete the table
+1
.
b) All of the
sub-a
(3)
corresponding tomic particles show
n in the table
have a
its antibaryon antiparticle. State one
e
example of
not shown
a bary
structures.
in the table
and state their on and
quark
c) The elect
ron and
(4)
particle–antipa the positron are an
exam
rticle pair.
the same as
State one prop ple of a lepton
an electron,
and one prop erty of positron that
4 Figure 2.21
is
erty that is
shows two
different.
particles inter
been drawn
(2)
acting. An
to represent
offset line has
the exchange
a) State the
particle.
name of the
inter
of the exch
ange particle action shown and give
draw
the
name
n on the diagr
b) In this inter
am.
actio
(2)
Give the name n, momentum and
e
energ
of another
quantity that y are conserved.
Figure 2.21
is conserved
Feynm
.
(1) showing two partic an diagr am
38
les interacting
469884_02_AQ
Figure 2.22
show
has been draw s another type of inter
actio
n to represent
the exchange n. Another offset line
c) Name the
particle.
A
particles label
led A, B and
d) State the
C.
B
name of this
type of parti
(3)
e) Momentum
cle interactio
n.
and energy
(1)
are
interaction.
Name two otheralso conserved in this
particle
for this inter
quan
tities
action to occu
that must be
conserved
r and show
f) The Stand
how they are
ard Model
conserved.
predicted the
C
(4)
particle invol
existence of
ved
the
observed expe in this interaction
before it had exchange
rimentally.
been
the prediction
Expl
of a scientific ain why it is important
model expe
to test
5 The K − kaon
rimentally.
is a meson
with strangene
muon, μ −,
(3)
and a muo
p
e
n-antineutrino ss −1. It can decay
equation below
into a
, vμ , as show
:
Figure 2.22
n in the
Another type
K− → μ− +
of interaction
vμ
.
a) State and
explain whic
h particle inter
this decay.
action is respo
nsibl
e for
b) Energy
and momentu
m are cons
other quan
(1)
erved in this
tities. State
deca
the name of
c) State one
these two quan y, as are two
prop
tities.
makes it differ erty of a strange parti
(2)
cle, such as
ent to a nonthe K −
strange parti
The K + kaon
cle, such as −kaon, that
is
a π pion.
following deca also strange. It coul
(1)
d
deca
y
equa
y
via either of
tions, how
is not poss
ever one of
the
ible:
these deca
y equations
K+ → μ+ +
vμ
K+ → π0 + +
π
d) State and
explain whic
h of these deca
The K − kaon
y equations
can
is not poss
muon-neutrin also decay in the follo
ible. (2)
o and a third
wing way, prod
particle, X:
ucing a muo
K− → X + μ−
n, a
+ vμ
e) State whic
h interactio
n is responsibl
f) Particle
e for this deca
X is identified
y.
as a pion.
a meson.
(1)
Explain why
X must be
g) State the
charge on this
particle.
(2)
6 During β −
deca
antineutrino. y a neutron decays into
(1)
a proton and
an electron
a) Write an
39
equation descr
ibing this deca
b) Using cons
y.
ervation laws,
is produced
explain why
(1)
rather than
an anti-elect
an electron-n
ron-neutrino
c) Draw a
eutrino.
Feynman diagr
am for this
(2)
decay.
ss
0
-1
e
e
.
A_A-Level_Phy
sics_018-041.in
dd 38-39
tions
S
1 The grou
p of particles
know
are two sub-g
roups of hadr n as hadrons are comp
osed
ons: mesons
a) What is
and baryons. of quarks. There
the name of
the property
b) State the
that defines
quark struc
a hadron?
ture of a meso
(1)
c) State the
n.
quark struc
ture of a bary
(1)
d) The antip
on.
roton
that the proto is the antiparticle of
(1)
the
n
they are differ and the antiproton are proton. State one way
similar, and
ent.
one way that
e) What are
the following
values of an
(2)
antiproton:
i) its charg
e
ii) its bary
on number
(1)
iii) its quar
k structure?
(1)
2 In the Stand
ard Model
of Matter, parti
exchange parti
(1)
cles
cles. Here is
a list of parti can be leptons, hadrons
electron
cles:
or
proton
neutron
a) From the
muon
list state the
name of one
pion
b) State one
lepton and
difference betw
one hadron.
een leptons
(2)
c) State the
and hadrons.
difference in
structure betw
(1)
d) From the
een baryons
list state the
and mesons.
name of one
(1)
3 The table
baryon and
below show
one meson.
s some basic
(2)
information
Table 2.9
about three
hadrons.
Sub-
Practice ques
You will find Practice questions
at the end of every chapter. These
follow the style of the different types
of questions you might see in your
examination, including multiplechoice questions, and are colour
coded to highlight the level of
difficulty. Test your understanding
even further with Stretch and
challenge questions.
(3)
06/03/19
9:25 PM
Questions are colour-coded, to help target your practice:
Green – Basic questions that everyone should be able to answer
without difficulty.
Orange – Questions that are a regular feature of exams and that all
competent candidates should be able to handle.
Purple – More demanding questions which the best candidates should
be able to do.
Stretch and challenge – Questions for the most able candidates to test
their full understanding and sometimes their ability to use ideas in a
novel situation.
Key terms and formulae
0.2 m
–10–8 C
H
E
F
The electric field at a distance r from the centre of an isolated charged
sphere carrying a charge +Q is
K
E=
I
Y
X
E=−
O
A
B
C
D
Z
V =−
=
b) i) Calculate the field strength, at Y, if the charge of −10 −8 C on sphere Z is
replaced with a charge of +10 −8 C.
ii) Calculate the force between the spheres in this case. Is the force
attractive or repulsive?
●● Electrical potential
+Q
Figure 19.21
96
r
Q
0
∫ 4πε r
2
dr
Q
4πε 0r
Absolute electric potential The potential
difference between a point and a point at
zero potential, which is infinitely far away.
The formula V =
Q
4πε0r
helps us to define absolute electric potential. The
absolute electric potential at a point r from a charge +Q is the work done
per unit positive charge in moving it from ∞ to that point. Note that if that
charge is −Q, then the potential is negative and the electric field does work
in moving a positive charge closer to the point r.
The potential at a distance r from the centre of a sphere carrying a charge Q
is given by
EXAMPLE
Electric field near a charged sphere
Q
4πε 0r
Note that if the charge is negative, then the potential near to the charged
sphere is also negative.
Figure 19.22 shows a metal sphere of radius 10 cm,
charged to a potential of 1000 V. The electric field
strength at C is 10 000 V m−1.
These formulae are very similar to the formula for gravitational potential –
except that gravitational potential must always be negative, whereas an
electrical potential, close to a charge, can be positive or negative. In the
case of gravitational potential, the zero point of potential is defined as a
point infinitely away from any planet or star. In a similar way, the zero
point of electrical potential is defined as a point infinitely far away from
the charged sphere. In practice, the surface of the Earth is our reference
point of zero potential, and by suspending our sphere a long way from
anything else, the Earth can be treated as being infinitely far away. The
Maths box shows how the formula for potential can be derived from the
formula for electric field.
r
∞
Note that the limits of the integration set the potential as zero at an
infinite distance.
Figure 19.21 shows a small charged isolated sphere. In theory, an isolated
sphere should be in contact with nothing and infinitely far away from
anything else. In practice, in a laboratory, the best we can do is to suspend a
sphere by a fine, insulating thread and make sure the sphere is a few metres
from everything else.
V=
dV
dr
so
Figure 19.20
insulating thread
supporting sphere
Q
4πε 0r 2
But
G
Figure 19.19
5 ELECTRIC FIELDS

vi
0.3 m
+5 × 10–9 C
These provide additional material
for the more mathematical
physicists.
Examples of questions and
calculations feature full workings
and sample answers.
J
Calculate the magnitude of the electric field strength for each of the
points B to K. (You will need to use a combination of Pythagoras’s
theorem and the inverse square law.)
15 a) Figure 19.20 shows two charged spheres, X and Z. Calculate the
electric field strength at point Y, which lies along the line XZ joining
the centres of the two charged spheres. Sphere X has a charge of
+5 × 10 −9 C, and sphere Z a charge of −10 −8 C.
Maths boxes
Examples
MATHS BOX
14 Figure 19.19 shows a grid marked with squares. A positive charge
placed at O produces an electric field of strength 3600 N C −1 at point A.
Electrical potential
These are highlighted in the text
and definitions are given in the
margin to help you pick out and
learn these important concepts.
V = 1000 V
Sketch graphs to show how
A
B
C
–50 cm
–10 cm
0
D
10 cm
50 cm
Figure 19.22
1 the potential
2 the field strength
vary along the line A to B and then from
C to D.
V/V E/V m–1
1000
10000
Answer
Figure 19.23 shows the answer. These are
the points to note.
● V is a scalar and is always positive.
1
● V obeys a
law and falls from 1000 V
r
500
–50
–40
●
dr
5000
–10
–500
from the sphere.
E is a vector, so must change direction.
E is connected to the potential by the
equation E = − d V . On the right-hand
–20
10
97
20
30
40
50
r/cm
at 10 cm to 200 V at a distance of 50 cm
●
–30
–1000
–5000
–10000
E/V m–1
Figure 19.23
AQA has provided five optional topics as part of the full A-level course so students can focus on their areas of
interest: Astrophysics, Medical physics, Engineering physics, Turning points in physics, and Electronics. A chapter covering
the first optional topic, Astrophysics, has been included in this book (Chapter 27).
Dedicated chapters for developing your Maths and Practical skills and Preparing for your exam can be found at
the back of this book.
469884_FM_AQA_A-Level_Physics_i-vi.indd 6
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1
Particles and nuclides
PRIOR KNOWLEDGE
●
●
●
●
●
●
Matter is made of atoms. Atoms are made up of a very small central
nucleus containing particles called protons and neutrons, surrounded
by orbiting electrons. The number of protons contained in the nucleus
of an atom is called the proton or atomic number and the total number
of protons and neutrons is called the nucleon or mass number.
Protons are positively charged particles with a relative charge of +1;
electrons are negatively charged particles with a relative charge of −1;
neutrons are electrically neutral with a relative charge of zero.
Atoms are electrically neutral overall, which means that they have
the same number of protons and electrons. When atoms lose or
gain electrons they become ions. An excess of electrons produces a
negatively charged ion whereas a shortage of electrons produces a
positively charged ion.
The diameter of an atomic nucleus is of the order of 10 000 times
smaller than the diameter of an atom, but contains the vast majority
of the atomic mass. Protons and neutrons have a relative mass of 1,
with electrons about 1800 times less massive.
Elements are made up of atoms with the same proton number. Atoms
can have the same number of protons but different numbers of
neutrons: these are called isotopes.
The relative atomic mass of an element compares the mass of atoms
of the element with the carbon-12 isotope.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 Naturally occurring hydrogen has three isotopes: (‘normal’) hydrogen,
deuterium and tritium. What three properties do atoms of each of
these isotopes have in common? How do the isotopes differ?
2 What is the relative charge of a copper atom that has lost two
electrons?
3 Chlorine has two naturally occurring isotopes that exist in an almost
75 : 25 abundance ratio. The average relative atomic mass of chlorine
is 35.5. What are the relative atomic masses of these two isotopes?
Find out their proton and nucleon numbers.
4 As at spring 2019, the element with the highest atomic number so far
discovered is ognasson (Og, atomic number 118), first synthesised in
2002, of which only five or six atoms have been observed, each with a
relative atomic mass of 294. Determine its:
a) proton number
b) neutron number
c) electron number
d) proton : electron mass ratio.
469884_01_AQA_A-Level_Physics_001-017.indd 1
1
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Figure 1.1 Peter Higgs and the Higgs
Boson Theory.
Figure 1.2 A more recent image taken
by a transmission electron microscope
showing individual gold atoms each
separated from its neighbour by a
distance of 2.3 nm (2.3 x 10 –9 m).
On 8th October 2013 the Nobel Prize in Physics was awarded jointly to
François Englert and Peter Higgs for the theoretical discovery of the particle
known as the Higgs Boson. (This followed its experimental discovery by
the Large Hadron Collider team at CERN on 4th July 2012). This discovery
completes our current best model of what the Universe is built from, called
the Standard Model, a model that started development nearly two and a half
thousand years ago by a Greek philosopher called Democritus.
It is said that Democritus observed that the sand on a beach was once part
of the rocks of the cliffs and he questioned whether the sand could be cut
into ever smaller pieces by a succession of sharper and smaller knives. It
was Democritus who first coined the word ‘atom’, meaning the smallest
indivisible piece that the sand could be cut up into. Nearly 2500 years later,
JJ Thomson was able to extend Democritus’ thought experiment when he
discovered the electron, splitting up atoms and discovering the first subatomic particle.
In the early 1980s, researchers at IBM in Zurich produced a machine
called an atomic force microscope that was able to image individual atoms.
Figure 1.2 shows a more recent image taken by a transmission electron
microscope showing individual gold atoms. Each individual gold atom is
separated from its neighbour by a distance of 2.3 nm (2.3 × 10−9 m).
●● What are the building blocks
1 PARTICLES AND NUCLIDES
of the Universe?
2
What is the material of the Universe made of? Like so many things in
physics, it depends on which model you are using to explain things. Physics
is a series of evermore complex, layered models designed to explain how
the Universe behaves as we see it now. Over the years, as our observations
of the Universe – particularly on the sub-atomic and cosmological scales –
have become more and more detailed, so the physical models have had to
adapt to the new observations and measurements. It is rare now for a theory
to be developed as the result of a ‘thought’ experiment such as Democritus’.
Models in physics
In science, and in physics in particular, we rely on models to explain how
the Universe around us works. Models make complex, often invisible
things or processes easier to visualise. Models take many forms. Some,
like the models of atoms and nuclei used in this chapter, are visual models
used as analogies of the real thing. The Rutherford-Bohr model used in the
next section is a good example of this type of model, where the nucleus is
modelled as a small ball containing neutrons and protons, with electrons
whizzing round the outside. Atoms don’t actually ‘look’ like this, but it is a
convenient, easy first model to use because it uses the analogy of everyday
objects that we are very familiar with.
Other types of model in physics are more mathematically based. A good
example of this is the kinetic theory of gases, where simple physical rules
expressed mathematically are applied to a model of gas molecules behaving
as hard, bouncy elastic balls. These rules allow the model to predict the
behaviour of real gases.
469884_01_AQA_A-Level_Physics_001-017.indd 2
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Shell
Positive
charged
nucleus
Figure 1.3 The Rutherford-Bohr model
of the atom.
Some models are very good analogies and others not so. Models and
analogies have their limits. The best physical models are those that are
adaptable to take in new experimental discoveries. Good physicists always
state and explain the limitations of the models that they are using.
JJ Thomson was the first person to create a model of the structure of the
atom, which he called the plum pudding model. Thomson produced this
model just after he had discovered the electron in 1897, and it was named
after a popular steamed dessert. His model consisted of a sphere of positive
charge with the electrons embedded throughout the positive charge, rather
like the small pieces of plum (or currants) inside the steamed pudding.
The electrons were allowed to move through the structure in ringed orbits.
Thomson’s model, however, was short-lived. The nucleus of the atom was
discovered by Rutherford, Geiger and Marsden in 1909.
The Rutherford-Bohr atom
The Rutherford-Bohr model of the atom consists of a
tiny, central, positively charged nucleus, containing
protons and neutrons, surrounded by orbiting negatively
charged electrons (analogous to the Solar System). This
model is particularly useful for chemists as it can be used
to explain much of the chemical behaviour of atoms. The
Rutherford-Bohr model is a useful model for visualising the atom, and is
used a great deal by the popular media to describe atoms.
What are the building blocks of the Universe?
Electrons
Rutherford scattering
Figure 1.4 If the atom was the size of
the Wembley Stadium complex, then the
nucleus would be the size of a pea on
the centre spot.
Experiments carried out by Ernest Rutherford and his research assistants
Hans Geiger and Ernest Marsden in 1911 proved that the nuclear radius
is about 5000 times smaller than the atomic radius. Geiger and Marsden
performed alpha-particle scattering experiments on thin films of gold,
which has an atomic radius of 134 pm (picometres) – where 1 pm = 1 ×
10−12 m. They found that the radius of the nucleus of gold is about 27 fm
(femtometres or fermi) – where 1 fm = 1 × 10−15 m. However, modern
measurements of the radius of a gold nucleus produce a value of about 5 fm.
So a more accurate ratio of the radius of a gold atom to the radius of a gold
nucleus is about 25 000.
As a way of visualising the scale of the atom and the nucleus, if the atom
had a radius equal to that of the Wembley Stadium complex, then the
nucleus would be about the size of a small pea on the centre spot.
Charges, masses and specific charges
The charges and masses of the proton, neutron and electron are shown in
Table 1.1 below:
3
Table 1.1
Sub-atomic particle
Proton
Neutron
Electron
469884_01_AQA_A-Level_Physics_001-017.indd 3
Charge/C
10 −19
+1.60 ×
0
−1.60 × 10 −19
Relative charge
Mass/kg
+1
0
−1
1.673 × 10 −27
1.675 × 10 −27
9.11 × 10 −31
17/04/19 8:57 AM
TIP
Next year you will meet the
concept of binding energy,
where some of the mass of the
particles involved with forming
nuclei is transferred into nuclear
energy holding the nuclei
together. So the calculations in
the examples are approximations
based on the separate masses of
the protons and neutrons.
The specific charge of a particle is defined as the charge per unit mass, and
its units are C kg−1. Specific charge is calculated using the formula:
charge of the particle Q
=
specific charge of a particle =
mass of the particle m
So the specific charge of the proton is calculated by:
specific charge of a proton =
+1.60 × 10−19
1.673 × 10 −27
= 9.56 × 107 C kg−1
The specific charges of the neutron and the electron are 0 and
−1.76 × 1011 C kg−1 respectively.
EXAMPLE
a) Calculate the specific charge of a beryllium-9
nucleus, containing 5 neutrons and 4 protons.
b) Calculate the specific charge of a lithium-7 (+1)
ion, containing 4 neutrons, 3 protons and
2 electrons.
1 PARTICLES AND NUCLIDES
Answer
4
a) The total charge, Q, of the nucleus is
4 × (+1.60 × 10−19)
= +6.40 × 10−19 C
The total mass, m, of the nucleus is (mass of
protons) + (mass of neutrons)
m = (4 × 1.673 × 10−27) + (5 × 1.675 × 10−27)
= 1.507 × 10−26 kg
The specific charge of the nucleus is:
specific charge = Q
m
+6.40 × 10−19
=
1.507 × 10−26
= +4.25 ×107 C kg−1
Nucleon is the word used to describe
protons or neutrons. These are particles that
exist in the nucleus.
The proton number of a nucleus is the
number of protons in the nucleus.
The nucleon number of a nucleus is the
number of protons plus the number of
neutrons.
nucleon
number
A
Z
proton
number
b) The relative total charge, Q, of the ion is +1
(3 protons and 2 electrons) so the charge is
1 × (+1.60 × 10−19)
= +1.60 × 10−19 C
The total mass, m, of the ion is (mass of protons)
+ (mass of neutrons) + (mass of electrons)
m = (3 × 1.673 × 10−27) + (4 × 1.675 × 10−27)
+ (2 × 9.11 × 10−31)
= 1.17 × 10−26 kg
The specific charge of the ion is:
−19
specific charge = Q = +1.60 × 10−26
m 1.17 × 10
= +1.37 × 107 C kg−1
Describing nuclei and isotopes
The number of protons in any given nucleus is called the proton number, Z.
The total number of protons and neutrons in the nucleus is called the
nucleon number, A. (Protons and neutrons are both found in the nucleus
and are therefore called nucleons). These two numbers completely describe
any nucleus, as the number of neutrons in a nucleus can be calculated by
subtracting the proton number from the nucleon number. The two nuclear
numbers and the chemical symbol are used together as a shorthand way of
describing any nucleus. This is called the AZ X notation.
The nucleon number, A, is always written as a superscripted prefix to
the chemical symbol, and the proton number, Z, is always a subscripted
prefix.
X
chemical
symbol
Figure 1.5 The ZA X notation.
469884_01_AQA_A-Level_Physics_001-017.indd 4
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TIP
Note: chemists use different
names for A and Z. They call
A the mass number and Z the
atomic number; this is because
chemists generally use this
notation for describing atoms
on the Periodic Table whereas,
as physicists, we are using the
notation to very specifically
describe nuclei.
This notation allows all nuclei to be described uniquely, including isotopes,
which are nuclei (and atoms) of the same element having the same proton
number, Z, and therefore chemical symbol, X, but different numbers of
neutrons and hence different nucleon numbers, A.
A good example of how the AZ X notation is used is when describing the
two main isotopes of uranium used in nuclear fission reactors. About
99.3% of all naturally occurring uranium atoms (and therefore nuclei)
are uranium-238, meaning 238 nucleons (the nucleon number, A),
comprising of 92 protons (the proton number, Z) and 146 neutrons
(238 − 92 = 146). Only about 0.7% of naturally occurring uranium is
the more useful uranium-235, which can be used in nuclear reactors as
nuclear fuel. Uranium-235 has the same proton number, 92, but only
143 neutrons (235 − 92 = 143). Using the AZX notation, the two nuclei are
written as:
238
92
hydrogen
hydrogen-1
1
1
H
deuterium
hydrogen-2
2
1
H
tritium
hydrogen-3
3
1
235
92
U
uranium-238
U
uranium-235
Figure 1.6 The two most abundant isotopes of uranium.
H
= proton
= neutron
Figure 1.7 The isotopes of hydrogen.
Naturally occurring hydrogen has three isotopes – ‘normal’ hydrogen,
deuterium and tritium – 99.98% of all hydrogen is normal hydrogen-1; less
than 0.02% is deuterium, hydrogen-2; and there are only trace amounts of
tritium, hydrogen-3.
TEST YOURSELF
Data for test yourself questions is generally given in
the questions, but you can find extra information on
specific nuclides using an online database such as
Kaye and Laby from the National Physical laboratory
(NPL), kayelaby.npl.co.uk.
1 Explain what is meant by the specific charge of an
electron.
2 What are isotopes?
3 Using the data on p. 3 obtained by Rutherford,
Geiger and Marsden, calculate the (atomic
radius : nuclear radius) ratio for gold.
4 Rutherford realised that the nuclear radius that he
had calculated from Geiger and Marsden’s data from
positively charged alpha particle scattering would be
a maximum value. (This value is now called the charge
radius.) More modern methods of determining the
nuclear radius of gold, using electron scattering, put
the radius at about 5 fm (~ 5 × 10−15 m). Why is there
such a difference between the two values?
5 This question is about the element silicon-28, 28
14 Si,
which is commonly used as a substrate for the
manufacture of integrated circuits.
a) How many protons, neutrons and electrons are
there in an atom of 28
14 Si?
28
b) The 14 Si atom loses two electrons and forms an
ion.
i) Calculate the charge of the ion.
469884_01_AQA_A-Level_Physics_001-017.indd 5
What are the building blocks of the Universe?
Isotopes are nuclei with the same number of
protons, but different numbers of neutrons.
6
7
8
9
ii) State the number of nucleons in the ion.
iii) Calculate the specific charge of the silicon ion.
a) Germanium atoms are frequently laid onto the
top of silicon substrates. About 37% of naturally
occurring germanium is germanium-74.
Determine the charge of a germanium-74 nucleus.
b) A positive ion with a germanium-74 nucleus has
a charge of 4.80 × 10 −19 C. Calculate the number
of electrons in this ion.
Calculate the specific charges of the following:
a) a deuterium nucleus (heavy hydrogen – one
proton + one neutron)
b) a carbon-12 nucleus
c) an (oxygen-16)2− ion.
Radon-222 is a colourless, naturally occurring
radioactive gas.
a) How many protons are there in a nucleus of
radon-222?
b) How many neutrons are there in a nucleus of
radon-222?
c) Write the ZAX notation for radon-222.
There are two naturally occurring isotopes of
copper: 69% is copper-63, 31% is copper-65.
a) Write the ZAX notations for each of the two isotopes.
b) Calculate the number of neutrons in each of the
two isotopes.
c) Calculate the average atomic mass of these two
isotopes.
5
17/04/19 8:57 AM
●● Stable and unstable nuclei
Why are most nuclei stable? Protons are positively charged, so when they
are confined close together in a nucleus they should all repel each other
and break up the nucleus, so why don’t they? If nuclei are stable then
there must be another attractive, short-range force that is stronger than the
force of electromagnetic repulsion between the protons. In addition, this
stronger force must also act between neutrons as well as protons, otherwise
it would be extremely easy to remove neutrons from nuclei; at extremely
short-ranges the force must be repulsive, otherwise the nucleons inside the
nucleus would implode.
force (N)
The strong nuclear force is one of the four
fundamental forces of nature (including
electromagnetic, weak and gravitational
forces). It is a very short-range force and
acts between nucleons (protons and
neutrons) holding nuclei together.
Repulsive
force
1
2
3
nucleon
separation
(fm)
Attractive
force
1 PARTICLES AND NUCLIDES
Figure 1.8 The strong nuclear force.
6
The neutrino, ν, is a neutral, almost massless fundamental sub-atomic particle that
rarely interacts with matter. Antineutrinos
are the antiparticles of the neutrino. There
are three forms of neutrino: the electronneutrino, νe; the muon-neutrino, νμ; and the
tau-neutrino, ντ.
TIP
The strong nuclear force is one of the four fundamental forces (the other
three are electromagnetic, gravitational and weak nuclear) and it acts between
nucleons. It is the force that holds nuclei together and keeps them stable.
The strong force is attractive up to distances of about 3 fm (3 × 10−15 m) and
repulsive below very short-range distances of about 0.4 fm. The graph of
force against nucleon separation is shown in Figure 1.8.
You can see from Figure 1.8, that at separations of about 0.4 fm, the
magnitude of the strong nuclear force is zero. The force is neither
attractive nor repulsive. This is the equilibrium position for the nucleons
(protons or neutrons), where the resultant force on each nucleon is zero.
In a stable nucleus, this is the separation of each nucleon. The graph also
shows how short range the force is: at a separation of just 3 fm, the strong
force has dropped to zero.
Alpha and beta radioactive decay
Not all nuclei are stable. In fact the vast majority of known isotopes
are unstable. Unstable nuclei can become more stable by the process
of radioactive nuclear decay. Although there are many different decay
mechanisms, three types are far more common than all the others. These
are alpha, beta and gamma decay. Alpha decay involves the emission
of two protons and two neutrons joined together (identical to a helium
nucleus). Beta decay involves the decay of a neutron into a proton, and an
electron and an antineutrino, which are subsequently emitted from the
nucleus. Gamma emission involves the protons and neutrons inside the
nucleus losing energy (in a similar way to electrons changing energy level)
and emitting a gamma ray photon as part of the process.
The antineutrino emitted during beta decay was first
suggested to be emitted by the Austrian physicist
Wolfgang Pauli in 1930. Beta particles are emitted
from a parent nucleus with a range of different
energies, up to a maximum value. Pauli suggested
that when beta particles are emitted, another particle
is emitted at the same time, taking up the balance of
the energy. Pauli called these particles ‘neutrinos’,
symbol υ, meaning ‘little neutral ones’. He proposed
469884_01_AQA_A-Level_Physics_001-017.indd 6
that neutrinos were electrically neutral and almost
mass-less (their mass is so small that it has never
been determined accurately); they also only feel
the weak nuclear and gravitational forces. The antineutrino is the antiparticle of the neutrino. It has
nuclear properties opposite to that of its particle
‘sister’. Antiparticles have the same nuclear symbol
as their particle equivalents, but they have a bar over
them, hence the symbol for the antineutrino is −ν.
17/04/19 8:57 AM
Number of electrons emitted
E
ν
The observed beta particle spectrum graph (Figure 1.9)
shows that beta particles are emitted with a range of energies,
Eβ, from just over zero up to a maximum value ETOT. ETOT is
the expected energy determined by analysing the energies of
the parent and daughter nuclides. ETOT is constant for any
one radioactive beta-emitting nuclide, so another particle,
the antineutrino, must be emitted with an energy E −,
ν which
can be any value from just above zero up to a maximum
value ETOT. The combined energy of the beta particle and the
antineutrino must equal ETOT, so:
Expected beta
electron
energy, ETOT
ETOT Eβ E
ν
Kinetic energy of
antineutrino
Kinetic energy of beta
particles (MeV)
Endpoint
of spectrum
Figure 1.9 A typical beta particle emission spectrum.
The total emission energy ETOT is made up of the
kinetic energy of the beta particle plus the kinetic
energy of the antineutrino.
ETOT = Eβ + E−.
ν
Detecting nuclear radiation
Kinetic
energy
of one beta
electron
Eβ
Observed
spectrum
of energies
●● Detecting nuclear radiation
The cloud chamber
glass plate
ionising
nuclear
radiation
source
One of the earliest ways of detecting ionising nuclear radiation was
invented by Charles Wilson in 1911. He came up with the idea of
creating clouds artificially in the laboratory, by rapidly expanding
light
beam
saturated air air saturated with water vapour inside a sealed chamber. Wilson
suspected that clouds form on charged particles in the atmosphere,
piston
and he experimented by passing X-rays through his chamber.
He discovered that the X-rays left wide cloudy tracks inside the
metal
chamber. Wilson then passed alpha, beta and gamma radiation
cylinder
through the chamber and found that the highly ionising alpha
particles left broad, straight, definite length tracks; the ionising
beta particles left thin, straight or curved tracks (depending on
Figure 1.10 The Wilson cloud chamber. (If you
are conducting this experiment yourself, follow how high their energy was); but the weakly ionising gamma rays
the CLEAPSS L93 risk assessment.)
left no tracks at all.
The spark counter
Radiation
source
0V
+ 5000 V
gauze
thin wire
sparking
here
Figure 1.11 A spark counter. (If you are
conducting this experiment yourself
use a current-limited (5 mA max) EHT
and follow the CLEAPSS L93 risk
assessment.)
469884_01_AQA_A-Level_Physics_001-017.indd 7
Spark counters only detect highly ionising alpha particles. Beta and gamma
radiation do not ionise enough of the air between the metal gauze and the
thin wire underneath. When the air particles are ionised by the alpha particles
the charged particles produced cause a spark to be formed. That spark jumps
across the 5000 V gap between the gauze and the wire. The spark can be
seen, heard and counted by an observer or with a microphone. This method
of detecting alpha particles is particularly useful for showing that they have a
very short range in air.
7
The Geiger-Muller counter
Figure 1.12 shows a modern Geiger-Muller counter, which is designed to detect
and count radioactive emisions automatically, without having to count sparks or
scintillations one at a time.
Ionising radiation enters the GM tube through the thin mica window
(alpha or beta radiation), or through the window or sides of the counter
17/04/19 8:57 AM
Mica
window
Cathode
Ionising
radiation
Atom
Electron
Ion
500 V R
(gamma radiation). The low-pressure inert gas inside the detector
(helium, neon or argon) is ionised producing a cascade of charged
particles that are attracted to oppositely charged electrodes. The small
pulse of current produced by the moving charge is detected by the
electronic counter, which registers a ‘count’. Geiger counters have a
distinct advantage over spark counters and cloud chambers in that they
can detect all three types of radiation.
Counter
Anode
Figure 1.12 A Geiger-Muller (GM) counter.
8
10 T
he pictures to the right show
the paths of alpha particles,
beta particles and X-rays inside
a cloud chamber:
a) Which cloud chamber
picture(s) show the
following?
i) alpha particle tracks
ii) an X-ray track
iii) a beta particle track
b) Explain why cloud chamber
picture C show two types of
particles?
11 E xplain why a spark counter
can easily detect alpha particles
but usually cannot detect beta
particles or gamma rays.
12 A Geiger counter is used to
detect the intensity of alpha
particles being emitted from
a radium source containing
two alpha-emitting isotopes,
radium-223 and radium-226.
A graph of intensity (counts per
minute) against range in air
(cm) for this source is shown in
Figure 1.17.
Radium-223 emits alpha
particles with higher energy
than radium-226. Use the graph
to determine the range of alpha
particles in air emitted by each
isotope. Explain your reasoning.
Figure 1.13 Cloud
chamber Picture A.
Figure 1.14 Cloud
chamber Picture B.
Figure 1.15 Cloud chamber Picture C.
Figure 1.16 Cloud chamber Picture D.
5000
intensity, counts per minute (cpm)
1 PARTICLES AND NUCLIDES
TEST YOURSELF
4000
3000
2000
1000
0
0
1
2
3
4
range in air (cm)
5
6
7
Figure 1.17 Graph of intensity versus range in air.
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Nuclear equations
ACTIVITY
The beta particle emission spectrum of bismuth-210
A student carried out an experiment measuring the kinetic energy of beta particles
emitted by the radioisotope bismuth-210. Her measurements are shown below.
Table 1.2
Kinetic energy of
beta particles,
E (× 10 −15 J)
Intensity of beta
particles (arbitrary
units)
0
16
32
48
64
80
10.0 11.5 12.0 11.5 10.0 8.0
96
112
128
144
160
176
186
6.0
4.4
2.9
1.5
0.6
0.2
0.0
1 Plot a graph of beta particle intensity (y-axis) against beta particle energy (x-axis).
2 A second student measures beta particles being emitted with a second bismuth-210
source with an intensity of 11 arbitrary units. Use the graph to determine the possible
kinetic energies of the emitted beta particles from the second source.
3 The maximum kinetic energy of beta particles emitted from bismuth-210 is 186 fJ (186 × 10−15J).
Calculate the energies of the antineutrinos emitted at this intensity.
4
2
0
–1
He – alpha particle (α)
–
e – beta particle (β )
γ – gamma ray (strictly 00 γ)
ν – neutrino (strictly 00 ν)
ν – antineutrino (strictly 00 ν)
●● Nuclear equations
Figure 1.18 The ZAX notation for
ionising radiation (Note: values of
o
oX are usually written without the
zeros).
TIP
Some nuclei that are unstable and
decay via gamma ray emission
have the letter ‘m’ added after
their nucleon number, meaning
‘metastable’. A good example
of this is technetium-99m, a
common gamma ray emitter used
in nuclear medical imaging. This
is covered in next year’s book.
The ZAX notation can be used to describe the nuclear reactions that take
place during radioactive decay.
During alpha, beta and gamma radioactive decay, two simple conservation
rules come into play:
●
●
nucleon number, A, is always conserved
proton number, Z, is always conserved.
In practice this means that the total (addition) value for A before the decay
is equal to the total (addition) value of A after the decay (and, similarly,
the totals for Z remain the same before and after the decay). Look at the
following two examples:
1 A common alpha decay is the decay of radium-226 in granite rocks,
producing an alpha particle and the radioactive gas, radon-222. This
nuclear decay can be written as:
226Ra
88
→ 222
+4
86Rn 2H e
Total nucleon number before = 226
226
Ra
88
Total proton number before = 88
Total nucleon number after 222 + 4 = 226
+
222
Rn + 4 He
86
2
+
Total proton number after 86 + 2 = 88
9
Figure 1.19 Decay of radium.
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2 The same rules apply to beta decay. A good example is the decay of
carbon-14, a naturally occurring radio-nuclide used in carbon dating
techniques. Carbon-14 decays via beta emission to nitrogen-14, emitting
an antineutrino in the process. This decay can be summarised by:
TIP
Total nucleon number before = 14
It is handy to have access to
a database of the different
nuclides. The database should
give you the values of A and Z, as
well as other key data, such as
the half-life, abundance and its
decay mechanisms. You will find
such databases on the web.
14
C
6
Total proton number before = 6
Total nucleon number after 14 + 0 = 14
+
14
N + 0β+ν
7
–1
+
Total proton number after 7 + (–1) = 6
Figure 1.20 Decay of carbon-14.
In both these examples, both the nucleon number, A, and the proton
number, Z, are conserved during the decay.
TEST YOURSELF
13 P
lutonium was the first man-made element. The
element is made by bombarding uranium-238
with deuterons (nuclei of deuterium, 21H),
producing neptunium-238 and two neutrons, (10n);
the neptunium-238 then decays via beta decay
(where the nucleus emits an electron, −10e) to
plutonium-238.
1 PARTICLES AND NUCLIDES
Write nuclear equations summarising the formation
of plutonium-238. Uranium has a proton number
of 92.
Use a nuclide database, or a periodic table, to write
nuclear equations to summarise the following
radioactive decays.
10
14 A
mericium-241 is an alpha emitter commonly
used in smoke detectors; its daughter nuclide is
neptunium-237.
15 Strontium-90 is a beta emitter used as a
fuel source in radioisotope thermoelectric
generators (RTGs) on space probes and
lighthouses; it decays to yttrium-90, emitting an
antineutrino in the process.
16 Phosphorous-32 is a beta emitter used as a
tracer in DNA research.
17 Plutonium-238 is an alpha emitter also used as
the fuel source for RTGs.
18 Tritium (or hydrogen-3) is a beta emitter used in
some ‘glow-in-the dark’ paints.
●● The photon model of electromagnetic
radiation
Around the turn of the twentieth century, physicists such as Albert Einstein
and Max Planck developed a particle model of electromagnetic radiation to
account for two experimental observations that were impossible to explain
using classical physics.
Planck developed the concept of a fundamental unit of energy, which became
known as a quantum. He proposed that atoms absorb and emit radiation in
multiples of discrete amounts that are given by the Planck equation:
E = hf
where:
f is the frequency of the radiation absorbed or emitted, and
h is a constant, now called the Planck constant.
It was Planck who called these discrete units of energy ‘quanta’ and the
small ‘packets’ of electromagnetic radiation making up these quanta became
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TEST YOURSELF
Antiparticles
known as photons. Measurements of the energy of photons has produced a
value for the Planck constant of 6.63 × 10−34 J s. This theory is discussed in
more detail in Chapters 3 and 4.
19 A
light-emitting diode (LED) emits light photons of
energy 3.6 × 10−19 J.
a) Calculate the frequency of this
electromagnetic radiation.
b) Calculate the energy emitted per second
(called the output power) of the LED, when it
emits 0.9 × 1017 photons each second.
20 LEDs are now used extensively as the light source
in torches.
The LEDs used in the torch shown in the diagram
above emit blue photons of light. The wavelength
of the blue light is 420 nm.
a) Show that the energy of a photon from this
LED is about 5 × 10 –19 J.
Figure 1.21 LED torch.
LEDs need to be connected to a certain minimum
voltage, called the activation voltage, before they
emit light. A blue LED needs a higher voltage than
a longer wavelength red LED.
b) Explain why a red LED needs a lower voltage
than a blue LED.
●● Antiparticles
Rest mass-energy is the amount of energy
released by converting all of the mass of a
particle at rest into energy using Einstein’s
famous mass-energy equation,
E = mc2
where m is the rest mass of the particle and
c is the speed of light.
Mega electron-volts The energy of nuclear
particles is usually given in MeV, mega
electron-volts. One electron-volt is a very
small amount of energy, equivalent to 1.6 ×
10 −19 J. This is the same numerical value as
the charge on an electron and is defined as
the amount of energy needed to accelerate
an electron of charge ‘e’, (1.6 × 10 −19 C)
through a potential difference of 1 volt. One
MeV is a million electron-volts, equivalent
to 1.6 × 10 −13 J. This unit comes from the
definition of the volt. A volt is the amount of
energy per unit charge or:
energy
volts =
charge
so
energy = charge × volts
469884_01_AQA_A-Level_Physics_001-017.indd 11
In 1928, Paul Dirac proposed the existence of the positron, the antiparticle
of the electron. He suggested that the positron had the same mass as an
electron, but that most of its other physical properties (such as its charge)
were the opposite to that of the electron. Dirac thought that positrons were
positively charged electrons.
All particles have a corresponding antiparticle. Each particle–antiparticle
pair has the same mass and therefore the same rest mass-energy (the
amount of energy released by converting all of the mass into energy), but
they have opposite properties such as their charge. The positron is one
of the few antiparticles that has its own different name; most other antiparticles are described by the word ‘anti’ in front of the particle name. The
other more common antiparticles are the antiproton, the antineutron and
the antineutrino. Antiparticles generally have the same symbol as their
particle but have a bar drawn over the top of the symbol; for example, the
proton has the symbol p and the antiproton has the symbol p−. An exception
to this is the positron. It was the first antiparticle to be discovered and is
given the symbol e+.
Antiparticles are not common. The Universe appears to be overwhelmingly
dominated by matter. Antimatter only becomes apparent in high-energy
particle interactions, for example in the interaction of high-energy cosmic
rays with the atmosphere, or in particle accelerator experiments such as the
Large Hadron Collider (LHC) at CERN.
11
17/04/19 8:57 AM
Particle–antiparticle interactions
TIP
Other particle–antiparticle
pairs with ‘non-bar’ symbols
are: muon+ (μ+) and muon− (μ−);
pion+ (π+) and pion− (π−); kaon+
(K+) and kaon− (K−). The photon
and the pion0 (π0) are their own
antiparticles. You will meet these
particles again in Chapter 2.
When a particle meets its corresponding antiparticle they will annihilate
each other. This means that the total mass of the particle pair is converted
into energy, in the form of two gamma ray photons.
i. Proton–antiproton annihilation
γ
p antiproton
proton p
ii. Electron–positron annihilation
γ
e+ positron
electron e–
γ
γ
Figure 1.22 Diagram showing annihilation.
When a particle meets its antiparticle and
their total mass is converted to energy in
the form of two gamma ray photons, the
particles annihilate each other.
Two gamma photons are always produced during particle–antiparticle
annihilation in order to conserve momentum. The total energy of the
gamma photons is equal to the total rest energies of the particle–antiparticle
pair, in order to conserve energy.
EXAMPLE
The rest energy of an electron is the same as that of a positron and is
0.51 MeV. Calculate the wavelength (in picometres, pm) of the two gamma
ray photons produced when an electron and a positron annihilate each
other at rest.
1 PARTICLES AND NUCLIDES
Answer
12
As the electron and the positron are initially at rest, their total momentum
is zero. This means that the total momentum of the photons produced when
the electron/positron annihilates must also be zero. Both the photons will
have the same energy and will be moving at the same speed (the speed of
light), so they must be moving in opposite directions for their momenta to
cancel out to zero.
The total energy of the annihilation will be 1.02 MeV. The two photons that
are produced during the annihilation must therefore have an energy equal
to 0.51 MeV.
0.51 MeV = 0.51 × 1.6 × 10−13 J = 8.2 × 10−14 J
To calculate the wavelength of the photons:
E = hc
λ
hc
λ=
E
6.6 × 10−34 Js × 3 × 108 ms−1
λ=
8.2 × 10−14 J
= 2.4 × 10−12 m
λ = 2.4 pm
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The opposite process to annihilation is called pair production. In this
process, a photon with enough energy can interact with a large nucleus
and be converted directly into a particle–antiparticle pair. The rest energy
of an electron (and therefore a positron) is 0.51 MeV, or 8.2 × 10−14 J.
In order to create this particle–antiparticle pair, the photon must have
enough energy to create both particles, in other words, (2 × 0.51) =
1.02 MeV, or 16.4 × 10−14 J. The wavelength of the photon needed to do
this can be calculated by:
Antiparticles
Pair production
E = hc
λ
λ = hc
E
−34
× 108 m s−1 = 1.2 × 10−12 m
λ = 6.6 × 10 J s × 3−14
16.4 × 10 J
λ = 1.2 pm
TEST YOURSELF
21 A
proton and an antiproton annihilate at rest
producing two high-energy photons, each with
an energy of 1.5 × 10 −10 J. Calculate the frequency
and the wavelength of the photons. Compare
the wavelength of these photons to the photons
produced by electron–positron annihilation. The
rest mass-energy of the proton and antiproton is
938 MeV.
22 Every type of particle has its corresponding
antiparticle.
a) Write down the name of one particle and its
corresponding particle.
b) State one property that the particle and its
antiparticle share.
c) State one property that is different for the
particle and its antiparticle.
23 Under certain circumstances it is possible for a
very high energy photon to be converted into a
proton and an antiproton, each with a rest energy
of 1.50 × 10 −10 J.
a) State the name of this process.
The photon in this process must have a minimum
energy in order to create a proton and an antiproton.
b) Calculate the minimum energy of the photon
in joules, giving your answer to an appropriate
number of significant figures.
c) A photon of even higher energy than that
calculated in part (b) is also converted into a
proton/antiproton pair. State what happens to
the excess energy.
d) Explain why the photon required to produce
an electron/positron pair would not be able to
produce a proton/antiproton pair.
e) The antiprotons produced during this process
have very short lifetimes. Describe what is
likely to happen to the antiproton soon after it
is formed.
f) Explain why a single photon could not produce
a single proton during this process rather than
a proton/antiproton pair.
13
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Practice questions
1 Which line represents the correct number of protons, neutrons and
electrons in an atom of one of the isotopes of lead 208
82 Pb ?
protons
82
126
208
82
A
B
C
D
neutrons
126
82
82
126
electrons
126
126
208
82
2 What is the specific charge of a gold 197
79 Au nucleus? The mass of the gold
−25
nucleus is 3.29 × 10 kg; the charge on the electron is 1.6 × 10−19 C.
A 3.63 × 107 C kg−1
C 3.92 × 107 C kg−1
B 3.84 × 107 C kg−1
D 9.56 × 107 C g−1
3 Uranium-236 may split into a caesium nucleus, a rubidium nucleus and
four neutrons as shown below in the following nuclear equation. What is
the value of X for the rubidium nucleus?
236U
92
→ 137
+ X + 410n
55Cs 37Rb
A 92
B 95
C 98
D 99
1 PARTICLES AND NUCLIDES
4 What is the charge, in C, of an atom of 157N from which a single electron
has been removed?
A – 9.6 × 10−19 C
C +1.6 × 10−19 C
B –1.6 × 10−19 C
D + 9.6 × 10−19 C
5 In a radioactive decay a gamma photon of wavelength 8.3 × 10−13 m
is emitted. What is the energy of the photon? The speed of light is
3 × 108 m s−1.
A 5.4 × 10−46 J
C 57 J
B 2.4 × 10−13 J
D 2.0 × 1043 J
6 Thorium decays by the emission of an alpha particle as shown in the
equation:
229Th
90
→ XY Ra + α
What are the correct values for X and Y?
X
14
7
Y
A
225
88
B
88
225
C
229
91
D
227
86
14C
6
is a radioactive isotope of carbon. It can form an ion when two
electrons are removed from the atom. What is the charge on this ion in
coulombs?
A −9.6 × 10−19 C
C 3.2 × 10−19 C
B −3.2 × 10−19 C
D 9.6 × 10−19 C
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A 3.89 × 10−40 J
C 1.13 × 10−27 J
B 1.17 × 10−31 J
D 3.38 × 10−19 J
9 An alpha particle has a kinetic energy of 9.6 × 10−13 J. The mass of the
alpha particle is 6.6 × 10−27 kg. What is the speed of the particle?
10
A 1.2 × 107 m s−1
C 1.5 × 1014 m s−1
B 1.7 × 107 m s−1
D 2.9 × 1014 m s−1
Practice questions
8 The line spectrum from helium includes a yellow line with a wavelength
of 587.6 nm. What is the energy of a photon with this wavelength?
238U
92
decays by emitting α and β− particles in a number of stages to
−
form 206
82 Pb . How many β decays are involved in this decay chain?
A 2
B 4
C 6
D 8
11 a) Name the constituent of an atom which
i)
has zero charge
(1)
ii) has the largest specific charge
(1)
iii) when removed leaves a different isotope of the element.
(1)
b) The equation
→ AZ Ru + −10 β + X
99
43 Tc
represents the decay of technetium-99 by the emission of a
β− particle.
i)
Identify the particle X. ii) Determine the values of A and Z.
(1)
(2)
12 Alpha decay is a process by which an unstable isotope of an
element may decay.
a) State what is meant by an isotope.
b) Copy and complete this equation for alpha decay:
...
AZ X → ...Y + 42 H e
c) Explain why the alpha particle, once outside the nucleus, is
unaffected by the strong nuclear force of the parent nucleus.
(2)
(2)
(2)
13 An atom of calcium, 48
20 Ca, is ionised by removing two electrons.
a) State the number of protons, neutrons and electrons in the ion
formed.
(1)
b) Calculate the charge of the ion.
(1)
c) Calculate the specific charge of the ion.
(2)
14 a)Describe how the strong nuclear force between two nucleons varies
with the separation of the nucleons, quoting suitable values for the
separation.
469884_01_AQA_A-Level_Physics_001-017.indd 15
15
(3)
17/04/19 8:57 AM
b) An unstable nucleus can decay by the emission of an alpha
particle. State the nature of an alpha particle.
(1)
c) Copy and complete the equation below to represent the
emission of an α particle by a 238
92 U nucleus.
...
→ ...Th + ... α
238
92U
(2)
15 a) Explain what is meant by the specific charge of a nucleus.
(1)
The incomplete table shows information for two isotopes
of uranium.
Number of
protons
Number of
neutrons
92
143
First isotope
Second isotope
Specific charge of
nucleus...
3.7 × 107
b) Copy the table and add the unit for specific charge in the
heading of the last column of the table.
(1)
c) Add the number of protons in the second isotope to the second
row of the table.
(1)
d) Calculate the specific charge of the first isotope and write this
in the table.
(3)
e) Calculate the number of neutrons in the second isotope and
put this number in the table.
(4)
Stretch and challenge
1 PARTICLES AND NUCLIDES
16 Figure 1.23 shows an arrangement to measure the Planck constant.
16
flying lead
1
2
3
4
5
V
Figure 1.23 Circuit diagram for measuring the Planck constant.
The data obtained from this experiment is shown below.
LED number
1 violet
2 blue
3 green
4 yellow
5 red
Wavelength of
emitted photons,
λ /nm
413
470
545
592
625
Frequency
of emitted
photons, f /Hz
Activation
voltage, VA /V
3.01
2.65
2.28
2.09
1.98
a) Write a method as a numbered list for this experiment.
b) Copy and complete the table, calculating the frequency of
the emitted photons.
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d) Draw a line of best fit on the graph and calculate the gradient.
e) Use the equation eVA = hf, together with the gradient of the best
fit line, to determine a value of the Planck constant. Show your
working.
Practice questions
c) Plot a graph of activation voltage (y-axis) against photon
frequency (x-axis).
17 How does the proton number, Z, and the nucleon number, A, of a
nucleus change due to:
a) the emission of an alpha particle
b) the emission of a beta particle
c) the fusion with a deuterium nucleus?
42
18 The potassium isotope 42
19 K disintegrates into 20 Ca.
a) What are the likely type/s of radiation produced?
b) How many protons, neutrons and electrons are present in an
atom of the daughter nucleus 42
20 Ca ?
19 A muon and an antimuon annihilate each other to produce two
γ-rays. Research the data that you need for this question and use
it to calculate the minimum energy of the photons.
17
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2
Fundamental particles
PRIOR KNOWLEDGE
●
●
●
●
Matter is made of atoms.
Atoms are made up of a very small central nucleus containing particles
called protons and neutrons, surrounded by orbiting electrons.
Atoms are electrically neutral. Protons carry a positive charge and
electrons carry the same magnitude, negative charge. In atoms, the
number of protons equals the number of electrons, so their overall
charge is zero.
Protons and neutrons have a mass about 1800 times larger than
electrons. Most of the mass of an atom is concentrated in the nucleus
due to the protons and neutrons.
TEST YOURSELF ON PRIOR KNOWLEDGE
Data for Test Yourself questions is generally given in the questions
but you can find extra information on specific nuclides using an online
database such as Kaye and Laby from the National Physical Laboratory
(NPL), kayelaby npl.co.uko.
1 a) Calculate the total charge on a sodium-23 nucleus.
b) Calculate the approximate mass of the sodium-23 nucleus.
2 The atomic radius of an element can be estimated using the Avogadro
Number, NA = 6.02 × 10 23 (the number of particles in 1 mole), the
density of the element and the molar mass of the element. The molar
mass of carbon-12 is 12.0 × 10 −3 kg mol−1, and carbon-12 (diamond
allotrope) has a density of 3500 kg m −3.
a) Calculate the mass of one atom of carbon-12.
b) Calculate the volume of one atom of carbon-12.
c) Calculate the radius of one atom of carbon-12 (volume of a sphere
is given by V = 4 π r 3).
3
d) The nuclear radius of an element is given by the following formula:
r = r 0A
where r 0 is a constant equal to 1.25 × 10 −15 m, and A is the nucleon
number (number of protons + number of neutrons). Use this
information to calculate the radius of a carbon-12 nucleus.
e) Calculate the ratio atomic radius for carbon-12.
1
3
nuclear radius
18
●● The particle garden
The discovery of the three basic sub-atomic particles: the electron (in 1897
by JJ Thomson); the proton (in 1917 by Ernest Rutherford) and the neutron
(in 1932 by James Chadwick), seemed to complete the structure of the
atom. However, in the 1930s, physicists started to discover a range of new subatomic particles, such as the positron and the muon, primarily as the result of
experiments on high-energy cosmic radiation from space. The Second World
War forced a break in the discovery of new particles as physicists were put to
use working on new warfare technology, such as the atomic bomb and radar.
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everyday matter
u
2/3
up
d
c
charm
-1/3
104M
s
171.2G 2/3
t
-1/3
4.2G
b
-1/3
strange
bottom
0.511M -1
105.7M -1
1.777G -1
electron
νe
muon
0
<0.17M
νμ
τ
tau
0
<15.5M
νt
γ
photon
0
e-neutrino μ-neutrino t-neutrino
80.4G +/-1
w+-
91.2G
z
H
higgs
graviton
gravitational force
<2.2
μ
g
gluon
125-6G
weak nuclear force
6 leptons (+6 antimatter)
down
e
charge
mass
(eV/c2)
top
electromagnectic (charge)
4.8M
2/3
1.27G
mass
outside of
giving standard model
strong nuclear force (color
charge)
6 quarks (+6 antimatter)
2.4M
exchange
particles
exotic matter
Figure 2.1 The Standard Model of matter showing the 12
fundamental particles and the exchange particles, together
with their masses and charges.
TIP
The masses of fundamental particles are nearly always
given as mass-energy equivalents. Einstein’s energymass equation, E = mc2, can be rearranged to m = cE2.
As c2 is a constant value then mass also has the unit
eV/c2. The mass of the electron is 0.51 MeV/c2, which is
It became obvious that particles were being discovered
in ‘families’, rather like the groups of elements on the
periodic table, and new names such as leptons, hadrons
and mesons were invented to classify these groups.
In 1967, the Standard Model of particle physics was first
described by Steven Weinberg and Abdus Salam, and it
is from this that the current picture of the structure and
behaviour of matter was constructed. In 2012, the final
piece of the Standard Model jigsaw, the Higgs Boson,
was ‘discovered’ by a large team of physicists working on
data from the Large Hadron Collider at CERN in Geneva.
(CERN stands for Conseil Européen pour la Recherche
Nucléaire or European Council for Nuclear Research.)
Fundamental particles
After the end of the Second World War, physicists returned to their research
into sub-atomic particles. The huge advance in technology made during the war
years led to the discoveries of many new particles and the production of more
sophisticated particle accelerators and detectors.
The Standard Model
The Standard Model of matter has proved to be very
powerful. It successfully describes the huge proliferation of
particles discovered during the latter part of the twentieth
century. The model is surprisingly simple, consisting of
two families of fundamental particles and a set of forces
that bind them together.
equivalent to 9.11 × 10−31 kg. The heaviest fundamental
particle, the top quark, has a mass of 171.2 GeV/c2
or 3.1 × 10−25 kg. Rest mass-energies can therefore
be given as a mass (usually in MeV/c2) or an energy
equivalent (usually in MeV).
●● Fundamental particles
Antiparticles are fundamental particles
with the same mass and energy as their
particle counterparts, but have opposite
properties such as charge. When a particle
meets its antiparticle they annihilate
(see Chapter 1) converting to gamma ray
photons.
469884_02_AQA_A-Level_Physics_018-041.indd 19
The word ‘fundamental’ in physics has profound meaning. Fundamental
particles are particles that appear to have no structure. Fundamental
particles cannot be broken down into smaller pieces and they are the basic
building blocks of the Universe. When Democritus performed his thought
experiment on the nature of matter on a beach in Ancient Greece, his idea
was simple – the smallest building block of matter was called an atom,
and for about two and a half millennia, this concept dominated science.
JJ Thomson’s discovery of the electron started the hunt for sub-atomic
particles, and the Standard Model appears to finish the hunt with a group
of 12 fundamental particles (and their antiparticles). The 12 fundamental
particles (and the four exchange particles that hold them together) of the
Standard Model can be arranged in many different ways to make up any
observed composite particles in the Universe, with the overwhelming
proportion of the observed Universe appearing to be made of just three
19
17/04/19 9:14 AM
fundamental particles: the electron, the up quark and the down quark.
Up and down quarks combine together in threes to make protons and
neutrons, and these combine together to form nuclei. The addition of
electrons to surround the nuclei forms atoms.
TEST YOURSELF
1 What is the ‘Standard Model’?
2 State one similarity and one difference between
a ‘particle’ and its ‘antiparticle’ (for example, an
electron and a positron).
3 What is the conversion factor from MeV into joules?
These questions refer to the Standard Model shown
in Figure 2.1.
4 State the six leptons and the six quarks that make up
the Standard Model.
5 What is the charge on the following particles?
a) electron
b) muon-neutrino
c) up quark
d) positron
e) down quark
f) strange quark.
6 What are the four most common particles shown in
Figure 2.1?
7 What is meant by a fundamental particle?
8 How many quarks are there in a proton?
●● Quarks
2 FUNDAMENTAL PARTICLES
Quarks are fundamental particles that make
up particles such as protons and neutrons.
They exert the strong nuclear force on one
another.
20
Although you have already met the electron as a fundamental particle, you
may not have come across quarks. The concept of quarks as fundamental
building blocks of matter was first proposed by Murray Gell-Mann and
George Zweig in 1964, as part of the original Standard Model, which
initially only contained three ‘flavours’ of quarks (up, u; down, d; and
strange, s). The other three flavours (charm, c; bottom, b; and top, t) were
added later. (The bottom quark is also known as the beauty quark.) A
basic difference between these quarks is their mass. Figure 2.2 shows the
masses of each quark drawn to scale (with the area representing mass).
The proton and electron masses are shown for comparison to the left.
electron
proton
up
down strange
quark quark quark
charm
quark
bottom
quark
top quark
Figure 2.2 The relative masses of the six quarks.
Gluons are one of the four exchange
particles of the Standard Model. Gluons
act between quarks holding them together.
Gluons have an extremely short range of
action of about 10 −15 m.
469884_02_AQA_A-Level_Physics_018-041.indd 20
Most of the mass of a proton is due to the energy of the interactions
between the quarks and the gluons that hold the quarks together via
the strong nuclear force. These gluons constantly come into and go out
of existence as they exchange between the quarks. The energy required
to do this is included in the mass of the proton or the neutron. The
up and down quarks are much less massive than the other quarks.
The strange quark is observed in particles at high altitude due to the
interaction of high-energy cosmic rays within the upper atmosphere.
The charm, bottom and top quarks are only observed in particle
accelerator detectors at extremely high energy.
17/04/19 9:14 AM
The first direct observation of quarks was carried out using deep-inelastic
scattering of electrons by protons at the Stanford Linear Accelerator (SLAC)
facility in 1968. Protons were observed to be made up of two up quarks
and a down quark. The strange quark was observed as a result of further
experiments at SLAC and the charm quark was observed in 1974. The
bottom quark was discovered three years later in 1977, but it took until 1995
for a particle accelerator with enough energy, the Tevatron at Fermilab, to
produce interactions involving the top quark.
jet of particles
formed by the
electron-quark
interaction
Fundamental forces
Deep-inelastic scattering involves firing
electrons at protons at very high energies
(hence the word ‘deep’). Elastic scattering,
for example, involves two particles such as
two protons colliding with each other and
rebounding off each other with the same
kinetic energy – rather like two snooker balls
hitting each other head-on and rebounding
back. The word ‘elastic’ in this context
means that no kinetic energy is lost. Inelastic
scattering involves the conversion of kinetic
energy into other forms. In this case the
electrons penetrate into the proton and
interact with the quarks (via exchange of
photons); kinetic energy is converted into
mass as the proton shatters, producing a
shower of other particles. Using the snooker
analogy, it would be as if one snooker ball
entered the second ball causing it to break
into other pieces as the first snooker ball
scattered away from it.
electron interacts
with quark
proton recoil
incident
electron
other jets
proton
recoil electron
Figure 2.3 Deep-inelastic scattering.
TIP
Make sure you know about the up, u; down, d; and strange, s, quarks (and
their antiquarks). You do not need to know about the charm, top or bottom
quark, although they could be set as examples in the exam, where all factual
knowledge would be provided in the question.
TEST YOURSELF
9 W
hat are the six ‘flavours’ of quarks and what is
the main difference between them?
10 The original deep-inelastic scattering
experiments carried out at SLAC in 1968 involved
which of the following particle interactions?
A protons and neutrons
B protons and positrons
C protons and electrons
D protons and muons
11 W
hy are quarks believed to be ‘fundamental
particles’?
12 Arrange the following particles in order of their
masses (heaviest first):
proton, electron, bottom quark, down quark, up
quark.
13 Explain what is meant by ‘deep-inelastic scattering’.
21
●● Fundamental forces
In the same way that the Standard Model describes the 12 fundamental
particles, it also describes the four fundamental forces that allow
the particles to interact with each other. Indeed the words force and
interaction are interchangeable in this context and mean the same
469884_02_AQA_A-Level_Physics_018-041.indd 21
17/04/19 9:14 AM
thing. The word interaction is used to describe these forces on a
microscopic, quantum scale, because the mechanism used to explain
how the forces work is different from the classical ‘Newtonian’ field
theory that is used to describe how forces work in the large-scale,
macroscopic world.
In the macroscopic world, charges and masses exert forces on each other
due to the electric and gravitational force fields that extend out into space,
away from the charge or mass. The patterns of the force field lines show
the action of the forces. On the microscopic, quantum scale, classical field
theory cannot explain some of the ways that particles interact with each
other, and a different ‘quantum’ interaction approach is used.
Exchange particles are particles involved
with the interaction of particles via the
four fundamental forces of nature on the
quantum scale. Exchange particles are only
created, emitted, absorbed and destroyed
between the interacting particles.
The Standard Model describes particles interacting by transferring
exchange particles. There are three fundamental forces described by
the Standard Model, each one with its own distinct exchange particle or
particles. Table 2.1 outlines the three fundamental forces of the Standard
Model and gravity (which falls outside the Standard Model, but is included
in Grand Unification Theories) and their exchange particles.
Table 2.1 Fundamental forces and their exchange particles.
Fundamental force
Nature
Electromagnetic
Acts between charged particles
Photon
Strong
Acts between quarks
Gluon
Weak
Responsible for radioactive decay
(with the exception of alpha decay)
and nuclear fusion
Acts between particles with mass
W+, W− , Z 0
Gravity
Graviton
On the quantum scale, two protons can interact with each other by
exchanging virtual photons via the electromagnetic interaction. The
photons are exchanged in both directions but exist for only minute amounts
of time – hence the word virtual. A simple model of this interaction could
be constructed by two basketball players throwing a basketball back and
forward to each other as shown in Figure 2.4. Momentum is exchanged
between the players and the ball, and a force is produced.
2 FUNDAMENTAL PARTICLES
22
Exchange particle(s)
F
P
Direction of
motion
Exchange
particle - photon
P
F
Figure 2.4 Two basketball players
‘exchanging’ a basketball. This is
analogous to two protons exchanging
a photon.
This model works quite well for illustrating the repulsive force that occurs
between two particles with the same charge, but it does not work very
well with the attractive forces, such as the strong, weak and gravitational
forces, and the electromagnetic attraction between particles with opposite
charge. In this case, if you visualise the basketballs attached to elastic
cords that stretch tight just before being caught by the catcher, this means
that as the basketball is caught it pulls the catcher back towards the
thrower as shown in Figure 2.5. To improve the model further, make the
elastic cord stiffer as the distance between the catcher and thrower gets
smaller, increasing the force of attraction. This makes it more realistic to
the quantum world where the attractive forces decrease with increasing
separation. Trying to visualise the quantum world is very difficult from
our viewpoint.
The electromagnetic force
We are very used to the everyday effects of the electromagnetic force. Not
only do we see and feel the effects of static charge, but the electromagnetic
469884_02_AQA_A-Level_Physics_018-041.indd 22
17/04/19 9:14 AM
Ball thrown
attached to
elastic cord
Fundamental forces
force is the source of the contact forces between everyday objects. On the
quantum scale the electromagnetic interaction occurs between charged
particles, most commonly electrons and protons.
Stretched cord
exerts pull on
catcher
P
F
Direction of
motion
Exchange
particle - photon
F
P
Figure 2.5 A basketball on a cord
provides an attractive force.
Figure 2.6 An electromagnetic interaction.
distance r
(1)
e
(2)
e
Force F
p
Force F/4
p
distance 2r
Figure 2.7 As the distance between two oppositely charged
particles doubles, then the force reduces by a factor of four.
The electromagnetic interaction exchange particle, the
photon, acts over infinite distances. However, the strength
of the force decreases with an inverse-square relationship
to distance, r12 . This means that if the distance between the
particles is doubled then the force reduces by a factor of
four as shown in Figure 2.7.
The photons created during electromagnetic interactions
are called virtual photons, because they only exist
during the time of the interaction. They are created,
interact and decay all within the time of the interaction.
The electromagnetic interaction is responsible for most of the behaviour of
matter on an atomic and molecular scale. (In fact the other three forces are
almost insignificant on this scale.)
23
The strong nuclear force
The strong interaction is an extremely short-range force (typically acting on
the scale of 10−14 to 10−15 m – the scale of the nucleus) and it acts between
quarks. The exchange particle is the gluon.
Although the strong force is very short range, it has a very high
magnitude, 137 times larger than the electromagnetic force, hence the
name ‘strong’. The strong force acts between quarks, so it is the force
that holds nucleons such as protons and neutrons together, and it is
469884_02_AQA_A-Level_Physics_018-041.indd 23
17/04/19 9:14 AM
also the force that holds nuclei together. If the strongest of the
fundamental forces, the strong nuclear force, has a magnitude of
1, then the corresponding electromagnetic force would have a
1
magnitude of 137.
The weak force
A version of the weak force was first described in 1933 by Enrico
Fermi while he was trying to explain beta decay. The weak force
is now known to be responsible for β− and β+ radioactive decay,
electron capture and electron–proton collisions. There are three
exchange particles involved with the weak force, the W+, W− and
the Z0, but these were not experimentally verified until 1983.
The weak force is about a million times weaker than the strong
nuclear force (hence the name), and it acts over an even shorter
range, typically 10−18 m, which is about 0.1% of the diameter of a
proton.
2 FUNDAMENTAL PARTICLES
Gravity
The force of gravity is well known to us. We feel its effect at all
times. It is the fundamental force that drives the macroscopic
behaviour of the Universe as it acts over infinite distances
and it acts between masses. On the quantum scale, gravity is
the weakest of all the four fundamental forces (typically 6 ×
Figure 2.8 Murray Gell-Mann, originator of the
10−39 of the magnitude of the strong force). Although classical,
Standard Model, quarks and gluons.
macroscopic, gravitational field theory works very well, the
quantum nature of gravity is not very well understood. The
proposed
name of the exchange particle is the graviton but, on the
Macroscopic means ‘large scale’, as
opposed to the microscopic, quantum scale.
quantum scale, the graviton would be almost impossible to observe due to
Quantum effects operate at distances less
the extremely small magnitude of the force of interaction. Huge, planetthan about 100 nm, so anything above
scale, detectors would be needed to capture the extremely rare effects of
this scale is considered macroscopic, and
graviton interactions.
classical (sometimes called Newtonian)
physics is applied.
TEST YOURSELF
TIP
In the examination, you will not
be required to recall information
regarding the gluon, Z0 or
the graviton.
24
469884_02_AQA_A-Level_Physics_018-041.indd 24
14 W
hat are the four fundamental forces? For each one, state its
exchange particle.
15 Why are the fundamental forces called ‘interactions’ on the
quantum scale?
16 Why are exchange particles sometimes called ‘virtual particles’?
17 W
hich of the following particle events is not an example of the weak
interaction?
beta− decay
alpha emission
beta+ decay
electron capture
17/04/19 9:14 AM
Feynman diagrams are pictorial ways of representing the interactions of
quantum particles. They were first introduced by Richard Feynman in
1948. Feynman realised that the interactions of particles on the quantum
scale could be represented on paper by a series of arrows and wavy lines,
following a set of simple rules. As an example, Figure 2.9 shows the
standard Feynman diagram illustrating β− radioactive decay.
p
e
νe
W
n
Figure 2.9 Feynman diagram illustrating
β− decay.
Feynman diagrams
●● Feynman diagrams
Feynman diagrams are generally read from left to right. Figure 2.9 shows
a neutron decaying into a proton and a W− exchange particle, which
subsequently decays into an electron and an electron antineutrino. This is
an example of the weak interaction and can be written as an equation:
1
n
0
→ 11 p + −01 e + ve
You can see immediately the advantage of the Feynman diagram over the
symbol equation. The Feynman diagram summarises all the parts of the
interaction, whereas the equation only tells us what goes into the interaction
and what comes out. It tells us nothing about what goes on during the
interaction.
Feynman diagram rules
●
●
●
●
●
e
e
●
Particles are represented by straight lines with arrow heads drawn on them.
Exchange particles are represented by wavy lines.
Time generally moves on the x-axis from left to right (although this is
not a hard and fast rule, and many Feynman diagrams have time running
vertically).
Particles are created and annihilated at the vertices between the lines.
Particles made up of quarks have the quark lines draw parallel and next
to each other.
Exchange particles generally transfer from left to right unless indicated
by an arrow above the wavy line.
Feynman diagram examples
γ
Two electrons scattering off each other (Figure 2.10)
e
e
Figure 2.10 Feynman diagram illustrating
two electrons meeting.
Two electrons meet, exchange photons and scatter away from each
other. The photon symbol γ indicates that this is an example of an
electromagnetic interaction.
β+ (positron) radioactive decay (Figure 2.11)
n
e
W
νe
p
Figure 2.11 Feynman diagram illustrating
β+ (positron) radioactive decay.
469884_02_AQA_A-Level_Physics_018-041.indd 25
In this case, a proton decays into a neutron and a W+ exchange particle,
which subsequently decays into a positron and an electron neutrino.
This is another example of the weak interaction, (like β− decay), and is
summarised by the equation:
1
p
1
25
→ 10 n + +01 e + ve
Electron capture (Figure 2.12)
Electron capture is another example of the weak interaction. An electron is
absorbed by a proton within a nucleus. The proton decays into a neutron
and a W+ exchange particle, which interacts with the electron forming an
17/04/19 9:14 AM
electron neutrino (as the proton acts on the electron the arrow above the
exchange particle moves left to right):
νe
n
1
p
1
W
p
+ −01 e → 10 n + ve
Electron–proton collision (Figure 2.13)
e
Figure 2.12 Feynman diagram illustrating
electron capture.
An electron and a proton collide transferring a W− exchange particle,
indicating the weak interaction; the proton decays into a neutron and the
electron decays into an electron neutrino (as the electron collides with the
proton the arrow above the exchange particle moves right to left):
1
p
1
+ −01 e → 10 n + ve
νe
n
TIP
β− decay is negative so involves e−, an antineutrino and W−, whereas β+
decay is positive so involves e+, a neutrino and W+.
W
p
e
Figure 2.13 Feynman diagram
illustrating electron–proton collision.
p
n
Proton–neutron bound by a gluon (Figure 2.14)
A gluon is exchanged between a neutron and a proton binding the two
particles together (the process repeats over and over again). Notice that the
Feynman diagram symbol for a gluon is a different wavy line from that of
the photon or the W±/ Z exchange particles. This is an example of the strong
interaction.
g
2 FUNDAMENTAL PARTICLES
n
ACTIVITY
Drawing Feynman diagrams
p
Figure 2.14 Feynman diagram illustrating
proton–neutron binding via gluon
exchange.
26
Write decay equations and draw Feynman diagrams for the
following decays.
1 A muon plus (µ+) decays into a W+ exchange particle and a muonantineutrino (v µ). The W+ then decays into a positron (e+) and an
electron neutrino (ve).
2 A kaon zero (K0) decays into a pion minus (π−) and a W+ exchange
particle, which subsequently decays into a pion plus (π+).
3 An electron neutrino and an electron antineutrino annihilate into a
gamma ray photon.
4 An electron neutrino decays into an electron and a W+ exchange
particle, which subsequently collides with a neutron producing a proton.
5 An electron and a positron annihilate to produce a gamma ray photons,
one of which then pair produces an electron and a positron pair.
●● Classification of particles
Although the Standard Model is used to describe the nature of the
matter in the observed Universe, many of the fundamental particles that
are part of the model are rarely observed on their own and, even when
they are, it is only at extremely high energy. Most of the fundamental
particles are seen in combination with others, forming particles that can
exist on their own at lower energies. Most of these composite particles
were ‘discovered’ before their constituent fundamental particles and,
469884_02_AQA_A-Level_Physics_018-041.indd 26
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Classification of particles
in the years after the Second World War, names were given to these
composite particles and the groups that they seemed to belong to.
The particles were arranged into three groups: hadrons, leptons and
exchange particles.
Matter
Leptons
Do NOT ‘feel’ the
strong force.
Are subject to the
weak force.
Are ‘fundamental’
particles, i.e. cannot
be split up into
anything else.
Exchange particles
Transmit the four forces of
matter:
Electromagnetic – Photon
Strong – Gluon
Weak – W±/Z
Gravity – Graviton
Electron – e–
Electron neutrino – νe
Muon – μ–
Muon neutrino – νμ
Tau – τ –
Tau neutrino – ντ
(plus antiparticles)
Hadrons
Are made up of
‘QUARKS’.
Feel the strong force.
Quarks
Up (u)
Down (d)
Strange (s) Charm (c)
Top (t)
Bottom (b)
(plus antiparticles)
Are fundamental
particles.
Baryons
Three quarks – qqq
Antibaryons – qqq
Mesons
Quark – Antiquark pair – qq
Examples:
Pion – �+ (ud)
Kaon – K+ (us)
(plus antiparticles)
Examples:
Proton – p (uud)
Neutron – n (udd)
(plus antiparticles)
Figure 2.15 The particle garden.
TIP
All the leptons, exchange particles and quarks make up the particles of the
Standard Model. However, the tau and tau-neutrino leptons and the charm,
top and bottom quarks are not part of the examination specification. Any
questions set on the examination papers involving these particles would
give you all the information that you need to answer the question.
TEST YOURSELF
18 Use the following list to answer parts (a) to (e).
proton pion muon photon neutron
a) Which particle is a lepton?
b) Which particles are hadrons?
c) Which particles are fundamental particles?
d) Which particle is a meson?
e) Which particle is an exchange particle?
19 State the difference between a baryon and a meson.
20 Which particles feel the weak interaction?
21 Use the following list of particle groups to answer parts (a) to (d).
lepton hadron meson baryon exchange particle.
27
Which group(s) do the following particles belong to?
a) neutron
b) kaon minus
c) W+
d) electron antineutrino
469884_02_AQA_A-Level_Physics_018-041.indd 27
17/04/19 9:14 AM
Leptons
TIP
All leptons have a lepton number
L = +1, but a charge Q = −1(e)
or 0. All antileptons have a lepton
number L = −1 and a charge
Q = +1(e) or 0.
TIP
In the examination, you will only
be set questions involving the
conservation of lepton number
in terms of electron leptons
(e−, e+, νe, νe) and muon leptons
(μ−, μ+, νμ, νμ). You do not need to
know about tau leptons.
Exchange particles aside, matter is arranged into two broad groups with
very different properties: leptons and hadrons. Leptons are fundamental
particles (and are described as part of the Standard Model). The lepton
group is made up of the electron, the muon and the tau particles, their
respective neutrinos and all their antiparticles (12 particles in total).
Leptons do not feel the strong force, but they are subject to the weak
force. All leptons are assigned a quantum number, called a lepton
number, L, which distinguishes them as leptons. All the leptons (like
the electron) have a lepton number L = +1, all the antileptons (like the
positron) have a lepton number L = −1, and all other (non-leptonic)
particles have a lepton number L = 0 (zero). In any particle interaction,
the law of conservation of lepton number holds. The total lepton
number before an interaction must be equal to the total lepton number
after the interaction.
For example, during β− radioactive decay, lepton number, L, is conserved.
1
n → 11 p + −01 e + ve
1
L: 0 → 0 + (+1) + (−1)
L: 0 → 0 + 1 − 1
✓
conserved
Remember – protons and neutrons are not leptons.
Muon decay
2 FUNDAMENTAL PARTICLES
Particle lifetime is the average time that a
particle exists from its creation to its decay.
Table 2.2 shows some examples.
28
Muons are unstable particles with a mass of about 200 times the mass of an
electron. Muons have unusually long lifetimes, of the order of 2.2 μs and
only the neutron, proton and atomic nuclei have higher lifetimes. All muons
decay via the weak interaction into three particles, one of which has to be
an electron (or a positron) and the other two particles are neutrinos. The
decay equations for the muon and the antimuon are:
μ− → e− + ve + vμ
μ+ → e+ + ve + vμ
Table 2.2 Mean lifetimes for some particles.
νμ
e
W
e
μ
Figure 2.16 Feynman diagram showing
muon decay.
Particle
proton
electron
(free) neutron
muon
π0
π+
W+
Mean lifetime
>1 × 1029 years
>4.6 × 1026 years
885.7 seconds
2.2 × 10 −6 seconds
8.4 × 10 −17 seconds
2.6 × 10 −8 seconds
1 × 10 −25 seconds
The Feynman diagram for the decay of the muon is given in Figure 2.16.
469884_02_AQA_A-Level_Physics_018-041.indd 28
17/04/19 9:14 AM
The following information is taken by kind permission
from the COSMUS website and Sergio Sciutto from
the AIRES software package:
http://astro.uchicago.edu/cosmus/projects/auger/
and
http://astro.uchicago.edu/cosmus/projects/aires/
‘The Pierre Auger Observatory in Malargue,
Argentina, is a multinational collaboration of
physicists trying to detect powerful cosmic rays
from outer space. The energy of the particles
here is above 1019 eV, or over a million times more
powerful than the most energetic particles in any
human-made accelerator. This value is about 1 J
and, as such, would warm up one cubic centimetre
of water by 0.2 °C. No one knows where these rays
come from.
Such cosmic rays are very rare, hitting an area the
size of a football pitch once every 10 000 years. This
means you need an enormous “net” to catch these
mysterious ultra high energy particles. The Auger
Classification of particles
ACTIVITY
The Pierre Auger Observatory and Project AIRES Cosmic
Ray Shower Simulations
project will have, when completed, about 1600
detectors.
Each detector is a tank filled with 11 000 litres of
pure water and sitting about 1.5 km away from the
next tank. This array on the Argentinian Pampas
will cover an area of about 3000 km2, which is about
the size of the state of Rhode Island or ten times
the size of Paris. A second detection system sits
on hills overlooking the Pampas and, on dark nights,
captures a faint light or fluorescence caused by the
shower particles colliding with the atmosphere’.
The cosmic rays that hit the atmosphere create huge
showers of particles, particularly leptons; the paths
of these particles through the atmosphere has been
modelled by Sergio Sciutto’s AIRES software and
you can investigate these particle showers using
the software and the download movie files. A good
place to start is the “Five showers” animation on the
AIRES page. Investigate the interactive software and
the movie files to find out more about how cosmic
rays produce particle interactions in the upper
atmosphere.
TEST YOURSELF
22 What is the law of conservation of lepton number?
23 The rest-mass of an electron is 9.11 × 10 −31 kg. Use this value to
estimate the rest-mass of a muon.
24 Use the law of conservation of lepton number to explain why an
interaction involving an electron and a positron annihilating and
producing two muons is not possible.
25 E xplain why an electron antineutrino is always produced during
β− emission.
Hadrons
TIP
Remember – you need to know
about the following quarks:
up (u) charge, + 23 e
down (d) charge, −
1
3
e
1
3
e
(and their antiquarks, with opposite
signs of charge).
strange (s) charge, −
469884_02_AQA_A-Level_Physics_018-041.indd 29
29
Hadrons are particles that are made up of quarks and are therefore subject
to the strong nuclear interaction. There are two sub-classes of hadrons.
Baryons, such as the proton and the neutron (and their antiparticles), are
made up of three quarks (or three antiquarks). The proton comprises uud,
with a total charge of +1e and the antiproton comprises u u d, with a total
charge of −1e.
Mesons, such as the pion and the kaon (and their antiparticles), are made
up of a quark–antiquark pair.
17/04/19 9:14 AM
TIP
Make sure you know the quark structure of the following particles:
● Proton, p, uud (and antiproton, u u d)
● Neutron, n, udd (and antineutron, u d d)
● Pion+, π+, u d
● Pion−, π−, d u
● Pion0, π0, u u or d d
(The π+ and π− are antiparticles of each other, π0 is its own antiparticle.)
● Kaon+, K+, u s
● Kaon−, K−, s u
● Kaon0, K0, d s (and antiparticle, K 0, s d)
(The K+ and K− are antiparticles of each other.)
Baryons
2 FUNDAMENTAL PARTICLES
As baryons are made up of three quarks, and there are six flavours of quark,
there are many different possible baryons. Two of these baryons, the proton and
the neutron, are well known and make up most of the mass of the Universe.
The other baryons, containing the more massive quarks, are more exotic and
are only observed at high energy inside particle detectors such as the LHC,
or high up in the atmosphere as the result of interactions of cosmic rays with
particles in the upper atmosphere. Each baryon has its own antibaryon – which
is made up of the corresponding antiquarks. For example, the sigma baryon,
Σ+, is made up of two up quarks and a strange quark, uus, and the anti-sigma
baryon, Σ + , is made up of two anti-up quarks and an anti-strange quark, u u s.
30
TIP
Remember charge is also a
quantum number like baryon
number and is also conserved in
particle interactions.
For example, during electron
capture a proton inside a nucleus
can interact with one of the
electrons surrounding the nucleus
and it can decay into a neutron
and W+ exchange particle, which
then interacts with the electron
producing an electron neutrino:
1
p
1
+ −01 e → 10 n + v e
B: + 1 + 0 → + 1 + 0 ✓ conserved
Q: + 1 + (–1) → + 0 + 0
✓ conserved
469884_02_AQA_A-Level_Physics_018-041.indd 30
The proton is the most stable and abundant baryon. Spontaneous free
proton decay has never been observed and, although some non-Standard
Model theories predict that it can happen, the predicted lifetime of the
proton is of the order of 1034 years. The current measurement of the
age of the Universe is only 13.8 billion years (13.8 × 109 years). As the
Standard Model has proved to be remarkably robust, then it seems that free
protons are stable, and all other baryons will eventually decay into protons.
Neutrons also appear to be stable within most nuclei (unless they are β−
radioactive decay emitters), but when they are isolated on their own (free)
they have a mean lifetime of 882 seconds (about 15 minutes). The vast
majority of all the other baryons have vanishingly short lifetimes, between
10−10 and 10−24 seconds.
All baryons are assigned a baryon quantum number, B. All baryons have
baryon number B = +1, all anti-baryons have a baryon number B = −1 and all
non-baryons have B = 0. Like lepton number, baryon number is also conserved
in particle interactions. The total baryon number of all particles before an
interaction must equal the total baryon number after the interaction.
As baryons have integer values of baryon number, quarks must have a baryon
number of + 13 and antiquarks have a baryon number of − 13. Protons are
baryons with a quark structure of uud, so they must have a baryon number
of + 13 + 13 + 13 = +1, and antiprotons with a quark structure of u u d must have a
baryon number of − 13 − 13 − 13 = −1.
Feynman diagrams can be drawn involving composite particles such as the
proton and the neutron (containing quarks), and they also show how the
17/04/19 9:14 AM
n
udu
νe
e
W
ud
p d
Classification of particles
quarks change during an interaction. The quarks making up the composite
particle are shown by arrowed lines drawn parallel and next to each other.
The Feynman diagram for β− decay then becomes:
Figure 2.17 Feynman diagram for β− decay.
In this example, the down quark decays into the W− exchange particle and
an up quark.
TEST YOURSELF
26 Explain why a proton can be
a hadron and a baryon.
27 What is the quark structure
of a neutron?
28 During positron emission,
the quark structure of a
proton changes. Describe
this change.
29 Write a nuclear reaction
equation for positron
emission and use the
equation to show that baryon
number is conserved.
Mesons are hadron particles made up of a
quark–antiquark pair.
The Feynman diagrams for positron emission by protons in terms of quarks
is also shown below:
n
udd
νe
e
W
ud
p u
Figure 2.18 Feynman diagram illustrating β+ (positron) radioactive decay involving
quarks.
Mesons
Meson particles were first proposed by the Japanese physicist Hideki Yukawa
in 1934 as a way of explaining the strong force holding protons and neutrons
together to make nuclei. Yukawa suggested that the strong force was due to
the proton and the neutron exchanging mesons. We now know that pions
(pi-mesons) are exchanged between protons and neutrons, but that the strong
interaction is actually due to the interaction between quarks that make up the
protons and neutrons. Pions, being made up of quarks, also feel the strong
force and are able to exist outside nucleons and so the strong interaction
between the proton and the neutron is due to the pion exchange with the
pions acting as a ‘force carrier’.
_
Mesons are made up of quark–antiquark pairs, qq, and as with baryons,
because there are six quarks and six antiquarks, there are many different
possible mesons. Most mesons are high-energy particles and are only seen
to exist inside the detectors of particle accelerators, but the pion and the
kaon are produced when high-energy cosmic rays interact with the upper
atmosphere and can be observed by high-altitude particle detectors. Mesons
have a lepton number, L = 0 (they are not leptons) and a baryon number,
B = 0 (they are not baryons).
31
TIP
Although mesons are hadrons, they are not baryons and hence their
baryon number is 0.
469884_02_AQA_A-Level_Physics_018-041.indd 31
17/04/19 9:14 AM
Pions are combinations of the up, u, and down, d, quarks and their
antiparticles. There are three types of pion:
π+ = ud
π− = d u
π0 = uu or dd
The π+/π− mesons are antiparticles of each other and the π0 is its own
antiparticle. All the pions are unstable and decay with lifetimes of the order
of 10−8 s for the π+/π− mesons and 10−16 s for the π0 meson. The π+/π− mesons
decay into muons and electron neutrinos via the weak interaction:
π+ → μ+ + vμ
π− → μ− + vμ
The Feynman diagram for π− decay is shown in Figure 2.19.
�
μ
d
W
ˉˉ
u
ν
Figure 2.19 Feynman diagram
showing pion minus decay.
The π0 meson decays into two gamma rays.
The other common meson produced by cosmic ray interactions is the
kaon. Kaons contain the strange quark (or antiquark), s, which has a
charge of − 13e. The quantum number strangeness, symbol S, is a property
possessed by particles containing the strange quark and was coined by
Murray Gell-Mann to describe the ‘strange’ behaviour of particles that are
always produced in pairs by the strong interaction but decay via the weak
interaction. To explain this, Gell-Mann suggested that strangeness was
conserved in the production of strange particles, but not in their decay.
There are four different kaons, (with their strangeness values, S):
K + = u s (S = +1)
2 FUNDAMENTAL PARTICLES
TIP
32
The K+ particle has a charge of
+1 because of the addition of the
charges on the u and s quarks,
+ 23 e + + 13 e = +1e. The same
is true for the charge on the K−
( )
( )
particle, − 23 e + − 13 e = −1e.
K − = s u (S = −1)
K 0 = d s (S = +1) and K 0 = s d (S = −1)
Kaon pair production occurs via the strong interaction (where strangeness is
conserved).
For example, during the high-energy collision of two protons, a K+/K− pair
is produced:
p + p → p + p + K+ + K−
Strangeness check:
TIP
The strange quark has
strangeness –1. The anti-strange
quark has strangeness +1.
0 + 0 = 0 + 0 + (+1) + (−1) = 0
✓
conserved
Kaons are unstable, decaying via the weak interaction with lifetimes of
about 10−8 s to 10−10 s. There are several processes by which the charged
kaons can decay:
K + → μ+ + vμ
K + → π + + π0
K + → π + + π + + π−
K − → μ− + vμ
K − → π− + π0
K − → π 0 + μ− + vμ
None of the decay products of these decays contain a strange quark, so
strangeness is not conserved.
469884_02_AQA_A-Level_Physics_018-041.indd 32
17/04/19 9:14 AM
30 W
hat is the charge and baryon number of the
following particles:
a) proton
f) positron
b) antiproton
g) uds baryon
c) neutron
h) u u s baryon
d) electron
i) dss baryon
e) antineutron
31 The Standard Model and the particle garden
arranges matter into the following groups:
A leptons
D baryons
B quarks
E mesons
C exchange particles
hich groups do the following particles
W
(represented by their symbols only) belong to?
a) p
e) μ+
b) n
f) π −
c) e −
g) u
d) γ
h) uds
Conservation laws
TEST YOURSELF
32 W
hich of the following particles are possible
baryons?
_
_
a) duu
e) d u s
b) duu
f) sdu
c) sss
g) ddd
d) dud
h) u s d
33Which of the following are impossible mesons?
a) ud
d) sd
b) d s
e) dd
c) u u
f) dd
34 What is the baryon and lepton structure of the
following atoms?
a) 21 H
d) 23
11 Na
b) 42 He
c)
14
C
6
e)
238
92 U
f)
294
118 Og
ACTIVITY
The Lancaster Particle Physics Package (LPPP)
The LPPP is an online resource for studying the
interactions of particles. The package consists of a
series of guided computer simulations that take you
through the way that particle physicists study particle
interactions experimentally. The computer simulations
are backed up with basic physics explanations of
what is going on inside each simulation. The package
contains some material studied in this book. The
package can be accessed by using the following link:
www.lppp.lancs.ac.uk
Work your way through the package; you can dip in and
out of it throughout the whole of the A-level course.
●● Conservation laws
TIP
Remember – Einstein’s energymass equation, E = mc2, shows
that, on the quantum scale, mass
and energy are interchangeable –
mass can convert to energy and
energy can convert back to mass.
It is better to talk about massenergy being conserved on a
quantum scale rather than the
conservation of mass and the
conservation of energy.
469884_02_AQA_A-Level_Physics_018-041.indd 33
Throughout this chapter you have met several different quantum number
conservation laws – properties or physical quantities that are the same after
an interaction as they are before the interaction. Three further quantities are
also always conserved in any interaction, these are:
●
●
●
charge, Q
momentum, p
mass-energy, E = mc2.
33
To these we add the quantum number conservation laws:
●
●
●
lepton number, L
baryon number, B
strangeness, S.
17/04/19 9:14 AM
With the exception of strangeness, all the other quantities are always
conserved in any interaction. Strangeness is conserved in strong interactions
but not in weak interactions (as the strange quark changes flavour).
Some examination questions ask you to decide if a given particle interaction,
usually given to you in equation form, can happen or not. Momentum and
mass-energy will always be conserved, so all you have to do is to determine
if charge, Q; lepton number, L; baryon number, B; and strangeness, S, are
conserved. (If strangeness is not conserved this could indicate that the
interaction is a weak interaction). Consider the examples shown below.
EXAMPLE
Is this particle interaction possible?
p + ve → e+ + n
Answer
A great way to do this is to construct a table similar to the one below:
Table 2.3
Conservation
quantity
Q
B
L
S
Quantity
After interaction
e+
n
Total conserved?
✓
+1
0
+1
✓
0
+1
+1
✓
−1
0
−1
✓
0
0
0
Before interaction
p
ve
Total
+1
0
+1
+1
0
+1
0
−1
−1
0
0
0
2 FUNDAMENTAL PARTICLES
In this example all the quantities are conserved, so the interaction is possible.
34
EXAMPLE
Is this interaction possible?
p + e+→ e− + Σ0 + K+
Answer
The Σ0 baryon has the following properties: Q = 0; B = +1; L = 0 and S = −1.
Using the same table as in the previous example, but adding an extra product
column:
Table 2.4
Conservation Before interaction
Quantity
After interaction
quantity
+
−
0
+
p
e
Total
e
Σ
K
Total conserved?
✗
Q
+1
+1
+2
−1
0
+1
0
✓
B
+1
0
+1
0
+1
0
+1
L
0
−1
−1
+1
0
0
+1
✗
S
0
0
0
0
−1
+1
0
✓
In this case charge, Q, and lepton number, L, are not conserved, so this
interaction is not possible.
469884_02_AQA_A-Level_Physics_018-041.indd 34
17/04/19 9:14 AM
What you need to know
TIP
ACTIVITY
The format of the table can
be modified depending on the
number of particles involved –
you can add extra columns or
take them away.
Use the conservation laws to decide if the following particle interactions
can occur. A table of some of properties of particles is also shown.
Table 2.5
Particle Charge, Q Baryon number, B Lepton number, L Strangeness, S
p
+1
+1
0
0
n
0
+1
0
0
−1
0
+1
0
e−
+1
0
−1
0
e+
0
0
+1
0
νe
νe
0
0
−1
0
0
+1
0
−1
Σ0
−1
+1
0
−1
Σ−
+1
0
0
+1
K+
1 p + π − → Σ − + K+
2 p + νe → e+ + Σ 0
3 n → p + e+ + νe
4 p + e+ → e− + Σ 0 + K+
5 n → p + e− + νe
6 π − + p → n + π 0 + νe
●● What you need to know
●
●
●
●
●
●
●
●
●
●
●
●
●
●
469884_02_AQA_A-Level_Physics_018-041.indd 35
For every type of particle there is a corresponding antiparticle.
Particles and antiparticles have: rest-mass (in MeV/c2); charge (in C) and
rest-energy (in MeV).
The positron, the antiproton, the antineutron and the electron antineutrino
are the antiparticles of the electron, the proton, the neutron and the
electron neutrino respectively.
The four fundamental interactions are: gravity, electromagnetic, weak and
strong. (The strong nuclear force is also known as the strong interaction.)
Exchange particles are used to explain forces between elementary
particles on the quantum scale.
The virtual photon is the exchange particle for the electromagnetic
interaction.
Examples of the weak interaction are β− decay, β+ decay, electron capture
and electron–proton collisions.
The W+ and W− are the exchange particles of the weak interaction.
Feynman diagrams are used to represent reactions or interactions in
terms of particles going in and out, and exchange particles.
Hadrons are particles that are subject to the strong interaction.
There are two classes of hadrons:
– baryons (proton, neutron) and antibaryons (antiproton and antineutron)
– mesons (pion, kaon)
Baryon number, B, is a quantum number that describes baryons. Baryons
have B = +1; antibaryons, B = −1; non-baryons, B = 0.
Baryon number is always conserved in particle interactions.
The proton is the only stable baryon and all other baryons will eventually
decay into protons.
35
17/04/19 9:14 AM
●
●
●
●
●
●
●
●
●
●
●
●
2 FUNDAMENTAL PARTICLES
●
●
Free neutrons are unstable and decay via the weak interaction forming a
proton, β− particle and an electron antineutrino.
Pions and kaons are examples of mesons. The pion is the exchange
particle of the strong nuclear force between baryons. The kaon is a
particle that can decay into pions.
Leptons are particles that are subject to the weak interaction.
Leptons include: electron, muon, neutrino (electron and muon types)
and their antiparticles.
Lepton number, L, is a quantum number used to describe leptons;
leptons have L = +1; antileptons, L = −1; non-leptons, L = 0.
Lepton number is always conserved in particle interactions.
Muons are particles that decay into electrons.
Strange particles are particles that are produced through the strong
interaction and decay through the weak interaction (e.g. kaons).
Strangeness (symbol S) is a quantum number to describe strange
particles. Strange particles are always created in pairs by the strong
interaction (to conserve strangeness).
Strangeness is conserved in strong interactions. In weak interactions the
strangeness can change by −1, 0 or +1.
Quarks have: charge + 13 , − 13 , + 23 , − 23 , baryon number, + 13 , − 13 and
strangeness (+1, 0 or −1).
Hadrons have the following quark structures:
__
– baryons (proton, uud; neutron, udd), antiproton, u u d, and
antineutron, u d d)
– mesons:
_
_
- pions - Pion+, π+, u d;_ Pion−, π−, d u;_Pion0, π0, u u or
_ dd
- kaons - Kaon+, K+, u s ; Kaon−, K−, su; Kaon0, K0, d s or s d.
During β− decay a d quark changes into a u quark, and during β+ decay a
u quark changes into a d quark.
Conservation laws for charge, baryon number, lepton number and
strangeness can be applied to particle interactions.
TEST YOURSELF
35 Match the pions to their correct quark structures:
π0 π+ π−
d u u u ud, u d
36 W
hich of the following pion decays is not possible?
a) π + → e+ + νe
d) π − → μ− + μ+
−
−
b) π → μ + νμ
e) π + → μ+ + νμ
0
c) π → γ + γ
36
469884_02_AQA_A-Level_Physics_018-041.indd 36
37 Identify the quark structure of the following
strange particles from the quark list below.
P uus
Q uds
_
R us
S su_
T ds
a) K+
b) K−
c) Λ0 (lamda0) baryon
17/04/19 9:14 AM
Table 2.6
Quark
Baryon number Charge/e
Strangeness
s
1
3
1
−3
−1
s
−3
1
+3
1
+1
d
1
3
−3
1
0
d
−1
+1
0
3
Identify particles X and Y.
‘Jet’ of
particles
ˉˉ
b
b
Table 2.7
ds
ˉˉt
t
Collision
happens
here
e
a) Quarks and antiquarks can combine to form
four possible mesons. Copy and complete the
table, calculating the baryon number, charge
and strangeness of the four mesons.
Quark pair
ss
Name
phi kaon0
kaon0
rho0
(anti-symmetric) (symmetric)
dd
Baryon
number
Charge/e
Strangeness
b) The phi and rho0 mesons have the same
properties in the table. Suggest a way that
these two mesons could be distinguished
from each other.
he last quark to be identified experimentally was
39 T
the top, t, quark in 1995, 18 years after it was first
added as a concept to the Standard Model.
a) Suggest why physicists were able to predict
the existence of the top quark even though it
took 18 years to observe it experimentally.
b) The top quark was observed experimentally by
the Tevatron particle accelerator at Fermilab
in the USA. The tevatron collided protons
and antiprotons with an energy of 1.8 TeV
(1.8 × 1012 eV). Although the protons and antiprotons have a high momentum, the sum of the
momentums of a colliding pair of particles – a
proton and an antiproton – is zero, because
469884_02_AQA_A-Level_Physics_018-041.indd 37
‘Jet’ of
particles
X
3
sd
they are travelling in opposite directions with
the same speed. The resulting collisions
produced top quark–antitop quark pairs that
were then observed in the particle detectors.
Explain why top quark–antitop quark pairs
move in opposite directions after they are
produced by the collision.
c) The top quark (and antiquark) has an
extremely short lifetime (5 × 10 −25 s) and
decays into a bottom quark and a W± exchange
particle. This is shown in Figure 2.20.
What you need to know
38 S
trange quarks, responsible for the long halflives of strange particles during the weak
interaction, were first proposed by Murray
Gell-Mann and George Zweig in 1964. As baryon
number had already been defined as +1 for
protons and neutrons the subsequent discovery of
quarks required them to have + 13 and − 13 baryon
numbers. Table 2.6 compares some of properties
of strange and down quarks and their antiquarks:
W
μ
νμ
Y
Figure 2.20 Feynman diagram of top–antitop quark
pair production.
40 T
he charm quark was observed experimentally
by two particle physics teams at almost the
same time in 1974. One team was based at the
Brookhaven National Laboratory (BNL) and the
other at the Stanford Linear Accelerator (SLAC).
Both teams observed a cc meson. The BNL team
called it the psi (ψ) meson and the SLAC team
named it the J meson – since then it has been
known as the J/ψ meson. The J/ψ meson has the
following properties:
Table 2.8
Charge
Baryon number
Lepton number
0
0
0
The J/ψ meson can decay in many ways; two of these
are shown below:
J/ψ → e+ + e−
J/ψ → μ+ + μ −
37
Explain, using conservation laws, how both of these
decay equations are correct. You need to consider:
● energy
● momentum
● charge
● baryon number
● lepton number.
17/04/19 9:14 AM
Practice questions
1 The group of particles known as hadrons are composed of quarks. There
are two sub-groups of hadrons: mesons and baryons.
a) What is the name of the property that defines a hadron?
(1)
b) State the quark structure of a meson.
(1)
c) State the quark structure of a baryon.
(1)
d) The antiproton is the antiparticle of the proton. State one way
that the proton and the antiproton are similar, and one way that
they are different.
(2)
e) What are the following values of an antiproton:
i)
(1)
its charge
ii) its baryon number
(1)
iii) its quark structure?
(1)
2 In the Standard Model of Matter, particles can be leptons, hadrons or
exchange particles. Here is a list of particles:
electron
proton
neutron
muon
pion
a) From the list state the name of one lepton and one hadron.
(2)
b) State one difference between leptons and hadrons.
(1)
c) State the difference in structure between baryons and mesons.
(1)
d) From the list state the name of one baryon and one meson.
(2)
2 FUNDAMENTAL PARTICLES
3 The table below shows some basic information about three hadrons.
38
Table 2.9
Sub-atomic particle Quark structure Baryon OR Meson
ud
meson
udd
Relative charge
Σ+, sigma+
a) Copy and complete the table.
+1
(3)
b) All of the sub-atomic particles shown in the table have a
corresponding antiparticle. State one example of a baryon and
its antibaryon not shown in the table and state their quark
structures.
(4)
Baryon number Strangeness
0
0
-1
e
e
c) The electron and the positron are an example of a lepton
particle–antiparticle pair. State one property of positron that is
the same as an electron, and one property that is different. (2)
4 Figure 2.21 shows two particles interacting. An offset line has
been drawn to represent the exchange particle.
a) State the name of the interaction shown and give the name
of the exchange particle drawn on the diagram.
(2)
b) In this interaction, momentum and energy are conserved.
Give the name of another quantity that is conserved.
469884_02_AQA_A-Level_Physics_018-041.indd 38
(1)
e
e
Figure 2.21 Feynman diagram
showing two particles interacting.
17/04/19 9:14 AM
A
c) Name the particles labelled A, B and C.
(3)
d) State the name of this type of particle interaction.
(1)
e) Momentum and energy are also conserved in this particle
interaction. Name two other quantities that must be conserved
for this interaction to occur and show how they are conserved.(4)
C
f) The Standard Model predicted the existence of the exchange
particle involved in this interaction before it had been
observed experimentally. Explain why it is important to test
the prediction of a scientific model experimentally.
(3)
K−
5 The
kaon is a meson with strangeness −1. It can decay into a
muon, μ−, and a muon-antineutrino, vμ, as shown in the
equation below:
B
Practice questions
Figure 2.22 shows another type of interaction. Another offset line
has been drawn to represent the exchange particle.
e
p
Figure 2.22 Another type of interaction.
K− → μ− + vμ
a) State and explain which particle interaction is responsible for
this decay.
b) Energy and momentum are conserved in this decay, as are two
other quantities. State the name of these two quantities.
(1)
(2)
c) State one property of a strange particle, such as the K− kaon, that
(1)
makes it different to a non-strange particle, such as a π− pion.
The K+ kaon is also strange. It could decay via either of the
following decay equations, however one of these decay equations
is not possible:
K+ → μ+ + vμ
K+ → π0 + π+
d) State and explain which of these decay equations is not possible. (2)
The K− kaon can also decay in the following way, producing a muon, a
muon-neutrino and a third particle, X:
K− → X + μ−+ vμ
e) State which interaction is responsible for this decay.
(1)
f) Particle X is identified as a pion. Explain why X must be
a meson.
(2)
g) State the charge on this particle.
(1)
39
6 During β− decay a neutron decays into a proton and an electron
antineutrino.
a) Write an equation describing this decay.
(1)
b) Using conservation laws, explain why an anti-electron-neutrino
is produced rather than an electron-neutrino.
(2)
c) Draw a Feynman diagram for this decay.
(3)
469884_02_AQA_A-Level_Physics_018-041.indd 39
17/04/19 9:14 AM
β+ decay involves the emission of positrons. 23
12Mg is a positron emitter,
Na
nucleus,
a
positron
and
another particle, X.
decaying into a 23
11
d) State the name of the other particle, X.
(1)
e) State whether each decay product is a baryon or a lepton.
(3)
During β+ decay, an up quark decays to a down quark and
an exchange particle, which subsequently decays into the positron
and the other particle, X.
f) State the quark structure of a neutron and a proton.
(2)
g) Draw a Feynman diagram for this decay.
(3)
νe
n
7 Figure 2.23 illustrates the particle interaction known as
electron capture.
W
a) During electron capture, charge, baryon number and lepton
number are all conserved. Show how these three quantities
are conserved.
(3)
b) The isotope potassium-40, 40
19K , is an extremely unusual
nuclide because it can decay by all three types of beta decay.
It can decay via electron capture or positron emission or β−
emission. Write an equation summarising each decay.
p
e
Figure 2.23 Feynman diagram showing
electron capture.
(6)
2 FUNDAMENTAL PARTICLES
8 The proton is an example of a hadron. The positron is an example of lepton.
40
a) State one similarity and one difference between protons and
positrons.
(2)
b) There are four forces that act between particles. Exchange
particles can be used to explain these forces. Match the correct
exchange particle to its force.
(3)
Table 2.10
Gravity
Force
Exchange particle Graviton
Strong
Photon
Weak
Gluon
Electromagnetic
W ±, Z
c) Describe how the force between a proton and a neutron varies
with the separation distance between the two particles and quote
suitable values for separation distance.
(3)
d) Positrons and protons can interact via three of the above forces.
Identify the force that cannot exist between the proton and the
positron, and explain why they cannot interact in this way.
(2)
9 Pions and muons are produced when high-energy cosmic rays interact
with gases high up in the Earth’s atmosphere. Write an account of how
the Standard Model is used to classify pions and muons into particle
groups. Your account should include the following:
●
the names of the groups of particles that pions and muons belong to
●
other examples of particles in these groups
●
details of any properties that the particles have in common
●
description of the ways that each particle can interact with
other particles.
469884_02_AQA_A-Level_Physics_018-041.indd 40
(6)
17/04/19 9:14 AM
Practice questions
Stretch and challenge
10 The Σ+ baryon has a strangeness of −1.
a) State the quark composition of this particle.
The Σ+ is unstable and can decay into a meson and another baryon
as shown in the equation below:
Σ+ → π+ + n
b) State the names and the quark structures of the two decay particles.
c) Apart from momentum and energy, which two other quantities are
conserved in this decay?
d) Name a quantity that is not conserved in this decay.
e) State the name of the particle interaction shown by the equation.
f) Both of the decay product particles of the Σ+ baryon are unstable.
Write the decay equation for the decay of the n particle.
1
11 Figure 2.24 shows the decay of a strange quark, s.
2
a) State which interaction is responsible for the decay of the strange quark.
b) Give names for particles 1, 2 and 3.
12 A π + meson has a rest mass of 139.6 MeV/c2 and a mean lifetime of
2.6 × 10−8 s. This meson decays into an antimuon and a muon neutrino.
s
a) What sort of particle interaction is involved with this decay?
Figure 2.24 Feynman
diagram showing the decay
of a strange quark.
b) Which exchange particle is involved in this decay?
c) Draw a Feynman diagram for this decay.
A particle of rest mass M decays at rest into two high-speed particles,
m1 and m2, as shown in Figure 2.25.
The relativistic version of Einstein’s energy-mass equation is given by:
E2 = c2p2 + m02 c4
d) Use this relationship to show:
3
W
m1
M
m2
momentum, p1
momentum, p2
total energy, E1
total energy, E2
Figure 2.25 Particle decay diagram.
E1 = (M2 + m12 − m22) c2 and cp1 = √(E12 − m12c4)
2M
e) Experiments have shown that the rest mass of an antimuon is
105.7 MeV/c2, and the Standard Model requires a muon neutrino to
be massless. Using this data, and the data for the pion, show that the
kinetic energy of the antimuon emitted during the decay is 4.1 MeV.
41
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3
Electrons and
energy levels
PRIOR KNOWLEDGE
●
●
●
●
●
Energy is measured in joules, J. A potential difference of 1 volt
transfers 1 joule per coulomb, C, of charge: 1 V = 1JC-1.
In the nuclear model of the atom, the atom has a central, massive
nucleus, which is positively charged. The nucleus contains protons
and neutrons and is surrounded by a cloud of orbiting electrons,
which are negatively charged.
A neutral atom has equal numbers of protons and electrons. When an
atom loses or gains an electron, it is ionised. The atom is left with an
overall positive charge if it loses electrons, or a negative charge if it
gains electrons.
Light is one form of electromagnetic radiation. Higher frequency
radiation, for example gamma rays, has a shorter wavelength
compared with lower frequency radiation, for example microwaves.
In order of decreasing wavelength, the members of the
electromagnetic spectrum are: radio waves, microwaves, infrared,
visible light, ultraviolet, X-rays, gamma rays.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 How much energy does a potential difference of 6 V transfer to 2 C
of charge?
2 A sodium atom has 11 protons. When a particular sodium atom is
ionised, it loses electrons.
a) How many electrons does a sodium atom have?
b) Is this sodium ion positively or negatively charged?
3 Explain one difference between a sodium atom and a sodium ion.
4 Put these members of the electromagnetic spectrum in order of
increasing frequency: gamma rays, infrared, visible light, microwaves.
5 Which colour of visible light has the longest wavelength?
42
Electromagnetic radiation carries information throughout the Universe to
Earth at the speed of light. Under the right conditions, all elements formed
since the Universe began, wherever they are in the Universe, can give out
the same unique patterns of light, called a spectrum. The spectrum from
a star can be analysed in detail to reveal which elements are present in the
star. Light from very distant galaxies takes billions of years to reach Earth.
When it arrives, the spectrum from each galaxy has been red shifted. When
the red shift of the light is measured, the speed and distance of each galaxy
from Earth can be calculated, helping us estimate of the age of the Universe.
There are many other uses of spectra. Using line spectra from samples
collected at the scene, forensic scientists can identify samples of illegal
drugs, or match paint chips from vehicles thought to be involved in
469884_03_AQA_A-Level_Physics_042-056.indd 42
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Photons
hit-and-run road accidents. One lucky lottery winner was able to collect
her winnings after spectral analysis proved her ticket was genuine; a
computer error had originally suggested it was fake. In this topic you will
learn how line spectra are caused, and the role of photons and electrons
in atoms.
●● Photons
Figure 3.1 The spectrum of our Sun
compared with the spectrum from
BAS11, a galaxy about 1 billion light
years away.
A quantum of energy is a small packet of
energy. The word quantum means discrete
or separate.
A photon is the name given to a discrete
packet (quantum) of electromagnetic
energy.
We often see electromagnetic radiation behaving as a wave. For example,
waves change wavelength and speed at the boundary between different
materials (refraction) and spread through gaps, or around obstacles
(diffraction). When electromagnetic waves in the same region overlap,
their amplitudes add (superposition). Waves of similar amplitude and
wavelength, in the same region, produce interference patterns, which can
be detected (see Chapter 6).
Other observations can only be explained by thinking of electromagnetic
radiation as a stream of packets (or quanta) of energy called photons. A photon
has no mass or charge and is described by its energy, wavelength or frequency.
For example, an X-ray photon can have a wavelength of 1 nm and frequency
of 3 × 1017 Hz, and a photon of yellow light has a wavelength of 600 nm and
frequency of 5 × 1014 Hz. The energy carried by a photon is discussed below.
Energy of photons
The energy of a photon is proportional to its frequency. A photon of light
carries less energy than an X-ray photon because the photon of light has a
lower frequency (this is covered further in Chapter 4).
The energy carried by each photon is calculated using:
E = hf
where E is the energy in joules
h is Planck’s constant
f is the frequency of the photon in hertz.
Planck’s constant is 6.63 × 10−34 J s.
EXAMPLE
Calculate the energy carried by a photon that has a frequency of
5.60 × 1013 Hz.
Answer
E = hf
43
= 6.63 × 10−34 J s × 5.60 × 1013 s−1
= 3.71 × 10−20 J
Since frequency × wavelength equals wave speed, you can rewrite the
equation above as E = hcλ .
469884_03_AQA_A-Level_Physics_042-056.indd 43
17/04/19 9:15 AM
TIP
EXAMPLE
When working with very large
and very small numbers, use
the brackets on your calculator
wisely. Use the powers of ten to
work out the order of magnitude
of the answer you should expect.
Calculate the energy carried by a photon that has a wavelength
of 630 nm.
Answer
E = hc
λ
−34
× 108 m s−1
= 6.63 × 10 J s × 3.0
−9
630 × 10 m
= 3.16 × 10−19 J
Intensity
The intensity of electromagnetic radiation is the energy transferred per unit
time per unit area. The intensity depends on the energy carried by photons,
the number of photons transferred each second and the
area on which they are incident. The intensity increases:
S
r
●
2r
3r
3 ELECTRONS AND ENERGY LEVELS
Figure 3.2 Doubling the distance from
the light source spreads the energy
over four times the area.
Inverse square law applies when a
quantity, such as intensity, is inversely
proportional to the square of the distance
from the source.
44
●
●
when the light source is made more powerful, so more photons are
transferred per second; for example, the intensity of light from a 100 W
bulb is greater than the intensity from a 10 W bulb if all other factors
remain the same
when each photon transfers more energy; for example, a beam of
ultraviolet photons is more powerful than a beam transferring the same
number of infrared photons per second
when the light is incident on a smaller area; for example, when you move
closer to a light source, more light energy enters your eye each second,
and you sense a greater intensity of light. Intensity follows an inverse
square law, so the intensity of light measured from a light source
quadruples if you half the distance away from the same light source.
EXAMPLE
An 11 W light bulb emits light of average wavelength 550 nm. The bulb
is 20% efficient.
Calculate the energy carried by one photon, and hence the number
of visible photons emitted each second.
Answer
Energy per photon = hc
λ
6.63
× 10−34 J s × 3 × 108 m s−1
=
550 × 10−9 m
−19
= 3.6 × 10 J
The bulb is 20% efficient.
Power output = power input × efficiency = 11 W × 0.2 = 2.2 W
Power output = number of photons emitted per second × energy per photon
2.2 W
Number of photons per second =
3.6 × 10−19 J
= 6 × 1018 photons per second.
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Electrons inside atoms can absorb photons and gain energy. They may gain
enough energy to move further from the nucleus into a higher energy level or
leave the atom altogether.
An electron volt (eV) is a unit of energy
equal to 1.6 × 10 −19 J. It is the energy
gained by an electron when it is accelerated
through a potential difference of 1 volt.
Photons
The electron volt
The energy carried by a photon and the energy gained by electrons are so
small that we use a different unit of energy for these changes. The unit of
energy is called the electron volt, eV. An electron volt is the energy needed
to move an electron through a potential difference of one volt. When charge
flows around an electrical circuit, 1 joule of work is done moving
1 coulomb of charge through a potential difference of 1 volt. In the same
way, work is done moving electrons in an electric field.
The work done in electron volts is calculated using:
W = VQ
where
W is the energy transferred in electron volts
V is the potential difference in volts
Q is the electron charge, 1.6 × 10−19 C.
Since the charge on an electron is 1.6 × 10−19 C, the energy transferred per
electron volt is 1.6 × 10−19 J.
●
●
To convert joules to electron volts, divide the energy in joules by
1.6 × 10−19 J/eV.
To convert electron volts to joules, multiply the energy in electron volts
by 1.6 × 10−19 J/eV.
EXAMPLE
Calculate the energy of an electron when it moves through a potential
difference of 6 V in electron volts and in joules.
Answer
W = VQ
=6×e
= 6 eV
One electron volt equals 1.6 × 10−19 J so the energy in joules is:
= 6 × 1.6 × 10−19
= 9.6 × 10−19 J
45
TEST YOURSELF
TIP
Check the maths chapter
(Chapter 28) if you are not
confident working with very small
or very large numbers.
469884_03_AQA_A-Level_Physics_042-056.indd 45
1 State the formula that links the energy of a photon with its frequency.
Calculate the energy carried by these photons in joules. Planck’s
constant is 6.63 × 10 −34 J s.
a) an infrared photon with a frequency of 2 × 1013 Hz
b) a visible light photon with a frequency 6 × 1014 Hz
c) an ultraviolet photon with a frequency 9 × 1015 Hz
17/04/19 9:15 AM
2 State how to convert joules into electron volts. Convert your answers
for question 1 into electron volts.
3 Calculate the frequency of a photon with the energies below.
Remember to convert electron volts to joules.
a) 5.0 × 10 −19 J
b) 2.5 eV
c) 1975 eV
4 State the formula which relates the energy of a photon to its
wavelength. Calculate the wavelength of a photon with these
energies.
a) 8.0 × 10 −19 J
b) 16 eV
c) 254 eV
5 a) Explain what is meant by an electron volt, and how to convert
between electron volts and joules.
b) Calculate the energy needed to move these charged particles
through a potential difference of 600 V:
i) an electron
ii) a helium nucleus (this contains two protons and two neutrons).
e−
Emission
hf
Photon
3 ELECTRONS AND ENERGY LEVELS
Absorption
hf
Absorbing and emitting photons
Figure 3.3 Atoms emit photons when
electrons move from a higher energy
level to a lower energy level. They move
from a lower energy level to a higher
energy level when they absorb photons.
Electrons can only occupy a small number
of separate energy levels within an atom.
Electrons in an atom move between
different energy states or levels when they
absorb or emit a photon.
Potential energy
0
Step 5
Step 4
Step 3
46
Step 2
Step 1
Figure 3.4 There are fixed values of
allowed energy when you climb stairs,
just as electrons in an atom have fixed
values of allowed energy.
469884_03_AQA_A-Level_Physics_042-056.indd 46
An electron in an atom gains and loses energy as it moves within the atom.
The electron has a combination of kinetic energy and electrostatic potential
energy. Because the electron has a negative charge and the nucleus has a
positive charge, the electron is attracted to the nucleus and work must be
done to move the electron away from the nucleus. This is why the electron
has less energy close to the nucleus.
The electron moves further from the nucleus if it gains the right amount
of energy by absorbing a photon. We say the electron moves to a higher
energy level.
If the electron drops from a higher energy level to a lower energy level,
it loses its surplus energy by emitting a photon and moves closer to the
nucleus.
Quantised energy levels
An electron can only absorb a specific amount of energy because
the possible, or allowed, energies for electrons in an atom are not
continuous. Only certain, fixed separate energy levels are allowed. These
energy levels are called quantised energy levels because they have fixed
energy values.
You come across this idea when you use stairs. You cannot climb part of a
step, but must gain or lose gravitational potential energy in precise amounts
that match the energy difference between each stair. You can gain or lose
energy in larger amounts to move between stairs further apart, but you
are still only allowed fixed values of energy. However, while stairs all have
the same spacing, the arrangements of energy levels in atoms are more
complicated.
17/04/19 9:15 AM
When electrons in an atom are in their
lowest energy state, they are in the ground
state.
When an electron in an atom moves to a
higher energy level (above the ground state)
after it has absorbed energy, it is in an
excited state. Excitation has occurred.
If all electrons in an atom are in the lowest energy state, we say the atom is
in its ground state.
If an electron, or electrons, in an atom have absorbed energy and moved to
higher energy states, we say the atom is excited. Excitation occurs when
electrons absorb exactly the right amount of energy to move to higher
energy levels. This occurs either by:
●
●
absorbing a photon with the exactly the right amount of energy to move
between two levels. A photon is not absorbed if its energy is different
from the amount needed for the electron to move between two levels
absorbing exactly the right amount of energy to move between two levels
after colliding with a free electron that has energy equal to or greater
than the energy required. The energy gained by the electron in the atom
equals the energy lost by the colliding electron. The free electron’s kinetic
energy after the collision is equal to its kinetic energy before the collision
minus the energy transferred to the excited electron in the atom.
Excitation and ionisation
●● Excitation and ionisation
Ionisation
Ionisation occurs when an atom gains or
loses an electron and becomes charged. It
has been ionised.
Ionisation energy is the energy required to
remove an electron from its ground state to
infinity, i.e. to become detached from the
atom.
TEST YOURSELF
ionisation
0.544 eV
0.85 eV
1.51 eV
3.4 eV
13.6 eV
n5
n4
second excited state
n3
first excited state
ground state
If an electron in an atom absorbs enough energy to escape the atom
completely, we say the atom is ionised. Ionisation occurs when an atom
gains or loses an electron and becomes a charged particle called an ion. The
ionisation energy is the minimum energy needed to remove the electron
from the atom completely.
n2
n1
Figure 3.5 When an electron absorbs a
photon, it moves into higher excited states
or escapes completely (ionisation). This
diagram shows a number of possible
excited states for a hydrogen atom.
6 Describe the processes by which electrons gain and lose energy in
the atom.
7 Explain the difference between the ground state and an excited state
in an atom.
8 Explain the difference between excitation and ionisation.
9 This question refers to the energy level diagram in Figure 3.5. A
free electron collides with an electron in a hydrogen atom, which
is originally in the ground state, n = 1. The ground state electron is
moved to the exited state, n = 2.
a) How much energy has been transferred to the excited electron?
Give your answer in electron volts, and in joules.
The electron now falls to the ground state by emitting a photon.
b) Calculate the wavelength of this photon.
47
ACTIVITY
Calculating the Planck constant using lightemitting diodes
A student has been asked to measure Planck’s
constant by investigating the equation:
E = hc.
λ
469884_03_AQA_A-Level_Physics_042-056.indd 47
She sets up the series circuit shown in Figure 3.6
using a variable power supply, light-emitting diode
(LED) and a 300 Ω resistor. The potential difference
across the LED is measured using a voltmeter.
She increases the potential difference across the
LED until the LED is just lit; this potential difference
17/04/19 9:15 AM
is measured by the voltmeter. She changes the LED
and repeats the experiment using different colours
of LEDs; her results are shown in Table 3.1, which
compares the wavelength of the light emitted, λ , with
the operating p.d., V.
Voltmeter
LED
300-ohm
resistor
Table 3.1
DC power
supply
Colour of LED
Figure 3.6 Circuit to measure
Planck’s constant.
The principle behind the operation of an LED is
illustrated in Figure 3.7.
Conduction
band
Electron
Emitted energy
in the form
of light
3 ELECTRONS AND ENERGY LEVELS
Valence
band
Hole
Figure 3.7 The principle behind
the operation of an LED.
A semiconductor allows electrons to occupy different
energy levels. Energy is needed for electrons to
move from the valence band to the conduction
band. This energy is supplied by applying a potential
difference to the LED. When an electron moves to
the conduction band, it leaves a positively charged
‘hole’ in the valence band. When an electron in the
conduction band drops down to the valence band and
recombines with a hole, the electron releases surplus
energy as light. The energy of these photons equals
the difference in energy between the valence band
and the conduction band.
Red
Orange
Yellow
Green
Blue
Wavelength/nm
624
602
590
525
470
p.d. when LED is
just lit /V
1.99
2.06
2.11
2.36
2.64
1 Explain why the blue LED needs a higher p.d. to
light than the red LED.
2 Using the data in Table 3.1, plot a graph of the
operating p.d., V, across the diodes against 1 . Use
λ
the gradient of the graph to calculate the Planck
constant.
Line spectra and continuous spectra
A diffraction grating is a piece of
transparent material ruled with very closely
spaced lines, used to see the diffraction of
light.
A continuous spectrum is a spectrum
where all frequencies of radiation or colours
of light are possible.
A line spectrum is a spectrum of discrete
coloured lines of light.
Light passing through a fine gap is diffracted, which means that it spreads
out after passing through the gap. A diffraction grating is made from a
transparent material with many gaps between very closely ruled lines, or
ridges. Light passing between each gap in the grating spreads and interferes
with light spreading through neighbouring lines. This process splits the
light into a spectrum of the colours it contains. The pattern of the coloured
light carries information about the light source. You will meet a more
detailed explanation of the diffraction grating in Chapter 6.
Sunlight and light from a filament bulb (or incandescent bulb) give
a continuous spread of colours that merge into each other. This is a
continuous spectrum. A bulb has a solid filament that is heated. Energy
levels in a solid overlap, so all energy changes for the electrons are allowed.
This means the electrons can emit photons with any energy, producing a
continuous spectrum when a solid is heated.
48
Figure 3.8 Comparing the emission
spectra of different light sources.
469884_03_AQA_A-Level_Physics_042-056.indd 48
Light from a fluorescent tube produces a spectrum of coloured lines
on a dark background. This pattern of lines is called a line spectrum.
The pattern of lines is characteristic for certain elements. All fluorescent
bulbs have a similar line spectrum because electrons in the mercury
vapour inside them have the same excited states regardless of the shape
of the bulb.
17/04/19 3:22 PM
An emission spectrum is a bright spectrum
seen when photons are emitted by atoms.
An emission spectrum is seen when electrons in atoms fall from higher energy
levels to lower energy levels, releasing photons. In a gas, mercury vapour for
example, electrons can only occupy a small number of discrete energy levels.
When a gas is heated, electrons absorb energy and reach higher energy levels.
However, electrons stay in these higher states (or excited states) for a short
time, before falling back to a lower energy level. When the electron falls to its
lower state, it emits a photon whose energy is equal to the difference in energy
between the two energy levels. Since only a small number of possible energy
levels exist, there is only a small number of possible transitions between them.
The photons that are emitted have specific energies, and therefore specific
wavelengths and colours. It is these specific colours that we see as lines when
the light is diffracted through a diffraction grating.
Excitation and ionisation
Emission spectra
When you heat salts in a Bunsen flame, sometimes you see different
colours. For example, compounds with copper in them emit green light,
and sodium compounds a bright yellow light. These colours are determined
by electrons falling from one energy level to another, emitting the particular
colour of light specific to that element.
Absorption spectra
4
3
2
4
3
2 1
An absorption spectrum can be seen when light shines through a gas,
and electrons in the atoms absorb photons corresponding to the possible
energy transitions. All other photons pass through, as they cannot be
absorbed. The dark lines of the spectrum correspond to the wavelengths
of the possible energy transitions for the electrons of the gas atoms. The
electrons become excited, moving from lower energy levels to higher
energy levels. In fact, the electrons then fall back to
their original energy state, releasing photons as they
Emission
do so, but the spectrum still appears to have black
lines as these photons are emitted in all directions.
Absorption
Energy
An absorption spectrum is a spectrum of
dark lines seen on a coloured background
produced when a gas absorbs photons.
n 1
Fraunhofer lines: an absorption spectrum
A schematic
representation of the
orbital energy levels
Figure 3.9 An energy level diagram.
Joseph von Fraunhofer realised that the continuous
spectrum from the Sun was overlaid with about 570
dark absorption lines, now known as Fraunhofer lines.
Processes within the Sun create a continuous
spectrum of radiation that passes through cooler
gases in the Sun’s outer layers. Electrons, which are bound to atoms in the
cooler gases, absorb photons at specific frequencies and then re-emit the
photons in all directions when returning to their ground state. This is why
we see a dark line surrounded by the rest of the continuous spectrum. The
continuous spectrum is caused by radiation emitted by free electrons.
49
TEST YOURSELF
10 A
beam of sunlight passes through the
Sun’s atmosphere. Explain why dark lines are
seen in the continuous spectrum from the
sunlight. (These lines are called Fraunhofer
lines.)
469884_03_AQA_A-Level_Physics_042-056.indd 49
11 D
escribe the difference between an emission
spectrum and an absorption spectrum.
12 E xplain whether an emission spectrum is caused
by atoms in their ground state or by atoms in
excited states.
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The energy of spectral lines
EXAMPLE
Use Figure 3.10 to calculate the
frequency of a photon released
when an electron moves from
energy level n = 6 to n = 2 in a
hydrogen atom.
Answer
The energy of the photon is the
difference in energy between level
6 and level 2, which is −0.38 eV −
(−3.41 eV) = 3.03 eV.
3 ELECTRONS AND ENERGY LEVELS
To convert energy in electron
volts to energy in joules, multiply
by the electron charge, e, 1.6 ×
10−19 C. This gives an energy of
4.85 × 10−19 J.
To find f, rearrange the equation
E = hf
E
f=
h
4.85 × 10−19 J
=
6.63 × 10−34 J s
= 7.3 × 1014 Hz
When an electron moves to a lower energy level, it emits a photon. The
energy lost by the electron equals the energy of the photon. This energy is
calculated as:
E2 − E1 = hf
where E1 and E2 are the energies of energy levels 1 and 2 (in J), h is the
Planck constant (in J s) and f is the frequency of radiation (in Hz).
Many spectral line diagrams show the energy in electron volts. One electron
volt is 1.6 × 10−19 J.
Figure 3.10 shows energy levels in a hydrogen atom.
ground state is labelled as n = 1, and the first excited energy level is
n = 2 and so on.
●The ionisation energy is 13.6 eV (moving the electron from its ground
state up to n = ∞).
●The arrows indicate the electron moves to a lower energy level, emitting
photons. An emission spectrum would be seen.
●The energy is always negative as we define the zero point of energy is
when the electron is at infinity. This energy gets lower as the electron
gets closer to the nucleus.
●The energy of the green line is the difference in energy between n = 4 and
n = 2, that is −0.85 eV − (−3.41 eV), or 2.56 eV.
● 2.56 eV is equivalent to 4.1 × 10−19 J.
●The
n=∞
n=6
n=5
n=4
n=3
–1.51 eV
n=2
–3.41 eV
n=1
–13.6 eV
Figure 3.10 Spectral lines in a hydrogen atom.
TEST YOURSELF
Use Figure 3.10 to answer these questions.
50
0 (ionisation)
–0.38 eV
–0.54 eV
–0.85 eV
13 W
hat is the ionisation energy of hydrogen in
electron volts?
14 How would the diagram be different if photons
were absorbed, not emitted?
15 Which energy levels does an electron move between
if it emits a photon with each of these energies?
a) 0.31 eV b) 1.9 eV c) 1.13 eV
469884_03_AQA_A-Level_Physics_042-056.indd 50
16 C
alculate the frequency of a photon emitted
when an electrons falls from:
a) energy level n = 5 to n = 2
b) energy level n = 6 to n = 5.
17 E xplain which of these photon energies can be
absorbed by the electron: 10.1 eV, 0.16 eV, 1.20 eV
and 0.47 eV.
17/04/19 9:15 AM
Fluorescence is when a substance absorbs
short wavelength electromagnetic radiation
and emits it as longer wavelength radiation.
If you are at school or college, the room you are in probably uses fluorescent
lights. Fluorescence in these lights occurs when electrons absorb photons
of ultraviolet radiation, and move to a higher energy level. When the excited
electrons fall back to the lower energy level, energy is released as visible light.
anode
mercury vapour
cathode
The fluorescent tube
●● The fluorescent tube
phosphor coating
Figure 3.11 The main components of a fluorescent tube.
A fluorescent tube is a type of light bulb
that gives out light when its inner coating
fluoresces.
Thermionic emission happens when
free electrons are released from a heated
filament.
Plasma is a mixture of ions and electrons
in a gas.
A fluorescent tube is a glass tube filled with mercury vapour and coated
inside with fluorescent materials called phosphors. When the light
is switched on, the cathode is heated causing thermionic emission.
Thermionic emission occurs when a heated cathode releases free electrons
from its surface. The free electrons have a range of energies. A potential
difference of 500 V, applied across ends of the glass tube, accelerates the
electrons from the cathode to the anode through the mercury vapour. If the
free electrons collide with mercury atoms inelastically, some energy may be
transferred from the free electrons to the mercury atoms. These atoms may
be ionised or excited, provided the free electrons transfer enough kinetic
energy. High-energy electrons cause ionisation, and lower energy electrons
cause excitation. As the mercury atoms in the vapour become ionised (lose
electrons), a mixture of ions and free electrons is created; this is called a
plasma. When the electrons in the excited mercury atoms return to their
ground state, they release photons of ultraviolet radiation. These photons
strike the phospors in the coating and are absorbed. The energy is re-emitted
as visible light, and some energy is transferred as heat.
TEST YOURSELF
ionised
18 D
escribe the purpose of these parts of a
fluorescent tube: electrode, mercury vapour,
phospor coating.
19 E xplain why a person viewing a fluorescent bulb
though a diffraction grating observes specific
peaks in the spectrum.
20 E xplain why the spectrum from a fluorescent
lamp and an incandescent bulb are different.
21 The diagram shows some allowed energy levels
for mercury. Use the diagram to calculate:
a) the energy of photons emitted for each of the
three electron transitions from excited states
to the ground state
b) the wavelength of the emitted photons.
10.44
8.85
8.84
7.93
7.73
6.70
51
5.46
4.89
4.67
ground state
0.00 eV
Figure 3.12 Some allowed energy levels for mercury.
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Practice questions
1 Figure 3.13 shows some energy levels, in eV, of an atom.
Energy/eV
0.1
level
n4
3.1
n3
12.4
n2
18.6
n 1 (ground state)
Figure 3.13
An emitted photon has an energy of 9.92 × 10−19 J. Which transition was
responsible for this photon?
A n = 4 to n = 1
B n = 3 to n = 1
C n = 2 to n = 1
D n = 3 to n = 2
2An atom emits light of wavelength 2.0 × 10−7 m. What is the energy, in J,
of a photon with this wavelength?
3 ELECTRONS AND ENERGY LEVELS
A 5.0 × 10−19 J
52
B 9.9 × 10−19 J
C 1.2 × 10−18 J
D 9.9 × 10−18 J
3 Figure 3.14 shows part of the energy level diagram for a hydrogen atom.
n4
n3
n2
0.85 eV
1.50 eV
3.40 eV
n1
13.60 eV
Figure 3.14
What is the ionisation energy of the atom in J?
A 3.04 × 10−19 J
B 4.08 × 10−19 J
C 2.04 × 10−18 J
D 2.18 × 10−18 J
4 The electron in a hydrogen atom (as shown above) is excited to the n = 4
level. How many different frequencies of photons would this produce in
the hydrogen emission spectrum?
A3
C 6
B4
D 7
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5 Read the following statements.
2 As the energy of electromagnetic radiation increases, its frequency
decreases.
3 An atom can be excited by emitting light energy.
4 An excited atom can return to a lower energy level by emitting light
energy.
Practice questions
1 The frequency and wavelength of light are inversely proportional.
Which of the statements are true?
A 2 and 3
C 1 and 3
B 1 and 4
D 3 and 4
6 A student is using LEDs in an electronics project. The wavelength of
the LEDs are given in the table. What is the energy of the lowest energy
photon emitted by the LEDs in the project?
Wavelength / × 10 −9 m
530
600
680
A 1.5 × 10−19 J
C 2.9 × 10−19 J
B 2.2 × 10−19 J
D 3.7 × 10−19 J
7 An electron is accelerated through a large potential difference and gains a
kinetic energy of 36 keV. What is this energy expressed in joules?
A 2.88 × 10−18 J
C 2.88 × 10−15 J
B 5.76 × 10−18 J
D 5.76 × 10−15 J
8 A photon has an energy of 4.14 keV. What is the frequency of the
photon?
A 5.86 × 1013 Hz
C 1.06 × 1018 Hz
B 1.00 × 1015 Hz
D 6.27 × 1033 Hz
9 Which of the following graphs best describes how photon energy varies
with the wavelength of the photon?
A E/eV
B E/eV
0
0
53
0
λ/nm
C E/eV
0
0
λ/nm
0
λ/nm
D E/eV
0
λ/nm
0
Figure 3.15
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10 Figure 3.16 represents the three lowest energy levels of an atom.
Which of the diagrams in Figure 3.17 best represents the emission spectrum
that could arise from electron transitions between these energy levels?
Energy
A
n3
C
Increasing wavelength
B
n2
n1
Increasing wavelength
D
Increasing wavelength
Figure 3.16
Increasing wavelength
Figure 3.17
11 Figure 3.18 shows the lowest four energy levels of a hydrogen atom.
ionisation
0.544 eV
0.85 eV
1.51 eV
3.4 eV
3 ELECTRONS AND ENERGY LEVELS
13.6 eV
54
second excited state
first excited state
ground state
n5
n4
n3
n2
n1
Figure 3.18 Energy levels of a hydrogen atom.
An electron in a hydrogen atom is excited from the ground state to the n = 2
energy level. The atom then emits a photon when the electron returns to the
ground state.
a) Explain what must happen to the electron for it to move from
the ground state to its excited state. (2)
b) Calculate the frequency of the photon and state which part of
the electromagnetic spectrum it belongs to. (3)
12 A fluorescent tube contains mercury vapour at low pressure. When
the mercury atoms are excited then fall back to their ground state
the atoms emit electromagnetic radiation.
a) What is meant by the ground state of an atom?
(1)
b) Explain how mercury atoms can become excited in a
fluorescent tube.
(3)
c) What is the purpose of the cathode in a fluorescent tube? (2)
d) Explain why only specific peaks in the spectrum are observed.
(2)
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0
Ionisation
–1.8
–3.0
–11.4
Practice questions
13 Some of the energy levels of a tungsten atom are shown in Figure 3.19.
–69.6
Energy/keV
Figure 3.19 Some of the energy levels of a tungsten atom.
a) Calculate the wavelength of a photon emitted as an electron
moves from the energy level −1.8 keV to energy level −11.4 keV. (3)
b) Calculate the frequency of the photons required to ionise the
tungsten atoms.
(3)
c) State one transition which emits a photon of a shorter wavelength
than that emitted in the transition from level n = 4 to level n = 3. (2)
d) i)
n electron collided with an atom, exciting the atom from
A
the ground state to the level −11.4 keV. The initial kinetic
energy of the incident electron is 3.2 × 10−15 J. Calculate its
kinetic energy after the collision. ii) Show whether the incident electron can ionise the atom
from its ground state. e) State the ionisation energy of tungsten in joules. Give your
answer to an appropriate number of significant figures.
(2)
(2)
(3)
14 Photons of more than one frequency may be released when electrons
in an atom are excited above the n = 2 energy level.
a) Sketch a diagram showing the lowest three energy levels in an
atom to explain why photons of more than one frequency can
be released. You do not need to include specific energy values.
(3)
b) Use your diagram to explain how many different frequencies
can be produced.
(2)
15 Fluorescent tubes are filled with low-pressure mercury vapour. Describe
the purpose of the main features of a fluorescent tube and use this to
explain how visible light is generated.
The quality of your written communication will
be assessed in this question.
469884_03_AQA_A-Level_Physics_042-056.indd 55
55
(6)
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16 Figure 3.20 shows the energy levels in a sodium atom. The
arrows show some of the transitions that occur.
energy/eV
a) State the ionisation energy of the sodium atom.
b)The most common transitions are 589.6 nm and
589 nm, giving the strongest spectral lines. What
colour do you expect a sodium discharge tube to be?
c)Use the diagram to determine the the wavelength of
light emitted for these transitions: 4s to 3p; 3d to 3p;
3p to 3s.
1.0
4d
5s
4p
4p
3d
4d
4f
3d
4s
2.0
3.0
3p
3p
d) Which transitions gives rise to a spectral line of
wavelength 620 nm?
Stretch and challenge
4.0
17 X-ray machines produce photons with wavelengths
5.0
between 0.01nm and 10 nm by colliding electrons with
3s
a tungsten metal target. X-rays with energies above 5 keV
Figure 3.20 The energy levels in a sodium atom.
are categorised as hard X-rays. These X-rays pass through
most tissues. Soft X- rays usually have energies in the range
100 eV to 5 keV, and are totally absorbed in the body. Mammograms
are routine X-rays images used to detect tumours in the breast before
symptoms are observed.
3 ELECTRONS AND ENERGY LEVELS
a) Explain whether hard X-rays or soft X-rays are most useful
for medical diagnosis, giving two reasons
18 A successful mammogram uses the lowest x-ray dose that can pass
through tissues, while still producing clear contrast between tumours
and surrounding tissue. Low energy x-rays increase contrast but are less
penetrating; higher energy x-rays reduce contrast but pass well through
tissues. Many mammograms uses energies of about 24 keV
a) Calculate the wavelength of a typical mammogram X-ray
b) Explain whether bone X-rays should have a higher or lower
wavelength than for mammograms.
56
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4
Particles of light
PRIOR KNOWLEDGE
●
●
●
●
●
●
●
●
●
●
●
Electromagnetic waves are transverse waves with a range of
wavelengths. In order of increasing frequency, members of the
electromagnetic spectrum are radio waves, microwaves, infrared,
visible, ultraviolet, X-rays, gamma rays.
When atoms are ionised, they lose or gain electrons. Electromagnetic
waves can ionise atoms when an electron absorbs a photon.
Electromagnetic waves with a higher frequency and shorter
wavelength are more ionising.
Diffraction occurs when waves spread through a gap or around an
obstacle. The wavefront is more curved when the wavelength is about
the same size as the gap, and diffraction is greatest.
Electrons are negatively charged particles. The electron’s mass is
about 2000 times less than that of a proton or neutron.
Electrical current is a flow of charged particles.
In classical mechanics, momentum is calculated using:
momentum = mass × velocity.
Velocity is a vector, so the direction of motion must be specified.
Kinetic energy = 12 mass × velocity2.
Energy is conserved. It cannot be created or destroyed, but must be
accounted for at all stages of a process.
1 electron volt = 1.6 × 10 −19 J
The energy of a photon is given by: E = hf, where h is Planck’s constant
and f the frequency of the electromagnetic radiation.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 List the members of the electromagnetic spectrum in order of
increasing frequency, giving one use for each member.
2 Sketch the shape of wavefronts diffracting through a gap when the gap is:
a) much larger than the wavelength
b) approximately equal in size to one wavelength.
3 a) Compare the kinetic energy and momentum of a proton and an
electron travelling at the same velocity.
b) Compare the kinetic energy of a proton and an electron with the
same momentum.
4 Describe the change in momentum of an atom when it bounces off a
surface and travels in the opposite direction at the same speed.
5 Describe the energy transfers that occur in a torch circuit when it is
turned on.
57
Have you ever wondered how photovoltaic (solar) cells work? Lightsensitive material in photovoltaic cells absorbs photons from sunlight.
When electrons in the material absorb this energy, the electrons are released
and can generate electrical power. Other devices, such as CD players and
bar code scanners, work in a similar way using light sensitive cells.
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Charge coupled devices (CCD) are digital picture sensors that also use
photodiodes. They are found in digital cameras, astronomical telescopes
and digital X-ray machines. A CCD device is a grid of thousands of
photodiodes. Each photodiode absorbs photons from the light shining on
it and emits electrons. More electrons are released if brighter light shines on
the photodiode. The signals from all the photodiodes combine to produce
signals that change in real-time.
All of these applications make use of quantum effects, which you will learn
about in this chapter.
–
ultraviolet
light
●● The photoelectric effect
The photoelectric effect was first observed in the 1880s. This was a highly
significant discovery because it caused scientists to think about light in an
entirely new way. At that time light was thought to be a wave, and wave
theory proved to be sufficient to explain many properties of light, such as
reflection, refraction and diffraction, which you will meet in Chapter 5.
leaf falls
immediately
Figure 4.1 Demonstrating the
photoelectric effect of ultraviolet light on a
negatively charged gold leaf electroscope.
+
ultraviolet
light
4 PARTICLES OF LIGHT
no effect
Figure 4.2 Demonstrating the
photoelectric effect of ultraviolet light on a
positively charged gold leaf electroscope.
+
or
–
bright source
of white light
The photoelectric effect can be demonstrated using the apparatus shown in
Figure 4.1. A freshly cleaned sheet of zinc is placed on top of a negatively
charged gold leaf electroscope. When a weak source of ultraviolet light is
directed towards the zinc sheet, the electroscope discharges immediately.
However, when the gold leaf electroscope is charged positively, as shown
in Figure 4.2, the electroscope remains charged. This is evidence that the
ultraviolet light knocks electrons out of the metal surface. A zinc plate with
a negative charge can release electrons but this is not possible when the
plate is positively charged. Electrons which are emitted from the surface of
a metal are called photoelectrons.
Figure 4.3 shows that a very bright white light shining on the zinc sheet
does not discharge the electroscope, nor does a very intense red laser. This
result surprised the physicists of the 1880s. They expected the electroscope
to discharge when a strong light source was incident on it. At that time,
physicists were familiar with the thermionic emission of electrons from
a heated metal wire. The electrons in such a wire gain enough energy for
them to escape from the attractive forces of the atoms in the wire. This is a
similar process to evaporation. When water molecules have enough kinetic
energy, they are able to escape from the body of the fluid.
Using wave theory, physicists had argued that shining a bright light on the zinc
sheet would be rather like heating it. After a small delay, the light falling on the
zinc would heat it, and some electrons would be emitted from the surface.
The problem was that the bright white light did not discharge the electroscope,
while the weak ultraviolet light began to discharge the electroscope immediately.
58
no effect
Figure 4.3 Demonstrating the
photoelectric effect–bright source of
white light.
A quantum of light is a small indivisible
package of energy. This is the smallest unit
of energy for a particular frequency of light.
The total energy emitted from a source of
light is a multiple of this quantum of energy.
469884_04_AQA_A-Level_Physics_057-069.indd 58
(Note: if you are conducting an experiment like this yourself this yourself,
you will need a 280 nm (UVC) lamp. Do not shine UV light on skin or eyes.)
An explanation with quantum theory
In 1900 Max Planck suggested a solution that would solve the problem
with his quantum theory. His idea was that light and other types of
electromagnetic radiation are emitted in indivisible small packets of energy,
which he called quanta (a single packet of energy is called a quantum).
You met this idea in Chapter 3, and you will recall that the quanta are also
17/04/19 9:16 AM
E = hf
where E is the energy of the photon in joules, f is the frequency of the light
in Hz, and h is the Planck constant, which has a value of 6.6 × 10 −34 J s.
This theory provides a solution to the problem of the electroscope. For
the electron to be removed from the surface of the metal, it must absorb
a certain minimum amount of energy. Planck suggested that when light
falls on to the surface of the metal, one photon interacts with one electron,
and that the photon’s energy is absorbed by the electron. The ultraviolet
photons have enough energy to knock an electron out of the metal; so, as
soon as the ultraviolet photons hit the zinc sheet in Figure 4.1, electrons
are removed from the metal and the electroscope immediately begins to
discharge. However, the photons of visible light have a lower frequency
and, when they are absorbed by an electron, the electron does not have
sufficient energy to escape from the metal surface, so the electroscope does
not discharge, as shown in Figure 4.3.
The photoelectric effect
known as photons. Planck argued that the energy of a quantum of light is
proportional to its frequency, and is given by the equation:
Further photoelectric experiments
The previous section described a simple demonstration of the photoelectric
effect using ultraviolet light and an electroscope. There are many more
experiments that can be done using different metals and a wide range of
frequencies of incident light. It is also possible to measure the kinetic energies
of the photoelectrons after they have been emitted from the surface of the
metal. Here is a summary of these experimental observations.
Threshold frequency is the minimum
frequency of light required to cause the
photoelectric effect.
●
●
Intensity is rate of energy transfer per
unit area.
A photoelectron is an electron emitted
during the photoelectric effect.
●
●
No photoelectrons are emitted if the frequency of light is below a certain
value. This is called the threshold frequency
The threshold frequency varies for different materials. Visible light
removes photoelectrons from alkali metals, and calcium and barium.
Other metals require ultraviolet radiation, which has a higher frequency
than visible light.
More photoelectrons are emitted as the intensity of light increases, but
only if the frequency of the light used is above the threshold value.
Photoelectrons, emitted from a particular metal, have a range of energies.
Their maximum kinetic energy depends on the frequency of the incident
light so long as the frequency of this light is above the threshold frequency.
TEST YOURSELF
1 a) Describe the main features of the photoelectric
effect.
b) State two experimental observations of the
photoelectric effect that were evidence that light
does not behave as a wave.
2 A 2.5 mW laser pointer is powered by one 1.5 V cell
delivering a current of 6.0 mA.
a) Calculate the efficiency of the laser pointer.
b) If the wavelength of light produced is 532 nm,
calculate the number of photons released per
second.
469884_04_AQA_A-Level_Physics_057-069.indd 59
3 Explain why this statement is incorrect: ‘The
photoelectric effect is only seen using ultraviolet
light.’
4 The threshold frequency of copper is 1.1 × 1015 Hz.
Describe what is observed if a sheet of copper,
which is placed on the top of a negatively charged
electroscope, is exposed to light of frequency:
a) 1.0 × 1015 Hz
b) 1.2 × 1015 Hz
5 a) Why was the photoelectric effect significant
when it was first observed?
59
17/04/19 9:16 AM
b) The photoelectric effect is only seen with zinc
using ultraviolet light, but can be seen when
visible light shines on sodium. Compare the
energy needed to release photoelectrons from
the different metals.
c) Explain how the kinetic energy of
photoelectrons released from sodium changes
when ultraviolet light is used instead of visible
light.
6 Explain whether a photon of ultraviolet radiation
carries more or less energy than a photon of
gamma radiation.
7 A laser pen emits red light of frequency 4.3 × 1014 Hz.
In terms of photons, explain the changes if:
a) a laser pen emitting more intense light is used
b) a laser pen emitting green light (frequency 5.5 ×
1014 Hz) with the same intensity is used.
8 Give two examples of quantities that are quantised.
A model to explain the photoelectric effect
Threshold energy is the minimum
amount of energy required to cause the
photoelectric effect.
●
Potential well is a model to help us
understand why electrons require energy to
remove them from a metal. An electron has
its least potential energy in the potential well.
●
4 PARTICLES OF LIGHT
high energy violet photon
potential well
●
electron leaves metal
electron
kinetic energy
●
60
quantum energy
work function
●
Figure 4.4 An electron can leave a
potential energy well if it absorbs
enough energy by absorbing a photon.
469884_04_AQA_A-Level_Physics_057-069.indd 60
To explain threshold energy, we can describe electrons in a metal
as being trapped in a potential well. This type of well is a well of
electrostatic attraction. The ‘depth’ of the well represents the least
amount of energy an electron requires to escape from the metal. For
photoelectrons, this amount is the threshold energy. It is a bit like a
person trying to jump out of a hole. The person can only jump out if
they make one jump that is high enough. They cannot escape from the
hole by making several small jumps. In the same way, an electron also
needs enough energy to escape from the metal, which is gained when
the electron absorbs a photon. If the energy from the photon matches or
exceeds the threshold energy then the electron escapes from the potential
well and is released as a photoelectron.
The energy needed to escape from the potential well is different for
different metals, so the threshold energy varies for these metals. For
example, in alkali metals the potential well is small, so the threshold
energy is small and photoelectrons are released using visible light. The
potential well also varies within the metal. Electrons at the surface
require less energy to escape compared to electrons deeper within the
metal, which may be bound more strongly to atoms. The threshold
energy is the minimum energy required to remove an electron from
the metal.
Planck’s formula for the energy of a photon, E = hf, explains the
threshold frequency. Below the threshold frequency, photons do not
carry enough energy to release a photoelectron so the photoelectric
effect is not observed. Electrons cannot be released by absorbing
two photons, just as the person cannot escape from a hole with two
smaller jumps. Each photon must carry at least the threshold energy
in order to release an electron.
As light intensity increases, the number of photons increases but the
energy per photon stays the same. Because there are more photons,
more electrons can be released, so long as the frequency of the
radiation is above the threshold frequency.
The photon’s energy does work to release the photoelectron and give it
kinetic energy. The maximum kinetic energy of a photoelectron is the
difference between the photon’s energy and the threshold energy. This
is why the maximum kinetic energy of photoelectrons increases with
frequency.
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9 D
escribe how threshold energy can be explained
using the potential well model for a metal. 10Use Table 4.1 to calculate the energy carried by a
photon of each colour of light in joules and in eV.
Table 4.1 Frequencies and wavelengths for photons of
different colours.
Colour
Yellow
Green
Blue
Violet
Ultraviolet
The photoelectric effect
TEST YOURSELF
11 T
he threshold energy for sodium is 2.29 eV and for
selenium is 5.11 eV. Calculate the threshold energy
of each metal in joules.
12 E xplain why the maximum kinetic energy
of photoelectrons is not proportional to the
frequency of radiation.
Frequency (1014 Hz) Wavelength (nm)
5.187
578
5.490
546
6.879
436
7.409
405
8.203
365
Einstein’s photoelectric equation
Energy from a photon is used to release a photoelectron and give the
photoelectron kinetic energy. Conservation of energy means that energy
from the photon equals the threshold energy plus kinetic energy of the
photoelectron.
Work function is the least energy needed
to release a photoelectron from a material.
This equals the threshold energy.
The work function of a material is the least energy needed to release a
photoelectron from a material. The work function has the symbol ϕ.
Applying the conservation of energy when energy, hf, is transferred from the
photon to a photoelectron:
hf = ϕ + 1 mvmax2
2
where h is Planck’s constant
f is the frequency of the radiation in hertz
ϕ is the work function of a material in joules
1 mv
2
max
2
is the maximum kinetic energy of photoelectrons, with mass m and
maximum velocity vmax.
This equation is called Einstein’s photoelectric equation. Electrons emitted
from the material’s surface have the maximum kinetic energy because energy
is not used moving to the surface. Electrons from deeper in the material have
less kinetic energy as some energy is used moving to the surface.
700 nm
1.77 eV
KEmax 0.81 eV
max velocity 5.3 105 ms1
500 nm
2.50 eV
KEmax 0.21 eV
max velocity 2.7 105 ms1
400 nm
3.1 eV
61
no
electrons
Potassium
Figure 4.5 Potassium’s work function is 2.29 eV. Surplus energy (above the work function) becomes the kinetic
energy of emitted electrons.
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TIP
Energy (joules) = energy (eV) x charge (C)
For a photon:
● E = hf, or
● E = hc/ λ
A photon of energy 1 eV has a frequency of about 2 x 1014 Hz and a
wavelength of about 10−6 m.
EXAMPLE
a) Calculate the minimum frequency of radiation that
causes the photoelectric effect for aluminium. The
work function for aluminium is 4.08 electron volts.
b) Calculate the maximum kinetic energy (in electron
volts) of a photoelectron if light of wavelength
240 nm shines on the surface.
Answer
a) Convert this energy in electron volts to joules:
4.08 eV × 1.6 × 10−19 C = 6.53 × 10−19 J
se E = hf to calculate the frequency that matches
U
this energy:
6.53 × 10−19 J
f= E =
h 6.63 × 10−34 J s
f = 9.8 × 1014 Hz
b) Calculate the energy of a photon if its wavelength is
240 nm:
hc
E = hf =
λ
6.63
× 10−34 J s × 3 × 108 m s−1
=
240 × 10−9 m
= 8.28 × 10−19 J
Convert joules to electronvolts:
8.28 × 10−19 J
E=
= 5.18 eV
1.6 × 10−19 J/eV
Using Einstein’s photoelectric equation, KEmax = E − ϕ
Subtract the work function from energy carried by the
photon to find KEmax
KEmax = 5.18 eV − 4.08 eV
= 1.10 eV
4 PARTICLES OF LIGHT
Using the work function
62
When a material’s work function is less than 3.1 eV, visible light can release
electrons. For example, the work function for calcium is 2.87 eV, potassium
is 2.29 eV, and caesium is 1.95 eV.
A semiconductor is a material with
conductivity between a metal and an
insulator.
When a material’s work function is less than 1.77 eV, infrared light can
release electrons. Semiconductors are materials that have been treated so
they have a low work function and respond to visible light and infrared
radiation. Semiconductors are found in many applications, for example in
CCD devices in digital cameras, and in photovoltaic cells.
TEST YOURSELF
13 Using your answers to questions 10 and 11:
a) For each of the metals, sodium and selenium,
explain which colours of light release
photoelectrons corresponding to the threshold
energy for each metal.
b) For each colour of light, calculate the maximum
kinetic energy in eV of photoelectrons. Refer to
the table in question 10, p. 61.
14 V isible light releases electrons from the
surface of caesium but not from iron.
Ultraviolet light releases electrons from
both metals. Explain which metal has the larger
work function.
469884_04_AQA_A-Level_Physics_057-069.indd 62
15 The work function of silver is 4.26 eV:
a) Calculate the threshold frequency for silver.
b) Ultraviolet light of frequency 1.50 × 1015 Hz
shines on a silver surface. Explain whether
photoelectrons are emitted from this metal’s
surface.
c) Calculate the maximum kinetic energy of the
photoelectrons. (Hint: calculate the energy
from incident photons first.)
16 The work function of calcium is 2.87 eV.
a) Calculate the threshold wavelength of light
that matches the work function of calcium.
17/04/19 9:16 AM
c) L
ight of wavelength 400 nm now shines on the
surface of the calcium. Calculate the maximum
kinetic energy of electrons emitted from the
surface.
Stopping potential
light
negatively
charged plate
positively
charged plate
evacuated tube
Figure 4.6 shows a photoelectric cell. When light strikes the metal surface
on the left photoelectrons are emitted for some wavelengths of light. The
photoelectrons that are emitted can be detected by a sensitive ammeter
when they reach the electrode on the right of the tube.
Electron diffraction
b) L
ight of wavelength 700 nm falls on the
surface of a piece of calcium. Discuss
whether electrons are emitted from the
surface.
By making the potential of the right-hand electrode negative with respect to
the left-hand electrode, the photoelectrons can be turned back, so that they do
not reach the right-hand electrode. At this point, the current is zero, and we
say that we have applied a stopping potential to the electrons. The stopping
potential gives a measure of the electrons’ kinetic energy, because the work
done by the electric field, eV, is equal to the photoelectron’s kinetic energy.
So the electron’s kinetic energy may be calculated from the equation:
1 mv2 = eV
stop
2
Figure 4.6 Photoelectric cell.
EXAMPLE
Use Figure 4.7 to answer these questions:
8
Vstop (V)
6
b) Calculate the work function for this material.
c) Calculate the maximum kinetic energy of
photoelectrons when the incident radiation is of
frequency 2.5 × 1015 Hz.
Answer
4
2
0
0
1
2
3
f ( 1015 Hz)
4
Figure 4.7 Graph to show how stopping
potential varies with frequency.
a) Explain why the stopping potential changes with
frequency.
a) Higher frequency photons transfer more energy to
each photoelectron. The surplus energy above the
work function is changed to kinetic energy of the
photoelectron
b) The work function equals hf when the line intercepts
the x-axis. hf = 6.63 × 10-34 J s-1 × 1 × 1015 Hz
= 6.63 × 10-19 J
c) Using the graph, the stopping potential is 6V when the
incident radiation is of frequency 2.5 × 1015 Hz. Using
KEmax = eVstop gives KEmax= e × 6V = 6 eV or 9.6 × 10-19 J
●● Electron diffraction
Since waves behave as particles, scientists wondered if particles could
behave as waves. Diffraction can be seen when waves spread round an
obstacle or through a gap and is a defining wave property. If scientists could
observe diffraction patterns using particles such as electrons, this would
prove that particles also behave as waves.
63
The wavelength of particles
If particles behave as waves, they must have a wavelength. In 1923, Louis
de Broglie suggested how to calculate the wavelength of particles using the
particle’s momentum. The momentum of a photon,
p = mc.
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The de Broglie wavelength for a particle is
the Planck constant divided by a particle’s
momentum; it represents the wavelength of
h
a moving particle, or λ = p .
Substituting p = mc in the equation E = mc2 gives
E = pc or p = Ec
Since E = hf then
hf h
p=
=
c λ
De Broglie proposed that this relationship would hold for an electron, or any
particle. He calculated that the de Broglie wavelength of electrons travelling
about 0.1 to 1% of the speed of light is similar to the wavelength of X-rays.
Diffraction using crystals
Diffraction is greatest when the wavelength of the wave is roughly equal to the
size of the gap it passes through. The wavelength of X-rays is of the same order
of magnitude as the spacing between ions in many crystals. Diffraction images
were first produced using X-rays travelling through crystals in 1912. Since the
wavelength of electrons is similar to the wavelength of X-rays, the same crystals
should diffract X-rays and electrons. In 1927, electron diffraction was seen
in experiments by Davisson and Germer, then repeated in other laboratories
around the world.
Figure 4.8 X-ray diffraction.
Figure 4.9 Electron diffraction.
4 PARTICLES OF LIGHT
ACTIVITY
Diffraction rings
In the diffraction rings experiment, a beam of
electrons is fired from an electron gun. The electrons
pass through a thin graphite screen in which the
carbon atoms are arranged in a lattice, acting as a
diffraction grating. As the electrons pass through the
lattice, they diffract and interfere. Many individual
diffraction patterns are produced, which combine to
form a single pattern of diffraction rings. A voltage
across the electron gun accelerates the electrons.
evacuated tube
64
thin graphite
target
electron gun
electrons show
particle properties
low
accelerating voltage
diffraction
rings
high
accelerating voltage
Figure 4.10 Diffraction rings apparatus.
469884_04_AQA_A-Level_Physics_057-069.indd 64
The screen glows where the electrons strike it and the
diameter of the rings can be measured. When the rings
are closer together, electrons have deviated less from
their original path so less diffraction has occurred. 1 Explain why a diffraction pattern is produced.
2 What does the diffraction pattern prove about the
spacing in the graphite lattice and the wavelength
of the electrons? 3 Calculate the kinetic energy of the electron when the
accelerating voltage is:
a) 2000 V
b) 4000 V
4 Calculate the speed of the electron for each
voltage.
5 Calculate the wavelength of the electron for
each voltage.
6 Use your answer to question 5 to explain why there
is less diffraction when the accelerating voltage is
higher.
17/04/19 9:16 AM
EXAMPLE
Calculate the wavelength of an
electron moving at 0.1% of the
speed of light. The mass of the
electron is 9.1 × 10−31 kg.
Answer
λ= h
mv
6.63 × 10−34 J s
(9.1 × 10−31kg × 3 × 105 m s−1)
= 2.4 × 10−9 m
=
The de Broglie wavelength is the wavelength of a moving particle. The
wavelength is calculated as Planck’s constant divided by the momentum of
the particle using this equation:
h
λ = mv
where λ is the de Broglie wavelength in metres
h is Planck’s constant
m is the mass of the particle in kg, v is the velocity of the particle in m s−1.
This equation shows that faster moving particles have a shorter wavelength.
This equation can also be written as:
h
λ= p
Wave–particle duality
Calculating the de Broglie wavelength
where p is the momentum of the particle, mass × velocity.
●● Wave–particle duality
Wave–particle duality is the idea that
matter and radiation can be described best
by sometimes using a wave model and
sometimes using a particle model.
Wave–particle duality is the idea that particles such as electrons and
electromagnetic radiation can behave as particles as well as waves. Electron
diffraction proves that electrons do diffract and interfere, just as waves do.
The photoelectric effect can only be explained using a particle model of
electromagnetic radiation. It was no longer possible to use the same model
to explain the behaviour of electrons in an electric field and the behaviour
of electrons when they diffract. The idea of wave–particle duality and
the quantum theory were radically different to conventional thinking of
the time. The quantum theory completely changed our understanding of
physics and led us to a better understanding of the microscopic world.
TEST YOURSELF
17 A
crystal of rock salt diffracts X-rays but not
visible light. a) State the conditions for diffraction.
b) X-rays have a wavelength of about 1 nm and
visible light has a wavelength of about 400 nm
to 700 nm. Suggest the approximate spacing in
the crystal lattice.
18 a) Compare the de Broglie wavelength for a
neutron with the de Broglie wavelength of an
electron moving at the same speed.
b) Use your answer to (a) to suggest why neutrons
do not produce diffraction patterns from the
same crystals that electron diffraction patterns
are clearly seen with, if the neutrons and
electrons are travelling at the same speed.
469884_04_AQA_A-Level_Physics_057-069.indd 65
19 Calculate the de Broglie wavelength of:
a) a 70 kg person running at 5 m s−1
b) a proton travelling at 1 × 106 m s−1. The proton
mass is 1.67 × 10 −27 kg
c) Comment on your answer to (a).
d) Suppose Planck’s constant suddenly became
100 J s. Explain what would happen to a class of
students as they left a classroom to go for lunch. 20 E xplain which of these are evidence for wave–
particle duality:
a) the diffraction of neutrons, and the knowledge
that a neutron can have kinetic energy and
momentum
b) polarisation of X-rays
c) diffraction of microwaves.
65
17/04/19 9:16 AM
Practice questions
1 Which of the following is not true about the photoelectric effect?
A For most metals UV light is needed for the photoelectric effect to occur.
B A bright light causes more electrons to be emitted than a faint light provided
the frequency is above the threshold frequency.
C Higher frequency light emits electrons with higher kinetic energies.
D A faint light contains very little energy so it takes a few minutes before
electrons are emitted from the metal it is shining on.
2 Monochromatic light of wavelength 4.80 × 10−7 m is shone on to a metal surface.
The work function of the metal surface is 1.20 × 10−19 J. What is the maximum
kinetic energy, in J, of an electron emitted from the surface?
A 2.94 × 10−19 J
C 5.34 × 10−19 J
B 4.14 × 10−19 J
D 6.25 × 1014 J
3 Electrons travelling at a speed of 5.00 × 105 m s−1 exhibit wave properties. What is
the wavelength of these electrons?
A 7.94 × 10−13 m
C 1.46 × 10−9 m
B 2.42 × 10−12 m
D 6.85 × 108 m
4 PARTICLES OF LIGHT
4 Radiation from a mercury lamp has an energy of 2.845 eV. What is the wavelength
of the radiation?
66
A 355.0 nm
C 453.5 nm
B 435.9 nm
D 554.2 nm
5 Ultraviolet light shines on a metal surface and photoelectrons are released. The
effect of doubling the intensity of the ultraviolet light while keeping the frequency
constant is to release what?
A Twice as many photoelectrons, and double their maximum kinetic energy.
B The same number of photoelectrons with double the maximum kinetic energy.
C Twice as many photoelectrons with the same maximum kinetic energy.
D The same number of photoelectrons with the same maximum kinetic energy.
6 Some electrons have a de Broglie wavelength of 1.80 × 10−10 m. At what speed are
they travelling?
A 2.20 × 103 m s−1
C 8.04 × 106 m s−1
B 4.04 × 106 m s−1
D 1.02 × 108 m s−1
7 The work function of caesium is 1.95 eV. Electrons are produced by the
photoelectric effect when a caesium plate is irradiated with electromagnetic
radiation of wavelength 434 nm. What is the kinetic energy of the emitted
photons?
A 1.46 × 10−19 J
C 2.92 × 10−19 J
B 2.35 × 10−19 J
D 4.16 × 10−19 J
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A 1.2 × 10−26 m
C 5.2 × 10−21 m
B 1.7 × 10−22 m
D 1.3 × 10−11 m
9 Monochromatic ultraviolet radiation with a photon energy of 3.2 × 10−19 J
falls on the surface of a metal. Photoelectrons are emitted from the metal
with a maximum kinetic energy of 2.1 × 10−19 J.
Practice questions
8 What is the wavelength of an electron that has been accelerated through a
potential difference of 9.0 kV?
What is the work function of the metal?
A 0.7 × 10−19 J
C 1.5 × 10−19 J
B 1.1 × 10−19 J
D 5.3 × 10−19 J
10 A proton and an electron have the same velocity. The de Broglie wavelength
of the electron is 3.2 × 10−8 m. What is the de Broglie wavelength of the
proton?
A 1.7 × 10−11 m
C 6.4 × 10−8 m
B 5.5 × 10−9 m
D 2.3 × 104 m
11 When light shines on a clean copper surface, photoelectrons with a range
of kinetic energies up to a maximum of 2.4 × 10–20 J are released. The work
function of copper is 4.70 eV.
a) State what is meant by a photoelectron.
(2)
b) Explain why the kinetic energy of the photoelectrons has a
maximum value.
(1)
c) Calculate the wavelength of the light. Give your answer to an
appropriate number of significant figures.
(4)
12 As a result of work on the photoelectric effect and electron diffraction,
a model of wave–particle duality was suggested.
a) State one effect that can only be explained if electrons have wave
properties. Details of apparatus used are not required.
(1)
b) The theory of wave–particle duality modified existing ideas of the
behaviour of particles and waves. Explain, with a relevant example,
the stages involved when an existing scientific theory can be
modified or replaced with a new theory. The quality of your written
communication will be assessed in this question.
(6)
13 When a clean metal surface in a vacuum is irradiated with ultraviolet
radiation of a high frequency, electrons are emitted from the metal.
a) Explain what is meant by the threshold frequency.
67
(2)
b) Explain why electrons may be emitted from the surface of one metal
but not from the surface of a different metal when light of the same
frequency shines on each surface.
(2)
c) Explain what is meant by the term work function, and state its unit.
469884_04_AQA_A-Level_Physics_057-069.indd 67
(1)
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14 Electrons can exhibit wave properties.
(2)
a) Explain what is meant by wave–particle duality.
b) Calculate the de Broglie wavelength of electrons travelling at a speed
(3)
of 3.2 × 104 m s–1. The mass of an electron is 9.11 × 10−31 kg.
c) The proton has a mass equal to 1836 times the mass of an electron.
Calculate the speed of protons with the same de Broglie wavelength
(2)
as the electrons in part (b).
15 Light energy of frequency 1.4 × 1015 Hz is incident on each square
centimetre of an aluminium surface at a rate of 3.0 × 10–7 J s−1.
a) Calculate the energy of an incident photon. (2)
b) Calculate the number of photons incident per second per square
centimetre of the aluminium surface.
(2)
16 Describe the main features of the photoelectric effect and explain how
these provided evidence for the particle nature of light. The quality of
your written communication will be assessed in this question.
(6)
17 A freshly cleaned zinc plate has a work function of 3.6 eV.
(2)
a) i) Explain what is meant by the term work function.
ii) Ultraviolet light of wavelength 280 nm is now shone on to
the metal surface. Show that photoelectron will be emitted
from the zinc surface.
(3)
b) The zinc plate is now charged to a potential of +2 V relative to
the surroundings. What is the minimum wavelength of light
which will cause photoelectrons to be emitted now?
(3)
4 PARTICLES OF LIGHT
Stretch and challenge
18 A teacher demonstrates an experiment to show electron diffraction (see
Figure 4.11). A metal filament is heated, releasing electrons. A high voltage
accelerates electrons through a vacuum in an evacuated tube. As the electrons
pass through the carbon disc, the carbon lattice forming the disc diffracts
them. The diffraction rings seen on the phosphor layer are caused by different
layers of atoms.
a) Explain why the electrons are diffracted as they pass
through the carbon target.
b) The electrons are accelerated through 5000 V. Calculate:
i)
68
their kinetic energy
ii) their velocity
iii) use your answer to calculate the de
Broglie wavelength of the electrons.
c) The potential difference is halved. Predict how these
quantities change:
i)
filament
carbon
disc
6.3 V ac
supply
to the
filament
phosphor
0V
cathode
5000 V
electron
beam
diffraction
rings
Figure 4.11 Using a diffraction tube to diffract
electrons. For safety, the 5000 v current is
limited to 5 mA.
kinetic energy
ii) momentum of electrons
iii) de Broglie wavelength.
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(a)
e) Suggest why a magnet held near the screen changes
the pattern seen.
19 Young’s double slit experiment was a famous experiment you will
learn about in more detail in Chapter 6. It demonstrated that
waves passing through two very narrow slits interfere with each
other, causing a pattern of alternating regions of darkness and
brightness called fringes. These patterns can only be seen
with waves.
a) Compare the images in Figure 4.12 which show the
pattern obtained by passing protons, electrons and
very low intensity light, one photon at a time, through
double slits.
(b)
Practice questions
d) When the potential difference decreases, the diameter
of the rings increases. Suggest why.
(c)
b) Suggest why the patterns look similar and how you
could use this as evidence for wave–particle duality.
c) Explain whether the slit separation would be the same
in all experiments.
Figure 4.12 Images showing the
patterns obtained when a) electrons
b) protons and c) single photons pass
through double slits.
69
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5
Waves
PRIOR KNOWLEDGE
●
●
●
●
●
●
●
●
●
Wave motion transfers energy and information without transferring
matter.
Vibrations or oscillations in transverse waves are perpendicular
to the direction of energy transfer, for example electromagnetic
waves.
Vibrations or oscillations in longitudinal waves are in the same
direction as the energy transfer; for example sound waves.
Frequency (in Hz) is the number of cycles per second; wavelength
(in m) is the distance between equivalent points in successive cycles.
Amplitude is the maximum displacement from the equilibrium
position.
The wave equation states that the speed of a wave (in m s−1) equals its
frequency (in Hz) multiplied by its wavelength (in m).
v=f×λ
Refraction is the change of wave speed and direction at a boundary.
Reflection occurs when a wave bounces off a surface. The angle of
reflection equals the angle of incidence.
Total internal reflection occurs when all wave energy reflects off the
inside surface of a material.
Diffraction is the spreading of waves around an obstacle or through a gap.
TEST YOURSELF ON PRIOR KNOWLEDGE
70
469884_05_AQA_A-Level_Physics_070-094.indd 70
1 Two transverse waves travel on a slinky. Both waves have the same
frequency but one wave has double the amplitude of the other
wave. Make a sketch to show these waves, using the same axes:
the y-axis shows the transverse displacement of the waves, and
the x-axis the distance along the slinky. Label the amplitude for
each wave.
2 Sketch two waves with the same amplitude but one wave has
double the frequency of the other wave. Label the wavelength for
each wave.
3 Write down one similarity and one difference between reflection and
refraction.
4 a) Calculate the speed of a sound wave: its wavelength is 2 m and its
frequency is 170 Hz; give your answer in m s−1.
b) Sound waves from the same source now travel in water at 1500 m s−1.
The frequency does not change. Calculate the new wavelength.
5 Draw a diagram to show how a mirror can be used to see around a
corner. Include the incident ray and the reflected ray, and label the
angle of incidence and the angle of reflection.
6 Write down two similarities and one difference between transverse
waves and longitudinal waves.
17/04/19 9:16 AM
Figure 5.1 An endoscope.
●● What are progressive waves?
A progressive wave is a wave that travels
through a substance or space, transferring
energy.
What are progressive waves?
Optical fibres transmit information using an electromagnetic wave, which
travels down a fibre by means of repeated reflections off the surface of
the fibre. This is called total internal reflection. Cable TV networks and
communications networks transmit information through optical fibres by
converting electrical signals to digital pulses of light or infrared. These travel
through the fibre by total internal reflection. Endoscopes also use optical
fibres to examine inside a patient without needing to operate. Optical fibres
inside a light cable transmit light into the patient; reflections are transmitted
back through the fibre optic cable and sent to a display screen to show a
real-time image.
Waves are caused by a vibrating source. A progressive wave is an oscillation
or vibration that transfers energy and information. Only energy is transferred,
although the substance is disturbed as the waves pass through. The particles
oscillate about their fixed positions but do not move to a different place. As
you hear your teacher speak, a sound wave travels through the air from your
teacher to your ear. The air does not move from your teacher’s mouth to your
ear, but the air oscillates backwards and forwards.
Waves are represented in two different ways:
●
●
Figure 5.2 The energy spreads out from
the centre.
ray
●
wavefront
A wave front joins points on a wave that are at the same point of the
cycle as their neighbours.
For waves caused by a stone dropped in water, the ripples are wave
fronts, and the rays radiate outwards from where the stone fell into
the water. You can see wave fronts spreading out from a point in
Figure 5.3.
A ray is an imaginary line showing the direction the wave travels in. It
joins the position of the wave source to the wave fronts.
How do particles move in waves?
Mechanical waves on a string cause particles to vibrate perpendicular to the
direction energy is transferred. The amplitude, is the maximum displacement
from the particle’s undisturbed position and the larger the amplitude is,
the more energy is transferred. The distance between wave peaks is the
wavelength, λ. λ is the distance between two equivalent points in successive
cycles.
wavelength, λ
Amplitude is the maximum displacement
from the undisturbed position (for a wave).
The distance between wave peaks is the
wavelength.
displacement/m
Figure 5.3 Water spreading out from
a ripple.
A
71
distance/m
A
wavelength, λ
Figure 5.4 Wave terminology.
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Frequency is the number of cycles per
second, measured in Hz.
Hertz is the unit of frequency, equivalent
to s−1.
Period is the time for one complete cycle,
measured in seconds.
The frequency of a wave, measured in hertz, is the number of cycles or
vibrations per second. The time per cycle is the period of a wave and is
measured in seconds. A wave with a frequency of 10 Hz has 10 cycles per
second, each one taking 0.1 seconds. Period and frequency are related using:
T=
1
f
where T = period in s
f = frequency in Hz
Since 1 Hz is a very low frequency we also measure hertz in kHz
(103 Hz), MHz (106 Hz) and GHz (109 Hz).
displacement/cm
2
1
0
0.5
1.0
1.5
2.0
2.5
time/s
–1
–2
Figure 5.5 The time period for one oscillation is 0.8 seconds.
The wave equation
TIP
Be careful when you are using the
wave equation as c can be used
to represent wave speed but it is
also often used to represent the
speed of light, 3 × 108 m s−1.
Different waves travel at very different speeds, but in each case the speed is
calculated the same way using the equation:
distance
speed =
time
For a wave, the wavelength, λ, is the distance travelled in one cycle and the
time to complete one cycle is the period, T.
This means that:
wavelength
wave speed =
period
The period is the time (in seconds) for one complete cycle, and frequency
1
is the number of cycles per second. This gives frequency, f = . Substituting
T
for 1 in the equation above gives:
5 WAVES
T
wave speed = frequency (Hz) × wavelength (m)
c (m s−1 ) = f (Hz) × λ (m)
This is known as the wave equation.
72
EXAMPLE
A microwave oven operates at a frequency of 2450 MHz and produces
waves of wavelength 12.2 cm. Use this information to calculate the
speed of microwaves.
Answer
TIP
Remember to convert units to
metres, seconds, hertz, m s−1.
469884_05_AQA_A-Level_Physics_070-094.indd 72
wave speed = frequency (Hz) × wavelength (m)
= 2450 × 106 Hz × 0.122 m
= 2.9 × 108 m s−1
17/04/19 9:16 AM
Particles along a wave that move
out of phase are at different points in their
cycle at a particular time.
Particles along a wave that move in
antiphase move in opposite directions at
the same speed. The particles have opposite
displacements from their mean position.
Particles moving in antiphase are separated
by a distance of a whole number, n, of
wavelengths plus an extra half wavelength,
nλ + λ .
2
A wave repeats its cycle at regular time intervals. The phase of a wave
describes the fraction of a cycle completed compared to the start of the
cycle. Particles in parts of a wave that are moving at the same speed in the
same direction are in phase. They are out of phase if they are at different
points in their cycle at a particular time. Particles in parts of a wave that
move in opposite directions and at exactly the same speed are moving in
antiphase, or completely out of phase.
direction of wave travel
C
20
displacement/cm
Particles along a wave that move in phase
move in the same direction with the
same speed. The particles have the same
displacement from their mean position.
Particles in phase are separated by a whole
number, n, of wavelengths, nλ.
What are progressive waves?
Phase difference
10
B
A
10
20
K undisturbed position
of slinky
D
J
E
I
F
H
G
L
M
N
P
Q
distance
along
slinky
/metres
O
Figure 5.6 The displacement of a slinky along its length. The arrows on the graph
show the direction of motion of the slinky, the wave is travelling from right to left.
One wavelength is the shortest distance between two points that are
moving in phase. Points that are a whole number of wavelengths apart
move in phase; points that are half a wavelength or 1.5 wavelengths or
2.5 wavelengths, etc. move exactly out of phase (or in antiphase) with
each other.
Figure 5.6 shows a transverse wave travelling on a slinky. In this diagram,
●
●
Phase difference is measured as a fraction
of the wave cycle between two points along
a wave, separated by a distance x.
Φ = 2πx rad
λ
points E and M are in phase as they move in the same direction at the
same speed and are at the same point in their cycle. These three points
are out of phase with all other points
points E and I are moving in antiphase, because they are moving in
opposite directions with the same speed.
Any phase difference can be measured as an angle in degrees or
radians. The motion of many waves can be described using a sine
function. Since a sine function repeats itself over an angle of 360o or
2π radians, we say the phase change of a wave during one complete
cycle is 360o or 2π radians. You can read more about radian and degree
measure in Chapter 28.
The maths of phase differences
One complete rotation involves turning
through 2π radians, so 2π (or 6.28) radians
is equivalent to 360o.
73
One complete rotation of a circle involves turning through an angle of 360o.
You can also measure this angle in radians. One complete rotation involves
turning through 2π radians, so 2π radians is equivalent to 360o.
The motion of the wave in Figure 5.6 is sinusoidal with a time period of T.
When time t, equals T, one cycle has been completed, so the value of the
angle in the sine function must be 2π radians. This happens when t = T for
the angle, 2πt.
T
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We can describe the vertical displacement of the particles in the wave using
an equation of the form:
2πt
T
Since f = T1 , this can also be written as
y = A sin
y = A sin (2πft)
where y is the vertical displacement at a time t, A is the amplitude of the
wave, and f is the frequency of the oscillations.
At a different point, the oscillations will be out of phase. We describe this
phase difference as a fraction of the angle 2π. So points that are half a
wavelength apart have a phase difference of π (points B and F, for example,
in Figure 5.6), and points three-quarters of a wavelength apart have a phase
(points C and I, for example, in Figure 5.6). In general, two
difference of 3π
2
points separated by a distance x on a wave have a phase difference of:
Φ
=
2πx
λ
where Φ is the phase difference, measured in radians.
We also talk about phase differences in terms of a fraction of a cycle. For
example, a phase difference of 14 of a cycle is a phase difference of 2π
radians, and a phase difference of 1 a cycle is a difference of π radians.
2
EXAMPLE
What is the phase difference between two points along a wave separated
by a distance of 3.5λ?
Answer
Φ=
2πx
λ
= 2π
5 WAVES
= 7π
74
(a) in phase
(b) out of phase
(c) completely out of phase
Figure 5.7
469884_05_AQA_A-Level_Physics_070-094.indd 74
3.5λ
λ
But this phase difference can be reduce to π (or half a cycle), because a
phase difference of 6π is the same as a phase difference of 0. Points that
are three wavelengths apart, on a wave train, move in phase.
Figure 5.7 show two waves travelling in the same direction. They are in
phase when, at all places, the particles in each wave move up and down
together at the same speed and in the same direction, as in Figure 5.7a.
The waves are out of phase when the particles at the same distance along
the wave train, do not move together in the same direction and with the
same speed. This is shown in Figure 5.7b.
In Figure 5.7c the particles in the each wave, at the same position on the
wave train, move exactly out of phase, or in antiphase. Now the particles on
each wave move in the opposite direction but with the same speed.
17/04/19 9:16 AM
Phase change on reflection
reflective surface
TEST YOURSELF
1 A student uses a rope, fixed at one end, to demonstrate transverse
incident wave
reflection
Figure 5.8 When a wave reflects from
a denser material, there is a phase
change of 180°.
waves. A knot is tied at the midpoint of the rope. One end of the rope
is fixed and the student can move the other end of the rope. Describe
how the displacement of the knot changes during one complete cycle,
starting from its equilibrium position.
2 a) A student reads that some radio navigation systems use waves
of frequency 15 kHz. Calculate the period of a radio wave of this
frequency.
b) A pulse of these radio waves is emitted for 0.03 s. How many
complete waves are emitted during this time?
3 A signal generator, attached to a loudspeaker, generates a sound
wave with a wavelength of 3 m. The sound wave travels at 330 m s−1
in air.
a) Calculate the frequency of the sound wave.
b) The loudspeaker from the signal generator is now placed
underwater. The frequency of the sound wave does not change.
What is the wavelength of this sound wave underwater? The speed
of sound in water is 1500 m s−1.
4 Calculate the smallest phase difference in degrees and radians
a) for two points along a wave that are 1 of a cycle out of phase
b) for two points along a wave that are
4
3
4
Longitudinal waves and transverse waves
When a wave is reflected off the surface of a denser medium, it undergoes a
phase change of 180°. This is illustrated in Figure 5.8.
of a cycle out of phase.
5 Calculate the smallest distances, in terms of wavelength, between
two points along a wave, which are:
a) π out of phase
b)
c)
π
out of phase
3
2π
out of phase.
3
6 Some students are using a slinky spring to create transverse waves.
The amplitude (maximum displacement) of the transverse wave at
point X on a slinky is 3 cm when the time is 0 s. The period of the wave
cycle is 4 s.
a) State the displacement at the point X when the time t = 1 s, 2 s,
3 s and 4 s.
b) Calculate the displacement at the point at a time of 0.5 s.
●● Longitudinal waves and transverse waves
In longitudinal waves particles vibrate
parallel to the direction that energy travels in.
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75
Sound waves, primary seismic waves and pulses moving along a slinky
spring are all examples of longitudinal waves. Longitudinal waves are
waves where the wave source vibrates parallel to the direction in which the
wave travels, so particles in the medium also vibrate parallel to the direction
of energy travel.
17/04/19 9:16 AM
In transverse waves vibrations are at right
angles to the direction that energy travels in.
Electromagnetic waves are transverse
waves all of which travel at the speed
3 × 108 m s−1 in a vacuum.
Electromagnetic waves and secondary seismic waves are examples of
transverse waves. Transverse waves are created when the wave source
vibrates at right angles to the direction the wave travels in, so vibrations
in the medium are at right angles to the direction that energy travels in.
Electromagnetic waves are transverse waves all of which travel at the
speed 3 × 108 m s−1 in a vacuum. The electric field and magnetic fields of
the electromagnetic wave vibrate at right angles to each other and to the
direction of transmission. The wavelength of an electromagnetic wave is the
shortest distance between two points where the electric field (or magnetic
field) is in phase.
E
B
electric field oscillation
(horizontal)
distance
magnetic field oscillation
(vertical)
Figure 5.9 The energy in electromagnetic waves is carried by oscillating
electric and magnetic fields.
Mechanical waves cannot travel through
a vacuum, but need a medium to travel
through.
Mechanical waves need a medium to travel through and can travel as
transverse or longitudinal waves. Seismic waves, sound waves and waves on
a string are examples of mechanical waves.
ACTIVITY
Slinky springs
5 WAVES
You can investigate longitudinal and transverse mechanical waves using
a slinky spring.
vibration at the end
direction of wave motion
oscillation
of loop
compression
76
rarefaction
‘slinky’ spring
oscillation
of loop
vibration at the end
direction of wave motion
Figure 5.10 A slinky spring.
469884_05_AQA_A-Level_Physics_070-094.indd 76
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Longitudinal waves and transverse waves
More about longitudinal waves
Sound waves are longitudinal waves produced by vibrations that move
backwards and forwards along the line of wave progression. These
vibrations produce regions of high and low pressure. The regions of high
pressure are called compressions and the regions of lower pressure are
called rarefactions. The wavelength of a longitudinal wave is the distance
between successive compressions, or the distance between successive
rarefactions.
TIP
A longitudinal wave shown on
an oscilloscope looks similar
to a transverse wave although
the processes creating it are
different. Compressions and
rarefactions in the sound wave
vibrate a microphone, which
generates an electrical signal.
The signal on the oscilloscope
shows the form of these
compressions and rarefactions.
trace shown on an oscilloscope
Figure 5.11 This diagram shows the changes in air pressure at a microphone
diaphragham, measured over a period of time.
When sound travels through a solid, energy is transferred through intermolecular or interatomic bonds. Sound travels quickly through solids
because the bonds are stiff and the atoms are close together. In gases the
energy is transferred by molecules colliding, so the speed of the sound
depends on the speed of the molecules. Sound waves travel fastest in solids
(5100 m s−1 in aluminium), less quickly in liquids (1500 m s−1 in water) and
even slower in gases (340 m s−1 in air). Sound waves are mechanical waves,
so they cannot travel through a vacuum.
Electromagnetic waves
All electromagnetic waves are similar in that they all travel at the speed
of light, and their energy is carried by oscillating electric and magnetic
fields. However, the properties of the waves vary considerably with their
wavelength so we normally consider the spectrum as seven main groups
shown in Figure 5.12.
10–12m
10–9m
10–6m
10–3m
103m
1m
77
X-rays
gamma rays
infra-red
ultra-violet
radio waves
microwaves
visible light
Figure 5.12 The electromagnetic spectrum.
Hodder Science: Physics Matters 3rd Edition
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Text: Helvetica 8/9
HODDER & STOUGHTON EDUCATIONAL
Studio: Peters & Zabransky
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Effect of electromagnetic radiation on living cells
The effect of electromagnetic radiation on living cells depends on the
intensity, duration of exposure and frequency of the radiation. Radio
waves, microwaves and infrared radiation do not cause ionisation, but
may have a heating effect. Ionising radiation includes shorter wavelength
electromagnetic radiation: ultraviolet radiation, X-rays or gamma rays.
If living cells absorb ionising radiation, DNA molecules in the cells may
be damaged, which can lead to mutations or cancer. (The ionising effect
of short wavelength radiations can be understood by the particle nature
of waves; photons of ultraviolet light, X-rays and gamma rays have
sufficient energy to remove electrons from atoms. This is explained in
Chapter 3.)
●
●
●
●
●
5 WAVES
●
78
Communication uses of electromagnetic radiation
Electromagnetic waves are used in many different situations to
transmit information. Our radios, televisions and phones all depend on
electromagnetic waves carrying programmes and messages. In deciding
which wavelength to use for a particular form of communication,
consideration must be given to how much a wave will be absorbed by
the atmosphere or how much a wave will spread out due the effects of
diffraction.
●
469884_05_AQA_A-Level_Physics_070-094.indd 78
Radio waves have little effect on living cells.
In microwave ovens, water and fat molecules in food absorb microwaves
effectively. The molecules are forced to vibrate, heating the food.
Microwave ovens used in many homes and businesses produce the
frequencies most strongly absorbed by water, 2450 MHz.
Molecules that absorb infrared radiation vibrate more, increasing their
internal energy. The heating effect is used for cooking and heating.
We can see because light-sensitive cells in the retina at the back of our
eyes absorb visible light.
Ultraviolet radiation absorbed by skin causes a tan as a molecule called
melanin develops in these cells. Exposure to ultraviolet radiation raises
the risk of skin cancer but, if a person is tanned, the melanin absorbs
some UV radiation, reducing the amount of UV reaching cells deeper
in the tissues. Skin cells exposed to ultraviolet radiation also produce
vitamin D.
Gamma rays and X-rays have similar properties, but gamma rays
come from radioactive decay while X-rays are created when electrons
strike a metal target. X-rays are used for medical imaging as they can
penetrate soft tissues but are absorbed by dense cells such as bones.
High doses of radiation kills cells, for example from exposure to
high-intensity, high-energy radiation or exposure for a long duration.
Gamma rays are used in radiotherapy and to sterilise surgical
instruments.
Radio waves transmit TV and radio programmes by superimposing
information from the programme on to a carrier wave, changing
(modulating) either its amplitude or its frequency. The receiver is tuned
to the frequency of the carrier wave and converts the signal into sound
and images. Radio telescopes are massive, land-based telescopes that
receive radio signals from bodies in space that can penetrate through
the atmosphere.
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Absorbtion of tm waves
100%
Most of the
infrared spectrum
absorbed by
atmospheric
gases (best
observed
from space)
Radio waves
observable
from Earth
Longitudinal waves and transverse waves
Gamma rays, X-rays and ultraviolet
light blocked by the upper atmosphere
(best observed from space)
Visible light
observable
from Earth,
with some
atmospheric
distortion
Long-wavelength
radio waves
blocked
50%
0%
0.1 nm
1 nm
10 nm 100 nm
1μm
10μm
100μm 1 mm
Wavelength
1 cm
10 cm
1m
10 m
100 m
1 km
Figure 5.13 Astronomical telescopes can be based on earth or in space.
●
●
●
transmitting aerial
E
Some microwave frequencies penetrate the atmosphere and are used for
satellite communication and in mobile phone networks. Microwaves
have a shorter wavelength than radio waves and diffract less, so
transmitters and receivers need a straight line of sight.
Infrared and visible radiation can carry data in optical fibres. Remote
controls often use infrared radiation as it only travels a short distance
in air.
Gamma rays, X-rays and short-wavelength ultraviolet radiation do not
penetrate the atmosphere, so space-based telescopes are needed to
investigate these wavelengths in the Universe.
Polarisation
Figure 5.14 shows a simple way to demonstrate polarisation
using laboratory equipment. In the diagram you can see
M
E
a radio aerial transmitting 30 cm radio waves, which are
M
received by another aerial placed a metre away. A highE
M
frequency (1 GHz) signal is applied to the transmitting
M
E
aerial, which causes electrons to vibrate up and down.
M
Therefore, the electromagnetic waves that are transmitted
E
only have their electric fields oscillating vertically (and their
magnetic fields oscillating horizontally). When the fields
receiving aerial
of an electromagnetic wave only oscillate in one direction,
the wave is said to be a polarised wave. In this case we
say the wave is vertically polarised, because the electric
coil, acting as receiving aerial
fields are confined to the vertical plane. (By convention
Figure 5.14 A simple way to demonstrate polarisation.
the direction of polarisation is defined as the direction of
Hodder Science: Physics Matters 3rd Edition
the
electric
field
oscillation.) We can show the polarisation of the waves by
G09.01
When
the oscillations of the wave are
Mac/eps/Illustrator 8.0/4-col & B/W s/s
turning
the
receiving
aerial: when the receiving aerial is placed vertically, as
confined
to one plane
Text: Helvetica
8/9 the wave is a
in
the
diagram,
radio
waves are detected, because they cause electrons in
polarised
wave.
For example,EDUCATIONAL
in an
HODDER
& STOUGHTON
Studio: Peters &
Zabransky
electromagnetic
wave
the electric field might the vertical aerial to oscillate up and down; when the aerial is turned to a
be confined just to the vertical plane. These
horizontal direction, no signal is detected, because the electric field is in the
waves are said to be vertically polarised.
wrong direction to make the electrons move along the aerial.
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79
17/04/19 9:16 AM
(a)
Unpolarised
(b)
Plane-polarised
vertically
(c)
Plane-polarised
horizontally
Figure 5.15 Unpolarised and
polarised light. These lines show
the direction of the oscillating
electric fields.
Light is an example of an electromagnetic wave that is usually
unpolarised when it is transmitted. Light waves are produced when
electrons oscillate in atoms producing electromagnetic waves of
frequency about 5 × 1014 Hz. Since electrons in atoms can vibrate in
any direction, the electric and magnetic fields of light oscillate in any
direction – so such light is unpolarised. Figure 5.15 shows the electric
fields of unpolarised and polarised light (the magnetic fields are omitted
for clarity): in part (a) light is unpolarised because the electric field
oscillating in any direction. In part (b) the light is vertically polarised,
and in part (c) the light is horizontally polarised.
Figure 5.16 shows unpolarised microwaves incident on a metal
grille. The electric fields of the waves have vertical and horizontal
components. The horizontal electric field components of the waves
are absorbed by the wires of the grille because they cause electrons to
oscillate along this direction. The vertical components of the electric
field are not absorbed by the grille, so the waves that emerge from
the grille are polarised with their electric field vertical. Polarisation
is a property of transverse waves only. Longitudinal waves cannot be
polarised as the particles in a longitudinal wave always oscillate along
the direction of energy transfer.
Figure 5.16 A polariser.
5 WAVES
Polarisation effects
80
If a second polarising filter is held at right
angles to the original filter, this is called
crossing the polarisers.
469884_05_AQA_A-Level_Physics_070-094.indd 80
Light is polarised when it passes through a polarising filter. The filter
only allows through electric field oscillations in one plane because the
filter absorbs energy from oscillations in all other planes. The metal
grille shown in Figure 5.16 is a good model for a polarising light
filter, such as that shown in Figure 5.17a. Long molecules of quinine
iodosulphate are lined up, and electrons in these molecules affect the
light in the same way as the microwaves are affected by the grille.
Polarised light is less intense than unpolarised light because only half
the energy is transmitted through the filter. If a second polarising filter
is held at right angles to the original filter, all oscillations are blocked
and no light is transmitted. This is called crossing the polarisers.
17/04/19 9:16 AM
polarising filters
Longitudinal waves and transverse waves
polarised light
no light
a)
polarised
reflected ray
unpolarised light
unpolarised ray
b)
Figure 5.17 a) A polarising filter and b) reflected light is sometimes polarized.
Light reflecting from a surface can be polarised by reflection. At some angles
of incidence, the only reflected rays are rays whose electric fields oscillate in
one direction.
Figure 5.18 A scene without polarising sunglasses (left) and with polarising
sunglasses (right). Polarised sunglasses do not transmit reflections from the water
surface, making it easier to see into the water.
Horizontally mounted aerial
Vertically mounted aerial
Figure 5.19 A horizontally mounted
aerial absorbs horizontally polarised
signals efficiently.
469884_05_AQA_A-Level_Physics_070-094.indd 81
Polarisation applications
Polarising sunglasses include lenses with polarising filters that are orientated
so that they reduce the glare of reflected light. When light has been reflected
it is partially polarised. The filters are orientated so that they cut out light
reflecting from horizontal surfaces, such as water and snow. This reduces
eyestrain for skiers and makes it much safer for drivers. Using polarising
filters to reduce the glare of reflected light also makes it far easier to see
into water.
81
Outdoor television and FM radio aerials must be correctly aligned for
the best reception. Transmitters generate plane-polarised electromagnetic
waves, which are picked up most effectively by a receiver with the same
alignment. It is possible to reduce interference between nearby transmitters
if one of the two transmitters is aligned vertically, and the other is aligned
horizontally as shown in Figure 5.19.
17/04/19 9:16 AM
TEST YOURSELF
7 a) Describe how you could use a
displacement
slinky spring to demonstrate the
nature of
i) transverse waves
ii) longitudinal waves.
b) Describe how you could show
that light is a transverse wave.
8 Explain how the properties of
microwaves make them suitable for
a) cooking food
b) satellite communication.
9 Explain why the alignment of a TV aerial affects
the strength of the signal received.
10 State one everyday use of a polarising filter,
explaining why it is useful.
11 Earthquakes produce transverse and longitudinal
waves that travel through the Earth. The diagram
shows the displacement of particles of rock at a
particular time for different positions of a transverse
wave.
direction of wave travel
P
Q
R
position
along wave
a) i) State the phase difference between points
P and R.
ii) Describe the difference in the motion of the
rock at points Q and R.
b) Describe the motion of the rock and point
P over the next complete cycle.
c) A seismologist detects a wave that is
polarised. Explain what the seismologist can
deduce from this.
d) The frequency of the seismic wave is 0.65 Hz and
its speed is 4.8 km s−1.
i) Calculate the time period of the wave.
ii) Calculate the wavelength of the wave.
ACTIVITY
Polarisation of glucose solution
5 WAVES
A student investigates how glucose solution affects the plane of polarisation
of light passing through it. The student is told that glucose solution rotates the
plane of polarisation of polarised light. The angle of rotation is measured using
two polarising filters, one placed below the analyser and one placed above.
Both polarising filters are aligned so transmitted light has maximum intensity
when there is no solution. Glucose solution is poured into the analyser and one
polarising filter is rotated until the light intensity is a maximum again.
82
eyepiece
rotating
analyser
scale
The student uses the same concentration of glucose each time but changes the
depth of the solution.
glass
container
liquid
Here is a table of her results.
polariser
Depth (cm)
2
4
6
8
10
Angle of rotation (1)/°
8
21
30
42
54
Angle of rotation (2)/°
11
25
29
38
52
LED
Figure 5.20 Polarisation of
glucose solution.
1 Find the mean of the student’s results
2 The student is told that the equation linking these results is:
measured angle of rotation = specific angle of rotation per
decimetre × solution depth in decimetres
Plot a suitable graph to show if her results obey this relationship.
3 Use the graph to calculate the specific angle of rotation.
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Refraction is the change in direction at a
boundary when a wave travels from one
medium to another.
incident ray
θi
glass block
θr
What is refraction?
Waves change speed as they travel from one medium into another.
When light travels at any angle other than along the normal, the change
in speed causes a change in direction. This is called refraction. All
waves refract, including light, sound and seismic waves. The frequency
of a wave does not change when it travels from one medium to another
but its wavelength does. When light travels from air into a medium
such as glass or water, the wave slows down and the wavelength gets
shorter.
●
refracted ray
●
●
Figure 5.21 Light in air travels faster
than light in the glass block. Light is
refracted at each boundary.
Angle of incidence is the angle between the
incident ray and the normal (θi in Figure 5.22).
The normal is an imaginary line at right
angles to the boundary between two
materials.
Angle of refraction is the angle between
the refracted ray and the normal
(θr in Figure 5.22).
Refraction
●● Refraction
Waves arriving along the normal do not change direction because all
parts of the wave front reach the boundary simultaneously.
Waves arriving at any other angle change direction as some parts of
the wave front arrive earlier and slow down before the rest of the
wave front.
Waves change direction towards the normal when they slow down on
entering a medium. Waves change direction away from the normal when
they speed up on entering a medium.
Figure 5.22 shows light entering three different mediums from air, all at
the same angle of incidence (the angle between the incident ray and the
normal). The light changes direction least when it enters water, and the
most when it enters diamond. This is because the light travels fastest in
water, slower in glass and, of the three mediums, the slowest in diamond.
Light refracts more when it enters diamond from air than it does when
it enters glass from air. The angle of refraction is least in diamond.
Diamond is said to have a higher refractive index than glass.
θi 45
θi 45
Air
Water
θr 32
Air
Glass
θi 45
θr 28°
Air
Diamond
θr 17
Figure 5.22 Angle of refraction for different materials. In each case the angle of
incidence is 45°, but the angle of refraction is different for each material.
Refractive index
Refractive index is the ratio of a wave’s
speed between two materials. It is normally
quoted for light travelling from a vacuum
into a material.
The refractive index, n, is the ratio of the wave’s speed between two
materials. Because refractive index is a ratio it has no units. A refractive
index is calculated using:
n = cc
s
where n is the refractive index
83
c is the speed of light in a vacuum
cs is the speed of light in the material.
469884_05_AQA_A-Level_Physics_070-094.indd 83
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c1
However, the speed of light in a vacuum is almost exactly the same as the
speed of light in air. This leads to two useful and accurate approximations:
θ1
θ2
medium 1
●
medium 2
●
the refractive index of air ≈ 1
the refractive index of a material
n≈
c2
Figure 5.23 Light refracted between
two materials.
speed of light in air
speed of light in the material
The refractive index of window glass is about 1.5, which means that the speed
of light in air is about 1.5 times faster than it is in glass. The refractive index
of water is 1.33, so light travels 1.33 times faster in air than it does in water.
When light travels from one material to another (other than air) we can
define the relative refractive index between them as follows:
c1
1n2 = c
2
where 1n2 = refractive index between materials 1 and 2
c1 is the speed of light in material 1
c2 is the speed of light in material 2.
However, if we divide the top and bottom of the equation above by the
speed of light in air, c, we get:
1n2
Since
= (c1 / c)
(c2 / c)
c1 1
c2 1
c = n1 and c = n2
it follows that:
(1 / n1) n2
=
1n2 =
(1 / n2) n1
EXAMPLE
Calculate the refractive index
when light passes from:
5 WAVES
a) water to glass
b) glass to water.
n2
n1
b) 1n2 =
1.33
1.5
= 0.89
=
84
This last example leads to an important result, which shows that the
refractive index travelling from water to glass is the reciprocal of the
refractive index when light travels from glass to water.
In general:
Answer
a) 1n2 =
The advantage of this formula is that we only need to know one refractive
index for a material, and we can calculate a new refractive index when light
passes between any pairs of materials other than air.
n1
n2
1.5
1.33
= 1.125
=
1n2
=
1
2n1
Law of refraction
For many hundreds of years scientists have studied refraction and have
been able to predict the angles of refraction inside transparent materials.
Willebrord Snellius was the first person to realise that the following ratio is
always constant for all materials:
sin θi
= constant
sin θr
where θi is the angle of incidence and θr is the angle of refraction. This is
now known as Snell’s law.
At a later date it was understood that this constant ratio is the refractive
index between the two materials that the light passes between.
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Refraction
More generally, Snell’s law of refraction is stated as:
sin θ1
1n2 =
sin θ2
where
1n2 = refractive index between materials 1 and 2
θ1 = angle of incidence in material 1
θ2 = angle of refraction in material 2
or, since
1n2 =
n2
n1
it follows that:
n2 sin θ1
=
n1 sin θ2
and
n1 sin θ1 = n2 sin θ2
where
n1 = refractive index of material 1
n2 = refractive index of material 2
EXAMPLE
The speed of light is 3.00 × 108 m s−1 in air, and 2.29 × 108 m s−1 in ice.
a) Calculate the refractive index of ice.
b) Calculate the angle of refraction for a ray of light that is incident on
the air–ice boundary at 35°.
Answer
a) refractive index =
=
c1
c2
3.00 × 108
2.29 × 108
= 1.31
b) Snell’s law states that:
sin θ1
1n2 =
sin θ2
sin 35
1.31 =
sin θ2
sin 35
θ2 = sin−1
= 26°
1.31
(
)
EXAMPLE
Light travelling in water is incident on a glass surface at an angle of 37°.
What is the angle of refraction of the light in the glass of refractive index 1.5?
Answer
85
n1 sin θ1 = n2 sin θ2
1.33 × sin 37° = 1.5 sin θ2
1.33
sin θ2 =
× sin 37°
1.5
( )
and
sin θ2 = 0.533
θ2 = 32°
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Total internal reflection
Total internal reflection of light is the
complete reflection of light at a boundary
within a material that has a higher refractive
index than its surroundings.
Critical angle is the angle of refraction for
which the angle of incidence is 90°.
EXAMPLE
The refractive index of ice is 1.31.
Calculate the critical angle for
light striking a boundary between
ice and air.
Answer
n2
n1
1
sin θc =
= 0.763
1.31
Total internal reflection is the complete reflection of waves back inside
a medium at a boundary with a second material in which the wave travels
faster. For example, light can be totally internally reflected off the inside of
a glass block in air. Light travels more slowly in glass than air. (Air has a
lower refractive index than glass.)
In the example of light passing from glass to air, the angle of refraction is
larger than the angle of incidence. When the angle of refraction reaches
90°, refracted light travels along the boundary and the angle is called the
critical angle, θc. When the angle of refraction reaches 90° all light is
refracted internally. At this point the refracted angle is 90° and sin θ = 1.
When light is in a material with refractive index n1 and is incident on a
boundary with medium which has refractive index n2, we can use Snell’s
law to predict the critical angle as follows:
n1 sin θ1 = n2 sin θ2
n1 sin θc = n2
since θ2 = 90°
sin θc =
So
sin θ2 = 1
so
sin θc =
θc = 50°
n2
n1
incident
ray
semi-circular
glass blocks
reflected
ray
reflected ray
5 WAVES
EXAMPLE
Light travelling inside flint glass
with a refractive index of 1.58 is
incident at boundary with a layer
of paraffin, which has a refractive
index of 1.44. What is the critical
angle for the glass–paraffin
boundary?
Answer
n2
n1
1.44
sin θc =
= 0.91
1.58
sin θc =
86
So
θ1
θ1
θ2
θ1
θ1
θ2 = 90 °
refracted
ray
a θ1 less than the critical
angle. Some light is
reflected
θ1
b θ1 equals the critical
angle. Light has been
refracted to 90 °. Total
internal reflection just
begins.
c θ1 greater than the
critical angle. Total
internal reflection
takes place.
Figure 5.24 Total internal reflection.
θc = 66°
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12 a) C
alculate the angle of refraction for a ray of
light travelling from water into air, when the
angle of incidence is 24°. The refractive index
of water is 1.33.
b) Calculate the critical angle for water.
13 Light travels from water into oil, approaching the
boundary at an angle of 30°. The refractive index
of the oil is 1.52 and the refractive index of water
is 1.33.
a) Calculate the angle the refracted ray makes
with the boundary.
b) Calculate the ratio speed of light in water
(cw) : speed of light in oil (co).
14 W hen light passes from air into a crown glass
prism, it disperses into a visible spectrum from
red to violet light. The refractive index of red
light in crown glass is 1.509 and the refractive
index of violet light is 1.521. Calculate the
difference in refracted angle for each colour
if the light originally approaches the boundary
at 40°.
15 Diamond has a refractive index of 2.4.
a) Calculate the critical angle of diamond.
b) Explain what a jeweller must do to a diamond
to make it sparkle.
c) Calculate the critical angle for rays leaving a
diamond when it is submerged in water with a
refractive index of 1.33.
16 A light ray leaves a fish in a fish tank and is
incident on the glass aquarium wall at an angle
of 40°. At what angle of refraction does the ray
emerge into the air on the other side of the glass?
The refractive index of water is 1.33.
17 Read the following passage then answer the
questions below.
Figure 5.25 shows a cross-section of the
Earth with seismic waves spreading out from
an earthquake at the top of the diagram. Two
types of wave are shown: p-waves (primary
or pressure waves) are longitudinal waves;
s-waves (secondary or shear waves) are
transverse waves. The Earth has three major
layers: the solid mantle is its outer layer; below
that is the outer core, which is molten or liquid
rock; and the deepest layer is the solid inner
core. Waves are refracted as they pass into the
Earth’s centre, and there is marked change of
direction when the waves pass from the mantle
into the outer core.
469884_05_AQA_A-Level_Physics_070-094.indd 87
a) Explain why pressure waves can travel through
the outer core but shear waves cannot.
b) i) Explain how the diagram shows that
p-waves slow down when they pass from
the mantle to the outer core.
ii) Explain how the diagram shows that both
p-waves and s-waves travel more quickly
as they travel deeper into the mantle.
c) P-waves travel with a velocity of 12.5 km s−1 as
they reach the mantle/outer-core boundary.
Use the information in the diagram to calculate
the speed of the p-waves as they enter the
outer core from the mantle.
d) Calculate the critical angle for p-waves
as they pass from the outer core into the
mantle. Comment on the angles you see in the
diagram.
e) A p-wave with a frequency of 0.2 Hz reaches
the mantle/outer-core boundary. Use the
information in part (c) to calculate the
wavelength of the wave.
f) The diagram shows a region that is called
the ‘shadow zone’. The theory of refraction
predicts that no seismic waves will reach
this zone. However, weak seismic waves are
detected in this zone. A seismologist suggests
waves may reach the shadow zone due to
diffraction. Explain how this might happen.
Refraction
TEST YOURSELF
epicentre
(the point on the crust immediately
above the focus of the earthquake)
P and S
waves
P and S
waves
42˚
90˚
23˚
37˚
23˚
103˚
103˚
37˚
87
42˚
90˚
140˚
140˚
Figure 5.25 Seismic waves.
17/04/19 9:17 AM
Optical fibres
Optical fibre is a thin glass (or plastic) fibre
that transmits light.
Step index optical fibre is an optical fibre
with a uniform refractive index in the core
and a smaller uniform refractive index for
the cladding.
An optical fibre is a thin glass (or plastic) fibre that transmits light or
infrared radiation. These waves travel through the glass but are trapped
inside by repeated total internal reflection. This is possible if each reflection
has an angle of incidence larger than the critical angle. The critical angle
depends on the ratio between the refractive index of the optical fibre and its
cladding (coating).
A step index optical fibre has a central core with a uniform refractive
index, while the cladding has a different, smaller, refractive index. By
choosing a suitable material for the core and cladding, only certain
wavelengths of light or infrared radiation can travel through the fibre by
total internal reflection.
Material and modal dispersion
A spectrum seen using a prism is caused by dispersion. Different colours
of light travelling through the glass slow down by different amounts. The
refractive index varies with wavelength. A similar effect happens inside
optical fibres. As a signal travels within an optical fibre, it disperses. Two
types of dispersion occur in step index optical fibres:
low refractive
index cladding
●
high refractive
index core
material dispersion occurs because the refractive index of the optical
fibre varies with frequency. Different wavelengths of light in the signal
travel at different speeds. This causes a sharp pulse to spread into a
broader signal. Therefore, the duration of each pulse increases. This is
called pulse broadening. Pulse broadening is a problem because it limits
the maximum frequency of pulses and therefore the bandwidth available.
Figure 5.26 Optical fibre.
5 WAVES
Material dispersion is the spreading of a
signal caused by the variation of refractive
index with wavelength.
Pulse broadening occurs when the duration
of a pulse increases as a result of dispersion
in an optical fibre.
Figure 5.27 Material dispersion: different frequencies have a different
refractive index.
Modal dispersion is the spreading of a
signal caused by rays taking slightly different
paths in the fibre.
●
88
modal dispersion occurs when rays inside an optical fibre take
slightly different paths. Rays taking longer paths take longer to travel
through the fibre, so the duration of the pulse increases and the pulse
broadens. Modal dispersion is significant in multimode fibres, because
these fibres are broad enough to allow rays to take different paths. For
communications, monomode fibres are used. These have a very narrow
core, so that light is very nearly confined to one single path along the
axis of the cable.
travels further
travels less far
Figure 5.28 Modal dispersion: rays can take more than one path in the fibre.
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Absorption occurs when energy from a
signal is absorbed by the optical fibre in
which it travels.
Some wavelengths of light are absorbed strongly in materials that are used
to make optical fibres, so the signal strength falls. It is important to make
an optical fibre from a material with low absorption at the wavelength used
to send signals. The wavelengths commonly used are 650 nm, 850 nm and
1300 nm. It may also be necessary to amplify the signal if it travels long
distances through the optical fibre.
Refraction
Absorption
TEST YOURSELF
18 O
ptical fibres have a core surrounded by cladding made from a different material.
a) Describe the function of the core and of the cladding for an optical fibre.
b) Suggest at least one suitable property for each material.
normal
19 The refractive index of the cladding of an optical fibre is 1.52,
and the core has a refractive index of 1.62.
Calculate the critical angle for light incident on the core–
75°
cladding boundary.
glass Y
glass X (1.73)
20 Describe one similarity and one difference between modal
θ
dispersion and material dispersion.
21 a) Describe one way an optical fibre can be designed to
minimise multipath dispersion.
normal
b) Explain why it is important to minimise pulse broadening.
normal
20°
22 Figure 5.29 shows three clear pieces of glass, X, Y and Z, which
are all joined together. Each piece of glass has a different
refractive index. A ray of light passes through the blocks as
shown.
a) State two conditions necessary for light to undergo total
glass Z (1.36)
internal reflection at a boundary between two transparent
media.
Figure 5.29 Multiple refractions.
b) The refractive index of glass X is 1.73. Calculate the speed of
light in this glass.
c) Show that angle θ is about 34°.
d) The refractive index of glass Z is 1.36. Calculate the critical
angle between glass X and glass Z.
e) Explain what happens to the ray once it strikes glass Z.
89
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Practice questions
1 A prism for which all angles are 60° has a refractive index of 1.5. What
angle of incidence at the first face will just allow total internal refraction
at the second face?
A 48°
C 28°
B 42°
D 60°
2 Two polarising filters are aligned to transmit vertically polarised light.
They are held in front of a source of horizontally polarised light. The
filter closest to the light source is rotated by 45°. The intensity of light
passing through the filters
A does not change
C increases to maximum intensity
B increases
D decreases
3 Electromagnetic radiation and sound waves are two types of waves.
Which statement below correctly describes both of them?
A progressive waves that can be polarised and reflected
B transverse waves that can be polarised and refracted
C longitudinal waves that can be refracted and reflected
air
40 °
D progressive waves that can be refracted and reflected
4 Figure 5.30 shows the path of light travelling from air, through
water and into glass. The refractive index of glass is 1.50.
θw
5 WAVES
The refracted angle in glass, θg, is
90
A 25°
C 40°
B 30°
D 66°
5 A ship is stationary on the ocean. As the ocean waves move past
the ship, the front of the ship rises and falls with a period of 10 s.
When the crest of a wave is under the front of the ship, an adjacent
trough is under the back of the ship. The ship is 60 m long. What
is the speed of the ocean waves relative to the sea bed?
A 6 m s−1
C 70 m s−1
B 12 m s−1
D 120 m s−1
water
θg
glass
Figure 5.30 Light travelling from air,
through water and into glass.
6 A wave of frequency 2.5 Hz travels along a string with a speed of
20 m s−1. What is the phase difference between the oscillations at points
2 m apart along the string?
A
π
4
rads
C π rads
B
π
2
rads
D 2π rads
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displacement
direction of wave travel
Which line shows the correct phase difference between A and B, and B
and C?
Phase difference / rads
between A and B
between B and C
A
π
4
3π
2
B
π
2
π
C
π
2
3π
2
D
3π
4
π
A
B
C
position
along
wave
Figure 5.31 The position of particles
in the rock at an instant as the wave
passes.
Practice questions
7 A transverse seismic wave is produced by an earthquake. The
wave travels through rock. Figure 5.31 shows the position of
particles in the rock at an instant as the wave passes.
8 The frequency of the seismic wave shown in question 7 is measured
and found to be 6.0 Hz. The wave speed is 4.5 × 103 m s−1. What is the
wavelength of the seismic wave?
A 750 m
C 13 500 m
B 7 500 m
D 27 000 m
light ray
9 Figure 5.32 shows a typical glass step index optical fibre used for
communications. What is the refractive index of the core?
cladding
core
normal line
A −3.1
C 1.47
14.4°
normal
line
air
B 0.04
D 1.49
Figure 5.32 An optical fibre.
9.74°
cladding
X
10 A light ray travels from a slab of air into amber. The refractive index of
amber is 1.56. Which of the following pairs gives possible values for the
angle of incidence and the angle of refraction?
A
B
C
D
Angle of incidence
20°
40°
60°
80°
Angle of refraction
16°
20°
34°
40°
11 The diagram shows an a.c. supply. This is a three-phase supply, so
three voltages are supplied, running at the same time but out of
phase with each other.
1.0
Phase 2
Phase 3
0.5
a) The supply has a frequency of 50 Hz. Calculate
its time period.
(1)
b) State the phase difference between phase 2
and phase 3 in terms of radians.
(1)
0
c) Calculate the time between phase 1 having a maximum value
and phase 3 having a maximum value.
(1)
469884_05_AQA_A-Level_Physics_070-094.indd 91
Phase 1
91
90°
180°
270°
360°
0.5
1.0
120°
120°
Figure 5.33 An a.c. supply.
17/04/19 9:17 AM
12 Submarines communicate underwater using very low-frequency
radio waves. These waves travel at approximately 3.33 × 107 m s−1
in seawater. One underwater system uses a frequency of 82 Hz.
a) Calculate the wavelength of radio waves of this frequency,
giving the unit.
(2)
A dolphin near the submarine sends out a series of clicking
sounds of frequency 100 kHz. The speed of sound waves in
water is 1500 m s−1. The time for the dolphin to detect the
sounds reflecting off the submarine is 100 ms.
b) How far away is the submarine?
(3)
c) Compare the two methods of communicating underwater.
(4)
5 WAVES
13 Figure 5.34 shows a glass sculpture made by Colin Reid. In one
picture, a reflection of the internal surface is visible; from a
slightly different angle the reflection is not visible.
92
a) State the effect that makes this possible.
(1)
b) Explain what has changed to change the appearance of the
sculpture.
(4)
Figure 5.34 Two pictures of a glass sculpture made by Colin Reid.
14 A prism (not drawn to scale) has a refractive index of 1.50, and
internal angles of 60° and 120°. Light is incident on the face
shown at an angle of 30° to the normal, which is shown as a
dotted line. Use calculations to show the path of the light
through the prism.
15 Figure 5.36 shows an optical fibre.
a) Name the process by which light is trapped in the
optical fibre.
normal
Figure 5.35 A prism.
(1)
b) The refractive index of the core is 1.60 and the refractive
index of the cladding is 1.50. Calculate the critical angle
for light travelling in the optical fibre.
(3)
core
cladding
coating
Figure 5.36 An optical fibre.
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i)
Explain what is meant by pulse broadening.
(2)
ii) Explain why a multimode optical fibre is more likely
to suffer from pulse broadening than a single mode
optical fibre.
(2)
iii) Explain one implication of pulse broadening.
(2)
d) Explain why it is important to select the material for the core
very carefully.
Practice questions
c) A multimode optical fibre has a wider diameter core than a single
mode fibre, and carries more information.
(5)
16 A student measures the refractive index of glass using Snell’s law. The
student measured the angle of incidence and angle of refraction for rays
travelling through the glass block.
Angle of incidence Angle of refraction
0
0, 0
10
5, 6
15
10, 10
20
12, 14
25
15, 18
30
20, 20
35
23, 24
40
25, 25
45
29, 30
50
32, 32
55
34, 33
Angle of refraction (mean)
0
5.5
10
13
16.5
20
23.5
25
29.5
32
33.5
a) The student kept the same number of decimal places in the
raw data. Explain why this is acceptable.
(1)
b) The student has calculated the mean. Explain why the
calculated data has not been presented correctly in the table.
(1)
c) Calculate the values of sin i and sin r, and plot these on a graph.
(5)
d) The ratio sin i : sin r equals the refractive index, n, for the material.
Use your graph to calculate the refractive index for the material. (4)
e) Calculate the maximum uncertainty in the readings.
(2)
f) Describe two difficulties with carrying out this experiment, and
how the student could overcome these.
(4)
g) The angles are measured using a protractor to a precision of ±0.5°.
For the angle θ, measured to be 20°, calculate the range of the values
of sin θ and the percentage uncertainty in sin θ.
93
17 The Fermi gamma ray space telescope is a satellite that orbits Earth
at a height of 550 km. Communications with the satellite from
Earth use electromagnetic radiation.
a) Calculate the minimum time it takes a signal to reach
the satellite.
(3)
b) Suggest, with reasons, a suitable type of electromagnetic
radiation to communicate with the satellite.
(2)
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c) Explain two advantages of using a space-based telescope orbiting
Earth to observe gamma ray bursts from distant galaxies.
(2)
The Arecino Observatory includes a telescope with a 305 m
diameter dish. It detects wavelengths between 3 cm and 1 m.
(2)
d) Describe the two advantages of basing the telescope on Earth.
Stretch and challenge
18 A fibre optic cable has a core of refractive index nco, and is surrounded
by cladding of refractive index ncl.
a) Write an expression for the critical angle, θc, for a pulse of red light
travelling inside the core.
The fibre optic cable is made from glass whose refractive index varies
with wavelength. The table gives information about the refractive index
of the cladding and the core for light of different wavelengths.
Wavelength
400 nm
700 nm
Refractive index (core)
1.635
1.602
Refractive index (cladding)
1.470
1.456
b) Calculate the difference in critical angle for red light and for
green light.
0.2 mm
An endoscope is a device that uses optical fibres to look inside
the body. One endoscope uses the material described in part a)
for a light guide used to illuminate inside a patient’s body. The
diagram, which is not to scale, shows the core of the optical fibre
of diameter 0.20 mm.
A ray of white light travelling along the axis of the fibre reaches
the region where the fibre is coiled into an arc of radius R.
R
axis
Figure 5.37
c) Write down the condition needed for all light to remain in the fibre.
5 WAVES
d) Calculate the smallest value of R, which allows the ray to remain
within the fibre
94
e) Explain why it is important to use a cladding material for optical
fibres used in an endoscope
19 The speed of longitudinal waves, c on a spring is given by
c = √(kl/µ)
where k is the spring constant, l is the stretched length of the spring and
µ is the mass per unit length.
a) By substituting the dimensions of the quantities in this equation,
show that the dimensions of the spring constant are MT−2
b) The slinky has a mass of 0.6 kg and is stretched by 3 m using a
force of 9 N. Calculate the speed of longitudinal waves in the slinky.
Assume the spring had negligible length initially.
c) Show that doubling the length of the stretched slinky does not affect
the time for the wave to travel from one end of the slinky to the other.
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6
Combining waves
PRIOR KNOWLEDGE
●
●
●
●
●
●
●
●
●
The wave equation states that the speed, v, of a wave (in m s −1) equals
its frequency, f (in Hz), multiplied by its wavelength, λ (in m).
v=f×λ
The frequency of a wave, f = 1T, where T is the period of the wave.
The phase of a wave describes the fraction of a cycle completed
compared to the start of the cycle.
Phase difference for a single wave compares different points along
the wave at the same time.
i Parts of a wave are in phase when particles in the medium move at
the same speed in the same direction.
ii Parts of a wave that move in different directions and speeds are out
of phase.
iii Parts of a wave moving in opposite directions at the same speed
are in antiphase.
In one complete cycle, waves travel a distance of one wavelength.
The shortest distance between two points moving in phase is one
wavelength, λ.
The phase difference between two points a wavelength apart is 2π or
360°.
The phase difference, Φ, between two points along a wave, separated
by a distance x, is given by:
Φ = 2πx rad
λ
Waves diffract when they go through a small gap. The angle of
diffraction is small when the wavelength is small in comparison with
the gap width; the angle of diffraction is large when the wavelength is
comparable in size to the gap width.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 Calculate the time period of a wave of frequency 50 Hz.
2 Calculate the frequency of a wave if its time period is 2 × 10 −3 s.
3 What is the phase difference between two points along a wave
separated by a distance of 2.75 wavelengths?
4 When two points of a wave are in phase, what can you say about their
motion?
5 Explain why a radio receiver can detect radio waves (wavelength 30 m
to 1000 m) emitted by a transmitter on the opposite side of a hill that is
between the receiver and transmitter.
6 Explain why we can hear but not see around the corner of buildings
(wavelength of sound is 0.3 m to 10 m; wavelength of light is 400 nm to
700 nm).
7 What is the speed of radio waves of wavelength 3 km and frequency
100 kHz?
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95
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Figure 6.1 Permanent changes to your
vision are possible using laser surgery.
Lasers have impacted our lives since they were developed in the 1960s. The
word laser is an acronym standing for Light Amplification by Stimulated
Emission of Radiation. You learned about emission of photons in Chapter 3.
The great intensity of laser beams has several practical uses: laser surgery
allows corrections to eyesight by reshaping the cornea on the outer surface
of the eye; surgeons can cut through flesh using a laser instead of a scalpel.
The accuracy and focus of the beam makes lasers useful in DVD players,
scanners and optical disc drives. Much scientific research relies on the
precise, polarised, coherent beam of radiation from a laser, which is created
using two mirrors in an optical cavity to produce stationary waves. You will
learn about stationary waves in this chapter, and other effects caused when
two waves interact.
●● Superposition
When two waves of the same type meet at the same point and overlap, the
resultant displacement of the oscillations is the vector sum of the displacements
of each wave. This phenomenon is called the superposition of waves. You can
see water waves superpose when you watch waves on the seashore: when two
peaks of waves meet, they combine to make a larger wave.
Superposition is when two waves of the
same type (e.g. sound waves) overlap and
interact. The displacement of the medium,
where the waves overlap, is the vector sum
of the two wave displacements.
A
= 2A
Figure 6.2 shows the superposition of two water waves of the same amplitude,
A, in phase and in antiphase. The resultant amplitudes are 2A when the waves
are in phase, and zero when the waves are in antiphase. When water waves
overlap in phase, larger displacements of the water occur; when waves overlap in
antiphase, the water can remain still. The idea of wave superposition is used to
reduce unwanted noise. Figure 6.3 shows the principle. Unwanted sound, from
some nearby source, is received by a microphone and fed into noise-cancellation
circuitry. This circuitry inverts the noise (which is the same as changing its phase
by 180°), so that the original sound and the newly added waveform cancel each
other out. A worker can be protected from the sound by wearing headphones
that soften a loud noise, which has the potential to damage their hearing.
6 COMBINING WAVES
+
A
4: Original background noise and newly
added waveform cancel each other out
1: Incoming background noise
3: Noise-cancellation circuitry
inverts wave and sends it
back to headphone
2: Noise is picked up by microphone
andsent to noise- cancellation
circuitry
constructive superposition
+
=
96
destructive superposition
Figure 6.2 The superposition of
two water waves.
469884_06_AQA_A-Level_Physics_095-115.indd 96
Figure 6.3 How noise cancellation works.
Superposition of waves only occurs between the same types of wave. Two
water waves can superpose, but light and sound waves cannot superpose.
We can notice the superposition of water waves as the process takes
place relatively slowly. Often when two sources of waves overlap, we
do not notice the superposition, as the frequency of the waves might be
high and any pattern of superposition lasts for a very short time. Under
some circumstances, two sources of waves can overlap in such a way
17/04/19 9:47 AM
that positions of high and low intensity are produced in fixed positions.
Here fixed patterns of wave superposition occur in the same place over
long periods of time, so that we can observe them. A fixed pattern of
superposition is called interference.
Interference
Interference is the name given to
the superposition of waves from two
coherent sources of waves. Interference
is constructive if waves are in phase,
or destructive if waves are in
antiphase (out of phase by 180°).
●● Interference
Two waves are coherent when they have
a fixed phase difference and have the same
frequency.
Two sources of waves can overlap and produce constructive and destructive
regions of superposition. However, a stable pattern or superposition
(or interference) will only occur if the two sources of waves are coherent.
Two sources are coherent when the waves from each source have the
same wavelength (and, therefore, the same frequency), and there is also a
constant phase relationship between the two sources. For example, light
which meets at a point from two coherent sources can produce regions of
bright light (or maxima) where light interferes constructively, and regions of
darkness (or minima) where light interferes destructively.
Interference patterns from sound waves
A single sustained note from a signal generator played through two
loudspeakers creates interference patterns. The loudness of the sound
heard by a person walking in front of the speakers varies from loud to quiet
on a regular spacing due to the pattern of constructive and destructive
interference. When coherent sound waves are in phase, the sound is louder
because of constructive interference.
The path difference is the difference in distance travelled by the two waves
produced by the loud speakers. Path difference is usually measured in
multiples of wavelength.
If the waves are in phase when they leave the speakers, they are in phase
at any point where they have travelled the same distance or where their
path difference is a whole number of wavelengths, nλ. These sound waves,
initially in phase, are out of phase at any point where the path difference
is a whole number of wavelengths plus a half wavelength, (n + 12 )λ. Points
along a wave are in phase if they are separated by a whole number of
wavelengths, and in antiphase when the distance between them is a half
wavelength, one and a half wavelengths, and so on.
path difference = λ
constructive
interference
Figure 6.4 Path difference.
P2
λ
path difference = −
2
destructive
interference
97
P1
S2
S1
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EXAMPLE
A student walks between two loudspeakers
each playing a note of frequency 300 Hz.
The waves are in phase as they leave the
loudspeakers. The speed of sound in air is
340 m s−1.
a) Calculate the smallest path difference for
waves from each loud speaker when the
student hears
i) constructive interference
ii) destructive interference.
b) Now calculate the next smallest path
difference for
i) constructive interference
ii) destructive interference.
signal generator
QUIET
speaker
LOUD
QUIET
LOUD
QUIET
LOUD
Figure 6.5 An experiment demonstrating interference of sound waves.
Answers
a i) C
ondition for constructive interference is a path difference of nλ.
The shortest path differences occur when n = 0 so the shortest path
difference is 0 m.
ii) Condition for destructive interference is a path difference of (n + 12)λ.
For the minimum path difference, n = 0.
The path difference is 2λ = 1.13
2
6 COMBINING WAVES
= 0.57 m.
98
b i) T
he next smallest path difference for constructive interference, is
n=1
So the path difference is
λ= c
f
(340
m s−1)
=
300 Hz
= 1.13 m
ii) T
he next path difference for the first minimum occurs at a distance of
1.5 λ = 1.70 m. (n = 1)
EXAMPLE
The student in the previous example stands in a region
of destructive interference. Explain what the student
would expect to hear
a) if one loudspeaker is covered with a cushion
b) if the connections to one loud speaker are reversed.
b) Changing the connections round makes sound
from the two speakers move out of phase with each
other. A path difference of 2λ now makes the waves
in phase again. So there is constructive interference
and the sound becomes louder.
Answers
a) The student will hear the sound get louder as there
is no destructive interference, but only the sound
from one speaker.
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destructive
interference
(crest on trough)
S1
S2
constructive
destructive
constructive
Figure 6.6 Interference in a ripple tank.
Young’s double slit experiment–Required practical
number 2
Young’s double slit experiment demonstrates interference between coherent
light sources, thus showing the wave nature of light. The experiment uses two
coherent sources of light waves, produced from a single source of light, which
then pass through two very narrow, parallel slits. The light diffracts (spreads)
through the slits, producing an interference pattern of fringes on a screen.
Interference occurs because the waves overlap and superpose in a stable pattern.
Interference
constructive
interference
(crest on crest)
Light has such a short wavelength it is difficult to see its interference
patterns. This is why Young’s double slit experiment works best using a
blacked out room and a very bright white light source, or a laser as a source
of intense single-wavelength (monochromatic) light. The slits must be very
narrow and less than 1 mm apart. An interference pattern can be seen where
patches of bright light alternate with regions of darkness. These correspond
to areas of constructive and destructive interference. These patterns are
called fringes. The interference fringes are visible on a screen placed at least
1 m away from the slits. If the screen is further away, the fringe separation
increases but their appearance becomes fainter.
P
x
s1
θ
θ
s
s2
central axis
O
w
nλ
D
Figure 6.7 Young’s double slit experiment.
Light from each slit travels a slightly different route to the screen, creating
a path difference as shown in Figure 6.7. Dark fringes occur where there
is destructive interference (the path difference between the two slits is
(n + 12 )λ). Bright fringes occur where there is constructive interference
(the path difference between the two slits is nλ).
Fringes are ordered. The order of the central bright fringe is n = 0. The
order of the fringes closest to the central fringe is n = 1; the order of the
next pair of fringes is n = 2, etc.
Referring to Figure 6.7, the condition for constructive interference (bright
fringes) is:
s sin θ = nλ
99
where s = spacing between the slits
θ = angle from the beam towards the screen
n = a whole number
λ = wavelength of light.
The extra distance travelled by the waves leaving S2 is s sin θ, and for
constructive interference this distance (or path difference) must be a whole
number of wavelengths, nλ.
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We can also express the sepration of the fringes in terms of the angle θ.
Refering again to Figure 6.7, we can see that:
X = tan θ
D
where X is the distance to the nth fringe from the midpoint, and
D is the distance from the slits to the screen on which we see the
interference fringes.
However, since D X, the angle θ is small.
Then, the small angle approximation gives:
»
sin θ ≈ tan θ
since
and
sin θ = nλ
s
6 COMBINING WAVES
tan θ = X
D
it follows that (for small angles):
nλ = X
s D
We can use this formula to predict the fringe spacing, w, between two
adjacent bright fringes. When n = 1 the distance X becomes the spacing, w,
between adjacent bright fringes.
So
λ w
=
s D
and
λD
w= s
100
EXAMPLE
A student shines monochromatic light of wavelength 580 nm through
two slits 0.25 mm apart. A screen is positioned 2.3 m away. What is the
separation of the bright interference fringes seen on the screen?
Answer
λD
s
(580 × 10−9 m) × (2.3 m)
w=
(2.5 × 10−4 m)
= 5.3 × 10−3m or 5.3 mm
w=
LASER DANGER
Lasers produce a very intense light. In schools, the most powerful lasers
that are allowed are class 2 lasers which must have a power output of less
than 1 mw. You should never look directly at a laser or its reflection, or allow
it to reflect from shiny surfaces into anyone’s eyes.
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Stationary waves
TEST YOURSELF
displacement
1 a) Radiowaves are emitted from two aerials in
5 A Young’s slit experiment uses light of wavelength
phase. At a position x, the path difference
600 nm.
between the two aerials is 3 wavelengths.
a) Explain why laser light is often used for Young’s
Explain whether there will be constructive or
slit experiments.
destructive interference of the waves at x.
b) If the two slits are 0.2 mm apart, and the screen
b) Explain how your answer to part (a) would
is 3.0 m away, calculate the separation of the
change if the waves were out of phase as they
fringes.
left the aerial.
c) What changes could be made to increase the
c) Coherent soundwaves are emitted from two
separation of the fringes?
loudspeakers in phase. A microphone is placed
6 A microwave transmitter transmits waves with a
at a position where the path difference between
wavelength of 2.8 cm through slits in a metal plate
the two loudspeakers is 5.5 wavelengths. What
that are 5.8 cm apart. A probe is placed 30 cm
type of interference is detected?
from the slits and detects regions of high and low
d) Explain how your answer to part (d) would
intensity. Calculate the separation of the regions of
change if the waves were out of phase as they
high intensity.
left the loudspeaker.
7 In a Young’s slit experiment, a student measures
2 Compare the different conditions needed for
the separation between six bright fringes to
superposition and for interference.
be 3.0 mm. The light used in the experiment
3 Explain what is meant by two sources of coherent
has a wavelength of 600 nm, and the slits are
waves, giving an example.
placed 36 cm from the screen. Calculate the slit
separation.
4 Two radio wave transmitters, that are separated by a
distance of 6.0 m, emit waves of frequency 1 GHz. The
8 The diagram below shows two waves approaching
waves leave the transmitters in phase. The waves
each other on a string. Describe the displacement
are directed into a field where they are detected by a
of the string at points A and B over the next 5
man holding an aerial at a distance of 200 m.
seconds.
a) Describe what the man detects on his
aerial as he walks along a line XY, which
1
1 ms1
1 ms1
is parallel to the line joining the two
A B
0
transmitters.
2
3
4
5
6
7
8
9
10 11
1
b) The man stands at a point of maximum
position/m
–1
intensity, when the phase of one of the
transmitters is advanced by 14 of a cycle.
Figure 6.8 Two waves approaching each other on a string.
How far does he have to walk along XY to
detect a maximum again? Does it matter
which way he walks?
●● Stationary waves
A stationary (or standing) wave is a
wave formed by the superposition of two
progressive waves of the same frequency and
amplitude travelling in opposite directions.
If you watch a plucked guitar string closely, you will see that it appears to be in
two fixed positions at once. In fact, the string is vibrating rapidly between two
positions. What you see is called a stationary (or standing) wave. Stationary
waves can be set up in many situations including using a guitar string.
101
Creating stationary waves
Stationary waves are created when two progressive waves of the same
frequency and amplitude moving in opposite directions superpose. This
often occurs when reflections of a progressive wave superpose with the
original wave. When waves reflect off a rigid boundary, the reflected waves
are out of phase with the original wave by 180° and travelling in the
opposite direction, but have the same amplitude and frequency.
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TIP
The maximum displacement
of all the particles along a
stationary wave is not the same
for stationary waves. Particles
at nodes do not move at all, but
particles at antinodes vibrate at
the maximum amplitude.
Nodes
Antinodes
Figure 6.9 A stationary wave.
Stationary waves only form on a guitar string at specific frequencies. Where
the waves are in phase, the displacements add to form a peak or a trough
of double the original amplitude. Where the waves are in anti-phase, their
displacements cancel out. For stationary waves, the positions of maximum and
minimum amplitude remain in the same places, with particles, at antinodes,
vibrating rapidly between their positions of maximum displacement. This is
why the waves don’t appear to progress along the string.
A stationary wave alternates between the positions shown by the red wave
and the blue wave in Figure 6.9.
Nodes and antinodes
Nodes and antinodes are at fixed points along the guitar string. At a node,
the amplitudes of the two progressive waves moving in opposite directions
always cancel out so the particles do not oscillate at all. At an antinode, the
amplitudes add together and the guitar string is displaced between the peak
and trough during a cycle. The amplitude of the peak or trough is double the
amplitude of the two progressive waves. Nodes are always separated by half
a wavelength, as are antinodes. The displacements of particles in positions
between the nodes and antinodes vary but do not remain zero or reach the
amplitude of an antinode during the cycle according to their positions.
How stationary waves form
A node is a point of zero amplitude on a
stationary wave.
An antinode is a point of maximum
amplitude on a stationary wave.
Figures 6.10a to d show how stationary waves form on a guitar string. Wave
1 and wave 2 are progressive waves travelling in opposite directions along
the string with the same frequency and amplitude. The amplitude of the
wave 3 is the amplitude of the two waves superimposed.
The diagrams show a sequence of snapshots of the wave at different times
during one complete cycle.
6 COMBINING WAVES
(1)
102
(2)
λ
2
λ
2
λ
2
(3)
(a)
t0
(b)
1
t4 T
(c)
1
t2 T
(d)
t
3
T
4
Figure 6.10(a) Waves 1 and 2 are in phase at the start of the cycle but move in opposite
directions. There is constructive superposition and their displacements add. The
combined displacement of both waves is double their original displacement.
Figure 6.10(b) Waves 1 and 2 are in antiphase after quarter of a cycle. The particles
at all points are displaced in opposite directions. Their displacements cancel out as
the superposition is destructive. The combined displacement of both waves is zero.
Figure 6.10(c) Waves 1 and 2 are in phase halfway through their cycle. Compared
to the start of the cycle, the position of peaks and troughs is reversed. There
is constructive superposition and their displacements add. The combined
displacement of both waves is double their original displacement.
Figure 6.10(d) Waves 1 and 2 are in antiphase three-quarters of the way through
a cycle. Their displacements cancel out as the superposition is destructive. The
combined displacement of both waves is zero.
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Harmonics
Stationary waves
A harmonic is a mode of vibration that is a
multiple of the first harmonic.
A guitar string can support different modes of vibration for stationary
waves. The different modes of vibration are called harmonics. Harmonics
are numbered: the first harmonic is the mode of vibration with the longest
wavelength. The second harmonic is the mode of vibration with the next
longest wave.
The frequency of the vibration is found using
f= v
λ
where f = frequency of the harmonic in Hz
v = speed of wave in m s−1
Figure 6.11 The three lowest harmonics
for a guitar string.
λ = wavelength of harmonic in m
In Figure 6.11,
the first harmonic on a string includes one antinode and two nodes.
Stringed instruments have nodes at each end of the string as these points
are fixed. For a guitar string of length l, the wavelength of the lowest
harmonic is 2l; this is because there is one loop only of the stationary
wave, which is a half wavelength. Therefore the frequency is:
f1 = v = v
λ 2l
●the second harmonic has three nodes and two antinodes; the wavelength
is l and frequency is:
f2 = v = v , or 2f1
λ l
●the third harmonic has four nodes and three antinodes; the wavelength is
2L/3 and frequency is:
f3 = v = 3v or 3f1
λ 2l
●
The first harmonic on a string
For waves travelling along a string in tension, the speed of a wave is given by:
T
v=
μ
where:
T = tension in the string in N
μ = mass per unit length of the string in kg m−1
Using Figure 6.11, you can see that the first harmonic for a stationary wave
on a string is half a wavelength. The wavelength of the first harmonic is 2l
for a string of length l.
103
We can combine the equation above and the wave equation as follows:
c= f
so T = f
μ
= f × 2l
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Therefore:
f=
1
2l
T
μ
This equation gives the frequency of the first harmonic where:
f = frequency in Hz
T = tension in the string in N
μ = mass per unit length of the string in kg m−1
l = the length of the string length in m
EXAMPLE
A string is stretched between two points 0.45 m apart.
a) Calculate the frequency of the first harmonic when
the tension in the string is 80 N and the mass per
unit length is 3.2 g m−1.
b) Predict the effect on frequency of:
i) doubling the length of the string
ii) doubling the tension in the string.
Answers
a) f =
=
1
80 N
× 2 × 0.45 m
3.2 × 10−3 kg m−1
= 175.7 Hz or 176 Hz to 3 s.f.
b) i) F
requency is proportional to 1l, so doubling the
length halves the frequency (to 88 Hz).
ii) Frequency is proportional to √T, so doubling the
tension increases the frequency by a factor of √2
(to 248 Hz).
1 T
2l μ
6 COMBINING WAVES
Stationary waves: Sound waves and musical
instruments
104
Longitudinal waves such as sound waves also form stationary waves.
Sound waves are pressure waves with particles vibrating in the direction
the wave travels. Stationary wave diagrams for sound waves show the
amplitude for particles vibrating longitudinally in the air column. The
amplitude is greatest at the open end of pipes, where there is an antinode.
The particles cannot vibrate at a closed end and so there is a node. If the
pipe has two open ends, the stationary wave has at least two antinodes,
at either end of the pipe.
Figure 6.12 Stationary waves for sound waves in open-ended tubes. When there is one
open end, there is an antinode at the open end and a node at the closed end. In a pipe
that is open at both ends, there is a node in the middle and antinodes at both ends.
Musical instruments include stringed instruments, where stationary waves
form on the string when it is plucked or bowed. Stationary waves also
form when the air column in organ pipes or wind instruments is forced
to vibrate. The harmonic frequencies available to players will vary with
different instruments, even if the air column is the same length, depending
on whether the instrument has one or both ends open.
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A
A
N
λ
l= —
2
N
A
f = f0
N
A
motion of
air particles
A
A
A
l=λ
A
N
f = 2f0
A
3λ
l=—
2
f = 3f0
N
A
N = node
A = antinode
λ
l=—
N
4
l
A
N
Stationary waves
Figure 6.13 Different notes are created
from the different harmonics of
stationary waves in air columns.
A
f = f0
A
N
3λ
l=—
A
4
N
f = 3f0
A
N
N
5λ
l=—
4
f = 5f0
N
REQUIRED PRACTICAL 1
Investigation into the variation of the frequency of stationary waves on a string with length,
tension and mass per unit length of the string
Note: This is just one example of how you might tackle this required practical.
A student used the equipment shown in Figure 6.14 to
investigate how the frequency of stationary waves on a
wire varied with tension in the wire. The student fixed
a length of wire at one end and passed the other end
over a pulley. A hanger was attached to the end of the
wire. The tension in the wire was changed by adding
masses to the hanger.
The student was told the mass per unit length of wire
was 0.16 g/m.
1. Suggest why the string is passed over a pulley, and
not over the edge of the table. The student obtained these results with a length, l, of
0.4m:
l
T
Figure 6.14
The student placed a microphone connected to a
frequency analyzer near the wire. When the wire was
469884_06_AQA_A-Level_Physics_095-115.indd 105
plucked, the frequency of the sound could be read
off the frequency analyzer. The student sometimes
observed several frequencies, but only recorded the
lowest frequency.
Mass on
hanger
(kg)
Frequency
(Hz)
1st reading
Frequency
(Hz)
2nd reading
0.20
140
135
0.40
195
200
0.60
240
240
0.80
280
278
1.00
310
310
1.20
340
340
1.40
365
368
1.60
392
390
Frequency
(Hz)
mean
105
2) Calculate the average frequency using the data
3) Calculate the frequency using the equation
1 T
f = 2l μ
17/04/19 9:47 AM
4) Suggest, with reasons, the maximum percentage
uncertainty in this experiment. You will need to
suggest the resolution of the equipment used in the
experiment.
Extension
Measure the change in frequency of standing waves
if wires of different length are used or if a wire with a
different mass per unit length is used.
ACTIVITY
Measuring the wavelength of sound waves
A student uses a set-up called a
Kundt’s tube to measure the speed
of sound in air. The Kundt’s tube is
placed horizontally, and a fine powder
is sprinkled along its length. One end
of the tube is closed; the other end has
a loudspeaker connected to a signal
generator to produce notes of different
frequency inside the tube.
6 COMBINING WAVES
The pitch of the note is changed until
the sound from the tube increases to
a maximum, and a stationary wave
is set up inside the tube. The powder
moves with air and settles into piles
corresponding to nodes.
closed end
signal generator
power pack
microphone
adjustable
position
Figure 6.15 Kundt’s tube apparatus.
1 Use the pattern of powder to work
out which frequency matched a
stationary wave with a wavelength of 1.5 m. You may
Pitch of Length of Where powder Where there was
find it helpful to sketch the pattern. note/Hz
tube/m
piled up/cm
no powder/cm
2 Calculate the number of stationary waves in the
75
0, 150
118
1.5
tube for each frequency.
235
1.5
38
and
112
0,
75, 150
3 Calculate the wavelength of the stationary wave and
357
1.5
25, 75, 125
0, 50, 100, 150
speed of sound for each frequency.
−1
4 The speed of sound in air is about 340 m s . Comment
179
1.5
0, 100
50, 150
on your answer and any reason for differences.
106
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Stationary waves: microwaves
Microwaves and other electromagnetic waves can interact and form
stationary waves. Microwaves in microwave ovens are generated using a
device called a magnetron. Since microwaves reflect off metals, they are
directed from the magnetron and reflect off metallic inner surfaces to
ensure they spread evenly throughout the oven. However, stationary waves
still tend to develop, resulting in over-cooked food at the antinodes and
undercooked food at the nodes. This is why most microwave ovens have
a rotating turntable. Grated cheese or chocolate covering a plate placed in
a microwave oven with the turntable disabled melt in places about 6 cm
apart. These places correspond to antinodes for the first harmonic. This
information tells us that the wavelength of the microwaves is about 12 cm.
17/04/19 9:47 AM
Electromagnetic waves are carried by oscillating
electric and magnetic fields. For clarity of explanation
we will consider only the electric fields in this chapter.
When two microwaves overlap and superpose,
interference can take place. Where the electric fields
of the two waves are in phase, there is constructive
interference and regions of high intensity. The
intensity is low where the electric components are out
of phase. The changes in intensity are detected using
a probe and this information can be used to calculate
the wavelength of the microwaves.
after it has been reflected from the metal plate,
deduce the wavelength of the microwaves.
2 The student takes measurements to check the
wavelength. Describe how she could reduce
uncertainties in her measurements.
3 Explain the student's could explain her observations,
using ideas about stationary waves.
receiving
aerial
A student sets up a microwave transmitter,
microwave detector and aluminum plate placed
about 40 cm from the transmitter as shown in
Figure 6.16. Microwaves that reflect off the metal
plate interfere with the transmitted microwaves.
The interference pattern has regions of low intensity
where the waves are out of phase, and regions of
high intensity where the waves are in phase. She
uses a receiving aerial, or probe, to detect nodes.
The student discovers that the distance between
successive regions of destructive interference is 1.4 cm.
1 By considering the path difference between the
wave reaching the receiver directly from the
transmitter, and the wave reaching the receiver
Stationary waves
ACTIVITY
Interference patterns from microwaves
reflecting
board
mircowave
transmitter
Figure 6.16 An experiment demonstrating interference of
microwaves.
TEST YOURSELF
9 State and explain one difference between:
a) a stationary wave and a progressive wave
b) a node and an antinode.
10 State three conditions that are needed for a
stationary wave to be et up.
11 Two progressive waves travelling in opposite
directions originally have the same amplitude.
Explain why the amplitude of an antinode in a
stationary wave is double the amplitude of the
progressive waves.
12 Describe how the displacement of a particle at an
antinode changes during one complete wave cycle.
13 A stationary wave is set up on a rope of length
3 m. Write down the wavelength of:
469884_06_AQA_A-Level_Physics_095-115.indd 107
a) the first harmonic
b) the third harmonic.
14 a) A stationary sound wave is set up in a hollow
pipe with both ends open of length 1.4 m. Write
down the wavelength of:
i) the first harmonic
ii) the second harmonic.
b) One end of the pipe is now closed. Write down
the wavelengths of:
i) the first harmonic
ii) the second harmonic.
15 A student blows across the top of a bottle, setting
up a stationary wave, and creating a note. The
bottle is 20 cm high.
107
17/04/19 9:47 AM
a) W
hat is the wavelength of the first harmonic?
b) The speed in air is 340 m s−1. What is the
frequency of the note?
c) T he student fills half the bottle with water.
Describe the note that the student hears now
when he blows across the top of the bottle.
16 a) A string has a mass per unit length of
5 g m−1. It has a length of 0.9 m and has a first
harmonic frequency of 170 Hz. Calculate the
tension in the string.
b) The tension is increased by a factor of 4.
Calculate the frequency of the first and second
harmonics now.
17 It is possible to calculate the diameter of a metal wire
by using information about its harmonic frequencies.
Work through this question to find out how.
steel wire is fixed at one end and the other
A
end is stretched over a pulley, with a mass of
4 kg hung on the end. The wire is set vibrating
at a frequency of 83 Hz, and the length of wire
between the fixed end and the pulley is 63 cm.
At this frequency a stationary wave is set up,
which has two nodes between the pulley and
the fixed end.
a) Calculate the wavelength of the waves.
b) Calculate the speed of the waves.
c) Calculate the tension in the wire.
d) Use your answers to parts (b) and (c) to
calculate the mass per unit length of the wire.
e) Given that the density of the steel wire is
7800 kg m −3 , calculate the diameter of
the wire.
●● Diffraction
When waves pass through a gap or
move past an obstacle, they spread out.
This is called diffraction.
A single source of light can also create interference patterns. This effect
occurs because of diffraction. Diffraction occurs when waves spread around
an obstacle or through a gap. It affects all waves, and is why we can hear
conversations around corners even if we cannot see the people speaking.
Diffraction can be shown easily using a ripple tank, and is most pronounced
if the wavelength is of a similar order of magnitude as the obstacle or gap.
The wavelength of diffracted waves does not change, but the waves become
curved and spread into shadow regions around the obstacle or gap.
6 COMBINING WAVES
Diffraction of light
Figure 6.17 Diffraction of water through a gap.
108
Visible light has a very short wavelength (400 × 10−9 m
to 700 × 10−9 m) so diffraction is only significant if
the slit is very narrow. When light diffracts through a
narrow slit, a fringe pattern is seen as a bright central
fringe surrounded by other less-bright fringes either side.
These fringes, called maxima, are due to constructive
interference of light. The dark regions in between
(minima) are due to destructive interference of light. The
interference pattern is due to light from one part of the
slit diffracting, overlapping and then interfering with light
that has diffracted from other parts of the slit. When the
path difference between the top half of the slit and the
bottom half of the slit is half a wavelength, destructive
interference occurs and the light intensity is zero.
The width of the central diffraction maximum depends on the wavelength of
light and the slit width. At the edges of the central maxima, light leaving the
top half of the slit interferes destructively with light leaving the bottom half, so:
λ a
= sin θ
2 2
or the first minimum occurs at an angle given by:
λ
sin θ =
a
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monochromatic
light
Diffraction
intensity
of light
slit
fringe pattern
on screen
Figure 6.18 Diffraction fringes from a single slit.
where λ is the wavelength of light, a is the slit width and θ is the angle of
diffraction.
As the wavelength of light increases, the central maximum becomes wider
because, as λ increases, sin θ = aλ increases.
As the slit width increases, the central maximum becomes narrower because,
as a increases, sin θ = aλ decreases.
If a source of white light is used, the fringes are coloured because white
light is a mixture of different colours of light. Since the central maximum
depends on the ratio aλ , the central maximum is broader for red light than it
is for blue light. Red light has a longer wavelength than blue light.
Figure 6.19 Comparing diffraction of monochromatic light
and white light.
109
Diffraction gratings–Required practical number 2
A diffraction grating has thousands of slits
spaced very closely together.
We can see interference patterns using a diffraction grating. A diffraction
grating has thousands of slits spaced very closely together. The slits are
very narrow so that the light diffracts through a wide angle. The pattern is
a result of light overlapping (or superposing) and interfering from a great
number of slits.
When light passes straight through the slits, the path difference between
each slit is zero, so the light from each slit is in phase and the overlapping
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a
θ
θ
d
b
c
e
waves combine to form a maximum intensity. This maximum is called the
central maximum. It can also be called the zeroth order maximum because
there is 0 path difference between each slit. Such a maximum is easily seen
on a screen when a laser is shone through a grating. A diffraction grating also
produces other maxima on either side of the central maximum. These occur
when the waves leaving the slits arrive in phase on the viewing screen. The
condition for constructive interference is that the path difference for waves
leaving adjacent slits must be a whole number of wavelengths. The first
order maximum is seen when the path difference for waves leaving adjacent
slits is one wavelength; the second order maximum is seen when the path
difference for waves leaving adjacent slits is two wavelengths, and so on.
For the first order maxima:
interference is constructive
● the path difference of light from adjacent slits is λ
● light travels at angle θ1 to the direction of incident light.
●
d
g
f
Figure 6.20 Finding the path difference
for a diffraction grating.
Applying this information to the triangle abc,
λ
sin θ1 = bc =
ab d
This means that
λ
sin θ1 =
d
where λ is the wavelength of light in m
θ1 is the angle between the 0th and 1st order maxima
d is the spacing between slits in m.
For second order fringes, the path difference is 2λ and light travels at angle
θ2 to the direction of incident light. The equation becomes:
2λ
d
In general, the condition for maxima to occur is given by:
nλ
sin θn =
d
where n is a whole number, and the order of the maximum.
6 COMBINING WAVES
sin θ2 =
EXAMPLE
Calculate the angle of a third order maximum for light of wavelength
500 nm passing through a diffraction grating with 600 lines per mm.
Answer
110
−3 m
10
The spacing of the grating, d, is 1 × 600
= 1.67 × 10−6 m
The order, n, is 3; the wavelength, λ = 500 × 10−9 m
nλ
sin θn =
d
−9 m
= 3 × 500 × 10
−6
1.67 × 10 m
= 0.898
sin−1 0.898
=θ
So θ = 64°
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Applications of diffraction gratings
Diffraction
Diffraction gratings are used to separate light of different wavelengths in
great detail (high resolution). This is useful when the diffraction grating
is part of a spectrometer and used for investigating atomic spectra in
laboratory measurements. Diffraction gratings are also used in telescopes to
analyse light from galaxies. CDs and DVDs use diffraction gratings but they
reflect light from a grating rather than transmitting it. The light source is a
laser installed inside the CD or DVD drive.
TEST YOURSELF
18 W
hy is diffraction of light best seen using very
narrow slit sizes?
19 Refer to the photograph in Figure 6.19, which
shows a diffraction pattern for green light when
it passes through a narrow slit. Describe how the
fringe pattern changes as the angle increases
away from the central maximum.
20 A student is looking at a diffraction pattern from a
single slit. He makes the slit narrower. How does
this affect the fringe pattern he sees?
21 Light of wavelength 500 nm passes through
a diffraction grating with 500 lines per nm.
Calculate the angle for the second order
maximum seen for the light.
22 C
alculate the spacing of a diffraction grating
when the angle of a second order maximum for
light (λ = 400 nm) is 20°.
23 A diffraction grating has 400 lines per mm. White
light is shone through the grating.
a) Describe the diffraction pattern produced by
this grating.
There is a second order maximum for red light
of wavelength 690 nm at the same angle as a
third order maximum for blue light of wavelength
460 nm.
b) Calculate this angle.
c)How many orders of the blue light and how many
orders of the red light does this grating produce?
111
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Practice questions
1 When two waves are initially in phase, their path difference at any point
where the waves are in antiphase is:
λ
A nλ
C nλ +
4
λ
D 0
B nλ +
2
2 The separation between a node and the adjacent antinode in a stationary
transverse wave on a rope is:
λ
C λ
A
4
λ
D 2λ
B
2
3 Which statement is not always true about coherent waves?
A Coherent waves have the same frequency.
B Coherent waves have the same wavelength.
C Coherent waves have a constant phase difference.
D Coherent waves have the same amplitude.
4 In an experiment to investigate nodes and antinodes produced by
stationary microwaves, a microwave probe detects antinodes 3.0 cm
apart. What is the wavelength of the microwaves?
A 3.0 cm
C 6.0 cm
B 1.5 cm
D 12.0 cm
6 COMBINING WAVES
5 A pipe, open at both ends, has a first harmonic of 30 Hz. The frequency
of the second harmonic is:
112
A 60 Hz
C 120 Hz
B 15 Hz
D 40 Hz
6 When a wave reflects off a denser medium, its phase change is:
A 90°
C 360°
B 180°
D 45°
7 Which of these statements is incorrect about interference patterns for light?
A Dark fringes are caused by the interference of waves with a phase
difference of 180°.
B Dark fringes are the result of constructive interference.
C Light fringes are caused by the interference of waves with a path
difference of a whole number of wavelengths.
D Dark fringes are the result of destructive interference.
8 Light of wavelength 500 × 10−9 m shines through a pair of narrow slits
0.5 mm apart. The separation of fringes on a screen 2 m away is:
A 2 × 10−4 m
C 2 × 10−3 m
B 1.25 × 10−3 m
D 0.125 m
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a) Calculate the spacing of the lines on the grating.
(1)
b) Calculate the angle of the 3rd order diffraction lines when light of
wavelength 633 nm shines through the grating.
(3)
c) Explain if it is possible to see the 4th order diffraction lines.
(2)
10 Young’s double slit experiment is set up using slits set 0.3 mm apart
and a screen 2 m from the slits. A student shines light of wavelength
557 nm from a krypton source through the slits and measures the
fringe pattern.
a) Calculate the width of ten fringe spacings.
(3)
b) Describe the changes the student could make to increase the
separation of the fringes.
(3)
Practice questions
9 A diffraction grating has 500 lines per mm.
11 A few people can see a spectrum caused by diffraction through their
eyelashes.
a) Explain the conditions needed for a diffraction pattern like this
to be seen.
(2)
b) The wavelength of visible light is 400 nm to 700 nm. Your eyelashes
have a separation of about 0.1 mm. Use this information to estimate
the angles at which the 4th order spectrum will be seen.
(4)
c) Comment on your answer.
12 a) Explain what is meant by a stationary wave.
b) i)
The frequency of the microwaves generated in a microwave
oven is 2.4 GHz. Calculate the wavelength of the waves.
(2)
(2)
(2)
ii) Using your answer to part (i), calculate the distance between
(1)
antinodes inside the oven.
c) A person placed a plate of grated cheese inside the microwave oven,
switched off the turntable and turned the oven on for about 30
seconds. State and explain what would happen, and how
a person could use this to verify the frequency of the microwaves. (2)
13 Discuss the formation of stationary waves in an open pipe. You should:
●
include a diagram of the first harmonic, explaining where nodes and
antinodes are found
●
describe how nodes and antinodes form
●
explain what conditions are needed for the formation of stationary waves.
The quality of written communication will be assessed in your answer. (8)
14 Coherent light of wavelength 460 nm shines on a pair of parallel slits;
a pattern of dark and bright fringes is seen on a screen some distance
away.
a) State one reason why the experiment was important when it was
first performed 200 years ago.
(1)
b) Explain why coherent waves are necessary for this experiment.
(2)
c) Calculate the ratio of slit width to screen distance if the fringes
are 1.8 mm apart.
(3)
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113
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15 a) A laser emits monochromatic light. Explain what is meant by
monochromatic light. (1)
Blue light from a laser travels through a single slit. The intensity
graph for the diffracted blue light is shown.
b) Explain how the intensity graph changes if a different laser
that emits red light is used.
(2)
c) State and explain one risk of working with lasers, describing
how to reduce this risk.
(2)
d) The red laser light is replaced by a non-laser source emitting
white light. Describe the new appearance of the pattern. (2)
16 Monochromatic light of wavelength λ shines on a diffraction
grating, which has lines separated by a distance d.
Intensity
Single-slit diffraction pattern
0
Figure 6.21
a) Derive the formula: d sin θn = nλ for light of wavelength λ
shining through a diffraction grating of spacing d where θn is
the angle at which the nth order maximum occurs. (4)
b) Describe and explain two ways that the pattern of maximum
intensities changes as the wavelength increases. (2)
c) Calculate the wavelength of light for 3rd order fringes. The
spacing of the grating is 1 × 10−6 m and one sharp maximum
is seen at a diffraction angle of 70°.
(2)
10−6 m
17 A grating with slit separation of 2 ×
is used analyse a
helium spectrum. A blue line and a red line are observed very
close to each other at 42°. What order are these lines? (Range of
red light is 645 nm to 700 nm; range of blue light is 440 nm
to 490 nm.) (5)
6 COMBINING WAVES
18 A grating has 300 lines per mm and is used with a laser that
emits light of wavelength 7 × 10−7 m.
114
a) What is the slit spacing of the grating?
(1)
b) If the light is incident at right angles, how many orders of
diffraction are visible and in what directions do they occur?
(4)
Stretch and challenge
19 Compare the three ways of observing interference patterns:
Young’s double slits, diffraction gratings and diffraction through
a single slit, including a discussion of why the patterns form.
20 White light travels through a diffraction grating. Use calculations
to work out the lowest order spectrum when colours overlap with
the next order spectrum. The wavelength of red light is 700 nm
and violet light is 400 nm.
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S1
B
rays from Sun
θ
S2
θ
θ
Sea
R
receiver
θ
Practice questions
21 A stellar interferometer is located on a cliff top, 100 m high. It is
used to study 1.5 m waves from the sun.
100 m
AA
Figure 6.22
a) Using the letters on the diagram, name the path difference that may
give rise to interference.
b) The path difference is 200 sin θ. Write down the condition for the
receiver to detect an intensity maximum. Remember that there is a
phase change for waves reflecting off the sea. (Mathematicians might
like to prove this result for the path difference.)
c) Explain why the intensity goes through a series of maxima and
minima as the sun sets.
d) Will the changes in maxima and minima be most rapid when the sun
is directly overhead or when it is nearly set? Explain your answer.
22 Explain why a diffraction grating with 100 lines per mm produces a few
sharply defined maxima when laser light shines through it, whereas the
same laser shining through just two slits, produces equal areas of light
and dark. (Hint: think of part of the grating – perhaps 100 lines. If the
λ
, what is the path
path difference between the first and second slit is 100
difference between the first and 51st slit?)
23 Two microscope slides, length L, are placed flat with one slide on top
of the other.
One end of the top slide is raised through a small distance D, enclosing
a wedge of air between the two slides.
Monochromatic light shines on to the slides and an interference pattern
of parallel, equally spaced fringes can be seen.
a) Light reflects in two places: from the bottom face of the upper slide
and from the upper face of the lower slide. Which reflected ray
undergoes a phase change of 180°?
b) Write down the general condition for dark fringes to be seen in this
interference pattern.
115
c) The reflection takes place where the thickness of the air wedge is a
distance, d. Write down an expression for the path difference when
dark fringes can be seen in terms of distance d, and wavelength λ.
d) Use your answer to part (c) to predict how the fringe pattern would change if:
i)
the light source does not change but the air gap is filled with a
transparent liquid
ii) the only change is to increase the wavelength of light.
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7
Introduction to
mechanics
PRIOR KNOWLEDGE
●
●
●
●
●
●
A scalar quantity has only size, for example a mass of 2 kg.
A vector quantity has both size and direction, for example a 5N force
acting downwards.
A force is a push or a pull. Forces are measured in newtons, N.
The size of a turning effect is called a turning moment or torque.
Turning moment = force applied × perpendicular distance from
the pivot.
Newton’s first law of motion states that when no force acts, or
balanced forces act on an object, that object will remain at rest or
continue to move in a straight line at a constant speed.
Figure 7.1 Formula 1 racing cars
moving at speed around a corner.
116
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Figure 7.2 A modern bridge with central
pillar chords.
Engineers who build bridges, or who design Formula 1 cars, need a
profound understanding of mechanics. Engineers use the laws of physics,
and their knowledge of the strength and flexibility of materials, to calculate
the size and shape of material required at each stage of design and
construction. In this chapter you will meet ideas about the vector nature of
forces, and also the concept of turning moments. Both these ideas are used
by design and construction engineers.
17/04/19 9:48 AM
1 Explain why speed is a scalar quantity and velocity is a vector quantity.
2 Make an estimate of the size of the following forces:
a) The weight of an apple.
b) The push on a pedal when you cycle.
c) The drag on a car moving at 20 m s–1.
3 A parachutist falls at a constant speed of 5 m s–1. Which of the
following statements is true?
a) The pull of gravity on the parachutist is greater than the drag.
b) The pull of gravity on the parachutist is less than the drag.
c) The pull of gravity on the parachutist is the same size as the drag.
4 a) Explain why it is easier to undo a nut using a spanner with a long
handle.
b) Calculate the turning moment when a force of 30 N is applied at
right angles to the handle of a spanner with a length of 0.25 m.
Scalar and vector quantities
TEST YOURSELF ON PRIOR KNOWLEDGE
●● Scalar and vector quantities
A scalar quantity is one which has size only.
A vector quantity is one which has size and
direction.
A scalar quantity is one that only has size. Scalar quantities include:
mass, temperature, energy, distance, speed and time. We say that ‘room
temperature is 19°C’; it makes no sense to say ‘19°C westwards’.
A vector quantity is one that has magnitude (size) and direction. Vector
quantities include: force, acceleration, velocity and displacement. Velocity
and speed seem like similar quantities, as they both can be measured in
m s–1. But, if you write that a car is travelling at 30 m s–1, you are talking
about a speed; when you write that a car is travelling at 30 m s–1 due
east, you have used a vector quantity, which is called velocity. In a similar
way, the word displacement is used to describe a distance travelled in a
particular direction.
A resultant vector is the vector that results
from adding two or more vectors in a vector
sum (e.g. resultant force or resultant
displacement).
You have learnt to add or subtract simple vectors. In Figure 7.3 a two forces
acting in the same direction add up to make a larger resultant force of
200 N, and in Figure 7.3 b two forces acting in opposite directions make a
smaller resultant force of 300 N.
100 N
800 N
100 N
500 N
117
800 N
100 N
100 N
200 N
500 N
300 N
Figure 7.3 (a) two forces acting in the same direction and (b) two forces acting in
opposite directions.
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TEST YOURSELF
1 Name three scalar and three vector quantities,
including at least one of each that do not appear
on this page or on page 123.
2 Which of the following are vector quantities and
which are scalars: density, volume, momentum
(mass × velocity), weight, gravitational field
strength, electrical resistance, potential
difference?
3 A motorist drives from Abinton to Barmouth via
Chine Roundabout (see Figure 7.4). The journey
takes 0.8 hours. Calculate:
a) the distance travelled and the displacement
of the car.
b) the average speed of the car during the
journey and the average velocity of the car
during the journey.
N
Chine Roundabout
30 km
10 km
Abinton
25 km
Barmouth
Figure 7.4 Route map of journey from Abinton to
Barmouth via Chine Roundabout.
●● The addition of vectors
The method of adding vectors is illustrated by the following example.
EXAMPLE
7 INTRODUCTION TO MECHANICS
A rambler walks a distance of 8 km travelling due east, before
walking 6 km due north. Calculate their displacement.
118
C
θ
Answer
Displacement is a vector quantity, so we must calculate its
magnitude and direction. This calculation, for vectors at right
angles to each other, can be done using Pythagoras’ theorem
and the laws of trigonometry.
10 km
6 km
N
θ
A
(AC)2 = (8 km)2 + (6 km)2
N
8 km
B
Figure 7.5
(AC)2 = (64 + 36) km2
⇒ AC = 10 km
Directions can be calculated on a bearing from due north; this is the angle θ,
in Figure 7.5, which is also the angle θ, in the triangle.
8 km
tan θ =
6 km
⇒ θ = 53°
The displacement is 10 km on a bearing of 53°.
You are expected to be able to calculate vector magnitudes when two
vectors are at right angles, but the calculations are harder when two vectors
are separated by a different angle. Now the magnitude can be calculated
using a scale drawing.
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The addition of vectors
EXAMPLE
In Figure 7.6a, a liner is being pulled by two tugs. What
is the magnitude and direction of the resultant force?
Answer
two tug boats; these can be added by simply adding
the vector BC (which is parallel to line AD) to vector
AB. The resultant force is represented by the vector
AC, which can be measured to be 870 000 N.
B
500 000 N
500 000N
500 000N
tug
30
30
A
30°
30°
870 000N
C
tug
passenger liner
500 000N
500 000 N
D
Figure 7.6a A liner being pulled by two tugs.
100 000N
scale
Figure 7.6b shows you how to make the scale drawing.
The two vectors AB and AD show the pulls from the
Figure 7.6b Prepare a scale drawing.
TEST YOURSELF
4 A
plane flies due north with a speed of 420 km h–1;
a cross wind blows in a westerly direction with a
speed of 90 km h–1.
6 A
light aeroplane, whose air speed is 60 m s–1, is
flying in a westerly gale of 40 m s–1. (Westerly means
that the wind is coming from the west.)
a) Calculate the plane’s velocity relative to the
ground.
b) Calculate the time taken for a 600 km flight.
On another day, the plane flies due west with a
velocity of 400 km h–1, and the wind blows from
a bearing of 45° with a speed of 100 km h–1.
Draw vector diagrams to calculate the plane’s
ground speed, when:
a) The pilot keeps the plane heading east.
b) The pilot keeps the plane heading north
(the plane’s nose is pointing north, but it
does not travel north because the wind blows
it sideways).
c) The pilot now keeps the plane flying north.
In which direction is the plane headed in
this case?
c) Make a scale drawing to calculate the plane’s
velocity.
d) Calculate the time taken for a 600 km flight.
5 Calculate the resultant force on the plane shown
in Figure 7.7.
lift 22 000 N
119
drag
10 000 N
thrust from
engines 15 000 N
weight 20 000 N
Figure 7.7
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●● The resolution of vectors
Figure 7.8a shows a passenger pulling his wheelie bag at the airport. He
pulls the bag with a force, F, part of which helps to pull the bag forward
and part of which pulls the bag upwards, which is useful when the bag
hits a step.
Resolving a vector Split it into two
mutually perpendicular components that
add up to the original vector.
The force, F, can be resolved into two components: a horizontal component
Fh and a vertical component Fv. A vector can be resolved into any two
components that are perpendicular, but resolving a vector into vertical and
horizontal components is often useful, due to the action of gravity.
F
F
Fv
θ
Fh
Figure 7.8a and Figure 7.8b
Using the laws of trigonometry:
sin θ = Fv
F
and
cos θ = Fh
F
The vertical component of the force is:
Fv = F sin θ
The horizontal component of the force is: Fh = F cos θ
EXAMPLE
In Figure 7.9, a ball is travelling at a velocity of
25 m s–1 at an angle of 30 ° to the horizontal.
Calculate the horizontal and vertical components
of the ball’s velocity.
7 INTRODUCTION TO MECHANICS
TIP
You must make sure you
understand your trigonometrical
formulas to resolve vectors.
Answer
25 m s1
30°
Figure 7.9
vh = (25 m s–1) × cos 30°
= 21.7 m s–1
vv = (25 m s–1) × sin 30°
= 12.5 m s–1
An inclined plane
Figure 7.10 shows a car at rest on a sloping road. The weight of the car
acts vertically downwards, but here it is useful to resolve the weight in
directions parallel (||) and perpendicular (⊥) to the
road. The component of the weight parallel to the
θ
road provides a force to accelerate the car downhill.
W cos θ
There are other forces acting on the car, which are
W
dealt with in Figure 7.12.
120
θ
θ
W sin θ
The component of the weight acting along the slope is:
W|| = W sin θ
The component of the weight acting perpendicular to
the slope is:
W
Figure 7.10
469884_07_AQA_A-Level_Physics_116-132.indd 120
W⊥ = W cos θ
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You met the idea of balanced forces in the prior knowledge
section. In Figure 7.11 the motorcyclist has a weight of
800 N and the bike has a weight of 2500 N – a total of
3300 N. Those forces are balanced by the two contact forces
exerted by the road, so the bike remains at rest. This is an
example of the application of Newton’s first law of motion.
1100 N
2500 N
800 N
The example in Figure 7.11 is easy to solve as all of the forces
act in one direction. Figure 7.12 shows all the forces acting
on the stationary car first shown in Figure 7.10. Newton’s
first law of motion applies to this situation too, although the
forces do not all act in a straight line. However, the vector
sum of the forces must still be zero. Three forces act on the
car: its weight, a normal reaction perpendicular to the road,
and a frictional force parallel to the road.
2200 N
Figure 7.11
θ
N
N
W
F
F
The resolution of vectors
Forces in equilibrium
The triangle of forces shows that the three forces – W, N
and F – add up to zero, so there is no resultant force to
accelerate the car. We can also explain why the car has no
resultant force acting on it by resolving the forces along
and perpendicular to the plane.
Along the plane: F = W sin θ
θ
At right angles to the plane: N = W cos θ
W
So the two components of the weight are balanced by
friction and the normal reaction.
Figure 7.12
ACTIVITY
The apparatus shown in Figure 7.13 provides a
practical way for you to test the theory discussed
in the last section. Three weights are set up in
equilibrium. A central weight, W3, is suspended by two
light strings that pass over two frictionless pulleys.
These strings are attached to the weights W1 and W2.
Since point O is stationary, the three forces acting on
that point, T1, T2 and W3, must balance. The tensions
T1 and T2 are equal to W1 and W2 respectively.
We can demonstrate the balance of forces by
resolving the forces acting on point O vertically
and horizontally. The forces must balance in each
direction.
wooden
board
T1 sin θ = 3 N sin 60 = 2.6 N
pulley
T2
T1
W1
θ1
(3 N)
W3
θ2
O
(6 N)
W2
(5 N)
So, resolving horizontally:
T1 sin θ1 = T2 sin θ2
And resolving vertically:
T1 cos θ1 + T2 cos θ2 = W3
We can check using the example numbers in the
diagram:
pulley
T2 sin θ = 5 N sin 30 = 2.5 N
string
So the horizontal components balance to within a
reasonable experimental error.
slotted
metal
weights
121
Resolving vertically:
3 N cos 60 + 5 N cos 30 = 5.8 N
Figure 7.13 In this example θ1 = 60° and θ2 = 30°.
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And, to within a reasonable experimental error, the
vertical components balance too; this is close to the
weight of 6 N shown in the diagram.
a) Discuss how you would measure the angles θ1
and θ2 accurately.
b) Discuss what other sources of error might
contribute to the size of errors seen in these
checks.
c) A student now carries out some experiments
and in each case tries to calculate the unknown
weights or angles shown in Table 7.1. Copy and
complete the table, filling in the gaps. Suggest
what weight the student actually used; in each
case she used weights that were a whole number
of newtons.
Table 7.1
W1/N
6
W2 /N
6
8
7
6
W3/N
8
8
14
10
θ1
60°
θ2
60°
40°
42°
53°
40°
28°
7 INTRODUCTION TO MECHANICS
TEST YOURSELF
122
7 An oil tanker is pulled into harbour by two tugs.
Each tug applies a force of 180 000 N at an angle
of 37 ° to the forwards direction of the tanker.
Calculate the resultant force on the tug. You may
calculate this or use a scale drawing.
8 Three men are trying to move a piano. One pushes
southwards with a force of 100 N; the second
pushes westward with a force of 173 N; the third
pushes in a direction 60 ° east of north with a
force of 200 N. Calculate the resultant force on
the piano.
9 A helicopter rotor provides a lift of 180 000 N when
the blades are tilted 10 ° from the horizontal.
a) Resolve the lift into horizontal and vertical
components.
b) The helicopter is flying level. Calculate its mass.
Lift 180 000 N
10°
10
Figure 7.14
10 A
cyclist climbs a 1 in 5 slope as shown in Figure
7.15. The combined weights of the cyclist and the
bike are 950 N. The frictional forces acting on the
cyclist are 40 N.
469884_07_AQA_A-Level_Physics_116-132.indd 122
B
1m
θ
A
Figure 7.15
a) By resolving the weights of the cyclist and
the bike, calculate the force acting down the
slope.
b) Calculate the force he must exert on the road
to climb at a constant speed.
11 A massive ball is suspended on a string 1 m long,
and pulled sideways by a force of 1 N through a
distance of 30 cm.
a) Calculate the angle θ
shown in Figure 7.16.
b) Make a scale drawing
(in the form of a vector
triangle )of the three
forces that act on the
ball: its weight W, the
tension in the string T,
and the sideways pull of
1 N. Use the triangle to
calculate the magnitude
of W and T.
c) Check your answers by
calculation.
θ
1m
T
1N
0.3 m
Figure 7.16
17/04/19 9:48 AM
Turning moments
●● Turning moments
Introducing moments
(a)
You know from experience that it takes a large force to turn the lid on an
unopened jam jar, and that it is easier to turn a screw using a screwdriver
with a broad handle. So, to turn something effectively it is necessary to
apply a large force and to apply that force at a large distance from the
central point, or pivot, of the rotation.
100 N
P
(b)
The size of the turning effect is called a turning moment or torque.
0.3 m
Turning moment = force applied × perpendicular distance from the pivot
100 N
Figure 7.17a and Figure 7.17b illustrate the importance of the
perpendicular distance. In figure 7.17a there is no turning effect when the
force acts through the pivot. However, in Figure 7.17b:
Figure 7.17
Moment The force × perpendicular distance
from the pivot. The unit of a moment is N m
or N cm.
Pivot The point about which an object
rotates. In Figure 7.18 the support of the
see-saw is the pivot.
Equilibrium An object is in equilibrium,
when the sum of the forces = 0 and the sum
of the turning moments = 0.
Turning moment = 100 N × 0.3 m
= 30 N m
You know from Newton’s first law of motion that an object will remain
at rest if the forces on it balance. However, if the body is to remain at
rest without translational movement, or rotation, then the sum of the
forces on it must balance, and the sum of the moments on the object
must also balance.
Another way of expressing Newton’s first law is to say that when the vector
sum of the forces adds to zero, a body will remain at rest or move at a
constant velocity.
Figure 7.18 shows two children on a see-saw. The see-saw has a weight of
200 N, which can be taken to act through its support. We can show that the
see-saw is in equilibrium as follows.
1m
1.5 m
anticlockwise
450 N
clockwise
200 N
300 N
Figure 7.18
123
The anticlockwise turning moment of the boy = 450 N × 1.0 m = 450 N m
The clockwise turning moment of the girl = 300 N × 1.5 m = 450 N m
Since the clockwise and anticlockwise moments balance (or add up to
zero), one condition for equilibrium is met.
The total weight of the see-saw and the children, which acts on its support
is: 450 N + 300 N + 200 N = 950 N. The support provides an upwards force
of 950 N on the see-saw, so the resultant force acting on it is zero and the
see-saw is in equilibrium.
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TEST YOURSELF
12 State the unit of turning moment.
13 Figure 7.19 shows three children on a see -saw.
F
30
1.8 m
1.0 m
cm
1.6 m
P
B A
5 cm
250 N
350 N
F
500 N
Figure 7.19
Figure 7.20
a) Use the principle of moments to show that the
see -saw is in equilibrium.
b) State the size of the upward force exerted by
the pivot on the see -saw. Assume the see saw itself has negligible weight.
14 Figure 7.20 shows a pair of wire cutters.
b) You want to cut some thick wire. Use
your knowledge of moments to establish
whether it is better to cut the wire at point
A or B.
c) A wire, which needs a force of 210 N to cut it,
is placed at B, 5 cm from the pivot. How big a
force will you need to exert at the ends of the
handle to cut the wire?
a) Explain why the cutters have long handles.
7 INTRODUCTION TO MECHANICS
Centre of mass
The centre of mass is the point in a body
around which the resultant torque due to
the pull of gravity is zero.
Figure 7.21a shows a rod in equilibrium on top of a pivot. Gravity acts
equally on both sides of the rod, so that the clockwise and anticlockwise
turning moments balance. This rod is equivalent (mathematically) to
another rod that has all of the mass concentrated into the midpoint; this
is known as the centre of mass. In Figure 7.21b the weight acts down
through the pivot, and the turning moment is zero.
R
(a)
weight
pivot
(b)
R
(c)
pivot
weight
W
Figure 7.21a, Figure 7.21b and Figure 7.21c
124
centre of
gravity
W
The centre of mass is the point in a body around which the resultant torque
due to the pull of gravity is zero.
This means that you can always balance a body by supporting it under its
centre of mass. In figure 7.21c the tapered block of wood lies nearer to the
thicker end.
Figure 7.22 The rocking toy is stable
because its centre of gravity lies below
the pivot.
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0.3 m
P
0.21 m
45
Turning moments
Moments in action
When we look at problems in real situations they are not always as
straightforward as the examples above. Figure 7.23 shows the action of a
force to turn a spanner, but the line of action lies at an angle of 45° to the
spanner. How do we determine the turning moment now?
This can be done in two ways.
First, a scale drawing shows that the perpendicular distance between the
line of the force and the pivot is 0.21 m, so the turning moment is:
100 N × 0.21 m = 21 N m.
100 N
Figure 7.23
Secondly, the turning moment can be calculated using trigonometry,
because the perpendicular distance is (the length of the spanner) × sin θ.
⇒Moment = F × l × sin θ
= 100 N × 0.3 m × sin 45
= 21 N m
EXAMPLE
Figure 7.24 shows a man standing on a small bridge; he has a weight of
800 N and the bridge has a weight of 1200 N, which acts through the midpoint
(centre of mass) of the bridge. How can we use the principle of moments to
calculate the reaction forces, R1 and R2, which support the bridge?
Answer
If we calculate the moments
about A, then the clockwise
moments of the two weights are
balanced by the anticlockwise
moment of R2.
A
B
R1
R2
It follows that:
R2 × 5 m = 800 N × 1 m +
1200 N × 2.5 m
800 N
1200 N
R2 × 5 m = 3800 N m
R2 = 760 N
0
1m
2m
3m
4m
5m
Figure 7.24
But the sum of the forces, in the
vertical direction, must also balance:
R1 + R2 = 800 N + 1200 N = 2000 N
R1= 2000 N – 760 N = 1240 N
R1 is larger than R2 because the man is standing closer to A than to B.
469884_07_AQA_A-Level_Physics_116-132.indd 125
125
17/04/19 9:49 AM
Couples
F
Figure 7.25
Couple A pair of forces that provide a turning
effect but no translational movement. They
act in opposite directions, are parallel, but do
not act along the same line.
In Figure 7.25 a force is applied to an object. This will have two effects. The
first is to set the object off moving in the direction of the force (translation);
the second is to start the object rotating about its centre of gravity. If the
intention of applying the force is just to rotate the object, applying one
force is inefficient because it sets the object moving along too. To avoid
this problem we often apply two forces to the object that are parallel but in
opposite directions; this is called a couple.
A couple is shown in Figure 7.26a. You apply a couple when you turn a
steering wheel; the two forces turn the wheel but exert no translational
force. If you take one hand off the wheel (which you should do only to
change gear), you can still exert a couple, but the second force is applied by
the reaction from the steering wheel (see Figure 7.26b).
20 cm
(a)
(b)
20 N
10 N
10 N
20 N
7 INTRODUCTION TO MECHANICS
Figure 7.26a and Figure 7.26b
126
TEST YOURSELF
The bucket is now moved to the other end of the
ladder, 1 m away from her shoulder.
c) Calculate the reaction force now.
d) Explain how the woman can arrange the bucket
and ladder so that the reaction force is only
100 N.
16 The man shown in Figure 7.24 now stands 2 m
away from A. By taking moments about B,
calculate the reaction forces R1 and R2.
15 F
igure 7.27 shows a window cleaner carrying a
ladder and bucket.
0.5 m
3m
1m
17 C
alculate the two couples in Figures 7.26a and
7.26b; comment on the answers.
F
80 N
20 N
1m
Figure 7.27
a) Calculate the force, F, she must exert to
balance the ladder.
b) Now calculate the reaction from her shoulder
to support the ladder.
469884_07_AQA_A-Level_Physics_116-132.indd 126
18 T
he unit of a turning moment is N m. The joule,
which is the unit of work, can also be written as N m.
Explain why a turning moment is never described
as a joule. Hint: how do you calculate the work done
when a turning moment moves an object?
19 F
igure 7.28 shows a windsurfer in action.
The sail experiences a lift, L, which can be
represented by a force of 500 N as shown; this
acts at an angle of 30° and at a height of 1.5 m
17/04/19 9:49 AM
L
30
1.5 m
W
d
Figure 7.28
20 A
uniform beam AC is 6 m long. It is supported at
two points, B and C, where AB = 1 m, BC = 5 m.
The weight of the beam is 250 N.
a) Calculate the reaction forces from the
supports on the beam at points B and C. To
start solving this problem, take moments
about point B.
A man of mass 850 N stands on the beam at a
point D. The reaction forces from the supports at
B and C are now equal.
b) State the size of the two reaction forces now.
c) Calculate the distance AD.
21 The Royal Engineers build a small bridge to
cross a river. The bridge is 18 m long and is
supported at each end by two pillars of concrete.
The bridge is of uniform cross section along
its length. The weight of the bridge span is
32 000 N. A jeep with its occupants, with a weight
of 12 000 N, is 6 m from one end. Calculate the
forces on the pillars.
Turning moments
above the board. The surfer has a weight of 650 N.
How far out does he have to lean to balance the
board?
127
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Practice questions
1 Which of the following quantities is a scalar?
A acceleration
C kinetic energy
B momentum
D force
C
2 The direction and size of two forces acting on a body are shown
in Figure 7.29. The sum of these forces acts along the line AC.
Which of the following pairs gives the correct magnitude of the
sum of the forces, and the angle θ?
A 10.0 N, 34°
C 7.2 N, 48°
B 7.2 N, 34°
D 8.4 N, 42°
4N
A
θ
Figure 7.29
3 In Figure 7.30, a car accelerating down a slope under the action of
gravity; its engine is turned off. The mass of the car is 800 kg, and
the frictional force acting up the slope against the car is 720 N.
Which of the answers below gives the correct acceleration of the
car down the slope?
A 8.5 m s–2
C 4.9 m s–2
7.6 m s–2
4.0 m s–2
B
D
B
6N
N
F
W
30°
7 INTRODUCTION TO MECHANICS
4 Which of the following values gives the value for the normal force N, Figure 7.30
that the road exerts on the car in the previous question?
128
A 7800 N
C 3900 N
B 6800 N
D 3400 N
5 Figure 7.31 shows a metal ball supported by a string and held
at an angle of 20° to the vertical by a horizontal force of 1.5 N.
Which of the following possible answers is the correct value for
the weight of the ball?
A 4.1 N
C 1.4 N
B 0.6 N
D 4.4 N
20°
1.5 N
6 A plank of wood with a weight of 120 N rests symmetrically on two Figure 7.31
supports, A and B, which are 3.0 m apart as shown in the diagram
below. A box of weight 300 N is placed on the plank a distance
0.75 m from B. By taking moments about A, determine which of the
answers is the correct reaction force, R2, exerted on the plank at point B.
A 285 N
C 195 N
B 225 N
D 180 N
7 The weight is now moved to the right of B,
to a point where the plank begins to tip up
about B. Which of the answers below gives the
correct distance between the box and point B?
A 0.50 m
C 0.75 m
B 0.60 m
D 1.20 m
469884_07_AQA_A-Level_Physics_116-132.indd 128
3.0 m
1.5 m
R2
R1
A
B
120 N
300 N
0.75 m
Figure 7.32
17/04/19 9:49 AM
8 A piece of wood of weight 2.4 N is balanced on a support as
shown in Figure 7.33. Which of the answers gives the correct
position of the wood’s centre of gravity?
Practice questions
0.2 m
A 0.8 m to the right of the support
B 0.5 m to the right of the support
support
C 0.4 m to the right of the support
6N
D 0.1 m to the left of the support
9 Figure 7.34 shows a uniform ruler which is supported at its
midpoint. Two forces act on either side of the ruler as shown: a
force, F, acting at an angle of 37° and a weight of 3 N on the other
side. The ruler is in equilibrium. Which of the following answers is
the closest to the value of the force, F?
A 12 N
C 6N
B 9N
D 4N
C 9N
B 10 N
D 6N
42 cm
35 cm
37°
3N
10 The ruler has a weight of 3 N. Which of the following answers is
the closest to the vertical cimponent of the reaction force exerted
on the ruler by the pivot?
A 12 N
Figure 7.33
F
Figure 7.34
11 Copy and complete the table below to show the fundamental SI units of
the listed quantities, and to show whether each quantity is a vector or
a scalar quantity.
Table 2.10
Quantity
Force
Kinetic energy
Acceleration
Displacement
Power
Fundamental SI units
kg m s–2
Type of quantity
vector
12 A sprinter is set for the start of a race. His weight, W, is 780 N and the
reaction from the track on his hands, R1, is 320 N.
0.18 m
R1
R2
W
R3
129
0.54 m
A
B
C
0.81 m
Figure 7.35
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By taking moments about B, calculate:
a) the reaction R3
(3)
b) the reaction R2.
(1)
13 A truck has a weight of 16 000 N and a load of 8000 N.
0.75 m
1.5 m
8000 N R
2
R1
A
16 000 N
3.0 m
B
7 INTRODUCTION TO MECHANICS
Figure 7.36
130
a) Calculate the combined moments of the load and the weight
of the truck about point B.
(2)
b) State the principle of moments.
(1)
c) Use the principle of moments to calculate the reaction force R1,
and then calculate R2.
(3)
d) State the values of R1 and R2 when the truck carries no load.
(1)
14 A golfer decides to determine the position of the centre of
gravity of one of his clubs. He balances it as shown in the
diagram. The mass of the club is 350 g.
60 cm
28 cm
14 cm
head
a) Explain why this balancing arrangement produces a stable
position of equilibrium.
(1)
b) Calculate the weight of the club.
(1)
handle
pivot
5.4 N
Figure 7.37
c) Use the information to determine the position of the centre
of gravity from the head end of the club.
(2)
The golfer now hits the ball with a speed of 45 m s–1 at an angle
of 40° to the horizontal.
L
d) Calculate:
i) the vertical component of its velocity
(2)
ii) the horizontal component of its velocity.
(2)
D 7.4 N
20
T
W 1.7 N
The ball is in flight for 5.9 s.
e) Determine the distance the ball travels before it lands
on the ground.
(2)
15 Figure 7.38 shows the forces acting on a kite that is in
equilibrium.
a) Use the information in the diagram to calculate the magnitude
of the tension, T, in the string.
(3)
b) Now calculate the magnitude of the lift on the kite, L.
469884_07_AQA_A-Level_Physics_116-132.indd 130
(3)
Figure 7.38
17/04/19 9:49 AM
a) Draw a diagram for the point A (where the
car is suspended from the cable) to show
how the three forces are in equilibrium. (3)
T
Practice questions
16 Here is a cable car which is stationary on a wire.
T
9
9
A
b) Either by scale drawing, or by calculation,
determine the tension, T, in the cable.
(4)
17 A shelf with a weight of 16 N is placed (without
being fixed) on two supports B and D (see
Figure 7.40).
a) State the size of the reaction contact forces
at B and D.
(1)
A box of weight 40 N is now placed on the shelf
as shown in Figure 7.41.
W 13 500 N
Figure 7.39
b) Use the principle of moments to calculate the
reaction contact forces at B and D now. (4)
75 cm
The box is now moved to the left of B.
25 cm
c) Calculate the position of the box so that
the contact force at D is zero.
(2)
d) Calculate what additional weight you would
have to place above C to ensure that the
box will not tip the shelf when it is placed
at A or E.
(3)
C
A
D
B
E
Figure 7.40
15 cm
R1
18 A pedal bin is a labour-saving device in the
kitchen. Pressing on the pedal with your foot
causes a series of pivots and levers to lift the lid
so that you can throw rubbish into the bin.
When the pedal is depressed, rod R lifts the lid
(see Figure 7.42).
R2
C
A
B
D
E
40 N
Figure 7.41
a) When a force of 20 N is applied to the
pedal, as shown, calculate the force acting
(2)
upwards at A, to lift the rod R.
b) If 20 N is the minimum necessary force
applied to the pedal to lift the lid,
calculate the weight of the lid.
25 cm
3 cm
15 cm
Pivot 2
Lid
W
R
(2)
c) Calculate the contact force between pivot 2
and the lid as it is lifted by the 20 N force
from the pedal.
(2)
d) Explain how a small displacement of the
pedal lifts the lid a much larger distance. (2)
40 cm
20 N
131
Pivot 1
A
Pedal
9 cm
18 cm
Figure 7.42
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Stretch and challenge
A
19 The figure below shows a big wheel at a funfair.
The wheel’s centre of mass is at its centre of
rotation, point O. In the following calculations,
you may ignore any effects of friction.
At present only two of the seats on the wheel
are occupied: at A two people sit with a
combined weight of 1600 N; one person sits at
B with a weight of 800 N.
a) Calculate the vertical force required at the
embarking point C to prevent the wheel
rotating:
O
B
C
i) in the position shown
ii) after the wheel has been rotated
clockwise by 60°.
b) In what position will the wheel be in stable
equilibrium?
Figure 7.43
7 INTRODUCTION TO MECHANICS
c) In which position would you place two
more passengers, and what should their combined
weight be, so that the wheel has no tendency to rotate
in any position?
20 A light, rigid straight beam ABC is smoothly hinged at end
A, on a vertical wall, and supported in a horizontal position
by a wire, which is attached to the wall and the midpoint of
the beam, B, at an angle of 60° to the beam.
Calculate the tension in the wire and the direction
and magnitude of the force on the hinge when a mass M is
suspended from point C.
132
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8
Motion and its
measurement
PRIOR KNOWLEDGE
●
●
●
●
●
●
●
●
●
●
●
average speed = distance
time
Speed is measured in m s−1 or km h−1.
When we deal with the vector quantities, displacement and velocity,
we write a similar equation:
displacement
average velocity =
time
When an object speeds up it accelerates, and when it slows down it
decelerates.
Acceleration is a vector quantity as it has a direction. It is defined by
this equation:
change of velocity
acceleration =
time
The units of acceleration are m/s per second, usually written as m s−2.
The gravitational field strength, g, near a planet is defined as the
gravitational force acting on each kilogram:
On Earth, g = 9.81 N kg−1
Newton’s first law of motion: an object will continue to move in a
straight line with a constant speed (or remain at rest) unless acted on
by an unbalanced force.
Newton’s second law of motion: resultant force = mass × acceleration
A velocity vector may be resolved into horizontal and vertical
components.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 A car travels at 40 km h−1 for 2 hours then 60 km h−1 for 4 hours.
Calculate the car’s average speed.
2 A car slows down from a speed of 30 m s−1 to 12 m s−1 in a time of 6 s.
Calculate the car’s acceleration.
3 Convert a speed of 54 km h−1 to m s−1.
4 A ball is falling in a direction at 35° to the horizontal with a speed
of 12 m s−1.
Calculate the vertical and horizontal components of its velocity.
5 An astronaut throws a ball horizontally when standing on the surface
of the Moon, where there is no atmosphere.
Draw a diagram to show the direction of the force(s) acting on it.
(There is no need to calculate the size of any force.)
469884_08_AQA_A-Level_Physics_133-150.indd 133
133
17/04/19 9:51 AM
●● Motion in a straight line
In October 2012 Felix Baumgartner broke the existing records for the
highest manned balloon flight, the highest altitude for a parachute jump
and the greatest free fall velocity. He jumped from a height of nearly
39 km and reached a speed of 377 m s−1 (1357 km h−1). The successful
completion of such a challenge depends on careful planning and design.
Felix needed a balloon inflated to exactly the right pressure at sea level so
that it could expand as the atmospheric pressure reduced at high altitude;
he needed a pressurised space suit, capable of withstanding the shock of
free fall; and he relied on the calculations of physicists using the equations
of motion to predict his time of fall and knowing when it was safe to open
his parachute. You will learn about these equations in the chapter that
follows.
Displacement-time graphs
In the work which follows we study motion in a straight line, but since
the motion might be from left to right, or upwards then downwards, the
direction is important. Therefore we shall use the vector displacement, s,
and velocity, v.
134
displacement/m
B
10
D
A
3
Figure 8.2 shows a displacement–time (s–t) graph for a walker. In the first
three seconds she walks to the right (which is chosen to be positive); then
she stands still for two seconds, so her displacement does not change;
then she moves back to the left for five seconds and gets back to where
she started. So, after ten seconds, her displacement is zero.
The velocity of the walker can be calculated from the gradient of the
graph.
C
5
time/s
10
Figure 8.2 A displacement–time (s–t)
graph for a walker.
displacement/m
8 MOTION AND ITS MEASUREMENT
Figure 8.1 Alan Eustace on his way up
to break Felix Baumgartner’s record. In
October 2014, Alan fell from a height of
41.5 km.
P
Δs
Δt
time/s
Figure 8.3 A displacement–time graph
for a vehicle that is accelerating.
TIP
The gradient of a displacement–
time graph is velocity.
469884_08_AQA_A-Level_Physics_133-150.indd 134
Over the region AB:
v = (10 – 0) m = 3.3 m s−1
3s
Over the region BC:
v=0
Over the region CD:
v = (0 – 10) m = −2 m s−1
5s
Because the gradient is negative over the region CD the velocity is negative;
the walker is moving to the left.
In Figure 8.2 there are three constant velocities. This means that the
gradients of the graph are constant, so the graphs are ‘straight line’.
Figure 8.3 is a displacement–time graph for a vehicle that is accelerating.
As the velocity increases, the gradient increases.
To calculate the velocity at point P, we measure the gradient, which we do
by drawing a tangent to the curve at that point. We measure a small change
in displacement, Δs, which occurred over a small time interval, Δt.
Then
v=
Δs
Δt
17/04/19 9:51 AM
Motion in a straight line
TEST YOURSELF
1 Figures 8.4a, b and c show displacement–time
graphs for three moving objects. In each case
describe the motion in as much detail as possible.
2 Draw displacement–time graphs to illustrate the
motion in each of the following cases.
a) A walker travels at a constant velocity of 3 m s−1
for 30 s.
b) An express train slows down as it comes into a
station and stops; five minutes later it sets off
and accelerates up to its original velocity.
(a)
c) A parachutist jumps out of a plane, accelerates
to a high constant velocity before opening
the parachute and coming to ground. Draw
a graph with the ground as the point of zero
displacement.
d) A rubber ball falls from a height of 1 m and
hits the ground. It bounces back to height of
0.5 m. Sketch a graph to show the displacement
against time up to the second bounce.
(c)
(b)
60
40
s
s
s/m
0
t
20
0
0
5
10
15
t/s
t
Figure 8.4 Displacement–time graphs for three moving objects.
velocity/m s–1
160
120
Velocity–time graphs
Δt
80
D
C
Δv
P
B
40
Over the part AB, you can see from the graph that the acceleration is
4 m s−2. This is the gradient of the graph: (80 m s−1)/(20 s). But, over the
region BC, the gradient is changing, so we now use the formula:
A
20
40 60
time/s
80
Figure 8.5 shows a velocity–time graph for an aeroplane as it takes off.
Initially, the plane accelerates at a constant rate, shown on part AB of the
graph. Then the acceleration slows until the plane reaches a constant
velocity at point D.
100
a = Δv
Δt
Figure 8.5 A velocity–time graph for
an airplane at take off.
velocity/m s–1
20
A
A2 = 1 × 20 ms–1 × 8 s
2
15
10
where Δv means a small change in velocity, and Δt means
a small interval of time.
B
= 80 m
A1 = 20 ms–1 × 16 s
= 320 m
5
C
4
8
12
time/s
16
20
24
Figure 8.6 How to use a velocity–time graph to calculate the
distance travelled by a motorbike.
469884_08_AQA_A-Level_Physics_133-150.indd 135
At point P:
Δv 60 m s–1
= 1.25 m s−2
=
a=
Δt
48 s
Figure 8.6 shows how we can use a velocity–time
graph to calculate the distance travelled by a
motorbike. In this graph, the motorbike travels
at a constant velocity over the period AB, before
decelerating over the period BC. While the velocity is
constant, the distance travelled is represented by area A1,
which is 320 m.
135
17/04/19 9:51 AM
While the bike decelerates, we could use the formula;
distance = average velocity × time
But the average velocity is 10 m s−1, which is the average of 0 m s−1 and 20 m s−1.
So the distance can also be calculated using the area:
1
A2 = × 20 m s−1 × 8 s = 80 m
2
Figure 8.6 also shows that when the gradient of a velocity–time graph is
zero, then the velocity is constant; when the gradient is negative, the bike is
decelerating.
TIP
The area under a velocity–time
graph is the distance travelled.
The gradient of a velocity–time
graph is the acceleration.
TEST YOURSELF
B
136
C
A1
B
A
A2
D
t/s
C
100
velocity/m s–1
8 MOTION AND ITS MEASUREMENT
120
5 Figure 8.8 is a velocity–time graph for a firework
rocket which takes off vertically and, when the fuel
runs out, falls back to where it took off.
v/m s–1
3 a) Draw a velocity–time graph for a train that
slows down and stops at a station, and then sets
off again in the same direction, reaching the
same speed.
b) Draw a second velocity–time graph for a train
that comes into a terminus, stops and then
returns back in the opposite direction, reaching
the same speed.
4 Figure 8.7 is a velocity–time graph for the Maglev
train, which travels from Longyang Station in
Shanghai to Pudong International Airport.
E
F
80
Figure 8.8 A velocity–time graph for a
firework rocket.
60
40
20
A
0
0
D
100
200
300
time/s
400
Figure 8.7 A velocity–time graph for the
Maglev train.
a) Calculate the train’s acceleration over the
region AB.
b) Calculate the distance travelled when the train
travelled at a constant velocity.
c) Calculate the distance from Longyang Station to
Pudong Airport.
d) Calculate the train’s average speed for the journey.
a) Explain why the gradient gets steeper over the
region AB.
b) At which point does the rocket stop burning?
c) Explain why the gradient of the graph should be
about 10 m s−2 over the region CD.
d) Why is the gradient zero over the region EF?
e) Explain why the areas A1 and A 2 are the same.
6 A rubber ball falls for a time of 0.6 s before hitting
the ground. It is in contact with the ground for 0.02 s,
before bouncing upwards for 0.4 s; it then falls to
the ground in 0.4 s. Sketch a velocity-time graph for
the ball up to the second bounce.
●● Equations for uniform acceleration
So far you have learnt to analyse motion using graphs. We can use a
velocity–time graph, where the acceleration is constant, to derive formulas
to help us calculate velocity and displacement changes.
469884_08_AQA_A-Level_Physics_133-150.indd 136
17/04/19 9:51 AM
velocity/m s–1
v+u
2
v–u
Area = 1 (v – u)t
2
u
Area = ut
u
t
time/s
The first equation is:
acceleration = change of velocity
time
or
a=v–u
t
Rearranging the equation to make v the subject of the
formula gives:
v = u + at
(Equation 1)
The second equation is:
Equations for uniform acceleration
Figure 8.9 is a velocity–time graph showing a vehicle
increasing its velocity with a constant acceleration, a, from
an initial velocity, u, to a final velocity, v, over a time, t.
v
displacement = average velocity × time
Figure 8.9 A velocity–time graph showing a vehicle
increasing its velocity with a constant acceleration.
The average velocity is half way between the initial and
final velocities.
Substituting for average velocity using:
(u + v)
2
gives:
(Equation 2)
s = u + v t
2
The displacement can also be calculated from the area under the velocitytime graph, which is the sum of the two areas shown in Figure 8.9.
( )
The third equation is:
1
s = ut + (v − u)t
2
(which is the area under the graph).
Therefore
1
s = ut + at2
2
because
(v − u) = at
(Equation 3)
(from Equation 1)
The fourth equation, which links velocity, acceleration and displacement, is:
v2 = u2 + 2as
(Equation 4)
There is no need to be able to derive these equations and you will find them on
your formula sheet. The derivation of Equation 4 is shown in the Maths box.
137
MATHS BOX
The fourth equation can be derived as follows:
v = u + at
→ v2 = (u + at)2
→ v2 = u2 + 2uat + a2 t2
→ v2 = u2 + 2a ut + 1 at2
2
→ v2 = u2 + 2as
469884_08_AQA_A-Level_Physics_133-150.indd 137
(using Equation 3)
17/04/19 9:51 AM
EXAMPLE
You drop a stone off the edge of a high cliff (making sure there is no-one
below). How far does it fall after 2 seconds?
Answer
Use
s = ut + 1 at2
2
The initial velocity is zero, and the acceleration due to gravity is 9.8 m s−2.
s = 0 + 1 × 9.8 m s−2 × 22s2
2
s = 19.6 m
EXAMPLE
What is the displacement after 2 seconds if you throw
the stone:
a) with an downward velocity of 5.0 m s−1
b) with a upward velocity of 5.0 m s−1?
Answer
8 MOTION AND ITS MEASUREMENT
We need to remember that velocity and displacement
are vector quantities, so we need to define a direction.
We will call down a positive direction, and up a
negative direction. (The choice is arbitrary; you can
reverse the direction signs if you wish.)
138
a) Use
s = ut + 1 at2
2
s = 5.0 m s−1 × 2 s + 1 × 9.8 m s−2 × 22s2
2
s = 29.6 m
b) Use
s = ut + 1 at2
2
s = −5.0 m s−1 × 2 s + 1 × 9.8 m s−2 × 22s2
2
s = 9.6 m
EXAMPLE
A ball is thrown upwards with an initial velocity of
15 m s−1 from a height of 1 m above the ground. When
does it reach a height of 7 m above the ground?
Answer
The first point to appreciate about this example is that
there are two times when the ball is 7 m above the
ground: when it is on the way up, and when it is on the
way down again.
We use the same equation again, using s = −6 m (for
the distance travelled from the start); u = −15 m s−1;
g = 9.8 m s−2. Here the downward direction has been
defined as positive.
Use
−6 = −15t + 1 × 9.8 × t2
2
2
4.9t − 15t + 6 = 0
This can be solved using the standard formula for the
solution of quadratics:
2
t = 15 ± √15 − 4 × 4.9 × 6
2 × 4.9
15 ± 10.3
t=
9.8
t = 2.6 s or 0.5 s
So the ball reaches a height of 7 m after 0.5 s on the
way up, and passes the same height on the way down
after 2.6 s.
s = ut + 1 at2
2
469884_08_AQA_A-Level_Physics_133-150.indd 138
17/04/19 9:51 AM
7 A
plane accelerates from rest along a runway at
a constant rate of 1.8 m s−2. The plane takes off
after 40 s.
a) Calculate the velocity of the plane at take off.
b) Calculate the minimum length of runway
required for the plane to take off.
8 A car slows down from a speed of 30 m s−1 to
22 m s−1 over a distance of 120 m. Calculate the
deceleration of the car.
9 A Formula 1 car accelerates off the grid reaching
a speed of 55 m s−1 in 3.6 s.
a) Calculate the acceleration of the car over this
time.
b) Calculate the distance travelled by the car in
this time.
10 An express train travelling at 50 m s−1 applies its
brakes for 3 minutes before reaching a station,
Equations for uniform acceleration
TEST YOURSELF
which it passes at 30 m s−1. How far from the
station was the train when the brakes were first
applied?
11 A car starts from rest with a uniform acceleration
of 2 m s−2 along a long straight track. At the
moment the car starts, a second car passes it on
a parallel track travelling at a constant speed of
20 m s−1. Calculate:
a) the time at which the cars are level
b) the velocity of the first car at that time
c) the distance travelled by the first car when it
catches the second car.
12 A ball is launched upwards from ground level
with a speed of 20 m s−1. At what times is the ball
at a height of 10 m above the ground? (Hint: look
at the example above.)
Acceleration due to gravity, g
One of the first scientists to study the acceleration of objects due to gravity
was Galileo Galilei. There is story that he dropped two iron balls of different
masses from the top of the Leaning Tower of Pisa. Thereby, he demonstrated
that gravity accelerates all masses at the same rate, provided that air
resistance is negligibly small.
By such an experiment you could calculate the acceleration due to gravity,
using the equation:
1
s = gt2
2
which rearranges to give
2s
g= 2
t
For example, Galileo’s assistant could have timed such a fall (from the top of
the tower, which is 55 m high) as three and a quarter swings of his 1-second
pendulum.
2 × 55 m
3.252
g = 10.4 m s−2
g=
This answer illustrates one of the problems we have if we try to measure
g accurately. We need very accurate timing and, because the calculated
answer depends on the value of t2, small errors in the measurement of t
become significant.
469884_08_AQA_A-Level_Physics_133-150.indd 139
139
17/04/19 9:51 AM
REQUIRED PRACTICAL 3
Determination of g by a free-fall method
Note: This is just one example of how you might tackle this required
practical.
Figure 8.10 shows an experimental arrangement using two light gates to
calculate the gravitational acceleration. A steel ball is dropped through
the two gates, and the times to fall through the gates are measured
and recorded in the table. The ball is guided through the light gates by a
glass tube.
Experiment number
Diameter of the ball (cm)
Time for the ball to pass
through gate A (ms)
Time for the ball to pass
through gate B (ms)
Time for the ball to pass from
the centre of gate A to the
centre of gate B (ms)
1
1.32
8.3
2
1.32
6.7
3
1.32
9.7
2.8
3.1
5.2
318.0
219.7
121.2
8 MOTION AND ITS MEASUREMENT
Using the results from experiment 1, the acceleration g is calculated as follows.
140
The velocity of the ball as it falls through gate A:
0.0132 m
= 1.59 m s−1
vA =
0.0083 s
The velocity of the ball as it falls through gate B:
0.0132 m
= 4.71 m s−1
vB =
0.0028 s
VB − VA
g=
t
4.71 m s−1 − 1.59 m s−1
=
0.318 s
= 9.81 m s−2
steel ball
to computer
A
An advantage of using light gates and a computer is that the
experiment can be repeated several times, and the computer
can be programmed to calculate the acceleration directly.
1 The experiment is repeated using different separations of the
light gates. Use the data in experiments 2 and 3 to calculate
the acceleration due to gravity.
2 Use the equations of motion to calculate the separations of
the light gates in each of the experiments.
3 Explain why it is not necessary to know the separation of the
light gates to calculate to gravitational acceleration.
4 Does it matter how far above the first light gate you release
the steel ball?
5 Discuss the possible sources of error in this experiment.
6 Calculate the average value of the acceleration and estimate
the percentage error in the measurements.
469884_08_AQA_A-Level_Physics_133-150.indd 140
glass tube
to computer
B
Figure 8.10 A ball is timed as it falls between
two light gates.
17/04/19 9:51 AM
Drag is the name given to resistive forces
experienced by an object moving through a
fluid such as air or water.
Terminal speed is the speed reached when
the weight of an object in free fall is balanced
by drag forces acting upwards on the object.
1N
1N
In everyday life we are all aware of the effect of wind resistance, or drag,
when something falls to the ground. A piece of paper that has been rolled
up into a ball accelerates quickly downward, but the same piece of paper
left as a sheet will flutter from side to side as it falls due to a large drag
acting over the larger surface area.
The size of the drag on a falling object increases with:
●
●
10 N
1N
Figure 8.11 At this instant the two
balls fall at the same speed; the drag
is the same on each but the blue ball
continues to accelerate.
speed
surface area.
Figure 8.11 explains in more detail how drag affects the accelerations of two
similar balls. They are both of identical size and shape, but the blue ball
has a weight of 10 N and the red ball 1 N. In the diagram they are falling at
the same speed and they both have an upward drag of 1 N. The blue ball
continues to accelerate because there is a resultant downwards force acting
on it; but the red ball moves with a constant speed as the pull of gravity is
balanced by the drag. The red ball has reached its terminal speed.
Equations for uniform acceleration
Terminal speed
TEST YOURSELF
13 A parachutist together with his parachute has a weight of 850 N.
a) Draw a diagram to show his weight and the drag force acting on him
when he falls at his terminal speed without his parachute open.
b) Draw a second diagram to show the weight and drag forces when he
falls at his terminal speed with his parachute open. Explain why the
two terminal speeds are different.
14 A student drops a ping-pong ball and a steel ball of the same diameter
from a height of 5 m.
a) Explain why the steel ball reaches the ground first.
b) Sketch graphs to show how the velocity of each ball changes with time.
15 Figure 8.12 shows another method of measuring g. A small steel ball is
dropped from a height, h, and passes through a light gate. The data from
the gate goes into a computer, which calculates the ball’s speed, v.
Height fallen: h (m)
Speed: v (m s−1)
0.10
1.40
0.15
1.72
0.26
0.38
2.26
2.73
0.55
0.75
3.28
3.84
steel ball
v2 (m2 s−2)
glass tube
h
141
a) Copy the table and complete the column to calculate value of v 2.
b) The relationship between v and h is given by the equation:
v 2 = 2gh
This relationship can be deduced from the equation of motion:
v 2 = u2 + 2as.
Plot a suitable graph and use it to calculate the value of g.
469884_08_AQA_A-Level_Physics_133-150.indd 141
Figure 8.12 The speed is measured as
the ball falls from different heights.
17/04/19 9:51 AM
velocity/m s–1
16 F
igure 8.13 shows the velocity-time graph that described Felix
Baumgartner’s free fall from a height of 39 km.
a) Determine his acceleration over the region AB of the graph.
b) Explain why his acceleration decreases over the region BC of the graph.
c) Over the region CD of the graph, Felix was
400
falling at his terminal speed. Yet this
C
terminal speed kept decreasing. Explain why.
300
d) At time E Felix opened his parachute. Use the
graph to estimate the distance he
fell before he opened his parachute.
200
B
e) Calculate Felix’s average speed over the period
A to E.
100
D
f) One of the claims made about this descent was
E
F
that Felix reached ‘Mach 1.2’: that
A
he travelled at 1.2 times the speed of sound.
100
200
300
From the graph, deduce the speed of
time/s
sound at high altitude. Why is this speed less
Figure 8.13 The velocity–time graph that described Felix
than the speed of sound at sea level,
Baumgartner’s free fall from a height of 39 km.
at 20 °C, which is 340 m s−1?
●● Projectile motion – or falling sideways
Projectile motion builds on the following principles.
●Newton’s
8 MOTION AND ITS MEASUREMENT
●
●
●
first law of motion: an object will continue to move in a straight
line with a constant speed (or remain at rest) unless acted on by an
unbalanced force.
Newton’s second law of motion: an object accelerates in the direction of
a resultant force acting on it (resultant force = mass × acceleration),
see Chapter 9.
Projectile paths may be predicted using the equations of motion.
A vector velocity may be resolved into horizontal and vertical components.
Falling sideways
Figure 8.14 shows a diver jumping into the sea. Each
image is taken 0.2 seconds apart. You can see that the
diver is displaced sideways by a constant amount in each
frame of the picture. This is because there is no horizontal
force acting on him, so he keeps moving with a constant
speed in that direction. Yet his displacement downwards
increases with each frame of the picture. This is because
gravity acts to accelerate him, so his downwards velocity is
increasing.
142
Figure 8.14 A man jumping off a diving board into the sea.
469884_08_AQA_A-Level_Physics_133-150.indd 142
An important principle is that the vertical and horizontal
motions of the diver are independent of each other. When
the diver jumps horizontally off the diving board, he will
always reach the water in the same time because he has no
initial velocity in a vertical direction. However, the faster he
runs sideways, the further he will travel away from the board
as he falls.
17/04/19 9:51 AM
EXAMPLE
s = 1 gt2
2
12.5 m = 1 × 9.8 m s−2 × t2
2
Rearranging to find t2:
12.5 2
t2 =
s
4.9
Taking square roots of each side of the equation gives:
In Figure 8.14, the board is 12.5 m above the sea, and
the man jumps 7.0 m sideways, from the end of the
board, by the time he splashes into the sea.
a) How long does it take him to fall into the sea?
b) How fast was he running when he jumped off the
board?
Answer
a) Use the equation:
s = ut + 1 at2
2
This equation is being used to consider the vertical
displacement under the influence of gravity; the final
displacement is 12.5 m, the initial (downward) velocity
is zero, and the acceleration due to gravity, g, is
9.8 m s−2.
Projectile motion – or falling sideways
In all of the calculations that follow, we shall assume that we may ignore the
effects of air resistance. However, when something is moving very quickly –
a golf ball for example – the effects of air resistance need to be taken into
account. The effect of air resistance on the path of a golf ball is to reduce
its maximum height and to reduce the horizontal distance (range) that it
travels.
t = 1.6 s
b) Because his horizontal velocity, vh, remains constant:
vh = horizontal displacement
time
7m
vh =
1.6 s
= 4.4 m s−1
EXAMPLE
A basketball player passes the ball at
a speed of 11 m s−1 at an angle of 30°
as shown in Figure 8.15.
B
Calculate
a) the greatest height the ball reaches
b) the time the ball takes to reach this
height
c) the horizontal distance, x, it travels.
A
30
h
C
Answer
a) Figure 8.16 shows how we can
resolve the velocity of the ball into
vertical and horizontal components,
vv and vh.
vv = 11 sin 30° = 5.5 m s−1
vh = 11 cos 30° = 9.5 m s−1
469884_08_AQA_A-Level_Physics_133-150.indd 143
2m
143
x
Figure 8.15 A basketball player passing a ball.
17/04/19 9:51 AM
The height, h, to which the ball rises above the line AC
can be calculated using the equation of motion:
v = 11 m s–1
vv2 = 2 gh
(5.55 m s−1)2 = 2 × 9.81 m s−2 × h
30°
h = 1.5 m
v cos 30 = 9.5 m s–1
s o the greatest height the ball reaches above the ground
is 3.5 m.
b) The time taken to reach this height is calculated
using the equation of motion:
v = u + at
he final vertical velocity at the top of the ball’s path
T
(point B) is zero, the initial velocity is −5.5 m s−1, and
the acceleration is +9.8 ms−2.
Figure 8.16 Resolving the velocity of the ball
into vertical and horizontal components.
c) The time taken for the ball to travel from A to B is
0.56 s, so the time taken for it to travel from A to C
is 2 × 0.56 s = 1.12 s.
he horizontal displacement, x, is calculated using
T
the equation:
x = vht
0 = −5.5 m s−1 + 9.8 m s−2 × t
5.5 m s−1
t=
9.8 m s−2
= 0.56 s
x = 9.5 m s−1 × 1.12 s
x = 10.6 m
144
1 m/s
1 A student thinks that a ball which is thrown
sideways takes longer to reach the ground than
a similar ball dropped from the same height.
Look at Figure 8.17 and comment on
this observation.
2 Use the information in the diagram to show
that the time interval between each photograph
is 0.1 s.
3 Either copy the diagram and mark in the positions
B 3 and B4, or write the coordinates of B 3 and B4.
4 Calculate the velocity of ball A at position A 4
(g = 9.8 m s−2).
5 Calculate the magnitude and direction of the
velocity of ball B at position B5.
A1
0
Figure 8.17 shows a multiple exposure photograph
of two small balls that are allowed to fall freely
under gravity. Ball A is dropped from rest and falls
vertically from position A1 to position A5. Ball B is
released at the same time as A, but falls with an
initial sideways velocity of 1 m s−1; ball B falls from
position B1 to position B5. The grid behind the balls
allows you to calculate how far they travel.
B1
B2
A2
10
A3
20
30
40
y/cm
8 MOTION AND ITS MEASUREMENT
ACTIVITY
Projectile paths
All the positions of ball A – A1 to A5 – are shown, but
positions, B 3 and B4 of ball B have been removed.
v sin 30 = 5.5 m s–1
A4
50
60
70
A5
80
B5
90
100
0
10
20
30
40
50
60
70
x/cm
Figure 8.17 A multiple exposure photograph of two small balls.
469884_08_AQA_A-Level_Physics_133-150.indd 144
17/04/19 9:51 AM
260
A8
240
A7
A6
220
A5
200
180
A4
160
140
y/cm
6 Copy the diagram on to graph paper
and mark in the positions B4, B5, B6
and B7, or write their coordinates.
7 Position A 8 is the maximum height
reached by ball A.
a) State the velocity of ball A at this
highest position.
b) State the velocity of ball B at its
highest position, B8.
8 Draw diagrams to show the positions
of balls A and B at intervals of 0.1 s as
they fall back to the ground.
Projectile motion – or falling sideways
In a second experiment balls
A and B are photographed again with
multiple exposures separated by
0.1 s. This time A is thrown upwards
with an initial velocity of 7 m s−1, while
B has an initial upward velocity of
7 m s−1and a horizontal velocity of 2 m s−1
to the right.
A3
120
B3
100
80
B2
A2
60
40
20
A1
0
0
B1
20
40
60
80
100
x/cm
120
140
160
180
Figure 8.18 The second experiment.
TEST YOURSELF
17 E
xplain which of the following statements is/are
true.
a) A bullet fired horizontally takes longer to fall
to the ground than a bullet which is dropped
from the same height.
b) When you throw a ball sideways, there is a
force pushing it forwards while the ball is in
mid-air.
c) The pull of gravity on an object is independent
of the velocity of the object.
18 A marksman aims his rifle at a target 160 m away
from him. The velocity of the bullet as it leaves
the rifle horizontally is 800 m s−1.
469884_08_AQA_A-Level_Physics_133-150.indd 145
a) Calculate the time the bullet takes to reach
the target.
b) How far has the bullet fallen in this time?
c) The target is now placed 480 m away from the
marksman.
i) Calculate how far the bullet falls now before
reaching the target.
ii) Explain why the marksman must adjust his
sights before aiming at the second target.
19 A cricketer throws a ball sideways with an initial
velocity of 30 m s−1. She releases the ball from a
height of 1.3 m. Calculate how far the ball travels
before hitting the ground.
145
17/04/19 9:51 AM
Practice questions
1 A ball is thrown vertically upwards, with an initial speed of 20 m s−1 on
a planet where the acceleration due to gravity is 5 m s−2. Which of the
following statements about the ball’s velocity after 6 s is true?
A The ball’s velocity is 0.
B The ball’s velocity 10 m s−1 upwards.
C The ball’s velocity is 10 m s−1 downwards.
D The ball’s velocity is 5 m s−1 downwards.
2 A car accelerates from rest at a rate of 3.5 m s−2 for 6 s; it then travels at
a constant speed for a further 8 s. Which of the following is the correct
distance that the car has travelled in this time?
A 231 m
C 147 m
B 215 m
D 49 m
8 MOTION AND ITS MEASUREMENT
3 A plane accelerates from rest along a runway. It takes off after travelling a
distance of 1200 m along the runway at a velocity of 60 m s−1. Which of
the following is the average acceleration of the plane along the runway?
A 3.00 m s−2
C 0.67 m s−2
B 1.50 m s−2
D 0.33 m s−2
4 A ball P of mass 2m is thrown vertically upwards with an initial speed
u. At the same instant a ball Q of mass 3m is thrown at an angle of 30°
to the horizontal with an initial speed of 2u. Which of the following
statements about the two balls is true?
A Q reaches twice the vertical height as P.
B Q reaches the ground first.
C Q and P reach the ground at the same time.
D The horizontal component of Q’s velocity is u.
In each of the questions 5–7, select from the list A–D the relationship that
correctly describes the connection between y and x.
A y is proportional to x2
1
B y is proportional to x 2
C y is proportional to x
146
D y is proportional to 1x
Question
5
6
7
y
the speed of an object falling freely from rest
the speed of an object falling freely from rest
the distance of an object falling freely from rest
469884_08_AQA_A-Level_Physics_133-150.indd 146
x
time
height fallen
time
17/04/19 9:51 AM
y
x
Question
the speed of a car which decelerates at an increasing rate
time
8
the horizontal distance travelled by a projectile moving
time
9
without air resistance
the speed of an object falling when air resistance acts on it time
10
A
B
C
D
Practice questions
In each of the questions 8–10, select from the graphs A–D the one that
correctly shows the relationship between y and x.
11 Figure 8.20 shows the trajectory of a real golf ball (path A) and an
Figure 8.19 Four graphs.
idealised ball (path B) that is not affected by air resistance.
60
Ideal ball,
no air
resistance
B
40
Golf
ball
A
20
0
V
0
40
80
120
160
200
240
280
Figure 8.20 The trajectory of a real golf ball (path A)
and an idealised ball (path B) that is not affected by
air resistance.
Drag
(2)
a) State three differences between the two trajectories.
b) Figure 8.21 shows the forces acting on the ball just after it
has been struck by the club. Use this information to explain
your observations in part (a).
(3)
W
Figure 8.21 The forces
acting on the ball just after it
has been struck by the club.
12 A boy throws a ball vertically upwards; it rises to a maximum height of
19.0 m in 2.0 s.
a) Calculate the speed of the ball as it left the boy’s hand.
(2)
b) Calculate the height of the ball, above its position of release,
3.0 s after the boy threw it.
(2)
c) State any assumptions you have made in these calculations.
(1)
a) Use the graph to calculate the acceleration of the ball:
i)
while it is falling
ii) while the ball is in contact with the floor.
(2)
(3)
In each case state the magnitude and direction of the
acceleration.
b) i)
Calculate the distance from which the ball was dropped. (2)
ii) Calculate the height of the first rebound.
469884_08_AQA_A-Level_Physics_133-150.indd 147
velocity/m s–1
13 Figure 8.22 shows a velocity–time graph for a hard rubber ball
which is dropped on to the floor before the rebounding upwards.
10
8
6
4
2
–2
–4
–6
–8
1
2
time/s
147
Figure 8.22 Velocity – time graph for a
(2) hard rubber ball.
17/04/19 9:51 AM
V
14 Figure 8.23 is a velocity–time graph for a moving object.
a) Explain the significance of:
i)
(2)
the positive and negative gradients of the graph
ii) the areas under the graph, both above and below the time axis. (2)
b) Use your answer to part (a i) to help you draw an
acceleration–time graph for the object.
(2)
c) Use your answer to part (a ii) to help you to draw a
displacement–time graph for the moving object.
(3)
t
Figure 8.23 A velocity–time
graph for a moving object.
15 A bungee jumper takes a plunge off a bridge. Figure 8.24 shows
a velocity–time graph for the jumper as he falls from the bridge.
a) Use the graph to calculate his acceleration over the first
2 seconds.
15
At which time does the jumper reach the lowest
point of his fall?
(1)
ii) Use the graph to estimate approximately how far he
falls before being stopped by the bungee rope. Is the
(1)
distance fallen closest to: 20 m, 40 m, or 60 m?
iii) Explain how the graph shows you that the jumper
does not bounce back to his original height.
8 MOTION AND ITS MEASUREMENT
148
5
0
16 Figure 8.25 shows a fort that has been built to protect the coast
from enemy ships. A cannon ball, which is fired horizontally
from the fort, falls into the sea a distance of 1250 m from the
fort.
A
C
1.0
2.0
3.0
E
4.0
–5
(1)
(1)
c) Calculate the acceleration at point B on the graph.
10
velocity/m s–1
b) i)
B
20
(2)
5.0
6.0
time/s
–10
–15
D
Figure 8.24 A velocity–time graph for the
jumper as he falls from the bridge.
cannon
93 m
1250 m
Figure 8.25 A fort that has been built to protect the coast from enemy ships.
a) Use the information in the diagram to show that the ball takes
about 4.4 s to fall into the sea.
(3)
b) Calculate the horizontal velocity of the cannon ball during its flight. (2)
c) Calculate the vertical velocity of the cannon ball after it has fallen
for 4.4 s.
(2)
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Practice questions
d) Calculate, or determine by scale drawing, the magnitude
and direction of the ball’s velocity as it falls into the sea. (3)
e) Discuss how you would adjust the cannon to hit a ship
that is further away from the fort.
(1)
17 A student uses two light gates as shown in Figure 8.26 to
measure the time taken for a small ball bearing to fall through
different heights. The table below shows the results of the
experiments. The ball bearing was dropped three times from
each different height. To release the ball bearing the student
held it just above the higher light gate.
Height (mm)
100
200
300
400
500
600
t1 (s)
0.14
0.18
0.25
0.29
0.32
0.34
t2 (s)
0.15
0.20
0.25
0.28
0.31
0.35
t3 (s)
0.14
0.21
0.26
0.30
0.32
0.34
h
To data logger
Figure 8.26 Using light gates to measure
the time taken for a small ball bearing to
fall through different heights.
a) Calculate the average time of fall from each height.
(1)
b) Calculate the maximum percentage error in timing when the
ball bearing is dropped from a height of
i)
100 mm
(2)
ii) 600 mm.
c) Calculate the approximate percentage error in measuring
the height of 100 mm.
(1)
d) Comment of the errors that could be introduced by the
student’s method of releasing the ball bearing.
(2)
t2
e) Plot a graph of height on the y-axis against on the x-axis.
Use the graph to determine a value for g. Comment on the
accuracy of the result.
(5)
18 A plane travelling with a constant horizontal speed of 54 m s−1
drops a food parcel to people stranded in a desert. The parcel
takes 3.7 s to reach the ground.
a) Calculate the height of the plane above the ground.
(3)
b) Calculate the horizontal distance travelled by the parcel
before it hits the ground.
(2)
c) i) Calculate the parcel’s vertical component of velocity when
it hits the ground.
ii) Calculate the magnitude of the parcel’s velocity as it hits the
ground.
149
(2)
(2)
Stretch and challenge
19 A family enjoying an outing on a river are cruising upstream in their
boat. Just as they pass a bridge (point A) their sandwich box falls into
the river.
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They continue upstream for 10 minutes before they notice
their loss. They then turn round (point B in the diagram) and
travel downstream. They pick up the sandwich box 1.0 km
downstream from the bridge at point C.
Bridge
A
Calculate the velocity of the river, assuming that the boat has the
same velocity relative to the water up and downstream, and that
no time is wasted turning round the boat at point B.
(Hint: you do not need to know the velocity of the boat; you
do need to set up some equations to calculate the time taken to
catch up with the sandwich box.)
20 A swimmer at point A next to the side of a swimming pool
wishes to get to a friend at point B as quickly as possible. He can
walk next to the pool at 3 m s−1 and swim at a speed of 1 m s−1.
B
C
1 km
riverflow
Figure 8.27 A family enjoying an outing.
Show that he must swim at an angle θ = sin−1 13 to get there in the
shortest time.
(Hint: you need to set up an equation to calculate the time in terms of
distances L, x and y, and the speeds. Then differentiation gives you the
minimum time. You need to use your knowledge of Maths A-level to
solve this.)
Lx
A
8 MOTION AND ITS MEASUREMENT
θ
150
y
x
B
L
Figure 8.28 Swimming pool.
21 A supersonic aircraft leaves Heathrow to travel to Kennedy airport.
You are required to use the information below to calculate the plane’s
position between the airports at 12:30 p.m. (Hint: sketch a velocitytime graph to help you plan your solution.)
●
The plane departs from Heathrow at 10:30 a.m. sharp.
●
The flight time is exactly 3 hours and 40 minutes.
●
The distance between the airports is 5800 km.
●
The plane flies for the first 15 minutes to a point 235 km from
Heathrow. At that point it is flying at 990 km h−1. It then accelerates at
a constant rate for 11 minutes over a distance of 250 km to a cruising
speed of 2160 km h−1.
●
The plane flies in a straight line.
●
The deceleration into Kennedy airport is uniform.
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9
Newton’s laws of motion
PRIOR KNOWLEDGE
●
●
●
●
●
●
A force is a push or a pull. Forces can be contact or non-contact
forces. A contact force is exerted between bodies that touch each
other; non-contact forces are exerted over a distance by gravitational,
electric and magnetic fields.
The strength of a gravitational field is measured in N kg−1. On the
surface of the Earth the gravitational field strength is 9.8 N kg−1.
Weight = mass × gravitational field strength
Newton’s first law of motion: a body will remain at rest or continue to
move in a straight line with a constant velocity unless it is acted on by
an unbalanced force.
Newton’s second law of motion: when an unbalanced (or resultant)
force acts on a body, it will accelerate in the direction of that force. The
size of the acceleration may be determined by using the equation F = ma.
Newton’s third law of motion: when body A exerts a force, F, on body B,
body B exerts an equal and opposite force, F, on body A.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 Karen has a mass of 57 kg.
a) Calculate her weight on the Earth.
b) Calculate her weight on Mars, where the gravitational field
strength is 3.7 N kg−1.
2 a) Describe what will happen to the stage of the motion of each of the
bodies shown below. In each case use a suitable phrase such as:
remains at rest; continues to move at a constant speed; changes
direction; accelerates; decelerates.
(i)
(ii)
m 2 kg
10 N
v0
(iv)
10 N
(iii)
m 2 kg
v
3 ms1
v
v 3 ms1
10 N
10 N
3 ms1
(vi)
(v)
m 2 kg
m 2 kg
10 N
m 2 kg
m 2 kg
8N
v 3 ms1
v 3 ms1
151
Figure 9.1 What will happen to these bodies?
b) C
alculate the acceleration of the
bodies in each case.
10 N
Tony’s push
on the wall
300 N
3 In Figure 9.2 you can see Tony pushing
against a wall with a force of 300 N.
State the magnitude and direction of the
force that the wall exerts on Tony.
Figure 9.2 Tony pushing
against a wall.
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Figure 9.3 shows a man enjoying a ride with a water jet pack. The pack
on his back pumps water up through the yellow pipe and then forces the
water out in two high-velocity jets. This recreational toy relies on each of
Newton’s three laws of motion. The jet pack pushes water downwards, and
the escaping water pushes the jet pack back in an upwards direction. When
the upwards push from the water on the jet pack balances the weight of the
man and his jet pack, he remains at a constant height.
●● Newton’s first law – free body diagrams
Figure 9.3 This water jet pack
relies on each of Newton’s three
laws of motion.
A free body diagram shows all the forces
acting on a single body; no other body is
shown in the diagram.
9 NEWTON’S LAWS OF MOTION
D
A body may be subjected to a number of forces, and the body may also
exert forces on other bodies that come into contact with it. If we try to draw
forces on more than one body, on the same diagram, it is easy to become
confused over which force acts on which body. So, to understand the effect
of forces on a body, we draw a free body diagram, which only shows the
forces acting on a single body.
The diagrams below illustrate free body diagrams with balanced forces.
In Figure 9.4 a skydiver and parachute fall to the ground at a constant
speed. Here the free body diagram is easy as the skydiver and parachute
are not in contact with anything (except the air). The drag, D, upwards
balances the weight, W, of the skydiver and parachute downwards.
In Figure 9.5a, a climber is abseiling down a rock face; she has just
paused for a rest and is stationary. To draw a free body diagram, we must
remove the rock face. The three forces acting are shown in Figure 9.5b:
the weight of the climber, W; the tension of the rope, T; the reaction
from the rock face, R. Since the climber is stationary, these forces act
through the centre of gravity of the climber. They add up to zero, as
shown in Figure 9.5c.
T
T
W
W
centre of
gravity
Figure 9.4 Skydiver falling to the ground
at a constant speed.
152
R
R
W
Figure 9.5a A climber is
abseiling down a rock face.
Figure 9.5b The forces
acting on the climber.
Figure 9.5c The forces
add up to zero.
Figure 9.6a shows a man climbing a ladder that rests against a smooth wall.
Figure 9.6b shows a free body diagram for the ladder when the man stands
on it.
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Newton’s second law of motion
Rw
Rm
W
RF
F
Figure 9.6a A man climbing a ladder.
Figure 9.6b A free body diagram for
the man on the ladder.
The forces acting on the ladder are:
RW, a horizontal reaction force from the wall
RF, a vertical reaction force from the floor
F, a horizontal frictional force from the floor
W, the weight of the ladder
Rm, a contact force from the man that is equal in size to his weight (this is
not the man’s weight, which acts on him).
Since the ladder remains stationary, the forces on it balance.
TIP
So
To draw a free body diagram,
consider only one body. Draw the
forces that act on that body, not
the forces that it exerts on other
bodies.
RF = W + Rm
These are the forces acting vertically.
F = RW
These are the forces acting horizontally.
●● Newton’s second law of motion
When an unbalanced force is applied to an object, it experiences a change
in velocity. An object speeds up or slows down when a force is applied
along its direction of motion. When a force is applied at an angle to
an object’s direction of travel, it changes direction. When we calculate
accelerations we use the equation:
153
resultant force = mass × acceleration
F = ma
Provided the force is in newtons, N, and the mass in kilograms, kg, then the
acceleration is measured in m s−2.
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EXAMPLE
Figure 9.7 shows the forces on a cyclist accelerating along
the road. The forces in the vertical direction balance, but
the force pushing her along the road, F, is greater than the
drag forces, D, acting on her. The mass of the cyclist and
the bicycle is 100 kg. Calculate her acceleration.
D 180 N
R 1 400 N
R 2 580 N
Answer
Her acceleration can be calculated as follows.
F – D = ma
F 320 N
320 N – 180 N = 100 kg × a
a=
140
m s−2
100
W 980 N
Figure 9.7 The forces on a cyclist.
= 1.4 m s−2
EXAMPLE
A passenger travels in a lift that is
accelerating upwards at a rate of 1.5 m s−2.
The passenger has a mass of 62 kg. By
drawing a free body diagram to show the
forces acting on the passenger, calculate
the reaction force that the lift exerts on her.
9 NEWTON’S LAWS OF MOTION
Answer
154
T
R
a 1.5 ms2
R – W = ma
R − W = ma
R = W + ma
= 62 kg × 9.8 m s−2 + 62 kg × 1.5 m s−2
= 700 N (2 sig. figs)
Figure 9.8 A passenger in a lift.
W
TEST YOURSELF
1 Draw free body diagrams to show the forces acting in the following cases.
a) A car moving at a constant speed.
b) An aeroplane just at the point of take off from a runway.
c) A submarine moving at a constant speed and depth.
d) A skydiver, just after she has opened her parachute.
Illustrate your diagrams to explain whether the bodies move at constant
speed, or whether they are accelerating or decelerating.
2 a) Calculate the reaction force acting on the passenger in Figure 9.8 when
i) the lift moves up at a constant speed of 3.0 m s −1
ii) it accelerates downwards at a rate of 2.0 m s −2.
b) Explain why the passenger feels heavier just as the lift begins to
move upwards.
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Newton’s third law of motion
m 1 950 kg
D 1000 N
velocity/m s–1
1
3 Figure 9.9 shows the horizontal forces
m 2 450 kg
acting on a truck that is pulling a trailer.
D 2 600 N
a) Calculate the acceleration of the
truck and trailer.
F 2400 N
b) Calculate the tension, T, in the tow
bar that is pulling the trailer.
Tow bar
4 A driver tests his car’s acceleration.
Figure
9.9
The
horizontal
forces
on a truck.
The result of this trial is shown by the
velocity–time graph in Figure 9.10.
a) Use the graph to calculate the car’s acceleration
15
i) between 0 and 2 s
ii) between 6 s and 8 s.
10
The car and the driver have a mass of 1200 kg.
b) Calculate the initial force that accelerates the car
from rest.
5
c) Assuming that the same driving force acts on the car
between 6 s and 8 s, calculate the drag forces acting on
0
the car at this time using information from the graph.
0
2
4
6
8
10
d) Calculate the initial acceleration of the car when it is
time/s
loaded with passengers and luggage so that its mass
Figure 9.10 Velocity–time graph for a car.
is 1450 kg.
12
●● Newton’s third law of motion
stretched band
(a)
1 kg
0
10
20
1 kg
30
40
50
60
70
80
90 100 cm
scale
slack
rubberband
(b)
0
10
20
30
40
50
60
70
80
Figure 9.11 Demonstrating Newton’s third law.
469884_09_AQA_A-Level_Physics_151-167.indd 155
90 100 cm
scale
This law has already been stated in the introduction
but it is stated as follows in a slightly different way.
Every force has a paired equal and opposite force. This
law sounds easy to apply, but it requires some clear
thinking. It is important to appreciate that the pairs of
forces must act on two different bodies, and the forces
must be the same type of force.
An easy way to demonstrate Newton’s third law is
to connect two dynamics trolleys together with a
stretched rubber band, as shown in Figure 9.11.
When the trolleys are released, they travel the same
distance and meet in the middle. This is because
trolley A exerts a force on trolley B, and trolley
B exerts a force of the same size, in the opposite
direction, on trolley A. Some examples of paired forces
are given in Figure 9.12.
155
17/04/19 10:17 AM
100 N
100 N
Force from the
road on the car
Force from the
car on the road
Figure 9.12a If I push you with a
force of 100 N, you push me back
with a force of 100 N.
Figure 9.12b When the wheel of a car
turns, it pushes the road backwards.
The road pushes the wheel forwards
with an equal and opposite force.
Pull from the Earth
on the spacecraft
Pull from the spacecraft
on the Earth
θθ
B exerts a
force on A
9 NEWTON’S LAWS OF MOTION
Figure 9.12c A spacecraft orbiting the
Earth is pulled downwards by Earth’s
gravity. The spacecraft exerts an
equal and opposite gravitational force
on the Earth. So if the spacecraft
moves towards the Earth, the Earth
moves too, but because the Earth is so
massive its movement is very small.
A
B
A exerts a
force on B
Figure 9.12d Two balloons have been
charged positively. They each experience
a repulsive force from the other. These
forces are of the same size, so each
balloon (if of the same mass) is lifted
through the same angle.
All of these are examples of Newton’s third law pairs.
Newton’s third law pairs always have these properties:
●
they act on two separate bodies
are always of the same type, for example two electrostatic forces
or two contact forces
they are of the same magnitude
they act along the same line
they act in opposite directions.
●they
●
156
●
TIP
When applying Newton’s third
law, remember that the ‘equal
and opposite’ forces act on
different bodies. Do not confuse
this with Newton’s first law, when
balanced forces keep a body at
rest or at a constant speed.
469884_09_AQA_A-Level_Physics_151-167.indd 156
●
A tug of war
Alan and Ben are enjoying a tug of war, as this gives them an opportunity to
apply all of Newton’s laws of motion. In Figure 9.13a Alan, who is stronger
than Ben, is just beginning to win. Alan, Ben and the rope are accelerating
to the left at 2 m s−2. By drawing free body diagrams for Alan (mass 80 kg),
the rope (mass 2 kg) and Ben (mass 60 kg), we can analyse the forces acting
on each person. Figure 9.13b shows these diagrams.
17/04/19 10:18 AM
mass 80 kg
mass 60 kg
Newton’s third law of motion
(a)
accelerating at 2 m s2
Alan
(b)
Ben
mass 80 kg
mass 60 kg
T 1 240 N
T 2 236 N
mass 2 kg
240 N
(i)
236 N
(ii)
(iii)
R 2 116 N
R 1 400 N
Figure 9.13 Free body diagrams for the tug of war. Exercise caution if you are repeating this experiment
yourself.
For each body the resultant forces caused on acceleration of 2 m s−2. These
calculations use Newton’s second law.
Table 9.1
Resultant force (N)
=
Mass (kg)
×
Acceleration ( m s−2)
Alan
(400 − 240)
=
80
×
2
Ben
(236 − 116)
=
60
×
2
Rope
(240 − 236)
=
2
×
2
Newton’s third law pairs
●
●
●
●
The Earth pushes Alan with 400 N to the left.
Alan pushes the Earth with 400 N to the right.
The Earth pushes Ben with 116 N to the right.
Ben pushes the Earth with 116 N to the left.
The rope’s tension pulls Alan to the right with 240 N.
Alan pulls the rope to the left with 240 N.
The rope’s tension pulls Ben to the left with 236 N.
Ben pulls the rope to the left with 236 N.
157
TIP
To work out the acceleration of a body you must draw a free body diagram
showing all the forces that act on that body. Alan and Ben accelerate
because unbalanced forces act on them.
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9 NEWTON’S LAWS OF MOTION
TEST YOURSELF
5 Figure 9.14 shows a dog sitting still on the ground.
a) Draw a free body diagram to show the two forces acting on it to keep
it in equilibrium.
b) For the two forces above, state and explain the two forces which
are their Newton third law pairs.
6 This question refers to the tug of war in Figure 9.13. Just before Alan
started to win, the two men were in equilibrium. The force from the
ground on Alan was 300 N.
a) Draw free body diagrams for
i) Alan
ii) the rope
iii) Ben
to show all the forces acting on them, both horizontally and vertically.
b) Count as many Newton third law pairs as you can. (Remember to
include forces acting on
the Earth.)
7 Use Newton’s third law to explain the following.
a) You cannot walk easily on icy ground.
b) You have to push water backwards so that you can swim forwards.
8 If you blow up a balloon and then release it, the balloon flies around
the room. Draw a free body diagram and explain why the balloon
travels at a constant speed when air resistance balances the
forwards push from the air escaping from it backwards. You need
Newton’s first law to explain this.
9 a) Under what circumstances would a constant force acting on a body
produce zero acceleration?
b) Under what circumstances would a constant force acting on a body
produce a decreasing acceleration?
c) Under what circumstances would a constant force acting on a body
produce an increasing acceleration?
[Hint: for all of these three examples, there are other forces acting
on the body. It is possible that the mass of a body changes while
it accelerates.]
10 A body of mass 5 kg has three forces acting on it of magnitude
10 N, 7 N and 5 N.
a) Sketch a diagram to show how the body may be in equilibrium.
b) The 7 N force is removed. Calculate the body’s acceleration.
11 A man, whose mass is 90 kg, jumps off a wall of height 1.5 m.
While he is in free fall, what force does he exert on the Earth?
Figure 9.14 A dog sitting still
on the ground.
158
●● Practical investigations
Here are several practical exercises that you can try for yourself, or you
can use the data and diagrams to discuss or calculate the outcomes of
these experiments. In these activities it will make your calculations more
straightforward if you assume g = 10 N kg−1.
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Practical investigations
ACTIVITY
Investigating tension
Look at Figures 9.15a, 9.15b and 9.15c. In each case predict the reading
on the forcemeter. Then check your predictions by reading the meters.
Comment on your results.
(a)
pulley
F
pulley
(b)
pulley
(c)
F
F
5N
5N
5N
5N
Figure 9.15 Investigating tension.
ACTIVITY
Balanced forces
Look at Figure 9.16a and predict the weight, W, you will
have to put on the string to balance the 4 N weight that
is suspended by the lower pulley block.
pulley
1kg
W
Now look at Figure 9.16b. Calculate the angle θ, to
which you will have to tip the slope so that the 3 N
weight balances the pull of gravity down the slope
on the trolley.
Figure 9.16a What weight
θ
will you have to put on the
string?
4N
3N
Figure 9.16b To what angle will you have
to tilt the slope?
ACTIVITY
Acceleration
Refer to Figure 9.17a. A small ball is suspended
from a stick attached to the trolley; it is free to
move. The mass of the trolley, stick and ball is
1 kg. The trolley is held stationary by hand while
a stretched spring exerts a force of 2 N on the
trolley. The trolley is then released so that it
accelerates to the right.
Predict what happens to the ball.
Calculate its maximum angle of deflection.
159
2N
mass 1 kg
2N
Figure 9.17a A small ball is suspended from a stick
attached to the trolley. It is free to move.
Finally, the trolley stops abruptly, when it hits a barrier.
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Predict what happens to the ball.
Now try the experiment and check your predictions,
as far as possible.
In Figure 9.17b a trolley is held in a stationary
position by a weight of 2.5 N. The string holding the
weight is then cut and the trolley accelerates down
the slope covering the distance of 0.5 m in 0.41s.
Calculate the mass of the trolley.
Now check the mass of a trolley in this way with your own
practical measurements.
0.50 m
2.5 N
Figure 9.17b A trolley is held in a stationary position by a
weight of 2.5 N.
In Figure 9.17c, three trolleys (with a mass of about 1 kg each), are held in
a stationary position. The 3 N pull of the weight is being balanced by a
student’s hand. At the front of each trolley is a spring that has been extended
by 12 cm due to the pull of the 3 N force.
The hand now releases the last trolley. Predict and explain what will happen
to the extension of each spring, before you check by trying the experiment out.
3N
A
B
C
3N
9 NEWTON’S LAWS OF MOTION
Figure 9.17c Three trolleys (with a mass of about 1 kg each), are held in a stationary
position. The 3 N pull of the weight is being balanced by a student’s hand.
160
ACTIVITY
F = ma
Light gates
0.9 kg
Figure 9.18 shows an arrangement of light gates
that can be used to calculate accelerations
directly. Two light gates measure the time it takes
for a piece of card on an accelerating trolley to
timers
pass through the gates, and the time interval
between each of the timed measurements.
Explain how the three time measurements allow
links to computer
the acceleration to be calculated. What further
information is needed?
1 N weight
Predict the acceleration that would be
Figure 9.18 An arrangement of light gates that can be used to
calculated in this experiment.
calculate accelerations directly.
Design an experiment to help you prove the general relationship: F = ma.
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Practical investigations
ACTIVITY
Newton’s Third Law
In Figure 9.19a a weight is suspended from a
forcemeter. It is then lowered into a beaker of water,
which sits on top of an electronic balance. Use the
following information to draw free body diagrams for
F
F
a) the water in the beaker
b) the suspended weight when the weight has been
lowered into the water.
●
●
When the weight is suspended above the water, the
forcemeter reads 2.7 N and the balance reads 6.8 N
When the weight is in the water the forcemeter
reads 2.1 N and the balance reads 7.4 N
Name as many Newton’s third law pairs as you
can in this experiment.
Figure 9.19b shows a similar idea to the previous
experiment. Predict what you expect to happen to
the forcemeter reading and the balance reading as
the magnet hanging from the forcemeter is lowered
over the second magnet on the balance. Explain the
significance of Newton’s third law to this experiment.
beaker
of water
electronic
balance
Figure 9.19a A weight is
suspended from a forcemeter
and is then lowered into a
beaker of water, which sits on
top of an electronic balance.
electronic
balance
Figure 9.19b What do you expect
to happen to the forcemeter
reading and the balance reading
as the magnet hanging from the
forcemeter is lowered over the
second magnet on the balance?
161
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Practice questions
T
1.11 ms2
1 A lift and its passengers have a total mass of 2000 kg. It accelerates
upwards with an acceleration of 1.1 m s−2. Which of the following is
the correct value for the tension in the lift cable?
A 2200 N
C 19 600 N
B 17 400 N
D 21 800 N
2 Two masses hang at rest as shown in Figure 9.21. The
thread separating the two masses is burnt through.
Which of the following gives the magnitude of the
accelerations of the masses after the thread burns
through?
Acceleration of
A ( m s−2)
Figure 9.20 A lift and its
passengers.
Acceleration of
B ( m s−2)
A
0.5 g
g
B
g
g
C
1.5 g
1.5 g
D
1.5 g
g
3 Figure 9.22 shows a box at rest on the floor. The
reaction force from the floor on the box equals the
weight of the box. Which law predicts that these two
forces are equal?
2m
A
m
B
Figure 9.21 Two masses hang at rest.
R
9 NEWTON’S LAWS OF MOTION
A The principle of conservation of momentum.
162
B The principle of the conservation of energy.
W
C Newton’s First Law of Motion.
D Newton’s Third Law of Motion.
Figure 9.22 A box at rest on the floor.
4 Figure 9.23 shows a truck accelerating from rest down a
slope that makes an angle of 30° to the horizontal. Which
of the following is the distance travelled by the truck after
4 seconds?
A 78.4 m
C 39.2 m
B 58.8 m
D 19.6 m
5 Figure 9.24 shows two boxes, A and B, which are connected by
a light thread X. The boxes are moving at a constant speed of
2.0 m s−1. A frictional force of 10 N acts on each box, and a force
F is applied to a second light thread Y. Which of the following
statements is true?
A F is slightly greater than 20 N.
30
Figure 9.23 A truck accelerating from
rest down a slope.
2.0 m s–1
A
2 kg
B
X
3 kg
Y
F
B String X exerts a force of 10 N on B.
C X exerts a greater force on A than it does on B.
D X exerts a greater force on B than it does on A.
469884_09_AQA_A-Level_Physics_151-167.indd 162
F 10 N
F 10 N
Figure 9.24 Two boxes, A and B, which
are connected by a light thread X.
17/04/19 10:18 AM
Motion of A
Motion of B
A
travels 0.4 m before stopping
accelerates at a rate of 3 m s−2
B
stops moving in a time of 0.4 s
accelerates at a rate of 5 m s−2
5 m s−2
Practice questions
6 The force F in Figure 9.24 is suddenly increased to 25 N and, at the same
instant, the string X breaks. Which of the following pairs of answers best
describes the subsequent motion of boxes A and B?
accelerates at a rate of 8 m s−2
C
decelerates at
D
Stops moving in a time of 2 s
accelerates at a rate of 5 m s−2
7 A car accelerates from rest and travels a distance of 20 m in 4.0 s.
The driving force acting on the car remains constant at 1750 N.
Which of the following is the mass of the car?
A 900 kg
C 700 kg
B 800 kg
D 600 kg
8 A skydiver is falling at a terminal speed of 39.8 m s−1 when he opens his
parachute. For the next 2 seconds the parachute exerts an average drag
force of twice the skydiver’s weight in an upwards direction. Which of
the following statements about the skydiver’s motion over the next two
seconds is correct?
A The skydiver moves upwards for 2 seconds.
B After two seconds the skydiver moves at a constant speed.
C Over two seconds the skydiver falls 40 m.
D Over two seconds the skydiver falls 60 m.
15 N
9 A child pulls a trolley with a force of 15 N in the
direction shown in Figure 9.25. A frictional force of
3 N acts on the trolley. Which of the following gives
the acceleration of the trolley along the ground?
A
3.0 m s−2
B 2.4 m s−2
C
1.8 m s−2
37
5 kg
a
3N
D 1.2 m s−2
The information below is needed for both questions
10 and 11.
Figure 9.25 A child pulls a trolley with a force of
15 N in the direction shown.
A four-engine aeroplane of mass 80 000 kg takes off along a runway of length
1.0 km. The aeroplane’s take-off speed is 50 m s−1. The average drag forces
acting against the plane during take-off are 20 kN.
10What is the plane’s minimum acceleration if it is to take-off from the
runway?
A 4.00 m s−2
C 1.50 m s−2
B 2.50 m s−2
D 1.25 m s−2
163
11 What is the minimum thrust required from each of the four engines to
achieve an acceleration of 2.0 m s−2?
A 35 000 N
C 140 000 N
B 45 000 N
D 160 000 N
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12 Figure 9.26a shows a cylinder of wood. Its density is 720 kg m−3.
a) Use the information in the diagram to show that the weight of the
cylinder is about 12 N.
(3)
(1)
b) State Newton’s first law of motion.
The cylinder, Y, is now placed on a table with a second identical
cylinder, X, placed on top of it. (Figure 9.26b)
Figure 9.26c is a free body diagram that shows the forces that act on
cylinder Y.
(a)
A
(c)(c)
(b)
cylinder Y
120 mm
B
X
Y
centre of gravity
150 mm
Y
table
C
Figure 9.26a A cylinder of wood.
Figure 9.26c A free body diagram that
shows the forces that act on cylinder Y.
Figure 9.26b Two cylinders of wood.
c) Name and state the magnitude of the forces A, B and C.
(3)
d) A force, F, forms a Newton’s third law pair with force B.
State the following:
iii) the type of force that F is
ii) the direction of F
iv) the object on which F acts.
(4)
A
13 A student with a weight of 580 N stands on a set of weighing
scales. She holds a mass of 2 kg in her hand by her side.
a) Explain why the scales record a weight of about
600 N.
(1)
b) Using Newton’s laws of motion, explain the readings
on the graph at times:
A
ii) B
iii) C.
B
400
200
(6)
14 Two boxes are pulled with a force of 50 N, which causes
them to accelerate to the right. Frictional forces of 3 N and
2 N, respectively, act on each box as shown in
Figure 9.28.
a) Calculate the acceleration of the two boxes. (3)
C
600
She moves her arm upwards rapidly to lift the weight
about her head. The scales change as shown in Figure
9.27.
i)
164
the magnitude of F
weight measured on scales/N
9 NEWTON’S LAWS OF MOTION
i)
2N
t
Figure 9.27 How the reading on the
scales changes.
10 kg
T
15 kg
50 N
3N
b) Calculate the tension, T, acting in the string
that connects the two boxes.
(2) Figure 9.28 Frictional forces acting on boxes.
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17/04/19 10:18 AM
P
250 N
Q
Figure 9.29 Crates in a warehouse.
a) The crates accelerate at 1.2 m s−2.
Determine the resultant force acting on the crates.
(2)
ii) Determine the total resistive force acting on both crates.
(1)
i)
Practice questions
15 Two large crates rest on the floor of a warehouse.
Each crate has a mass of 48 kg. A forklift truck
provides a force of 250 N and the crates accelerate to
the left is shown in Figure 9.29.
b) Assuming that the resistive force is spread equally between the two
crates, calculate the force that crate P exerts on crate Q when the
(4)
force of 250 N is applied to crate Q. Explain your answer.
16 A toy plane is powered by a pressured water bottle. The
principle is illustrated in Figure 9.30. A water bottle is
pressurised with a pump and inserted into the plane. When
the plane is launched, water is pushed backwards out of the
plane and the plane accelerates forwards. The plane flies
horizontally after it is thrown to launch it.
bottle of water
in the plane
a) Use Newton’s second and third laws of motion to explain
how the water accelerates the plane.
(4)
The plane starts with a mass of 1.3 kg, of which 0.4 kg is the
initial mass of the water in the bottle.
Figure 9.30 Principle of how a toy plane works.
Figure 9.31 shows a velocity-time graph for the plane’s flight.
6
E
D
velocity/m s–1
5
4
C
3
B
2
A
F
1
G
0
0
1
2
3
4
5
time/s
6
7
8
9
10
Figure 9.31 A velocity-time graph for the plane.
b) Calculate the plane’s acceleration:
i)
over the first second
165
(2)
ii) over the time 1 second to 2 seconds
on the graph.
(2)
iii) Explain why the plane’s acceleration increases during
the period of 1 to 2 seconds.
(2)
c)Calculate the resultant force on the plane at the start of its flight. (2)
d) At which point does the water run out? Explain your answer.
469884_09_AQA_A-Level_Physics_151-167.indd 165
(2)
17/04/19 10:18 AM
e) Explain the shape of the velocity-time graph over the following
periods of time:
C to D
(2)
ii) D to E
(1)
iii) E to F
(2)
iv) F to G.
(1)
i)
f) The distance flown by the plane is closest to which of these values:
15 m
iii) 55 m
ii) 35 m
iv) 75 m
i)
(2)
When the plane increases its speed, its kinetic energy increases.
Explain what source provided this kinetic energy.
(2)
g) i)
(1)
ii) What else gains kinetic energy in this flight?
17 A student designs an experiment to investigate the resistive forces that
act on a small mass as it falls through a column of water. The apparatus
is shown in Figure 9.32. The mass is immersed in the water and released
from rest. A ticker tape attached to the mass records the motion of the
mass as it falls.
9 NEWTON’S LAWS OF MOTION
The ticker timer marks the tape every 0.02 s with a small dot. The
separation of the dots may be used to calculate the speed of the mass
over a small interval of time. The student analyses the tape from one of
their experiments and records their results in the table shown below.
166
Interval
number
Gap between successive
dots (mm)
Interval
number
Gap between successive
dots (mm)
1
2.0
10
21.5
2
5.0
11
22.5
3
8.0
12
23.5
4
11.0
13
24.0
5
13.5
14
25.0
6
15.5
15
26.0
7
17.0
16
25.0
8
18.5
17
26.0
9
20.0
ticker timer
weight
water
Figure 9.32 An experiment
to investigate the resistive
forces that act on a small
mass as it falls through a
column of water.
a) Estimate the accuracy of these measurements. Could the results have
been recorded more accurately?
(2)
b) Show that the acceleration of the mass over the
first four intervals is 7.5 m s−2.
(4)
c) Explain how the data shows that the acceleration
of the mass decreases with time.
(2)
d) Calculate the terminal speed of the falling mass.
(2)
e) Sketch graphs to show:
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17/04/19 10:18 AM
ii) how the resistive forces on the mass vary
with speed.
f) Criticise this experiment, explaining how you would
improve this method to investigate the size of the drag
force on a falling object.
Practice questions
i) how the velocity of the mass varies with time
(4)
(4)
30 m s1
Stretch and challenge
18 A firework rocket reaches its highest point at 100 m above
the ground, where it is stationary. It then explodes into four
equal parts as shown in Figure 9.33a. Each fragment has
a mass of 0.1 kg.
Q
C
30 m s1
a) Using Newton’s laws explain why there should be
symmetry to the explosion.
P
R
S
b) Calculate how long each of the fragments P, Q, R and S
take to reach the ground. Ignore the effects of air
resistance in this and all subsequent calculations.
c) Calculate where each fragment reaches the ground
relative to the centre of gravity, C, which is 100 m above
the ground.
30 m s1
30 m s1
Figure 9.33a A rocket explodes into four
equal parts.
30 m s1
30 m s1
30 m s1
30 m s1
d) Calculate the kinetic energy of each fragment as it reaches
the ground.
[Hint: you might be able to solve this and the next part
using your knowledge from GCSE, or you might have
to read ahead into the next chapter and come back
to solve this.]
e) A second rocket explodes with the four fragments moving
as shown in Figure 9.33b. State the kinetic energy of each
fragment when they hit the ground.
19 A man with a mass of 80 kg has designed a lift in which
he can use his own body power to lift himself. The mass
of the lift is 40 kg.
Figure 9.33b A rocket explodes into four
equal parts.
a) In Figure 9.34 he is stationary. By drawing free body
diagrams for the man and the lift, determine the tension
in the rope and the reaction between the man and the
lift floor.
b) He pulls on the rope with a force of 648 N. Determine the
acceleration of the lift now.
c) Comment on any safety issues you might have identified
with this lift.
167
mass of lift – 40 kg
mass of man – 80 kg
Figure 9.34 A lift in which a man can use
his own body power to lift himself.
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10
Work, energy
and power
PRIOR KNOWLEDGE
●
●
●
Work is done when a force moves an object in the direction of the force.
Work (J) = force (N) × distance moved in the direction of the force (m)
This equation defines the joule:
1J = 1N × 1m
●
●
●
Energy is the capacity to do work or to heat something up – these
quantities are measured in joules.
Power is the rate of doing work or transferring energy.
energy
Power =
time
This equation defines the watt:
1 W = 1 J s−1
●
Energy is conserved. This means that energy cannot be created
or destroyed, but it can be transferred from one form to another.
TEST YOURSELF ON PRIOR KNOWLEDGE
168
1 Calculate the work done when you lift a mass of 24 kg through a
height of 1.3 m.
2 You leave a 2 kW electrical heater on for 312 hours. How much
electrical energy have you used?
3 In which of these examples is work being done?
a) A weightlifter holds a weight stationary above his head.
b) A car moves at a constant speed along a road.
c) A spacecraft moves at a constant speed in outer space in ‘zero g’.
d) A spacecraft moves in a circular orbit around the Earth.
e) A magnetic attraction keeps a magnet attached to a fridge door.
f) You walk down the stairs at a steady speed.
4 Explain the energy transfers that occur in these cases.
a) You throw a ball into the air and it lands on the ground.
b) A car drives along a road.
c) You use a rubber band to launch a paper pellet.
Typhoon Haiyan, which hit the Philippines in November 2013, was the
most powerful storm in history to make landfall. Communities were
devastated by the energy of the wind. Consider the energy transfers, which
are responsible for the building up of the storm, then consider how the
storm dissipates its energy when it hits land.
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We all know that if we expect to earn money doing a job, we must do
something useful. Nobody wants to pay an employee who is not working
effectively. The same idea applies to the scientific definition of work. It is
possible to use energy but to do no work. Figures 10.2 and 10.3 show two
examples of people using energy but not working.
10 N
10 N
Work
●● Work
In Figure 10.1 Tracy is doing some weight training. She is holding two
weights but she is not moving them. She gets tired holding them out, and her
arms convert chemical energy into heat energy, but no work is done as the
weights do not move. Tracy does work when she lifts the weights.
In Figure 10.2 three people are helping to push-start a friend’s car. Salim
and Anne are pushing at the back of the car. Here Salim and Anne are
working, because they are pushing along the direction in which the car is
travelling. Jim does no work as he is pushing at right angles to the direction
of motion.
Figure 10.1 Tracy is using energy but
not working.
Often an object moves in a different direction to the applied force. To
calculate the work done, we resolve the force into components parallel
and perpendicular to the direction of motion. Only the component parallel
to the motion does work. Figure 10.4 shows a passenger at an airport
wheeling her luggage. She pulls the bag with a force, F, at an angle θ to the
ground.
push
direction of
motion
Anne
work is done
push
Salim
Jim
direction of
motion
no work is done
F
Figure 10.2 Salim and Anne are working but Jim is not working as he is pushing at
right angles to the car.
θ
The component of force F sin θ does no work, but supports the bag. The
component F cos θ does work against frictional forces.
d
The general formula for work is
Figure 10.3 A passenger at an airport
wheeling her luggage.
F sin θ
θ
F cos θ
Figure 10.4 She pulls the bag with a
force, F, at an angle θ to the ground.
469884_10_AQA_A-Level_Physics_168-182.indd 169
169
work done = F s cos θ
where F is the applied force, s the displacement of the object, and θ the
angle between the force and displacement.
There are times when the force we are working against does not remain
constant, so the calculation of the work done is slightly more complicated.
Figure 10.5 shows an example of the force increasing as an object is
displaced. We calculate the work done by considering a small displacement,
∆s, when the applied force is F.
17/04/19 10:19 AM
Then the small amount of work done is:
300
∆W = F × ∆s
force/N
250
200
However, ∆W is the small area marked under the graph. So to
calculate the work done over a bigger distance we calculate the area
under the graph.
F
150
100
50
0
10 s 20
30
displacement/m
EXAMPLE
40
Figure 10.5 We calculate the work done
by considering a small displacement,
∆s, when the applied force is F.
Calculate the work done from Figure 10.5 when the object is displaced
40 m from its original position.
Answer
Work done = the area under a force-displacement graph
= 1 (150 N + 300 N) × 40 m
2
= 9000 J
10 Work, energy and power
TEST YOURSELF
4 A bag of mass 23 kg is pulled along a floor a
1 A passenger pulls her suitcase, similar to the one in
Figure 10.4, a distance of 400 m through an airport.
distance of 50 m; the frictional forces acting on
The case has a mass of 20 kg, and she pulls it along
it are 40 N. It is then lifted on to a desk of height
with a force of 60 N at an angle of 50° to the ground.
0.9 m. Calculate the total work done in the process.
a) Calculate the work done against resistive
forces.
b) Explain why it is an advantage to pull a
25 suitcase rather than to push it.
400 N
2 Three people help to push-start a car.
The strength and direction of the forces
that they apply are shown in Figure 10.6. 300 N
direction of
They push the car along the road for 60 m.
movement
250 N
Calculate the work done by the three
25 people in total.
3 A satellite, with a mass of 75 kg, is in a
Figure 10.6 The strength and direction of applied forces.
circular orbit of radius 7000 km around a
planet. At this height the gravitational field strength
is 6 N kg−1. Calculate the work done by the
gravitational field during one complete orbit of
the planet.
170
●● Calculating the energy
When work is done against resistive forces, for example when the suitcase
is moved at an airport, that work is transferred to heat energy. Therefore the
wheels of the suitcase and their surroundings warm up by a small amount.
Work can also be done to transfer energy into other forms. We can use this
idea to derive formulas for gravitational potential energy, elastic potential
energy and kinetic energy.
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In Figure 10.7 a load with a weight W, has been lifted through a height Δh.
The work done = increase in gravitational potential energy
work = W × Δh
W
But
Δh
weight W = mg
So the increase in potential energy, ΔEp, is given by:
Calculating the energy
Gravitational potential energy
ΔEp = mgΔh
Figure 10.7 A load with a weight W lifted
through a height Δh.
Note that we use the symbol Δ to indicate a change in energy and height.
There is no defined zero point of gravitational potential energy on the Earth,
so we talk about changes.
Elastic potential energy
Elastic potential energy (Eep) is stored in a stretched wire or rubber band.
Figure 10.8 shows how a rubber band may be stretched.
runway
rubber band
puck
TIP
The area under a forcedisplacement graph is equal to
the work done.
s
∆x
F
Figure 10.8 How a rubber band may be stretched.
When it is released it can convert its stored energy into the kinetic
energy of a puck, which slides along a runway.
5
force/N
4
Figure 10.9 shows how the force required to stretch the band depends
on the distance it is pulled back. Because the force changes we have to
use the average force to calculate the energy stored.
3
2
Δ(Eep) = Fav × s
1
In the case of Figure 10.9, where the force to stretch the band is
proportional to the distance moved:
2
4
6
8
10
12
average force = 1 × final force
2
Δx/cm
So
Figure 10.9 How the force required to
stretch the band depends on the distance
ΔEep = 1 Fs
2
it is pulled back.
The stored elastic potential energy can also be calculated more generally
using the area under the force-extension graph.
velocity 0
velocity v
s
Figure 10.10 A constant force accelerates a car.
469884_10_AQA_A-Level_Physics_168-182.indd 171
171
Kinetic energy
In Figure 10.10 a constant force F accelerates a car, starting
at rest, over a distance s. Work is done to increase the
kinetic energy of the car. We can use this idea to find a
formula for kinetic energy, Ek, in terms of the car’s speed
and mass.
17/04/19 10:19 AM
ΔEk = Fs
but from Newton’s Second Law:
F = ma
and from the equations of motion:
s = 1 at2
2
which gives
ΔEk = ma × 1 at2
2
1
= ma2 t2
2
since
v = at
ΔEk = 1 mv2
2
So the kinetic energy of a body of a mass m, moving at a velocity v, is given by:
Ek = 1 mv2.
2
Note this is a scalar quantity because v2 has no direction.
TEST YOURSELF
5 a) i) A crane lifts a container of mass 25 000 kg
10 Work, energy and power
though a height of 12 m above the quay next to
a ship. Calculate the container’s increase in
gravitational potential energy.
Figure 10.11 A container being loaded on to a ship.
172
ii) The crane then rotates, moving the container
sideways 20 m over the ship. The average
frictional force acting on the crane as it
moves sideways is 150 N. Calculate the work
done in this movement.
iii) The container is then lowered into the ship’s
hold so that the container lies 8 m below
the level of the quay. Calculate its potential
energy now, when compared to its original
potential energy.
b) A spacecraft of mass 18 000 kg falls into a black
hole. The decrease in the spacecraft’s potential
energy as it falls 100 m near the event horizon
is 5 × 1018 J. Calculate the average gravitational
field strength in this region.
6 a) A Formula 1 car with a mass of 720 kg
accelerates from a speed of 42 m s−1 to 88 m s−1.
Calculate its increase in kinetic energy.
b) A bullet with mass of 0.05 kg has a velocity of
300 m s−1. Calculate its kinetic energy.
7 In an experiment similar to that shown in Figure 10.8,
a puck of mass 0.15 kg is used. The rubber band is
stretched by 0.12 m with an average force of 4.0 N.
a) Calculate the elastic potential energy stored in
the rubber band when it has been pulled back a
distance of 0.12 m.
b) The puck is released and it slides along
the runway a distance of 0.96 m past the
unstretched position of the band (the distance s
in the diagram). Estimate the average frictional
force acting on the puck to slow it down.
●● The principle of conservation of energy
You are already familiar with the principle of conservation of energy, which
may be stated as follows: energy cannot be created or destroyed, but may be
transferred from one form of energy to another.
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Efficiency
In the last section you met formulas for gravitational potential energy,
kinetic energy and elastic potential energy. In this section you will learn
to use these formulas to calculate and predict the motion of bodies as
one form of energy is transferred to another.
EXAMPLE
A boy throws a ball upwards with a speed of 16 m s−1. It
leaves his hand at a height of 1.5 m above the ground.
Calculate the maximum height to which it rises.
Answer
The kinetic energy of the ball as it leaves the boy’s
hand is transferred to gravitational potential energy in
the ball as it rises. At the ball’s highest point it stops
moving, so all its kinetic energy has been transferred
to potential energy at that point.
So
ΔEk = ΔEp
1
⇒ mv2 – 0 = mgΔh
2
2
⇒ Δh = v
2g
2
−1 2
= 16 (m s )−2
2 × 9.8 m s
= 13 m
but the total height gained is 13 m + 1.5 m = 14.5 m.
EXAMPLE
EXAMPLE
A stretched catapult stores 0.7 J
of elastic potential. The catapult
is used to launch a marble of
mass 0.01 kg. Calculate the initial
speed of the marble.
A car of mass 1300 kg is travelling at 9 m s−1. The driver applies the
brakes, which exert a braking force of 4400 N on the car. Calculate the
braking distance of the car.
Answer
The stored elastic potential energy
in the catapult is transferred to
the marble’s kinetic energy.
So
ΔEep = ΔEk
0.7 J = 1 × 0.01 kg × v2
2
2
v = 1.4 (m s−1)2
0.01
⇒ v = 12 m s−1
Answer
The work done by the brakes is equal to the change of kinetic energy
in the car. (The kinetic energy is transferred to heat energy in the
brake discs).
work done = ΔEk
F × s = 1 mv2
2
4400 N × s = 1 × 1300 kg × (9 m s−1)2
2
650
kg × 81 (m s–1)2
s=
4400 N
= 12 m
●● Efficiency
electrical supply
In Figure 10.12 an electric motor is being used to lift a weight. Electrical
energy is being transferred to gravitational potential energy. However, the
motor also transfers electrical energy into heat energy and sound energy.
173
In Figure 10.13 a Sankey diagram shows how 100 J of electrical energy
are transferred to other types of energy. The majority of the energy is
transferred into heat and sound energy, with only 30 J being transferred into
useful gravitational potential energy.
Figure 10.12 An electric motor being
used to lift a weight.
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Sankey diagram A specific flow diagram
in which the width of the arrows shown is
proportional to the quantity of the flow. In
this chapter Sankey diagrams are used to show
how energy is transferred in various processes.
30 J gravitational
potential energy
100 J electrical energy
12 J heat energy
in the pulley
8 J sound
energy
50 J of heat
energy in
the motor
Figure 10.13 A Sankey diagram.
The efficiency of a machine is defined as follows:
efficiency = useful energy transferred
energy supplied
For the motor:
efficiency = 30 J
100 J
= 0.3 or 30%
10 Work, energy and power
TEST YOURSELF
8 A motor similar to the one shown in Figure 10.12
uses 280 J of electrical energy to lift a mass of
2.3 kg through a vertical height
of 3.5 m. Calculate the efficiency of the motor.
9 The efficiency of a car’s petrol engine is 27%. This
means that only 27% of the chemical energy stored
in the car’s fuel does useful work to drive the car
forwards against the drag and frictional forces.
174
469884_10_AQA_A-Level_Physics_168-182.indd 174
a) One litre of petrol in a fuel tank stores 30 MJ
of chemical potential energy. Calculate the
energy available in 1 litre of fuel to work against
frictional forces.
b) A car travelling at a constant speed of 20 m s−1
uses 1 litre of petrol to drive 7 km. Calculate the
frictional forces acting on the car.
●● Power
Power is defined by the equation:
energy transferred work done
power =
or
time
time
The unit of power is the watt, W, or J s−1. This definition can be used to
produce a useful formula to calculate the power transferred by moving
vehicles.
power = work done
time
F×s
=
t
⇒ power = F × v
17/04/19 10:19 AM
18 m s−1.
A car is moving at a constant speed of
The
frictional forces acting against the car are 800 N in
total. The car has a mass of 1200 kg.
a) Calculate the power transferred by the car on a
level road.
b) The car maintains its constant speed while climbing
a hill of vertical height 30 m in 16 s. Calculate the
power transferred by the car now.
a)
Power
Answer
EXAMPLE
P=F×v
= 800 N × 18 m s−1
= 14 kW
b)
P = 14 kW + mgΔh
t
1200
kg × 9.8 N kg–1 × 30 m
= 14 kW +
16 s
= 14 kW + 22 kW
= 36 kW
ACTIVITY
How high can you climb using the energy from
a chocolate bar?
Estimate the height of a mountain that you can climb using the energy from
a chocolate bar. You will need to research the efficiency of the body in turning
food energy to work, and the calorific value of your chocolate bar.
The purpose of the following group of experiments is to account for energy as
far as is possible. At the start of each experiment you will calculate the amount of
energy in one form, and see whether you can explain how the original energy has
been transformed into other forms of energy. Preferably you should do your own
experiments based on these, but you can use the data provided to guide you.
ACTIVITY
Elastic potential energy transferred
to kinetic energy
light gate
Here you investigate how elastic potential energy
is transferred into the kinetic energy of a trolley,
which is tethered by two springs. In Figure 10.14
the trolley is in equilibrium. Pull the trolley a
suitable distance sideways, and measure the force
required to do this. Then release the trolley and
use the light gate to measure the speed directly.
Here is some sample data obtained with a
trolley of mass 0.85 kg.
Figure 10.14 The speed of the trolley is measured as it passes through
the light gate.
Extension of spring (cm) Force to extend the spring (N) Speed of trolley at its midpoint (cm s–1)
10.0
4.6
69
Experiment 1
20.0
9.3
131
Experiment 2
175
Use these results to calculate the elastic potential energy stored in the
springs before the release of the trolley. Use the formula:
∆Eep = 1 force × extension
2
Then calculate the trolley’s kinetic energy at its fastest point of motion.
Comment on your results.
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ACTIVITY
Gravitational potential energy to kinetic energy
This experiment monitors the transfer of gravitational
potential energy to kinetic energy. In Figure 10.15 a ball
bearing is dropped through two light gates which measure
the ball’s speed in each position.
Use the data provided in Figure 10.15 to calculate the change
of gravitational potential energy as the ball falls from A to
B, and the change in kinetic energy over the same distance.
Comment on your results.
A
1.56 ms1
light gates
0.35 m
B
The tube is now tilted so that the ball bearing rolls down the
tube rather than falling (see Figure 10.16). It falls through the
same vertical height as before, 0.35 m.
3.04 ms1
mass of ball
bearing 15.7 g
When the ball bearing rolls it has two types of kinetic
energy: translation kinetic energy, which is calculated
by the formula:
10 Work, energy and power
1 mv 2
2
and rotational kinetic energy due to the rotation of the ball
about its own axis. The rotational kinetic energy can be
calculated using the formula:
176
1 mv 2.
5
Use this theory to predict the speed of the ball bearing
after is has rolled down a vertical height of 0.35 m, to light
gate B after it travelled past light gate A with a speed of
0.91 m s−1. Explain which other energy transformations
may take place during this process.
Check by doing the experiment for yourself.
Figure 10.15 The light gates read the
speed directly as the ball bearing passes
through them.
A
0.91ms1
light gates
mass of ball
bearing 15.7 g
0.35 m
B
Figure 10.16 The tube is now tilted.
TEST YOURSELF
10 A
ball is dropped from a height of 15 m.
a) Calculate its speed when it hits the ground
below.
b) The same ball is thrown downwards from a
height of 15 m with an initial speed of 10 m s−1.
Calculate its speed when it hits the ground.
11 A ball has a mass of 0.1 kg. It is thrown sideways
with a speed of 10 m s−1 at a height of 20 m above
the ground.
469884_10_AQA_A-Level_Physics_168-182.indd 176
a) Calculate its kinetic energy when it is first
thrown.
b) Calculate the ball’s potential energy at a height
20 m above the ground.
c) Calculate the ball’s kinetic energy when it hits
the ground.
d) Calculate the ball’s speed when it hits the
ground.
17/04/19 10:19 AM
8
7
force/mN
6
5
4
3
2
1
4
8
12
16
20
extension of web/mm
Figure 10.17 How the centre of the web
is displaced as it slows the fly down.
a) The maximum stretch of the web when the fly
hits it is 14 mm. Estimate the elastic potential
energy stored in the web for this displacement.
Give your answer in μJ.
b) Estimate the speed of the fly when it hit the web.
State any assumptions you make.
16 Read this extract from a scientific article.
‘ The hind legs of a locust are extremely
powerful. The insect takes off with a speed
of 3 m s −1. The jump is fast and occurs in a
time of 25 milliseconds. The locust’s mass is
about 2.5 g.’
Use the information to answer the questions below.
a) Calculate the locust’s average acceleration
during take-off.
b) Calculate the average force exerted by the
locust’s hind legs.
469884_10_AQA_A-Level_Physics_168-182.indd 177
c) The locust’s legs extend by 5 cm during the
jump. Calculate the work done by the legs.
d) Calculate the power developed in the locust’s
muscles.
e) Compare the power/mass ratio for the locust,
with that of a high jumper. Assume the high
jumper has a mass of 70 kg; he lifts his centre
of gravity by 1.3 m in the jump; his take-off foot
is in contact with the ground for 0.2 s.
17 A power station turns the chemical energy stored
in coal into electrical energy with an efficiency
of 36%. Some of the electrical energy is used
to drive a tram. The tram’s motor transfers
electrical energy to work against resistive forces
with an efficiency of 25%. What fraction of the
coal’s original chemical energy is used to drive
the tram? What happens to the rest of the energy?
18 A hydroelectric turbo-generator is driven by water
that descends 160 m from a high-level reservoir.
Water passes through the turbines at a rate of
700 m3 s−1. The density of water is 1000 kg m−3. The
efficiency of the turbo-generator is 0.8.
Calculate the electrical power output from the
generator. Express your answer in GW.
19 It has been suggested that a tidal barrage could
be built across the River Severn to generate
electricity. The idea is that, as the tide rises, water
flows in through gates in the barrage into a lake.
When the tide begins to turn, the gates are closed
and the water is stored behind the barrage. At low
tide water in the lake is then allowed to flow out
through turbo-generators.
You waill be asked to carry out some calculations
on the generation of electrical power. You will
need the following information.
● The surface of the lake is at a height of 13 m
above the turbo-generators at the start of the
generation process and falls to a height of 6 m
above the turbo-generators.
● The surface area of the lake is 200 km2.
● The electricity is generated over a period of 5.5
hours.
● The density of sea water is 1020 kg m −3.
● g = 9.81 N kg−1.
● The efficiency of the turbo-generators is 78 %.
Power
12 A
car of mass 1800 kg slows from a speed of
20 m s−1 to 15 m s−1 over a distance of 200 m.
Calculate the average resistive forces acting on
the car over this distance.
13 A force of F is used to accelerate, from rest, two
cars over a distance, d. One car has a mass m, the
second car a mass of 2 m. Which car now has the
greater kinetic energy?
14 An express train is powered by diesel engines that
have a maximum output of mechanical power to
drive the train of 2.0 MW. Using this power the train
travels at a top speed of 55 m s−1.
a) Calculate the resistive forces acting on the
train at its top speed.
b) The diesel fuel stores 40 MJ per litre of
chemical energy. The engines have an efficiency
of 32%. Calculate the volume of fuel required
for a journey of 200 km at the train’s top speed.
15 A fly of mass 16 mg flies into a spider’s web. The
kinetic energy of the fly is converted to elastic
potential energy in the web as it stretches.
Figure 10.17 shows how the centre of the web is
displaced as it slows the fly down.
177
a) Calculate the gravitational potential energy
available in the lake at high tide. Express your
answer in TJ.
b) Calculate the average power produced by the
turbo-generators. Express you answer in MW.
c) Assuming there are two high tides a day,
estimate the total electrical energy production
a year.
17/04/19 10:19 AM
Practice questions
1 A car exerts a tractive force of 750 N when travelling at 108 km h−1.
What is the work done by this force in 10 minutes?
A 10.8 MJ
C 15.6 MJ
B 13.5 MJ
D 19.6 MJ
2 A horse pulls a barge along a canal with a
force of 160 N. The barge moves at a steady
speed of 2.0 m s−1. How much work does
the horse do in pulling the barge forwards
in 1 hour?
A 1.15 MJ
C 0.49 MJ
B 1.04 MJ
D 0.05 MJ
160 N
25
3 An electric motor uses 100 W to raise 10 kg
vertically at a steady speed of 0.4 m s−1.
What is the efficiency of the system?
A 10%
C 40%
B 25%
D 100%
Figure 10.18 A horse pulling a barge.
10 Work, energy and power
4 An electric motor has an efficiency of 22%. It lifts a 6.6 kg load through
a height of 2 m in 4 s. What power is the motor using?
178
A 75 W
C 150 W
B 100 W
D 225 W
5 A car of mass 1200 kg brakes from a speed of 30 m s−1 to a speed of 18 m s−1
over a distance 240 m. What is the average braking force over this distance?
A 360 N
C 1080 N
B 720 N
D 1440 N
6 A ball is falling at a speed of 8 m s−1 when it is 6 m above the ground.
What is the speed when it hits the ground?
A 13 m s−1
C 17 m s−1
B 15 m s−1
D 19 m s−1
The information here refers to both questions 7 and 8.
A long bow, when fully flexed, stores 75 J of elastic potential energy. The
bow transfers this energy with 80% efficiency to an arrow of mass 60 g.
The arrow has a range of 200 m and penetrates a target to a depth of 12 cm.
7 The speed of the arrow as it leaves the bow is:
A 50 m s−1
C 40 m s−1
B 45 m s−1
D 35 m s−1
8 The average retarding force exerted by the target on the arrow is:
A 500 N
C 50 N
B 200 N
D 5N
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17/04/19 10:19 AM
A 90 m
C 900 m
B 180 m
D 1800 m
Practice questions
9 A manufacturer states that a 50 g chocolate bar provides your body with
1080 kJ of energy. Our bodies can convert energy from food to work
with an efficiency of 10%. How far up a mountain can a 60 kg student
climb using the energy from this chocolate bar?
10 A runner loses heat from her body at an average rate of 90 W when
at rest. She plans to run a race that will require her to use energy at a
rate of 600 W for three hours. What is the athlete’s minimum energy
requirement for the day?
A 6.8 MJ
C 16.7 MJ
B 13.3 MJ
D 18.0 MJ
pile driver
11 A student investigates the action of a pile driver using the apparatus
shown in Figure 10.19.
guide tube
A steel rod is dropped on to a nail, which is embedded 4 mm into a
block of wood. The student then measures the additional depth to
which the nail is knocked in for each drop of the model pile driver.
Use the data below, collected by the student, to answer the following
questions:
●
Mass of the pile driver: 0.31 kg
●
Height of the pile driver above the nail for each drop: 25 cm
Length of nail above the
wood (mm)
41
35
31
28
25
23
Number of drops of the
pile driver
0
1
2
3
4
5
Figure 10.19 Apparatus for
investigating the action of a pile
driver.
a) Calculate the average force to drive the nail into the wood
for the first drop of the pile driver.
(4)
b) Use your knowledge of forces, work and energy to explain the
pattern of the student’s results.
(2)
12 A conveyor belt in a power station lifts
9.6 tonnes of coal per minute to tip into
the furnace.
a) Use the information in Figure 10.20 to
calculate the increase in gravitational
potential energy of the coal per second.(3)
179
112
m
36 m
b) The conveyor belt is powered by a 80 kW
electric motor. Calculate the efficiency of
the motor, ignoring frictional effects. (2)
Figure 10.20 Conveyor belt lifting coal.
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13 The following is an extract from a flying manual.
‘The typical drag on a jet aircraft of mass 110 000 kg, flying at 240 m s−1
at a constant height of 12 000 m, is 92 kN. On a four-engine aircraft this
means that each engine must be delivering 23 kN of thrust; and under
these conditions 23 kN of thrust requires a power of just over 5.5 MW.’
a) How do you know from this information that the plane is
travelling a constant speed?
(2)
b) Calculate the lift on the aircraft.
(1)
c) Show how the author calculated the power delivered by the engines. (3)
(2)
14 a) State the principle of conservation of energy.
b) A boy of mass 32 kg travels down a slide
as shown below. The distance he travels
is 12 m.
Calculate his loss of gravitational
potential energy.
platform
rails for
safety
steps
(3)
12 m
c) The boy reaches a speed of 9.3 m s−1 at
the bottom. Calculate his increase in
kinetic energy.
(2)
h
10 Work, energy and power
d) The increase in the boy’s kinetic energy
30 is less than his decrease in potential
Figure 10.21 Boy sliding down slide.
energy. Explain how the principle of
conservation of energy still applies here.
(2)
180
e) A student suggests that the frictional force acting on the boy as he
descends the slide will generate heat energy. Use this idea to calculate
(3)
an average value for this frictional force.
15 A golfer hits a golf ball so that it travels with a speed of 41 m s−1. The golf
club is in contact with the ball for 0.49 ms. The golf ball has a mass of 46 g.
a) i)
Calculate the average acceleration of the ball while it is in
contact with the club.
(2)
(2)
ii) Calculate the force of contact that accelerated the ball.
b) Calculate the ball’s kinetic energy as it leaves the club.
(2)
c) Calculate the distance that the club was in contact with the
ball, assuming that all of the work done by the club was used to
accelerate the ball.
(3)
d) Explain in practice why the contact distance might be a little more
than the distance you calculated in part (c).
(2)
16 An attraction at a theme park is the water plunge as
shown in Figure 10.22. The riders fall down a slope of
17 m in height and splash into a pool at the bottom,
which slows them down to rest over a distance d.
d
17 m
The mass of a boat and its three occupants is 1300 kg.
a) Calculate the change in gravitational potential energy
as the boat falls through 17 m.
(2)
469884_10_AQA_A-Level_Physics_168-182.indd 180
Figure 10.22 The water plunge.
17/04/19 10:19 AM
(2)
ii) Explain why the speed of the boat may be less than your
calculated values.
(1)
Practice questions
Calculate the maximum speed of the boat at the bottom.
b) i)
The boat is now loaded with more passengers so that its mass is 1500 kg.
c) Explain carefully the effect of the larger mass on:
i)
the final speed at the bottom
ii) the distance d over which the boat slows to rest.
(4)
17 Figure 10.23 shows how the drag forces on a car depend on its speed.
A student observes that it will save petrol to drive the car at 20 m s−1
rather than 30 m s−1.
b) Calculate the power generated by the car to drive at
(2)
a steady speed of 30 m s−1.
The car engine is 27% efficient when driving at a speed
of 30 m s−1.
c) i)
Calculate the power input to the engine at
30 m s−1.
500
(3)
(2)
400
drag force/N
a) Explain how the graph backs up the student’s
comment.
300
200
100
ii) Explain what happens to the wasted power. (1)
The petrol in the car stores 32 MJ of chemical potential
energy per litre.
d) Calculate how much petrol is used in 1 hour if the
car drives at a constant speed of 30 m s−1.
5
10
15
20
25
30
velocity/m s1
Figure 10.23 How the drag forces on a car
depend on its speed.
(3)
18 The extract below is taken from a scientific article. Read it and answer
the questions that follow.
‘A crater in the Arizonian desert is thought to have been caused
by a meteor impact about 50 000 years ago. The crater measures
1200 m across and is about 200 m deep; the volume of the crater is
about 350 000 000 m3, and the density of rock in the region is about
3000 kg m−3. There is evidence from the surrounding region that the
debris from the impact must have been thrown 5 km into the air before it
settled back to earth. Meteors entering the Earth’s atmosphere can travel as
fast as 14 km s−1, therefore they possess a great amount of kinetic energy.’
a) Use the data in the article to calculate the mass of rock and earth
removed from the crater.
(2)
b) Calculate the gravitational potential energy gained by
the material as it rose 5 km above the plain.
(2)
c) Calculate the minimum mass of the meteor that made
this crater.
(2)
d) Suggest why the meteor was probably bigger than the
mass you calculated in part (c).
(1)
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181
17/04/19 10:19 AM
2800
19 A sprinter uses blocks to help him accelerate quickly
after the starting gun goes. Figure 10.24 shows how
the horizontal force from the blocks acts on the
sprinter as he starts the race. The sprinter has a mass
of 86 kg.
b) Estimate the velocity of the sprinter just after he
left the blocks.
(3)
Stretch and challenge
2000
force/N
a) Calculate the work done by the sprinter while he
is in contact with the blocks.
(3)
2400
1600
1200
800
400
20 A car accelerates using constant power. It takes 15
seconds to reach a speed of 40 km h−1. How long does
it take the car to reach a speed of 80 km h−1?
21 Figure 10.25 shows how the force of attraction varies
between bar magnets of length 6 cm and mass 50 g.
0.1
6.0
force/N
b) Since work is done to pull the magnets apart, energy has been
transferred. Where has this energy been transferred to?
c) While separated by 5 cm, the two magnets are released and
they move back together. Each magnet experiences a constant
frictional force of 0.4 N. What is the speed with which each
magnet is moving, relative to the table, when they collide?
10 Work, energy and power
0.3
Figure 10.24 How the horizontal force from the
blocks acts on the sprinter as he starts the race.
a) Two magnets are placed in contact. Use the graph to show
that the work done to move the magnets apart to a separation
of 5 cm is about 0.1 J.
182
0.2
displacement/m
d) One magnet is now turned round and the magnets are again
put into contact, so that two north poles touch each other. By
using the graph and the information in the text, estimate how
far the magnets move apart before they stop moving.
4.0
2.0
1.0 2.0 3.0 4.0 5.0 6.0 7.0
separation of magnets/m
Figure 10.25 How the force of attraction
varies between the bar magnets.
e) The magnets are placed 10 cm apart with a north pole facing a south
pole. What is the closest distance they can be moved towards each
other without the magnetic forces pulling the magnets together?
22 a) The unit of pressure is the pascal, which is equivalent to 1 N m 2.
Show that the unit 1 N m−2 is also equivalent to 1 J m−2.
b) When a champagne cork is released from a bottle, the cork (with
a mass of 15 g) is launched 10 m into the air. Use the idea above
to calculate the excess pressure in a 75 cl bottle.
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11
Momentum
PRIOR KNOWLEDGE
●
●
●
●
●
●
●
Momentum is defined as mass multiplied by velocity: p = m v.
Momentum is a vector quantity.
Newton’s second law of motion states that resultant force is the mass
multiplied by the acceleration: F = m a.
Newton’s third law of motion states that to every force there is an
equal and opposite force. Such paired forces are of the same type, act
along the same line, and act on separate bodies.
Acceleration is the change of velocity divided by time: a = Δv/Δt
Kinetic energy = 12 mass × (velocity)2
The principle of conservation of energy states that energy cannot
be created or destroyed, but it can be transferred from one type of
energy to another.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 Explain why the definition of momentum shows it must be a
vector quantity.
2 Calculate the momentum of a person of mass 80 kg running with a
velocity of 8 m s–1. State the units of momentum.
3 A car experiences a resultant forwards force of 1470 N. It accelerates
from a speed of 6 m s–1 to 14 m s–1 in 4 seconds. Calculate the
car’s mass.
4 You hold a book with a weight of 2 N in your hand. How big is the equal
and opposite force to that weight, and on what body does that force act?
5 A car with a mass of 1200 kg travelling at a speed of 25 m s–1 applies
its brakes and comes to a halt.
a) Calculate the kinetic energy that is transferred to other types
of energy.
b) Describe the energy transfers in this process.
●● Introducing momentum
Momentum is a useful quantity in physics because the amount of
momentum in a system always remains the same provided no external
forces act on that system. This principle allows us to predict what will
happen in a collision or an explosion.
183
In the example of the exploding firework, chemical potential energy
is transferred to thermal energy, light energy and kinetic energy of the
exploding fragments. However, during the explosion the momentum
remains the same. Momentum is a vector quantity; the momentum of
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Momentum The product of mass and
velocity. The unit of momentum is kg m s–1.
The symbol p is used for momentum.
p = mv
a fragment travelling in one direction is balanced by the momentum of
a fragment travelling in the opposite direction. The same laws of physics
apply equally to all masses whether they are planets, objects we meet
every day or the sub-atomic particles studied by nuclear physicists.
TIP
When a ball hits a wall with
momentum p and bounces back
in the opposite direction with
momentum –p, the change of
momentum is 2p.
EXAMPLE
A ball of mass 0.1 kg hits
the ground with a velocity
of 6 m s–1 and sticks to the
ground. Calculate its change of
momentum.
Answer
Using the formula for momentum
change:
Δp = m Δ v
= 0.1 kg × 6 m s–1
= 0.6 kg m s–1
Figure 11.1 An understanding of momentum and energy
enables us to explain what is going on here.
11 MOMENTUM
EXAMPLE
A ball of mass 0.1 kg hits the ground with a velocity
of 6 m s–1 and bounces back with a velocity of 4 m s–1.
Calculate its change of momentum.
Answer
Using the formula for momentum change, where mu
is the momentum of the ball when it hits the ground
and mv is its momentum when it begins to bounce
back upwards:
184
Δp = mv – mu
= 0.1 kg × 4 m s–1 – (–0.1 kg × 6 m s–1)
= 1.0 kg m s–1
In this case the momentum before and the momentum
after are in opposite directions so one of them must
be defined as a negative quantity; we have defined
upwards as positive and downwards as negative.
●● Momentum and impulse
Newton’s second law can be used to link an applied force to a change of
momentum:
F = ma
Substituting
a = Δv
Δt
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17/04/19 10:20 AM
gives
Momentum and impulse
or
F = mΔv
Δt
F = Δ(mv)
Δt
TIP
Note that momentum can be
expressed in kg m s–1 or N s.
This can be put into words as follows: force equals the rate of change of
momentum. This is a more general statement of Newton’s second law
of motion.
The last equation may also be written in the form:
F Δt = Δ(mv)
or
F Δt = mv2 – mv1
Impulse The product of force and time. The
unit of impulse is N s.
where v2 is the velocity after a force has been applied and v1 the velocity
before the force was applied. The quantity F Δt is called the impulse.
EXAMPLE
gymnast
lands
at 8 m s–1
gymnast
bounces
back at 8 m s–1
A gymnast is practising her skills on a trampoline.
She lands on the trampoline travelling at 8 m s–1 and leaves the trampoline
at the same speed. Her mass is 45 kg and she is in contact with the
trampoline for 0.6 s. Calculate the average force acting on the gymnast
while she is in contact with the trampoline.
Answer
F=
Figure 11.2
mv2 − mv1
Δt
−1
−1
= 45 kg × 8 m s − (45 kg × −8 m s )
0.6s
= 1200 N
Gymnasts use trampolines to reach and then fall from considerable heights,
for example 5 m. If you jumped from 5 m and landed on a hard floor, you
would hurt yourself and might even break a bone in your foot, ankle or leg.
The equation
185
F = Δ(mv)/Δt
helps you to understand why.
When you fall you have an amount of momentum that is determined by
how far you fall. The force on you when you stop moving depends on the
time interval, Δt, in which you stop. On a trampoline Δt is long, so the force
is small; on a hard floor Δt is short and the force much larger.
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17/04/19 10:20 AM
TEST YOURSELF
1 Use the ideas of impulse and change of momentum to explain the
following:
a) why you bend your legs when you jump from a wall on to the ground
b) why hockey players wear shin pads to protect their legs
c) why you move your hands backwards when you catch a fast-moving
ball coming towards you.
2 A gymnast of mass 45 kg jumps from a wall of height 3 m. When she
lands, her legs stop her moving in 0.2 s.
a) Calculate her momentum on landing.
b) Calculate the force on her legs when she lands.
Figure 11.3 An impulsive force: the
momentum of the tennis ball is changed
by a large force acting for a short time.
Car safety
The idea of impulse is vital in designing cars safely. Figure 11.4 shows two
force–time graphs for passengers A and B in a high-speed car crash. Marked
on the graph is a small area F Δt; this is equal to the change in momentum
in that time interval, m Δv.
80
A
70
60
force/kN
50
40
B
30
20
F
area F Δt
11 MOMENTUM
10
0.0
0.05
0.10
0.15
0.20
time/s
0.25
0.30
0.35
Figure 11.4 Force–time graphs for two passengers in a car crash.
So the total change of momentum of one of the passengers in the crash is
the sum of all the small areas: ∑ F Δt. Thus:
186
change of momentum = area under the force–time graph
TIP
It is useful to remember that the
area under a force–time graph
equals the change of momentum.
469884_11_AQA_A-Level_Physics_183-199.indd 186
The two passengers have different masses, so the areas under each graph
are different. However, passenger B was strapped in by their safety belt
and was stopped in the time it took the crumple zones at the front of the
car to buckle. Passenger A, in the back of the car, was not restrained and
was stopped as they hit the seat in front of them. Passenger A stopped in a
shorter time, so the maximum force on them was much greater.
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Momentum and impulse
Figure 11.5 In this test the dummy is protected by an airbag, and the crumple
zone at the front of the car allows time for the passengers to slow down.
TEST YOURSELF
3 a) Explain why crumple zones and seat belts are
vital safety measures for passengers in cars.
b) If a helicopter crashes it is most likely that the
impact will take place on the bottom of the craft.
Explain what safety features should be built into
the seats in the helicopter.
4 In an impact, a person who experiences an
acceleration of 300 m s –2 , of short duration, will
receive moderately serious injuries. However,
a much larger acceleration is likely to inflict
gravely serious injuries.
a) i) Explain why rapid accelerations cause injury.
ii) Explain why injuries are likely to be more
severe if a very high acceleration or
deceleration acts over a longer period of time.
In the car crash described in Figure 11.4, passenger A
has a mass of 82 kg and passenger B a mass of 100 kg.
b) Describe the likely injuries to both passengers A
and B.
c) By using the area under graph B, show that the
car crashes at a speed of about 35 m s–1.
5 A karate expert can break a brick by hitting it with
the side of their hand.
a) They move their hand down at a velocity of
12 m s–1 and the hand bounces back off the brick
with a velocity of 4 m s–1. Calculate the impulse
delivered to the brick if the hand and forearm
have a mass of 2.2 kg.
b) The time of contact between the hand and brick
is found to be 64 ms. Calculate the average force
exerted by the hand on the brick.
c) Calculate the average force exerted by the brick
on the person’s hand.
You can investigate how forces vary with time by
using a ‘force plate’ that is connected to a data
logger. Such plates use piezoelectric crystals, which
produce a p.d. that depends on the pressure applied
to the crystal.
A student investigates his reaction force on the
ground in various activities. He uses a force plate
and a data logger, which records the changes in force
with time.
In his first activity, the student steps on to the plate
in a crouching position; then he stands up, before he
steps off the plate again. See Figure 11.6.
469884_11_AQA_A-Level_Physics_183-199.indd 187
reading on force plate/kN
ACTIVITY
Investigating varying forces
A1
1
A2
B
0.5
0
E
A
0
1
187
D
C
2
3
4
5
6
time/s
Figure 11.6
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2.5
1 a) Explain what is happening over each of the
following times.
AB, BC, CD, DE
b) Explain why the areas A1 and A2 on the graph
are equal.
c) The student now jumps on to the force plate.
Sketch a graph to show how the reaction force
changes now.
2 In the next activity the student measures his
reaction force on the plate while running on the
spot. This is recorded in Figure 11.7.
a) Use the graph to show that the student’s
change of momentum on each footfall is about
250 kg m s–1.
b) Explain why his upward momentum as he takes
off on one foot is 125 kg m s–1.
c) The student’s mass is 80 kg. Show that his
centre of mass rises by a height of about 12 cm
between each stride. (Here you will need to
reading on force plate/kN
Answer the following questions, based on the
student’s investigation
2.0
R
L
1.5
1.0
0.5
0
0.1
0.2
0.3
time/s
0.4
0.5
0.6
Figure 11.7
use the equations of motion that you learned in
Chapter 8.)
d) Sketch a graph to show how the horizontal
component of the foot’s reaction on the ground
varies with time when the runner is moving at a
constant speed.
●● Conservation of linear momentum
11 MOMENTUM
Linear momentum The momentum of an
object that moves only in one dimension.
We use the term linear momentum when we refer to collisions (or
explosions) that take place in one dimension, i.e. along a straight line.
Figure 11.8 shows a demonstration of a small one-dimensional explosion.
The head of a match has been wrapped tightly in aluminium foil. A second
match is used to heat the foil and the head of the first match, which ignites
and explodes inside the foil. The gases produced cause the foil to fly rapidly
one way, and the matchstick to fly in the opposite direction. (It is necessary to
use ‘strike-anywhere’ matches, the heads of which contain phosphorous. The
experiment is safe because the match blows out as it flies through the air, but
common-sense safety precautions should be taken – do this in the laboratory,
not in a carpeted room at home, and dispose of the matches afterwards.)
matchstick
moving to
the left
(a)
(c)
(b)
aluminium foil
moving to the
right
188
– F ∆t
+ F ∆t
– ∆(mv)
+ ∆(mv)
Figure 11.8 A simple experiment that demonstrates the conservation of linear momentum.
In all collisions and explosions, both total energy and momentum are
conserved, but kinetic energy is not always conserved.
In this case, the chemical potential energy in the match head is transferred
to the kinetic energy of the foil and matchstick, and also into thermal, light
and sound energy.
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Since F Δt = Δ(mv), it follows that the foil gains exactly the same positive
momentum as the matchstick gains negative momentum.
We can now do a vector sum to find the total momentum after the
explosion (see Figure 11.8c):
+Δ(mv) + [–Δ(mv)] = 0
Conservation The total momentum of two
bodies in a collision (or explosion) is the
same after the collision (or explosion) as it
was before.
So there is conservation of momentum: the total momentum of the foil and
matchstick was zero before the explosion, and the combined momentum of
the foil and matchstick is zero after the explosion.
Conservation of linear momentum
As the match head explodes, Newton’s third law of motion tells us that both
the matchstick and foil experience equal and opposite forces, F. Since the
forces act for the same interval of time, Δt, both the matchstick and foil
experience equal and opposite impulses, F Δt.
Collisions on an air track
Collision experiments can be carried out using gliders on linear air tracks.
These can demonstrate the conservation of linear momentum. The air
blowing out of small holes in the track lifts the gliders so that frictional
forces are very small.
glider A
mass 0.63 kg
velocity 1.35 m s –1
TIP
Momentum is always conserved
in bodies involved in collisions
and explosions provided no
external forces act. This is the
principle of conservation of
momentum.
Blu Tack
lamps
air
input
+ momentum
holes for
air escape
light
gates
stationary glider B
mass 0.42 kg
data
logger
Figure 11.9 A laboratory air track with gliders.
EXAMPLE
In one air-track experiment a moving glider collides
with a stationary glider. To get the gliders to stick
together, a bit of BluTack can be stuck to each glider
at the point of impact. We can use the principle of
conservation of momentum to predict the combined
velocity of the gliders after impact.
momentum after = (mA + mB)v
In Figure 11.9 we define positive momentum to the
right.
So, from conservation of momentum:
Answer
momentum before = mAvA + mBvB
= 0.63 kg × 1.35 m s–1 + 0
= 0.85 kg m s–1
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The two gliders have the same velocity after the
collision because they have stuck together.
= (0.63 kg + 0.42 kg) × v
189
= 1.05v kg m s–1
1.05v kg m s–1 = 0.85 kg m s–1
v = 0.81 m s–1
Both gliders move to the right at 0.81 m s–1. A result
such as this can be confirmed by speed measurements
using the light gates and data logger.
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EXAMPLE
In a second experiment, glider B is pushed to the left
with a velocity of 2.7 m s–1 and glider A is pushed to the
right with a velocity of 1.5 m s–1. What happens after
this collision?
Answer
Note that B has negative momentum, and that the
total momentum of both gliders together is negative.
momentum after = (mA + mB)v
So
– 0.19 kg m s–1 = 1.05v kg m s–1
momentum before = mAvA + mBvB
= 0.63 kg × 1.5 m s–1 – 0.42 kg × 2.7 m s–1
v = – 0.18 m s–1
So both trolleys move to the left with a velocity of
0.18 m s–1.
= – 0.19 kg m s–1
11 MOMENTUM
TEST YOURSELF
6 A student decides to investigate the dynamics of
the exploding match as shown in Figure 11.8. Using
a data logger she discovers that the match starts to
move at 7.1 m s–1 immediately after the explosion.
a) The mass of the matchstick is 0.15 g and the
mass of the aluminium foil is 0.06 g. Calculate
the initial speed of the foil.
b) Calculate the kinetic energy of:
i) the aluminium foil
ii) the matchstick.
c) Calculate the minimum chemical potential
energy stored in the match head before it
explodes. Explain why the stored energy is likely
to be greater than your answer.
d) The student discovers that the matchstick travels
further than the foil. She expected the foil to travel
further. What factors affect the distance travelled
by the matchstick and the foil? (You may need to
refer back to the work on drag in Chapter 8.)
7 This question refers to the collisions between the
two gliders on an air track as shown in, Figure
11.9. In each of the following cases, calculate the
velocity of the gliders after they have collided.
a) Before the collision glider A is moving to the
right at 2.0 m s–1 and glider B is moving to the
right at 1.0 m s–1.
b) Before the collision glider A is stationary and
glider B is moving to the left at 2.0 m s–1.
c) Before the collision glider A is moving to the
right at 1.0 m s–1 and glider B is moving to the
left at 1.5 m s–1.
8 A student now investigates some more collisions
between gliders on an air track but he makes
changes to the apparatus. He replaces the
BluTack in Figure 11.9 with two magnets that
repel each other. Now when a collision takes
place, the two gliders move independently after
the collision.
Calculate the velocity of glider B after each of the
following collisions.
a) Before the collision glider A is moving to the
right at 2.0 m s–1 and glider B is moving to the
right at 1.0 m s–1. After the collision glider A is
moving at 1.2 m s–1 to the right.
b) Before the collision glider A is stationary and
glider B is moving to the left at 2.0 m s–1. After
the collision glider A is moving at 1.6 m s–1 to the
left.
c) Before the collision glider A is moving to the
right at 1.0 m s–1 and glider B is moving to the
left at 1.5 m s–1. After the collision glider A
is moving at 1.0 m s–1 to the left.
●● Momentum and energy
190
In elastic collisions kinetic energy is
conserved.
In inelastic collisions kinetic energy is not
conserved. Some or all of the kinetic energy
is transferred to heat or other types of
energy.
469884_11_AQA_A-Level_Physics_183-199.indd 190
In collisions between two or more bodies, both momentum and energy are
conserved. However, the total kinetic energy of the bodies does not always
stay the same because the kinetic energy can be transferred to other types
of energy. Collisions in which the kinetic energy of the particles is the same
after the collision as it was before are described as elastic. Collisions in
which kinetic energy is transferred to other forms of energy are described
as inelastic. Most collisions on a large scale are inelastic, but collisions
between atomic particles are often elastic.
17/04/19 10:20 AM
To answer Questions 9 and 10
you may need to refer back to
definitions from Chapters 4 and 6.
9 Define the volt.
10 a) Calculate the speed of an
electron that has been
accelerated through
a potential difference of
5000 V.
b) Which has the larger
momentum, a proton that
has been accelerated
through a p.d. of 20 kV, or
an electron that has been
accelerated through
20 kV?
−27
mp = 1.67 × 10 kg
−31
me = 9.1 × 10 kg
−19
e = 1.6 × 10 C
Although momentum and kinetic energy are very different quantities, they
are linked by some useful equations.
Momentum is given the symbol p.
p = mv
Kinetic energy is given the symbol Ek.
Ek = 12 mv2
But Ek can also be written in this form:
2 2
Ek = m v
2m
giving
Elastic and inelastic collisions
Some useful equations
TEST YOURSELF
p2
2m
It is important to remember that momentum is a vector quantity and
kinetic energy is a scalar quantity. We give momentum a direction: for
example, +p to the right and −p to the left. A vehicle travelling with
momentum p has as much kinetic energy when travelling to the right as
it does to the left.
Ek =
To the right:
(+p)2 p2
=
Ek =
2m 2m
To the left:
Ek =
(−p)2 p2
=
2m 2m
When a vector is squared, it becomes a scalar quantity.
●● Elastic and inelastic collisions
Figure 11.10 shows an unfortunate situation: a van just fails to stop in a line
of traffic and hits a stationary car; and they move forwards together. Is this an
elastic or inelastic collision? Without doing any calculations, we know that this
is an inelastic collision because the crash transfers kinetic energy to other forms –
the car is dented, so work must be done to deform the metal, and there is
noise. However, we can calculate the kinetic energy transferred as shown in the
example below.
3 m s–1
ν
18 000 kg
20 000 kg
191
at rest
2000 kg
Before the collision
After the collision
Figure 11.10 An inelastic collision.
In the world of atomic particles, collisions can be elastic because, for example,
electrostatic charges can repel two atoms or nuclei without a transfer of kinetic
energy to other forms.
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17/04/19 10:20 AM
EXAMPLE
EXAMPLE
In Figure 11.10 the momentum
before the crash is equal to the
momentum after the crash.
A helium nucleus of mass 4m collides head-on with a stationary proton
of mass m. Use the information in Figure 11.11 to show that this is an
elastic collision.
18 000 kg × 3 m s–1 + 0 = 20 000 kg × v
v=
u
0.6u
ν
4m
m
2.7 m s–1
Before the crash the kinetic
energy of the van was:
Ek = 12 mvvv2
at rest
m
4m
Before the collision
2
= 12 × 18 000 kg × (3 m s–1)
After the collision
Figure 11.11
= 81 kJ
Answer
After the collision the kinetic
energy of the van and car
together is:
The velocity of the proton can be calculated as follows.
The momentum before the collision equals the momentum after the collision.
Ek = 12 (mv + mc)v2
4mu + 0 = 4m × 0.6u + mv
mv = 4mu – 2.4mu
v = 1.6u
The total kinetic energy before the collision was:
2
= 12 × 20 000 kg × (2.7 m s–1)
= 73 kJ
So about 8 kJ of kinetic energy
is transferred to other forms of
energy.
Ek = 12 × 4mu2
= 2mu2
The total kinetic energy after the collision was:
Ek = 12 × 4m(0.6u)2 + 12 × m(1.6u)2
= 2m(0.36u2) + 12 m(2.56u2)
TIP
= 0.72mu2 + 1.28mu2
11 MOMENTUM
In elastic collisions, kinetic
energy is conserved. In inelastic
collisions energy is transferred to
other forms.
= 2mu2
So the total kinetic energy is conserved in this collision. It is a useful
rule to know that in an elastic collision the relative speeds of the two
particles is the same before and after the collision.
TEST YOURSELF
11 Look at the diagram of two dodgem cars in a collision at a funfair.
192
5 m s–1
3 m s–1
A
3.4 m s–1
B
mass 200 kg
A
mass 160 kg
Before the collision
ν
B
mass 200 kg
mass 160 kg
After the collision
Figure 11.12
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17/04/19 10:20 AM
the velocity of car B after the collision.
b) E xplain whether this is an elastic or an
inelastic collision.
12 Two ice skaters, each of mass 65 kg, skate
towards each other. Each skater has a velocity
of 1 m s–1 relative to the ice. The skaters collide;
after the collision each skater returns back
along their original path with a velocity of 2 m s–1.
a) Show that momentum is conserved in the
collision.
b) Calculate the gain in kinetic energy of each
skater during the collision.
c) E xplain where the extra energy has come
from.
13 Look again at Question 8. Show that each of
the collisions between the two gliders is elastic.
14 A
ball of mass 0.1 kg falls from a height of 5.0 m
and rebounds to a height of 3.2 m. For this
question assume g = 10 m s−2.
a) Calculate the velocity of the ball just before it
hits the ground.
b) Calculate the velocity of the ball just after it
has hit the ground.
c) The ball is in contact with the ground for
0.004 s. Calculate the average force that acts
on the ball in this time.
d) The momentum of the ball has changed during
this process. The principle of the conservation
of momentum states that ‘momentum is
always conserved’. Explain how momentum is
conserved in this case. [Hint: there is another
body involved in this collision – what is its
momentum?]
Elastic and inelastic collisions
a) Use the information in Figure 11.12 to calculate
More advanced problems in momentum
The generalised form of Newton’s second law of motion enables you to
solve some more complicated problems.
force = rate of change of momentum
F =
Δ(mv)
Δt
EXAMPLE
Calculate the force exerted by water leaving a fire hose
on the firefighter holding the hose. The water leaves
with a velocity of 17 m s–1; the radius of the hose is
3 cm; the density of water is 1000 kg m–3.
Answer
The mass of water leaving the hose each second is Δm .
Δt
This is equal to the density of water ( ρ) × the volume of
water, V, flowing per second:
Δm
ΔV
Δt = ρ Δt
But the volume flow per second is the cross-sectional
area of the hose (A) multiplied by the velocity of the
water (v). So:
469884_11_AQA_A-Level_Physics_183-199.indd 193
Δm
= ρ Av
Δt
Thus the force, or change of momentum per second,
is:
F=
Δp
= ρ Av × v
Δt
= ρ Av2
2
= 1000 kg m–3 × π × (0.03)2 m2 × (17 m s–1)
= 820 N
This large force explains why a hose is sometimes
held by two firefighters, and why large hoses have
handles.
193
17/04/19 10:20 AM
Collisions in two dimensions
Before the collision
stationary
p
A
An interesting special result occurs when two atomic particles of the
same mass collide elastically, when one of the particles is initially
stationary.
B
After the collision
stationary
A
B
p
Figure 11.13 Head-on collision of
two protons.
Before the collision
A
p
θ
B
p sin θ
After the collision
θ
A
90 – θ
What happens when two particles of the same mass collide, but the
collision is not head-on, as shown in Figure 11.14? The momentum
of particle A can be resolved into two components: p cos θ along the
line of collision and p sin θ perpendicular to the line of the collision.
As in Figure 11.13, all of the momentum of A along the line of the
collision (here p cos θ) is transferred to particle B. This leaves particle
A with the component p sin θ, which is at right angles to the line of
the collision.
Figure 11.15 shows how momentum is conserved as a vector quantity.
p cos θ
Figure 11.14 Non-head-on collision of
two protons.
p
θ
p cos θ
11 MOMENTUM
In this way both momentum and kinetic energy are conserved. This only
happens when the particles are of the same mass; in all other cases both
particles will be moving after the collision.
So, in a non-head-on elastic collision between two particles of the same
mass, they always move at right angles to each other.
B
194
If the two particles (protons for example) collide head-on, then the
momentum and kinetic energy of the moving proton (A) is transferred
completely to the stationary proton (B). See Figure 11.13.
p sin θ
Figure 11.15 In vector terms, p = p cos θ
+ p sin θ.
Kinetic energy is also conserved. Before the collision the kinetic energy is:
p2
2m
After the collision the kinetic energy of the two particles is:
p2
p2
cos2θ +
sin2θ
2m
2m
However, since cos2θ + sin2θ = 1, the kinetic energy after the collision is the
p2
.
same as before,
2m
TEST YOURSELF
15 A
n alpha particle with a mass of about 4 amu is
emitted from a uranium nucleus, mass about
238 amu, with a kinetic energy of 4.9 MeV.
−27
1 amu (atomic mass unit) = 1.66 × 10 kg;
−19
e = 1.6 × 10 C
a) Calculate the momentum of the alpha particle.
b) Calculate the momentum of the nucleus after
the alpha particle has been emitted.
c) Calculate the kinetic energy of the nucleus
after the alpha particle has been emitted.
d) Discuss where the kinetic energy has come
from in this process.
469884_11_AQA_A-Level_Physics_183-199.indd 194
16 A
rocket of mass 400 000 kg takes off on a voyage
to Mars. The rocket burns 1600 kg of fuel per
second ejecting it at a speed of 2600 m s–1 relative
to the rocket.
a) Calculate the force acting on the rocket due to
the ejection of the fuel.
b) i) Calculate the acceleration of the rocket at
take-off.
ii) Calculate the acceleration of the rocket 90
seconds after take-off.
17/04/19 10:20 AM
1 Which of the following is a correct unit for momentum?
A Ns
C N s–1
B kg m s–2
D kg m–1 s
2 A body is in free fall. Which of the following quantities is equal to the
rate of increase of the body’s momentum?
A kinetic energy
C weight
B velocity
D decrease in potential energy
Practice questions
Practice questions
3 A tennis ball of mass 60 g is travelling at 35 m s–1. A player hits the ball
back in the opposite direction, with a velocity of 45 m s–1. What is the
magnitude of the impulse which the tennis racket exert on the ball?
A 2.4 Ns
C 0.6 Ns
B 4.8 Ns
D 2.7 Ns
4 An alpha particle has kinetic energy of 8.0 MeV. The mass of the alpha
particle is 6.8 × 10–27 kg; the charge on an electron is 1.6 × 10–19 C.
Which of the following gives the correct value for the particle’s
momentum?
A 1.8 × 10–22 kg m s–1
C 2.6 × 10–22 kg m s–1
B 1.8 × 10–19 kg m s–1
D 1.3 × 10–19 kg m s–1
5 A lorry travelling at 5.5 m s–1 collides with a stationary car. After the
collision, the lorry and car move forwards together. Use the information in
the diagram below to calculate which of the following values is the correct
velocity of the car and lorry after the collision.
5.5 m s–1
at rest
mass 20 000 kg
mass 2000 kg
Figure 11.16
A 5.5 m s–1
C 5.0 m s–1
B 2.75 m s–1
D 1.5 m s–1
6 In the collision described in Question 5, some of the kinetic energy of
the lorry is transferred to other forms, such as heat and sound. Which
of the following gives the correct value for the change of kinetic energy
during the crash?
A 300 000 J
C 150 000 J
B 275 000 J
D 27 500 J
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7 A snooker ball of mass 160 g collides elastically with the cushion
of a snooker table. Use the information in the diagram below to
calculate which of the following is the correct value for the change
of momentum of the ball in the direction normal (perpendicular)
to the cushion.
3.0 m s–1
3.0 m s–1
30°
30°
cushion
Figure 11.17
C 0.42 kg m s–1
B 0.24 kg m s–1
D 0.48 kg m s–1
8 A driver passing through a town sees a hazard and brings his car
to an emergency stop. The driver has a mass of 80 kg. The graph
shows how the force exerted on him by his seat belt varies while
the car slows down. Use the graph to calculate which of the
following gives the correct value for the initial speed of the car
before the brakes were applied.
A 24 m s–1
C 12 m s–1
B 16 m s–1
D 6 m s–1
11 MOMENTUM
9 A rubber ball of mass 0.2 kg lands on a floor with a velocity of
6 m s–1; it bounces back up with a velocity of 3 m s–1. The ball is
in contact with the floor for 0.06 s. Which of the following is the
correct value for the average force that the floor exerts on the ball
during the bounce?
A 60 N
C 20 N
B 30 N
D 10 N
10 Two trolleys collide as shown in figure 11.19.
After the collision they stick together. Use the
information in the diagram to calculate which
of the velocity values below is the correct
value for the trolleys after the collision.
960
force/N
A 0
1.0
time/s
Figure 11.18
1.5 m s–1
3 kg
2.0
2.5 m s–1
1 kg
Figure 11.19
A 1.75 m s–1 to the right
B 1.75 m s–1 to the left
196
C 0.25 m s–1 to the left
D 0.50 m s–1 to the right
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300
Practice questions
11 A car travelling at speed collides with a wall and comes to a halt.
The force acting on the car during the crash is shown in
the diagram.
a) Use the graph to show that the change of momentum of the
car in the crash is approximately 26 000 N s.
(3)
force/kN
200
b) The car and its passengers have a mass of 1300 kg. Calculate
the speed of the car before the crash.
(2)
100
c) There were two passengers in the car. One was wearing a seat
belt and the second was not. Explain why the one without the
seat belt is more likely to receive serious injuries.
(3)
d) Sketch a graph to show how the velocity of the car changed
with time during the crash.
(3)
0.1
time/s
12 In the diagram to the right an alpha particle has just been emitted Figure 11.20
from a large nucleus. It is repelled by the large nuclear charge and
7
ν = 1.5 × 107 m s–1
leaves with a velocity of 1.5 × 10 m s–1.
–27
a) The alpha particle has a mass of 6.8 × 10 kg and the nucleus
a mass of 4.0 × 10–25 kg. Calculate the recoil velocity of the
(3)
nucleus.
alpha particle
b) Calculate the total kinetic energy of the alpha particle and
–19
nucleus. Express your answer in MeV. (e = 1.6 × 10 C)
0.2
nucleus
Figure 11.21
(5)
c) Explain how momentum and energy have been conserved in this
alpha decay process.
(2)
13 The diagram below illustrates part of a ride at a funfair. One of the vehicles,
with a mass of 240 kg, reaches the bottom of the slope reaching a speed of
14 m s–1. It is slowed down to a speed of 5 m s–1 by passing through a trough
of water. The vehicle slows down over a time of 0.6 s. When it hits the water,
a mass of water is thrown forwards with a velocity of 18 m s–1.
5 m s–1
14 m s–1
mass of vehicle = 240 kg
Figure 11.22
a) Explain what is meant by the principle of conservation of
momentum.
(2)
b) Show that the mass of water thrown forwards by the vehicle is
about 120 kg.
(3)
c) Calculate the change in kinetic energy of the vehicle, and the
increase in kinetic energy of the water, as the vehicle splashes
into the trough.
(4)
d) Explain how energy is conserved during this process.
(2)
e) Calculate the average deceleration of the vehicle in the water.
(2)
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197
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14 A soft rubber ball is allowed to fall on to the ground
and bounce back up again. A data logger measures
the speed as the ball falls. The graph shows how the
ball’s velocity changes with time.
6
(1)
The ball has a mass of 0.08 kg.
b) Determine the momentum change of the ball
as it hits the ground.
(3)
4
2
0
0.2
c) Use your answer to part (b) to calculate the force
exerted by the ground on the ball as it
is in contact with the ground.
(2)
–2
d) How big is the force that the ball exerts on the
ground while it is in contact?
(1)
–6
e) A hard rubber ball of the same mass, 0.08 kg,
is now dropped from the same height. When it
is in contact with the ground it exerts a larger
force than the soft ball. Explain two changes you
might see in the graph of velocity against time
(4)
measured by the data logger.
A
8
velocity/ m s–1
a) Use the graph to determine the time of the
ball’s bounce over the region AB.
10
0.4
0.6
0.8
time/s
1.0
1.2
1.4
1.6
–4
B
–8
Figure 11.23
15 A steel ball of diameter 10 cm is allowed to fall from a height of
2 m so that it collides with a steel spike that is embedded in
a piece of wood.
a) Steel has a density of 9000 kg m–3. Show that the ball has a
mass of about 4.7 kg.
(2)
b) Determine the ball’s speed as it comes into contact with the
steel spike.
(2)
11 MOMENTUM
The ball now moves forwards with the spike together. The spike
has a mass of 2.6 kg.
c) Calculate the velocity of the ball and spike as they move
forwards.
(2)
d) By calculating the change of kinetic energy in the collision, explain
whether this is an elastic or inelastic collision.
(4) Figure 11.24
The spike penetrates 3.5 cm into the wood before coming to rest.
e) Calculate the average force acting to slow the ball and spike.
198
2.0 m
(3)
16 a) A neutron travelling with a velocity of 1.2 × 107 m s–1 is absorbed by a
stationary uranium-238 nucleus. Calculate the velocity of the nucleus
after the neutron has been absorbed.
(3)
b) An alpha particle decays from a nucleus of polonium-208. The speed
of the alpha particle is 1.5 × 107 m s–1. Calculate the velocity of the
polonium nucleus. The alpha particle has a relative atomic mass of 4.(3)
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Practice questions
Stretch and challenge
(a)
17 The rotor blades of a helicopter push air vertically downwards
with a speed of 5.0 m s–1.
18 m
a) Use the information in Figure 11.25a and the value for the
density of air, 1.2 kg m–3, to calculate the momentum of the air
pushed downwards per second.
b) When the helicopter pushes air downwards with a speed of
5.0 m s–1, it hovers stationary. Calculate the helicopter’s mass.
(b)
c) The helicopter’s rotor blades are now tilted forwards at an
angle of 14° to the horizontal as shown in Figure 11.25b.
The speed of rotation is increased so that the helicopter flies
horizontally and accelerates forwards.
14°
v
Figure 11.25
Calculate:
i)
the speed of air being pushed away from the blades
ii) the initial horizontal acceleration of the helicopter.
18 A firework rocket is launched upwards. When it reaches its highest point,
it explodes symmetrically into a spherical ball with 100 fragments, each
of which has a mass of 20 g. The chemical energy stored in the explosives
is 2 kJ; 80% of this energy is transferred to the kinetic energy of the
fragments. Calculate the speed of each fragment.
19 The photo on the right shows a computer-enhanced image of a
cloud chamber photograph. The event shown is the radioactive
disintegration of a helium-6 nucleus, 62 He, into a lithium-6
nucleus, 63 Li, and a β-particle. The helium nucleus was originally
stationary at point O. The β-particle is ejected along the thin
broken red track OA, and the lithium-6 nucleus travels a short
distance along the thick green track OB. Both particles travel in
the plane of the paper. The β-particle’s track is curved due to the
presence of a strong magnetic field.
a) Deduce the direction of the magnetic field.
A
0
B
Figure 11.26
b) Explain why the lithium nucleus track is short and thick, and the
β-particle track longer and thinner.
c) Judging from the cloud chamber tracks, momentum is not conserved in
this decay process. Explain why this appears to be the case.
Since physicists accept the principle of conservation of momentum to be
a universal law, it was suggested that another particle must be emitted in
β-decay. This particle is the antineutrino, which is uncharged and leaves no
track in the cloud chamber.
199
d) The β-particle has momentum p1 and the lithium nucleus momentum
p2. Use the principle of momentum conservation to determine the
momentum of the antineutrino. Express your answer in terms of
p1 and p2, giving both a magnitude and direction.
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12
Properties of
materials
PRIOR KNOWLEDGE
●
●
●
A force acting on an object can cause it to change shape and stretch,
squash or bend.
Weight is the pull of gravity on an object and can be calculated using
the formula:
W = mg
where W is the weight in N
m is the mass of the object in kg
g is the gravitational field strength in N kg−1.
Work done (or energy transferred) by a force on an object depends
on the magnitude of the force in the direction of motion of the object.
This can be calculated using the formula:
W = F s cos θ
where W is the work done in J
F is the magnitude of the force in N
s is the distance moved in the direction of the force in m
θ is the angle between the force and the direction in which the
object moves.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 The Voyager 2 spacecraft is travelling through space at a steady speed
of approximately 23 km s−1 relative to Earth. Discuss whether work is
being done by the spacecraft to travel at this speed, assuming there
are no gravitational forces acting on Voyager.
200
Figure 12.1 Voyager 2, launched over 36 years ago and still
travelling through space.
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Density
2 The Mars lander, Curiosity, has a mass of 899 kg. Calculate the
lander’s weight on Earth and on Mars. The gravitational field strength
on Earth is 9.81 N kg−1 and on Mars is 3.7 N kg−1.
3 Alex is going on holiday. At the airport she has to lift her suitcase and
carry it up a flight of stairs which has 20 steps, each 18 cm tall and
22 cm deep. Her suitcase has a mass of 20 kg.
Calculate the total amount of work Alex did in moving her suitcase
(assume g = 9.81 N kg−1).
4 A girl is sledging on a snowy day. She pulls her sledge on to a path
where the snow has melted. The girl does 12 000 J of work to pull the
sledge 100 m along the path. Calculate the horizontal force she has
used to pull the sledge along the path.
5 Figure 12.2 shows an example of a penguin toy race. Wheeled
penguins are lifted by a motor up a moving staircase. When they get
to the top they roll down the long slide and back to the staircase. The
height of the staircase is 20 cm, and each penguin has a mass of 20 g.
a) Calculate the work done by the motor lifting one penguin from the
bottom to the top of the staircase.
b) Compare the value you calculated in part (a) with the change in
gravitational potential energy of the penguin as it goes down the
slide.
Figure 12.2 A penguin toy race.
●● Bulk properties of solids
People have been using, and altering, the properties of materials since the
Stone Age. Flint tools, samurai swords and glass beakers all made use
of the strength, flexibility or optical properties of the materials they were
created from.
Today, engineering materials researchers use their knowledge of the
properties of materials to develop novel materials, and find new uses
for old ones.
●● Density
Figure 12.3 The pole used in pole
vaulting must be lightweight, strong and
have some flexibility. Vaulting poles were
originally made from wood. Nowadays,
they are usually made from layers of
fibreglass and carbon-fibre materials.
The density of a material is a measure of the mass per unit volume. It is
given the symbol, ρ. Density is a useful quantity because it allows us to
compare different materials.
ρ=m
V
where ρ is density in kg m−3
m is mass in kg
201
V is volume in m3
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Figure 12.4 The large stainless steel globe is more massive than the small stainless
steel ball bearing. The globe floats; the ball bearing doesn’t. Why?
EXAMPLE
Calculate the density of the stainless steel globe
shown in Figure 12.4 using this data.
Mass of globe = 0.30 kg
Diameter of globe = 0.2 m
Volume of a sphere = 34 πr3
Answer
12 Properties of materials
Volume of the globe = 4.19 × 10−3 m3
density = mass
volume
0.3 kg
4.19 × 10−3 m3
= 72 kg m−3
=
The density of the globe depends on the density of the
steel and the density of the air that is within the globe.
Together, the combined (or average) density is much
less than the density of water, which is 1000 kg m−3;
this is why the globe floats. The ball bearing, which has
a density of 7800 kg m−3, sinks.
Aerogels have the lowest density of any solid. Table 12.1 compares the
density of aerogel with other common materials.
Table 12.1 The density of some common materials
Material
Density (kg m−3)
Aerogel
20
Air (at 20°C)
1.2
Water
1000
Silica glass
2200
Steel
7480–8000
Figure 12.5 Aerogel: a very low-density
material that is 99.8% air.
Aerogel has been used in aerospace industries as a very low-density
insulation material. Aerogel is a good insulator because it traps air and
prevents it from moving. If air can circulate it will transfer heat through
convection currents. NASA used aerogel on the Mars landers Spirit and
Opportunity to provide thermal insulation of key components.
202
Figure 12.6 The sample collector
from the Stardust space mission. Each
rectangular block is made from aerogel.
469884_12_AQA_A-Level_Physics_200-220.indd 202
The NASA Stardust mission used aerogel to capture particles from the
tail of comet Wild 2. The very low density of the aerogel ensured that
the comet dust particles were slowed down without being damaged. The
insulating properties of aerogel also meant that the dust particles were
protected from the heat generated during re-entry of the sample collector
into the Earth’s atmosphere.
17/04/19 10:21 AM
1 Suggest why different samples of steel can have
different densities.
2 a) The stainless steel ball bearing shown in Figure
12.4 has a mass of 32.1 g and a diameter of
19.98 mm. Calculate the density of the stainless
steel used to make the ball bearing.
b) The diameter of the ball bearing was measured
using a pair of vernier calipers. These measure
to an accuracy ±0.02 mm. Estimate the
percentage uncertainty for the value of the
diameter that was measured.
3 The globular cluster M13 contains approximately
300 000 stars. It has a diameter of 145 light years. A
light year is the distance travelled by light in a year.
Take the speed of light to be 3 × 108 m s−1.
a) If a typical star in the cluster has a mass of
2 × 10 30 kg calculate the average density of the
globular cluster.
b) Use your answer to part (a) to suggest why stars
within the globular cluster rarely collide with
each other.
Hooke’s law
TEST YOURSELF
4 A window made from silica glass has an area of
1.5 m × 2 m and is 3 mm thick. Calculate the mass
of the window using the value for density given in
Table 12.1.
5 A plumber uses a brass washer when fitting a
tap. The washer is a flat ring of metal that is
0.8 mm thick. It has an internal diameter of
4 mm and an external diameter of 8.5 mm.
The density of brass is 8.4 g cm −3 . Calculate
the mass of the washer.
6 The aerogel sample collector from the Stardust
mission used 132 aerogel blocks, each of which
was 3 cm thick. The dust collector had a total
surface area of 1039 cm2 of aerogel.
Calculate the mass of aerogel used to capture the
comet dust.
7 Estimate the density of a typical human body.
●● Hooke’s law
Compressive forces are forces that tend
to squeeze an object and reduce its size in
the direction that the forces are applied.
For example, a heavy weight placed on a
column, together with the upwards force on
the bottom of the column, will reduce the
height of the column.
Tensile forces are forces that act to pull or
stretch an object. The metal ropes holding
up a lift have tensiles force acting on them
due to the weight of the lift and the upwards
pull from the ceiling on the end of the rope.
Materials can be characterised by the
properties they show when forces are
applied to them. Different materials
will exhibit very different properties.
When scientists are testing materials
they may apply compressive forces
or tensile forces.
In 1678, Robert Hooke wrote about
his discovery of elasticity. He used
his investigations into springs to
develop a spring for use in the first
portable timepiece, or a watch as
they are now known.
Hooke realised that the extension
of some springs shows a linear
region for a range of applied forces.
In other words, the extension was
proportional to the force applied.
extension ∝ force
Figure 12.7 A bungee trampoline makes
use of the elastic properties of ropes.
When a tensile force is applied to the
ropes, they stretch. The greater the
force applied, the more they stretch.
203
Δl ∝ F
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Spring constant, k, is a measure of how
hard it is to bend or stretch a spring. A large
spring constant means that the spring is
stiff. k has units of N m−1.
Extension is the length a material has
stretched when a load is added. It is
calculated by subtracting the original
length of the material from the length
when stretched.
Limit of proportionality is the endpoint
of the linear section of a force–extension
graph.
Elastic limit is the load above which a
material is permanently deformed.
A material is said to be elastic when it
returns to its original dimensions once the
applied load is removed.
A material is said to be plastic when it is
permanently deformed and does not return
to its original dimensions once the applied
load is removed.
How much a spring extends will also depend on the spring constant of
the spring. This is a measure of how easy it is to stretch when a force is
applied. A spring that extends a large amount for a force of 1 N is not as stiff
as a spring that extends only a small amount for the same force. The spring
constant, k, defines the stiffness of a spring as the force required for a unit
extension of the spring.
We can therefore write Hooke’s law as:
F = kΔl
where
F is the applied force in N
k is the spring constant in N m−1
Δl is extension in m
Using the simple equipment shown in Figure 12.8, the spring constant of
a spring can be measured. The spring is stretched using a tensile load. This
is done by adding masses to a mass hanger. The extension of the spring is
measured by noting the change in the position of the pointer against the rule.
rule
10
8
F/N
spring
6
12 Properties of materials
4
2
0
weights
0
10 20 30 40 50 60 70 80
x/mm
Figure 12.8 Experimental set-up for Hooke’s law and graph.
204
When the results from the experiment are plotted, the graph initially has a
linear region, as shown in Figure 12.8. Here the spring is still coiled and the
extension is directly proportional to the load. When more masses are added,
the spring starts to stretch more for each increase of load: the extension is
no longer proportional to the load. The point at which this occurs is called
the limit of proportionality.
Figure 12.9 These two springs were
the same length. One of them has gone
beyond its elastic limit and will not
return to its original length.
469884_12_AQA_A-Level_Physics_200-220.indd 204
The elastic limit of a material is the point below which the spring will
return to its original length when the load is removed. The spring is
showing elastic deformation. Above the elastic limit, the spring will be
stretched out of shape and will not return to its original length. This is
known as plastic deformation.
17/04/19 10:21 AM
Hooke’s law
EXAMPLE
Springs are used in newton-meters to make
measurements of force. One particular newton-meter
is used to measure forces of up to 50 N. The length of
the scale on the newton-meter is 10 cm long.
Calculate the value of the spring constant for the
spring in the newton-meter.
Answer
so
F = kΔl
k=
F
Δl
k = 50 N
0.1m
k = 500 N m−1
ACTIVITY
Extension of springs
(a) the total extension 2Δl
Δl
(b) in each spring, extension 2
Figure 12.10 shows two different combinations of
springs: in series and in parallel. All the springs
are the same length and have the same spring
constant.
A student hangs a load from the bottom of each
spring combination and then measures the extension.
Her results are given in the table below.
Δl
Δl
2
load
Extension (cm)
Load (N)
Series combination Parallel combination
0
0
0
0.5
2.5
0.7
1.0
6.2
1.5
1.5
9.5
2.6
Figure 12.10 Springs in series and parallel.
2.0
13.6
3.4
2.5
17.5
4.4
3.0
21.4
5.3
2 Use your graph to calculate the effective spring
constant for each combination of springs.
3 The student predicted that the effective spring
constant of the series combination will be four
times smaller than the effective spring constant of
the parallel combination.
Use the data from your graph to show if this prediction
is correct.
4 Predict the spring constant when two identical
springs, which are placed in parallel, are then put
in series with another pair of springs, which are
also in parallel.
5 For the series combination of springs, calculate the
percentage uncertainty in each measurement if the
student measured the extension of the springs with
an accuracy of ±0.1 cm.
1 Plot a graph of load against extension for the series
and parallel combination of springs.
TIP
It is usual to plot the independent variable on
the x-axis and the dependent variable on the
y-axis. However, in this case we plot the load
(independent variable) on the y-axis so that the
spring constant, k, is given by the gradient.
469884_12_AQA_A-Level_Physics_200-220.indd 205
load
Δl
205
17/04/19 10:21 AM
TEST YOURSELF
8 The experiment shown in Figure 12.8 is repeated
with a spring having double the value of spring
constant. Describe, and explain, how the linear
part of the graph would change.
9 A spring has a spring constant of 1 N cm−1. The
scale on a newton-meter is 10 cm long. Would the
spring be suitable for a newton-meter used to
measure forces up to 10 N?
10 An acrobat is climbing up an aerial rope that is
suspended vertically from a high ceiling. The
unstretched length of the rope is 5 m, and it
stretches to 5.7 m when the acrobat is starts
climbing up. The mass of the acrobat is 55 kg.
Assuming that the rope obeys Hooke’s law,
calculate the spring constant of the rope.
11 Three identical springs with a spring constant k
are hung in series with each other. A force, F, is
applied. What is the effective spring constant, ke,
for this combination of springs in series.
12 a) A pair of springs in parallel are attached to
the bottom of three springs in series. All five
springs are identical and have a spring constant
k. A force, F, is suspended halfway between the
lower ends of the parallel springs. What is the
effective spring constant, ke, of this combination?
b) The three springs, originally in series, are now
put in parallel and then attached, in series, to the
pair of springs in parallel. Calculate the effective
spring constant of the new combination.
Investigating wires and fibres
12 Properties of materials
Nearly all materials show Hooke’s law behaviour up to a point. This includes
metals such as copper and steel, fibres such as cotton and silk, natural
rubber and polymers.
The applied force beyond which materials no longer obey Hooke’s law
will be different for each material. In a school laboratory it is possible to
investigate the properties of materials such as copper wire and nylon thread.
These materials produce measurable extensions for an easily obtainable
range of applied forces. Other materials, such as steel wire, require much
larger forces and so specialist equipment is required.
ACTIVITY
Investigating the properties of copper wire
wooden blocks
copper wire
A thin copper wire is clamped between two
wooden blocks at one end of a bench. The other
end of the wire is passed over a pulley.
A rule is placed on the bench next to the wire,
and a strip of sticky tape is attached to the wire
so that it touches the rule.
Figure 12.11 shows a simple experimental set-up that
can be used to measure the properties of copper wire.
To begin with, carry out a preliminary test to
determine what length of wire will produce
a reasonable amount of extension as you
increase the load. The tape marker must not
go past the pulley. Choose a length of wire,
and attach a mass hanger on to the end. Place
masses on to the mass hanger and note how far
the tape marker moves as you increase the load.
Wear safety glasses while carrying out the experiment.
Place padding beneath the masses to break their fall.
Once you have decided on the length of wire, set up the
experiment again. Before attaching the mass hanger,
G-clamp
rule
tape
206
load
Figure 12.11 Stretching copper wire.
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17/04/19 10:21 AM
Carefully add a 100 g mass to the hanger and read
the extension off the scale. Continue taking readings
until the extension obtained for each increase in
weight becomes much larger than before. At this
point the wire has passed the elastic limit and is
now behaving plastically. Continue to add masses,
but allow time for the wire to continue to extend
after each addition. Eventually the wire will break.
1 Plot a graph of force against extension for the
copper wire. On your graph label:
● the region where the wire obeyed Hooke’s law
● the elastic limit
● the breaking point of the wire.
2 Calculate the spring constant for your wire in the
region where the wire obeys Hooke’s law.
3 Explain why using a longer wire gives a larger range
of values for extension.
Ductile materials can be formed into wires
by stretching them. They show ductility.
A brittle material is one that shows little, or
no, plastic deformation before breaking.
F/N
B
A
0
l/mm
Figure 12.12 A typical force–extension
graph for a copper wire. The wire shows
plastic behaviour when it is stretched
beyond the elastic limit.
F/N
material breaks
0
l/mm
Hooke’s law
stick the tape marker on the wire so that it is at the
lower end of the metre scale on the rule.
4 Estimate how accurately you were able to read the
position of the marker on the rule. Explain what
effect this will have on the value of spring constant
you have calculated.
Extension
● Repeat the experiment for different thickness of
copper wire and compare the results.
Carry out a similar experiment for fibres such
as cotton and nylon thread, and compare their
behaviour with that of copper.
If you do not have data for this experiment then you
can use these results for a copper wire.
Mass (kg)
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Extension
(mm)
0
0.8
1.6
2.3
3.2
4.2
5.3
6.8
9.6
Wires obey Hooke’s law because the bonds between the metal atoms
act like springs. When the wire is stretched the bonds lengthen slightly.
When the force is removed, the bonds return to their original length.
However, if the force applied is too great, and the elastic limit exceeded,
then the metal atoms will be able to move past one another and the
wire lengthens. This is known as ductility, and is a very useful property
as it allows metals to be formed into thin wires. The wire formed
shows plastic behaviour and will not return to its original length when
the force is removed. Ductile behaviour is also an example of plastic
deformation.
In Figure 12.12 the dotted line represents the extension measured as the
force is removed from the loaded wire. It can be seen that the wire has
permanently lengthened because, even with no applied force, there is still a
measurable extension.
Some materials do not show plastic behaviour but are brittle and break
when the elastic limit is exceeded. Cast iron and glass are two examples of
brittle materials. Figure 12.13 shows a typical force–extension graph for
high-carbon steel, which is also a brittle material. The material fractures and
breaks. It does not show plastic behaviour.
The way in which ductile and brittle materials fracture is also different. In
a ductile material, the sample of material will elongate and ‘neck’ before it
breaks. On a force–extension graph, necking occurs in the plastic region
of the graph. In a brittle material there is no change in the shape of the
material because it does not undergo plastic behaviour. A straight break
in the material is seen. Figure 12.14 shows the difference between the two
types of fracture.
207
Figure 12.13 Force–extension curve for
high-carbon mild steel.
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Elastic strain energy
The exercise equipment shown in Figure 12.15 makes use of springs.
When you pull the two handles apart you increase the energy stored by
the springs. We can calculate how much energy is stored as elastic strain
energy in the springs.
The energy stored is equal to the work done stretching the springs. The
work done depends on the average force applied and the extension of the
springs.
work done = average force × extension
Ductile
Brittle
Figure 12.14 Ductile and brittle
fracture.
Elastic strain energy is the energy stored
by stretched materials.
For a material that obeys Hooke’s law, the average force is:
Fmax
2
We can express the elastic strain energy using the equation:
elastic strain energy = 1 × F × Δl
2
where:
J is elastic strain energy
F is force in N
Δl is extension in m.
We can also express the elastic strain energy in terms of the spring constant:
Elastic strain energy = 1 × F × Δl
2
and
F = kΔl
12 Properties of materials
substituting for F we obtain:
elastic strain energy = 1 × (kΔl) × Δl = 1 × k × Δl2
2
2
For a material that does not fully obey Hooke’s law, we can still calculate
the elastic strain energy by calculating the area under the load–extension
graph.
For any graph, we can choose small changes in extension, δ l, and calculate
the work done by the load to produce that small extension. The total work
done is then the sum of all these values.
WT = ΣFδ l
Figure 12.16 shows how this is done for a simplified force–extension graph.
208
Figure 12.15 This exercise equipment
uses parallel springs to strengthen arm
muscles.
469884_12_AQA_A-Level_Physics_200-220.indd 208
On Figure 12.16 the region of the graph marked as OA represents the
applied loads for which the material obeys Hooke’s law. For this region the
area under the graph is a triangle, so:
W1 = 1 Fmax Δl1
2
For the second region of the graph, AB, the material no longer obeys
Hooke’s law. However, energy is still required to stretch the material, so
work is still being done.
17/04/19 10:22 AM
A
Fmax
W2 = Fmax × Δl2
B
which means that the total work done in stretching this material is:
WT = W1 + W2
WT = 1 Fmax Δl1 + Fmax Δl2
2
or
0
Δ l2
W1 =
1
2
Hooke’s law
F/N
The area of this region is that of the rectangle below the line:
W2 = Fmax Δ l2
Δ l /mm
WT = Fmax 1 Δl1 + Δl2
2
The elastic strain energy stored is equivalent to the work done in stretching
the material.
Fmax Δ l1
Figure 12.16 Calculating elastic strain
energy.
EXAMPLE
F/N
For non-linear graphs elastic strain
energy is calculated by counting the
number of squares under the line and
multiplying this by the value of work
equivalent to each square. Find the elastic
strain energy in Figure 12.17.
Answer
Note: We could have estimated the elastic
strain energy by assuming that shape C
was a triangle and calculating the area of
that triangle. However, that would give a
less accurate answer than the method of
counting squares shown here.
In Figure 12.17, the work equivalent of
each small square
= 0.2 N × 0.1 × 10−3 m = 0.02 × 10−3 J
1
× 5 N × 2 × 10−3 m
2
= 5 × 10−3 J
8
7
C
6
5
4
3
B
2
A
1
0
0
1
2
3
4
Δ l/mm
5
6
7
Figure 12.17 Elastic strain energy is calculated by counting the number of
squares under the line and multiplying this by the value of work equivalent
to each square.
area of A =
rea of B = 5 N × 5 × 10−3 m
a
= 25 × 10−3 J
Number of small squares in region C ≈ 448
Work equivalent = 448 × 0.02 × 10−3 J = 9 × 10−3 J
209
So
elastic strain energy = 3.9 × 10−2 J
TIP
Convert all measurements to
standard units, for example,
cm to m.
469884_12_AQA_A-Level_Physics_200-220.indd 209
Energy and springs
When a material is stretched or compressed its elastic strain energy is
altered. Figure 12.18 shows two common toys that make use of stored
elastic strain energy.
17/04/19 10:22 AM
EXAMPLE
The spring jumper toy shown in Figure 12.18 is
compressed by pushing the suction cup on to the stand.
When the suction cup releases the toy ‘jumps’ to a
height of 65 cm. The toy has a mass of 16 g. The spring is
originally 3.6 cm long, and is compressed to 0.9 cm long.
Assuming that no energy is dissipated as heat and
sound when the toy jumps, calculate:
a) the elastic strain energy stored by the spring
b) the spring constant of the spring.
Answer
a) If there are no energy loses during the jump then
the amount of gravitational potential energy gained
by the toy will be equal to the amount of elastic
strain energy stored by the spring.
gpe = mass × gravitational field strength × height
= 0.016 kg × 9.81 N kg−1 × 0.65 m
= 0.102 J
instead of the amount the spring has extended we
need to know the amount it has compressed.
compression = initial length − final length
= 3.6 cm − 0.9 cm = 2.7 cm
Rearranging the equation for elastic strain energy
we get:
2 × elastic strain energy
Δl2
2
×
0.102
J
k=
0.027 m2
k = 280 N m−1
k=
The elastic properties of some materials, such as rubber, can be complex.
When a rubber band is stretched it will return to its original length.
However, the way in which it does this is very different from a metal
wire. Figure 12.19 shows an experimental set-up that can be used to
carry out this measurement. Figure 12.20 shows a typical force–extension
curve obtained from this experiment.
newtonmeter
rule
rubber
band
210
Figure 12.19 Experimental set-up for
measuring the extension of a rubber
band when a force is applied. Raising and
lowering the clamp stand boss alters the
applied force, which is measured using
the newton-meter. If you are conducting
this experiment, wear safety glasses.
469884_12_AQA_A-Level_Physics_200-220.indd 210
Initially, there is a small amount of extension as the force is applied.
Then, as more force is applied, the rubber band stretches easily. Finally,
just before it breaks (not shown in Figure 12.20) it becomes harder to
stretch again. If you have ever blown up a balloon, you will be familiar
with this changing behaviour. Initially, balloons are hard to inflate, but
become much easier once you’ve
blown some air inside. Just before
they burst, it becomes more difficult
to inflate them further.
Figure 12.20 also shows that the
extension for a given force is different
when the rubber band is being
loaded (top curve) or unloaded
(bottom curve). This means that the
strain energy stored when the rubber
band is being loaded is greater than
the strain energy released when the
rubber band is being unloaded.
F/N
12 Properties of materials
Therefore, the elastic strain energy = 0.102 J
b) Elastic strain energy = 1 kΔl2
2
In this case the spring has been compressed,
not stretched. The equation is the same but
Figure 12.18 Each of these toys stores elastic strain
energy by deforming an elastic material.
Loading
Unloading
0
Δl/cm
Figure 12.20 Force–extension curve for
a rubber band. Notice that the loading
and unloading characteristics are
different. What happens to the elastic
strain energy stored by the rubber band?
17/04/19 10:22 AM
TEST YOURSELF
13 A
weight of 35 N is attached to the bottom of a vertical wire. The wire
stretches by 0.8 mm. Calculate the elastic strain energy stored in the
wire.
14 A 2 m rope is suspended from a tree branch. A child of mass 30 kg
hangs on to the end of the rope, which stretches to 2.05 m. Calculate
the elastic strain energy stored in the rope. Take g = 9.81 N kg−1.
15 Guitar strings are stretched using a tension key to tune them to the
correct frequency. A 750 mm long nylon guitar string is stretched so that
it experiences a tension force of 27.5 N. At this tension the guitar string is
785 mm long. Calculate the elastic strain energy stored in the string.
16 The exercise equipment shown in Figure 12.15 is designed to provide
a total resistive force of 1000 N when pulled to an extension of
40 cm. The effective spring constant of the five springs in parallel is
2500 N m−1. Calculate the minimum energy that the person who is
exercising needs to use to stretch the springs by this extension and
explain why this is a minimum value.
17 A toy uses a compression spring to launch a small ball at a target. The
ball moves horizontally. The spring constant of the spring is 1250 N m−1.
The spring is compressed by 3 cm before the ball is launched. Calculate
the elastic strain energy stored by the spring before launch.
Stress and strain: the Young modulus
However, the law of energy conservation states that energy cannot be
created or destroyed in a closed system. The difference in strain energy
must be accounted for. In the case of the rubber band, it will become warm
as it is stretched and relaxed. This is why there is a difference in energy
between loading and unloading.
●● Stress and strain: the Young modulus
Look back at Figure 12.10 showing springs joined in parallel and in series.
When two identical springs were joined together in series, we saw
that the springs extended twice as much as a single spring on its own.
The extension depends on the original length. Similarly, if we join two
springs in parallel, then the springs extend half as much as a single
spring on its own. The extension depends on the area over which the
force is acting.
In each case, the property of the object depends on the dimensions of the
object itself.
If we want to compare materials fairly, rather than compare two different
objects, then it is better to have a measurement that doesn’t depend on the
shape or size of the object. This comparison is made by using stress and
strain instead of force and extension.
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17/04/19 10:22 AM
Tensile stress =
Tensile stress is a measurement of the force applied over the cross-sectional
area of sample of material.
force
tensile stress =
cross-sectional area
F
tensile stress =
A
where tensile stress is measured in Pa, or N m−2
force
cross-sectional area
F is applied force in N
A is area in m2
F
A
Tensile strain is the ratio of the extension and original length of the sample.
Stress is sometimes given the symbol σ, so σ =
Tensile strain =
extension
original length
extension
original length
tensile strain = Δl
l
where tensile strain is dimensionless because it is a ratio
tensile strain =
Δl is the extension in m
l is the original length in m.
Strain may also be expressed as a percentage. It is sometimes given the
symbol ε.
212
We can now use these quantities to calculate a measure of the stiffness of an
elastic material that is independent of the shape of the sample of material.
This is called the Young modulus, E, of the material.
stress
strain
E = σε
Young modulus =
E=
(AF(
(Δl l(
E = Fl
AΔl
Young modulus is measured in Pa (or N m−2).
UTS
stress σ/Nm2
12 Properties of materials
Young modulus, E, is a measure of the
stiffness of an elastic material. It does not
depend on the dimensions of the sample
being tested. It is measured in Pa or N m−2.
EY
P
2.0 102
1.0 Figure 12.21 shows a stress–strain graph for copper. It looks like the force–
extension graph shown earlier but this graph will be valid for any sample of
copper.
O–P on the graph represents the range of tensile stress for which the copper
obeys Hooke’s law. The gradient of this section is the Young modulus for
copper. This value will be the same no matter what the size or shape of the
sample of copper being used. Point P represents the limit of proportionality
for the material.
102
0
0
5
10
15
20
strain/%
Figure 12.21 A simplified stress–strain
graph for copper.
Point E on the graph represents the elastic limit. Up to point E, if the stress
is removed, the sample of copper will return to its original length. Beyond
this point, copper behaves plastically. It does not return to its original length.
The yield point of the material is given by Y. This is the value of stress beyond
which the strain increases rapidly for small increases in stress.
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Make sure that you convert all the
measurements to standard units
before attempting the question.
Some metals will also show a phenomenon known as creep. This is when
the sample continues to extend over a period of time, even though the
stress applied is not increased.
EXAMPLE
Rock climbers often rely on their ropes. Some ropes
are chosen so that they will stretch if the climber
loses their footing on the rock and falls a short
distance. This reduces the danger from the fall.
To test the suitability of ropes for climbing a weight of
800 N is suspended from the rope and the extension
measured.
This test is carried out on 50 m of a rope that has a
diameter of 10 mm. If the rope extends 2 m during the
test, calculate the Young modulus of this rope.
Answer
First we calculate tensile stress.
force
tensile stress = area
The area of the rope is the area of a circle = πr2
The radius of the rope is half the diameter (in m)
= 5 × 10−3 m
so
800 N
7.85 × 10−5 m2
= 10.19 × 106 N m−2
tensile stress =
Stress and strain: the Young modulus
TIP
The ultimate tensile stress (UTS) of copper is sometimes called the
maximum strength or strength of the wire. Just before this point the copper
becomes narrower at the weakest point, known as necking. This can be
seen in Figure 12.14 for a ductile sample. The UTS is the stress at which the
material breaks.
Next we calculate tensile strain, making sure that both
measurements are in the same units.
extension
original length
= 2
50
= 0.04
tensile strain =
Young modulus =
=
tensile stress
tensile strain
10.19 × 106 N m−2
0.04
Young modulus of the rope = 2.5 × 108 N m−2
area = π (5 × 10−3)2 m2
= 7.85 × 10−5 m2
TIP
Sometimes the tensile stress and
Young modulus will be given with
the SI units of Pascal, Pa; 1 Pa is
equal to 1 N m−2.
REQUIRED PRACTICAL 4
Determination of the Young modulus by a simple method
Note: This is just one example of how you might tackle this required
practical.
A student carried out an experiment to measure the extension of a copper
wire as weights were applied. The original length of the wire was 2.5 m.
The student used a micrometer screw gauge to take three measurements
of the diameter of the wire at different places along the wire. The mean
diameter of the wire was 0.52 mm.
469884_12_AQA_A-Level_Physics_200-220.indd 213
213
17/04/19 10:22 AM
The data obtained are shown in the table.
reference wire
mm scale
sample wire
vernier scale
Mass (kg)
Extension (mm)
0.5
0.5
1.0
1.0
1.5
1.5
2.0
2.0
2.5
2.5
3.0
3.0
3.5
4.0
4.0
6.0
Stress (N m−2)
Strain
1 Suggest why the student took three measurements of the diameter of
the wire.
2 Calculate the stress and strain values for the results.
3 Plot a graph of stress against strain and calculate the Young modulus
of copper from your graph.
4 Explain whether the wire returned to its original length when the
weights are removed.
214
A simple experimental set-up for copper can be used to measure its Young
modulus. However, we can use a more accurate experimental set-up as
shown in Figure 12.22. This is particularly useful for materials such as steel,
which generally give smaller values of strain.
reference weight
Figure 12.22 Experimental set-up
to measure the Young modulus of a
metal wire. If you are conducting this
experiment, wear safety glasses. (Not
drawn to scale.)
1000
stress/MPa
12 Properties of materials
Extension
Measure the Young modulus of a number of different diameter copper
wires and compare the values obtained.
In Figure 12.22, the left-hand wire is a reference wire that holds the main
scale of the vernier calipers, usually calibrated in millimetres. The reference
wire has a mass hung from it to keep it taut. The sample wire is hung close to
the reference wire and holds the smaller scale of the vernier calipers. The two
wires are made from the same material.
As weights are added to the sample wire it extends, and the small scale moves
relative to the main scale. This allows the extension of the wire to be measured.
Interpreting stress–strain graphs
ceramic
Stress–strain graphs allow us to describe the properties of
materials, and also to predict the stresses at which changes in
those properties might occur.
high-carbon steel
glass
500
copper
0 0.01
0.1
strain
0.2
Figure 12.23 Stress–strain graphs for four different
materials.
469884_12_AQA_A-Level_Physics_200-220.indd 214
Figure 12.23 compares the stress–strain graphs for four different
materials: ceramic, steel, glass and copper.
Ceramics are extremely strong and have very high UTS values.
However, they are show very little (if any) plastic behaviour
before they fracture, so they are also very brittle. Glasses have
lower UTS values than ceramics and so are less strong, but
they are also brittle, generally showing no plastic behaviour
before they break.
17/04/19 10:22 AM
The high-carbon steel shown in the graph is a strong but brittle material.
It shows elastic behaviour at higher values of stress but fractures with very
little plastic behaviour. This type of steel is often used in cutting tools and
drill bits because it has a higher UTS value. Other types of steel may show
plastic behaviour but have a lower UTS value.
Copper has a long plastic region because it is a ductile material, and this
makes it ideal for forming into wires for use in electrical circuits.
Stress and strain: the Young modulus
Steel is made by adding different elements to iron to form an alloy.
Common elements used in steel-making are carbon, manganese and
chromium. Types of steel differ in the percentage composition of
the various elements added to iron to create them. This affects the
properties of the steels and they are generally much stiffer than ductile
metals such as copper. This can be seen on the graph from the shallower
gradient of copper giving a lower value of Young modulus.
Strain energy density
The strain energy density is the strain energy per unit volume of a sample.
Earlier we showed that the spring constant, k, depends on the dimensions
of the material sample being tested, but the Young modulus depends only
on the material used. In the same way we can calculate the strain energy
density, which is a measure of the energy stored in a material that does not
depend on the dimensions of the sample being tested.
Strain energy density is the strain energy per unit volume of the material.
From earlier:
1
strain energy = FΔl.
2
If l is the original length of the wire, and A its cross-section, then the
volume of the wire = Al.
Therefore:
Δl
strain energy per unit volume = 1 F
2 Al
Δl
1
= F×
l
2 A
but AF = stress and
Δl
l
= strain.
Therefore:
strain energy per unit volume, or strain energy density = 1 (stress × strain)
2
1
(stress
×
strain)
is
the
area
under
a
linear
stress–strain
graph.
Therefore, the
2
area under any stress–strain graph is equal to the energy per unit volume.
469884_12_AQA_A-Level_Physics_200-220.indd 215
215
17/04/19 10:22 AM
EXAMPLE
Why do all animals jump the same height?
Answer
Obviously all animals don’t jump exactly the same
height. However, they nearly all jump somewhere
between 0 and 1 m.
If all animals are the same shape and they are made
of the same stuff then we can say that:
1
strain energy per unit volume = (stress × strain)
2
Therefore:
1
energy stored in muscles = (stress × strain ×
2
volume of animal muscle)
gravitational potential energy (Ep) gained in jumping
= mgh
We can estimate the mass of the animal (muscle)
using:
mass = density × volume
so
Ep = density × volume × g × h
Assuming that all the energy stored in the animal is
used to increase the height of the animal, we can say:
energy stored in muscles = increase in gpe
1
(stress × strain × volume of animal muscle)
2
= density × volume × g × h
therefore
1
height jumped = (stress × strain)/density × g
2
So, if all animals are the same shape and made out
of the same stuff, i.e. they all have the same density
and muscle properties, then the height jumped is
independent of the size of the animal and they will all
jump the same height.
Although real animals do jump different heights,
the range of heights is much smaller than the
corresponding range of masses. For example, from
a standing start a flea can jump about 20 cm and a
human 60 cm. However, a typical flea has a mass of
approximately 0.5 mg and a typical human has a mass
of approximately 70 kg.
12 Properties of materials
TEST YOURSELF
216
18 A
brass wire of length 1.5 m and diameter 0.4 mm
is extended by 3 mm when a tensile force of 32 N
is applied. Calculate:
a) the applied stress
b) the strain on the wire
c) the Young modulus of the brass.
19 A towbar on a car is used to pull a trailer. The
trailer exerts a tensile force of 23 kN on the towbar.
The towbar can be modelled as a cylindrical bar of
steel with a diameter of 6 cm and a length of 25 cm.
The high-strength steel has a Young modulus
of 200 GPa. Calculate how much the towbar will
extend when it is in use.
20 T he high-strength steel used in the towbar in
question 19 is a brittle material with an ultimate
tensile strength of 500 MPa. Estimate the
maximum force the towbar can sustain before it
fractures and breaks.
21 A metal component has a length of 0.5 m. The
strain must not exceed 0.1% (1 × 10 −3). Calculate
the maximum extension allowed.
469884_12_AQA_A-Level_Physics_200-220.indd 216
22 A
steel strut has a cross-sectional area of
0.025 m2 and is 2.0 m long. Calculate the
compressive force that will cause the strut
to shorten by 0.3 mm. Assume that the Young
modulus for this steel is 200 GPa.
23 E stimate the stress on the lower leg of a person
who is standing still. Suggest why your value is
likely to be much less than the ultimate compressive
stress of bone.
24 A n architect makes a scale model of a bridge
she is going to build. It is made 20 times smaller
than the real bridge in every dimension. The
model is made from exactly the same material
as the real bridge. The bridge is supported
by four steel pillars. Calculate the ratio of the
stresses in the support pillars, in the real bridge
to model bridge.
25 L ions are approximately ten times larger than
cats. Use your understanding of stress to
explain why the legs of lions and cats are not the
same shape.
17/04/19 10:22 AM
stress/MPa
Figure 12.24 shows the stress–strain graph for two materials up to
their breaking points. You need this graph to answer questions 1 to 3.
1 Which word best describes the behaviour shown for material M?
A brittle
C fragile
B ductile
D plastic
2 Which word best describes the behaviour shown for material N?
A brittle
C plastic
B fragile
D strong
M
N
Practice questions
Practice questions
strain
Figure 12.24 A stress–strain
graph for two materials.
3 Which combination of materials could be best represented by the graph?
A
B
C
D
M
glass
high-carbon steel
cast iron
cast iron
N
cast iron
glass
high-carbon steel
rubber
4 A superbounce bouncy ball has a diameter of 2.5 cm and a mass of 35 g.
What is the density of the ball in kg m−3?
A 4.3 × 10−3
C 4.3 × 103
B 4.3
D 4.3 × 106
5 A 10 cm spring is stretched until it is 16 cm long. The force needed to
stretch the spring by this amount is 30 N. What is the spring constant
of the spring in N m−1?
A 1.9
C 480
B 50
D 500
You need the following information for questions 6 and 7.
A 2 m long square metal bar has sides of 40 mm width. A force of 80 kN
is applied to the bar and it extends by 0.046 mm.
6 What is the stress in the metal bar?
A 5.0 N m−2
C 5.0 × 107 N m−2
B 5000 N m−2
D 5.0 × 109 N m−2
7 What is the strain in the metal bar?
A 2.3 ×
10−5
B 0.23
217
C 23
D 2.3 × 105
You need the following information for questions 8 and 9.
A vertical steel wire of length 0.8 m and radius 1.0 mm has a mass of 0.2 kg
attached to its lower end. Assume that the Young modulus of steel is
2.0 × 1011 N m−2 and g = 10 N kg−1.
469884_12_AQA_A-Level_Physics_200-220.indd 217
17/04/19 10:22 AM
8 What is the extension of the wire?
A 0.128 × 10−6 m
C 0.255 × 10−3 m
B 2.55 × 10−6 m
D 2.55 m
9 What is the energy stored in the wire when stretched?
A 2.6 × 10−6 J
C 2.6 J
B 2.6 × 10−4 J
D 26 J
You need the following information for questions 10 and 11.
0.80
10 What is the Young modulus of the metal?
0.60
A 16 GPa
C 74 GPa
B 53 GPa
D 120 GPa
11 What does the area under the graph represent?
A the total stress on the wire
stress/GPa
Figure 12.25 shows the stress–strain graph for a metal
wire that is stretched until it breaks.
0.40
0.20
B the breaking strain
0
C the energy stored in the wire per unit volume
D the elasticity of the wire
0
1.0
12 Properties of materials
3.0
strain/%
4.0
5.0
Figure 12.25 The stress–strain graph for a metal wire
that is stretched until it breaks.
12 A sample of wood is tested in a tensile testing
machine. The wood sample breaks when the applied force is
840 N. The cross-sectional area of the sample is 1.3 × 10−5 m2.
Calculate the ultimate tensile stress for the sample.
218
2.0
(1)
13 A stone paving slab has a mass of 13 kg. Its dimensions are
600 mm × 300 mm × 35 mm.
a) Calculate the density of the stone.
(1)
b) Calculate the maximum compressive stress that the slab could
exert on the ground when resting on one of its edges.
(2)
14 A cord made from natural rubber is initially 20 cm long. A load of
30 N is attached to it. The new length is measured as 1.0 m.
a) Calculate the strain in the rubber cord.
(1)
b) The stress on the cord is 14 MPa. Calculate the cross-sectional
area of the cord.
(1)
15 An experiment is carried out to determine the spring constant of a
metal in the form of a wire. Weights are added to the wire in steps of
5.0 N up to 25.0 N.
Load (N)
0.0
5.0
10.0
15.0
20.0
25.0
Loading
Extension (mm)
0.00
0.12
0.23
0.35
0.48
0.61
469884_12_AQA_A-Level_Physics_200-220.indd 218
Unloading
Extension (mm)
0.00
0.12
0.23
0.34
0.49
0.61
17/04/19 10:22 AM
Practice questions
a) Use the results in the table (without plotting a graph) to state and
explain if the deformation of the wire is plastic or elastic. (1)
b) Describe how the length and extension of the wire could be
measured experimentally and explain what safety precautions
should be taken when carrying out the measurements.
(4)
c) Plot a graph using the results for loading the wire and use
your graph to calculate the spring constant of the wire.
(3)
8
d) What additional measurements would need to be made to
allow calculation of the Young modulus of the wire?
(1)
(1)
a) Calculate the mass of the 30 m cable.
b) The lift is stationary on the ground floor of the building. Show
that the tensile stress in the cable due to the lift and the mass
of the cable is approximately 30 MPa.
(3)
17 A student measures the extension of an elastic band when a force
is applied. She then continues to measure the extension as the
force is progressively removed from the elastic band. A graph of
her results is shown in Figure 12.26.
a) Describe a simple experiment that would allow measurement
of the force applied and the extension of the elastic band. (4)
b) The student did not take repeat measurements. Suggest
what effect this might have on her results.
(1)
c) Does the elastic band show Hooke’s law behaviour?
Explain your answer.
(2)
loading
unloading
6
5
force/N
16 A lift has a mass of 500 kg. It is designed to carry five passengers
with a maximum combined mass of 450 kg. The lift is moved by
means of a steel cable of diameter 20 mm. When the lift is on the
ground floor of the building the cable is at its maximum length of
30 m. The density of the cable is 9550 kg m−3.
Key:
7
4
3
2
1
0
0
5
10
15
20
25
extension/cm
Figure 12.26 Results from loading and
unloading an elastic band.
d) Explain how the student could estimate the amount of energy
dissipated by the elastic band during the experiment.
(2)
200
e) What physical change would occur in the
elastic band if it was repeatedly stretched
and released in a short space of time? (1)
a) Explain why no units are given for strain
on the x-axis.
(1)
b) Describe the behaviour of copper up to a
(1)
strain of 1.0 × 10−3.
c) State the breaking stress of this copper
wire.
(1)
d) Calculate the Young modulus for copper.(2)
e) A similar copper wire is loaded up to
a strain of 3.5 × 10−3. The load is then
469884_12_AQA_A-Level_Physics_200-220.indd 219
160
140
stress/106Pa
18 Figure 12.27 shows a stress–strain graph for
copper wire.
180
120
100
219
80
60
40
20
0
0
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
strain/103
Figure 12.27 A stress–strain graph for copper wire.
17/04/19 10:22 AM
removed from the wire. Sketch on the graph the stress–strain
measured as the load is removed and explain the shape of the
graph you have drawn.
(2)
Stretch and challenge
19 Figure 12.28 shows an experiment in which two wires, each 2.0 m long,
are hung vertically from a fixed bar. The ends of the wires are attached
to a lightweight horizontal bar 0.1 m apart.
0.1 m
wire 1
Wire 1 is made from steel and has a diameter of 0.8 mm. Wire 2 is made
from brass and has a diameter of 0.68 mm. A force of 100 N is applied
vertically downwards to the centre of the platform. The wires remain
vertical but the brass wire extends more than the steel wire and the bar
tilts at 1° to the horizontal.
wire 2
lightweight platform
Figure 12.28 An experiment
in which two wires, each
2.0 m long, are hung
vertically from a fixed bar.
a) Calculate the difference between the extensions of the two wires.
b) The Young modulus for steel is 2.0 × 1011 N m−2; show that the
extension of the steel wire is 1.0 mm.
c) Calculate the Young modulus for the brass wire.
d) Calculate the energy stored in the steel wire due to its extension.
12 Properties of materials
20 A pinball machine firing mechanism is used to launch a
small ball bearing, of mass 100 g, into the playing area.
The design of the mechanism is shown in Figure 12.29.
The playing area and the mechanism are at 8.5° to the
horizontal.
Initially the spring is relaxed. To fire the ball, the plunger is
pulled back and compresses the spring by 5.0 cm. When the
plunger is released the spring extends and the plunger applies
a force to the ball. The dashed circle shows the position of the
ball when it has a maximum speed, just as it loses contact with
the plunger.
spring
8.5°
vertical
height, h
Figure 12.29 A pinball machine firing
mechanism.
The ball leaves the plunger at a speed of 0.68 m s−1. Use ideas about
conservation of energy to calculate the spring constant of the spring.
Assume that the mechanism is frictionless.
21 The table below shows the Young modulus and breaking strength of
spider silk and steel.
Young modulus (Pa)
Spider silk
High-tensile steel
220
5.0 cm
plunger
Breaking strength (Pa)
1×
1010
1 × 109
2×
1011
2 × 109
Using data from the table, and considering the strain and elastic energy,
suggest why spiders silk is more effective at catching a fly than steel
(of the same thickness) would be.
22 A long rod of metal is suspended and stretches under its own weight. Its
length is l, its density ρ, and Young Modulus E. Show that its extension
under its own weight is ρgl2/2E.
469884_12_AQA_A-Level_Physics_200-220.indd 220
17/04/19 10:22 AM
13
Current electricity
PRIOR KNOWLEDGE
●
●
●
●
●
●
●
●
Electric current is a flow of electric charge.
The size of the electric current, I, is the rate of flow of electric charge,
which is given by the equation:
Q
I=
t
where Q is the charge in coulombs, C, and t is the time in seconds.
Electric current is measured in amperes, A, by an ammeter
connected in series with components.
The potential difference or voltage, V, between two points in an
electric circuit is the work done (energy transferred), W, per
coulomb of charge that passes between the points and is given by
the equation:
W
V=
Q
Potential difference is measured in volts, V, by a voltmeter connected
in parallel with components.
The resistance of a component, R, can be found by measuring the
current, I, through and potential difference, V, across the component;
resistance is defined by the equation:
V
R=
I
Resistance is measured in ohms, Ω.
The current flowing though, the potential difference across and
the resistance of a component are related by the equation:
V = I × R.
For example, if the current in Figure 13.1 is 4.5 A and the
resistance is 4.0 Ω, the potential difference across the resistor
is 18 V.
V = 18 V
A
I = 4.5 A
R = 4.0 Ω
V
221
Figure 13.1 Simple electric circuit.
●
469884_13_AQA_A-Level_Physics_221-243.indd 221
The potential difference provided by cells connected in series is
the sum of the potential difference of each cell (depending on the
direction in which they are connected).
17/04/19 10:22 AM
TEST YOURSELF ON PRIOR KNOWLEDGE
1 Draw a circuit showing how you would measure the electric current
flowing through a bulb powered from a 12 V battery.
2 Write down the names and symbols of the units of:
a) current
c) resistance
b) potential difference
d) charge
3 0.6 C of charge flows through a bulb during 40 s. Calculate the current
flowing through the bulb.
4 The battery provides a potential difference of 12 V across the bulb.
Calculate the electrical energy transferred into heat and light by the
bulb when 0.25 C of charge flows through the bulb.
●● Moving charge and electric current
Electric current is the rate of flow of
electric charge. The unit of current is the
ampere.
A lightning bolt is an extreme example of an electric current. During the
largest cloud-to-ground lightning bolts 2 × 1021 electrons might jump from the
bottom of a cloud to Earth in 0.003 s, delivering 5000 MJ of electrical energy.
An electric current is formed when the electrons move. Electric current, I,
is the rate of flow of charge:
amount of charge flowing, ΔQ (C)
time to flow, Δ t (s)
ΔQ
I=
Δt
The SI unit of current is the ampere (symbol, A), which is nearly always
abbreviated to amps. 1 amp is equal to a charge of 1 coulomb flowing in
1 second (so 1 A = 1 C s−1).
Note that here we have used the notation ΔQ and Δt, rather than Q and t. In
GCSE work, you performed calculations in which the current was a constant
value. At A-level you will meet questions in which the current is changing.
Under those circumstances we can calculate a current by considering a small
flow of charge, ΔQ, in a small time, Δt. So the equation above is a more
general expression of the equation you met during your GCSE course.
13 Current electricity
electric current, I (A) =
222
Figure 13.2 A cloud-to-ground lightning
strike.
EXAMPLE
The electrons moved by friction in a cloud accumulate
at the bottom of the cloud producing a large negative
charge. Each electron has a charge, e, where e = −1.6 ×
10−19 coulombs (symbol, C). The total magnitude of the
charge (symbol, Q) of all the electrons is about 320 C
(Q = ne = 2 × 1021 × 1.6 × 10−19 C = 320 C). The bottom
of the cloud is ‘near’ to the ground, which is ‘earthed’,
and the electrons flow from the cloud to Earth through
the air in 0.003 s, which provides a conducting pathway.
469884_13_AQA_A-Level_Physics_221-243.indd 222
How much current is produced?
Answer
Using the current relationship for a large lightning
bolt:
320 C
I=
= 1 × 105 A (1 s.f.)
0.003 s
17/04/19 10:22 AM
EXAMPLE
Answer
Some electric currents involve beams of electrons.
Any device containing a cathode ray tube (CRT),
such as an oscilloscope, involves electron beam
currents.
∆Q
= n × e = ne
∆t
1
(The charge, ∆Q, is the charge carried by each
electron, e, multiplied by the number of electrons).
I=
A current of 30 μA flows in a CRT. Calculate the
number of electrons flowing in the beam per
second, n.
Moving charge and electric current
Electronic circuits, such as those that control household appliances,
operate with much smaller currents, typically milliamps (mA, 10−3 A), and
many microelectronic circuits, such as the printed circuit boards inside
many computer devices, operate with currents of the order of microamps
(μA, 10−6 A). Even currents of the order of microamps still involve the
movement of about 6 × 1012 electrons per second.
I
30 × 10−6A
⇒n= e =
= 1.9 × 1014 electrons
1.6 × 10−19C
per second
Y plates
electron gun
cathode
fluorescent screen
electron beam
anode
X plates
cathode ray tube
Figure 13.3 Electron beam in a cathode ray tube.
An electrolyte is a conducting solution,
usually containing positive and negative salt
ions dissolved in water.
Not all electric currents involve the flow of electrons. Charged ions in
solution (called an electrolyte) can also flow and create a current. Car
batteries work due to the flow of hydrogen (H+) ions and sulfate (SO42−)
ions and can deliver currents of up to 450 A for 2.5 seconds.
load
resistance
e–
–
+
–
+
–
–
–
+
+
+
Positive ions (cations) move
toward the negative electrode.
Negative ions (anions) move
towards the positive electrode.
Electrons flow in the external
circuit from the positive electrode
to the negative electrode.
223
Figure 13.4 Electric current can also be due to the flow of ions.
469884_13_AQA_A-Level_Physics_221-243.indd 223
17/04/19 10:22 AM
Both of the ions in the solution contribute to the current, which can be
measured externally by an ammeter connected in series with both electrodes.
TIP
In the examination you are
assessed on your understanding
of the significant figures of
the numbers that you use in
calculations. In the example on
the right, both the pieces of data
used in the calculation are to
2 significant figures (s.f.). As a
general rule, you should state your
answer to the same number of
sf as the least significant piece of
data used – 2 s.f. in this case.
EXAMPLE
If a car battery delivers 450 A for 2.5 seconds, calculate the total charge
flowing, ΔQ.
Answer
ΔQ = I × Δt
= 450 A × 2.5 s
= 1125 C
= 1100 C or 1.1 × 103 C (2 s.f.)
Variations of current with time
Figure 13.5 shows the variation of the current drawn from a cell with time. The
current is drawn for a total of 10 seconds, but 5 s after turning on switch A,
switch B is closed allowing current to travel through the second identical resistor:
0.9
0.8
A
During the second 5 s the
current is constant at 0.8 A.
The total charge, Q5–10,
transferred is equal to the
area under the graph
= 0.8 × 5
= 4.0 C
0.7
I
R
0.6
13 Current electricity
R
current, I/A
B
0.5
0.4
During the first 5 s the
current is constant at 0.4 A.
The total charge, Q0–5,
transferred is equal to the
area under the graph
= 0.4 × 5
= 2.0 C
0.3
0.2
0.1
TIP
The area under a current-time
graph is the charge transferred.
224
0
0
1
2
3
4
5
6
time, t/s
7
8
9
10
11
Figure 13.5 Current–time graph.
As
∆Q = I × ∆t
the charge transferred during the first 5 s is the area under the current–time
graph from t = 0 s to t = 5 s. During the second 5 s, the current doubles. The
charge transferred during this time is the area under the graph from t = 5 s
to t = 10 s. The total charge ∆Q transferred during the whole 10 s period is
the total area under the graph
∆Q = Q0–5 + Q5–10 = 2.0 C + 4.0 C = 6.0 C
469884_13_AQA_A-Level_Physics_221-243.indd 224
17/04/19 10:22 AM
0.7
current, I/A
Figure 13.6 shows the variation of current with time through
part of the timing circuit for a set of traffic lights.
∆Q = 1 × 10 s × 0.6 A = 3 C
0.6
2
0.5
In this case, the current varies continuously with time. The total
charge transferred is still the area under the current–time graph,
which is calculated to be 3 C in this case.
0.4
0.3
0.2
0.1
0
0
2
4
6
time, t/s
8
10
12
In a similar way, the gradient of a charge–time graph (Q – t) is the
current. Figure 13.7 shows the current sparking off the top of a
school van der Graaff generator.
The current, I, during this discharge is determined by calculating
Figure 13.6 Current–time graph for a traffic
light circuit.
the gradient of the Q – t graph as I =
∆Q
∆t .
Moving charge and electric current
The total charge transferred is the area under a current–time
graph.
Total charge transferred, ∆Q
= area under graph
In this case,
0.5 × 10 −6C
I=
= 5 × 10−6A = 5μA.
0.1 s
0.6
charge, Q/µC
0.5
I = ∆Q = gradient of graph
∆t
–6
I = 0.5 × 10 C = 5 × 10–6 A = 5 µA
0.4
0.1 s
0.3
0.2
0.1
0
0
0.02
0.04
0.06
time, t/s
0.08
0.10
0.12
Figure 13.7 The current sparking off the top of a school van der Graaff generator.
TEST YOURSELF
1 Here is a list of electrical units:
A s A V−1
C s−1
J s−1
J C−1
Choose the unit for:
a) electric charge
b) potential difference
c) electric current.
2 An iPhone can be charged three ways:
● via a PC USB port, which delivers 0.5 A
● via the iPhone charger, which delivers 1.0 A
● via an iPad charger, which delivers 2.1 A.
469884_13_AQA_A-Level_Physics_221-243.indd 225
The iPhone battery capacity is 1420 mAh,
which equates to a total electric charge of
5112 C.
a) Calculate the time needed to charge an empty
iPhone battery from each of these charging
methods.
b) Explain why 1420 mAh equals 5112 C. (A charge
of 1 mAh is transferred when a current of 1 mA
flows for 1 hour.)
225
17/04/19 10:22 AM
3 Sprites are rare, large scale, very high altitude
lightning bolts that often take the shape of a
jellyfish.
Figure 13.8 High-altitude sprite lightning.
13 Current electricity
A sprite delivers a total charge of 285 C in 3.5 ms.
Calculate the current flowing in this type of sprite.
4 A blue light emitting diode (LED) has a current
of 25 mA flowing through it. Calculate the total
number of electrons passing through the blue
LED per second. (The charge on an electron is
1.6 × 10 −19 C.)
5 A scanning tunnelling microscope works by bringing
a very fine metal needle tip up very close to a
surface; electrons then flow between the tip and the
surface. Tiny piezoelectric crystals in the tip move
the tip up and down in response to the size of the
current flowing between the tip and the sample.
tunnelling
electrons
produce
tunnelling
current
sample
226
tip
Figure 13.9 A scanning tunnelling microscope.
15 million electrons per second flow across the gap
between the tip and the surface. Calculate the current
in the circuit.
6 An ion beam used to construct integrated circuits
delivers a charge of 80 nC in a time of 46 s.
Calculate the current of the beam and the number
469884_13_AQA_A-Level_Physics_221-243.indd 226
of ions passing per second. Assume that each
ion has a charge equivalent to the charge on the
electron.
7 This question requires you to apply your knowledge
of currents in a more complicated situation. A car
battery has a low voltage, typically of the order of a
few volts, but it can produce a very large current.
The large current results from rapid chemical
reactions involving large numbers of ions. You need
to use the following data:
● magnitude of e = 1.6 × 10 −19 C
● Avogadro number, the number of particles in one
mole, NA = 6.0 × 1023 particles/mole
● molar mass of lead, MPb = 207.2 g.
● In a lead chemical cell, the energy transferred
by the chemical reactions between the lead
electrodes and the electrolyte is about 24 kJ per
mole of lead.
a) Calculate the charge carried by a mole of
electrons (known as a Faraday of charge).
b) Show that the electrical potential energy
transferred to each coulomb of charge (i.e. the
potential difference, V) is 0.125 V. (Each lead
atom loses two electrons to become a Pb2+ ion.)
c) How many lead chemical cells would be
required to produce the same potential
difference as a 12 V car battery?
d) When driving at night the battery delivers a
current of 15 A to power the lights; how many
electrons flow from the battery per second?
e) Calculate the number of moles of electrons
flowing per second.
f) Chemical cells can produce large charge flows
for substantial times (i.e large currents) before
their chemicals have all reacted and the cell
‘runs out’. In a lead chemical cell, what mass of
lead is reacting per second to produce a current
of 15 A? (Remember that each lead atom loses
two electrons to become a Pb2+ ion.)
g) If the cell contains 640 g of reactable lead,
for how long could it generate this current
before running out? (Note that we protect a
car’s battery by running the lights off the car’s
alternator when the engine is running; the
alternator generates electricity using energy
from the engine.)
8 A mobile phone charger normally operates at
0.70 A. During a particular charging cycle, the
charger operates normally for 1.5 hours before it
suddenly develops a fault and outputs a current of
0.12 A for a further two hours.
a) Plot an I–t graph for this charging cycle.
b) Calculate the total charge transferred during
the whole cycle.
17/04/19 10:23 AM
Electric current only measures the rate at which charged particles (usually
electrons) flow around a circuit. It tells us nothing about the electrical energy
involved with circuits. The quantity used to describe the electrical energy
in circuits is potential difference (p.d.), symbol V, measured in volts, V.
A potential difference across an electrical component is measured by putting
a voltmeter across the component, in parallel with it.
Potential difference is the amount of
electrical work done per unit charge flowing
through a component.
Electromotive force is the amount of
electrical work done per unit charge by a
power supply such as a cell in an electrical
circuit. The power supply transfers the other
forms of energy, such as chemical energy
into electrical energy. The unit is the volt, V.
Electrical energy in circuits is defined in terms of the electrical work done
by the electric charge flowing through the circuit. Potential difference
(p.d.) is defined as the electrical work done per unit (coulomb) of charge
flowing through components such as bulbs, motors, resistors, etc. This
electrical energy is transferred into heat, light and other more useful forms
of energy by the components.
electrical work done by the charge, W (J)
potential difference, V (V) =
charge flow, Q (C)
W
V=
Q
1 V = 1 J C−1
Potential difference, however, cannot be used to describe the energy changes
involved with power supplies such as cells, generators and mains power supply
units. These devices transfer other forms of energy, such as chemical energy
into electrical energy. To make a distinction between these different energy
transfers we define another quantity, electromotive force (emf), symbol ε.
Both emf and p.d. are measured in volts, symbol V, using a voltmeter.
Potential difference and electromotive force
●● Potential difference and electromotive force
Electromotive force (emf) is defined as the electrical work done per unit
(coulomb) of charge as it flows through a source of electrical energy such
as a cell, generator or power supply unit (psu). The sources of electrical
energy transfer other forms of energy such as kinetic or light energy into
electrical energy.
electrical work done on the charge, E (J)
electromotive force, ε (V) =
charge flow, Q (C)
E
ε=
Q
The law of conservation of energy can now be written in terms of emf
and p.d. In a series circuit, where the components are connected one
after another in a complete loop, the total electrical energy per coulomb
transferring into the circuit (the sum of the emfs in the circuit) must equal
the energy per coulomb transferring into other forms of energy (the sum of
the p.ds). (There is more about this in Chapter 14 on electrical circuits.)
Here is an example. The following circuit is set up:
227
e = 6V
0.5 A
8Ω
4Ω
V1
V2
Figure 13.10 Circuit diagram.
469884_13_AQA_A-Level_Physics_221-243.indd 227
17/04/19 10:23 AM
The emf of the cell, ε, transfers 6 J C−1 (V) of chemical energy into electrical
energy (if the cell is 100% efficient).
The 6 J C−1 of electrical energy is shared between the two resistors. This energy
is shared in the same ratio as the resistance of the resistors (8 : 4 or 2 : 1).
The potential difference, V1, across the 8 Ω resistor is therefore 4 J C−1 (V) and
the potential difference, V2, across the 4 Ω resistor is therefore 2 J C−1 (V). Note
that the law of conservation of energy still holds here as 6 J C−1 (V) is transferred
from chemical energy into electrical energy in the cell and 6 J C−1 (V) in total is
transferred from electrical energy into heat energy within the resistors.
ACTIVITY
Investigating the ability of a travel mug heating element to boil water
13 Current electricity
In an experiment to investigate the ability of a travel
mug heating element to boil water in a mug for a cup
of tea, a student connected the heating element to a
car battery.
228
Figure 13.11 Travel heating element.
The student put 180 g of water into a mug and
measured the temperature of the water every minute
for 8 minutes. Here are the results:
Time, t (minutes)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
469884_13_AQA_A-Level_Physics_221-243.indd 228
Temperature, θ (°C)
15
26
37
48
59
70
81
92
100
One gram of water requires 4200 J of heat energy
to raise the temperature of the water by 1 °C. This
is called the specific heat capacity of water. An
ammeter measures the current in the heater to be
11.1 A during the heating.
1 Convert the times to seconds and plot a graph of
temperature, θ /°C against time, t/s.
2 Plot a best-fit line and calculate the gradient of
the line.
3 What quantity is represented by the gradient of
the line?
4 The heat energy, ∆ Q, required to heat the water is
given by the equation:
∆ Q = mc∆θ
where m is the mass of the water, c is the specific
heat capacity of the water and ∆ θ is the change in
temperature. How much energy heats 180 g by
one degree? (Note: The heat energy ∆ Q should not
be confused with electrical charge.)
5 How much heat has been supplied therefore after
two minutes?
6 How much charge is transferred in 2 minutes?
7 Assuming that the electrical energy supplied to
the heater is transferred into heat energy of the
water, use your answers to questions 5 and 6 to
calculate V, the potential difference across the
heater.
8 State two hazards involved with this experiment
and describe the risks and control measures that
you would use.
9 This experiment involves three measurements,
the volume and temperature of water, and the
time. What are the sources of uncertainty in these
measurements? For each one, state a way of
reducing this uncertainty?
17/04/19 10:23 AM
Positive ions from metal
–
–
–
+
–
–
–
–
+
–
–
–
–
–
–
–
–
+
–
+
–
–
+
–
–
+
–
+
–
+
–
–
+
–
–
‘Gas’ of electrons,
free to move
throughout the
structure
–
Figure 13.12 The structure of a metal.
The resistance of a conductor is the
opposition of the conductor to electric current
flowing through it. Unit is the ohm, Ω.
A superconductor is a material whose
resistance drops to zero below a specific
temperature, called the critical temperature.
When current flows through the material of a circuit, such as
the metal of the connecting wires, the material of the circuit
gets in the way of the flow of the charge. On a microscopic
level, as the electrons flow through the metal they collide
with the vibrating positive ion cores of the metal structure.
The collisions between the electrons and the positive ion
cores transfer electrical energy from the electrons to the
structure of the metal, causing the metal ion cores to vibrate
more, thus heating up the wire.
Resistance
●● Resistance
The opposition of a component to the flow of electric
current through it is called resistance, symbol R, unit
ohms, Ω.
The model of a metal shown above explains what happens to the
resistance of the metal when it is heated. The resistance increases as the
temperature increases. The vibrating, positively charged, ion cores move
around much more as the temperature increases, thus getting in the way
of electron flow. This opposes the flow of the gas of electrons moving
through the structure.
Components with very high resistances let very little current through
them, and are considered to be electrical insulators. Some materials, at
very low temperatures, have zero resistance. These materials are called
superconductors.
Superconductors
Figure 13.13 Magnetically levitating
superconductor.
Figure 13.14 One of the Shanghai Maglev
trains, where superconducting magnets
support the train, enabling it to travel at
speeds up to 270 mph (430 km/h).
469884_13_AQA_A-Level_Physics_221-243.indd 229
A current set up in a loop of superconducting material carries on flowing
indefinitely. Superconductors also exclude magnetic fields inside them.
This allows a strong permanent magnet to be repelled and held above
the superconductor. Superconducting materials completely lose their
resistance below a temperature called the critical temperature, Tc. Different
superconducting materials have different critical temperatures; the
metallic element tungsten, W, for example, has the lowest known critical
temperature of 0.015 K, (−273.135 °C); and a type of ceramic copper
oxide, called mercury barium thallium copper oxide, has the highest
known critical temperature of 139 K (−134.15 °C). This is much higher
than the boiling point of liquid nitrogen (77 K).
As new materials are developed, so the superconducting critical
temperature has risen substantially. The ultimate goal is to develop
superconductors with critical temperatures that are around room
temperature. Imagine the world of possibilities with zero-resistance
superconductors. There could be, for example, electronic devices and
computer units that don’t generate any heat and don’t need cooling
fans; batteries that last for an extremely long time on one charge; cheap
magnetic levitation; portable MRI scanners; super-strong electromagnets
and electrical power transmission lines that don’t waste any energy.
229
17/04/19 10:23 AM
●● Current/potential difference
characteristics and Ohm’s law
An electrical characteristic is a graph
(usually I – V) that illustrates the electrical
behaviour of the component.
TIP
13 Current electricity
I–V graphs can be drawn with
I or V on either axis to illustrate
the relationship between a
current through a component and
the potential difference across
a component. I–V graphs (with V
on the x-axis) are useful because
the current through a component
depends on the potential
difference across it. This type
of graph is particularly useful
for people who are designing
circuits for devices where they
need to know how a component
will behave with different
applied potential differences.
V–I graphs (with I on the x-axis)
are useful for illustrating the
resistance of a component, and
the characteristics of cells and
batteries are usually plotted
this way.
TIP
The equation:
V
R=
I
defines resistance. Some sources
on the web incorrectly call the
equation Ohm's Law. It is not.
230
TIP
Ohm's Law states that for some
conductors
IαV
469884_13_AQA_A-Level_Physics_221-243.indd 230
Most of the properties of an electrical component can be determined by
plotting the current – potential difference graph of the component. This is
called an electrical characteristic.
Electrical characteristics are graphs that show how the potential difference
and current vary when the component is connected in both forward
and reverse bias (when the current passes one way and then in the other
direction). The most straightforward electrical characteristic is that of a
fixed resistor.
You can see that the graph can be drawn with each variable on either axis.
current, I
potential
difference, V
IV characteristic
VI characteristic
potential
difference, V
OR
current, I
Figure 13.15 Electrical characteristic graphs of a fixed resistor.
Ohm’s law
In 1827, the German physicist Georg Ohm performed potential difference –
current experiments on metal wires at constant temperature. Ohm
discovered that the current, I, flowing through the wire was proportional to
the potential difference, V, across the wire, provided that the temperature
(and other physical variables) remained constant. If this is shown on an
electrical characteristic graph, then the graph will be linear. The relationship
between the current and the potential difference became known as Ohm’s
law, which can be written mathematically by:
current, I α potential difference, V
IαV
Ohm’s law applies to some conductors under some circumstances. It is
generally a special case applied to metal wires at constant temperature.
But, using the concept of resistance, a convenient, familiar mathematical
equation can be written that uses the relationship between I and V to define
the unit of resistance, the ohm (Ω):
V
R=
I
In other words, if a potential difference of 1 V produces a current of 1 A
flowing through a component, then its resistance must be 1 Ω. This defines
resistance in ohms. This equation applies in all cases and also makes it
clear that for a fixed p.d., V, a small measured current, I, implies that the
component has a large resistance, R, and vice versa. Similarly, for a fixed
current, I, we need a large p.d., V, to drive the current through a large
resistance, R.
17/04/19 10:23 AM
V
I
+
R
Fixed resistors and metal wires at constant temperature obey Ohm’s law
across their current range and their electrical characteristics are linear.
Components like this are said to be ohmic conductors – i.e. they obey
Ohm’s law.
–
I
Figure 13.16 An electrical
characteristic of a fixed resistor
showing Ohm’s law.
The current flowing through an ohmic
conductor is proportional to the p.d.
applied across it.
Other electrical characteristics
A standard tungsten filament lamp transfers electrical energy into light
and heat as the current flows through it. As the current increases, so the
frequency of electron collisions with the positive ion cores of the tungsten
lattice increases, transferring more kinetic energy. The positive ion cores
vibrate with greater amplitude and so the resistance increases. A higher
current leads to a higher temperature, which in turn leads to a higher
resistance. The electrical characteristic of a filament lamp is shown in
Figure 13.17:
Current/potential difference characteristics and Ohm’s law
gradient = R
One use of the electrical characteristic graph is that if the current is plotted
on the x-axis, for component’s following Ohm’s law then the gradient of the
line is the same value as the resistance of the component. As fixed resistors
and metal wires at constant temperature have a fixed resistance, then the
electrical characteristic graph will be a straight line. On this type of graph,
high-resistance components have large, steep gradients, and low-resistance
components have small, shallow gradients. Remember though, electrical
characteristics can be drawn with either variable on either axis.
current, I
potential
difference, V
Figure 13.17 Electrical characteristic
of a filament lamp.
A non-ohmic component is a component
that does not obey Ohm’s law; i.e. current is
not proportional to the potential difference
applied across it.
anode
(+)
cathode
(–)
Figure 13.18 A semiconductor diode –
symbol and picture. The current will
only flow in the direction of the arrow
(anode to cathode).
The electrical characteristic shows the ratio V/I increasing and therefore the
resistance increasing. Components like this, whose electrical characteristics
are non-linear, are said to be non-ohmic conductors.
Components such as fixed resistors and filament lamps have the same
characteristics independent of the direction of the current flowing through
them. Their V–I graphs produce the same shapes in forward and reverse
bias. Components such as semiconductor diodes do not behave in this way.
Diodes are generally found in electronic circuits where they act as one-way
gates, preventing the current from flowing back through the circuit. They
are particularly useful in mains power supplies where they can be used in
circuits to convert alternating current (ac) into direct current (dc).
231
Diodes only conduct in forward bias (in the direction of the arrow on the
symbol); their circuit symbol shows this direction using the arrow, and the
component itself normally has a different coloured ring at the forward bias end.
Diodes do not conduct in reverse bias; this means that the resistance of the
diode in reverse bias is infinite. Diodes have very low resistance in forward
469884_13_AQA_A-Level_Physics_221-243.indd 231
17/04/19 10:23 AM
bias. The electrical characteristics of diodes are usually drawn in current
(y-axis)–potential difference (x-axis) format:
current
TIP
Make sure that you are aware of
how the electrical characteristic
is plotted. Carefully check which
quantity is on each axis. Also, be
careful to ensure that you know
what the units of each axis are.
reverse bias
forward bias
potential
difference
Figure 13.19 Electrical characteristic of a semiconductor diode.
ACTIVITY
Plotting electrical characteristics
The electrical characteristics of a small bulb and a
fixed resistor were compared. The circuit diagrams
used are shown in Figure 13.20.
The table shows data taken from the experiment.
(When the components were connected in reverse
bias, the current and potential difference values were
numerically the same but with a negative sign.)
0–12 V
A
0–12 V
A
V
V
Figure 13.20 Electrical circuit diagrams.
13 Current electricity
Current, I (A)
232
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
Bulb potential difference, Vbulb (V)
Fixed resistor potential difference, Vresistor (V)
0.00
0.44
1.02
1.85
3.20
4.64
6.30
8.90
11.43
0.00
1.04
2.07
3.10
4.12
5.14
6.15
7.20
8.20
1 Plot an I–V electrical characteristic illustrating this
data (p.d. on the x-axis); show both sets of data on
the same graph.
2 Use your graph to determine and label which
component obeys Ohm’s law.
3 Plot a V–I electrical characteristic illustrating this
data (current on the x-axis); show both sets of data
on the same graph.
4 Explain why the gradient of the line showing the
results for the fixed resistor is the resistance (in
ohms) of the resistor.
5 Use the graph to calculate the resistance of the
fixed resistor.
6 At what current value is the resistance of each
component the same?
469884_13_AQA_A-Level_Physics_221-243.indd 232
7T
he resistance of the bulb is not constant.
Construct a copy of the table but only include the
data from the bulb (in forward bias only) and add
an extra column showing the resistance of the bulb
at each current value.
8 Plot a graph of bulb resistance against current.
9 E xplain why the resistance of the bulb varies in
this way.
10 W hat do you think happens to R as I tends to 0
(zero)?
11 Write a method detailing how you would collect
similar data for a diode. Include a circuit diagram
in your description.
17/04/19 10:23 AM
A thermistor is a component whose
resistance varies with temperature. Many
thermistors used in electrical circuits have
a negative temperature coefficient (ntc),
meaning that their resistance decreases with
increasing temperature.
Thermistors
●● Thermistors
Thermistors are a type of resistor that change their resistance with
temperature. Thermistors have the following circuit symbol:
resistance,
R (Ω)
temperature, T (°C)
Figure 13.21 Thermistor circuit symbol and resistance – temperature graph.
Most thermistors are negative temperature coefficient (ntc) components.
This means that their resistance decreases with increasing temperature. The
vast majority of ntc thermistors also have a non-linear response to changing
temperature; in other words, a graph of resistance against temperature is a curve.
Thermistors are widely used in circuits to sense temperature changes, and then
to control devices; the change in the resistance of the thermistor affects a current,
which can be used to switch devices on or off. They can be found in thermal
cut-out circuits to prevent devices from overheating (such as a hairdryer), central
heating circuits, digital thermometers and engine-management circuits.
80°C – High Temperature (large current)
Resistance = R = Ratio V/I
= 11/0.010 = 1100Ω
Low Resistance
0.014
0.012
current, I (A)
0.010
0.008
20°C – Low Temperature (small current)
Resistance = R = Ratio V/I
= 6/0.002 = 3000Ω
High Resistance
0.006
0.004
0.002
0.000
0
1
2
3
4
5
6
7
8
9
potential difference, V (V)
10
11
12
13
Figure 13.22 Thermistor electrical characteristic example.
At a low temperature of 20 °C, the resistance of the thermistor is high and
so the current (and the V/R ratio) is low:
R=V
I
= 6V
0.002 A
= 3000 Ω
A higher temperature of 80 °C, the resistance is lower and the V/R ratio is higher:
R=V
I
11 V
=
0.010 A
= 1100 Ω
the current flowing through the thermistor is therefore high.
469884_13_AQA_A-Level_Physics_221-243.indd 233
233
17/04/19 10:23 AM
ACTIVITY
The resistance of a thermistor
●
Design an experiment to determine how the
resistance of a thermistor varies with temperature.
You should include:
● a description of the independent, dependent and
control variables. For each variable you should
state appropriate ranges, intervals, values and
suitable measuring equipment
● a diagram of your experiment including a suitable
circuit diagram
●
●
●
●
●
a list of equipment including appropriate capacities,
ranges and settings
a statement of how you will minimise the errors on
the measurements that you will take
a numbered method stating how the experiment
will be performed
a suitable risk assessment
a table that could be used to record your primary data
a description of any calculations that need to be
carried out, including how you could calculate the
uncertainty in your measurements.
234
9 Sprites are a relatively newly discovered
phenomena. The first recorded visual observation
was in 1886, but the first photograph of a sprite
was only taken in 1989. Jellyfish sprites occur
at altitudes of up to 90 km and they can cover an
area of 50 × 50 km. If a sprite lightning strike is
generated by a potential difference of 200 000 000 V,
and 325 C of charge is transferred, calculate the
total energy transferred during the strike.
10 A small neon indicator lamp on a cooker circuit
requires 95 V in order to conduct. It then draws a
current of 0.8 mA.
a) Calculate the resistance of the lamp at this
current and voltage.
b) Calculate the number of electrons moving
through the lamp each second.
11 Figure 13.23 shows the electrical characteristic
graphs of three different electrical components:
A, B and C.
5
potential difference, V/V
13 Current electricity
TEST YOURSELF
4
A
3
B
C
2
1
0
0
50
150
100
current, I/mA
200
250
Figure 13.23 Electrical characteristic graphs of
three different electrical components.
469884_13_AQA_A-Level_Physics_221-243.indd 234
a) W
hich component, A, B or C, obeys Ohm’s law?
b) Use values from the graph to determine which
component has an increasing resistance at
higher current.
12 A red-coloured LED starts to conduct electrical
current when the potential difference across it is
greater than 1.5 V. Figure 13.24 shows the circuit
used to run the LED from a 6.0 V battery that has
negligible internal resistance.
6.0 V
R
Figure 13.24 Electrical circuit diagram.
a) E
xplain the reason for putting the resistor, R,
into the circuit.
b) W hen operating at its normal current of
20 mA the potential difference across the LED
is 2.2 V. Calculate the value of the resistor,
R, for operation of the LED at 20 mA from
the 6.0 V battery. (Remember, the sum of
the emfs is equal to the sum of the p.ds in a
series circuit.)
13 Figure 13.25 shows how the potential difference
varies with current for a red and a yellow LED.
a) Calculate the resistance of each LED at each
of the following currents:
i) 30 mA
ii) 10 mA
b) Describe one other difference between the
behaviour of the two LEDs.
17/04/19 10:23 AM
Thermistors
3
potential difference, V/V
2.5
2
Red LED
Yellow LED
1.5
1
0.5
0
0
5
10
15
20
current, I/mA
25
30
35
Figure 13.25 How the potential difference varies with current for a red and a
yellow LED.
14 Figure 13.26 illustrates how the resistance of
a negative temperature coefficient thermistor
varies with temperature.
a) A
voltmeter is connected to the circuit to
indicate an increasing potential difference
when the sensor detects an increasing
temperature. Copy and complete the
diagram showing the circuit connections
for a voltmeter to measure a potential
difference that increases with increasing
temperature.
b) The value of the fixed resistor in the
circuit diagram is 250 Ω. The thermistor
is at 25 °C. Using data from the graph in
Figure 13.26, calculate the current drawn
from the 6.0 V supply. See Question 12 on
the previous page.
1200
resistance, R/Ω
1000
800
600
400
200
0
0
20
40
60
temperature, T/°C
80
100
6.0 V
Figure 13.26 Resistance–temperature graph.
a) Use the graph to estimate the rate of change of
resistance with temperature at:
i) 24 °C
ii) 84 °C.
b) The thermistor is to be used as a resistance
thermometer. At which temperature is it more
sensitive, 84 °C or 24 °C? Explain your answer.
15 Figure 13.27 shows the thermistor from
Question 14 together with a resistor in a
temperature-sensing circuit, similar to circuits
found in electronic thermometers.
469884_13_AQA_A-Level_Physics_221-243.indd 235
resistor
thermistor
235
Figure 13.27 A thermistor
together with a resistor
in a temperature-sensing
circuit.
17/04/19 10:23 AM
●● Resistivity
An intrinsic property of a material is a
property of the material itself, independent
of other factors. The resistance of a piece
of wire, for example, depends on the
dimensions of the wire, in addition to the
material that it is made of – resistivity is
the intrinsic ‘resistance’ of the material,
independent of its dimensions.
How is the intrinsic ‘resistance’ of different materials compared? How
do you decide which material would be best for the internal connections
of a mains plug, or for using as an insulator on a high-voltage line? The
resistance of a metal wire, for example, depends on the length and area
of the wire, as well as the material that the wire is made from. Consider
Figure 13.28, showing a metal conductor.
length, l
cross-sectional area, A
resistance, R
Figure 13.28 The resistance of a wire depends on its dimensions, but
resistivity depends on what the material is made of.
The resistance of the conductor, R, increases with increasing length, l (there
are more positive ion cores in the way of the gas of electrons moving
through the conductor). In fact, if the length, l, doubles, then the resistance,
R, doubles. This means that R α l.
13 Current electricity
The resistance of the conductor decreases with increasing cross-sectional
area, A. (There are more conducting pathways through the conductor
for the electrons to move through.) In this case, if the cross-sectional area,
A, doubles,
The resistivity of a substance is the intrinsic
resistance of the material that the substance
is made from; it is independent of the
dimensions of the substance. Resistivity
also varies with temperature, although the
resistivity of metals changes only gradually
with temperature.
236
then the resistance, R, halves. This means that R α A1 . Combining both of
these proportionality statements together:
l
Rα
A
and replacing the proportionality sign and adding a constant of
proportionality, ρ, we have:
ρ×l
R=
A
where ρ (‘rho’) is called the electrical resistivity of the material.
Resistivity is the property that gives the intrinsic resistance of the material
independent of its physical dimensions, such as length and cross-sectional
area. Resistivity has the units of ohm metres, Ω m, and is defined by the
rearranged form of the equation:
ρ = RA
l
where R is the resistance (measured in ohms, Ω), A is the cross-sectional area
(measured in metres squared, m2) and l is the length (measured in metres, m).
The resistivity of a material depends on some intrinsic properties of the
material. In particular, it relates directly to the number of free, conducting
electrons that can flow through the structure and the mobility of these
electrons to flow through the structure. The arrangement of the atoms in
the conductor and any distribution of impurities affects this mobility, as
does the temperature of the material.
At room temperature (20 °C), good insulating materials, such as ABS plastic
(the material that most mains plugs are now made from), have extremely
high resistivity (ABS has an electrical resistivity of 1 × 1015 Ω m at 20 °C).
469884_13_AQA_A-Level_Physics_221-243.indd 236
17/04/19 10:23 AM
We can use a water analogy to help understand resistivity. A hose pipe full
of sand is like a high resistivity conductor, as water trickles through slowly
when pressure is applied. However, that same pressure pushes water quickly
through an empty hose pipe, which is like a low resistivity conductor.
Remember though that the ability of the water to pass through the hosepipe
also depends on the length of the hosepipe and its cross-sectional area.
Resistivity
Good metallic conductors have very low resistivity; the resistivity of copper,
for example, is 1.68 × 10−8 Ω m at 20 °C. Superconductors have zero
resistivity below their critical temperature.
Resistivity is also dependent on temperature. The resistivity of metals increases
with increasing temperature, and the resistivity of many semiconductors,
such as silicon and germanium, decreases with increasing temperature. The
resistivity of a superconducting material decreases with decreasing temperature
above its critical temperature (like a metal), but its resistivity drops to zero
below the critical temperature.
REQUIRED PRACTICAL 5
Determination of resistivity of a wire using a micrometer, ammeter and voltmeter
Note: This is only one example of how you might tackle this required practical.
A 2.0 m length of 36 swg (standard wire gauge)
6 The ammeter introduces an error of 5% on each
constantan is clamped to a laboratory bench. A suitable
measurement of the current and therefore an
measuring instrument is used to measure the diameter
error of 5% on each resistance measurement.
of the wire ten times at random distances along the
Use this information to plot error bars on your
length of the wire. A fixed 6.0 V power supply is then
graph.
connected to one end of the wire and an ammeter
7 Use a best-fit line technique to determine the
and a flying lead and then used to connect in series to
gradient of the graph.
different lengths of the wire. The distance between the
8 Use the gradient calculated in step 7 to determine
fixed end of the wire and the flying lead is then varied
the resistivity of the constantan wire.
from 0.00 m to 2.00 m in 0.25 m steps.
9 Kaye and Laby, the National Physical Laboratory
The diameter measurements are (in mm):
Tables of Physical and Chemical Constants (www.
kayelaby.npl.co.uk/general_physics/2_6/2_6_1.
0.19; 0.19; 0.20; 0.19; 0.18; 0.19; 0.20; 0.18; 0.20; 0.19
html) gives the electrical resistivity of constantan
1 Draw a circuit diagram for this experiment.
to be 49 × 10 −8 Ω m at 0 °C. Determine the
2 State the name of a suitable measuring instrument
percentage error in your answer compared to the
for measuring the diameter of the wire.
value given in Kaye and Laby.
3 Calculate the mean average diameter of the wire
10 The temperature coefficient of resistivity for
and determine an uncertainty for this measurement.
constantan is 8 × 10−6 °C−1 (this coefficient is defined
4 The following data was collected from this
as the fractional change in resistivity per °C). Explain
experiment. Copy this table and add a third
why the value from Kaye and Laby, quoted at 0 °C, is
row showing the resistance of each length of
likely to be very similar to the value at 20 °C.
constantan.
Extension
Length of
0.00
0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 Use the spread of the error bars and a suitable
wire (m)
graphical technique to calculate the uncertainty
Current Short
1.34 0.84 0.51 0.35 0.31 0.27 0.22 0.18 in the measurement of the resistivity of
circuit
(A)
constantan and compare it to the value from Kaye
and Laby.
5 Plot a graph of resistance, R/Ω (y-axis) against
length of wire, l/m (x-axis).
469884_13_AQA_A-Level_Physics_221-243.indd 237
237
17/04/19 10:23 AM
TEST YOURSELF
16 A square semiconductor chip made from selenium
is shown in Figure 13.29. The resistance of the
chip is about 100 Ω. Current is passed in and out of
the chip by the metallic connectors shown.
b) U
sing figure 13.12, describe and explain the
trend shown by the graph in terms of the
microscopic structure of metals.
18 The resistance, R, of a cylindrical wire is given by
the following equation.
ρL
ρL
R=
= 2
A
πr
Figure 13.29 Selenium chip.
a) D
raw a circuit diagram showing how you could
obtain the electrical measurements required
on this chip to determine the electrical
resistivity of selenium.
b) Make a list of all the measurements needed to
determine the resistivity, and explain how you
would make them.
c) Describe how all of the data can be used to
calculate a value for the resistivity of selenium.
d) Suggest one way in which you could improve
the uncertainty in the measurement of
resistivity by reducing the error in one of the
measurements.
17 Table 11.7 below shows the thermal conductivity
and electrical resistivity (at room temperature) of
five pure metals.
13 Current electricity
Table 11.7
Copper
Gold
Aluminium
Magnesium
Zinc
Thermal
conductivity, χ
(W m−1 K−1)
385
310
240
150
110
Electrical
resistivity, ρ
(Ω m)
1.7 × 10 −8
2.4 × 10 −8
2.7 × 10 −8
4.0 × 10 −8
5.9 × 10 −8
a) P
lot the data on a suitable graph with thermal
conductivity on the x-axis.
r
L
Figure 13.30 Cylinder showing
dimensions of a wire.
The dimensions are shown on Figure 13.30.
Here is a list of multiplying factors.
1
1
× × × 2 × 4
2
4
Which of the multiplying factors above best
describes the following changes:
a) The length L of the wire is doubled. The
resistance R will change by a factor of…
b) The radius r of the wire is halved. The
resistance R will change by a factor of…
19 Figure 13.31 shows a copper connector on the
surface of a mobile phone chip. The connector
has a cross-sectional area A = 4.0 × 10 −10 m2,
a length L = 17.0 × 10 −4 m and the resistivity, ρ,
of copper = 1.7 × 10 −8 Ω m. Calculate the
resistance R of this connector.
A
L
Figure 13.31 Copper connector.
238
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1 The resistivity of a copper wire at room temperature is 1.7 × 10−8 Ω m.
The wire has a cross-sectional area of 3.1 × 10−12 m2. The resistance of a
length of 0.1 m of this wire is:
A 0.002 Ω
C 548 Ω
B 14 Ω
D 5484 Ω
Practice questions
Practice questions
2 A platinum wire of length 1.6 m has a constant diameter of 0.001 mm.
The resistivity of platinum is 1.06 × 10−7 Ω m. The resistance of the
wire is:
A 220 kΩ
C 2 kΩ
B 200 Ω
D 2 MΩ
3 The current through a wire made from tin must not exceed 4.0 A. The
wire has a cross-sectional area of 7.8 × 10−9 m2. If a p.d. of 2.0 V is
connected across the wire, what is the minimum length of wire needed
to ensure the current remains below 4.0 A? The resistivity of tin is
1.1 × 10−7 Ω m.
A 0.035 m
C 0.140 m
B 0.070 m
D 0.350 m
4 How many electrons are passing through a wire each second if the
current flowing through it is 1.00 mA?
A 6.3 × 1013
B 6.3 × 1015
C 6.3 × 1018
D 6.3 × 1021
5 The heating element for a small electric heater is made from a wire of
resistance R. It is replaced by a wire made from the same material and
with the same diameter, but with half the length. The resistance of this
second wire is:
A
B
1
R
4
1
R
2
C 2R
D 4R
6 A resistor is connected to a cell. In a time, t, an amount of charge,
Q, passes through the resistor. During this time the electrical energy
dissipated by the resistor is W. What is the current in the resistor and
the emf of the cell?
Current
emf
A
Q/t
W/Q
B
Qt
WQ
C
Qt
W/Q
D
Q/t
WQ
469884_13_AQA_A-Level_Physics_221-243.indd 239
239
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7 The wires shown in Figure 13.32 are made from the same
material but have different dimensions. Which wire will have
the smallest value of resistance?
A
A
L
8 The unit of electric current, the ampere, is equivalent to:
A C s
B
J C−1
C Js
D
B
1 L
2
9 An electric eel can store charge in specialised cells in its body.
It then discharges these cells to protect itself from predators.
The discharge of an electric eel can transfer a charge of
2.0 mC in approximately 2.0 ms. What current does the eel
deliver?
A 1 mA
B 10 mA
C 100 mA
C
C 400 C
B 200 C
D 800 C
D
D 1A
Figure 13.32 These wires are made
from the same material but have
different dimensions.
13 Current electricity
b) This filament bulb is connected to a 6 V battery. Calculate
the normal operating resistance of the bulb when it has
reached its standard operating temperature.
(2)
240
c) Explain why the current through the lamp increases
rapidly between 0 and 50 ms before dropping to a
steady value.
d) Use the graph to estimate the total charge that has
passed through the filament bulb in the first 500 ms
of operation.
0.4
0.3
current, I/A
(1)
1 A
2
2L
11 When a tungsten filament lamp is switched on it takes about
500 ms for the filament to heat up and reach its normal
operating temperature. The variation of the current with time
is shown in Figure 13.33.
a) What is the maximum current passed through the
filament?
1 A
2
1 L
2
10 A car battery delivers a current of 400 A for 0.5 s. What is the charge
flowing?
A 100 C
A
C s−1
0.2
0.1
0
0
(2)
200
400
600
800
time, t/ms
Figure 13.33 Filament lamp current –
time graph.
(2)
12 The resistance of a metal wire changes with temperature. You are
given a length of nichrome wire and asked to determine how the
resistance of the nichrome wire changes between 0 °C and 100 °C.
a) Draw a labelled diagram of the experimental set-up that would
allow you to perform this experiment.
(2)
b) Write a suitable, numbered method that would allow you to
obtain accurate and reliable measurements of the resistance of
the wire over a range of temperatures between 0 °C and 100 °C.
(3)
c) The metallic element aluminium has a critical temperature of
1.2 K. Explain what is meant by the critical temperature of an
electrical conductor.
(2)
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a) State what is meant by the term ‘ohmic conductor’.
(1)
b) Draw a circuit diagram for this experiment.
(2)
Practice questions
13 A student wished to perform an experiment on an electrical component
to determine if the component was an ohmic conductor.
c) For the experimental circuit diagram that you have drawn, write an
account of a suitable experiment. Your account should include:
●
what measurements you would take
●
how you would use your measurements
●
how you would reach a conclusion.
(6)
14 A semiconducting diode and a filament lamp are both
examples of non-ohmic components.
current, I
a) Copy and complete a sketch on the axes shown in Figure
13.34 of the current–voltage characteristics for both
components.
(2)
potential difference, V
b) Describe, using the current–voltage characteristic that
you have drawn, how the resistance of the filament lamp
changes as the potential difference across it changes. (2)
c) Draw a suitable diagram of the circuit that would enable
you to collect data so that you could plot the I–V curve
(2)
for the semiconductor diode.
d) Write a method that you could follow in order to
obtain this data for a semiconductor diode.
Figure 13.34 Electrical characteristic axes.
6.0 V
(4)
15 Figure 13.35 shows a thermistor connected in series with a
resistor, R, and a 6.0 V battery.
When the temperature is 40 °C the resistance of the thermistor
is 1.4 kΩ. The voltmeter connected across R reads 1.8 V.
R
a) Calculate the potential difference across the thermistor. (1)
b) Calculate the current flowing through the thermistor.
(2)
c) Calculate the resistance of R.
(2)
V
Figure 13.35 Thermistor circuit.
d) The battery is now replaced with a power supply unit with an
internal resistance of 10 Ω. (This is equivalent to adding an extra
10 Ω to the circuit, but the resistance is inside the battery.) Without
calculation, state and explain the effect on the voltmeter reading. (2)
16 Carbon-conducting assembly paste is used by electronics manufaturers
to ensure good connections between components and printed circuit
boards. A cylindrical length of this paste is laid out, 8.0 × 10−2 m long
with a radius of 1.4 × 10−2 m. The resistivity of the paste is 0.82 Ω m.
a) Calculate the resistance between the ends of the cylinder of
the assembly paste. Give your answer to a suitable number of
significant figures.
241
(2)
b) The paste is now reshaped into a cylinder with half the radius and
a length that is four times as great. Calculate the new resistance. (3)
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17 A plastic-clad copper connecting wire used in school electricity
experiments is 0.60 m long and the copper inside the wire has a
cross-sectional area of 1.3 × 10–7 m2. The resistivity of copper is
= 1.7 × 10–8 Ω m.
(2)
a) Calculate the resistance of the connecting wire.
A filament lamp is connected to a power supply using two of the
connecting wires; an ammeter connected in series measures a current
of 2.2 A. The potential difference across the filament lamp is 12.0 V.
b) Calculate the potential difference across each of the wires.
(2)
c) Calculate the potential difference across the terminals of the power
supply.
(2)
Stretch and challenge
All of the following questions are provided with permission of
the British Physics Olympiad
E
18 A cell that produces a potential E (called an emf) is shown in Figure
13.36 and is connected to two resistors in series: a fixed resistor R1
and a variable resistor R2. The current, I, in the circuit is measured
by the ammeter, A, and the potential difference, V, across resistor R2
is measured with the voltmeter, V.
A
R1
R2
V
The relation between the potential E and the current I is given by:
Figure 13.36 Circuit diagram.
E = IR1 + IR2
13 Current electricity
Which of these graphs would produce a straight-line fit?
242
A V against I
C 1/V against 1/I
B V against 1/I
D I against 1/V
19 Two students decide to calibrate a thermistor in order to measure
variations in the temperature of a room. They connect a small
thermistor across the terminals of a 5 V power supply and in
series with a 1 A ammeter. The resistance of the thermistor is
120 Ω at room temperature. See figure 13.37.
a) Instead of showing small variations in room temperature, the
thermistor is likely to go up in smoke. Explain why.
b) In the light of this experience, they decide to redesign their simple
circuit as shown in Figure 13.37. They have a few values of resistor
R to choose from; 5 kΩ, 500 Ω, 50 Ω.
5V
R
V
0V
Figure 13.37 Circuit
diagram.
State which value of R would give the biggest variation of V with
temperature. Explain your choice.
c) State which value of R would be most likely to cause the same
problem as in part (a). Again, explain your choice.
469884_13_AQA_A-Level_Physics_221-243.indd 242
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Some metals are ductile, which means that they can be drawn into
long thin wires. In doing so, the volume V remains constant while
the length increases and the cross-sectional area of the wire decreases.
A wire of length 32 m has a resistance of 2.7 Ω. We wish to calculate
the resistance of a wire formed from the same volume of metal
but which has a length of 120 m instead.
Practice questions
20 The resistance of a wire is proportional to its length and inversely
proportional to its cross-sectional area. The resistance of a wire of
length l and cross-sectional area A is given by:
R = ρl
A
where ρ is a constant that depends on the material of the wire.
a) Write down the relationship between V, A and l. Obtain an
expression to show how R depends on the length l of the
wire and its volume V.
b) Rewrite the equation with the constants ρ and V on one side and
the variables we are changing, R and l, on the other.
c) Calculate the resistance of the longer wire.
243
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17/04/19 10:23 AM
14
Electrical circuits
PRIOR KNOWLEDGE
●
●
●
The rate at which energy is transferred by an appliance from one energy
store to another is called the power:
energy transferred
power =
time
or
E
P= t
Power, P, potential difference, V, and current, I, are related by the
equation:
P=I×V
The potential difference provided by cells connected in series is
the sum of the potential difference of each cell (depending on the
direction in which they are connected) – see Figure 14.1.
1.5 V
1.5 V
1.5 V
4.5 V
Figure 14.1 Circuit diagram showing cells and a battery.
●
For components connected in series:
• the total resistance is the sum of the resistance of each component
(see Figure 14.2).
10 Ω
20 Ω
30 Ω
60 Ω
Figure 14.2 Circuit diagram showing resistance in series.
• there is the same current through each component (see Figure 14.3).
0.5 A
0.5 A
0.5 A
0.5 A
Figure 14.3 Circuit diagram showing current in series circuits.
• the total potential difference of the supply is shared between the
components (see Figure 14.4).
6.0 V
1.0 V
244
2.0 V
3.0 V
Figure 14.4 Circuit diagram showing p.ds in series.
●
For components connected in parallel:
• the potential difference across each component is the same
(see Figure 14.5)
6.0 V
6.0 V
6.0 V
6.0 V
Figure 14.5 Circuit diagram sharing p.ds in parallel circuits.
469884_14_AQA_A-Level_Physics_244-273.indd 244
17/04/19 10:23 AM
Electrical circuits
• the total current through the whole circuit is the sum of the
currents through the separate components (see Figure 14.6).
0.6 A
0.1 A
0.2 A
0.3 A
0.1 A
0.2 A
0.3 A
0.6 A
Figure 14.6 Circuit diagram sharing current in parallel circuits.
●
The resistance of a light-dependent resistor (LDR) decreases as
light intensity increases.
●
The resistance of a thermistor decreases as the temperature
increases.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 An electric kettle takes 3 minutes to transfer 360 kJ of electrical
energy into thermal energy stored in the water. Calculate the power of
the kettle.
2 The kettle in the previous question is mains powered from a supply
with a potential difference of 230 V. Calculate the current flowing
through the kettle element.
3 A lab technician puts three used D cells into a battery pack. All three
are connected the same way around. The measured emfs of the cells
are: 1.45 V; 1.39 V and 1.47 V respectively. Calculate all the possible
emfs that can be delivered from this battery pack.
4 The circuit in Figure 14.7 uses a 1.5 V cell, with negligible internal
resistance, to feed current through two fixed resistors:
1.5 V
0.6 mA
A
R1
R 470 Ω
V1
V2
Figure 14.7 Circuit diagram for question 4.
a)
b)
c)
d)
469884_14_AQA_A-Level_Physics_244-273.indd 245
Calculate the value of V2.
Use the value of V2 to calculate the value of V1.
Now calculate the value of R1.
State the total resistance of the circuit.
245
17/04/19 10:23 AM
a) Calculate the total current,
I, delivered by the cell.
b) Now calculate the values of
R1, R2 and R3.
9.0 V
0.90 A
0.41 A
R3
R2
R1
5 In Figure 14.8 a 9.0 V cell of
negligible internal resistance
delivers current to three
resistors, R1, R2 and R3,
connected in parallel:
0.19 A
I
Figure 14.8 Circuit diagram for question 5.
●● Electrical power in circuits
The integrated circuit of an iPhone is an extremely complex piece of
electronic engineering. It involves the interconnection of many thousands
of miniature electronic components. These are designed to work together
controlling the many functions of the phone. Although the circuits are
intricate, they are all based on some simple basic circuit principles and
laws. Integrated circuits excel at doing simple, single-step things. They just
do them very quickly in predefined sequences.
14 Electrical circuits
One of the most fundamental properties that electronic engineers need
to consider when designing electrical devices is the power transferred
by the circuit in the device. High power consumption requires specialist
cooling systems and, if the device is battery powered, high battery capacity.
Minimising power consumption reduces costs and bulk.
The electrical power transferred by a circuit is the sum of all the power
transferred by the individual electrical components in the circuit. The electrical
power transferred by a component is related to the current flowing through it
and the potential difference across it. The general definition of power is:
power = rate of doing work or rate of energy transfer
or
P=
Figure 14.9 The iPhone 5s A7
processor.
246
469884_14_AQA_A-Level_Physics_244-273.indd 246
ΔW
Δt
So, in an electrical context, electrical power can be defined as the rate of
doing electrical work. The electrical power transferred by a component can
be determined by using the current, potential difference and resistance of
the component, and there are several equations that can be used to calculate
this power. These equations are produced by rearranging the basic equations
for electric current, potential difference and the definition of resistance.
charge
current =
time
in terms of symbols:
Q
I=
t
rearranging:
⇒ Q = I t
Equation 1
electrical energy
potential difference =
charge
17/04/19 10:23 AM
⇒ Q = W
V
Equating Equation 1 with Equation 2:
It = W
V
rearranging:
W = IV = P
t
or
P = IV
and using the definition of resistance
V = IR
Equation 2
Electrical power in circuits
in terms of symbols:
V=W
Q
rearranging:
P = I × (IR) = I2R
or
P= V ×V
R
2
V
=
R
The relationship P = IV is a really useful one as it allows the calculation
of electrical power using two easily measured and monitored quantities,
current and potential difference. Data-logging ammeters and voltmeters
allow for the real-time monitoring of the electrical power consumption of
a circuit, which then allows battery-powered devices such as smartphones,
tablets and laptops to display the amount of energy left in the battery and
predict the amount of usage time remaining before charging is required.
TEST YOURSELF
1 A mains electric iron operates at 230 V and is rated
at 2.2 kW. Calculate the resistance of the iron when
operating normally.
2 Look at this list of electrical units:
A As
D C s–1
–1
B VA
E VAs
C J s–1
Choose the unit for:
a) electric charge
c) electrical resistance
b) electrical power
d) electrical energy
3 A high-power resistor used inside electric
guitar amplifiers has a resistance of 330 Ω and a
maximum power of 25 W. Calculate the potential
difference across the power resistor when it is
running at full maximum power.
4 A mobile phone battery operates with a potential
difference of 3.8 V across its circuits and a
capacity (the product of the current and time
469884_14_AQA_A-Level_Physics_244-273.indd 247
that the current can be maintained) of 1560 mA h
(milliampere-hours).
a) While playing a high-graphic usage game
the power drawn from the battery is 0.30 W.
Calculate the current drawn from the battery.
b) Calculate the total charge that can be drawn
from the battery.
c) Calculate the total available game-play time.
5 Two relationships for electrical components are:
V = IR and P = IV.
a) Use these two relationships to produce an
equation for electrical power in terms of
V and R only.
b) The power dissipated from a 33 Ω resistor
operating at 1.5 V is 0.068 W. The p.d., V, is
now doubled and the resistance is halved.
What is the effect on the power dissipated by
the resistor?
247
17/04/19 10:23 AM
ACTIVITY
Comparing the power of a 12 V bulb and a rheostat
A student carried out a simple experiment to compare the power of a
12 V bulb and a rheostat (acting as a fixed resistor). The circuits that
she used are shown in Figure 14.10.
0–12 V dc
0–12 V dc
A
A
Here are her measurements.
Potential
difference, V (V)
±0.01 V
0.00
2.00
4.00
6.00
8.00
10.00
12.00
Current, I (A), ±0.01 A
Rheostat
Bulb
0.00
0.00
0.08
0.02
0.16
0.09
0.24
0.21
0.32
0.37
0.40
0.58
0.48
0.83
Power, P (W)
Rheostat
Bulb
14 Electrical circuits
1 Copy and complete the table, calculating the power
of each component at each potential difference.
2 Plot a graph of electrical power, P, against potential
difference, V, plotting both the rheostat and the
bulb on the same graph.
3 Describe in detail the pattern in the variation
of power with potential difference for both
components.
4 Use the data to determine the resistance of
the rheostat.
5 The student’s teacher suggests that the relationship
between the power, P, and the potential difference, V,
V
V
Figure 14.10 Circuit diagram.
for the bulb is given by P = kV 3, where k is a constant.
Use the data in the table to draw a suitable graph to
enable you to determine if the teacher was correct
and, if so, to calculate a value for k.
6 Explain why the power produced by the bulb varies
in a different way to the rheostat.
Extension
The potential difference is measured to the nearest
0.01 V, the current to the nearest 0.01 A. Use this
information to calculate a suitable maximum uncertainty
for the measurement of the resistance of the rheostat.
●● Circuit calculations
Electronic engineers also need to know the simple ways that current and
potential difference act in circuits. The ‘rules’ for these are universal and
apply to simple circuits as well as the intricate circuits found in integrated
circuit chips. These simple rules were first worked out by a German
physicist called Gustav Kirchoff in 1845 and have become known as
Kirchoff’s First and Second Circuit Laws.
Kirchoff’s First Circuit Law – the law of current
I1
248
I3
I2
I4
I5
Figure 14.11 A circuit junction.
At a circuit junction, the sum of the currents flowing into the junction equals the
sum of the currents flowing out of the junction.
Figure 14.11 shows a circuit junction with two currents (I1 and I2)
flowing into the junction and three currents flowing out of the junction
(I3, I4 and I5).
Kirchoff’s First Circuit Law states:
I1 + I2 = I3 + I4 + I5
469884_14_AQA_A-Level_Physics_244-273.indd 248
17/04/19 10:23 AM
A1
Conventional current flows from positive to negative, so the current flowing
into the junction indicated on the diagram is measured by ammeter A1.
Current splits or recombines at a junction, so the current flowing out of the
junction is measured by A2 and A3. Kirchoff’s First Circuit Law tells you that:
A2
A3
A1 = A2 + A3.
Figure 14.12 Circuit diagram showing
Kirchoff's First Circuit Law.
Circuit calculations
Figure 14.12 shows Kirchoff’s First Circuit Law in action in a real circuit
involving ammeters.
Written more generally in mathematical notation the law can be summarised
by:
∑ Iinto junction = ∑ Iout of junction
or, in other words, current is conserved at junctions.
Looked at from a slightly different perspective, as current is the rate of flow
of charge, or
I = ∆Q
∆t
then Kirchoff’s First Circuit Law could be written in terms of charge:
At a circuit junction, the sum of the charge flowing into the junction equals the
sum of the charge flowing out of the junction (per second).
∑ Qinto junction = ∑ Qout of junction
(per unit time).
Kirchoff’s Second Circuit Law – the law of voltages
In a closed circuit loop, the sum of the potential differences is equal to the sum
of the electromotive forces.
ε
Figure 14.13 shows a single closed loop series circuit. In this case, there is
one emf and two p.ds, and Kirchoff’s Second Circuit Law says that:
ε = V1 + V2
Or more generally, using mathematical notation, for any closed circuit loop:
V1
∑ε=∑V
V2
Figure 14.13 Circuit diagram
showing Kirchoff's Second
Circuit Law.
ε
A
If the circuit is extended to make it a parallel circuit such as Figure 14.14:
This parallel circuit is effectively made up of two series circuits: ABCD and
AEFD, so:
ε = V1 + V2
D
and
C
B
V1
V2
V3
V4
Figure 14.14 Circuit diagram
showing Kirchoff's Second
Circuit Law in a parallel circuit.
469884_14_AQA_A-Level_Physics_244-273.indd 249
or
(V1 + V2) = (V3 + V4)
F
E
249
ε = V3 + V4
Kirchoff’s Second Circuit Law applies to any circuit but, in the case of
parallel circuits, the circuit must be considered as a succession of individual
series circuits with the same power supply. This law is based on the
conservation of energy: the energy per coulomb transferred to the charge by
the battery, ε, is then transferred to other forms of energy by the charge as it
flows through the circuit components.
17/04/19 10:23 AM
Batteries and cells
The standard AA battery is a common battery for all manner of hand-held
devices such as games controllers and remote controls. The word battery is
actually a misnomer (a word or term that suggests a meaning that is known
to be wrong): the device shown in the picture is actually a cell. Several cells
joined in series or in parallel are called a battery. The standard electrical
symbols are shown in Figure 14.16.
Figure 14.15 An AA battery.
When identical cells are connected in parallel to form a battery, the emf
of the resultant battery is just the emf of the individual cells. Connecting
cells in parallel is generally not a good idea as they are rarely identical,
and one cell will force current back into the other, causing damage to the
cells. Cells in parallel have a higher capacity for storing and transferring
electrical energy than single cells, but is it is better to just buy and use a
bigger single cell.
Cell
Battery
Figure 14.16 Circuit symbols
for cell and battery.
TIP
●● Internal resistance and
electromotive force
Unless a question explicitly
states that the power supply
in the question has an internal
resistance, you should ignore it.
Real power supplies, such as batteries and laboratory power packs, always
have an internal resistance, r. As the current flows through the power
supply, the internal resistance creates a potential difference that leads to
electrical energy being transferred to thermal energy inside the power
supply. This is one of the reasons why hand-held devices such as tablets
warm up when they are in use for a long period of time.
14 Electrical circuits
The internal resistance of most batteries and power supplies is very low: a
typical AA battery has an internal resistance of about 0.2 Ω and, as such,
can effectively be ignored when measuring the electrical properties of
most circuits.
The internal resistance of a power supply cannot be measured directly
as it is ‘inside’ the power supply. The only way to measure it is to use its
electrical characteristic. A circuit that can be used to do this is shown
Figure 14.18.
battery or real power supply
r
ε
battery
I
A
250
ε
r
V
ideal battery
internal resistance
Figure 14.17 A real power supply
469884_14_AQA_A-Level_Physics_244-273.indd 250
Figure 14.18 A circuit used to measure
the internal resistance and electromotive
force of a real power supply.
17/04/19 10:23 AM
The electromotive force, ε, must be the same as the sum of the potential
differences in the circuit (Kirchoff’s Second Circuit Law). There are two
potential differences in this circuit: one is across the external variable
resistor, V, and the other is the p.d. across the internal resistor. This
cannot be measured directly, but the p.d. across this resistor is equal to Ir.
This means that:
ε = V + Ir
potential difference/V
y-intercept–
electromotive
force, ε
magnitude (size) of
gradient–internal
resistance, r
current/I
0
Figure 14.19 Electrical characteristic of a
real power supply, such as a battery.
The current, I, can be measured directly using an ammeter; ε and r are
both constants, so the equation can be rewritten as:
V = ε − Ir or V = −rI + ε
which is the equation of a straight line of the form, y = mx + c, where the
gradient is negative. If an electrical characteristic is drawn using values
of V and I from various values of R (the external load resistance) then the
y-intercept of the graph is the electromotive force, ε, and the gradient is
negative and equal to –r, the internal resistance.
Internal resistance and electromotive force
Circuit calculations can be used to determine the internal resistance and the
electromotive force.
TEST YOURSELF
ε 1.50 V
I 300 mA
r 0.20 Ω
mA
R 4.8 Ω
350
300
250
current, I/mA
6 Figure 14.20 shows an AAA battery with an emf, ε,
of 1.5 V and an internal resistance, r, of 0.20 Ω. The
cell delivers a current, I, of 300 mA into a fixed-load
resistor, R, of 4.8 Ω.
a) Use the data on the circuit diagram to calculate
the potential difference, V, across the fixed-load
resistor, R.
b) Explain why the potential difference, V, across
the load resistance is always less than the emf,
ε, of the cell.
200
150
100
50
0
V
Figure 14.20 Circuit diagram for question 6.
The AAA battery is left connected to the 4.8 Ω fixedload resistor for a few hours. Figure 14.21 shows how
the current from the AAA battery varies with time.
c) Using data from the graph, describe in detail
how the current varies with time.
469884_14_AQA_A-Level_Physics_244-273.indd 251
0
1
2
3
4
time, t/hours
5
6
251
Figure 14.21 Current–time graph.
d) Give reasons why the current varies with time in
this way.
e) Use the graph to estimate the total charge
delivered by the AAA battery during its
discharge.
17/04/19 10:23 AM
7 A radio-controlled model car can be modelled
with the circuit in Figure 14.22: a 9.0 V emf PP9
battery with an internal resistance r is connected
across a variable-load resistor R (representing
the motor).
ε 9.0 V
r
A
igure 14.23 shows the electrical characteristic for
F
this circuit.
Use the graph to calculate the internal resistance,
r, of the PP9 battery.
R
V
10.0
9.0
+
potential difference, V/V
8.0
Figure 14.22 Circuit diagram
for question 7.
+
+
7.0
+
6.0
5.0
4.0
+
3.0
2.0
+
1.0
0.0
0
2
4
6
8
10
current, I/A
14
12
16
18
20
Figure 14.23 Electrical characteristic of a PP9 battery.
252
X, Y or Z cell
R is varied and values of V and I are collected for
each cell; a composite electrical characteristic is
drawn and shown in Figure 14.25.
A
I
R
2.0
1.8
potential difference, V/V
14 Electrical circuits
8 Three cells, X, Y and Z, are connected consecutively
into the circuit shown in Figure 14.24.
X
1.6
V
1.4
Y
1.2
V
Z
1.0
Figure 14.24 Circuit diagram
for question 8.
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
current, I/A
Figure 14.25 Voltage–current graphs for cells X, Y and Z.
State which of the cells X, Y or Z:
a) has the largest emf
b) will deliver the most electrical power at a
current of 0.5 A
469884_14_AQA_A-Level_Physics_244-273.indd 252
c) has the largest internal resistance
d) can deliver an electrical power of 2.8 W.
17/04/19 10:23 AM
Internal resistance and electromotive force
REQUIRED PRACTICAL 6
Investigation of the emf and internal
resistance of electric cells and batteries by
measuring the variation of the terminal p.d.
of the cell with current in it
Note: This is just one example of how you might
tackle this required practical.
Lemons contain citric acid. This acid can undergo
electrochemical reactions when metal electrodes
made from copper and zinc (for example) are pushed
into the lemon. The electrochemical reactions
generate an electromotive force that produces
a current high enough to activate an LCD clock
screen. Lemons, however, have a very high internal
resistance which limits the current and makes
lemons a poor choice for most electronic circuits. It
would take about two and a half years, for example,
for a lemon to charge an iPhone battery from empty.
Although having a high internal resistance is
a problem for fruity powered real devices, it
makes them an ideal choice for school laboratory
experiments to measure emf and internal resistance.
A circuit that can be used to measure these values is
shown in Figure 14.27.
In this experiment, a variable resistor is used as an
external load; the current flowing through the circuit is
measured with a suitable milliammeter and the potential
difference across the variable resistor is measured with
a voltmeter for a range of resistance values. One such
experiment recorded the following data:
Resistance,
R (Ω)
100
220
470
1000
2200
4700
10 000
22 000
47 000
100 000
220 000
470 000
1
0.632
0.573
0.488
0.340
0.244
0.147
0.078
0.038
0.018
0.008
0.004
0.002
Current, I (mA)
2
3
0.623
0.644
0.566
0.581
0.482
0.500
0.336
0.343
0.242
0.246
0.146
0.148
0.077
0.078
0.038
0.038
0.018
0.018
0.008
0.008
0.004
0.004
0.002
0.002
Mean
1 Copy and complete the table by calculating the mean
average of the currents and the potential difference.
2 Plot an electrical characteristic graph of this data
with potential difference on the y-axis. Remember
to include a best-fit line.
469884_14_AQA_A-Level_Physics_244-273.indd 253
Figure 14.26 A lemon clock.
r
ε
A
I
R
V
Figure 14.27 A circuit to measure the emf
and internal resistance of a lemon.
1
0.059
0.125
0.229
0.419
0.544
0.676
0.774
0.833
0.865
0.881
0.890
0.894
Potential difference, V (V)
2
3
0.058
0.068
0.123
0.126
0.226
0.232
0.414
0.422
0.540
0.548
0.671
0.679
0.769
0.800
0.828
0.834
0.861
0.865
0.877
0.882
0.886
0.890
0.890
0.895
Mean
253
3 The formula for the internal resistance and emf
of a real power supply is given by the formula
ε = I(R + r).
17/04/19 10:23 AM
By using a suitable substitution, rearrange this
formula into the equation of a straight line with V,
the p.d. across the variable resistor, as the subject.
4 Use the graph to measure directly the emf, ε, of
the lemon.
5 Use the graph to calculate the internal resistance
of the lemon.
Extension
The ammeter measures with an uncertainty of
±0.001 mA, and the voltmeter measures with
an uncertainty of ±0.001 V. Use the instrument
uncertainties, the spread of the data and a suitable
graphical technique to determine uncertainty values
for the internal resistance, r, and the emf, ε.
Voltmeters and ammeters
Ammeters are always put into circuits in series with other components. This
is because they measure the flow of the charge (the electrons) through the
circuit. Older, analogue ammeters involve passing the current through a coil
of wire sitting inside a shaped permanent magnet.
soft steel ring
upper
control
spring
pointer
moving
coil
permanent
magnets
core
14 Electrical circuits
lower control spring
Figure 14.28 Moving coil ammeter.
high
resistance
254
Rs
low
resistance
I
Rs
The current in the coil generates a magnetic field that interacts with
a permanent magnetic field, causing the coil to turn; the current is then
measured by a pointer on an analogue scale. Modern, digital ammeters use
an integrated circuit within the meter to measure the current, which is then
displayed on a numerical display. Both designs, however, will always affect
the size of the current in some way, as any device put into the
circuit in series will have a resistance. The extra resistance of the
ammeter will therefore reduce the current in the circuit. This
effect cannot be overcome, so modern ammeters are designed to
have very low resistances and are calibrated to take into account
the reduction of the current due to the resistance of the meter.
component
I
ammeter
I
voltmeter
Figure 14.29 Ammeters have low resistance, and
voltmeters have high resistance.
469884_14_AQA_A-Level_Physics_244-273.indd 254
I
Voltmeters are always connected into circuits in parallel with
other components. Both analogue and digital voltmeters work
in very similar ways to ammeters, but a small current is drawn
from the circuit that passes through a set, known, very highresistance resistor so that the current is proportional to the
potential difference. High-quality voltmeters therefore have very
high resistance.
17/04/19 10:23 AM
Resistor networks
TEST YOURSELF
9 Look at the following list of units:
A Ω
D V
B W
E A
C J
Which of the above units is equivalent to the following:
a) J s−1
d) W A−1
2
−1
b) V W
e) V A−1
−1
c) C s
10 A
student is comparing the brightness of car bulbs when they are connected
to a 12 V car battery. She connects two bulbs, P and Q, in parallel with the battery
as shown in Figure 14.30.
Bulb Q is brighter than bulb P.
Copy and complete the following sentences
using one of the phrases listed below:
less than greater than the same as
a) The current in bulb P is………………………………that in bulb Q.
b) The p.d. across lamp P is……………………………...that in bulb Q.
c) The electrical power of bulb Q is………………………that in bulb P.
d) The electrical resistance of bulb Q is………………………that in bulb P.
P
Q
Figure 14.30 Circuit
diagram for question 10.
●● Resistor networks
When components are connected together to make useful circuits, each
component adds a certain resistance to the circuit. The effect of the added
resistance depends on how the component is connected into the circuit – in
series with other components or in parallel (or both in the case of circuits
containing many hundreds of components). There are some simple rules for
calculating the overall resistance of components connected in series or in
parallel.
ε
I
Resistors connected in series
Consider the resistor network shown in Figure 14.31.
R1
R2
R3
Figure 14.31b represents the single resistor that could replace the three
resistors in series in Figure 14.31a.
V1
V2
V3
Using Kirchoff’s Circuit laws and Ohm’s law leads to:
Figure 14.31a Circuit diagram showing
resistor combinations in series.
ε
I
ε = V1 + V2 + V3
and
ε = VT
255
where
RT
VT
Figure 14.31b Circuit diagram showing
equivalent resistor.
VT = V1 + V2 + V3
but because
V = IR
and, as the current is the same throughout a series circuit:
IRT = IR1 + IR2 + IR3 = I(R1 + R2 + R3)
469884_14_AQA_A-Level_Physics_244-273.indd 255
17/04/19 10:23 AM
so
RT = R1 + R2 + R3
and for a series network of n resistors:
RT = R1 + R2 + R3 + ... + Rn
or, using sigma notation
n
RT = ∑R i
i−1
Resistors connected in parallel
Consider the following circuit:
ε
ε
IT
R1
IT
I1
R2
RT
VT
I2
In the right-hand circuit one resistor, RT, has been used
to replace all three resistors arranged in parallel in the
left-hand circuit. Again, using Kirchoff’s Circuit laws and
Again, using Kirchoff’s Circuit laws and the definition of
resistance, V = IR:
Kirchoff’s First Circuit law says:
R3
IT = I1 + I2 + I3
I3
and
VT
Figure 14.32 Circuit diagram showing resistors
connected in parallel.
Rearranging gives:
IT
1 1 1
= + +
VT R1 R2 R3
So:
IT
1
=
VT RT
where:
1
1 1 1
= + +
RT R1 R2 R3
For a network of n resistors connected in parallel:
1
1
1 1
= + + ... +
Rn
RT R1 R2
or using sigma notation:
14 Electrical circuits
256
I=V
R
So as the potential difference, VT, is the same across all of
the resistors:
V
V
V
1 1 1
IT = T + T + T = VT( + + )
R1 R2 R3
R1 R2 R3
TIP
Resistors in parallel always have
a resistance less than any of the
individual resistors. This is a
useful check when you have done
a calculation.
469884_14_AQA_A-Level_Physics_244-273.indd 256
1 n 1
=∑
RT i=1 R i
In summary, for resistors in series the overall resistance is the sum of the
individual resistances. For resistors in parallel the reciprocal of the overall
resistance is the sum of all the reciprocal of the individual resistances.
Resistors connected in parallel always have a resistance less than any of the
resistors in that combination.
17/04/19 10:23 AM
Resistor networks
EXAMPLE
Three resistors are connected in parallel to a power supply as shown in Figure
14.33. Calculate the value of the one resistor that could replace all three resistors.
Answer
10 Ω
RT is the value of the one resistor that can replace all three so:
1 = 1 + 1 + 1
RT R1 R2 R3
substitute the numbers:
15 Ω
1 = 1 + 1 + 1
15 Ω
RT 10 Ω
20 Ω
1 = 0.10 + 0.07 + 0.05 = 0.217 Ω−1
RT
20 Ω
Figure 14.33 Circuit
diagram.
So
RT =
1
= 4.6 Ω (2 s.f.)
0.217 Ω−1
When calculating the overall resistance of a circuit containing a mixture of
resistors in series and in parallel, it is best to calculate the overall resistance
of each of the sections arranged in parallel first, and then calculate the
overall resistance by summing all the effective resistances in series.
EXAMPLE
Look at the combination circuit in Figure 14.34, containing resistors
connected in series and in parallel.
5Ω
What is the overall resistance of this circuit?
Answer
To calculate the overall resistance, RT, of this circuit, the first step is to
calculate the effective resistance, R, of the three resistors
in parallel:
1 + 1 + 1
1
=
R 5 Ω 10 Ω 15 Ω
1
= 0.20 + 0.10 + 0.067 = 0.367 Ω−1
R
1
R=
= 2.7 Ω
0.367 Ω−1
RT = 15 Ω + 2.7 Ω + 20 Ω
= 37.7 Ω
= 38 Ω (2 s.f.)
469884_14_AQA_A-Level_Physics_244-273.indd 257
15 Ω
10 Ω
20 Ω
15 Ω
Figure 14.34 Circuit diagram.
257
17/04/19 10:23 AM
14 Electrical circuits
TEST YOURSELF
258
11 T
hree resistors, each of resistance 3.3 kΩ, are
12 V
6.0 V
connected in parallel. Calculate the overall resistance of
this network.
47 kΩ
12 Two resistors are connected in parallel as shown in
1 kΩ
47 kΩ
Figure 14.35.
a) Calculate the overall resistance of this circuit.
b) The resistors are connected to a 12 V battery with
330 kΩ
47 kΩ
negligible internal resistance. Calculate the total
current drawn from the 12 V battery.
Figure 14.36 Circuit
13 Three 47 kΩ resistors are connected to a 6.0 V battery with Figure 14.35 Circuit
diagram for question 12. diagram for question 13.
negligible internal resistance as shown in Figure 14.36.
a) Calculate the total resistance of the circuit.
b) Calculate the current flowing through each resistor.
6.0 V
6.0 V
14 A teacher is thinking about the resistances of ammeters
and voltmeters. A battery with an emf of 6.0 V and
0.3 Ω
negligible internal resistance is connected first to a 2.2 Ω A
fixed resistor as shown in Circuit 1 in Figure 14.37.
2.2 Ω
2.2 Ω
a) Calculate the current flowing in Circuit 1.
Circuit 1
Circuit 2
An ammeter with a resistance of 0.3 Ω is connected into the
Figure 14.37 Circuit diagrams for questions 14 a) to f).
circuit in series with the resistor as shown by Circuit 2.
b) Calculate the total resistance of Circuit 2.
c) Calculate the current flowing in Circuit 2.
d) Explain why the current in Circuit 2 is lower than the
current flowing in Circuit 1.
e) Suggest how the manufacturer of the ammeter
could alter the design of the ammeter to take its
resistance into account.
f) The best quality ammeters, including most
6.0 V
6.0 V
digital ammeters, alter the current in the circuit
by as little as possible. Suggest a value for the
resistance of an ammeter to make it ‘perfect’.
The same battery is now connected across two fixed
X
Y
Z
X
Y
Z
resistors as shown in Figure 14.38 by Circuit 3.
47 k Ω
47 k Ω
47 k Ω
47 k Ω
g) Calculate the p.d. across points X and Y by Circuit 3.
Circuit 3
47 k Ω
A voltmeter with the same resistance as the fixed
V
resistors is now connected in parallel with X and Y as
shown by Circuit 4.
Circuit 4
h) Calculate the combined resistance across points
Figure 14.38 Circuit diagram for questions 14 g) to l).
X and Y in Circuit 4.
i) Calculate the p.d. across points X and Y in Circuit 4.
j) Explain why the p.d. across X and Y in Circuit 3 is
larger than the p.d. across X and Y in Circuit 4.
k) Explain why the best voltmeter to connect across X and
Y would be one with an extremely high resistance.
469884_14_AQA_A-Level_Physics_244-273.indd 258
17/04/19 10:23 AM
Potential dividers are simple, three-component circuits designed to control
the potential difference in a circuit.
They consist of a power supply (such as a cell); a fixed resistor; and a
third resistive component, whose resistance can be fixed (fixed resistor) or
variable (variable resistors, thermistors, light-dependent resistors, etc.).
ε
I
R1
R2
V1
V2
Figure 14.39 Circuit diagram of
a potential divider.
Potential dividers
●● Potential dividers
All of these components are connected in series and the emf of the power
supply is shared across the two resistive components.
The clue to how potential dividers work is in their name, potential dividers,
i.e. something that splits up potential difference. Look at the potential
divider circuit in Figure 14.39.
Generally we are trying to vary the potential difference V1 (across R1) by
varying R2.
Assuming that the cell has negligible internal resistance, then the total
resistance of the circuit RT is given by:
RT = R1 + R2
The current, I, flowing through the circuit is given by Ohm’s law:
ε
ε
I= =
RT (R1 + R2)
Then considering the resistor R1 we can write:
V1 = IR1
Substituting for I from above:
ε
× R1
V1 =
(R1 + R2)
so
εR1
V1 =
(R1 + R2)
From this equation it can be seen that if ε and R1 are fixed, then V1 only
depends on R2. In fact, as R2 increases, V1 decreases, and vice versa.
259
Figure 14.40 A wire potentiometer.
469884_14_AQA_A-Level_Physics_244-273.indd 259
17/04/19 10:23 AM
ACTIVITY
Analysing the results from a wire potentiometer
A potential divider can be made out of a length of high-resistance metal
wire such as the metal alloy nichrome. The wire is stretched out along
a ruler and a metal probe is used to make contact with the wire along
its length. This piece of apparatus is called a potentiometer, as shown in
Figure 14.40.
The circuit is shown in Figure 14.41:
A particular potentiometer has a maximum wire length of 1.0 m. The
potentiometer is connected to a 6.0 V battery of negligible internal
resistance. A student varies the distance, l1, between the fixed
connection end of the voltmeter and the sliding contact, and measures
the p.d. for a range of different wire lengths up to 1.0 m.
ε
l1
l2
V
Figure 14.41 Circuit for a potentiometer.
14 Electrical circuits
Here are her results:
260
p.d. across wire, V (V), ±0.01 V
Length of wire,
l1 (m), ±0.002 m
1
2
3
0.000
0.00
0.00
0.00
0.100
0.60
0.80
0.61
0.200
1.20
1.04
1.02
0.300
1.80
1.87
1.86
0.400
2.40
2.37
2.46
0.500
3.00
2.84
2.97
0.600
3.60
3.52
3.56
0.700
4.20
4.19
4.37
0.800
4.80
4.77
4.84
0.900
5.40
5.48
5.24
1.000
6.00
6.00
5.87
Mean
Uncertainty
1 Make a copy of the table.
2 Check the data for any anomalous results. In this experiment these are
normally due to pressing too hard on the wire with the slider. Deal with
these anomalies in the usual way (ignore them) and calculate the
mean p.d. across the wire for each length.
3 Use the spread of the data for each length to assign an uncertainty to
the p.d. of each length.
4 Plot a graph of the mean p.d. (y-axis) against length (x-axis), using the
uncertainties and the precisions to plot vertical and horizontal error
bars for each point.
5 Plot a best-fit line through the points. Should (0, 0) be a known fixed
point on this best-fit line?
6 Use the best-fit line to determine an equation for the relationship
between V and l.
7 The error bars that you have drawn will give you a confidence about the
best-fit line and, therefore, the relationship between the two variables,
V and I. Explain how the spread of the data and their error bars can help
you to decide the position of the best-fit line.
8 State and explain how her original results would differ if she used a
lower-resistance analogue voltmeter.
469884_14_AQA_A-Level_Physics_244-273.indd 260
17/04/19 10:23 AM
I
R1
R2
V1
V2
Another use of the potential divider is in sensor circuits. If the variable resistor,
R2, in the potential divider circuit shown in Figure 14.42 is replaced with a
component whose resistance varies with an external physical variable such
as temperature or light intensity, then the potential difference across the fixed
resistor, R1, will also vary with the external physical variable.
The voltmeter connected across R1 can then be calibrated in terms of the
value of the external physical variable. The circuit can then be made to act
as an electronic thermometer or an electronic light sensor.
Figure 14.42 Circuit diagram
for a potential divider.
With an electronic thermometer, a thermistor is connected in
place of R2. Most thermistors are ntc (negative temperature
coefficient) thermistors. This means that as the temperature
increases, their resistance decreases, as shown in Figure 14.43,
which also shows a circuit diagram showing how the components
are connected:
resistance
ε
I
Potential dividers
Potential dividers as sensors
ε
R1
temperature
V1
The effect on the potential difference across the fixed resistor R1 is that
as the temperature increases, so the resistance of the thermistor, R2,
Figure 14.43 A thermistor potential divider
drops and V1 rises. Thus increasing temperature produces increasing
circuit and the resistance–temperature graph
for an ntc thermistor.
p.d. If the voltmeter is connected across the thermistor, the opposite
will happen: increasing the temperature will produce a decrease in p.d. In
most cases, applications require the p.d. to increase with temperature, so the
voltmeter is usually connected across the fixed resistor.
EXAMPLE
The resistance–temperature graph for an ntc thermistor is shown in Figure 14.44.
100
90
thermistor resistance/kΩ
80
70
60
50
40
30
20
10
0
20
10
0
10
20
30
40
temperature/C
50
60
70
261
Figure 14.44 Resistance–temperature graph for an ntc thermistor.
The thermistor is connected into a potential divider circuit with a 6.0 V battery
of negligible internal resistance and 45 kΩ fixed resistor.
At −10 °C the resistance of the thermistor is 55 kΩ. At 40 °C the resistance of
the thermistor is 5 kΩ.
a) Calculate the voltage across the fixed resistor at −10 °C.
b) Calculate the voltage across the fixed resistor at 40 °C.
c) What effect does increasing the temperature from −10 °C to 40 °C have
on the voltage?
469884_14_AQA_A-Level_Physics_244-273.indd 261
17/04/19 10:23 AM
Answer
a) V1 =
εR1
6.0 V × 45 kΩ
= 2.7 V
=
(R1 + R2)
(45 + 55) kΩ
b) V1 =
εR1
= 6.0 V × 45 kΩ = 5.4 V
(45 + 5) kΩ
(R1 + R2)
c) Increasing the temperature of the thermistor from −10 °C to 40 °C
doubles the voltage (in this case).
A similar effect involves the use of a light-dependent resistor (LDR).
These are components that change their resistance with light intensity,
which makes them incredibly useful as light sensors. LDRs are made of
semi-conducting materials; as the light shines on the material, electrons
are freed from the structure and can flow through it, reducing the
resistance of the LDR.
In the dark, LDRs can have a very high resistance, of the order of megaohms, and yet in the light their resistance can drop as low as a few hundred
ohms. If an LDR is connected into a potential divider in place of R2 in the
potential divider circuit then, as the light intensity increases, then so will
the output p.d. V1 across the fixed resistor.
resistance
ε
I
R1
14 Electrical circuits
light intensity
262
(a)
(b)
(c)
V1
Figure 14.45 An LDR: its electrical circuit symbol (a) resistance–
light intensity graph (b) and its use in a potential divider circuit (c).
EXAMPLE
A particular LDR inside a digital camera has
a resistance of 150 kΩ in low light levels and a
resistance of 720 Ω in bright sunlight. The LDR is
connected to a 55 kΩ fixed resistor and a 3.0 V battery.
Calculate the p.d. across the fixed resistor in:
a) low light levels
b) bright sunlight.
Answer
b) In bright sunlight the p.d. across the fixed 55 kΩ resistor is:
εR1
(R1 + R2)
3.0 V × 55 kΩ
=
(0.72 + 55) kΩ
V1 =
= 3.0 V
(2 s.f.)
a) In low light levels the p.d. across the fixed 55 kΩ resistor is:
V1 =
εR1
3.0 V × 55 kΩ
=
= 0.80 V
(R1 + R2) (150 + 55) kΩ
469884_14_AQA_A-Level_Physics_244-273.indd 262
17/04/19 10:23 AM
15 Figure 14.46 shows part of the automatic light
sensor for a digital camera. The camera battery
has an emf of 6.0 V and has negligible internal
resistance. The voltmeter is a digital type with a
very high resistance.
6.0 V
330Ω
450Ω
V
Figure 14.46 Circuit diagram for question 15.
In bright daylight the resistance of the LDR is 450 Ω.
a) Calculate the ratio
p.d. across resistor
p.d. across LDR
b) Calculate the voltmeter reading in bright daylight.
At low light levels, the resistance of the LDR
increases to 470 kΩ.
16 An ntc thermistor is connected in series with
a 660 Ω fixed resistor and a 6.0 V battery of
negligible internal resistance forming a potential
divider circuit for an electric thermometer. A
high-resistance voltmeter is connected across
the fixed resistor.
a) Draw a circuit diagram for this circuit.
b) At 25 °C the thermistor has a resistance
of 1.5 kΩ. Calculate the current in the
circuit.
c) Calculate the p.d. across the 660 Ω resistor.
d) The thermistor is heated to 90 °C and
its resistance drops to 220 Ω. State and
explain what would happen to the voltmeter
reading as the temperature changes from
25 °C to 90 °C.
e) At 90 °C, when the thermistor has a
resistance of 220 Ω the thermometer
develops a fault and the current suddenly
rises to 0.4 A. Calculate the power loss in
the thermistor immediately after the fault.
f) State and explain what is likely to happen to
the thermistor in the circuit.
Potential dividers
TEST YOURSELF
c) Recalculate the ratio
p.d. across resistor
p.d. across LDR
at this lower light level.
d) Calculate the voltmeter reading in low light.
263
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Practice questions
1 A battery of emf 12 V and with negligible internal resistance is
connected to the resistor network shown in Figure 14.47.
What is the current through the 47 Ω resistor?
A 0.1 A
C 0.3 A
B 0.2 A
D 0.4 A
C 0.71 A
B 0.20 A
D 5.2 A
B
25
0.04
30 Ω
14 Electrical circuits
30 Ω
2.0 Ω
12 V
Figure 14.48 Circuit diagram for
question 2.
Q (Ω)
50
75.04
C
25
112.5
D
225
50
You need the following information for questions
4 and 5.
264
91 Ω
Figure 14.47 Circuit diagram for
question 1.
Each resistor has a resistance of 75 Ω. Which line represents the
equivalent resistance for each arrangement of resistors?
A
24 Ω
12 V
3 A student connects together three resistors together in the
arrangement shown in P and Q in Figure 14.49.
P (Ω)
180 Ω
47 Ω
2 A battery has an emf of 12 V and an internal resistance of 2.0 Ω.
Calculate the total current in the circuit shown in Figure 14.48.
A 0.19 A
27 Ω
P
Q
Figure 14.49 Circuit diagram for question 3.
A resistor and a thermistor are being used as part of a
temperature-sensing circuit. They are connected in series with a 12 V
battery of negligible internal resistance as shown in Figure 14.50.
B
A
4 At 200 °C the resistance of the thermistor is 18 Ω and
the resistance of the resistor is 130 Ω.
What is the voltage across the terminals AB at this temperature?
A 1.13 V
C 10.54 V
B 1.46 V
D 10.86 V
12 V
5 What is the power dissipated by the resistor at 200 °C?
A 0.08 W
C 0.85 W
B 0.12 W
D 0.97 W
Figure 14.50 Circuit diagram
for question 4.
3.0 Ω
4.0 Ω
6 Three resistors are connected to a battery of emf 12 V and a
negligible internal resistance as shown in Figure 14.51.
What is the voltage across the 6.0 Ω resistor?
A 2V
C 8V
B 4V
D 10 V
6.0 Ω
12 V
Figure 14.51 Circuit diagram for
question 6.
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Practice questions
7 A torch bulb is connected to a battery of negligible internal resistance.
The battery supplies a steady current of 0.25 A for 20 hours. In this time
the energy transferred by the bulb is 9.0 × 104 J.
What is the power of the bulb?
A 1.3 W
C 4500 W
B 21 W
D 5625 W
8 Electric blankets have wires in the centre. The energy dissipated by the
wires when a current flows through them heats the blanket and the
bed. One blanket has a power rating of 63 W and is plugged into mains
electricity (230 V).
What is the total resistance of the wire in the blanket?
A 4Ω
C 840 Ω
B 17 Ω
D 15 000 Ω
9 A cell of emf 1.5 V and internal resistance 1.0 Ω is connected to a 5.0 Ω
resistor to form a complete circuit. Calculate the current in this circuit.
A 0.25 A
C 0.60 A
B 0.30 A
D 1.5 A
10 An electrical generator produces 100 kW of power at a potential difference
of 10 kV. The power is transmitted through cables of total resistance 5 Ω.
What is the power loss in the cables?
A 50 W
C 500 W
B 250 W
D 1000 W
50 Ω
800 Ω
11 Car batteries have a typical emf of 12 V and a very low internal resistance
of 5.0 mΩ.
a) Explain what is meant by the terms ‘emf’ and ‘internal resistance’.
(2)
b) One such battery delivers a current of 500 A to the starter motor of
a car. Calculate the potential difference across the starter motor. (2)
12 Figure 14.52 shows three resistors connected to a battery of negligible
internal resistance.
a) Calculate the total resistance of the circuit.
(2)
b) The power dissipated by each of the 800 Ω resistors is
2.0 W. Calculate the p.d. across the two 800 Ω resistors.
(2)
c) Calculate the current through the 50 Ω resistor.
(2)
d) Calculate the emf of the battery.
(1)
800 Ω
Figure 14.52 Circuit
diagram for question 12.
P
R S
265
Q
13 Figure 14.53 shows four identical new cells, each with an emf of 1.2 V
and negligible internal resistance, connected to three identical resistors
each with a resistance of 5 Ω.
a) Calculate the total resistance of the circuit.
(2)
b) Calculate the total emf of the cells.
(1)
469884_14_AQA_A-Level_Physics_244-273.indd 265
Figure 14.53 Circuit
diagram for question 13.
17/04/19 10:23 AM
c) Calculate the current passing through cell Q.
(2)
d) Calculate the total charge passing through cell Q in 1 second.
(2)
e) Each of the cells shown in the circuit diagram stores the
same amount of chemical energy, and hence can transfer
the same amount of electrical energy. State and explain
which two cells in the circuit would transfer electrical
energy for the longest time period.
(2)
14 A car engine temperature-sensor circuit consists of a 12 V car battery
connected in series with a low-resistance ammeter, a 330 Ω resistor
and then connected to an ntc (negative temperature coefficient)
thermistor and a 1 kΩ resistor connected in parallel to each other
before returning to the battery as shown in Figure 14.54.
12 V
A
330 Ω
At its normal working temperature, the current reading on the
ammeter is 14.0 mA.
a) Calculate the p.d. across the 330 Ω resistor.
(2)
b) Calculate the p.d. across the 1 kΩ resistor.
(2)
c) Calculate the combined resistance of the thermistor and
1 kΩ resistor parallel combination.
(2)
d) Calculate the resistance of the thermistor.
(3)
1kΩ
Figure 14.54 Circuit diagram for
question 14.
e) The engine starts to overheat and the temperature of the thermistor
starts to rise. State and explain the effect that this has on the current
measured by the ammeter.
(2)
14 Electrical circuits
15 A thermistor is connected to a combination of fixed resistors
and a 6 V battery of negligible internal resistance, as shown in
Figure 14.55.
266
6V
a) At room temperature the resistance of the thermistor, T, is
2.5 kΩ. Calculate the total resistance of the circuit.
(3)
b) Calculate the current flowing through the 6 V battery.
(1)
c) A very high-resistance digital voltmeter is used to measure the
p.d. across different points in the circuit. Copy and complete
the table below by calculating the relevant p.d.s.
(3)
Position of the voltmeter
C-E
D-F
C-D
C-E
C
10 kΩ
E
p.d., V (V)
5 kΩ
B
d) The thermistor in the circuit is heated and its resistance
decreases. State and explain the effect that this has on the
voltmeter reading in the following positions:
i)
10 kΩ
A
D
T
F
Figure 14.55 Circuit diagram for question 15.
(4)
ii) D-F.
16 An LDR is connected in series with a fixed resistor, R, and a 3.0 V
battery with negligible internal resistance. A high-resistance voltmeter is
connected in parallel across the fixed resistor.
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Practice questions
(2)
a) Draw a circuit diagram of this circuit.
b) On a cloudy day when the light intensity is low, the resistance of
the LDR is 1.5 kΩ and the voltmeter reads 0.80 V. Calculate the p.d.
across the thermistor.
(2)
c) Calculate the current flowing through the battery.
(2)
d) Calculate the value of the resistance, R, of the fixed resistor.
(2)
e) On a bright sunny day the resistance of the LDR falls to 150 Ω. State
and explain what effect this has on the reading on the voltmeter. (2)
f) State and explain the effect on the current flowing though the battery
if the resistance of the voltmeter was the same as
the resistance of the fixed resistor.
(2)
17 Four resistors are connected into a circuit together
with a 12 V car battery of negligible internal resistance.
The current as measured by the low-resistance
ammeter is 2.2 A. The circuit is shown in Figure 14.56.
a) Calculate the total effective resistance of the
circuit.
A
Q
P
(2)
b) All the resistors have exactly the same resistance.
Calculate the resistance of resistor P.
(3)
c) Calculate the current flowing through resistor R. (3)
d) Calculate the power dissipated by resistor P.
R
S
Figure 14.56 Circuit diagram for question 17.
(2)
18 Two resistors are connected in series with a 12 V battery with
negligible internal resistance as shown in Figure 14.57.
12 V
a) The resistance R2 is 150 Ω and the voltmeter reads 7.5 V.
Calculate the current in the circuit.
(1)
b) Calculate the power dissipated by resistor R2.
(2)
c) Calculate the resistance R1.
(2)
R1
R2
d) Resistor R2 is replaced with an ntc (negative temperature
V
coefficient) thermistor. Explain why initially the voltmeter reading
remains constant, but slowly it changes and then reads another
Figure 14.57 Circuit diagram for
constant value. (2) question 18.
19 An LDR (light-dependent resistor), a fixed resistor R and a variable
resistor are connected in series with a 9.0 V battery with negligible
internal resistance. A high-resistance voltmeter is connected in parallel
with the fixed resistor, R, as shown in Figure 14.58 (overleaf).
267
a) In the dark, the resistance of the LDR is 120 kΩ, the resistance of the
fixed resistor is 8.0 kΩ and the variable resistor has a resistance of
42 kΩ. Calculate the current in the circuit flowing through the battery. (2)
b) Calculate the reading of the voltmeter.
(2)
c) The circuit is then taken out into the daylight. The resistance of the
LDR drops quickly to a small value. State and explain the effect on the
reading on the voltmeter.
(2)
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d) This light-sensing circuit forms part of an automatic
light cut-off for a greenhouse. At a predetermined light
intensity, the p.d. across the fixed resistor needs to be
0.90 V. At this light intensity, the resistance of the LDR is
12 kΩ. Calculate the required resistance of the variable
resistor for this particular light level.
(2)
9.0 V
R
20 The two sidelight bulbs of a car are rated as 12 V, 32 W.
The bulbs are connected in parallel to a 12 V car battery of
negligible internal resistance; an ammeter is connected in
series with the battery to measure the total current drawn
from the battery.
a) Draw a circuit diagram of this circuit.
(2)
b) Calculate the reading of the ammeter.
(2)
V
Figure 14.58 Circuit diagram for question 19.
c) Calculate the resistance of the bulbs.
(2)
d) State and explain the effect on the brightness of the bulbs if the
ammeter has a higher resistance.
(2)
Both of the bulbs are now connected to the battery in series
with the ammeter.
e) State and explain the change of brightness of the two bulbs in the
series circuit.
(2)
14 Electrical circuits
f) Apart from the brightness of the bulbs, suggest another reason why
connecting the bulbs in parallel would be preferable to connecting
them in series.
(1)
268
21 The internal lights on a helicopter consists of two
bulbs connected in parallel to a 24 V battery of
negligible internal resistance. The bulb in the cockpit
needs to be dimmer than the one in the cabin. Two
resistors are put into the circuit to ensure that the
bulbs are both operating at their working voltage, as
shown in Figure 14.59.
a) Bulb X is rated at 12 V, 36 W and bulb Y at 4.5 V,
2.0 W. Use this data to calculate the current
flowing though each bulb when it is operating at
(3)
its working p.d. b) Calculate the p.d. across resistor R1.
(2)
c) Calculate the current flowing through R1.
(1)
d) Calculate the resistance of R1.
(1)
24 V
R1
X
V
Figure 14.59 Circuit diagram for question 21.
e) The p.d. across bulb Y must be 4.5 V. Calculate
the p.d. across resistor R2.
(1)
f) Calculate the resistance of R2.
(2)
g) Bulb Y breaks. No current runs through bulb Y and resistor R2.
Explain, without using a calculation, the effect on the voltmeter
reading across R1.
(2)
h) State and explain what happens to bulb X.
(2)
469884_14_AQA_A-Level_Physics_244-273.indd 268
Y
R2
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ε
r
I
Figure 14.61 shows the results from an experiment involving
a D-type cell.
A
R
1.60
1.40
V
Figure 14.60 Circuit diagram for
question 22.
1.20
potential difference, V/V
Practice questions
22 The circuit in Figure 14.60 can be used to determine the emf and
internal resistance of a battery.
1.00
0.80
0.60
0.40
0.20
0.00
0.00
0.50
1.00
1.50
2.00
2.50
current, I/A
3.00
3.50
4.00
Figure 14.61 Electrical characteristic of a D-type cell.
a) Explain why the potential difference decreases as the current increases.(2)
b) Use the graph to determine the emf, ε, and the internal
resistance, r, of the D-type cell.
(3)
c) On a suitable sketch of the graph, draw on a line that illustrates the
results obtained from another battery with the same emf but double
the internal resistance of the first battery. Label this line X. (2)
d) On the same sketch draw another line that illustrates the results
obtained by a third battery with the same emf as the other two, but
with negligible internal resistance. Label this line Y.
(2)
B
Stretch and challenge
All of the following questions are
provided with permission of the
British Physics Olympiad
269
23 Twelve 1 ohm resistors are connected in a
cubic network, as shown in Figure 14.62.
a) Draw a two-dimensional circuit
diagram of this network.
b) Use your circuit diagram to determine
the overall resistance of this network
between points A and B.
469884_14_AQA_A-Level_Physics_244-273.indd 269
A
Figure 14.62 Circuit diagram for question 23.
17/04/19 10:23 AM
24 If the potential difference between A and B in Figure 14.63 is V,
then what is the current between A and B? All three resistors
are identical.
V
3V
C
A
3R
2R
R
A
2V
D 3VR
B
3R
2
25 A 10 V battery with negligible internal resistance is connected to
two resistors of resistance 250 Ω and 400 Ω, and to component Z,
as shown in Figure 14.64.
C 12.5 mA
B 7.5 mA
D 15.4 mA
26 In the circuit shown in Figure 14.65 the resistors have identical
resistances. If the power converted in R1 is P, what is the power
dissipated in R2?
A P/4
C P
B P/2
D 2P
Figure 14.63 Circuit diagram for
question 24 .
250 Ω
10 V
14 Electrical circuits
270
A 1V
C 3V
B 2.5 V
D 5V
28 A cell that produces a potential E (called an emf) is shown in
Figure 14.67 and is connected to two resistors in series: a fixed
resistor R1 and a variable resistor R2.
The current, I, in the circuit is measured by the ammeter, A,
and the potential difference, V, across resistor R2 is measured
with the voltmeter, V. The relationship between the
potential, E, and the current, I, is given by:
400 Ω
Z
Figure 14.64 Circuit diagram for
question 25.
27 Three identical voltmeters each have a fixed resistance, R, which
allows a small current to flow through them when they measure
a potential difference in a circuit. The voltmeters, V1, V2, V3 are
connected in the circuit as shown in Figure 14.66.
The voltage–current characteristics of the device D are
unknown. If V2 reads 2 V and V3 reads 3 V, what is the
reading on V1?
B
R
Z is a device that has the property of maintaining a
potential difference of 5 V across the 400 Ω resistor. The
current through Z is:
A 2.9 mA
R
R1
R2
R3
Figure 14.65 Circuit diagram for
question 26.
V1
V2
V3
D
Figure 14.66 Circuit diagram for
question 27.
E
ε = IR1 + IR2
Which of these graphs would produce a straight-line fit?
A V against I
C 1/V against 1/I
B V against 1/I
D I against 1/V
A
R2
R1
V
Figure 14.67 Circuit diagram for question 28.
469884_14_AQA_A-Level_Physics_244-273.indd 270
17/04/19 10:23 AM
R1
R2
A
C
R5
B
A potential difference of 1.5 V is connected in turn across
various points in the arrangement.
D
R3
R4
Figure 14.68 Circuit diagram for
question 29.
Practice questions
29 A combination of resistors shown in Figure 14.68
represents a pair of transmission lines with a fault in
the insulation between them. The wires have a uniform
resistance but do not have the same resistance as each
other. The following procedure is used to find the value
of the resistor R5.
With 1.5 V applied across terminals AC a current of 37.5 mA flows.
With 1.5 V applied across terminals BD a current of 25 mA flows.
With 1.5 V applied across terminals AB a current of 30 mA flows.
With 1.5 V applied across terminals CD a current of 15 mA flows.
a) Write down four equations relating the potential difference, the
resistor values and the currents.
b) Determine the value of resistor R5.
c) If the ends C and D are connected together, what would be the
resistance measured between A and B?
d) If the length AC (and also BD) is 60 metres, how far from A (or C)
does the fault occur?
30 A student decides to calibrate a thermistor in order to measure
variations in the temperature of a room. He connects a small beadsized thermistor across the terminals of a 5 V power supply and in
series with a 1 A ammeter. The resistance of the thermistor is 120 Ω
at room temperature.
5V
a) Instead of showing small variations in room temperature, the
thermistor is likely to go up in smoke. Explain why.
In the light of his experience, he decides to redesign his simple circuit as
shown in Figure 14.69. He has a few values of resistor R to choose from:
5 kΩ, 500 kΩ and 50 Ω.
b) State which value of R would give the biggest variation of V with
temperature. Explain your choice.
c) State which value of R would be most likely to cause the same
problem as in part (a). Again, explain your choice.
R
V
0V
Figure 14.69 Circuit
diagram for question 30.
271
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31 A single uniform underground cable linking A to B, 50 km
long, has a fault in it at distance d km from end A, as shown in
Figure 14.70. This is caused by a break in the insulation at X so
that there is a flow of current through a fixed resistance, R, into
the ground. The ground can be taken to be a very low-resistance
conductor. Potential differences are all measured with respect to the
ground, which is taken to be at 0 V.
d
A
X
50 – d
B
R
Figure 14.70 Circuit diagram.
In order to locate the fault, the following procedure is used.
A potential difference of 200 V is applied to end A of the cable.
End B is insulated from the ground, and it is measured to be at a
potential of 40 V.
a) What is the potential at X? Explain your reasoning.
b) What is:
i)
the potential difference between A and X?
14 Electrical circuits
ii) the potential gradient along the cable from A to X
(i.e. the volts/km)?
272
The potential applied to end A is now removed and A is
insulated from the ground instead. The potential at end B is
raised to 300 V, at which point the potential at A is measured
to be 40 V.
c) What is the potential at X now?
d) Having measured 40 V at end B initially, why is it that 40 V
has also been required at end A for the second measurement?
e) What is the potential gradient along the cable from B to X?
f) The potential gradient from A to X is equal to the potential
gradient from B to X.
i)
Explain why this is true.
ii) From the two potential gradients that you obtained earlier,
deduce the value of d.
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32 This question is about heating elements.
b) A student goes out to purchase an electric heater for their flat.
The sales person says that to get more heat they should purchase
a heater with a high resistance because P = I2R, but the student
2
thinks that a low resistance would be best because P = VR .
Explain who is correct.
c) Copper is a better conductor than iron. Equal lengths of copper
and iron wire, of the same diameter, are connected first in
parallel and then in series as shown by Figure 14.71. A potential
difference is applied across the ends of each arrangement in turn, and
the p.d. is gradually increased from a small value until, in each case,
one of the wires begins to glow. Explain this, and state which
wire will glow first in each case.
Case 1
Practice questions
a) The power dissipated as heat in a resistor in a circuit is given by
P = VI. Show that this may also be expressed as P = I2R
2
and P = VR .
Case 2
Figure 14.71 Copper and iron wire connected in different arrangements.
d) A surge suppressor is a device for preventing sudden excessive
flows of current in a circuit. It is made of a material whose
conducting properties are such that the current flowing through
it is directly proportional to the fourth power of the potential
difference across it. If the suppressor transfers energy at a rate of
6 W when the applied potential difference is 230 V, what is
the power dissipated when the potential difference rises to 1200 V?
273
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15
Circular motion
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
Acceleration =
●
●
●
●
●
change of velocity
time
Resultant force = mass × acceleration.
You need to recall that a vector quantity has magnitude and direction:
force, velocity and acceleration are vectors.
Resolving a vector into components.
Circumference of a circle = 2π × radius of circle, c = 2πr.
Newton’s first law of motion: a body remains at rest or continues
to move in a straight line at a constant speed unless acted on by an
unbalanced force.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 a) Explain the difference between speed and velocity.
b) Explain why acceleration is a vector quantity.
2
The three diagrams in Figure 15.1 show three separate examples
of how a vehicle’s velocity changes from v1 to v 2 over a time of 10 s.
15 CIRCULAR MOTION
Use the equation
274
469884_15_AQA_A-Level_Physics_274-284.indd 274
a=
v 2 −v 1
t
ν1 = 3 m s–1
ν2 = 4 m s–1
to calculate the magnitude
(a)
and direction of the
acceleration in each case.
ν1 = 3 m s–1
ν2 = 4 m s–1
3 The vehicle in question 2 has
(b)
a mass of 2 kg. In each case
shown in Figure 15.1, calculate
ν2 = 4 m s–1
the average resultant force that (c) ν1 = 3 m s–1
caused the acceleration of the
vehicle.
4 An unbalanced force acts on a
moving vehicle. Explain three
changes that could occur to the Figure 15.1
vehicle’s velocity.
5 You walk a quarter of the way round a circle of diameter 20 m.
a) Calculate the distance you have walked.
b) Calculate your displacement if you started at the north of the circle
and walked to the eastern side of the circle.
17/04/19 10:24 AM
s
●● Circular measure
B
You are used to measuring angles in degrees, but in physics problems
involving rotations we use a different measure.
r
In Figure 15.2, an arc AB is shown. The length of the arc is s, and the
radius of the circle is r. We define the angle θ as
θ
C
Figure 15.2
Radian The radian measure of a central
angle of a circle, θ, is defined as the ratio of
the arc length, s, subtended by the angle θ,
to the radius r:
θ = rs
θ = rs
The advantage of this measure is that θ is a ratio of lengths, so it has no
unit. However, to avoid the confusion that the angle might be measured in
degrees, we give this measure the unit radian, abbreviated to rad.
Circular measure
A
Since the circumference of a circle is 2πr, it follows that 2π radians is the
equivalent of 360°:
2π rad = 360°
so
360°
2π
= 57.3°
1 rad =
Equations of rotation
Angular displacement The angle
(measured in radians) through which a line
rotates about a fixed point.
Angular velocity The rate of change of
angular displacement (measured in radians
per second).
EXAMPLE
The ‘big wheel’
When something rotates about a fixed point we use the term angular
displacement to measure how far the object has rotated. For example, in
Figure 15.2, when an object rotates from A to B, its angular displacement is
θ radians.
The term angular velocity, ω, is used to measure the rate of angular
rotation. Angular velocity has units of radians per second or rad s−1:
ω=
θ
t
ω=
∆θ
∆t
or
A ‘big wheel’ at a funfair takes its
passengers for a ride, completing
six complete revolutions in 120 s.
1 Calculate the angular
displacement of the wheel.
Answer
where ∆θ is the small angle turned into a small time ∆t.
In general, there is a useful relationship connecting the time period of one
complete rotation, T, and angular velocity, ω, because after one full rotation
the angular displacement is 2π:
θ = 6 × 2π = 12π rad = 37.7 rad
ω=
2 Calculate the average angular
velocity during the ride.
275
or
Answer
ω=
2π
T
ω = 2πf
12 π
120 s
−1
= 0.1π = 0.31rad s
where f is the frequency of rotation. There is a further useful equation,
which connects angular velocity with the velocity of rotation. Since
s = θr
469884_15_AQA_A-Level_Physics_274-284.indd 275
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same
ω
and
1m
2m
v=
∆ s ∆ θ
r
=
∆ t ∆ t
then
v = ωr
This equation shows that the rotational speed of something is faster further
away from the centre. For example, all the children on a roundabout in a
playground have the same angular velocity ω, but the ones near the edge
are moving faster (Figure 15.3).
Figure 15.3
TEST YOURSELF
1 The Earth has a radius of 6400 km. The Shetland
Isles are at latitude of 60°.
a) Calculate the angular velocity of the Earth.
b) Calculate the velocity of rotation of a point on
the equator.
c) Calculate the velocity of rotation of the Shetland
Isles.
2 A proton in a synchrotron travels round a circular path
of radius 85 m at a speed of close to 3.0 × 108 m s−1.
a) Calculate the time taken for one revolution of
the synchrotron.
A
ν1
B
C
ν2
●● Centripetal acceleration
15 CIRCULAR MOTION
r
276
b) Calculate the frequency of rotation of the
protons.
c) Calculate the proton’s angular velocity.
3 The Sun rotates around the centre of our Galaxy,
the Milky Way, once every 220 million years, in an
orbit of about 30 000 light years.
a) Calculate the angular velocity of the Sun about
the centre of the Milky Way.
Calculate the velocity of the Sun relative to the
centre of the Galaxy.
[1 light year = 9.47 × 1015 m; 1 year = 3.16 × 107 s]
∆θ
In Figure 15.4 a particle is moving round a circular path at a constant
speed v, and because it is continually changing direction the particle is
always accelerating.
It is easier to understand the acceleration when you recall the formula:
O
acceleration =
change of velocity
time
Velocity is a vector quantity, so if the direction of the motion changes, even
though there is no change of speed, there must be an acceleration.
Figure 15.4 A particle moving round a
circular path with a constant speed is
always accelerating.
ν2
∆ν
∆θ
–ν1
Figure 15.5
469884_15_AQA_A-Level_Physics_274-284.indd 276
Figure 15.5 shows the direction of the acceleration. In going from position
A to position C, the particle’s velocity changes from v1 to v2. So the change
in velocity, Δv, is the vector sum v2−v1.
The diagram shows the change in velocity, Δv, which is directed along the
line BO, towards the centre of the circle. So, as the particle moves around
the circular path, there is an acceleration towards the centre of the circle.
This is called the centripetal acceleration. Because this acceleration is at
right angles to the motion, there is no speeding up of the particle, just a
change of direction.
17/04/19 10:25 AM
a=
The size of the acceleration, a, is calculated using this formula:
a=
v2
r
or because v = ωr
a = ω2r
v2
r
Here v is the constant speed of the particle, ω is its angular velocity, and r is
the radius of the path.
Centripetal force
Centripetal acceleration When a
particle moves in a circular path of radius
r, at a constant speed v, there must be a
centripetal acceleration towards the centre
of the circle, given by
MATHS BOX
You are not expected to be able to derive the formula for centripetal
acceleration, but it is given here for those who want to know where the
formula comes from.
In Figure 15.4, the particle moves from A to C in a small time Δt. We
now look at the instantaneous acceleration at the point B, by considering
a very small angle Δθ. The distance travelled round the arc AC, Δs, is
given by
Δs = v Δt
and
Δs = r Δθ
so
r Δθ = v Δt
curved track
and ∆ θ = v ∆ t
r
(i)
In Figure 15.5 the angle θ is given by
∆ v (ii)
∆θ = v
ball
v
provided Δθ is very small. Then by combining equations (i) and (ii), it
follows that
B
∆ v = v ∆t
v
r
or
2
a = ∆ v = vr
∆t
v
ball
A
straight track
Figure 15.6 A train carriage
turns a corner, but a ball on
the floor of the carriage keeps
on moving in a straight line.
469884_15_AQA_A-Level_Physics_274-284.indd 277
●● Centripetal force
277
Figure 15.6 illustrates the path of a railway carriage as it turns round a
corner (part of a circle), moving from A to B at a constant speed v. The
rails provide a force to change the direction of the carriage. However,
a ball that is placed on the floor behaves differently. The ball carries on
moving in a straight line until it meets the side of the carriage. The ball
experiences no force, so, as predicted by Newton’s first law of motion, it
carries on moving in a straight line at a constant speed, until the side of
the carriage exerts a force on it.
17/04/19 10:25 AM
(a)
(b)
(c)
W
T
R
T
T
W
W
Figure 15.7 Views of a ball suspended
from the back of the train carriage
looking forwards. (a) When the train
moves along a straight track, the ball
hangs straight down. (b) When the train
moves around the curved track, as in
Figure 15.6, the ball is displaced to the
right. (c) There is a resultant unbalanced
force R acting on the ball.
15 CIRCULAR MOTION
Centripetal force When an object moves
around a circular path, there must be a
centripetal force acting towards the centre
of the circle. Something must provide this
force, such as a pull from a string or a push
from the road.
278
Now suppose that the ball is suspended from the ceiling of the carriage and
the experiment is repeated. Figure 15.7 illustrates what happens now as the
carriage moves from a straight track to a curved track. In Figure 15.7a the
carriage moves along a straight track at a constant speed. The ball hangs
straight down and the forces acting on it balance: the tension of the string,
T, upwards, balances the ball’s weight, W, downwards.
In Figure 15.7b the train turns the corner. The ball keeps moving in a straight
line until tension in the string acts to pull the ball round the corner. Now the
forces acting on the ball do not balance. The vector sum of the tension T and
the weight W provides an unbalanced force R, which acts towards the centre
of the circle (Figure 15.7c).
This unbalanced force R provides the centripetal acceleration. So we can
write
mv2
R = r
where R is the unbalanced centripetal force, m is the mass of the ball, v
is the ball’s forward speed, and r is the radius of the (circular) bend it is
going round.
It is important to understand that a centripetal force does not exist
because something is moving round a curved path. It is the other way
around – according to Newton’s second law of motion, to make something
change direction a force is required to make the object accelerate. In
the example you have seen here, the tension in the string provides
the centripetal force, which is necessary to make the ball move in a
circular path. When a car turns a corner, the frictional force from the
road provides the centripetal force to change the car’s direction. When a
satellite orbits the Earth, the gravitational pull of the Earth provides the
centripetal force to make the satellite orbit the Earth – there is no force
acting on the satellite other than gravity.
A common misunderstanding
Figure 15.8 shows the same ball discussed earlier held hanging, at rest, at
an angle in the laboratory. Now it is kept in place by the balance of three
forces: the tension in the string, T, its weight, W, and a sideways push, P,
from a student’s finger.
θ
θ
T
W
(a)
(b)
P
W
Figure 15.8
469884_15_AQA_A-Level_Physics_274-284.indd 278
T
P
If the student removes his finger, the ball will accelerate and begin moving
to the left, because there is now an unbalanced force acting on it, exactly as
there was in Figure 15.7.
However, the situations are different. In Figure 15.8 the ball is
stationary until the finger is removed, and it begins to accelerate and
move in the direction of the unbalanced force. In Figure 15.7 the ball is
moving forwards and the action of the unbalanced force is to change the
direction of the ball.
17/04/19 10:25 AM
iii) Show that the angle at which the ball hangs
to the vertical is about 11.5°.
b) The train reaches a speed of 55 m s−1 and
travels round a curved piece of the track. At
this moment, the ball is deflected sideways by
about 11.5°.
i) State and explain the direction and
magnitude of the resultant force on the ball.
ii) Explain why the ball is accelerating. In which
direction is the acceleration? Calculate the
magnitude of the acceleration.
iii) Explain why the ball’s speed remains
constant.
iv) Calculate the radius of the bend the train is
going round.
c) The train carriage that carries the ball has a
mass of 40 tonnes.
i) Calculate the centripetal force that acts on
the carriage as it turns the corner. What
provides this force?
ii) Explain why trains go round tight bends at
reduced speeds.
4 Explain how a force can change the velocity of a
body without increasing its speed.
5 The force of gravity makes things fall towards the
ground. Explain why the Earth’s gravity does not
make the Moon fall towards the Earth.
6 A satellite is in orbit around the Earth, at a distance
of 7000 km from the Earth’s centre. The mass of
the satellite is 560 kg and the gravitational field
strength at that height is 8.2 N kg−1.
a) Draw a diagram to show the direction and
magnitude of the force (or forces) that act(s) on it.
b) Calculate the centripetal acceleration of the
satellite.
c) Calculate
i) the speed of the satellite
ii) the time period of its orbit.
7 This question refers to the suspended ball in the train,
illustrated in Figure 15.7. The ball has a mass of 0.15 kg.
a) The train accelerates forwards out of the station
along a straight track at a rate of 2 m s−2.
i) Explain why the ball is displaced backwards.
ii) Calculate the resultant force on the ball.
Centripetal force
TEST YOURSELF
ACTIVITY
Investigating centripetal forces
Figure 15.9 shows a way in which you can investigate
centripetal forces. The idea is that you whirl a rubber
bung around your head in a horizontal circle. The
bung is attached by a thin string to a plastic tube,
which is held vertically. A weight is hung on the
bottom of the string. This causes the tension to
provide the necessary centripetal force to keep the
bung moving in its circular path.
F
rubber bung
of mass m
thin string
plastic tube
Mg
Figure 15.9
●
●
Use a bung, of mass m, of about 50 g to 100 g.
Wear safety glasses (useful to protect yourself
from others doing the same experiment).
469884_15_AQA_A-Level_Physics_274-284.indd 279
●
●
●
r
F
●
●
A suitable plastic tube is an old case from a plastic
ballpoint pen.
The time of rotation, T, can be calculated by
measuring the time for 10 rotations, 10T.
The radius r can be measured after you have
finished 10 rotations by pinching the string with your
finger, then measuring the length from the
top of the tube to the centre of the bung.
We assume that there is no friction between the
plastic tube and the string.
It is assumed that the string is horizontal,
although this will not be entirely possible, so it is
important to try to meet this condition as far as
you can.
Table 15.1 shows some data measured by a student
doing this experiment.
279
1 Copy and complete Table 15.1 by filling in the gaps.
Comment on how well the results support the
hypothesis that the weight on the end of the string
causes the centripetal force to keep the bung in
its circular path. In this experiment the bung has a
mass of 0.09 kg.
17/04/19 10:25 AM
2 Discuss the sources of error in this experiment.
Suggest how the errors can be minimised.
3 To improve the reliability of the data, it might be
helpful to plot a graph.
a) Plot a graph of F against mω2r.
b) Explain why this should be a straight line. What
gradient do you expect to get when you measure it?
Table 15.1
F/N
M/kg
10T/s
ω/rad s−1
r/m
0.1
18.8
0.94
0.2
11.5
0.78
0.2
10.2
0.56
0.3
8.6
0.61
0.3
7.9
0.52
0.4
5.4
0.31
m ω2r/N
EXAMPLE
Round in circles
1 A physics teacher, shown in Figure 15.10,
demonstrates a well-known trick. She puts a beaker
of water on a tray, suspended by four strings at its
corners. Then she whirls the tray round in a vertical
circle, so that the beaker is upside down at the top.
She then asks why the water does not fall at the top
of the swing. A student (who has not been paying
attention) says ‘the pull of gravity is balanced by an
outwards force’. Explain why this is not correct.
15 CIRCULAR MOTION
Answer
280
B
A
E
0.95 m
The teacher gives this explanation: At the top,
the water and the beaker are falling together.
Look at Figure 15.10. At point A, the beaker is
travelling along the direction AB. The string pulls
the beaker down in the direction BC so at the top
it has fallen to point C.
The teacher repeated the demonstration and
asked the students to time the revolutions. The
students determined that the tray completed 10
revolutions in 8.3 s. They measure the radius of
the circle to be 0.95 m.
The speed of the tray is
2π r
T
2π × 0.95 m
=
0.83 s
v=
C
F
D
Figure 15.10
= 7.2 m s–1
So, while the beaker rotates, it has a centripetal
acceleration of
2
a=v
r
(7.2 m s–1)2
0.95 m
= 55 m s–2
=
469884_15_AQA_A-Level_Physics_274-284.indd 280
This tells us that the water is accelerating all
the time at 55 m s−2 (more than five times the
gravitational acceleration). So the water does
not fall out ofthe beaker at the top, because it is
already falling with an acceleration greater than
gravitational acceleration.
17/04/19 10:25 AM
40°
R
r = 25 m
40°
Answer
R
W
W
2
mv
r
110 kg × (14.5 m s–1)2
=
25 m
F=
Centripetal force
2 A cyclist is cycling at 14.5 m s−1 in a
velodrome where the track is banked
at an angle of 40° to the horizontal
(Figure 15.11). The track is curved so
that the cyclist is turning in a horizontal
circle of radius 25 m. The cyclist and
bicycle together have a mass of 110 kg.
a) Calculate the centripetal force acting
on the cyclist.
Unbalanced centripetal force
Figure 15.11
= 930 N (2 s.f.)
b) Calculate the contact force R from the track on
the bicycle.
Answer
Figure 15.11 shows the two forces acting on the
bicycle and cyclist: the contact force R and the
weight W. The forces combine to produce the
unbalanced centripetal force, which keeps the
cyclist moving round her horizontal circular path.
Force R may be resolved horizontally and vertically
as follows:
Rv = R cos 40°
Rh = R sin 40°
The vertical component Rv balances the weight, and
the horizontal component provides the unbalanced
centripetal force. (It is important to realise that the
cyclist can only lean her bicycle as shown because
she is accelerating towards the centre of the circle.
She would fall over if she were stationary.) So
R sin 40° = 930N
930N
sin 40°
=1440N
R =
TEST YOURSELF
8 This question refers to the teacher’s demonstration
with the beaker of water shown in Figure 15.10.
a) The beaker and water have a combined mass
of 0.1 kg. Use this information, together with
the information in the text, to calculate the
centripetal force required to keep the beaker in
the circular path that the teacher used.
b) The only two forces that act on the beaker and
water are their weight, W, and the contact force,
R, from the tray. Calculate the size and direction
of R at the following points shown in Figure 15.10:
i) C
ii) D
iii) E.
c) The water will fall out of the beaker at point C
if the beaker moves so slowly that the required
centripetal acceleration is less than g.
Assuming the teacher still rotates the beaker
469884_15_AQA_A-Level_Physics_274-284.indd 281
in a circle of 0.95 m radius, calculate the
minimum speed at which the water does not
fall out of the beaker at point C.
9 Formula 1 (F1) racing cars are designed to enable
them to corner at high speeds. Traction between
the tyres and the road surface is increased by
using soft rubber tyres, which provide a large
frictional force, and by using wings to increase the
down force on the car.
281
The tyres of an F1 car can provide a maximum
frictional force to resist sideways movement of
15 500 N. The car’s mass (including the driver) is
620 kg.
Calculate the maximum cornering speed of the car
going round a bend of
a) radius 30 m
b) radius 120 m.
17/04/19 10:25 AM
Practice questions
1 The orbit of an electron in a hydrogen atom may be considered to be
a circle of radius 5 × 10−11 m. The period of rotation of the electron
is 1.5 × 10−16 s. The speed of rotation of the electron is
A 2 × 105 m s−1
C 2 × 106 m s−1
B 4 × 105 m s−1
D 4 × 107 m s−1
2 From the information in question 1, the centripetal acceleration of
the electron is
A 3 × 1022 m s−2
C 12 × 1022 m s−2
B 9 × 1022 m s−2
D 30 × 1022 m s−2
3 The Moon orbits the Earth once every 29 days with a radius of orbit of
380 000 km. The angular velocity of the Moon is
A 2.5 × 10−6 rad s−1
C 8.5 × 10−6 rad s−1
B 5.0 × 10−6 rad s−1
D 25 × 10−6 rad s−1
4 From the information in question 3, the Moon’s centripetal acceleration is
A 2.4 mm s−2
C 7.6 mm s−2
B 4.0 mm s−2
D 24 mm s−2
15 CIRCULAR MOTION
5 The centripetal acceleration of a car moving at a speed of 30 m s−1 round
a bend of radius 0.45 km is
282
A 1.0 m s−2
C 100 m s−2
B 2.0 m s−2
D 200 m s−2
6 A satellite is in orbit around the Earth in a circular orbit of radius
10 000 km. The angular velocity of the satellite is 6.4 × 10−4 rad s−1.
The time of orbit of the satellite is
A 4800 s
C 8400 s
B 6800 s
D 9800 s
7 From the information in question 6, the centripetal acceleration of
the satellite is
A 2 m s−2
C 4 m s−2
B 3 m s−2
D 8 m s−2
8 A student swings a bucket of water in a vertical circle of radius 1.3 m. The
bucket and water have a mass of 2.5 kg. The bucket rotates once every 1.4 s.
When the bucket is upside down, the water does not fall out. Which of the
following gives a correct explanation of why the water stays in the bucket.
A The weight of the water is balanced by a centrifugal force.
B The centripetal force and the weight of the water balance.
C The water and bucket are falling at the same rate.
D The bucket moves so fast that the water has no time to fall.
469884_15_AQA_A-Level_Physics_274-284.indd 282
17/04/19 10:25 AM
B 36 N
Practice questions
9 From the information in question 8, the centripetal force on the
swinging bucket is
C 52 N
A 6N
D 65 N
10 From the information in question 8, for the swinging bucket, at the
bottom of the swing the rope exerts a force on the student’s hand of
A 90 N
C 52 N
B 63 N
D 36 N
11 An astronaut undergoes some training to test his tolerance to
acceleration. He is placed in a rotor, which carries him in a circle of
radius 7.0 m. The rotor completes 10 revolutions in 24.3 s, moving
at a constant speed.
a) Explain why the astronaut is accelerating, although his speed is
constant.
(2)
b) Calculate the size of the astronaut’s acceleration.
(3)
12 (Synoptic question: you need to think about energy transformations
to help solve this question.)
A large steel ball of mass 2100 kg is used to demolish buildings. The ball
is suspended on a cable of length 8 m, and is pulled back to a height of
4 m above its lowest point, before being released to hit a building.
a) Calculate the maximum speed of the ball just prior to hitting
the building.
(3)
b) Calculate the tension in the cable when the ball is at its
lowest point.
(4)
13 Figure 15.12 shows an aircraft propeller that is
undergoing tests in a laboratory.
A
0.6 m
The propeller is made out of high-strength, low-density
carbon-fibre-reinforced plastic (CFRP). In a test, it is
rotating at a rate of 960 times per minute.
a) Calculate the angular velocity of the propeller.
(2)
C
B
0.7 m
b) Calculate the speed of the propeller blade at these
two positions.(2)
i)
A
0.2 m
ii) B
c) Explain why the propeller blade is made of CFRP.(2)
d) At which point is the blade more likely to fracture, A or B?
Explain your answer.
283
Figure 15.12
(2)
e) Estimate the centripetal force required to keep a propeller blade
rotating at a rate of 960 times per second, if its centre of mass is
0.6 m from the centre of rotation and the mass of the blade is 3.5 kg. (3)
469884_15_AQA_A-Level_Physics_274-284.indd 283
17/04/19 10:25 AM
Stretch and challenge
14 This question is about apparent weight. Your weight is the pull of
gravity on you. But what gives you the sensation of weight is the
reaction force from the floor you are standing on.
a) A man has a mass of 80 kg. Calculate his apparent weight
(the reaction from the floor) when he is in a lift that is
i)
moving at a constant speed of 3 m s−1
ii) accelerating upwards at 1.5 m s−2
iii) accelerating downwards at 1.5 m s−2.
b) A designer plans the funfair ride shown in
Figure 15.13. A vehicle in an inverting roller coaster leaves point
A with a very low speed before reaching point B, the bottom of the
inverting circle. It then climbs to point C, 14 m above B, before
leaving the loop and travelling to point
D.
Assuming that no energy is transferred
to other forms due to frictional forces,
show that
i)
A
C
the speed of the vehicle at B is
20 m s−1
20 m
7m
ii) the speed of the vehicle at C is
11 m s−1.
15 CIRCULAR MOTION
c) Use your answers to (b) to calculate the
centripetal
acceleration required to keep the
vehicle in its circular path
284
i)
at B
D
B
Figure 15.13
ii) at C.
d) Now calculate the apparent weight of a passenger of mass
70 kg
i)
at B
rotation
ii) at C.
In the light of your answers, discuss whether or not this is a
safe ride.
e) Figure 15.14 shows the design of a space station. It rotates
so that it produces an artificial gravity. The reaction force
from the outer surface provides a force to keep people in
their circular path.
Use the information in the diagram to calculate the angular
velocity required to provide an apparent gravity of
9.8 m s−2.
469884_15_AQA_A-Level_Physics_274-284.indd 284
120 m
Figure 15.14
17/04/19 10:25 AM
16
Simple harmonic
motion
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
●
●
●
●
Displacement, velocity, force and acceleration are all vector
quantities.
change of velocity
Acceleration =
time
1
Time period =
frequency
Frequency = number of oscillations per second.
The natural measure of angle is the radian; 2π radians = 360°.
Resultant force = mass × acceleration.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 a) An electromagnetic wave has a frequency of 2.6 GHz. Calculate the
time period of the wave.
b) A boat that is anchored at sea lifts up six times in 30 s as waves
pass it. What is the frequency of the waves?
2 A car is travelling with a velocity of 20 m s−1 due north. Two minutes
later the car has travelled round a large bend and is travelling with a
velocity of 15 m s−1 due south. Calculate the car’s average acceleration
over this time.
3 Calculate the values of these trigonometric functions, where the
angle has been expressed in radians.
a) tan 0.01
b) sin π
c) cos π
●● Simple harmonic motion
285
Last year, when you studied wave motion, you learnt that all types of waves
require a vibrating source to produce them. For example, vibrating or
oscillating electric and magnetic fields are responsible for the production of
electromagnetic waves. There are also many examples of mechanical waves –
sound waves, water waves, waves on strings or wires, and shock waves from
earthquakes. All these waves are caused by a vibrating source.
469884_16_AQA_A-Level_Physics_285-307.indd 285
17/04/19 10:27 AM
In this chapter, you are going to be studying oscillations about a fixed point.
Figure 16.1 shows three examples of mechanical oscillations – a clamped
ruler, a mass on a spring and a pendulum. In each of these examples, we
observe that the motion is repetitive about a fixed point. The oscillating
object is stationary at each end of the motion, and is moving with its
maximum speed, in either direction, at the midpoint.
clamped rule
oscillations
pendulum
mass on
a spring
oscillations
oscillations
Figure 16.1
16 SIMPLE HARMONIC MOTION
To a good approximation, these objects have these features in common:
Simple harmonic motion A repetitive
motion about an equilibrium position. The
equation that describes this motion is
F = −kx.
286
469884_16_AQA_A-Level_Physics_285-307.indd 286
●
The force acting on the body always acts towards the equilibrium
position.
●
The force acting on the body is proportional to its displacement from the
equilibrium position.
An oscillating body that satisfies both these conditions is said to be moving
with simple harmonic motion or SHM. The two features of the motion
above may be summarised in the equation:
F = −k x
or
ma = −kx
or
a = − k x(i)
m
Here k is a constant (which can be called the spring constant or the force per
unit displacement). The significance of the minus sign is that it shows that
the force (and acceleration) are in the opposite direction to the displacement.
Force, acceleration and displacement are vectors, so we must define the
direction of the displacement and motion.
17/04/19 10:27 AM
equilibrium
position
F
x
A
Pull down a distance
A and release.
Mass accelerates upwards
due to resultant force F.
Figure 16.2 A mass on a spring is a simple harmonic oscillator.
Mathematical description of SHM
equilibrium
position
Figure 16.2 shows some important features of a simple harmonic oscillator.
When at rest, the mass hangs in its equilibrium position. A is the amplitude
of the oscillation – this is the greatest displacement of the oscillator from
its equilibrium position. When the mass is displaced downwards by x, the
force acts upwards on the mass towards the equilibrium position.
If you investigate the time period of a simple harmonic oscillator, you will
discover that the time period does not depend on the amplitude of the
oscillations, provided the amplitude is small. If you overstretch a spring or swing
a pendulum through a large angle, the motion ceases to be simple harmonic.
●● Mathematical description of SHM
The question we want to answer is this: How do the displacement, velocity
and acceleration of a simple harmonic oscillator vary with time?
Figure 16.3 gives us some insight. Here a mass is oscillating up and down
on a spring. The mass has been stroboscopically photographed by a camera,
moving horizontally at a constant speed. The shape of the curve we see is
sinusoidal. Figure 16.4 shows how the displacement of the mass varies with
time if it is released from rest with an amplitude A.
+A
displacement/m
Figure 16.3
287
0
�
T
2
2�
T
angle
time
–A
Figure 16.4
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The graph has the shape of a cosine function, which can be written as
TIP
NOTE
x = A cos θ
In the AQA specification the
symbol ω is used to represent
2π
2π f or
, so you will also
T
meet equations in this form:
x = A cos ω t
But the value of θ is 2π after one complete cycle so, at the end of the cycle,
x = A cos(2π)
However, we know that the oscillation is a function of t. The function that
fits the equation is
( (
x = A cos 2πt
T
2
a=–ω x
or
vmax = ωA
amax = ω 2 A
v = ±ω
2
A –x
2
x = A cos(2πft)(ii)
where T is the time period for one oscillation. Remember that T = 1 where f
f
is the frequency of the oscillation. This function solves the equation because
after one oscillation t = T, so the inside of the bracket has the value 2π.
Once we have an equation that connects displacement with time, we
can also produce equations that link velocity with time, and then also
acceleration with time. These are shown below:
x = A cos(2πft)
v = −2πfA sin(2πft)(iii)
a = −(2πf)2A cos(2πft)(iv)
16 SIMPLE HARMONIC MOTION
and since x = A cos(2πft)
a = −(2πf)2x(v)
We derive this assuming x = A when t = 0. However, the same equation
would have been obtained whatever the starting condition.
(Mathematicians will see that the velocity equation is the derivative of the
displacement equation, and that the acceleration equation is the derivative
of the velocity equation.)
Since the maximum value of a sine or cosine function is 1, we can write the
maximum values for x, v and a as follows:
xmax = A(vi)
vmax = 2πfA(vii)
288
amax = (2πf)2A(viii)
We also write down one further useful equation now, which allows us to
calculate the velocity v of an oscillating particle at any displacement x:
v = ±2π f A2 – x 2 (ix)
This will be proved later when we consider the energy of an oscillating system.
Figure 16.5 shows graphically the relationship between x, v and a. These
graphs are related to each other.
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0
(a)
2
3
4
5
6
7
8
time
displacement x
A
time t
–A
velocity v
2�ƒA
time t
–2�ƒA
acceleration a
(c)
(2�ƒ)2A
time t
–(2�ƒ)2A
Figure 16.5
graph of velocity v against time t links to the
gradient of the displacement–time (x–t) graph because
●The
∆x
∆t
For example, at time 0 (in Figure 16.5), the gradient of
the x–t graph (a) is zero, so the velocity is zero. At time
1, the gradient of the x–t graph (a) is at its highest
and is negative, so the velocity is at its maximum
negative value.
● The graph of acceleration a against time t (c) links
to the gradient of the velocity–time (v–t) graph (b)
because
∆v
a=
∆t
For example, at time 1 (in Figure 16.5), the gradient
of the v–t graph (b) is zero, so the acceleration is zero.
At time 2, the gradient of the v–t graph (b) is positive
and at its largest value, so the acceleration has its
largest value.
v=
Mathematical description of SHM
(b)
1
EXAMPLE
SHM of a mass on a spring
A mass hanging on a spring oscillates with simple
harmonic motion. The amplitude of the oscillation is
4.0 cm, and the frequency of the oscillation is 0.5 Hz.
The spring is released from rest at its lowest position,
4 cm below its equilibrium position (Figure 16.6).
1 Calculate the maximum velocity of the mass.
Answer
v max = 2πf A
= 2π × 0.5 s–1 × 0.04 m
= 0.13 m s–1
2 Calculate the maximum acceleration of the mass.
Answer
a
2
max
= (2πf ) A
2
= (2π × 0.5 s−1) × 0.04 m
= 0.39 m s –2
Equilibrium
position
4 cm
3 Calculate the acceleration of the mass 1.2 s after
release.
Answer
We need to define a direction before applying our
formulae. In Figure 16.6 we define positive as our
downwards direction. (This is arbitrary. You will get the
same answer if you choose this direction to be negative.)
289
2
m
Figure 16.6
a = –(2π f ) A cos(2π ft)
= –(2π × 0.5 s–1)2 × 0.04 m × cos(2π × 0.5 × 1.2)
= –9.9 s–2 × 0.04 m × (–0.81)
= 0.32 m s–2
Since this is positive, the acceleration is downwards.
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4 Calculate the velocity of the mass when it is
displaced 2 cm from itsequilibrium position.
2
v = ±2π f A – x
2
–1
2
1
2 2
= ±2π × 0.5 s (0.04 – 0.02 ) m
–1
= ± π s × 0.035 m
= ±0.11m s
–1
1 A pendulum is released from point A in Figure 16.7. It swings from
A to C and back with SHM. The distance AC is 24 cm, and the time
taken to travel from A to B is 0.8 s.
a) State the frequency of the oscillation.
b) State the amplitude of the oscillation.
c) i) Calculate the speed of the pendulum as it passes B.
ii) Calculate the velocity of the pendulum when it is displaced
4 cm from B.
d) Calculate the acceleration of the pendulum when it is displaced
6 cm to the right of B.
2 A ruler is clamped to a bench. When the free end is displaced, the
ruler oscillates with SHM, at a frequency of 100 Hz. The amplitude
of the oscillations is 1.8 mm.
a) Calculate the highest velocity of the ruler.
b) Calculate the highest acceleration of the ruler. State where
this is.
c) State the point in the oscillation where
i) the acceleration of the ruler is zero
ii) the velocity of the ruler is zero.
3 A marker buoy is oscillating in a vertical line with SHM. The buoy
takes 2.8 s for one oscillation and is seen to fall a distance of 1.8 m
from its highest to its lowest point.
a) Calculate the buoy’s maximum velocity.
b) Calculate the buoy’s acceleration when it is 1.4 m below its
highest point.
4 Figure 16.8 shows the displacement of a particle oscillating with
SHM. To represent its motion, the equation x = A sin(2π ft) is used,
where x = 0 when t = 0.
Copy the diagram and add sketches, using the same time axis, to
show the variations of v and a with time.
A
B
C
Figure 16.7
x/m
16 SIMPLE HARMONIC MOTION
TEST YOURSELF
time t
Figure 16.8
290
●● Time period of oscillations
We can combine two of the equations that we used in the previous
sections to produce a further equation that links the time period of
oscillation, T, to the mass of the oscillating particle, m, and the force per
unit displacement, k.
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1 A mass of 400 g hangs on
a steel spring, which has a
spring constant of 0.20 N cm−1.
Calculate the time period for
one oscillation.
Answer
T = 2π
= 2π
a=−kx
m
and
k = (2π f )2
m
or
k  2π 
m =  T 
m
k
0.4 kg
2
or
20 N m–1
m
k (x)
T = 2π
Once you recognise that a particle is oscillating with SHM, you can use this
general solution to calculate the time period of any oscillator.
1.4 N
250 g
The simple pendulum
3 cm
Figure 16.10a shows a pendulum held at rest by a small sideways force F.
Figure 16.10b shows the three forces acting on the pendulum bob to keep
it in equilibrium.
θ
Figure 16.9
2 In Figure 16.9, a light ruler
is clamped to a desk and a
mass of 250 g is attached
securely to the free end. When
a newtonmeter is attached to
the end of the ruler, a force
of 1.4 N displaces the ruler by
3 cm. Calculate the time period
of the oscillations, stating any
assumptions you make.
The force per unit displacement
is
1.4 N
k=
= 47 Nm–1
0.03 m
so
m
k
0.25 kg
47 N m–1
θ
θ
T
l
T
mg
F
mass m
Answer
= 2π
a = –(2π f )2 x
Combining these gives
= 0.89 s ≈ 0.9s
T = 2π
Time period of oscillations
EXAMPLE
Calculating the time period
The two equations that define the motion of a simple harmonic oscillator
that we need from the earlier sections are equations (i) and (v):
mg
Figure 16.10(a)
F
x
Figure 16.10(b)
Figure 16.10(c)
291
The force F = mg sin θ. For small angles we have θ ≈ sin θ, and therefore
F = mgθ.(xi)
Figure 16.10c shows that x can be related to the length of the pendulum, l, by
x = l tan θ.
= 0.46 s ≈ 0.5 s
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For small angles, we also have θ ≈ tan θ, and therefore
x = lθ
and
θ = x (xii)
l
Combining equations (xi) and (xii) gives
x
l
When the pendulum is released, the restoring force now acts in the
opposite direction. So
F = mg
ma = – mg x
l
and
g
a = –g x = – x
l
l
This is the defining equation for SHM because the acceleration is
proportional to, and in the opposite direction to, the displacement.
Therefore
(2πf )
2
g
l
=
 2π 2 g
  =
 T 
l
16 SIMPLE HARMONIC MOTION
and
292
T = 2π
l
g (xiii)
REQUIRED PRACTICAL 7
Investigation into simple harmonic motion
using a mass–spring system and a simple
pendulum.
Note: This is just one example of how you might
tackle this required practical.
Make a simple pendulum using a small mass hanging
on a piece of string about 1.5 m long.
l
l
l
Investigate whether the time period of the
pendulum depends on the amplitude of the swings.
Investigate how the time period, T, of the pendulum
varies with length between 1.5 m and 0.2 m.
Plot a suitable straight line graph to investigate
whether: T2 α l. Use the gradient of the graph to find
a value for g.
ACTIVITY
Oscillation of a tethered trolley
The purpose of this activity is to investigate how
the time period of a tethered trolley depends on its
mass. The mass of the trolley is changed by putting
additional weights on top of it.
469884_16_AQA_A-Level_Physics_285-307.indd 292
In Figure 16.11, the identical springs A and B are both
under tension, but in the trolley’s equilibrium position
the forces from the two springs balance.
17/04/19 10:29 AM
B
mass m
Figure 16.11 A tethered trolley oscillates with SHM. Each
spring has a spring constant k.
Each spring has a spring constant k. What is the
resultant force acting on the trolley when it is
displaced a distance x to the right?
l
l
Spring A exerts an extra force of kx to the left.
Spring B exerts a force reduced by kx to the right.
So the resultant force on the trolley is 2kx to the left.
Time period of oscillations
A
2 In an experiment, a student determines the spring
constant of her springs to be 17.8 N m −1. She then
recorded the set of data shown in Table 16.1 for the
oscillation of her trolley, as she varied its mass.
Table 16.1
Mass of
trolley/kg
1.0
1.5
2.0
2.5
3.0
4.0
Time for five
oscillations/s
5.3
6.4
7.4
8.3
9.1
10.5
Plot a graph of T2 against m.
a) Discuss whether or not your graph is consistent
with the formula quoted in part 1.
b) Determine the gradient of your graph. Comment
on this result.
1 Explain why the time period of the tethered trolley
is given by
T = 2π m
2k
ACTIVITY
Other systems that might show SHM
Figure 16.12 shows two systems to investigate. It is suggested that both are
simple harmonic oscillators.
1 A U-tube is partly filled with water, as shown in (a). The total length of
the water in the tube is L.
a) Investigate whether or not the amplitude of the oscillations affects
their time period.
b) Check to see if the time period of the oscillations agrees with the
formula:
T = 2π
x
L
L
2g
2 A weighted boiling tube is allowed to oscillate up and down in a large
beaker of water (b). In its equilibrium position, a length L of the tube is
submerged.
a) Investigate whether or not the amplitude of the oscillations affects
their time period.
b) Check to see if the time period of the oscillations agrees with the
formula:
T = 2π
x
equilibrium
position
(a)
equilibrium
position
293
L
L
g
lead shot
(b)
(Both formulae for the time periods are derived in the online material.)
Note: Make sure to wash your hands after handling lead shot.
469884_16_AQA_A-Level_Physics_285-307.indd 293
Figure 16.12
17/04/19 10:29 AM
TEST YOURSELF
5 A pendulum has a length of 2 m. Calculate its time period of oscillation
in each of these places:
a) on Earth
b) on the Moon, where g = 1.6 N kg−1
c) on a comet, where g = 0.0006 N kg−1.
6 A ‘baby bouncer’ is a harness that can be used to amuse and exercise
a baby before the baby can walk. Figure 16.13 shows a baby enjoying
this experience. This is a simplified model, in fact, the baby's feet will
touch the floor.
elastic ropes
suspended from
a door lintel
16 SIMPLE HARMONIC MOTION
displacement/m
The suspension ropes for a bouncer are 1.30 m long and stretch to
1.48 m when a baby of mass 9.5 kg is put in it.
a) i) Determine the spring constant for the baby bouncer.
Figure 16.13
ii) Determine the time period of the baby’s simple harmonic motion.
iii) Determine the baby’s maximum speed, when released from 10 cm
above the equilibrium position.
b) When the baby was bouncing three months later, the
0.1
baby’s father noticed that the time period of
the oscillations had increased to 1.0 s. He is
delighted that his baby has put on weight (or mass,
as baby’s mother correctly points out). What is
baby’s mass now?
A
c) Explain what is meant by the terms
i) weight
0.2
ii) mass.
7 Figure 16.14 shows a graph of displacement against
B
time for a mass of 0.5 kg oscillating on a spring.
a) Use the graph to estimate the speed of the mass
between points A and B.
b) Use your knowledge of SHM equations to calculate
–0.1
the theoretical maximum speed from the graph,
using the information shown on the axes.
Figure 16.14
c) Calculate the spring constant of the spring.
294
469884_16_AQA_A-Level_Physics_285-307.indd 294
0.4
time / s
●● Energy in simple harmonic motion
Figure 16.15 shows a pendulum swinging backwards and forwards from
A to B to C, and then back to B and A. As the pendulum moves, there is a
continuing transfer of energy from one form to another.
●At
A, the velocity of the pendulum bob is zero. Here the kinetic energy,
Ek, is zero, but the bob has its maximum potential energy, Ep.
●At B, the velocity of the pendulum is at its maximum value, and the
bob is at its lowest height. Therefore, Ek is at its maximum, and Ep is
at its minimum value. This can be defined as the system’s zero point of
potential energy.
17/04/19 10:29 AM
●At
The potential energy can be calculated as follows. The force
acting on the pendulum along its line of motion is −kx when
it has been displaced by x (where k is the force per unit
displacement).
A
C
v=0
v=0
Ek = 0
Ep = max
B
The work done to take the mass to x is
W = average force × distance
= 1 k x × (x)
2
1
= k x2
2
or the potential energy is given by
Ek = 0
Ep = max
v = max
Ek = max
Ep = min
Ep = 1 kx2(xiv)
2
So the maximum potential energy of any simple harmonic
oscillator is 12 kA2, where A is the amplitude of the
displacement.
Figure 16.15
energy
1
2
kA2
total energy
The kinetic energy of the oscillator at a velocity v is
potential
energy
kinetic
energy
–A
Figure 16.16
0
displacement x
Energy in simple harmonic motion
C, the velocity is once more zero. So the bob has
zero kinetic energy and its maximum value of potential
energy.
A
Ek = 1 mv2
2
Figure 16.16 shows how the potential energy Ep and the
kinetic energy Ek change with displacement for a simple
harmonic oscillator. The total energy of the system
remains constant (assuming there are no energy transfers
out of the system.)
Figure 16.17 shows how the potential, kinetic and total
energies change with time as the pendulum oscillates.
In one oscillation, the potential energy and the kinetic
energy both reach a maximum twice. The total energy
remains constant.
PE
1
kA2
2
KE
energy
Total
energy
295
0
T
4
T
2
3T
4
T
5T
4
3T
2
time
Figure 16.17
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We can now write an equation to link these three energies at a displacement x:
total energy = kinetic energy + potential energy
1k A2 = 1 mv2 + 1k x 2 (xv)
2
2
2
So
mv2 = k(A2 − x2)
and
1
 k 2
1
v = ±   ( A 2 – x 2 ) 2 (xvi)
 m 
(Remember that when you take the square root of a function, there is a
positive and a negative root.) But
T = 2π
m
k
or
1
= 2π m
k
f
and so
16 SIMPLE HARMONIC MOTION
2π f =
296
k
m
(xvii)
When equation (xvii) is substituted into equation (xvi), we get the familiar
equation for velocity:
1
v = ±2π f ( A2 – x 2 ) 2
or
v = ±2π f A2 – x 2
(xviii)
TEST YOURSELF
8 A simple model of a diatomic gas molecule
treats the two atoms as small masses, which are
connected by an atomic bond that behaves like a
tiny spring. The atoms in a particular molecule
vibrate with SHM at a frequency of 1013 Hz and
amplitude 2 × 10 −12 m. The mass of each atom is
about 10 −25 kg.
a) What fraction of a typical atomic separation
does this amplitude represent?
b) Calculate the approximate force constant of the
interatomic bond.
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c) Calculate the total energy of vibration of the two
atoms:
i) in joules (J)
ii) in electronvolts (eV).
9 Draw a sketch to show how the potential energy,
kinetic energy and total energy of a particle,
oscillating with SHM, vary with the particle’s
displacement.
10 A pendulum with a mass of 0.1 kg oscillates with an
amplitude of 0.2 m. When the displacement of the
pendulum is at its maximum from the equilibrium
17/04/19 10:29 AM
i) 0
ii) 0.1 m.
b) i) Use the maximum speed of the pendulum
and the amplitude of the swing to calculate
the time period of the pendulum.
ii) Now calculate the length of the pendulum.
●● Free, clamped and forced oscillations
So far, we have only dealt with free oscillations. These are oscillations
that (in theory) carry on indefinitely because there are no forces acting
to stop the oscillation. A close approximation to a free oscillator is a very
heavy pendulum supported by a very fine wire, which is attached to a
rigid support. Under these circumstances, the energy transfers from the
pendulum are very low, and the pendulum keeps swinging for a long time.
Of course, in the end any pendulum stops swinging because frictional forces
transfer the pendulum’s energy to the surroundings as heat. A pendulum
clock can run for a week, because energy from a slowly falling weight gives
the pendulum a little energy every time it swings.
Damped oscillation A damped oscillation
occurs when friction or wind resistance
takes energy out of the oscillation.
Free, clamped and forced oscillations
point, the pendulum has a potential energy of
0.08 J, and when the displacement is 0.1 m, the
potential energy is 0.02 J.
a) Calculate the speed of the pendulum at
displacements of
In practice, all mechanical oscillations are damped oscillations. In such
oscillations, the oscillator transfers energy to the surroundings. When
the damping is light, energy is transferred slowly. When the damping is
heavy, energy is transferred more quickly and the oscillations stop after a
few swings. The best way for you to investigate the effects of damping on
an oscillator is to use a motion sensor and a data logger. In this way, small
changes in amplitude can be recorded, which you could not do by eye.
(a)
(b)
rotary motion
sensor
computer
card
motion
sensor
interface/
data logger
interface/data logger
297
Figure 16.18
Figure 16.18 shows experimental set-ups to investigate damping in
two oscillating systems. In Figure 16.18a the motion sensor records the
displacement against time for a mass on a spring. The card on the bottom
has two functions: first, to act as a good reflector for the motion sensor;
second, to act as a ‘damper’. It causes drag to dampen the oscillations.
Increasing the size of the card will increase the damping of the oscillator.
In Figure 16.18b a rotary sensor records the motion of a pendulum. The
computer records how the angle of rotation varies with time. By attaching
469884_16_AQA_A-Level_Physics_285-307.indd 297
17/04/19 10:29 AM
cards to the pendulum, the motion of the pendulum can be damped.
Figure 16.19 shows how the angular displacement of a pendulum varies
with time, for different amounts of damping.
(a)
20
light damping
angular displacement (degree)
15
10
5
0
2.5
5
7.5
time/s10
2.5
5
7.5
time/s 10
2.5
5
7.5
time/s 10
–5
–10
–15
–20
(b)
medium damping
20
angular displacement (degree)
10
5
0
–5
–10
–15
–20
(c)
20
heavy damping
15
angular displacement (degree)
16 SIMPLE HARMONIC MOTION
15
298
10
5
0
–5
–10
–15
–20
Figure 16.19 Graphs of the angular displacement of a pendulum against time for different levels of damping:
(a) light damping, (b) medium damping and (c) heavier damping. Data provided by Data Harvest Group Ltd.
You may be able to use data logging equipment to investigate damping
for yourself.
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A teacher suggests that the amplitude of a damped
oscillator decays exponentially with time. This means
that the amplitude can be described by the following
equation:
2 To investigate if the amplitude decays exponentially,
we can plot a graph of the natural logarithm of the
amplitude against the number of swings. Since it is
suggested that the amplitude obeys the law
A = A 0 e–λt
where A is the amplitude at time t and A0 is the
amplitude at t = 0 (the start of the swings).
1 Investigate this relationship, using the graphs in
Figure 16.19. Work in teams of three, so that each
person can analyse one of the graphs. Rather than
working in seconds, work in ‘swings’. Then copy
and complete Table 16.2. Measure the amplitude
after each complete swing.
Table 16.2
Number Amplitude of
Amplitude of
Amplitude of
of swings graph
graph
graph
2.19(a)/degrees 2.19(b)/degrees 2.19(c)/degrees
0
A = A 0 e–λt (where t is the number of swings)
Forced oscillations
ACTIVITY
A damped oscillator
then
ln A = ln A 0 − λt
a) Using your data, plot a graph of In A against the
number of swings.
b) Discuss whether or not your data follows an
exponential law.
c) Determine the ‘half-life’ T 1 for your pendulum
2
using the expression
ln 2
2
λ
where you determine λ from your graph.
(You will find a similar expression derived in
Chapter 25.) Here T 1 is the number of swings it
2
takes your pendulum’s amplitude to reduce by a
half.
T1 =
1
2
3
4
5
6
7
8
9
10
●● Forced oscillations
When you let a pendulum swing freely, it swings at its natural frequency,
which is determined by its length. Gradually the pendulum transfers its
energy and slows down. It is also possible to force a pendulum to oscillate
at a different frequency by pushing it at regular time intervals. This is
demonstrated in Figure 16.20. In Figure 16.20a a pendulum is driven
by hand at a frequency below its natural frequency. The amplitude of the
oscillations is low, and the pendulum bob moves in phase with the hand. In
Figure 16.20b the hand moves backwards and forwards at a high frequency,
above the natural frequency of the pendulum. Now the pendulum bob
moves out of phase with the hand, and the amplitude of the oscillations
is still small. In Figure 16.20c the pendulum is given a push at its natural
469884_16_AQA_A-Level_Physics_285-307.indd 299
299
17/04/19 10:29 AM
(a)
(b)
Figure 16.20
Resonant frequency The resonant
frequency of a structure (or oscillator) is
the same as its natural frequency. When an
oscillator is driven or pushed at its natural
frequency, the amplitude of the oscillations
grows large.
frequency. Just as the pendulum stops moving, the
hand gives it a small nudge. Now the oscillations of the
pendulum become very large. The pendulum is said to
be driven at its resonant frequency. When you push a
child on a swing, you push at the resonant frequency.
Just as the child reaches the maximum displacement
from the centre, the swing momentarily stops. After that
point, you give the swing a push and the amplitude of
the swing builds up.
(c)
The idea of resonance is demonstrated by Barton’s
pendulums (see Figure 16.21). Here, a number of
light pendulums (A–E) are suspended from a string.
Also attached to the string is one heavy pendulum (X),
which is the ‘driving’ pendulum. When the driving
pendulum is released, it pushes the string as it swings.
These pushes then begin to drive the other pendulums. Most of them swing
with low amplitude, but the pendulum that has the same length (L) as the
driver swings with a large amplitude. This is because its natural frequency is
the same as the driving frequency.
Resonance An oscillator undergoes highamplitude oscillations (resonance) when the
driving frequency is the same as the natural
frequency.
L
L
E
D
16 SIMPLE HARMONIC MOTION
X
300
C
B
A
Figure 16.21 Barton’s pendulums.
ACTIVITY
Investigating resonance
You can use the apparatus in Figure 16.22 to help you
understand how the amplitude of a driven oscillator
changes with the driving frequency. The oscillator
pulls the string up and down. The string is attached
to a mass on a spring, which oscillates up and down
at the same frequency as the driving oscillator.
Proceed as follows.
l
l
Choose a spring (or springs) and masses so that
the natural frequency of the system is about 1–2 Hz.
Measure this frequency.
Vary the frequency of the oscillator in small steps
from about a third of the natural frequency to about
three times the natural frequency. Record the
amplitude of the oscillations in each case.
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natural
frequency
1
f0 = —
T0
forced
oscillations
to variable
frequency
signal
generator
large-amplitude
oscillations
when frequency
of signal
generator = f0
oscillator
(driving force)
Figure 16.22
17/04/19 10:29 AM
l
Repeat the experiment with a piece of card
attached to the masses to increase damping.
Figure 16.23 shows an idealised resonance curve that
you might get when you try the activity. The amplitude
of the driven oscillations peaks sharply at the natural
frequency of the system. The sharpness of the peak
depends on the amount of damping. When a system
is heavily damped, the peak is not so sharp because
energy is being lost from the system and the amplitude
does not build up so far.
amplitude
Forced oscillations
l
(This can be difficult because the oscillations do
not always settle into a steady pattern.)
Plot a graph of the amplitude of the oscillations
against frequency.
Figure 16.24 shows the effect of increasing damping on
a resonance curve:
applied frequency
amplitude resonance when
applied frequency = natural
frequency of oscillating body
Figure 16.23
the peak of the amplitude is lower
● the peak is broader
●the peak of the amplitude occurs at a frequency
slightly lower than the natural frequency of the
system.
●
amplitude
applied frequency
natural frequency
of oscillating body
resonating air
column
Figure 16.24
Examples of resonance
Musical instruments provide a good example of resonance. If the air in a
wind instrument oscillates at the natural frequency of the instrument, a
loud note is produced. In a stringed instrument, when a string is plucked it
vibrates at its natural frequency.
tuning fork forcing air
column into resonance
Figure 16.25
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301
Figure 16.25 shows how you can demonstrate resonance in the laboratory.
Here a tuning fork is held above a column of air. The length of the column
can be adjusted by moving the reservoir of water, on the right, up or down.
When the tuning fork is made to vibrate, the column of air vibrates. However,
the amplitude of the oscillations is small when the driving frequency does
not match the natural frequency of oscillations in the air column. When the
length of the column is adjusted so that its natural frequency is the same as
the tuning fork’s frequency, a loud sound is heard as the air resonates.
17/04/19 10:29 AM
A microwave oven takes advantage of the resonance of water molecules.
The frequency of the microwaves is matched to the natural frequency
of oscillation of water molecules. So when something is cooked in the
microwave oven, water molecules absorb energy from the microwaves. The
water molecules start to vibrate. This energy is then dissipated as random
vibrational energy among all the molecules in the food. Random vibrational
energy is heat energy.
Resonance can cause serious problems in any mechanical structure, because
all structures have a natural frequency of oscillation. Even a large structure
such as a chimney or a bridge can be set oscillating by eddies of wind. And
if the wind causes vortices of just the right frequency, large oscillations
can build up. Famous examples of bridges being made to oscillate by the
wind include the Millennium Bridge in London in 2000, and the Tacoma
Narrows Bridge in the USA in 1940. The decks of large boats can also be
made to oscillate if the boat hits waves with the same frequency as the
natural frequency of part of the deck. The Broughton Suspension Bridge
was an iron suspension bridge built in 1826 to span the River Irwell in
Manchester. In 1831, the bridge collapsed due to the mechanical resonance
caused by a troop of soldiers marching in step. Unfortunately for them,
the frequency of their steps caused the bridge to oscillate so much that it
collapsed. As a result of the accident, the British Army issued an order that
troops should ‘break step’ when crossing any bridge.
16 SIMPLE HARMONIC MOTION
Reducing resonance
The best way to avoid resonance in structures is to design them so that their
natural frequencies lie well outside the range of frequencies likely to be
caused by wind blowing across them. However, if that is not possible then
the amplitude of oscillations can be reduced by damping the motion. In the
case of the Millennium Bridge, the oscillations were reduced by applying
fluid dampers (see Figure 16.26).
302
Figure 16.26 Fluid dampers were used to reduce the lateral
oscillations of the Millennium Bridge.
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Forced oscillations
Pistons are able to move inside a cylinder containing fluid and this
causes the energy to be dissipated rapidly. Car suspension systems use
the same idea. A car uses springs and shock absorbers to make the ride
more comfortable for passengers. When a car goes into a pothole (for
example) a strong spring allows the wheel to drop into the hole. With no
shock absorber, the car would then oscillate up and down. But the shock
absorber removes the energy. It consists of a piston that moves inside an
oil-filled cylinder. The shock absorbers are critically damped. This means
that the wheel only oscillates once before returning to its normal position
relative to the car (see Figure 16.27).
Figure 16.27 The spring and the
piston-cylinder on a shock absorber
form the suspension system for a wheel
in a car.
TEST YOURSELF
11 Figure 16.28 shows a system of pendulums suspended
from a string. A heavy pendulum on the right is set in
motion, which then sets the other pendulums in motion
too. These light pendulums are made using small paper
cones.
a) Describe the motions of the eight light pendulums.
b) The small cones are replaced by larger paper cones.
Describe the changes you see in the motion of the
pendulums.
12 a) E xplain what is meant by the term ‘resonance’.
b) Give and explain an example of how resonance can be
useful in a mechanical system.
c) Give and explain an example of how resonance can
cause problems in a mechanical system.
heavy pendulum
Figure 16.28
303
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Practice questions
1 A metal ruler is clamped to the desk and a mass of 80 g is fixed
securely to the end (Figure 16.29). A force of 12 N is applied to
displace the end of the ruler by 3 mm upwards. When the ruler is
released, it oscillates with simple harmonic motion.
12 N
mass 80 g
3 mm
The frequency of the oscillation is
A 1 Hz
C 36 Hz
B 12 Hz
D 360 Hz
Figure 16.29
2 The maximum speed of the ruler in Figure 16.29 when released
with an amplitude of 3 mm is
A 0.36 m s−1
C 3.6 m s−1
B 0.67 m s−1
D 6.7 m s−1
3 In Figure 16.30a a mass m oscillates vertically on a spring with
time period T. The same mass is now attached to two springs, as
shown in Figure 16.30b. All the springs are identical.
The mass now oscillates with time period
T
A 2T
C
2
T
B 2T
D
2
16 SIMPLE HARMONIC MOTION
4 A pendulum of length 12 m is suspended from the ceiling of
a tall room. The time period of the pendulum is
304
A 2s
C 7s
B 5s
D 12 s
(a)
(b)
m
Figure 16.30
m
5 A mass m hangs on a spring with spring constant k. The mass
oscillates with simple harmonic motion. The amplitude of the
oscillations is A. The highest speed of the mass is
A
k
A
m
B 2π
k
A
m
C
1 k
A
2π m
k
D 2π A
m
6 An astronaut lands on a planet and decides to calculate the
gravitational field strength, g, using a pendulum of length 1.5 m.
For his pendulum, he discovers that the time period is 1.4 s. The
gravitational field strength is
A 15 N kg−1
C 45 N kg−1
B 30 N kg−1
D 60 N kg−1
7 A copper atom in a lattice can be modelled as a mass, m, tethered
by two springs, k, as shown in Figure 16.31, where m is 10−25 kg
and k for each spring is 40 N m−1.
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1 2k
by f = 2π m . The frequency of the oscillations is
A 3.5 × 1011 Hz
C 4.5 × 1012 Hz
B 6.6 × 1011 Hz
D 8.5 × 1012 Hz
k
C 0.5 eV
B 0.3 eV
D 0.6 eV
k
m
8 The oscillating copper atom in Figure 16.31 has oscillations with
an amplitude of 5 × 10−11 m. The energy of the oscillations is
A 0.1 eV
Practice questions
The frequency of the oscillations of the atom is given
Figure 16.31
9 A pendulum swings with an amplitude of 10 cm. The time period
of the swings is 0.8 s. When the pendulum is displaced 6 cm from its
equilibrium position, its speed is
A 0.6 m s−1
C 1.0 m s−1
B 0.8 m s−1
D 2.0 m s−1
10 A particle oscillates with simple harmonic motion. Its displacement
is given by the equation x = A cos(2πft), where A is the amplitude
of the oscillation, 2.5 cm, and f is the frequency of the oscillation,
10 Hz. At a time of 0.025 s, the displacement of the oscillator is
A −2.5 cm
C 0
B −1.25 cm
D +1.25 cm
11 a) What condition is necessary for a body to exist with simple
harmonic motion?(1)
b)A buoy of mass 65 kg oscillates up and down in the sea with
simple harmonic motion (Figure 16.32). It takes an additional
downwards force of 28 N to push the buoy an extra 5 cm into
the water.
(1)
,
c) The time period of the oscillation is given by T = 2π m
k
where m is the mass of the buoy and k is the force per unit
displacement.
Use the equation above to calculate the time period
of oscillations.
d) At t = 0 the buoy is at rest. Copy the axes of the graph in
Figure 16.33 and sketch how the kinetic energy of the
buoy changes over one complete oscillation, time T.
12 A girl sits on a swing. She takes 39 s to complete 14 swings.
The girl has a mass of 42 kg.
Figure 16.32
(2)
kinetic energy
Calculate the force per unit displacement, k, in N m−1,
for the buoy.
305
(2)
0
T
T
2
time
Figure 16.33
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a) Calculate the distance between the girl’s centre of gravity
and the suspension point of the swing.
(3)
The girl is travelling with a speed of 3.5 m s−1 at the lowest
point of her swing.
b) Calculate the height between the lowest and highest points
of her swing. Assume that she is swinging freely.
(2)
c) i)
Calculate the centripetal force that acts on the girl at
the bottom of her swing when she is travelling at
3.5 m s−1.
ii) Calculate the total upwards force exerted on her by
the swing at this point.
(2)
(1)
13 In Figure 16.34, a trolley of mass 0.7 kg is moving to the left with
velocity 0.1 m s−1. A stiff compressible spring is attached to the trolley,
which acts as a buffer. The trolley collides with a solid barrier and
rebounds elastically. The spring obeys Hooke’s law and has a
spring constant k.
a) Write an equation to describe how the force, F, that the spring
exerts on the trolley changes with the compression of the
spring, x. Hence explain why the motion of the trolley, while
the spring is in contact with the block, is simple harmonic.
Figure 16.34
(2)
0.1 ms–1
mass 0.7 kg
0.1
velocity
(i)
v/m s–1
16 SIMPLE HARMONIC MOTION
Figure 16.35 shows graphs of how (i) the velocity and (ii) the
acceleration of the trolley change with time while the trolley is in
contact with the block.
t
time
acceleration
(ii)
–0.1
306
Figure 16.35
b) Use your knowledge of the equations of motion to explain
the relationship between graphs (i) and (ii).(3)
c) The trolley has a mass of 0.7 kg and the spring has a spring
constant of 25 N m−1. Calculate the time, t, that the trolley is
in contact with the block.
(3)
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Calculate the maximum compression of the spring.(2)
ii) Calculate the maximum acceleration of the trolley.(2)
e) The trolley now approaches the block with a speed of 0.2 m s−1.
Describe what effect this has on the following:(2)
i)
the maximum acceleration of the trolley
ii) the time of contact with the block.
Practice questions
d) i)
f) Explain what would happen to the time of contact, t, if the
trolley was now made more massive by adding weight to it.(2)
14 Describe how you would design and set up some mechanical
apparatus to demonstrate resonance.(6)
Stretch and challenge
15 A particle of mass m is connected by two pieces of elastic, each of
unextended length l, to two fixed points A and B, one vertically
above the other, and separated by a distance 3l.
a) If the elastic is such that a force F causes an extension e given by
kkee
F=
l
find the equilibrium position of the particle.
b) The particle is now depressed a small distance z0 from its
equilibrium position, and then released. Assuming that it can
move only vertically up or down, and that damping effects can
be neglected, derive an equation describing its subsequent
motion. Then solve the equation to find the vertical motion z
of the particle as a function of time. Find the time period of the
subsequent oscillations, and the maximum velocity of the particle.
c) A piece of graph paper has x and y coordinates drawn on it. On
these axes is plotted a point whose x coordinate is the vertical
displacement of the particle at time t during its motion, and whose
y coordinate is the vertical velocity at the same instant. Explain
what would be seen if a whole series of points were drawn
corresponding to positions and velocities of the particle over an
extended time period.
d) Describe qualitatively the effects on the motion of the particle in
the following separate situations:
i)
the initial displacement z0 is not small
ii) the damping effects cannot be neglected.
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17
Gravitation
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
●
●
●
●
The Earth, other planets and stars produce gravitational fields, which
exert a force on other massive objects.
A gravitational field exerts a ‘non-contact’ force, which acts over very
long distances.
Gravitational field strength, g, is defined as the force that acts on a
mass of 1 kg. On the Earth, g = 9.81 N kg −1.
A planet’s or star’s gravitational field strength, at its surface, depends
on its mass and radius.
The gravitational potential energy, Ep (in J) gained by a mass m (in kg)
lifted through a height h (in m) in a uniform gravitational field g (in N kg−1)
is given by ΔEp = mgh.
The kinetic energy (Ek) of an object with mass m (in kg) and velocity v
1
(in m s −1) is given by Ek = 12 mv
mv12 2–. mv22 = m ∆V
2
TEST YOURSELF ON PRIOR KNOWLEDGE
308
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1 Listed below are four bodies in our Solar System, and four possible
values of gravitational field strengths at their surfaces. Match the
field strengths to the bodies, explaining your choices.
Four bodies in our Solar System:
Sun, Earth, Mercury, Ceres (a dwarf planet).
Possible values of surface gravity:
0.3 N kg−1, 3.7 N kg−1 , 9.8 N kg−1, 270 N kg−1.
2 Olympus Mons is the tallest mountain on Mars and stands a height
of 22.0 km above the Martian plain. Mount Everest stands at a
height of 8.8 km above sea level. Compare the energy expended by
a mountaineer climbing these two mountains. The gravitational
field strength on the surface of Mars is 3.7 N kg −1. (Assume that
the mountaineer has a mass of 120 kg, including equipment and/or
spacesuits.)
3 a) An object is dropped on Earth from a height of 5 m. Calculate its
speed when it hits the ground.
b) An object is dropped from a height of 31 m on the Moon, and it
reaches a speed of 10 m s−1 when it hits the surface of the Moon.
Calculate the Moon’s gravitational field strength.
c) On a planet, an object is dropped from a height h and it hits the
ground with a speed of 10 m s−1. Calculate the object’s speed when
it hits the ground if it is dropped from a height of 2h.
17/04/19 10:36 AM
We live in an age in which humans have travelled to the Moon and space
probes have been sent to investigate all the planets in our Solar System.
From our own experience and this exploration of space, we know that, on
Earth, gravity exerts a larger force on more massive objects. We have learnt
that the Moon has a smaller surface gravity than the Earth because it is a
less massive body. We have understood that a planet’s pull of gravity gets
weaker further away from that planet’s surface.
These facts seem obvious to us now, but that is because we have seen
experimental proof of them. It is a mark of Isaac Newton’s genius that he
had enough insight to produce a law of gravitation based
on the observations of astronomers before him.
(a)
r
m1
F
F
m2
Newton’s law of gravity
●● Newton’s law of gravity
Newton’s law of gravitation states that the gravitational
force of attraction between two point masses m1 and m2
(measured in kg) separated by a distance r (in m) is given by
F=
(b)
Gm1m2
r2
where G is the universal constant of gravitation:
r
G = 6.67 × 10−11 N m2 kg−2
F
F
This constant can only be accurately measured by
careful laboratory experiment.
m2
m1
(c)
r
F
m1
F
m2
Figure 17.1 In each of these examples, Newton’s law can be
used to calculate the gravitational force of attraction that
the objects exert on each other.
EXAMPLE
Gravitational force
1 Calculate the gravitational force between the Sun, mass
2 × 1030 kg, and Halley’s comet, mass 3 × 1014 kg, when
separated by a distance of 5 × 109 km. This is its furthest
distance from the Sun, which it will reach in 2023.
Although Newton’s law only applies to point masses,
it can also be used to calculate the force of attraction
between large spherical objects (such as planets and
stars) because a sphere behaves as if all the mass were
concentrated at its centre. So Newton’s law can be
correctly used to calculate the force of attraction in each
of the cases in Figure 17.1: (a) the force between two
point masses; (b) the force between a planet and a
small mass; and (c) the force between two planets or
stars. Newton’s law cannot be used to calculate the
force between two irregularly shaped objects, unless a
complicated summation of the forces is made.
Answer
F=
=
309
Gm1m2
r2
(6.67 × 10–11Nm2kg–2 ) × (2 × 1030 kg) × (3 × 1014 kg)
(5 × 1012 m)2
= 2 × 109 N(1s.f.)
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It is important to remember to convert the distance
into metres in the calculation.
This is what you would expect when you remember
that this is your weight, which can also be
calculated using:
2 Calculate the gravitational attraction between you
(assume you have a mass of 65 kg) and the Earth,
which has a mass of 6.0 × 1024 kg. The Earth’s radius
is 6400 km.
W = mg = 65 kg × 9.8 N kg−1 = 637 N
You should also remember, from Newton’s third
law of motion, that you exert a gravitational force of
637 N on the Earth too.
Answer
F=
=
Gm1m2
r2
(6.67 × 10–11Nm2kg–2 ) × (6.0 × 1024 kg) × (65kg)
(6.4 × 106 m)2
= 637N
light source
The inverse square law
When Newton formulated his law of gravity, he imagined that the
force of gravity spreads out in the same way as light spreads out
from a candle.
1m
Figure 17.2 shows his idea. When you hold a card at a distance of
1 m from the candle, you see a particular intensity of light, I. When
you move a distance of 2 m from the candle, the same amount of
light now spreads out over four cards of the same area. So the light
intensity is now a quarter of its original value, 14 I.
Light intensity obeys an inverse square law:
2m
17 GRAVITATION
I=
310
L
4πr 2
Here I is the intensity of light in W m−2, L is the luminosity of
the light source (the amount of energy emitted per second) in
W, and r is the distance away from the source in m. The factor
4π comes into the equation because the light spreads out into a
sphere, and the surface area of a sphere of radius r is 4πr2.
Figure 17.2 The intensity of light 2 m away
from a light source is a quarter of the intensity
1 m away from the source, because the light
spreads across four times the area.
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1 An astronaut of mass 100 kg (including his
spacesuit) stands on a planet of mass 4 × 1025 kg,
with a radius of 8400 km.
a) Calculate his weight.
b) Calculate the gravitational field strength at the
surface of the planet.
2 a) Make an estimate of the gravitational attraction
between two people, each of mass 80 kg,
standing about 10 m apart. What assumption do
you make in this calculation?
b) Explain why we do not notice gravitational
forces between objects on the Earth.
c) You can charge a balloon, by rubbing it, and get
it to stick to a wall. What does this tell you about
the size of electrostatic forces compared with
gravitational forces?
3 Jupiter is approximately five times further
from the Sun than the Earth is, and Jupiter has
Gravitational fields
TEST YOURSELF
a mass approximately 300 times larger than
that of the Earth. How many times larger is the
Sun’s gravitational pull on Jupiter than the Sun’s
gravitational pull on the Earth?
4 a) A spherical light bulb is fitted in a darkened
room. The light intensity at a distance 1 m away
from it is 0.2 W m−2. Calculate the light intensity
at a distance of 3 m away from it.
b) Calculate the electrical power of the light bulb
assuming it is 20% efficient in transferring
electrical energy into light energy.
5 Estimate the gravitational attraction between our
Galaxy, the Milky Way, and the Andromeda Galaxy.
The galaxies are separated by a distance of about
2.4 × 1019 km and each galaxy has a mass of about
7 × 1011 solar masses. The mass of the Sun
(one solar mass) is 2 × 10 30 kg.
●● Gravitational fields
Gravitational field A region in space
in which a massive object experiences a
gravitational force.
Field strength The strength of the
gravitational field measured in N kg−1. Field
lines represent the direction and strength of
the field.
A gravitational field is a region in which a massive object experiences a
gravitational force. Any object with mass produces a gravitational field,
but we usually use the term to describe the region of space around large
celestial objects such as galaxies, stars, planets and moons.
The gravitational field strength in a region of space is defined by
g=
F
m
where g is the field strength measured in N kg−1 and F is the gravitational
force in N acting on a mass m in kg.
(a)
g = 16 N kg−1
(b)
g = 8 N kg−1
Figure 17.3 a Field lines of equal spacing
in the same direction show a uniform
gravitational field. (b) This shows another
uniform field with a lower strength.
Uniform fields
Near the surface of a planet, the gravitational field is very nearly uniform,
which means that the field is of the same strength and direction everywhere.
Figure 17.3 a illustrates a uniform field. The field lines show the direction
of the gravitational force on an object, and the spacing of the lines gives
a measure of the strength of the field. Figure 17.3 b shows another
gravitational field, which is half the strength of the field in Figure 17.3 a.
You should remember that the spacing of the lines is chosen just for
illustrative purposes – another person might have represented the field
strengths in these diagrams with a different separation of the field lines.
311
Radial fields
Figure 17.4 shows the shape of a gravitational field near to Earth; this is a
radial field. Here the field lines all point towards the centre of the planet.
This is why we can use Newton’s law of gravity to calculate the gravitational
forces between two planets. The field is exactly the same shape as it would
have been if all of the mass of the planet were concentrated at its centre C.
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17/04/19 10:37 AM
r
C
planet
surface
You can also see from Figure 17.4 that the gravitational field decreases in
strength with increasing distance from the centre of the planet. The field
lines at a distance of 2r from the centre of the planet are further apart than
they are at distance r, which is the planet’s surface. (Remember that the
planet is a sphere, so that the lines spread out in three dimensions. The
diagram shows the field lines in just one plane.)
We can also produce a formula to describe the field strength close to a
planet. From Newton’s law of gravity, we know that
2r
F=
Figure 17.4 The radial gravitational
field decreases with distance away from
a planet.
Gm1m2
r2
Also if m2 is the mass of a small object close to the surface of the planet,
we know that
g=
F
m2
Therefore
g=
g/N kg–1
8
6
g=
GM
r2
Figure 17.5 shows how the gravitational field strength varies
with height above the Earth.
4
2
5
17 GRAVITATION
r2
Note that we often use a capital M to describe the mass of a
large object such as a star or planet. So using M as the mass of a
planet, and r as the distance away from the centre of the planet,
we have
10
312
Gm1
10
15
20
distance from the centre
of the Earth/km × 103
25
Figure 17.5 A graph of g as a function of distance
from the Earth’s centre.
30
For most planets, treating them as uniform spheres works as a
good approximation. For most objects with a mass larger than
1021 kg, the forces of gravity overcome the massive forces of the
rocks to turn the planet into a sphere. However, smaller moons
and minor planets can have irregular shapes, so Newton’s law of
gravity cannot be used to simply predict fields near them, but it
can be used accurately at large distances.
EXAMPLE
Planets and stars
Answer
1 A minor planet has a mass of 2 × 1022 kg, and it has
a radius of 1200 km. Calculate the gravitational field
strength at its surface.
g=
=
469884_17_AQA_A-Level_Physics_308-328.indd 312
GM
r2
(6.67 × 10–11Nm2kg–2 ) × (2 × 1022 kg)
(1.2 × 106 m)2
= 0.9 Nkg–1
17/04/19 10:37 AM
Answer
g1 =
g2 =
GM
So
 r 2
 1 2
g2 = g1  1  = 300 Nkg–1 ×  = 3 Nkg–1
10 
 r2 
3 Show that the gravitational field strength near to
the surface of a planet or star is given by g = 43 πGρr,
where ρ is the density of the body and r its radius.
r12
Answer
GM
g=
r22
r2
g2 GM
= 2 × 1
GM
g1 r2
 r 2
=  1 
 r2 
GM
Gravitational potential
2 A star has a gravitational field strength at its surface
of 300 N kg−1. Another star has the same mass but
10 times the radius of the first star. Calculate the
gravitational field strength at the surface of the
second star.
r2
G × 4 π ρr 3
3
=
r2
4
= πG ρ r
3
TIP
The volume V of a sphere of radius r is V = 4 πr 3.
3
TEST YOURSELF
In these questions use G = 6.7 × 10 −11 N m2 kg−2.
6 A planet has a mass of 4.6 × 1023 kg and a radius
of 3200 km.
a) Calculate the gravitational field strength at the
planet’s surface.
b) Calculate the gravitational field at a height of
6400 km above the planet’s surface.
7 A star has a gravitational field strength of
400 N kg−1 and a radius of 8.4 × 105 km. At the end
of its life, assume that it collapses to become a
neutron star, of the same mass but with a radius
of 14 km.
a) Calculate the ratio of the star’s initial radius
to the radius of the neutron star it eventually
becomes.
b) Calculate the gravitational field strength on
the surface of the neutron star.
8 Show that there is a negligible difference between
the Earth’s gravitational field strength at sea level
and at the top of Mount Everest. The Earth’s radius
at sea level is about 6400 km, and the height of
Mount Everest is 8.8 km. The mass of the Earth is
6.0 × 1024 kg.
9 This question requires you to think about
gravitational fields and also to recall last year’s
work on v–t and s–t graphs.
A spacecraft is flying away from the centre of
the Earth, at a height of 10 000 km. However, it is
travelling too slowly to escape from the Earth. At
a height of 20 000 km above the Earth’s surface, it
stops moving and begins to fall back to Earth. Use
Figure 17.5 to help you to sketch:
a) a velocity–time graph for the spacecraft
b) a displacement–time graph for the spacecraft.
In both cases, start the graph from the initial
height of 20 000 km, and finish the graphs as the
spacecraft crashes into the Earth.
10 Express the unit for G in SI base units.
313
●● Gravitational potential
You will be familiar with the equation that we use to calculate the increase
in gravitational potential energy when a mass is lifted in a gravitational
field. We calculate the change of gravitational potential energy, ΔEp, using
the equation
ΔEp = mgΔh
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17/04/19 10:37 AM
where m is the mass lifted in kg, g is the gravitational field strength in
N kg−1, and Δh is the increase in height.
Figure 17.6 shows the gravitational field lines (in green) close to the surface
of a planet, where the gravitational field strength is 5 N kg−1. When a 1 kg
mass is lifted through a height of 20 m in this field, the equation above tells
us that the increase in gravitational potential energy of the mass is 100 J
(ΔEp = 1 kg × 5 N kg−1 × 20 m = 100 J). When the mass is lifted through
40 m, the increase in potential energy becomes 200 J.
Gravitational
potential
Height
400 J kg−1
80 m
E
300 J kg−1
60 m
200 J kg−1
C
100 J kg−1
A
D
B
40 m
g = 5 N kg−1
20 m
Planet’s surface
Figure 17.6 Gravitational field lines and equipotentials close to the surface of a
planet, where g = 5 N kg −1.
Gravitational potential difference The
gravitational potential energy difference
per kilogram. Gravitational potential, and
potential difference, have units of J kg−1.
These calculations lead us to the idea of gravitational potential difference,
which can be defined as the change in gravitational potential energy per kg.
Gravitational potential is given the symbol V, and gravitational potential
difference is given the symbol ΔV. Since ΔEp = mgΔh, it follows that
∆V =
∆ Ep
m
= g∆ h
17 GRAVITATION
so
314
ΔV = gΔh
EXAMPLE
Gravitational potential
difference
Gravitational potential has units of J kg−1.
The equation in the main text
can be used to calculate the
magnitude of potential changes.
In Figure 17.6, what is the
gravitational potential difference
between being on the ground and
being at a height of 80 m?
Figure 17.6 also shows equipotentials close to the surface of the planet. In
the diagram these look like lines, but in three dimensions they are surfaces.
On the diagram, equipotential surfaces have been drawn at intervals of
100 J kg−1. When an object moves along an equipotential, it means that the
potential (and therefore the potential energy) stays the same.
Answer
ΔV = gΔh = 5 N kg−1 × 80 m
= 400 J kg−1
Gravitational field A gravitational field g is
linked to the gravitational potential gradient
by the equation
g =–
∆V
∆h
469884_17_AQA_A-Level_Physics_308-328.indd 314
Equipotential lines
Equipotential surfaces are always at right angles to the gravitational field.
When something moves at right angles to the field (and hence along an
equipotential), no work is done by or against the gravitational field, so there
is no potential energy change. When something moves along a field line,
there is a change of gravitational potential energy.
To be exact, in the link between gravitational field and potential, we
should link them with this equation:
g=–
∆V
∆h
17/04/19 10:38 AM
The significance of the minus sign is that the potential gradient
is in
a positive direction upwards, because the potential increases as the height
above the planet increases. The gravitational field direction is downwards.
EXAMPLE
Gravitational potential
energy change
Refer to Figure 17.6.
1 What is the gravitational
potential energy change in
moving a 2 kg mass from A to B?
Answer
The mass moves along an
equipotential, so the change is 0.
2 What is the gravitational
potential energy change in
moving a 2 kg mass from A to C?
Answer
It does not matter which path the
mass takes, the potential change
from A to C is 100 J kg−1. So the
potential energy change is
ΔEp = mΔV = 2 kg × 100 J kg−1
= 200 J
TEST YOURSELF
11 a) E xplain the term ‘gravitational potential difference’.
b) Give the unit of gravitational potential.
c) Explain the term equipotential.
12 Refer to Figure 17.6. Calculate the work done in moving a 5 kg mass
through these distances:
a) A to D
b) C to D
c) B to E.
13 Calculate the gravitational potential gradient in Figure 17.6.
Comment on your answer.
Gravitational potential
∆V
∆h
Gravitational potential in radial fields
In this section we calculate gravitational potential energy
changes over large distances in gravitational fields, which change in
strength. Figure 17.7 shows how the gravitational force on an object of
mass m changes near to a planet of mass M. (Note the common use of M for
the planet and m for a small mass near
the planet).
The graph shows that a force, F, acts on the object at a distance r from the
centre of the planet. If the object is moved a small distance, Δr,
further away from the planet, we can calculate the increase in the
object’s gravitational potential energy as
ΔEp = work done = FΔr
force/N
FΔr is more usually written as mgΔh because the force acting on
the mass is equal to its weight, mg.
How do we calculate the increase in gravitational potential energy
if the mass is moved from r1 to r2? This is more complicated
because the force changes as we move from r1 to r2.
area = F∆r
F
r1
∆r
r2
distance from centre of planet/m
Figure 17.7 Graph of gravitational force acting on
an object in the vicinity of a planet.
In the earlier calculation, FΔr represents a small area under
the graph. So the work done on the mass (or the increase in
gravitational potential energy) in moving from r1 to r2 is the area
under the graph.
The formula to calculate the increase in potential energy is given as
equation (i) (the Maths box shows how the formula is derived):
315
1 1
∆Ep = GMm  − 
 r1 r2  (i)
From this, we can also derive a formula for the increase in potential ΔV, because
∆V =
469884_17_AQA_A-Level_Physics_308-328.indd 315
 1 1
= GM  – 
m
 r1 r2 
∆ Ep
(ii)
17/04/19 10:38 AM
MATHS BOX
Using Newton’s law of
gravitation, the work done on m
is
work done = F ∆r =
GMm
r2
∆r
So the increase in gravitational
potential energy in moving m
from r1 to r2 is
r2
∆ Ep =
∫
r1
GMm
r2
dr
 GMm  r2

= –


r

 r1
 1 1
= GMm  – 
 r1 r2 
Equation (ii) allows us to think about defining gravitational potential close
to a planet. We can calculate the potential change in moving from a distance
r1 to a point infinitely far away from the planet. When r2 = ∞, 1 = 0. So the
r2
potential change in moving from r1 to ∞ is
∆V =
GM
r1 (iii)
However, we choose to define ∞ as the point of zero potential for all planets
and stars. If we chose any other point as zero, such as the surface of the
Earth, we would get a more complicated set of equations when we deal with
potentials near to other planets.
So, since we define the potential as zero at infinity, it means the potential
near to any planet is a negative quantity, because potential energy decreases
as something falls towards a planet. This leads to the following definition of
potential V at a distance r from the centre of a planet of mass M:
GM
r
Figure 17.8 shows the gravitational field lines and equipotentials near
to a planet. The equipotentials are shown in equal steps of 1 × 107 J kg−1
from the surface G, where the potential is −8 × 107 J kg−1, to A, where the
potential is −2 × 107 J kg−1.
V=–
17 GRAVITATION
A
Position
Potential (V) J kg−1
A
−2 × 107
B
−3 ×
107
C
−4 × 107
D
−5 × 107
E
−6 × 107
F
−7 × 107
G
−8 × 107
1
B
C
D
E
2
F G
3
4
316
5
radius of planet = 10000 km
Figure 17.8 This diagram shows gravitational field lines and equipotentials near to a planet.
469884_17_AQA_A-Level_Physics_308-328.indd 316
17/04/19 10:38 AM
This diagram shows two important linked points:
Using the information in
Figure 17.8, copy and complete
Table 17.1 linking potential and
distance from the centre of the
planet.You will need a ruler to
measure the distance of the
equipotentials from the planet
centre.
Table 17.1
1
Potential/ r/ 107 m r /
107J kg–1
10−7 m−1
−8
1.00
1.00
−7
1.14
0.88
−6
1.33
0.75
●
●
The potential gradient is steeper close to the planet’s surface, because the
equipotentials are closer together.
The field lines are closer together near the surface, because the
gravitational field strength g is stronger.
These two statements are linked through the equation you met earlier:
∆V
g=–
∆h
or
g=–
Escape velocity
ACTIVITY
Equipotentials and
variation of potential with
distance
∆V
∆r
These equations are exactly the same, except that Δh has been used for a
change in height, and Δr has been used for a change in distance from the
centre of a planet.
●● Escape velocity
−5
−4
−3
−2
The Earth has its atmosphere because the molecules of gas, moving in our
atmosphere, do not have enough kinetic energy to escape from the pull of
gravity at the Earth’s surface. So how fast does something have to move to
escape from the Earth’s surface?
When a fast-moving object leaves the surface of a planet, we can write that:
decrease in kinetic energy = increase in gravitational potential energy
ΔEk = ΔEp
1 Plot a graph of potential
against r.
2 Use the gradient of the graph
to determine the planet’s
gravitational field at a distance
of
a) 20 000 km from the centre
b) 40 000 km from the centre.
3 Plot a graph of potential
against r1. Use the graph to
determine the mass of the
planet.
Escape velocity The minimum velocity an
object must have at the surface of a planet
to escape the pull of gravity using its own
kinetic energy.
This assumes that the object is not affected by an atmosphere, and is in free
fall – this is not a spacecraft with a rocket.
The equation above can be written as
1 2 1 2
mv – mv = m ∆V
2 1 2 2
If the object is to just escape the pull of the planet, its speed, v2, will just
reach zero at an infinite distance from the planet. This leads to the idea of
escape velocity, which is the minimum velocity that an object must have
at the surface of a planet in order to escape the pull of gravity of the planet
using its own kinetic energy.
The gravitational potential at the surface of a planet is given by
V=–
GM
r
317
so the change in potential is
GM
r
So the escape velocity for a planet can be calculated using
∆V =
GMm
1 2
mv = m ∆V =
2
r
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17/04/19 10:38 AM
or
2GM r
For the Earth, M = 6 × 1024 kg and r = 6400 km, so
v2 =
v2 =
(iv)
2 × (6.67 × 10 –11 N m 2 kg –2 ) × (6 × 1024 kg)
6.4 × 106 m
v = 11 200 m s –1
Since air molecules travel at approximately 500 m s−1 on the surface of the
Earth, they travel well below the Earth’s escape velocity.
TEST YOURSELF
14 E xplain why we choose to define the zero point of
17 GRAVITATION
gravitational potential at an infinite distance from
any planet.
15 Calculate the gravitational potential at the
surface of Jupiter. The mass of the planet is
1.9 × 1027 kg and its radius is 70 000 km.
16 This question is based on the information in
Figure 17.8.
a) Calculate the work done in taking a 1200 kg
spacecraft from
i) point 1 to point 2
ii) point 2 to point 3
iii) point 3 to point 4
iv) point 4 to point 5.
b) Use your answer to part (a) to explain why a
spacecraft can stay in a circular orbit round a
planet indefinitely.
c) The spacecraft returns to the planet. It passes
point 5 travelling at a speed of 5200 m s−1, and
falls freely to point 2. How fast is it travelling as
it passes point 2?
17 The radius of the planet shown in Figure 17.8
is 10 000 km. Either by estimating the distance
between the equipotentials near the planet’s
surface, or otherwise, calculate the value of g
near the planet’s surface.
18 When a large star collapses at the end of its
life, it can collapse into a black hole. The pull
of gravity at its surface is so strong that not
even light can escape. Einstein’s theory of
general relativity predicts that black holes are
‘singularities’, which means they have collapsed
into a tiny space. However, we can use Newton’s
theory to calculate the maximum size of such
a hole.
a) Use the equation for escape velocity (marked
(iv) in the text) to calculate the maximum
radius for a black hole formed by a star of
mass 10 31 kg. The speed of light is 3 × 108 m s−1.
b) Calculate:
i) the gravitational field at this surface
ii) the gravitational potential at this surface.
●● Orbits
In Figure 17.9, a planet of mass m is in a circular orbit around a star of
mass M. You have already studied circular motion. Now we can combine
the equations of circular motion and gravitation to link the speed or time
period of a planet’s orbit to its distance from the Sun.
318
The pull of gravity provides the necessary centripetal force to keep the
planet in orbit. So we can write
M
m
r
Figure 17.9
469884_17_AQA_A-Level_Physics_308-328.indd 318
GMm mv2
=
r
r2
GM v2
=
r
r2
GM
v2 =
r
17/04/19 10:38 AM
time of orbit/years
150
2000
1500
Mercury
velocity
time
Venus
100
1000
Earth
Mars
50
500
Jupiter
Saturn
Uranus
orbit velocity/106 km year−1
200
Orbits
From this, you can see that the speed of the orbit is faster for small orbits.
Figure 17.10 shows the link between orbital speed and the distance of our
eight planets from the Sun.
Neptune
1000
2000
3000
4000
average distance from the Sun/106 km
Figure 17.10
We can also link the time period of the orbit (1 year for the Earth) to the
radius of the orbit. The speed of the orbit is linked to the circumference,
2πr, and the time period of the orbit, T, through the equation
v = ωr =
so
v2 =
r
3
T2
=
2πr
T
4π 2r 2
2
T
GM
=
GM
r
4π 2
 4π 2 
 r 3
T 2 = 
 GM 
Figure 17.10 also shows the relationship between time period and orbital
radius of the planet – the green curve.
Elliptical orbits
319
In the previous section, we treated all orbits as if they were circular – this
is because it is relatively easy to cope with the mathematics of circular
orbits. In practice, very few orbits are circular – most orbits have an
elliptical shape. Figure 17.11 shows a possible elliptical orbit for a comet
(black dot) travelling round the Sun. Most planetary orbits are nearly
circular. For example, the Earth’s closest approach to the Sun (perihelion)
is 147 × 106 km and its furthest distance (aphelion) is 152 × 106 km.
However, comets and some minor planets have elongated (or eccentric)
elliptical orbits.
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17/04/19 10:38 AM
D
FD
FA
A
Sun
FC
O
F2
F1
C
FB
B
Figure 17.11 Comets move in elongated or eccentric
elliptical orbits round the Sun. Note that the Sun’s solar
wind always blows the comet’s tail away from the Sun.
You should note that the relationship derived in the last section linking the
time period of an orbit to the radius of the orbit is still valid for elliptical
orbits, provided r is taken to be half the major axis of the ellipse. The major
axis in Figure 17.11 is the distance AC, and the minor axis is the distance
BD. So, for the elliptical orbits we can write
 4π 2 
( r )3
T 2 = 
 GM  sma
where rsma is the semi-major axis, equal to AO or OC.
When a planet or comet moves in an elliptical orbit, its speed changes, but
the total energy of the body stays the same.
At point A in Figure 17.11, the comet is moving at its fastest orbital
speed, but it has the smallest gravitational potential energy, because it is
closest to the Sun.
17 GRAVITATION
At point B, there is a component of the Sun’s gravitational pull on the
comet, which slows it down.
320
When the comet reaches point C, it is at its furthest point from the Sun.
It has its lowest kinetic energy at this point, but its highest gravitational
potential energy.
At point D, the comet is falling back towards the Sun. There is a component
of the Sun’s gravitational pull, which speeds it up. The comet’s potential
energy is being transferred into kinetic energy, and the comet reaches its
maximum speed again at A.
Satellite orbits
There are many satellites in orbit round the Earth, which are used for
a range of purposes. Two of the most common uses of satellites are
communications and observation.
Satellites placed in low orbits are able to take photographs of the world
below. We are used to seeing images of mountain ranges, lakes and cities
taken from space, and we use weather images on a daily basis.
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17/04/19 10:38 AM
low polar orbit
Figure 17.12
Satellites placed in higher orbits are useful for communications, because
messages may be sent from one part of the world to another via the satellite.
Different types of orbit are illustrated in Figure 17.12.
One of the most useful orbits for satellites is the geosynchronous orbit.
In this case the satellite is placed in an orbit above the Earth’s equator, at
such a height that it takes exactly one day to complete an orbit. The orbit is
synchronised with the Earth’s rotation, so that it remains in the same place
above the Earth’s surface. This means that our satellite dishes, for example,
can be aligned with a satellite, which always lies in the same position
relative to Earth.
Orbits
orbit over the
equator
EXAMPLE
Geosynchronous satellite
Calculate the height of a geosynchronous satellite in orbit above the
Earth’s surface.
Answer
We can use the equation we derived earlier in this chapter:
 4π2 

r 3
T 2 = 
 GM 
or
 GM 
r 3 =  2 T 2
 4π 
r3 =
(6.67 × 10–11 Nm2 kg–3 )×(6.0 × 1024 kg)×(24 × 3 600 s)2
4π 2
6
r = 42.4 ×10 m
Since the radius of the Earth is 6.4 × 106 m, the height of the orbit is
about 36 × 106 m or 36 000 km.
When a satellite is launched, it requires more energy to place it in a
higher orbit, even though it travels more slowly. This is because we
have to increase the potential energy of the satellite more to place it in
a high orbit.
MATHS BOX
In its orbit of radius r2, the satellite’s potential energy
How much energy has to be supplied to a satellite of
mass m to lift it into an orbit of radius r2 above the
Earth?
is
Answer
calculated as follows.
At the Earth’s surface, radius r1, the satellite’s
In a circular orbit we can write
and its kinetic
gravitational potential energy is – GMm
r1
energy is zero.
469884_17_AQA_A-Level_Physics_308-328.indd 321
–
GMm
.
r2
The satellite’s kinetic energy in orbit may be
321
mv2 GMm
= 2
r2
r2
17/04/19 10:39 AM
so
Ek =
So the work done to put a satellite in orbit is the
difference between its energy in its orbit and its
energy on the Earth’s surface:
1 2 GMm
mv =
2
2r2
The satellite’s total energy is the sum of its kinetic
energy and potential energy:
Ek + E p =
ACTIVITY
The moons of Saturn
Saturn is thought to have 62 moons in orbit around
it. Table 17.2 shows information about six of Saturn’s
inner moons.
17 GRAVITATION
Table 17.2
Moon
Orbital
radius/103 km
Orbital period/
days
Atlas
137
0.6
Mimas
185
0.9
Methone
194
Enceladus
238
1.4
Tethys
295
1.9
Dione
377
2.7
The orbital period and orbital radius are linked by the
equation:
 4π2 

r 3
T 2 = 
 GM 
1 Draw up a table of T2 and r 3 values (T should be
calculated in seconds, and r in metres). Plot a
graph of T2 against r 3. From your graph, determine:
a) Methone’s orbital period
b) Saturn’s mass.
2 Dione has the same orbital radius about Saturn as
our Moon does about Earth. Our Moon has an orbital
period of about 27 days, which is 10 times longer
469884_17_AQA_A-Level_Physics_308-328.indd 322
GMm  GMm 
– –

r1 
2r2 
1 1 

= GMm  –

r
2
r
 1
2 
GMm GMm
GMm
–
=–
2r2
r2
2r2
This shows us that the total energy approaches zero
as r2 tends to infinity, and it is more negative for
small value of r2. You will remember that we chose to
define the zero point of potential energy at infinity.
322
work done = –
In practice, the kinetic energy of the satellite at the
Earth’s surface is not zero, as the Earth is rotating. This
kinetic energy reduces the work done necessary to put
a satellite in orbit and is one reason why launch sites are
near the equator where the speed of rotation is greatest.
than Dione’s. Use the expression above to deduce
the ratio of Saturn’s mass to the Earth’s mass.
3 You could also check the relationship between T
and r by plotting a graph of log10 T against log10 r.
 4π2 

r 3
T 2 = 
 GM 
By taking the log of both sides we get
 2  
 4π  3 
logT 2 = log 
r 
 GM  
Therefore
 4π2 

 + log r 3
logT 2 = log
 GM 
and then
 4π2 

 + 3 log r
2 logT = log
 GM 
1  4π2  3
logT = log
 + log r
2  GM  2
When you plot a graph of log T against log r, you
should find
that its gradient is 3 , which confirms the relationship
2
that T 2 is proportional to r 3.
17/04/19 10:39 AM
19 A satellite is in a low orbit over the poles of the
Earth. It is 300 km above the surface of the Earth.
The gravitational field strength at the surface of
the Earth is 9.8 N kg −1, and the radius of the Earth
is 6400 km.
a) i) Explain how this satellite might be used.
ii) State another use of satellites, and explain
what orbit you would use for the satellite
you have chosen.
b) i) Calculate the gravitational field strength at
a height of 300 km.
ii) Calculate the orbital period of the satellite.
20 a) E xplain the difference between a planet’s orbit
and a comet’s orbit.
b) E xplain why a comet’s orbital speed changes
throughout its orbit.
21 A satellite takes 120 minutes to orbit the Earth.
a) Calculate the satellite’s angular velocity.
b) Calculate the radius of the satellite’s orbit.
Take GME to equal
4.0 × 1014 N kg−1 m2.
22 The Sun has a mass of 2 × 10 30 kg; the radius of
the Earth’s orbit is 1.5 × 108 km; the Earth’s mass
is 6 × 1024 kg.
a) i) Calculate the angular velocity of the Earth.
ii) Calculate the Earth’s orbital speed.
b) i) Calculate the centripetal force on the Earth.
ii) Show that the centripetal force on the Earth
is equal to the Sun’s gravitational pull on
the Earth.
c) Calculate the Sun’s gravitational field strength
at a distance equal to the Earth’s orbit round it.
Orbits
TEST YOURSELF
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Practice questions
1 The gravitational field strength at the surface of a planet, of radius
8 000 km, is 15 N kg−1. The gravitational field strength at a height of
4000 km above the planet is
A 10 N kg−1
C 6.7 N kg−1
B 8.0 N kg−1
D 4.4 N kg−1
2 The gravitational field strength on the surface of the Earth is g. The
gravitational field strength on a planet with twice the mass of the Earth
and twice the radius of the Earth is
A g/8
C g/2
B g/4
Dg
3 The gravitational potential on the surface of a planet with mass M and
radius R is −V. The potential on a second planet with mass 2M and
radius R/3 is
A −2V/3
C −2V
B −3V/2
D −6V
4 A satellite is in a circular orbit round a planet of radius 5200 km, at
a height of 1800 km. At this height the gravitational field strength is
4.2 N kg−1. The speed of the satellite is
A 5.4 km s−1
C 2.4 km s−1
B 4.5 km s−1
D 0.2 km s−1
5 The centres of two planets, each of mass M, are separated by
a distance r.
17 GRAVITATION
Which of the following correctly gives the gravitational field
strength and the gravitational potential, at a point halfway between
the centres of the planets?
324
A
B
C
D
Gravitational field
8 GM/r 2
4 GM/r 2
0
0
Gravitational potential
0
−2 GM/r
0
−4 GM/r
6 The gravitational field strength at the surface of the Earth is 9.8 N kg−1.
At the surface of the Moon the field strength is 1.7 N kg−1. The Earth has
a mass 81 times that of the Moon. The ratio of the Earth’s radius to the
Moon’s radius is
A 2.9
C 4.9
B 3.7
D 7.6
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7 Two stars of mass M and 4M are a distance r apart (Figure 17.13).
Star B
x
M
4M
r
Figure 17.13
The resultant gravitational field strength is zero along the line between
their centres at a distance x from the centre of the star with mass M. The
ratio of x/r is
A 3/4
C 1/2
B 2/3
D 1/3
Practice questions
Star A
8 The diameter of the Earth is twice that of Mars and the mass of the Earth is
10 times that of Mars. The gravitational potential at the surface of Mars is
−13 MJ kg−1. The gravitational potential at the surface of Earth is
A −290 MJ kg−1
C −65 MJ kg−1
B −120 MJ kg−1
D −28 MJ kg−1
9 A satellite is in orbit above the Earth at a distance of 9000 km from the
Earth’s centre. At this height the gravitational field strength is 5.0 N kg−1.
The time period of the orbit of the satellite is
A 90 minutes
C 180 minutes
B 140 minutes
D 270 minutes
10 The time period, T, of a body orbiting the Sun is given by the formula
4πR 3
T2 =
GM
where M is the mass of the Sun and R is the radius of the
orbit. Halley’s comet takes 76 years to orbit the Sun. The ratio
average radius of Halley’s comet orbit
average radius of the Earth’s orbit is
A 9
C 76
B 18
D 660
 2
g 
11 a) Work out the correct unit, expressed in SI base units, for  G  .(2)




b) The gravitational field strength at the surface of the Earth is six
times the gravitational field strength on the surface of the Moon.
The mean density of the Moon is 0.6 times the mean density
 radius of Earth 
of the Earth. Calculate the ratio:  radius of Moon  .(3)


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12 In Figure 17.14, the gravitational potential at A is −16 MJ kg−1 and the
gravitational field strength at A is 4 N kg−1.
planet
A
B
r
2r
Figure 17.14
a) Calculate the work done in taking a 120 kg mass from A to B.(3)
b) Calculate the gravitational field strength at B.(2)
13 The gravitational field strength at the surface of the Sun is
270 N kg−1. Betelgeuse is a red giant star, which has a density of
approximately 0.01 times that of the Sun and a radius about
1000 times that of the Sun.
Calculate the gravitational field strength on the surface of Betelgeuse.(3)
14 At point P in Figure 17.15, the gravitational field strength is zero,
and the gravitational potential is −8.0 × 107 J kg−1.
planet A
planet B
3r
P
r
Figure 17.15
17 GRAVITATION
a) Calculate the work done to remove a spacecraft of mass 600 kg
to a point infinitely far away from these planets.
326
b) Calculate the ratio: mass of planet A .
mass of planet B
15 Io is a moon of Jupiter. Io rotates around Jupiter once every
42 hours, in an orbit of radius of 420 000 km.
(3)
(3)
(3)
a) Calculate the angular velocity of Io.
b) Use the data above to calculate the mass of the planet Jupiter.
(3)
2
to T 3 ,
where
c) The radius of the orbit of a moon is proportional
T is the time period of the orbit. Ganymede is another moon of
Jupiter that takes 168 hours to rotate around the planet.
Calculate the radius of Ganymede’s orbit.
(4)
16 Two identical spheres exert a gravitational force F on each other.
a) What gravitational force do two spheres, each twice the mass
of the original spheres, exert on each other when the separation
of their centres is four times the original separation?
(2)
b) The gravitational force between two uniform spheres is
3.7 × 10−9 N when the distance between their centres is
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c) The gravitational potential difference between the surface of
a planet and a point 20 m above the surface is 800 J kg−1.
Calculate the gravitational field strength in the region close
to the planet’s surface.
(3)
(3)
17 a) Calculate the time period of the Earth’s rotation, if you were to
be made to feel weightless at the equator. The radius of the Earth
(3)
is 6.4 × 106 m.
Practice questions
200 mm. The mass of one sphere is 3.0 kg; calculate the mass
of the second sphere.
b) A satellite is in orbit, of radius r, around a planet of mass M.
Write down expressions for
i) its orbital speed
(2)
ii) the time period of its orbit.
(2)
18 Figure 17.16 shows a sketch of the Earth–Moon system. The
gravitational potential at the surface of the Earth is −62.8 MJ kg−1;
the gravitational potential at the surface of the Moon is −2.3 MJ kg−1; the
gravitational potential at point N is −1.3 MJ kg−1. Point N is the neutral
point between the Earth and Moon where the gravitational field is zero.
Earth
Moon
N
r1
r2
Figure 17.16
a)
The Earth is 81 times as massive as the Moon. Calculate
r
the ratio r1 .(2)
2
b) i) Calculate the minimum amount of energy required to
move a space probe of mass 2.0 × 104 kg from the Earth
to point N.
ii) Explain why no more fuel is required to take the space
probe from point N to the Moon.
c)
The amount of fuel required to take a spacecraft to the Moon
is much higher than that required to return it to Earth. Explain
why this is so referring to the forces involved – gravitational
field strength and gravitational potential.
Stretch and challenge
(3)
(1)
(6)
327
19 M87 (Messier Catalogue number 87), is a giant galaxy, and is about
6 × 1012 times as massive as our Sun. The gravitational pull of this
galaxy keeps star clusters in orbit around it. In the centre of this
galaxy is a giant black hole of about 5 × 109 solar masses.
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a) The event horizon of the black hole is the maximum radius from
which something can just escape the black hole travelling at the
speed of light.
i) Write down an expression for the gravitational potential at this
point, in terms of the mass of the black hole, M, and the radius
of the event horizon, r.
ii) Write an expression for the kinetic energy of a kilogram mass
travelling at the speed of light, c.
iii) Calculate the radius of the event horizon. Take the mass of the
Sun to be 2 × 1030 kg, the speed of light to be 3 × 108 m s−1, and
G = 6.7 × 10−11 N kg−2 m2.
b) i) A globular cluster of stars orbits M87 at a distance of 300 000
light years. Calculate the time period of the orbit.
[1 light year = 9.5 × 1015 m]
ii) What happens to any stars or clusters of stars near the galaxy if
they rotate too slowly to stay in an orbit?
20 Figure 17.17 shows the orbit of a comet as it falls in towards
the Sun and then leaves the inner Solar System again. The
gravitational potential due to the Sun has the following values
at these distances from the Sun: at Saturn’s orbit, −93 MJ kg−1; at
Jupiter’s orbit, −172 MJ kg−1; at the Earth’s orbit, −893 MJ kg−1.
D comet’s orbit
Saturn’s
orbit
Jupiter’s
orbit
A
a) From this information calculate this ratio:
radius of Saturn’s orbit
radius of Jupiter’s orbit
B
C
Earth’s
orbit
17 GRAVITATION
b) As the comet moves from point A to point B, it increases its
speed. Explain why.
c) i) The comet has a mass m. At point A its speed is vA and
at point B its speed is vB. Write an equation to link the
comet’s increase in kinetic energy to its decrease in
gravitational potential energy.
Figure 17.17
ii) At point A the comet’s speed is 3 × 104 m s−1. Calculate the
comet’s speed at point B.
iii) State the comet’s speed at points C and D.
328
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18
Thermal physics
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
●
●
●
The kinetic theory model of solids, liquids and gases assumes that
particles are incompressible spheres.
Solids have a close-packed, regular particle structure – the particles
vibrate about fixed points.
Liquids have a close-packed, random, irregular particle structure –
the particles are free to move.
Gases have a widely spaced, irregular particle structure – the
particles move at high speed in random directions.
Thermal energy can be transferred from somewhere hot (at a high
temperature) to somewhere cooler (at lower temperature) by the
processes of conduction, convection, radiation and evaporation.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 Draw simple diagrams showing the arrangements of particles inside
a solid, a liquid and a gas.
2 Explain the difference between the transfer of thermal energy, from a
hot body to a cold body, through conduction and through convection.
3 Explain how evaporation transfers thermal energy away from a hot
cup of tea.
●● Thermodynamics
During the late 18th and early 19th centuries scientists, inventors and
engineers began to develop steam engine technology, such as the giant
steam-powered beam-engine pumps used to pump water out of deep
Cornish tin mines.
Development of the engines required a systematic and fundamental
understanding of the nature of heat energy; its relationship to the behaviour
of steam and the other materials making up the engines; and the work done
by the engine. This study became known as thermodynamics and Britain
led the world, not only in the development of the new engines, but also in
the fundamental physics of thermodynamics.
329
Thermodynamics deals with the macroscopic (large-scale) behaviour of
a system, but it is complemented by the kinetic theory of matter, which
deals with the microscopic, particle-scale behaviour of matter. Some
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aspects of thermal physics are best explained in terms of macroscopic
thermodynamics, such as the behaviour of engines, but other aspects
are best explained using microscopic kinetic theory, such as Brownian
motion (the tiny random motion of pollen or smoke particles seen under
a microscope).
Internal energy
Internal energy The sum of the randomly
distributed kinetic and potential energies of
the particles in a body.
One of the most fundamental properties of thermodynamics is the concept
of internal energy, U, which is the sum (sometimes called an ensemble in
thermodynamics) of the randomly distributed kinetic energies and potential
energies of the particles in a body:
U = ∑(kinetic energies) + ∑(potential energies)
Consider a glass of water (Figure 18.1). The water particles have two types
of energy – kinetic energy associated with their movement (the faster they
move, or vibrate or rotate, the higher their kinetic energy) and potential
energy associated with any forces or interactions between the particles (such
as any electrostatic attraction or repulsion).
kinetic
energy
potential
energy
Figure 18.1 Glass of water, showing
internal energy.
Fluid A substance that can flow – i.e. a gas
or a liquid.
The kinetic energies of the particles depend on their temperature, and
the potential energies depend on any intermolecular forces between the
particles. For ideal gases, in which there are no intermolecular forces, the
internal energy is dependent on only the kinetic energies.
The first law of thermodynamics
The physicists working on the theories of how engines worked quickly
realised that there was an interplay between the changes in heat energy and
the work being done on or by the fluids in the engines. This was formalised
by the first law of thermodynamics, written by Rudolf Clausius in 1850.
A modern version of his law can be stated as follows:
18 THERMAL PHYSICS
The increase in internal energy of a system is equal to the heat added to
the system minus the work done by the system.
330
cylinder
frictionless piston
In terms of symbols, this can be written:
ΔU = ΔQ − ΔW
pressure, p
F = pA
piston area = A
piston moves
through a
distance ∆x
causing an
increasing
volume ∆V
where ΔU is the increase in internal energy of the system (usually a gas), ΔQ
is the thermal energy added to the system and ΔW is the work done by the
system.
Work done by an expanding gas
When a gas expands, it exerts a force on the surroundings, causing them to
move – the gas does work on the surroundings. We can use the first law of
thermodynamics to determine the work done, ΔW, by an expanding gas at
constant temperature (called an isothermal change). Consider a gas enclosed
Figure 18.2 A gas expanding in a cylinder. in a cylinder by a frictionless piston, as shown in Figure 18.2.
The gas of volume V exerts a pressure p on the walls of the cylinder. This in
turn exerts a force F on the frictionless piston of area A, where
F = pA
This causes an increase in the volume, ΔV. We assume that ΔV is very small
and that the force moves the piston at a slow but steady rate such that the
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1 What is thermodynamics?
2 State two ways in which the
internal energy of a gas inside
a bicycle tyre pump can be
increased.
3 Calculate the work done on a
gas when its internal energy
increases by 1864 kJ as it is
heated, causing its thermal
energy to increase by 1247 kJ.
ΔW = −FΔx
substituting for F = pA gives
ΔW = pAΔx
But AΔx = ΔV, the change in volume of the gas, so
ΔW = pΔV
●● Heating up substances and changes
of state
Heating up substances and changes of state
TEST YOURSELF
external force exerted on the piston is equal to the force exerted by the
pressure p of the gas in the cylinder. This effectively makes the pressure
exerted by the gas constant during the expansion. The gas does work, and
so ΔW is positive. The force on the piston moves it through a distance, Δx,
such that:
When substances are heated, thermal energy is supplied to the particles of
the substance, increasing their internal energy U, and therefore the average
kinetic energy of the particles. Increasing the average kinetic energy of
the particles causes the temperature of the particles to rise. The size of
temperature change, Δθ, is dictated by several macroscopic, measurable
factors: the amount of thermal heat energy supplied, Q; the mass of
the substance, m; and a quantity called the specific heat capacity of the
substance, c, which is unique to each substance; and its state. These factors
are related to each other by the equation:
Q = mcΔθ
The thermal energy Q is measured in joules (J), the mass m is measured in
kilograms (kg), and the temperature change Δθ is measured in kelvin (K),
so the units of specific heat capacity, c, are J kg−1 K−1.
The specific heat capacity of a material is a fundamental property of the
material and is particularly important to engineers and scientists designing
engines and insulation systems. A specific heat capacity dictates how easy it
is for a material to change its temperature. Materials with very high specific
heat capacities, such as water, cw = 4186 J kg−1 K−1 (usually rounded up
to 4200 J kg−1 K−1), require a great deal of thermal energy to increase the
temperature of 1 kg of the material by 1 K, whereas materials such as gold
with quite low specific heat capacities, cAu = 126 J kg−1 K−1, require only a
small quantity of thermal energy to increase the temperature of 1 kg of the
material by 1 K.
Water has a particularly high specific heat capacity. Other common
materials on Earth have substantially lower values: granite rock, for
example, has a specific heat capacity of 790 J kg−1 K−1, less than one fifth
that of water. Without this property, life may not have been possible on
Earth, because water would almost alway be in the gaseous state.
331
The specific heat capacity of a material enables us to measure the change in
temperature of a material following a change in thermal energy.
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EXAMPLE
Warming water
An aluminium saucepan is used to warm 1.50 kg
of tap water (at 18°C) for a hot-water bottle
by heating it on a 3.0 kW electric hob for 4.0
minutes. Assuming that 60% of the electrical
energy is absorbed by the water, and that there
are no subsequent heat losses, calculate the
final temperature of the warm water. The specific
heat capacity of water is cw = 4186 J kg−1 K−1.
Thermal energy supplied to the water is
Q=
= 4.32 × 105 J
But Q = mcΔθ so
∆ θ=
Answer
=
Total electrical energy produced by the electric
hob is
E = 3.0 × 103 W × 4.0 × 60 s
= 7.2 × 105 J
60
× 7.2 × 105 J
100
Q
mc
4.32 × 105 J
1.50 kg × 4186 Jkg–1 K–1
= 68.8 K= 69 K (2 s.f.)
Because a temperature change of 1 K is equal to a
temperature change of 1°C, the final temperature of the
water in the saucepan is 18°C + 69°C = 87°C.
EXAMPLE
Falling lead shot
The specific heat capacity of lead can be determined by letting
lead shot fall inside a long tube. The lead shot heats up as
gravitational potential energy is transferred to thermal energy
of the shot. The experiment is shown in Figure 18.3.
18 THERMAL PHYSICS
A student tipped some lead shot up and down in the tube and
found that after 20 turns the temperature of the lead had risen
by 1.5°C. Estimate the specific heat capacity of lead. You may
assume that the tube itself does not warm up.
Note: If you are handling lead shot, make sure to wash your
hands afterwards.
Answer
The gravitational potential energy of the falling lead is
transferred to heat in the lead. So
mgh = mcΔθ
and (because a temperature change of 1°C is equal to a
temperature change of 1 K) we obtain
c=
332
=
stopper
measure
temperature
before and
after falls
lead shot
allow shot to
fall 20 times
(turn tube
over and
over)
distance = 1 m
falls down
tube
closed
tube
stopper
Figure 18.3 Lead shot experiment.
gh
∆θ
9.8 Nkg–1 × 20 m
1.5 K
=130 Jkg–1 K–1
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If a hot liquid or a solid is placed into a cold liquid, the internal energy
transferred from the hot object when it cools down is equal to the thermal
energy gained by the cold liquid and its container, plus the thermal energy
lost to its surroundings. In the example shown in Figure 18.4, the thermal
energy lost to the surroundings is assumed to be negligible.
Mass m1 of liquid 1
at temperature T1
Test tubes
Mixed
Heating up substances and changes of state
Mixing hot with cold
Mass m1 + m2 of mixture
of liquids 1 and 2 and
test tube at
temperature T3
Mass m2 of liquid 2
at temperature T2
Figure 18.4 Specific heat capacity and mixtures.
EXAMPLE
Mixing hot and cold liquids
In an experiment, 20.0 g of hot seawater at 65°C
is mixed with 80.0 g of tap water at 12.0°C inside a
copper calorimeter of mass 75.0 g also at 12°C. If the
thermal energy lost to the surroundings is negligible,
calculate the new temperature of the mixture and the
calorimeter. The specific heat capacities of seawater,
tap water and copper are 3990 J kg−1 K−1, 4200 J kg−1 K−1
and 386 J kg−1 K−1, respectively.
Answer
The new (unknown) final temperature of the water
mixture and the calorimeter we will call T (°C). So
(because a temperature change in °C is equal to a
temperature change in K) the thermal energy Q lost by
the hot seawater is:
This is equal to the thermal energy Q gained by the tap
water and the copper calorimeter:
Q = [80.0 × 10−3 kg × 4200 J kg−1 K−1 × (T − 12) K] J
+ [75.0 × 10−3 kg × 386 J kg−1 K−1 × (T − 12) K] J
= [336T − 4032] J + [28.95T − 347.4] J
= (364.95T − 4379.4) J
Equating these two values and rearranging gives
5187 − 79.8T = 364.95T − 4379.4
444.75T = 9566.4
333
T = 21.5°C
Q = mseawater × cseawater × Δθseawater
= 20.0 × 10−3 kg × 3990 J kg−1 K−1 × (65 − T) K
= (5187 − 79.8T) J
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ACTIVITY
Measuring the specific heat capacity of
a metal block
The specific heat capacity of a solid material can be
measured (with reasonable certainty in the laboratory)
by heating a known mass of the material with a known
quantity of thermal energy, usually supplied via an
electrical heater. One such experiment involving a
copper block is shown in Figure 18.5.
The results of the experiment are shown in Table 18.1.
18 THERMAL PHYSICS
Table 18.1
334
Time, t /s
Temperature, T/°C
0
16.4
20
16.5
40
16.4
60
16.4
80
16.3
100
16.4
120
16.4
140
19.1
160
21.9
180
24.6
200
27.4
220
30.1
240
32.9
260
35.6
280
38.4
300
41.1
320
40.8
340
40.5
360
40.2
380
39.9
400
39.6
420
39.3
469884_18_AQA_A-Level_Physics_329-357.indd 334
15
stopwatch
30
V
thermal insulation
(e.g. plastic bubble wrap)
12 V dc electric heater
A
thermometer
material under test
(e.g. copper block)
12 V dc
Figure 18.5 Experiment to measure the specific heat
capacity of a metal block.
1 Plot a graph of the results and draw a smooth bestfitting line through the points.
2 A calorimetric technique is used to determine the
temperature change of the block. The cooling part of
the graph is used to take into account the heat still
present in the electric heater when it was turned off
but had not transferred into the block. A sketch of
how to use this technique is shown in Figure 18.6.
temperature T/K
In this experiment, the mass of the copper block was
measured with an electronic balance and was found
to be 0.814 kg. The block is heated using a stabilised
12.0 V dc power supply delivering 4.0 A of current to
the electric heater in the block. The temperature
of the block was measured every 20 s for 2 minutes
while all the apparatus came to thermal equilibrium.
After 2 minutes the heater was switched on and
the temperature recorded every 20 s again for a
further 3 minutes, before it was switched off. The
temperature of the block continued to be measured
every 20 s for a further 2 minutes, during which time
the block started to cool down.
60
45
∆T
time t/s
Figure 18.6 Graph showing how to calculate
the temperature change.
The change in temperature of the block, ΔT,
will always be slightly higher than the highest
temperature reached minus the starting
temperature – this accounts for the extra thermal
energy left in the heater when it is switched off. In
other words, ΔT gives the temperature the block
would have reached if all the energy could be
transferred instantly to the block, without any heat
being lost. Use this graphical calorimetric
17/04/19 10:41 AM
5 Explain how repeating the experiment would lead
to determining the uncertainty in the measurement
of the specific heat capacity of copper using this
technique.
6 This experiment suffers from a collection of
random and systematic errors. Identify these
errors, state whether they are random or
systematic in nature, and suggest ways in which
they could be minimised.
TEST YOURSELF
4 A student uses a microwave oven to warm up a cup
of cold tea.
a) Thermal energy is supplied to the tea at a rate
of 750 W. The tea has a mass of 0.42 kg and an
initial temperature of 17°C. Calculate the final
temperature of the tea. Assume that the specific
heat capacity of the tea is 4200 J kg−1 K−1.
b) In reality, some of the thermal energy goes into
the cup and some is used by (water) particles to
evaporate. What is the effect of this evaporation
on the final temperature of the tea?
c) Following re-heating the tea, the student decides
that the tea is too strong and adds milk from
the fridge at a temperature of 5.5°C, and the
temperature of the tea drops to 71.0°C. During
this time the student assumes that no thermal
energy is lost to the surroundings. Calculate the
decrease in thermal energy of the tea.
d) If all the thermal energy transferred by the tea is
used to heat up the milk, calculate the mass of
the milk added by the student to the tea. Take the
specific heat capacity of milk to be 4000 J kg−1 K−1.
5 A college sports dome has an internal air volume of
24 000 m3.
a) If the air inside the dome has a density of
1.2 kg m−3, calculate the mass of air inside the
dome.
b) During winter, the dome is kept inflated with
air pumped in from outside with a temperature
of 5.0°C. Overnight, the heater in the dome is
turned off, and the average temperature of the
air in the dome falls to 5.0°C. Calculate the
thermal energy required to heat the air in the
dome to a more pleasant 16°C in the morning.
The specific heat capacity of air is 1000 J kg−1 K−1.
c) The dome contains four industrial space heaters
rated at 14.7 kW. If the space heaters are 100%
469884_18_AQA_A-Level_Physics_329-357.indd 335
efficient, how long will it take for the heaters to
warm the air up in the dome to 16°C?
d) Explain why the actual energy value required
to heat up the dome will be larger than that
calculated in (b).
6 A large tropical fish tank has dimensions of
240 cm × 60 cm × 60 cm.
a) If the density of water is 1.0 g cm−3, calculate the
mass of water in the tank (in kg).
b) The tank is set up using water from an outside
water butt with a temperature of 9.5°C. The
thermostat on the heater is set to 25.5°C.
Calculate the thermal energy needed to
warm up the water in the tank to the desired
temperature. The specific heat capacity of water
is 4200 J kg−1 K−1.
c) The tank is kept at its optimum temperature of
25.5°C by a 100 W heater and thermostat. If the
heater should develop a fault and fail, show that
the initial rate of fall of temperature in the tank
will be about 0.1°C per hour.
d) The rate of fall of temperature of the water in
the fish tank, Δθ/Δt, can be described using
Newton’s law of cooling:
Heating up substances and changes of state
technique to determine the temperature change of
the copper block in this experiment.
3 Use the rest of the data to calculate the specific
heat capacity of the copper in the block.
4 The given specific heat capacity of copper at
room temperature is about 386 J kg−1 K−1. Suggest
reasons why your calculated value may be different
from the given value.
∆θ
= -k (θ w – θs )
∆t
where θw is the temperature of the water
in the tank and θs is the temperature of the
surroundings. If the temperature of the water
when the heater failed was 25.5°C and the
temperature of the room it was in was 15.0°C,
use your answer to (c) to determine the value of
the constant k.
e) Use your value of k to determine the rate of
cooling of the water in the fish tank if the
temperature of the room was to fall to 8°C.
335
17/04/19 10:41 AM
Measuring the specific heat capacity of water using
a continuous flow method
The specific heat capacity of a fluid can be measured using a continuous
flow method (Figure 18.7), where the fluid moves over an electric heater
at a constant rate. It is assumed that the thermal energy transferred
from the apparatus to the surroundings is constant. The experiment is
carried out and then the flow rate of the fluid is changed, and a second
set of readings is taken. The heat loss can then be eliminated from the
calculations.
liquid out
EXAMPLE
Continuous flow method
Work out the specific heat
capacity of water using the
following data measured during
one such experiment involving
water in a continuous flow
method:
●
●
●
●
●
18 THERMAL PHYSICS
●
336
●
time of each experiment,
t = 60 s
temperature difference in both
experiments, Δθ = 10.0°C
p.d. across the heater,
V1 = V2 = 12.0 V
current through heater in
experiment 1, I1 = 6.0 A
current through heater in
experiment 2, I2 = 2.0 A
mass of water flowing in
experiment 1 for 60 s,
m1 = 126.0 g
mass of water flowing in
experiment 2 for 60 s,
m2 = 56.0 g
Answer
The specific heat capacity is worked
out by substituting values in the
equation:
c=
(I1V1 – I2 V2 )t
(m1 – m2 )∆θ
((6 A × 12 V) – (2 A × 12 V)) × 60 s
=
(0.126 kg – 0.056 kg) × 10 K
= 4114 Jkg–1 K–1
469884_18_AQA_A-Level_Physics_329-357.indd 336
electric
thermometer
V
T2
T1
A
liquid in
heater
electric
thermometer
12 V
Figure 18.7 Measurement of specific heat capacity by the continuous flow method.
A fluid flows through an insulated tube containing an electric heating
wire, as shown in Figure 18.8. The rise in temperature of the fluid is
measured by the two electronic thermometers and calculated by Δθ =
T2 − T1. The mass of the fluid that flows through the apparatus in a time
t1 is m1, and is measured using a beaker on a balance and a stopwatch.
The flow rate of the fluid is then altered to give another value, m2, and
the heater controls are changed to give the same temperature difference
Δθ. The specific heat capacity of the fluid can then be determined by
assuming that the thermal losses to the surroundings are constant for
both flow rates.
For the first flow rate, the electrical energy supplied to the fluid in time t1
is given by
I1V1t1 = m1cΔθ + Elost(i)
where I1 and V1 are the initial current and p.d. of the heater and Elost is the
thermal energy lost to the surroundings. For the second flow rate:
I2V2t2 = m2cΔθ + Elost(ii)
Elost can be assumed to be the same in each experiment, so subtracting
equation (ii) from equation (i) gives
I1V1t1 − I2V2t2 = m1cΔθ − m2cΔθ
= cΔθ (m1 − m2)
If the experiments are both run for the same time t, then
c=
( I1V1 – I2V2 ) t
( m1 – m2 )∆ θ
17/04/19 10:41 AM
Specific latent heat The specific latent
heat of a material is the amount of thermal
energy required to change the state of
1 kg of material, without a change in
temperature, at a specified ambient pressure
(normally atmospheric pressure, p = 1 atm).
When liquids are heated up to their boiling point, the thermal energy is
used to increase the internal energy of the molecules of the liquid. We
measure this as a temperature change. However, at the boiling point, the
temperature change stops and all the thermal energy input is used to
overcome the intermolecular forces between the particles of the liquid,
converting it into a gas.
The amount of thermal energy required to change the state of a substance,
without a change in temperature, Q (in J), is given by
Q = ml
latent heat of vaporisation
liquid + gas
temperature
Tb
gas
condensing
latent heat of fusion
solid + liquid
boiling
liquid
Tf
freezing
melting
solid
time
Figure 18.8 Kinetic theory graph.
EXAMPLE
Evaporating water
A Bunsen burner delivers heat
energy at a rate of 900 W to water
inside a glass beaker. The water is
at its boiling point, and the 900 W
of thermal energy is used to
turn 0.50 kg of water into steam.
Calculate how long it will take for
the water to turn to steam. The
specific latent heat of vaporisation
of water is 2260 kJ kg−1.
Answer
Using the equation from the main
text
Q = mlv = 0.50 kg × 2260 ×
103 J kg−1 = 1.13 × 106 J
If the power supplied as thermal
heat to the water is 900 W, the
time required to boil the water is:
where m is the mass of the substance (in kg) and l is
the specific latent heat (‘latent’ means ‘hidden’) of
the substance (in J kg−1). This equation applies to all
the phase changes involved with changes of state.
So water, for example, has a specific latent heat of
vaporisation, lv, which deals with the phase change
from liquid to gas (and vice versa), and a specific
latent heat of fusion, lf, which deals with the phase
change from solid to liquid (and vice versa).
Heating up substances and changes of state
Changing state
The relationship between the kinetic theory models
of solids, liquids and gases and the concept of latent
heat is illustrated by Figure 18.8. Thermal energy
supplied to a substance that is changing state is
used to loosen the intermolecular bonds holding the particles together
(completely in the case of a liquid turning into a gas). The thermal
energy is called a latent heat because during the change of state the
temperature does not change, despite thermal energy being supplied to
the substance.
The values of lv and lf for a few selected materials are shown in Table 18.2.
Once again, the high values for water mean that a high proportion of the
water on planet Earth is in the liquid state, and our ambient temperature is
kept within a relatively small range.
Table 18.2
Material
Specific latent heat of
vaporisation, lv/kJ kg−1
Water
Specific latent heat of
fusion, lf /kJ kg−1
2260
334
Carbon dioxide
574
184
Nitrogen
200
26
Oxygen
213
14
Lead
871
23
337
Q 1.13 × 106 J
t= =
=1256 s ≈ 21minutes
P
900 W
Q 1.13 × 106 J
t= =
=1256 s ≈ 21minutes
P
900 W
469884_18_AQA_A-Level_Physics_329-357.indd 337
17/04/19 10:41 AM
TEST YOURSELF
7 A 12.5 g ice cube melts in the sunshine. Calculate
the thermal energy from the Sun absorbed by the
ice during melting. The specific latent heat of fusion
of water is 334 kJ kg−1.
8 Lead is a major component of the solder used to
construct integrated circuits. A soldering iron delivers
18 W of thermal energy to a small 4.2 g block of lead.
Calculate the time taken for all the lead to melt. The
specific latent heat of fusion of lead is 23 kJ kg−1.
9 A range oven rated at 3 kW actually delivers
2.7 kW of thermal power to 1.5 kg of water inside a
whistling kettle. The specific heat capacity of water,
c, is 4200 J kg−1 K−1, and the specific latent heat of
vaporisation of water, lv, is 2260 kJ kg−1.
a) How much thermal energy is required to heat
the water from 8°C to 100°C?
b) How long does it take the kettle to heat the
water from 8°C to 100°C?
c) The water starts to boil and the whistle on the
kettle starts to blow and keeps blowing for 25 s
until the kettle is removed from the heat. What
mass of water is converted into steam during
the 25 s of boiling?
10 The specific latent heat of vaporisation of a liquid
(such as water) can be measured using the
apparatus shown in Figure 18.9.
electrical
connections
18 THERMAL PHYSICS
holes
water
platinum
heater
condenser
The heat supplied to the water is given by
V1I1t = m1lv + E
where V1 is the p.d. across the heater, I1 is the
current supplied to the heater, t is the time that the
experiment is left to run, boiling a mass of water
m1 in the time t, lv is the specific latent heat of
vaporisation of water and E is the thermal energy
lost to the surroundings. The experiment is then
repeated with a different p.d., V2, across the heater
and a different current, I2, flowing through it,
boiling a different mass of water, m2, in the same
time t. In this case:
V 2 I 2 t = m 2 lv + E
(The thermal energy lost in each experiment, E, will
be the same.)
a) Use both equations to derive an expression for
the specific latent heat of vaporisation of water.
b) In one such experiment, the following data was
obtained in t = 600 s.
Quantity
Value
Quantity
Value
V1
8.00 V
V2
12.00 V
I1
2.41 A
I2
3.00 V
m1
5.8 g
m2
10.3 g
Use this data and your answer to (a) to calculate a
value for the specific heat capacity of water.
11 Stearic acid, a common chemical found in soaps,
is frequently used to show the phase change of a
material. A student set up an experiment using
a test tube with 4 g of initially liquid stearic acid
contained in a small water bath containing 25 g of
water set at 95°C. She puts thermometer probes
connected to a data logger into the stearic acid and
the water in the water bath and then she turns off
the temperature control. The data logger measures
and records the temperatures over the course of
ten minutes as the water and the stearic acid cool
down. Her results are shown in Figure 18.10.
100
90
stearic
acid
water
bath
338
water
temperature, T/°C
80
70
60
50
40
30
20
10
Figure 18.9 Apparatus for measuring the
specific latent heat of vaporisation of water.
469884_18_AQA_A-Level_Physics_329-357.indd 338
0
Figure 18.10
100
200
300
time, T/s
400
500
600
17/04/19 10:41 AM
c) Use the graph and the rate of thermal heat
transfer to estimate:
i) the specific heat capacity of solid stearic acid
ii) the specific latent heat of fusion of stearic
acid.
The gas laws
a) Draw a diagram of the experimental set-up.
b) The specific heat capacity of water is
4200 J kg −1 K −1. Neglecting the effect of the glass
test tube, estimate the rate of thermal energy
transfer from the water (and the stearic acid).
pressure/k Pa
●● The gas laws
120
Between 1662 and 1802 three laws were discovered by a collection of
European physicists that seemed to describe the behaviour of gases in
response to changes in their pressure, volume and temperature. The laws
themselves are all empirical, which means that they describe the mathematical
relationships between the three variables purely based on experiments.
100
Boyle’s law
The first gas law to be discovered was Boyle’s law, the relationship between
the pressure and volume of a gas. The experiments were carried out by
Robert Boyle and his research student Robert Hooke in 1662, involving
J-shaped tubes of sealed glass and mercury. Boyle quickly realised that
there was a relationship between the volume of the air trapped behind the
mercury and the weight of the mercury acting across the cross-sectional area
of the tube causing increased pressure.
80
60
40
20
10 15 20 25 30 35 40 45 50
volume/cm3
Figure 18.11 Boyle’s law graph.
Boyle realised that the pressure acting on the gas and the volume occupied by the
gas were inversely proportional to each other. Boyle would have obtained results
similar to those shown in Figure 18.11. A modern version of his law states:
For a fixed mass of an ideal gas at constant temperature, the pressure
of the gas is inversely proportional to its volume.
Writing this mathematically:
p ∝ 1 or p × V = constant
V
where p is the pressure acting on a gas of volume V. A more useful version
of this equation involves the same mass of gas at the same temperature, but
different pressures and volumes, where
p1V1 = p2V2
REQUIRED PRACTICAL 8
Investigating Boyle’s (constant temperature) law
Note: This is just one example of how you might tackle this required
practical.
339
A student uses the standard Boyle’s law apparatus
shown in Figure 18.12 to determine the value of the
constant involved in the equation
pV = constant
In the experiment, performed at 17°C, the foot pump
is used to pressurise the oil inside the cylinder, which
compresses the air column above it,reducing its volume.
The pump valve is closed when the pressure is at a
maximum and the air column volume is at a minimum.
469884_18_AQA_A-Level_Physics_329-357.indd 339
Figure 18.12 Apparatus used to
investigate Boyle’s law.
17/04/19 10:41 AM
The apparatus is then left to come to (thermal)
equilibrium as the oil drains back down the
sides of the column. The pressure and the volume of
the air column then
measured and recorded. The student then opens the
pump valve very slightly and the pressure is reduced
slightly, expanding the air in the column. The valve
is shut, the apparatus is allowed to come to thermal
equilibrium again and the pressure and volume are
measured and recorded. This process is repeated
until the pressure returns fully to its atmospheric
value. The student’s results are shown in Table 18.3.
Table 18.3
Pressure, p/105 Pa
(±0.01 × 105 Pa)
Volume, V/cm3 (±0.5 cm3)
3.5
9.0
3.0
10.0
2.5
12.0
2.0
15.5
1.5
20.0
1.0
30.5
pressure/kPa
1 Make a copy of the table and add two further
columns: 1/V (in cm−3) and p × V (in 105 Pa cm3) –
calculate the values and enter them in the table.
2 Plot graphs of the following:
l V against p
l 1/V against p.
3 For each graph, include error bars and a bestfitting line.
4 Use your graphs to measure a value for the
constant, where pV is constant. Use your graph
to estimate an uncertainty in the value of the
constant.
5 Explain why the student allowed the experiment to
come to thermal equilibrium before measuring the
pressure and volume in the apparatus.
The pressure–temperature law (Amonton’s law)
and absolute zero temperature
2.0
1.8
In 1702 Guillaume Amonton discovered the empirical relationship between
the pressure and temperature of a gas as a result of his efforts to design and
build air thermometers. Amonton realised empirically that there was a linear
relationship between the two variables, provided that the mass and the
volume of the gas were kept constant. Amonton struggled to build accurate
thermometers and, although his ideas were published, they lacked basic
quantitative data. A modern version of his graph is shown in Figure 18.13.
1.6
1.4
1.2
1.0
18 THERMAL PHYSICS
The student estimates that she can measure the
pressure readings from the pressure gauge with an
uncertainty of ±0.01 × 10 5 Pa, and the volume from the
measuring scale with an uncertainty of ±0.5 cm3.
0
300
400
500
temperature/K
600
A modern version of Amonton’s law can be written more formally as:
Figure 18.13 Amonton’s law (pressure–
temperature law).
340
The pressure of a fixed mass and fixed volume of gas is directly
proportional to the absolute temperature of the gas.
Writing this mathematically:
p∝T
or
p
= constant
T
This relationship has a third, more useful, form that is used to compare the
same gas under different pressures and temperatures. This can be written as:
p1 p2
=
T1 T2
Amonton realised at the time of his experiments that if he extrapolated
his data back through lower and lower temperatures there would be
a temperature where the pressure of a gas dropped to zero. At this
469884_18_AQA_A-Level_Physics_329-357.indd 340
17/04/19 10:41 AM
0K
−273.15 °C
pressure, p
absolute zero
273.15 K temperature, T
0 °C
Figure 18.14 Absolute zero.
TIP
Always use the absolute (Kelvin)
temperature scale when doing
problems and calculations
involving the gas laws.
temperature, the molecules would stop moving and so they could not
exert a pressure by hitting anything else. This temperature became
known as absolute zero and Amonton calculated it to be −240°C,
which, considering the thermometer technology available to him at the
time (air thermometers), was a pretty remarkable achievement – he was
only about 35°C out. In 1848 William Thomson (Lord Kelvin) used the
concept of absolute zero to construct a temperature scale with absolute
zero as the ‘zero’ of his scale. Using the better thermometer technology
of the day, Kelvin predicted that absolute zero would be at a temperature
of −273°C or 0 K – only 0.15°C away from today’s defined value of
−273.15°C (Figure 18.14).
The gas laws
Absolute zero The temperature when all
molecular motion ceases, and the pressure
of a gas drops to zero. The accepted value is
the zero of the Kelvin temperature scale and
is defined as −273.15°C.
Kelvin’s absolute scale of temperature became the SI unit of temperature,
and is defined in terms of two fixed temperature points – absolute zero
(0 K) and the triple point of water (0.01°C or 273.16 K – the temperature
and pressure values where ice, liquid water and water vapour can coexist).
Converting a temperature from °C into an absolute temperature measured
in K involves using the equation:
T (in K) = T (in °C) + 273.15
You must note that the magnitude (size) of 1°C is equivalent to 1 K. In other
words: |1°C| = |1 K|.
Charles’ law
volume V(m3)
In 1802, the French chemist Joseph Gay-Lussac published a paper
showing the experimental link between the volume and the temperature
for a gas. Gay-Lussac named the law after his balloonist friend, Jacques
Charles, who produced an unpublished version of the law following his
observations of the behaviour of balloons. The empirical law is illustrated
by the graph in Figure 18.15.
A modern version of Charles’ law states:
At constant pressure the volume of a fixed mass of an ideal gas is
directly proportional to its absolute temperature.
temperature T/K
Figure 18.15 Charles’ law.
Writing this mathematically:
V∝T
or
V
= constant
T
Once again, the temperature T is an absolute temperature using the Kelvin
scale (T/K = T/°C + 273.15). Another useful form of this relationship
(similar to the other gas laws) is
341
V1 V2
=
T1 T2
469884_18_AQA_A-Level_Physics_329-357.indd 341
17/04/19 10:42 AM
REQUIRED PRACTICAL 8
Investigation of Charles’ law for a gas
thermometer
Note: This is just one example of how you might tackle this
required practical.
gas syringe
Charles’ law can be used to make an estimate for the value of
absolute zero. This can be achieved by measuring the volume of
a gas at different temperatures and then extrapolating the graph
back to a volume of zero. In reality, non-ideal gases do not end
up with zero volume but, for the purposes of this experiment, the
difference is so small that it will not affect the outcome.
In this experiment, two small sealed gas syringes, one with a
total volume of 10 cm 3 and the second with a volume of 30 cm 3,
are put, one at a time, into a freezer cabinet at −15°C, and
then into a beaker of iced water at 0°C (Figure 18.16). The iced
water is then gradually warmed using a Bunsen burner. The
volume of the air trapped inside the gas syringes is recorded
at temperatures of −15, 0, 20, 40 and 80°C. The results of the
experiment are shown in Table 18.4.
water bath
stopper
Figure 18.16 Measuring absolute zero.
Table 18.4
Temperature, T/°C
Volume of air in syringe, V/ml (±0.2 cm3)
18 THERMAL PHYSICS
10 cm3 syringe
30 cm3 syringe
−15
4.3
13.5
0
4.6
14.2
20
4.9
15.1
40
5.2
16.4
80
5.9
18.3
1 Plot a graph of this data. Include error bars on the volume measurements
and best-fitting lines.
2 Extrapolate each best-fitting line back so that it crosses the temperature
axis. Use the temperature-axis intercepts to determine a range of values
for the absolute zero temperature in °C.
342
Combining the gas laws
The three gas laws – namely Boyle’s law, Amonton’s law and Charles’
law – can be combined into one expression linking the pressure, volume
and temperature of a gas, in a combined gas law, which is expressed
mathematically as:
pV
= constant
T
A more useful form of this equation can be written as
p1V1 p 2 V2
=
T1
T2
where p1, V1 and T1 describe the initial pressure, volume and temperature
of a gas, and p2, V2 and T2 describe the final pressure, volume and
temperature of the same gas after a change has been applied to it.
469884_18_AQA_A-Level_Physics_329-357.indd 342
17/04/19 10:42 AM
Remember that, when using
the combined gas law, absolute
temperature must be used and
temperatures in degrees Celsius
must be converted to kelvins.
Standard temperature and pressure This
refers to 0°C (273.15 K) and 1.01 × 105 Pa
(1 atm).
Room temperature and pressure This
refers to 25°C (278.15 K) and 1.01 × 105 Pa
(1 atm).
The gas laws
TIP
In questions on the combined gas law, pressure is usually measured in
pascals (Pa) or kilopascals (kPa), where a pressure of 1 Pa is equivalent
to a force of 1 N acting over an area of 1 m2. Pressure is also measured in
atmospheres (atm), where one standard atmosphere is defined as a pressure
equivalent to 101 325 Pa, and is the average value for atmospheric pressure
at sea level. The volume of a gas is normally measured in m3, but cm3, litres
and ml are also commonly used. Temperatures may be given in kelvin or
degrees Celsius. The latter must be converted to kelvin.
Which of these units should you use in your work? The answer is generally
to use the units included in the question, converting temperatures to kelvin.
State any calculated values using the units from the question.
Also, you should be aware that questions sometimes refer to standard
temperature and pressure (STP), which are 0°C (273.15 K) and 1.01 × 105 Pa
(1 atm), and room temperature and pressure (RTP), which are 25°C
(278.15 K) and 1.01 × 105 Pa (1 atm).
EXAMPLE
Changing volume of a balloon
The volume of a party balloon at a room temperature
of 15°C and an atmospheric pressure of 1.0 atm at sea
level is 2400 cm3. The balloon is taken up a mountain,
where the temperature is 1.4°C and the atmospheric
pressure has dropped to a value of 0.80 atm. Calculate
the new volume of the balloon at the top of the
mountain.
Answer
Use
pV
pV
11
= 2 2
T
T2
1
Substituting numbers:
1atm × 2400 cm3 0.80 atm × V2
=
(1.4 + 273.15)K
(15 + 273.15)K
Rearranging gives:
V2 =
1atm × 2400 cm3 × (1.4 + 273.15)K
(15 + 273.15)K × 0.80 atm
= 2858.4 cm3 ≈ 2900 cm3 (2 s.f.)
TEST YOURSELF
12 A
small patio heater gas bottle has a volume of
6.0 litres (6.0 × 10−3 m3) and contains butane gas at
a temperature of 5.0°C and a pressure of 2.5 MPa.
What would be the volume of the gas if it were let out
of the canister into an inflating balloon on a warm
day at 20°C and atmospheric pressure (0.1 MPa)?
13 A closed gas syringe contains a fixed mass of
air at 24°C. To what temperature must the gas
be heated so that its volume doubles, when the
pressure remains constant?
14 An empty syrup tin contains air at a temperature
of 16°C and a pressure of 1.5 × 105 Pa. The lid will
blow off the tin if the pressure inside the tin rises
beyond 2.4 × 105 Pa.
a) At what temperature will the top blow off if the
air is heated evenly with a Bunsen burner?
469884_18_AQA_A-Level_Physics_329-357.indd 343
b) W
hy will this trick work better if you put some
water in the tin?
15 A large car tyre has a volume of 22 × 10 −3 m3, and
the air inside is pumped to a pressure of 2.5 atm
above atmospheric pressure (1 atm). Calculate the
volume that the air inside the tyre would occupy
at atmospheric pressure. You should assume that
the temperature remains constant.
16 A steam cleaner has a steam tank with a
volume of 150 cm3 in which the steam is kept
at a temperature of 100°C and a pressure of
1.5 × 106 Pa. If the steam cleaner is used to clean
windows outside on a cold day where the air is
at atmospheric pressure and the temperature is
6.5°C, calculate the volume of steam generated if
all the steam is let out of the tank.
343
17/04/19 10:42 AM
●● Avogadro’s law, the ideal gas equation
and moles
A fourth experimental gas law was hypothesised and published by Amadeo
Avogadro in 1811. In this law he suggested that equal volumes of gases at the
same temperature and pressure contained the same number of molecules.
Mathematically, this means that:
V
V ∝ n or
= constant
n
where n is the number of moles of the gas. Once again, as with the other gas
laws, this is more usefully written as:
V1 V2
=
n1 n 2
In 1834, the French physicist Emile Clapeyron combined all four gas laws
and produced the ideal gas equation:
pV = nRT
where R is a constant now known as the molar gas constant, R =
8.31 J mol−1 K−1. The ideal gas equation is a tremendously powerful equation
that models the behaviour of gases extremely well, particularly when the
gases are at relatively low pressures and high temperatures. At conditions
close to liquefaction, it works less well because the gases behave much less
like ideal gases.
18 THERMAL PHYSICS
The Avogadro constant, NA, was proposed by Jean Baptiste Perrin in 1909,
to represent the number of particles (usually molecules or atoms) present
in one mole (1 mol) of a substance. Perrin named it in honour of Amadeo
Avogadro. Perrin went on to win the Nobel Prize in Physics for his attempts
to measure the Avogadro constant accurately. The Avogadro constant is
today defined as 6.022 141 29 × 1023 mol−1 (rounded to 6.02 × 1023 mol−1
on the AQA Physics Datasheet).
344
Dividing the molar gas constant, R, by the Avogadro constant, NA, yields
another fundamental constant in physics, the Boltzmann constant, k:
R
k=
NA
and it has the value:
R
8.31 J mol –1 K –1
= 1.38 × 10 –23 J K –1
=
N A 6.02 × 1023 mol –1
The Boltzmann constant is the fundamental constant that links the
macroscopic measurements of pressure, volume and temperature to the
microscopic behaviour of particles in a gas, and has a fundamental position
in the model of an ideal gas. The constant enables the microscopic model to
make predictions about the way that ideal gases behave on a macroscopic
scale where they can be measured empirically by the gas laws.
k=
The above equation can be rearranged to give:
R = kNA
which can be substituted into the ideal gas equation, giving:
pV = nNAkT
469884_18_AQA_A-Level_Physics_329-357.indd 344
17/04/19 10:42 AM
But nNA is the number of particles in the gas and is given the symbol, N, so:
(This implies that the pressure of an ideal gas is independent of the mass of
the particles.)
Molar mass and molecular mass
Counting particles is not a good way to gauge the amount of substance
present in a gas (or liquid or solid). It is incredibly difficult to observe
individual atoms or molecules, let alone to count them. A better way to work
out the amount of matter in a substance is to use the mass of particles and
weigh large collections of them using an electronic balance. If the mass of one
particle is known, then a measurement of the mass of a large number of them
will yield the number of particles present. We therefore define two quantities:
the molecular mass, m, which is the mass of one molecule of substance; and
the molar mass, Mm, which is the mass of one mole (NA) of molecules of the
substance. These two quantities are related to each other by:
Mm = NAm
If the mass of a known gas is measured, Mg, then dividing this value by the
molar mass gives the number of moles, n, and dividing it by the molecular
mass, m, gives the number of molecules, N:
Mg
Mg
n=
and N =
m
Mm
Avogadro’s law, the ideal gas equation and moles
pV = NkT
Both of these can then be substituted into the ideal gas equation, allowing
all quantities to be measured macroscopically:
pV =
Mg
Mm
RT and
pV =
Mg
m
kT
EXAMPLE
Propane gas cylinder
A propane gas cylinder has a volume of 0.14 m3 and the
pressure of the gas inside the cylinder is 2.0 × 106 Pa
above atmospheric pressure at 300 K.
1 Calculate the number of moles of propane gas
inside the cylinder.
Answer
Using the ideal gas equation, pV = nRT:
pV
n=
RT
(2.0 × 106 +1.0 × 105 )Pa × 0.14 m3
=
=118 mol
8.31Jmol–1 K–1 × 300 K
2 Calculate the number of propane molecules inside
the cylinder.
Answer
One mole of an ideal gas contains 6.02 × 1023
molecules of gas (the Avogadro constant, NA). The
469884_18_AQA_A-Level_Physics_329-357.indd 345
total number of molecules of gas in the container is
therefore 6.02 × 1023 × 118 = 7.1 × 1025 molecules.
3 Calculate the mass of gas inside the cylinder if the
molar mass of propane is 44.1 g mol−1.
Answer
If the molar mass of propane is 44.1 g mol−1, and
there are 118 mol of gas, then the mass of propane
inside the cylinder is 44.1 g mol−1 × 118 = 5204 g =
5.2 kg (2 s.f.).
345
4 Calculate the mass of one molecule of propane.
Answer
The mass of one molecule of propane is
molar mass
m=
NA
=
44.1× 10–3 kg mol–1
23
–1
6.02 × 10 mol
= 7.3 × 10–26 kg
17/04/19 10:42 AM
TEST YOURSELF
17 1.5 moles of an ideal gas at a temperature of 312 K
is kept at a pressure of 1.7 × 105 Pa. Calculate the
volume of the gas under these conditions.
18 A weather balloon contains helium gas and
occupies a volume of 0.85 m3. At a particular
weather station, the pressure is 1.2 × 105 Pa and
the temperature of the surrounding air is 18°C.
Assuming that helium behaves as an ideal gas,
show that the balloon contains about 42 mol of
helium gas.
19 A tyre on a cycle in the Tour de France contains
0.15 mol of air at a temperature of 293 K and has a
volume of 8.2 × 10 −4 m3. It is assumed that the air
behaves as an ideal gas.
a) Calculate the pressure of the air inside the tyre.
b) At the end of a stage in the race, the pressure
in the tyre has risen to 5.45 × 105 Pa. Use this
information to estimate the temperature of the
air in the tyre at the end of the stage. (Assume
that the volume does not change.)
20 Look at the graphs (labelled A, B, C and D) in
Figure 18.17 showing the behaviour of an ideal gas.
y
0
y
0
x
A
0
0
x
0
0
B
x
C
0
Quantity
Standard
temperature
and pressure
(STP)
Temperature,
T/K
273
Pressure,
p/10 5 Pa
1.01
Volume, V/m 3
y
y
a) Which graph shows the variation of the
volume of a fixed mass of the gas (y-axis)
at constant temperature with the pressure
(x-axis) of the gas?
b) Which graph shows the variation in pressure
(y-axis) of a fixed mass of gas at constant
volume with absolute temperature (x-axis)?
c) Which graph shows the variation of the product
(p × V) of a fixed mass of the gas (y-axis) at
constant temperature with the pressure
(x-axis) of the gas?
21 This question is about the two standard conditions of
an ideal gas – standard temperature and pressure
(STP) and room temperature and pressure (RTP).
The table below gives some data for 1 mol of an ideal
gas under each of these conditions.
Room temperature
and pressure
(RTP)
1.01
2.45 × 10 −2
Copy and complete the table.
0
x
D
18 THERMAL PHYSICS
Figure 18.17
346
ACTIVITY
The ideal gas equation and
Mount Kilimanjaro
Mount Kilimanjaro is the highest mountain in Africa
(Figure 18.18), and scientists have found that about
half of all the climbers who attempt to scale its height
suffer from altitude sickness before they reach the
summit as a result of ascending the mountain too
quickly. Every year approximately 1000 climbers are
evacuated from the mountain suffering from acute
altitude sickness, and on average 10 climbers die.
Kilimanjaro is a deceptively dangerous place.
Figure 18.18 Mt Kilimanjaro, the highest mountain
in Africa.
469884_18_AQA_A-Level_Physics_329-357.indd 346
17/04/19 10:42 AM
l
l
l
summit elevation 5895 m above sea level
summit air pressure 50 kPa
summit average air temperature −6.8°C
l
l
l
l
plains elevation 1018 m above sea level
plains air pressure 90 kPa
plains average air temperature 30°C
plains air density 1.03 kg m−3
1 Use this data to calculate the density of air at the
summit of Mount Kilimanjaro.
2 The proportions of oxygen and nitrogen in the
air on the surrounding plains and the summit
is constant (21% oxygen and 79% nitrogen). The
average adult lung capacity is about 6 litres.
Calculate the number of oxygen molecules in a
person’s lungs
a) on the surrounding plains
b) at the summit of Mount Kilimanjaro.
Ideal gases
Here is some data about Mount Kilimanjaro on the
mountain itself and on the plains below:
●● Ideal gases
There are many occasions in physics where we use a simplified model to explain
the behaviour of a system. Models use basic first principles and then usually
add in more complexity to fine-tune the behaviour of the model so that it better
reflects reality. One good example of a simple model is the model of an ideal gas,
which is used to explain the behaviour of gases subject to the changes in their
temperature, pressure and volume. Real gases do not behave exactly like ideal
gases but their general behaviour is sufficiently close that the model predicts and
explains most of the common patterns in the behaviour of the real thing.
The ideal gas model assumes the following:
●
●
●
●
●
●
●
●
●
●
An ideal gas consists of a large number of identical, small, hard spherical
molecules.
The volume of the molecules is very much smaller than the volume of
the container.
All the collisions between the molecules themselves and the container are
elastic and all motion is frictionless (i.e. no energy loss in motion or collision).
The movements of the molecules obey Newton’s laws of motion.
The average distance between molecules is very much larger than the size
of the molecules.
The molecules are constantly moving in random directions with a
distribution of velocities about a mean velocity.
There are no attractive or repulsive intermolecular forces apart from
those that occur during their collisions.
The only forces between the gas molecules and the surroundings are
those that determine the collisions of the molecules with the walls.
There are no long-range forces between the gas molecules and their
surroundings.
The time spent between collisions is very much larger than the time
spent colliding.
347
Molecular motion
One of the most important properties of an ideal gas is the idea that they
move in random directions. This is important because if this was not true
then gases would exert more pressure on one surface of their container than
they would on another, i.e. the direction that the particles travel in would
be important, and the theories would be different in different directions.
The fact that gases (and all fluid particles) have a random molecular motion was
first observed and described by the botanist Robert Brown in 1827, as a result of
his observations of pollen grains floating on water. Brown saw the grains moving
469884_18_AQA_A-Level_Physics_329-357.indd 347
17/04/19 10:42 AM
Figure 18.19 A modern version of
Jean Baptiste Perrin’s plot of the
Brownian motion of carbon particles.
in random directions as he observed them through a light microscope but he
was unable to explain why they moved – this was left to Albert Einstein during
his ‘annus mirabilis’ (miracle year) of 1905, during which he published his
ideas about the photoelectric effect, special relativity, mass–energy equivalence
(E = mc2) as well as Brownian motion. Einstein explained that the pollen grains
were moving in random directions as the result of the cumulative effect of the
water molecules randomly hitting the pollen grains. At different times the pollen
grains are hit by water molecules more on one side than they are on the other
sides, resulting in a motion in that direction that appears random in nature.
Einstein’s theory of Brownian motion was confirmed experimentally in 1908
by Jean Baptiste Perrin (of Avogadro constant fame). Perrin produced a series of
positional plots showing this random motion by observing the motion of 0.5 µm
carbon particles on a grid of 3 µm × 3 µm squares and recording their positions
every 30 s. His plots would have looked like the modern version in Figure 18.19.
Perrin analysed the motion of the particles and concluded that the motion
was truly random, in line with Einstein’s theory. Both Perrin and Einstein
were (separately) awarded Nobel Prizes partly because of their work on
Brownian motion.
Pressure, volume, temperature and
molecular motion
The importance of Brownian motion and the properties of an ideal gas
should not be underestimated. The observation of the random motion of
fluid particles and the subsequent theory proposed by Einstein provide a
way of explaining the macroscopic gas law quantities, and hence the gas
laws themselves in terms of a microscopic molecular model.
18 THERMAL PHYSICS
Pressure
348
Macroscopic pressure is defined in terms of a force acting over a given
area. The kinetic theory model of an ideal gas shows us that the force is
due to the collisions of the molecules with the walls of the container. The
molecules are moving in random directions with a mean average velocity.
The particles hit the walls of the container and rebound off at the same
speed (all the collisions are elastic). This produces a change of momentum,
and the cumulative effect of all the particles colliding over the total inside
surface area of the container per second causes a force per unit area, which
exerts a pressure acting in all directions (as the motion is random).
Volume
The motion of molecules inside a container is random in direction. This
means that there is no preferred direction, so the molecules will spread out
throughout the container filling its volume. Gases take the volume of their
container. If the dimensions of the container are changed, the motion of the
molecules will react to the change and will continue to fill the available volume.
The behaviour of real gases is closest to that of an ideal gas at low pressures,
well away from their phase boundary where they change into a liquid.
Temperature
For an ideal gas, because there are no intermolecular forces, increasing the
temperature of the gas only increases the kinetic energy of the particles.
This increases the average velocity of the particles. The particles still move
in random directions, and they fill the container. Increasing the temperature
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●● A molecular kinetic theory model
Although the gas laws are empirical in nature (they were developed as a result
of analysis of experimental data), the kinetic theory model is derived from
theoretical first principles. However, they both produce the same results when
observing the macroscopic behaviour of gases, but only the kinetic theory
version can explain the behaviour of gases on a microscopic, molecular scale.
z
L
L
c
y
L
x
Figure 18.20 Particle in a box.
A molecular kinetic theory model
for a fixed volume increases the pressure because the particles’ average
speed is higher and therefore the change of momentum during collisions
with the walls is greater, and the particles hit the walls more often. This
leads to higher forces and therefore higher pressures. Allowing the pressure
to remain constant requires the volume to change.
In 1860 James Clerk Maxwell and Ludwig Boltzmann both (independently)
used the assumptions of the ideal gas model to link the pressure and
density of a gas, connecting for the first time the molecular behaviour of
a gas to one of its mechanical properties. Their theories started with the
motion of one gas particle inside a cubic box (Figure 18.20).
The gas has a volume V and density ρ and is enclosed inside a cubic box
of side L. Inside the box there are N identical particles with the same mass,
m, and the gas particles have a range of different velocities c1, c2, c3…cN. It
is assumed that the volume of the particles is negligible compared to the
volume of the box.
Consider one particle moving parallel to the x-axis with a velocity c1. The
particle collides with the shaded wall in the diagram. The ideal gas theory
assumes that the collision is totally elastic and so the particle rebounds
back off the wall with a velocity of –c1. The particle therefore experiences a
total change in momentum equal to 2mc1, during the collision. If the totally
elastic collision assumption was untrue then the particles would gradually
loose energy during the collisions and the average velocity of the particles
in the box would decrease, resulting in a drop in overall gas pressure.
Experimental evidence tells us that this does not happen.
The particle then travels back across the box, collides with the opposite face
before returning to the shaded wall in a time interval ∆ t = 2c L . This means
1
that in the time interval Δt, the particle makes one collision with the wall and
exerts a force on it. If the particle obeys Newton’s Second Law of motion then,
F=
2mc1
mc2
change in momentum
=
= 1
L
time for change
(2L /c1 )
The shaded wall has an area, A = L2, so the pressure exerted by the one particle is:
2
F mc
p = = 31
A L
349
There are N particles in the box, and if they were all travelling parallel to
the x-axis: total pressure on the shaded wall would be
=
( ) × (c + c + c + … + c
m
L3
2
1
2
2
2
3
2
)
N
but in reality, particles are moving in random directions, with a velocity,
c, comprising components at right angles to each other in the x, y, and z
469884_18_AQA_A-Level_Physics_329-357.indd 349
17/04/19 10:43 AM
directions (cx, cy and cz). Using three dimensional Pythagoras Theorem,
c2 = cx2 + cy2 + cz2, but as on average, cx2 = cy2 = cz2, so cx2 = 13 c2. As there are N
particles in the box, the pressure, P, parallel to tae x axis is therefore
=1 ×
3
( ) × (c + c + c + … + c
m
L3
2
1
2
2
2
3
2
)
N
We now define a quantity called the root mean square velocity, (crms),
(the square root of the average of the square velocities) where:
crms =
( c12 + c22 + c32 + … + c2N )
so N ( crms )2 = ( c12 + c22 + c32 + … + c2N )
N
Substituting and replacing L3 = V, gives:
pV = 1 Nm( crms )2
3
Nm is the total mass of the gas inside the box so the density of the gas
inside the box is given by:
ρ = Nm
V
so:
p = 1 ρ ( crms )2
3
●● Comparing two models of the behaviour
of gases
We now have two models that describe the behaviour of a gas. The first
model, the ideal gas equation, describes the experimental, macroscopic
behaviour of the gas:
18 THERMAL PHYSICS
pV = nRT
350
The second model, involving the kinetic theory model, describes the
behaviour from a theoretical point of view in terms of a microscopic,
mechanical model of the particles:
1
1 Nm
p = ρ ( crms )2 =
( c )2
3
3 V rms
If the average molecular kinetic energy is Ek (the bar above the quantity
means ‘mean average’), then Ek = 1 m( crms )2 and so
2
2
2
1
N
2
p = × × m( crms ) or pV = × N × E k
3
3 V 2
If this equation is compared to the ideal gas equation, then
2
3 n
× N × E k or E k = × × RT
3
2 N
Because n is the number of moles of the gas and N is the number of
particles of the gas, then
n
1
N = n × N A or
=
N NA
nRT =
Substituting this for n/N gives
R
Ek = 3 ×
×T
2 NA
469884_18_AQA_A-Level_Physics_329-357.indd 350
17/04/19 10:43 AM
EXAMPLE
Molecules in a gas syringe
A collection of 50 ideal gas molecules are observed inside a gas syringe. At a particular time, the distribution of
their molecular speeds is:
Answer
Speed, c /km s−1
1.80
1.90
2.00
2.10
2.20
2.30
2.40
Number of particles
6
8
12
10
7
4
3
Comparing two models of the behaviour of gases
But R/NA has already been defined as equal to k, the Boltzmann constant,
which effectively is the gas constant per particle of gas. So
E k = 3 kkT
T
2
where Ek is the kinetic energy of one particle of the gas. This is truly a
remarkable end point. We started with three macroscopic, easily measureable
properties of a gas, and we end up with a simple equation that allows us to
measure the kinetic energy of a particle of gas by measuring only its temperature.
Calculate the root mean square speed of the particles.
Answer
The first step is to calculate the square speeds of the particles:
Speed, c /km s−1
1.80
1.90
2.00
2.10
2.20
2.30
2.40
c 2 /km2 s−2
3.24
3.61
4.00
4.41
4.84
5.29
5.76
Number of particles
6
8
12
10
7
4
3
The next step is to calculate the mean square speed, which is the average
of all the square speeds:
(3.24 × 6) + (3.61× 8) + (4.00 × 12) + (4.41× 10) + (4.84 × 7) + (5.29 × 4) + (5.76 × 3)
50
= 4.25 km2 s–2
c2 =
To calculate the root mean square (r.m.s.) speed crms, we need to take
the square root of this number:
(crms ) = c2 = 4.25 km2 s–2
= 2.06 km s–1
EXAMPLE
Air particles in a room
Calculate the r.m.s. speed of the air particles in a
room. The density of air is 1.3 kg m−3 and the
room pressure of the air is 1.01 × 105 Pa.
Answer
351
We start with
p = 1 ρ (crms )2
3
Rearrange to make crms the subject:
crms =
3p
3 × 1.01× 105 Pa
=
ρ
1.3 kg m–3
= 482.8 m s–1 = 480 m s–1 (2 s.f.)
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EXAMPLE
Neon particles in a bulb
A neon-filled lamp bulb used in advertising signs
contains neon particles at a pressure of 1.03 × 105 Pa
and a temperature of 60°C. The molar mass of neon
is 20.2 g mol−1. Calculate the density of the neon in the
bulb.
so
(crms )2 =
=
3
3
E k = kT = × 1.38 × 10–23 JK–1 × (273 + 60)K
2
2
= 6.89 × 10–21 J
This can then be used to calculate the value of (crms)2.
We have
1
E k = m (crms )2
2
m
=
2E k
Mm /N A
–21
2 × 6.89 × 10
=
2E kN A
Mm
J × 6.02 × 1023 mol–1
20.2 × 10–3 kg mol–1
= 408648 m2 s–2 = 4.1× 105 m2 s–2 (2 s.f. )
Answer
First calculate the kinetic energy of the individual neon
particles:
2E k
Substituting into
p = 1 ρ (crms )2
3
gives
ρ=
=
3p
(crms ) 2
3 × 1.03 × 105 Pa
408648 m2 s–2
= 0.756 kg m–3 = 0.76 kg m–3 (2 s.f.)
TEST YOURSELF
22 T
he sealed gas syringe shown in Figure 18.21 is
filled with argon gas.
18 THERMAL PHYSICS
argon gas
352
plunger area = 7.1 x 10–4 m2
Figure 18.21
When an argon atom makes an elastic collision
with the plunger, it undergoes a total momentum
change of 1.33 × 10 −22 kg m s −1.
a) If we assume that, every second, 2.6 × 10 24
argon atoms collide perpendicularly with the
plunger, calculate the force exerted by the
argon on the plunger.
b) Calculate the pressure of the argon gas inside
the syringe.
23 An official FIFA size 5 football has a circumference of
70 cm and an internal air pressure of
6.9 × 105 Pa at 300 K. The molar mass of air is
29 g mol−1.
a) Calculate the mass of air inside the football.
b) Calculate the root mean square velocity of the
air particles inside the football.
469884_18_AQA_A-Level_Physics_329-357.indd 352
24 A
t STP, 3 mol of an ideal gas occupy a volume of
0.067 m3. The Avogadro constant, NA = 6.02 × 1023.
Calculate:
a) a value for the molar gas constant, R
b) a value for the Boltzmann constant, k
c) the average kinetic energy of one molecule of
the ideal gas.
25 An ideal gas can be modelled as many molecules
in continuous motion enclosed in a container. Use
the ideas of the kinetic theory to explain why the
pressure of a fixed mass of an ideal gas at constant
volume increases as the temperature of the gas
rises. State any assumptions that you need to make.
26 A cylinder of helium gas, used to inflate party
balloons, has a volume of 3.0 × 10−3 m3 and contains
42 g of helium gas at a room temperature of 20°C.
The molar mass of helium is 4.0 g mol−1. Calculate:
a) the pressure inside the cylinder
b) the number of helium atoms inside the cylinder
c) the root mean square speed of the helium
atoms inside the cylinder.
The cylinder is now stored outside, where the
temperature is close to 0°C.
d) State and explain (without the aid of calculations)
the effect of this temperature change on the values
that you have calculated in parts (a), (b) and (c).
17/04/19 10:44 AM
1 Thermal energy is supplied at the rate of 2.5 kW for 140 s to 0.8 kg
of sunflower oil inside a saucepan with negligible heat capacity. This
produces a temperature change of 219 K. The specific heat capacity of
sunflower oil, in J kg−1 K−1, is
A 1500
C 2000
B 1800
D 2200
Practice questions
Practice questions
2 An ice sculpture of mass 25 kg at 0 °C absorbs thermal energy from its
surroundings at an average rate of 45 W. The specific latent heat of fusion
of ice is 334 kJ kg−1. The time, in days, for the sculpture to melt is
A 1.8
C 3.3
B 2.1
D 4.9
3 A 0.010 kg ice cube at 0°C is dropped into a glass containing 0.10 kg of
lemonade at 15°C. The ice cube melts, cooling the lemonade. What is
the new temperature of the drink in °C? The specific latent heat of fusion
of ice is 334 kJ kg−1, and the specific heat capacity of water (lemonade) is
4200 J kg−1 K−1.
A 6
C 10
B 8
D 12
4 A deep-sea diver is working at a depth where the pressure is 3.2 atm. She
is breathing out air bubbles. The volume of each bubble is 1.9 cm3. She
decompresses at a depth of 10 m where the pressure is 2.1 atm. What is
the volume of each bubble at this depth in cm3?
A 0.6
C 1.9
B 3.6
D 2.9
5 The helium in a sealed weather balloon at a temperature of 283 K has a
volume of 1.4 m3 and a pressure of 1.01 × 105 Pa. The balloon rises to a
height of 300 m, where the temperature is 274 K and the pressure is
0.98 × 105 Pa. The volume of the air in the balloon at 300 m, in m3, is
A 1.1
C 1.9
B 1.4
D 2.4
6 A mixture of helium and argon is used in a fire extinguisher system.
The molar masses are 4.0 g mol−1 and 40 g mol−1, respectively, and the
extinguisher contains one mole of each gas. The ratio of the pressure exerted
by the helium and the argon, respectively, on the inside of the extinguisher is
A 1:1
C 1 : 10
B 100 : 1
D 10 : 1
353
7 A deodorant can with a volume of 330 cm3 at 18°C contains deodorant
particles that exert a pressure of 3.2 × 105 Pa on the inside of the can.
The number of moles of deodorant particles in the can is
A 0.04
C 400
B 4
D 40 000
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17/04/19 10:44 AM
8 The density of air at 15°C and 1.01 × 105 Pa is 1.225 kg m−3. The r.m.s.
velocity of air particles, in m s−1, is
A 604
C 498
B 603
D 497
9 Five nitrogen gas molecules have the following velocities, in m s−1: 300,
450, 675, 700, 800. The root mean square velocity of the particles, in
m s−1, is
A 413
C 613
B 513
D 713
10 Carbon particles of mass 2.0 × 10−26 kg in the hottest part of a Bunsen
burner flame have a temperature of 1200°C. The r.m.s. velocity of these
particles, in m s−1, is
A 823
C 1746
B 1235
D 2143
11 A jewellery maker is making a gold pendant. She prepares a 3.0 kg iron
mould and then pours in 25.0 g of molten gold at a temperature of
1064°C. The mould’s temperature rises from 31°C up to 35°C when it is
then in thermal equilibrium with the solid gold.
Here is the thermal data about the gold and the iron:
●
mass of iron mould = 3.0 kg
●
specific heat capacity of iron = 440 J kg−1 K−1
●
specific latent heat of fusion of gold = 63 × 103 J kg−1
18 THERMAL PHYSICS
a) Calculate the thermal energy absorbed by the iron mould.(2)
b) Calculate the thermal energy given out by the gold as it
changes state from a liquid to a solid.(1)
c) Use the data to determine the specific heat capacity (c) of gold.(3)
d) State one assumption that you have made for your calculation of c. (1)
12 A student is making iced tea lollies using her family’s freezer. She
initially pours 0.050 kg of lukewarm tea at a temperature of 40.0°C
into a 0.12 kg aluminium mould at a temperature of 5.0°C. The specific
heat capacity of tea is 4250 J kg−1 K−1 and the specific heat capacity of
aluminium is 900 J kg−1 K−1.
a) Calculate the equilibrium temperature of the tea and the mould.(3)
354
b) The tea and the mould are then put into the freezer, which removes
thermal heat from the tea and the mould at a rate of 32 W.
Calculate how long it takes for the tea to freeze, if the specific latent
heat of fusion of tea is 3.38 × 105 J kg−1, stating any assumptions
that you make.(4)
13 A gas combi-boiler can heat water with a power of 15 kW. Cold water
with a temperature of 5°C flows into the heater at a rate of 0.24 kg s−1.
The specific heat capacity of water is 4200 J kg−1 K−1.
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17/04/19 10:44 AM
b) The water supply to the heater fails and 0.24 kg of water is trapped
inside the heating compartment of the heater. The water inside the
compartment has an average temperature of 35°C and the heater
continues to heat the water. How long will it take before the water
reaches 80°C, when the emergency cut-out valve turns off the
gas supply?(2)
Practice questions
a) Combi-boilers are highly efficient and you can assume that all
the thermal energy from the heater is transferred to the water.
Calculate the output temperature of the water.(2)
14 Formula 1 tyres have a volume of 0.09 m3 and are filled with
nitrogen to a pressure of 1.4 × 105 Pa at 285 K.
(1)
a) Calculate the number of moles of nitrogen in the tyre.
b) F1 tyres are designed to work at an optimum racing temperature
of 363 K. Calculate the racing pressure in the tyre. You can
assume that the tyre does not expand when heated.(2)
c) Calculate the root mean square (r.m.s.) velocity of the nitrogen
molecules in the tyre when it is at racing pressure. The molar
mass of nitrogen is 0.028 kg mol−1.(3)
d) Describe one similarity and one difference in the way that the
nitrogen molecules behave in the tyre at the different pressures.
(2)
15 A fixed mass of helium gas is enclosed in a container with a volume of
0.055 m3. The gas is cooled and a student measures and records the
pressure of the gas, in atm, for different temperatures. The table shows
the results:
Temperature, T/K
320
300
280
260
240
Pressure, p/atm
1.30
1.22
1.17
1.08
0.95
a) Use the data to plot a graph of the results, with temperature
on the x-axis and pressure on the y-axis. Start both axes at zero.
(3)
b) Use your graph to calculate the number of moles of helium gas
present in the container.
(3)
c) The pressure inside the container is reduced to 0.50 atm by
cooling the container. Use your graph to determine the
temperature of the gas at this pressure.
(1)
d) Use your answer to (c) to calculate the average kinetic energy
of a helium atom at a pressure of 0.5 atm.
(2)
e) Hence calculate the total internal energy, U, of the helium.
(2)
355
16 This question is about ideal gases.
a) State what is meant by an ‘ideal gas’.
(2)
b) An ideal gas at 300 K is enclosed inside a gas canister of
volume 3.3 × 10−4 m3 at a pressure of 2.02 × 105 Pa. Calculate
the number of moles of gas enclosed inside the canister.
(2)
c) The molar mass of the gas is 0.084 kg mol−1. Calculate the
density of the gas inside the canister.
(3)
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17/04/19 10:44 AM
d) The canister is taken to the top of Mount Kilimanjaro, where it is
used to inflate an air-mat. If the temperature at the top of the
mountain is 266 K and the air pressure is 0.50 × 105 Pa, what
is the combined volume of the canister and air-mat that could
be inflated by the gas in the canister at this pressure.
(1)
Stretch and challenge
The questions that follow here are British Physics Olympiad questions.
17 a) State Boyle’s law.
(a)
Figure 18.22a shows a length of capillary tubing in which
a column of air is trapped by a mercury column of length
100 mm. The length of the air column is 400 mm. The bottom
of the tubing is sealed and the top is open to the atmosphere.
b) The tubing is now inverted, as shown in Figure 18.22b,
and the air column is seen to increase in length to
520 mm. Use this observation to calculate a value for
atmospheric pressure, expressed in mm of mercury.
c) A typical value for atmospheric pressure, expressed in
SI units, is 101 kPa. The surface area, A, of the Earth is
related to its mean radius by the expression, A = 4πR2,
where R has the value 6400 km. Calculate:
i)
the sum of the magnitudes of the forces exerted by
the atmosphere on the surface of the Earth
(b)
100 mm
520 mm
air
400 mm
100 mm
Figure 18.22
ii) the mass of the Earth’s atmosphere, assuming that g
does not vary with height above the Earth’s surface
iii) the number of molecules in the atmosphere, assuming that
the molar mass of air is 30 g mol−1
18 THERMAL PHYSICS
iv) the height of the atmosphere if the density ρ = 1.2 kg m−3.
356
d) The height of the atmosphere calculated in c) iv) is less than the
height at which many aircraft fly. Explain why our calculation
gives a low result for the height.
e) The height of the atmosphere is typically given as 200 km. Does
this mean that our calculation of the mass is completely wrong
(by a significant factor)?
(BPhO A2-2005 Q2; and A2-2011 Q4)
18 An accurate thermometer, of heat capacity 20.0 J K−1, reads 18.0°C.
It is then placed in 0.250 kg of water and both reach the same final
temperature of 50°C. Calculate the temperature of the water before
the thermometer was placed in it. The specific heat capacity of
water is 4200 J kg−1 K−1.
(BPhO R1-2005 Q1(a))
19 Wet clothing at 0°C is hung out to dry. The air temperature is 0°C
and there is a dry wind blowing. After some time it is found that
some of the water has evaporated and the water remaining on the
clothes has frozen. The specific heat of fusion of ice is 333 kJ kg−1
and the specific latent heat of evaporation of water is 2500 kJ kg−1.
469884_18_AQA_A-Level_Physics_329-357.indd 356
17/04/19 10:44 AM
Practice questions
a) What is the source of energy required to evaporate the water? Explain
the mechanism of evaporation.
b) Estimate the fraction, by mass, of water originally in the clothes that
freezes.
(BPhO R1-2005 Q1(e))
20 A lead bullet at 320 K is stopped by a sheet of steel so that it reaches
its melting point of 600 K and completely melts. If 80% of the kinetic
energy of the bullet is converted into internal energy, calculate the speed
with which the bullet hit the steel sheet. The specific heat capacity of
lead is 0.12 kJ kg−1 K−1 and its specific latent heat of fusion is 21 kJ kg−1.
(BPhO R1-2007 Q1(f))
21 a) Water in an electric kettle is brought to
the boil in 180 s by raising its temperature
from 20°C to 100°C. It then takes a further
1200 s to boil the kettle dry. Calculate the
specific latent heat of vaporisation of water,
lv, at 100°C, stating any assumptions made.
b) A cylinder, with a weightless piston, has an
internal diameter of 0.24 m. The cylinder
contains water and steam at 100°C. It is
situated in a constant-temperature water
bath at 100°C, as shown in Figure 18.23.
Atmospheric pressure is 1.01 × 105 Pa. The
steam in the cylinder occupies a length of
0.20 m and has a mass of 0.37 g.
i)
atmospheric pressure
0.2 m
steam
constant temperature
bath at 100°C
water
Figure 18.23
What is the pressure p of the steam in the cylinder?
ii) If the piston moves very slowly down a distance 0.10 m, how
much work, W, will be done in reducing the volume of the
steam?
iii) What is the final temperature, Tf, in the cylinder?
iv) Determine the heat, Qc, produced in the cylinder.
c) A molecule of oxygen near the surface of the Earth has a velocity
vertically upwards equal in magnitude to the root mean square
(r.m.s.) value. If it does not encounter another molecule, calculate:
i)
the height H reached if the surface temperature is 283 K
ii) the surface temperature, Ts, required for the molecule to escape
from the Earth’s gravitational field if the potential energy per unit

M 
 –G E  .

RE 
−1
0.032 kg mol .
mass at the Earth’s surface is
molar mass of
469884_18_AQA_A-Level_Physics_329-357.indd 357
357
The oxygen molecule has a
(BPhO R1-2002 Q2)
17/04/19 10:44 AM
19
Electric fields
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
●
Atoms and molecules contain protons and electrons, which carry
positive and negative charges, respectively. These charges are equal
in size. An atom is neutral because there are as many positively
charged protons as there are negatively charged electrons.
Some materials, such as plastic, can become charged by rubbing with
a cloth. If the plastic is charged positively, then electrons have been
removed from the plastic and transferred to the cloth, which now
carries a negative charge. Another type of plastic might be charged
negatively when rubbed by a cloth – electrons have been transferred
to the plastic and the cloth will be charged positive.
Like charges repel each other, and unlike charges attract each other
(Figure 19.1).
support
thread
–
– – – –
– –
– F + +
+ ++
F
negatively
–
– –
–
charged rod
–
––
– ––
– –
–
positively
charged rod
negatively
charged rod
Figure 19.1
●
Electric charges exert a force on each other over a distance. For
example, a charged comb can pick up pieces of paper (Figure 19.2).
plastic comb
rising
paper
358
wooden surface
Figure 19.2
●
●
●
469884_19_AQA_A-Level_Physics_358-380.indd 358
Charges produce an electric field.
An electric field is a region in space where a charged object
experiences a force.
Forces between charges are stronger when they are closer together.
The forces are weaker when the charges are further apart.
17/04/19 10:45 AM
1 Explain why an atom of magnesium, which has
12 protons in its nucleus, is neutral.
2 Explain how ions are formed.
3 Explain what is meant by the term ‘electric field’.
Name two other types of force field.
Coulomb’s law
TEST YOURSELF ON PRIOR KNOWLEDGE
4 Draw a diagram to explain how a comb, which is
positively charged, can lift up a piece of paper,
which is neutral. (This takes some explaining,
in terms of electron movement in the paper and
attractive and repulsive forces.)
Volcanic ash thrust into the atmosphere produces ideal conditions for
lightning. The enormous quantity of pulverised material and gases ejected
into the atmosphere creates a dense plume of charged particles. The friction
of particles moving past each other transfers charge in the same way as a
balloon can be charged by rubbing. Potential differences of millions of volts
exist within the plume, which are sufficient to drive large currents, which
discharge the clouds of ash.
●● Coulomb’s law
The starting point for work in electrostatics is Coulomb’s law, which states
that the force between two point charges Q1 and Q2 separated by a distance
r (Figure 19.3a) is given by
F=
Q1Q2
4πε 0r 2
Here F is measured in newtons (N), r is measured in metres (m), Q1 and Q2
are measured in coulombs (C), and ε0 is a constant, the permittivity of free
space: ε0 = 8.85 × 10−12 F m−1. (The Maths box on p. 360 shows how the
units are derived, for the interested reader.)
1
Sometimes you will find the constant
quoted as 9.0 × 109 F−1 m. You
(4πε0)
might also see Coulomb’s law written in the form
F=
kQ1Q2
where k =
(a)
Q1
+
(b)
r
+Q1
Q2
+
+Q2
r
Figure 19.3
469884_19_AQA_A-Level_Physics_358-380.indd 359
r2
1 .
(4πε0)
The value of ε0 quoted refers to the permittivity of free space – which means
a vacuum. The value of permittivity varies from one medium to another.
However, the permittivity of air is very close to that of a vacuum, so we
shall use the value of ε0 quoted in those calculations.
359
The equation for Coulomb’s law is very similar to that for Newton’s law of
gravitation, except that the force between two charges can be repulsive if
the two charges have the same sign, or attractive if the two charges have
the opposite sign. The force between two masses is always attractive.
Coulomb’s law may also be used to calculate the electrostatic force
between two charged spheres carrying charges Q1 and Q2. In this case the
distance used is the separation of the centres of the two spheres, as shown
in Figure 19.3(b).
17/04/19 10:45 AM
MATHS BOX
Where does the unit of permittivity, ε0, quoted as
F m−1 or farads per metre, come from? The reasoning
below explains this.
Coulomb’s law states that
F=
so
Q1Q2
However, a joule = newton × metre, and
volt = joule . Therefore the units of ε0 are
coulomb
[C] × [C]
[J] × [m]
[C]
=
[V] × [m]
[ε 0 ] =
4πε 0r 2
ε0 =
Q1Q2
4πFr 2
But the definition of capacitance tells us that
farad = coulomb . Therefore ε0 has units F m−1.
volt
The units of ε0 are therefore
[C] × [C]
[ε 0 ] =
[N] × [m 2 ]
[C] × [C]
=
[N m] × [m]
EXAMPLE
Coulomb’s law
1 Calculate the force of attraction between two point charges A and
B separated by a distance of 0.2 m. The charge at A is +2 µC and
the charge at B is −1 µC.
(a)
(b)
q q
Answer
19 ELECTRIC FIELDS
F=
360
Q1Q2
15 cm
4πε 0r 2
=
q
mg
T
(2 × 10–6 C) × (–10–6 C)
(4π × 8.85 × 10–12 F m–1) × (0.2 m)2
= -0.45 N
The significance of the minus sign is to remind us that the force
is attractive, but it is not really necessary to include it.
3 cm
3 cm
F
Figure 19.4
2 Figure 19.4a shows two light polystyrene spheres, which have been
coated in a conducting metallic paint. Each has been charged positively
by a high-voltage supply to about 3 kV. They are suspended by pieces of cotton
15 cm long, and they are pushed apart by the repulsive electrostatic force
between them. The mass of each sphere is 0.08 g.
Use the information in the diagram to calculate the charge on each sphere.
Answer
Figure 19.4b shows the forces acting on the right-hand sphere. The tension
in the cotton, T, is balanced by the electrostatic force, F, and the weight of
the ball, mg. From the triangle of forces we can see that
tanθ = F
mg
469884_19_AQA_A-Level_Physics_358-380.indd 360
17/04/19 10:45 AM
F=
3 cm
= 0.2
15 cm
However, when θ is small, sin θ ≈ tan θ. Therefore
sinθ =
= 0.2 × (0.08 ×
= 1.6 ×
×
4πε 0r 2
=
Q2
4πε 0r 2
since Q1 = Q2. Therefore
Q2 = 4πε0r2 × F
tan θ ≈ sin θ and F = 0.2. So
mg
F = 0.2 mg
10−3 kg)
Q1Q2
Coulomb’s law
From Coulomb’s law,
From the triangle showing the displacement of the
two spheres we can see that
Q2 = 4π × (8.85 × 10−12 F m−1) × (0.06 m)2 × 1.6 × 10−4 N
Q = 8.0 × 10−9 C
9.8 N kg−1
10−4 N
ACTIVITY
Testing Coulomb’s law
Figure 19.5 shows an
experimental arrangement for
investigating Coulomb’s law.
The two polystyrene spheres
are charged by a high-voltage
supply. Sphere A is held in a
fixed position and sphere B is
free to move. A light bulb is
used to cast a shadow of the
spheres onto graph paper, so
that their separation, and also
the deflection of sphere B, can be
measured more easily.
insulating
suspension
threads
EHT (<5KV –
current limited
to <5mA)
vertical board
for shadows
support rod
graph paper
+ _
A
B
Insulating
handle
Table 19.1 shows the results for
six different separations of the
spheres’ shadows. Between each
set of measurements the spheres
were recharged.
shadow
light bulb
charged sphere of
metallised expanded
polystyrene
Figure 19.5
Table 19.1
Distance between centres of
spheres’ shadows/mm
Deflection of sphere B’s shadow/mm
100
70
50
40
30
20
1
3
5
8
15
29
361
1 Plot a graph of the deflection of sphere B’s shadow against 12 , where r is the separation of the centres of the
r
spheres’ shadows.
Discuss whether or not your graph supports Coulomb’s law.
2 Explain why the deflection of sphere B is proportional to the force between the balls.
3 Discuss the sources of error in this experiment. You should consider both systematic and random errors.
4 Discuss how you could calculate the real separation of the spheres, rather than their shadows.
469884_19_AQA_A-Level_Physics_358-380.indd 361
17/04/19 10:45 AM
TEST YOURSELF
For these questions, use the following values: mass
of an electron = 9.1 × 10 −31 kg, charge on an electron =
1.6 × 10 −19 C, ε0 = 8.85 × 10 −12 F m −1.
1 The centre of a small sphere carrying a charge of
+2.0 nC is placed at a distance of 240 mm from the
centre of a second small sphere carrying a charge
of −5.0 nC.
a) Calculate the force of attraction between them.
b) Calculate the size of the force between the
spheres when they are separated by each of the
following distances:
i) 120 mm
ii) 80 mm
iii) 60 mm
iv) 48 mm.
2 The electron and proton in a hydrogen atom are
on average about a distance of 5 × 10 −11 m apart.
Calculate the force the proton exerts on the
electron. How big a force does the electron exert
on the proton?
3 A uranium nucleus contains 92 protons; the
nucleus has a radius of 8.0 × 10 −15 m. Calculate
the force on an alpha particle at the surface of
the uranium nucleus. Comment on the size of this
force.
4 Two charges, one three times the size of the other,
experience a repulsive force of 80 mN when they
are separated by a distance of 10 cm. Calculate the
size of the larger charge.
5 A small polystyrene sphere of mass 1 g is
suspended near to another larger sphere so that
their centres lie along the same horizontal line.
Both spheres are charged negatively. The small
sphere is now deflected so that the thread holding
it is deflected to an angle of 36° to the vertical.
a) Draw a free-body diagram to show the forces
acting on the sphere.
b) Calculate the size of the repulsive force
between the spheres.
6 a) The average distance of electrons from the
nucleus, in the lowest energy state of a gold
atom, is 7 × 10 −13 m. Calculate the force between
such an electron and the gold nucleus. Gold has
an atomic number of 79.
b) By making the assumption that the electron
moves in a circular orbit of radius 7 × 10 −13 m.
calculate the speed of the electron. How does
this speed compare with the speed of light?
19 ELECTRIC FIELDS
●● Electric field strength
362
In the last section you met the idea of two point charges exerting a force
on each other. A charge produces an electric field around it, which exerts
a force on another charged object. This idea is similar to a magnetic field
close to a magnet, or a gravitational field around a planet.
An electric field strength is defined by the equation
E= F
Q
where F is the force, in newtons, which acts on a charge, Q, in coulombs.
So electric field strength is measured in newtons per coulomb, N C−1.
The direction of the electric field is defined as the direction of the force
on a positive charge. Electric field is a vector quantity because it has both
magnitude and direction.
We represent electric fields by drawing lines. Figure 19.6 shows two uniform
electric fields. The stronger field in Figure 19.6a is represented by field lines
that are closer together. The fields are uniform because in all places the field
has the same strength and the same direction. Note that the field lines start
on a positive charge and end on a negative charge. The positive charge in
Figure 19.6a experiences an electrostatic force downwards.
469884_19_AQA_A-Level_Physics_358-380.indd 362
17/04/19 10:45 AM
(a)
+
+
+
+
+
+
+
+
+
–
–
–
Electric field strength
+
+
400 N C–1
–
–
–
–
–
–
–
(b)
+
+
+
+
+
+
E = 200 N C–1
–
–
–
–
–
–
–
Figure 19.6
Electric field lines are a model to help us visualise a field, but a direct way
of showing an electric field is shown in Figure 19.7. In this photograph, a
potential difference has been applied to two metal plates, which have been
placed into an insulating liquid. Then short pieces of a fine thread have
been sprinkled on top of the liquid. When the electric field is applied, the
pieces of thread line up along the field lines, in the same way that iron
filings follow magnetic field lines.
Figure 19.7
EXAMPLE
Force on a small charge
Electric field strength and potential gradient
A small charge of +2 µC is
placed in the electric field in
Figure 19.7a. What force does it
experience?
Figure 19.8 shows a charge +Q placed in an electric field between two
parallel plates. The plates have a potential difference of V between them,
and their separation is d (in m). How much work is done if the charge is
moved from A to B? This question can be answered in two ways.
Answer
First, the work done = F × d, but the force to move the charge must be equal
in magnitude to the force on the charge, due to the electric field, E × Q. So
E=F
Q
work done = EQd
So
F = EQ
Secondly, the work done is also equal to the energy gained by the charge
in moving through a potential difference V. This is VQ – you should
remember that a volt is defined as a joule per coulomb. Therefore
= 400 N C−1 × 2 × 10−6 C
= 8 × 10−4 N downwards
EQd = VQ
and
+
+
+ B +
+
+
+Q
d
V
F = EQ
–
–
–
A
–
Figure 19.8
469884_19_AQA_A-Level_Physics_358-380.indd 363
–
–
363
E=
V
d
This equation allows us to calculate the magnitude of a uniform electric
field between two parallel plates. Note that the electric field strength can
also be measured in V m−1.
Figure 19.9 allows us to produce a more general formula to link electric
field and potential gradient. In the formula above, we just concentrated
on the magnitude of the field, but strictly speaking we should have
included its direction.
17/04/19 10:45 AM
+
In Figure 19.9 the charge Q is moved through a potential
ΔV through a small distance Δr by the force F = −EQ. (This
is marked in blue in the diagram and is in the opposite
direction to the force from the electric field.)
F = –EQ
Δr
ΔV
So we now get
+Q
work done = −EQ Δr = ΔV Q
or
increasing
potential
E=–
F = EQ
–
So the electric field strength is equal in magnitude to
the potential gradient, but it is in the opposite direction.
Figure 19.9
Figure 19.10 shows a graph of electric field strength
in a region. The area under the graph may be used to
calculate the potential difference between two points.
A shaded section (green) under the graph has an area
E Δr, which has a value ΔV.
2000
Δr
E/V m–1
E
1500
∆V
∆r
In general, the area under a graph of E against r gives
the change of potential ΔV.
1000
500
0.1
0.2
r/m
0.3
0.4
Figure 19.10
19 ELECTRIC FIELDS
Electric field strength is defined by
364
E=F
Q
In a uniform field
E = V
d
Use Figure 19.10 to calculate the change in potential in moving from
position r = 0 to r = 0.2 m.
Answer
and generally
E =–
EXAMPLE
Change in electric potential
∆V
∆r
ΔV = area under the graph
1
= (2000 V m −1 + 1500 V m −1) × 0.2 m
2
= 350 V
Units of electric field strength are N C−1 or
V m−1.
The area under an E against r graph is ΔV,
the change in potential.
Deflection of charged particles by electric fields
Any charged particle experiences a force in an electric field. So when a
moving charged particle enters an electric field, it will change direction.
(Only when a charged particle moves parallel to an electric field does it
keep moving in the same direction.) Figure 19.11 shows a photograph of an
electron beam tube, which can be used to deflect electrons.
Figure 19.12 shows the principle behind the electron beam tube.
Electrons are accelerated by an ‘electron gun’ on the left-hand side. They
469884_19_AQA_A-Level_Physics_358-380.indd 364
17/04/19 10:46 AM
Electric field strength
then travel across a fluorescent screen, which shows the electron path.
The electrons travel from P to Q. The electrons are deflected by applying
a potential difference between A and B. When a potential difference
is applied so that the top plate is positive, the electrons are deflected
upwards along a path such as PR.
Figure 19.11 Electrons can be deflected and
observed inside this evacuated tube.
metal plate
A
fluorescent screen marked
with squares 1 cm x 1 cm
+
R
Electron gun
path of
Q electron
beam
P
6V~
–
metal plate
–
V1
B
+
Figure 19.12
365
The path PR is a parabola, which can be explained as follows. The electrons are
travelling in a vacuum, so their velocity in the direction PQ remains unchanged.
While the electrons are in the electric field, they experience a constant
acceleration upwards due to the electric field. This is rather like throwing a ball
sideways – the ball’s horizontal velocity remains constant, but gravity gives the
ball a constant downwards acceleration. The ball falls along a parabolic path.
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EXAMPLE
Electron beam tube
This example refers to Figure 19.12. In an experiment,
a beam of electrons is directed along the line PQ. The
electrons arrive at P with a velocity of 4.0 × 107 m s −1
travelling in the direction PQ. The squares on the grid
measure 1 cm × 1 cm.
The acceleration is given by
1 Calculate the time taken for the electrons to travel
from P to Q.
where Q is the charge on an electron and m is its
mass. This gives
F
a=m
EQ
= m
Answer
a =
d =v × t
t =d
v
=
3.66 × 104 V m–1 × 1.6 × 10–19 C
9.1× 10–31 kg
= 6.4 × 1015 m s–2
0.1m
3 Show that the electron beam is deflected upwards
to point R, which is about 2 cm above point Q.
4.0 × 107 s
Answer
= 2.5 × 10–9 s
Now a potential difference of 2200 V is applied between
A and B, so that the beam deflects upwards.
2 Calculate the acceleration of an electron in this
electric field.
Answer
The electric field strength is
E =V
d
2200V
=
0.06 m
To calculate the upwards displacement of the beam,
we use the equation of motion:
s = ut + 1 at 2 = 1 at2
2
2
since the initial upward velocity u = 0. So


s =  1 ×6.4 ×1015 m s–2 ×(2.5 ×10–9 s) 2

 2
= 0.02 mor 2 cm
19 ELECTRIC FIELDS
= 36.6 kV m–1
366
TEST YOURSELF
7 T
he strength of an electric field may be expressed
in units of either N C−1 or V m−1. By considering
the definitions of the volt and the joule, show that
these two quantities are the same.
8 Figure 19.13 shows a charged particle placed
between two charged parallel plates. The
potential difference between the plates is 1500 V
and their separation is 7.5 cm.
P
Figure 19.13
a) C
alculate the magnitude of the electric field
strength between the plates.
b) A particle P experiences a force, from the
469884_19_AQA_A-Level_Physics_358-380.indd 366
electric field, of 1.5 × 10 −7 N in the direction
shown. Calculate the charge on the particle.
c) Calculate the work done by the electric field
on the particle in taking it from the top plate to
the bottom place.
9 This question refers to the deflection of electrons
shown in Figure 19.12. Explain what will happen to
the path of the electrons when, separately,
a) the potential difference between A and B is
increased
b) the potential difference V1 is increased so that
the electrons travel faster as they enter the
deflecting area.
10 A small polystyrene ball is dropped between a pair
of parallel plates as shown in Figure 19.14. As it
enters the plates, it has reached its terminal speed,
which is 1.0 m s−1. Initially, the plates are uncharged.
17/04/19 10:46 AM
f) Calculate the ball’s maximum sideways
deflection in the field. Why might the deflection
be less than this?
11 Calculate the potential difference between the
points r = 0 and r = 0.4 m in Figure 19.10.
12 A small oil drop, of mass 1.7 × 10 −15 kg, carries
a negative charge. It is suspended between two
parallel plates as shown in Figure 19.15. The p.d.
between the plates is adjusted to 312 V so that the
drop remains stationary.
+
1 m s–1
10 cm
+
–
Radial electric fields
2.5 cm
+
5 cm
Figure 19.14
312 V
a) Explain why a falling object has a terminal
speed. (You may need to refer back to
Chapter 8.)
b) Calculate the time taken for the ball to fall
through the plates.
The experiment is repeated when the plates
are charged as shown in Figure 19.14, and the
potential difference between them is 3000 V. In
this second experiment, the ball has been given a
positive charge of 1.2 × 10−8 C. It enters the plates
at the top at its terminal velocity of 1.0 m s−1.
c) Sketch the path of the ball as it falls between
the plates.
d) i) Calculate the electric field between the
plates.
ii) Calculate the force on the ball due to the
electric field.
e) The ball has a mass of 0.2 g. Calculate its
sideways acceleration as it enters the field.
–
6 mm
–
Figure 19.15
a) Draw a free-body diagram to show the forces on
the drop.
b) Use the information above and in the diagram to
calculate the charge on the drop.
c) The charge on the drop is changed by the use of
ionising radiation. A p.d. of 208 V is now used to
balance the drop. Calculate the charge on the
drop now.
d) What is the smallest charge the drop can carry?
For a drop that carries this charge, calculate
the p.d. that would hold the drop in a stationary
position.
e) Find out who first did this experiment to
determine the charge on the electron.
●● Radial electric fields
Figure 19.16 shows a photograph of the shape of an electric field close to
a small point charge. This photograph is obtained in the same way as that
shown in Figure 19.7. The electric field has a symmetrical radial shape near
to a small point charge.
Figure 19.17 shows how we can represent the electric field lines close to
a positively charged sphere. The lines point outwards symmetrically from
the sphere as if they had come from the centre of the sphere. You can also
see that the lines spread out. This means that the field gets weaker as the
distance increases from the sphere. This is very similar to the shape of the
gravitational field near to a planet, except that the gravitational field lines
must always point towards the planet. (Under what circumstances do the
field lines point towards a charged sphere?)
367
Figure 19.16
469884_19_AQA_A-Level_Physics_358-380.indd 367
17/04/19 10:46 AM
We can produce a formula for the electric field close to a sphere as follows.
We know from Coulomb’s law that the force between a sphere, carrying
charge Q, and a small charge q at a distance r from the centre is
+q
r
+Q
F=
Qq
4πε 0r 2
We also know that F = Eq. It follows that the electric field close to the
sphere is given by the formula
E=
Q
4πε 0r 2
So the strength of the electric field obeys an inverse square law.
Figure 19.17
As you can see from Figure 19.17, the electric field is a vector quantity. So
when we consider the field close to two or more point charges, we must
take account of the direction of the electric field.
EXAMPLE
Resultant electric fields
Figure 19.18 shows two small charges, one with
a charge of +4Q, the other with charge +Q. The
magnitude of the electric field at C due to the charge
+4Q is 40 N C−1.
+4Q
6 cm
A
6 cm
C
+Q
B
Figure 19.18
1 Calculate the magnitude of the electric field at C
due to the charge +Q alone.
19 ELECTRIC FIELDS
Answer
368
The field at C due to the charge +Q is 10 N C−1,
because the charge is 41 . But the field is in the
opposite direction (right to left).
2 Calculate the magnitude of the electric field at C
due to the two charges together.
Answer
The resultant field is 40 N C−1 − 10 N C−1 = 30 N C−1 to
the right.
3 Show that the position of point O, where the electric
field is zero along the line AB, lies 4 cm from B.
Answer
The field due to A is
EA =
k × 4Q
kQ
1
=
where k =
2
0.0016
4πε0
0.08
The field due to B is
kQ
kQ
=
2
0.0016
0.04
So at a point 4 cm from B the two fields cancel each
other out.
EB =
TEST YOURSELF
13 a) The electric field due to a point charge is 300 N C−1 at a distance
of 100 mm away from it. Calculate the strength of the field at
distances of
i) 50 mm
ii) 200 mm
iii) 250 mm.
b) Sketch a graph to show how the field strength varies with distance away from the charge.
469884_19_AQA_A-Level_Physics_358-380.indd 368
17/04/19 10:46 AM
J
Calculate the magnitude of the electric field strength for each of the
points B to K. (You will need to use a combination of Pythagoras’s
theorem and the inverse square law.)
15 a) Figure 19.20 shows two charged spheres, X and Z. Calculate the
electric field strength at point Y, which lies along the line XZ joining
the centres of the two charged spheres. Sphere X has a charge of
+5 × 10 −9 C, and sphere Z a charge of −10 −8 C.
0.3 m
+5 × 10–9 C
0.2 m
–10–8 C
K
H
E
O
A
F
B
Electrical potential
14 Figure 19.19 shows a grid marked with squares. A positive charge
placed at O produces an electric field of strength 3600 N C −1 at point A.
I
G
C
D
Figure 19.19
Y
X
Z
Figure 19.20
b) i) Calculate the field strength, at Y, if the charge of −10 −8 C on sphere Z is
replaced with a charge of +10 −8 C.
ii) Calculate the force between the spheres in this case. Is the force
attractive or repulsive?
●● Electrical potential
Figure 19.21 shows a small charged isolated sphere. In theory, an isolated
sphere should be in contact with nothing and infinitely far away from
anything else. In practice, in a laboratory, the best we can do is to suspend a
sphere by a fine, insulating thread and make sure the sphere is a few metres
from everything else.
The potential at a distance r from the centre of a sphere carrying a charge Q
is given by
insulating thread
supporting sphere
+Q
r
Figure 19.21
469884_19_AQA_A-Level_Physics_358-380.indd 369
Q
4πε 0r
Note that if the charge is negative, then the potential near to the charged
sphere is also negative.
V=
These formulae are very similar to the formula for gravitational potential –
except that gravitational potential must always be negative, whereas an
electrical potential, close to a charge, can be positive or negative. In the
case of gravitational potential, the zero point of potential is defined as a
point infinitely away from any planet or star. In a similar way, the zero
point of electrical potential is defined as a point infinitely far away from
the charged sphere. In practice, the surface of the Earth is our reference
point of zero potential, and by suspending our sphere a long way from
anything else, the Earth can be treated as being infinitely far away. The
Maths box shows how the formula for potential can be derived from the
formula for electric field.
369
17/04/19 10:46 AM
MATHS BOX
The electric field at a distance r from the centre of an isolated charged
sphere carrying a charge +Q is
E=
Q
4πε 0r 2
But
E=−
dV
dr
so
V=−
r
Q
∞
0
∫ 4πε r
2
dr
Q
4πε 0r
Note that the limits of the integration set the potential as zero at an
infinite distance.
=
Absolute electric potential The potential
difference between a point and a point at
zero potential, which is infinitely far away.
The formula V =
Q
4πε 0r
helps us to define absolute electric potential. The
absolute electric potential at a point r from a charge +Q is the work done
per unit positive charge in moving it from ∞ to that point. Note that if that
charge is −Q, then the potential is negative and the electric field does work
in moving a positive charge closer to the point r.
19 ELECTRIC FIELDS
EXAMPLE
Electric field near a charged sphere
370
V = 1000 V
Figure 19.22 shows a metal sphere of radius 10 cm,
charged to a potential of 1000 V. The electric field
strength at C is 10 000 V m−1.
Sketch graphs to show how
A
B
C
–50 cm
–10 cm
0
D
10 cm
50 cm
Figure 19.22
1 the potential
2 the field strength
vary along the line A to B and then from
C to D.
V/V E/V m–1
1000
10000
500
5000
Answer
Figure 19.23 shows the answer. These are
the points to note.
● V is a scalar and is always positive.
1
● V obeys a
law and falls from 1000 V
r
●
●
–50
–40
–30
–20
–10
10
–500
–5000
from the sphere.
E is a vector, so must change direction.
E is connected to the potential by the
–1000
–10000
dV
dr
. On the right-hand
469884_19_AQA_A-Level_Physics_358-380.indd 370
30
40
50
r/cm
at 10 cm to 200 V at a distance of 50 cm
equation E = −
20
E/V m–1
Figure 19.23
17/04/19 10:47 AM
sphere, because a positive charge is always repelled
by it.)
●
E obeys a
1
r2
law and falls from 10 000 V m−1 at
10 cm to 400 V m−1 at a distance
of 50 cm from the sphere.
Potential difference
Potential difference The work done,
against an electric field, in moving unit
charge from one point to a second point
at a higher potential. If a charge moves
from a point of higher potential to a lower
potential, work is done by the electric field.
Electrical potential
side of the sphere, the potential gradient is negative,
so E is a positive quantity. On the left-hand side of
the sphere, the potential gradient is positive, so E is
a negative quantity.
(More simply, the electric field must be in opposite
directions on either side of the
Electric potential difference is the difference in electrical potential between
two points. When the potential difference is ΔV, the work done in moving a
charge Q between the two points is
ΔW = Q ΔV
EXAMPLE
Work done in moving a charge
How much work is done in taking a charge of 1.00 × 10−9 C from the position r2
to r1, in Figure 19.24?
+1.00 × 10−7 C
1.0 m
r2
r1
4.0 m
Figure 19.24
Answer
The potential at a point is
V =
Q
4πε 0r
so the potential difference between r1 and r2 is
1 1
 − 
r r 
2
1
 1
1.00 ×10−7 C
1 

=
−
−12
−1 
4π × 8.85 ×10 F m  1m 4 m 
∆V =
Q
4πε0
= 675 V
So the work done is
∆W = Q ∆V
= 1.00 × 10−9 C × 675 V
= 675 nJ
371
Equipotential surfaces
Figure 19.25 shows a metal sphere that is charged to a potential of 1000 V.
The red circles drawn round the sphere show equipotential surfaces, where
the potential is the same, e.g. 900 V and 800 V. Although the diagram shows
circles, this is because the diagram can only be drawn in two dimensions.
The sphere is surrounded by spherical equipotential surfaces.
469884_19_AQA_A-Level_Physics_358-380.indd 371
17/04/19 10:47 AM
+
b
c
6
+ + +
5
+
4
+
+
fg +
+
d e
hi + + +
a
Position Potential
a
200 V
b
300 V
c
400 V
d
500 V
e
600 V
f
700 V
g
800 V
h
900 V
i
1000 V
3
2
1
0
10 cm
20 cm
30 cm
Figure 19.25
The equipotential surfaces may be linked to the electric field strength
through the equation
19 ELECTRIC FIELDS
E=–
372
∆V
∆r
Figure 19.25 shows that, near the surface of the sphere, the equipotential
surfaces are closer together. This means that both the potential gradient and
the electric field strength are higher near the sphere’s surface than they are
further away. The green lines in the diagram represent electric field lines,
which point radially away from the positive sphere. These lines get further
apart with distance from the sphere, which also shows a field diminishing
with distance away from the sphere’s centre.
The field lines are always at right angles to the equipotential surfaces. So,
when a charged particle moves along an equipotential surface, no work is
done by the electric field. This can be explained using two ideas.
First, work done is defined by
ΔW = ΔV Q
or, in words, the work done on a charge is the change in potential
multiplied by the charge. When ΔV = 0 (moving along an equipotential
surface) the work done is zero.
469884_19_AQA_A-Level_Physics_358-380.indd 372
17/04/19 10:47 AM
Secondly, work is also defined by
or, in words, the work done on an object is the force acting on the object
multiplied by the distance moved in the direction of the force. When a charge
is moved along an equipotential, it does not move in the direction of the
force or the electric field, which acts on it, but at right angles to the force.
Electrical potential
ΔW = F Δr
EXAMPLE
Moving a charge on equipotential surfaces
In Figure 19.25, a charge of 1.0 × 10−6 C is moved first from point 1 to point 2,
and then from point 2 to point 3. How much work is done in each case?
Answer
No work is done in moving the charge from point 1 to point 2, that is ΔV = 0.
In moving from point 2 to point 3
∆W = Q ∆V
= 1.0 × 10−6 C × (300 V − 200 V)
= 1.0 × 10−4 J
A
+Q
B
O
V = 25 V
D
Figure 19.26
C
Further examples of field and potential
Figure 19.26 shows a square ABCD. At point A there is a charge +Q; the
potential at point O (the centre of the square) due to this charge is 25 V.
What is the potential at O when a charge of +Q is placed at each of the
points A, B, C and D?
The answer to this is 100 V. To move a positive charge to point O, work has
to be done against each of the four charges. Therefore the potential at O is
four times larger. (The potential at a point is the work done per unit charge
to take it from infinity to that point.)
Since electric potential is a scalar quantity, we can calculate the potential
at a point close to two or more charges by adding the potential due to
each charge. Figure 19.27 shows the electric potential (calculated by
computer) near to a positive and negative ion, which are separated by a
distance of 10−9 m.
373
469884_19_AQA_A-Level_Physics_358-380.indd 373
17/04/19 10:47 AM
E
0V
E field
D
A
–1.0 V
–1.7 V –2.5 V
–5.0 V
+5.0 V
–Q
B
+2.5 V
+1.0 V
+1.7 V
+Q
F
C
10–9 m
0
Figure 19.27
19 ELECTRIC FIELDS
TEST YOURSELF
374
16 A
n isolated metal sphere is charged with a
positive charge of
1.5 × 10 −7 C.
a) i) Calculate the potential at point A, a distance
of 0.25 m from the sphere’s centre.
ii) Calculate the potential at point B, a distance
of 0.75 m from the sphere’s centre.
b)Calculate the work done in moving a charge of
2.0 × 10 −8 C from B to A.
17 This question refers to Figure 19.25. Calculate
the work done in moving a positive charge of 2.0 ×
10 −7 C from
a) point 3 to point 4
b) point 4 to point 5
c) point 3 to point 6.
18 This question refers to possible arrangements of
charges in Figure 19.26.
a) Calculate the potential at O, for each of the
following arrangements of charges:
469884_19_AQA_A-Level_Physics_358-380.indd 374
i) A, +Q; B, +Q; C, −Q; D, +Q
ii) A, +Q; B, −Q; C, +Q; D, −Q
iii) A, +2Q; B, −3Q; C, +Q; D, −Q.
b) For which of the above arrangements is the
electric field strength zero at O? Explain your
answer.
19 A n isolated charge of +Q is placed at point A in
Figure 19.28. The potential at point B is 120 V.
F
A
B
C
D
E
Figure 19.28
17/04/19 10:47 AM
20 T
his question refers to the equipotential surfaces
shown near to the ions in Figure 19.27.
a) Which of the following statements is/are true?
Explain your reasoning.
i) The direction of the electric field at point A
is correctly drawn.
ii) The strength of the electric field is
stronger at D than at E.
b) Calculate the change in electrical potential
energy when an electron is moved from
i) B to C
ii) B to F.
Express your answers in eV.
c) Sketch the shape of the electric field close to
the two ions.
Similarities between electricity and gravitation
a) Calculate the potential at points
i) C
ii) D
iii) E
iv) F.
b) A second charge of +Q is now placed at E.
Calculate the potential at points
i) B
ii) C
iii) D
iv) F.
c) The charge at E is now replaced with a charge
of −Q. Calculate the potential at points
i) B
ii) C
iii) D
iv) F.
●● Similarities between electricity and
gravitation
A comparison between Newton’s law of gravitation and Coulomb’s law
shows that there are many similarities between the actions of the two types
of force. There are also some important differences.
The main similarities are as follows:
l
l
l
Both electric and gravitational forces are non-contact; forces are exerted
over a distance without direct contact.
Both forces are of infinite range.
Both forces obey inverse square laws.
The main differences are as follows:
l
l
l
l
Gravitational forces between masses are always attractive; electric forces
can be attractive or repulsive.
An electric force is much stronger than the gravitational force.
It is possible to shield an electric force, but the gravitational force acts on
all objects.
An electric force only acts on charged objects; the gravitational force acts
on all objects.
375
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17/04/19 10:47 AM
Table 19.2 shows a useful summary of the similarities and differences.
Table 19.2
Characteristic
Gravitation
Electricity
Acts on
Mass (positive only)
Charge (positive or negative)
Force (N)
F=
GMm
r
2
attractive or repulsive
infinite range
infinite range
relative strength
1
10 36
Field
g=F
m
E =F
Q
units
N kg−1
N C−1 or V m−1
Potential difference
units
Potential gradient
Potential in radial fields
g=
GM
r2
∆W
m
∆V =
J kg−1
19 ELECTRIC FIELDS
Potential energy
E=
Q
4πε0r 2
∆V =
g =−
∆V
∆r
E =−
V =−
GM
r
V=
Ep =
∆W
Q
J C−1
V = 0 at ∞
469884_19_AQA_A-Level_Physics_358-380.indd 376
4πε0r 2
attractive only
radial field
376
Q1Q2
F=
−GMm
r
∆V
∆r
Q
4πε0r
V = 0 at ∞
Ep =
Q1Q2
4πε0r
TEST YOURSELF
21 a) C
alculate the gravitational force between two protons separated
by a distance of 10 −10 m.
b) Calculate the electrostatic force between the protons separated
by a distance of 10 −10 m.
c) Calculate the ratio of the two forces calculated above.
d) What will the ratio of the forces be when the protons are
separated by 10 −12 m?
(Look up data for these calculations.)
22 The strong nuclear force binds nucleons together in a nucleus. It is
thought that the force acts over a range of about 10 −15 m, and that the
force is 137 times stronger than the electric force. Comment on the
information in the previous sentence. Does it make sense?
23 E xplain how it is possible to shield a region from an electric field.
17/04/19 10:48 AM
Practice questions
Practice questions
1 A proton and an electron are separated by a distance of 5 × 10−11 m.
The size of the electrostatic force between them is
A 18 × 10−8 N
C 6 × 10−8 N
B 9 × 10−8 N
D 1.8 × 10−8 N
2 The potential energy of a proton–electron pair separated
by 5 × 10−11 m is
A 288 eV
C 28.8 eV
B 92.2 eV
D 5.6 eV
3 An electron starts at rest and is accelerated through a potential difference
of 1200 V. The speed of the electron after it has accelerated though this
p.d. is
A 8 × 106 m s−1
C 4 × 107 m s−1
B 2 × 106 m s−1
D 2 × 107 m s−1
4 On a dry day the electric field near the surface of the
Earth is 140 V m−1 downwards. A drop of water of
mass 4.2 mg is suspended in the electric field. The
charge that the drop carries is
A 3×
10−4 C
B −3 × 10−7 C
C 6×
A
+
+
B
C
A
D
C
10−7 C
A
D −3 × 10−4 C
B
C
D
A
B
C
D
A
B
C
D
Use this information to answer questions 5 and 6:
Figure 19.29 shows two equal positive charges that are
placed at B and C, along a line AD. Four graphs show the
possible variation of quantities along the line AD.
5 Which graph shows the variation of electric field
along AD?
B
D
A
B
C
D
Figure 19.29
6 Which graph shows the variation of electric potential along AD?
Use this information to answer questions 7 and 8:
The charge at C is now replaced with a negative charge of the same size.
7 Which graph now shows the variation of electric field
along AD?
160 V
D
120 V
8 Which graph now shows the variation of electric potential
along AD?
9 Figure 19.30 shows a series of equipotentials. Which of the
following statements is not true?
C
A
B
80 V
40 V
377
0V
Figure 19.30
A The work done in taking a charge of 0.1 C from A to B is zero.
B The work done in taking a charge of 0.1 C from A to C is 4 J.
C The electric field at D is stronger than the electric field at B.
D The direction of the electric field is downwards.
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17/04/19 10:48 AM
10 The electric field strength at a distance of 10 cm from the surface of a metal
sphere is 900 N C−1. The sphere has a radius of 20 cm. What is the electric
field strength at a distance of 70 cm from the surface of the sphere?
A 450 N C−1
C 100 N C−1
B 225 N C−1
D 50 N C−1
11 In Figure 19.31a an electron is placed at P in an electric field, which is
represented by the field lines shown.
a) i)
In which direction will the electron accelerate?
(1)
(a)
ii) Describe how the electron’s acceleration changes
with its position in the field. Explain your answer.(2)
b) An electron is now placed in another electric field, at Q,
as shown in Figure 19.31b.
i)
Describe how the electron’s acceleration changes
with its position in the field now.
–P
(b)
(1)
–Q
ii) The electron, at Q, is replaced by a proton. Compare the
proton’s acceleration with the electron’s acceleration. (3)
12 Figure 19.32 shows two parallel plates, which are connected to
a low-voltage supply. The plates are in a region where there is a
vacuum. A small polystyrene sphere is placed at X between the
plates. The sphere carries an electric charge of +4.0 × 10−18 C,
and it has a mass of 2.6 × 10−15 kg.
Figure 19.31
6 cm
a) Calculate the size of the electric force acting on the sphere.(3)
b) Draw a free-body diagram to show the forces acting on
the sphere.
(2)
19 ELECTRIC FIELDS
c) Calculate the magnitude and direction of the sphere’s
acceleration after its release.
300 V
+
(4)
d) A different sphere is now introduced into the field at point X.
It carries twice as much charge as the first sphere, and it is
twice as massive. Compare the magnitude and direction of
this second sphere’s acceleration with the first sphere.
(2)
+
X
–
0V
–
Figure 19.32
13 A small plastic ball is suspended on a fine glass spring as shown in
Figure 19.33. It carries a negative electric charge. When a potential
difference of 500 V is applied to the plates, the ball moves upwards by
a deflection of 9 mm. The spring constant of the spring is 0.12 N m−1.
378
20 cm
_
_
_
_
_
_
_
+
_
500 V
Figure 19.33
a) Use the deflection of the ball to calculate the electrostatic the
force acting on it.
(2)
b) Use the information in the diagram to calculate the electric
field strength.
469884_19_AQA_A-Level_Physics_358-380.indd 378
(2)
17/04/19 10:48 AM
c) Deduce the charge on the sphere.(2)
(You will only be able to do this part of the question if you
have studied simple harmonic motion in Chapter 16.)
0V
14 A small sphere of mass 2.4 g is charged and suspended in an electric
field (Figure 19.34). It is deflected from the vertical at an angle of 10°.
a) Use the information in Figure 19.34 to calculate the
strength of the electric field between the plates.
–
+
12 cm
Figure 19.34
b) Show that the force due to the electric field acting on
the sphere is about 4 × 10−3 N.
(3)
c) Calculate the charge on the sphere.
(2)
+
15 Figure 19.35 shows the arrangement of two protons that form a
hydrogen molecule. They are separated by a distance of 3.0 × 10−10 m.
A
(1)
b) i)Show that the electric potential at point C, due to the
proton at A only, is 9.6 V.
1.5 × 10–10 m
+
C
B
3.0 × 10–10 m
Figure 19.35
(3)
ii)State the potential at point C due to both protons at A and B.
(1)
c) An electron, in its ground state, has 3.7 eV of kinetic energy at point
C. Show that the total energy of the electron at
this point is −15.5 eV.
(2)
d) State the ionisation energy of the hydrogen
molecule.
D
(1)
16 In Figure 19.36 two charges are placed at points A
and B, which are 1.0 m apart. At A there is a charge
of +6 nC, and at B a charge of −6 nC.
A
C
B
+6 nC
–6 nC
a) Calculate the strength of the electric field at
point C, which is halfway between the charges. (3) Figure 19.36
b) i)Draw a vector diagram to show the two electric
fields due to the charges A and B at point D.
(2)
ii)Use the diagram to show the direction of
the electric field at D.
(1)
0.5 m
0.5 m
300 V
17 Figure 19.37 shows equipotentials around a positive charge.
379
+
a) Explain how the equipotentials show that the electric
field is stronger at point A than it is at point C.(2)
b) A charge of −2.0 nC is moved from
i)
B to C
A
400 V
B
C
ii) C to D.
Calculate the work done against the electric field in each case.(3)
469884_19_AQA_A-Level_Physics_358-380.indd 379
+600 V
+
(1)
a) What is the electric field strength at a point C, midway
between the two protons?
Practice questions
10º
d) The electric field is now switched off. Explain why the sphere
oscillates with simple harmonic motion. Calculate the time
period of the motion; the mass of the sphere is 15 g.
(3)
Figure 19.37
200 V
D 100 V
17/04/19 10:48 AM
18 When the electric field strength reaches about 3.0 × 106 V m−1,
air can become ionised. In strong electric fields, free electrons
gain sufficient energy to ionise air molecules.
Figure 19.38 shows electric potentials close to an isolated tree.
a) Explain why the electric field strength is stronger over the
top of the tree.
(2)
b) A free electron can ionise a molecule if it has sufficient
Figure 19.38
energy to dislodge an electron that is attached to an atom
or molecule. In air, at atmospheric pressure, an electron travels an
average distance of 0.5 µm between collisions (this is called the mean
free path).
i) Calculate the energy gained by an electron that accelerates a
distance of 0.5 µm through an electric field of strength 46 MV m−1.
Express your answer in eV.(3)
ii) Explain why gases may be ionised with weaker electric field when
the gas pressure is low.(2)
c) A thundercloud at a height of 300 m above the ground is charged to a
potential of −7 × 108 V relative to the Earth.
i) Sketch a diagram to show the electric field
between the cloud and the ground.
(1)
ii) Calculate the field strength under the cloud.
(2)
The cloud is discharged by a flash of lightning, which carries a charge
of 4.5 C, in a time of 0.024 s.
iii) Calculate the average current during the discharge.
(2)
iv) Calculate the energy dissipated during the lightning strike.
(2)
19 ELECTRIC FIELDS
Stretch and challenge
380
19 Figure 19.39 shows two positive charges, +q, separated by a
distance 2a.
C
q
a) Show that the magnitude of the electric field, EC, at C is given by
Ec =
qx
2
2
2πε0 ( x + a )
3
2
r
and that the direction of the field is along the line OC.
A
a
a
+q
.
b) Show that the electric field is a maximum for x = ±
2
20 This question also refers to Figure 19.39. A particle with charge +q
Figure 19.39
(the same size as the two charges at A and B) is directed along the
line CO. The particle starts a very long way from the charges. What is the
minimum initial speed the particle must have if it is just to reach point O?
469884_19_AQA_A-Level_Physics_358-380.indd 380
q
x
0
r
a
B
+q
17/04/19 10:48 AM
20
Capacitance
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
Electric current, I, is the rate of flow of charge, I = ∆Q .
∆t
Potential difference (p.d.), V, is the amount of electrical work done per
W
unit charge, V = .
Q
●
●
●
●
Electrical resistance, R, is defined by R = V .
I
V2
Electrical power, P, is the rate of doing electrical work, P = VI = I2R = .
R
Electrical energy, E = VIt.
Kirchhoff’s second circuit law states that (in a complete loop of the
circuit) sum of emfs = sum of p.d.s.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 A current of 25 mA flows through a fixed resistor for 1 minute.
Calculate the total charge that flows through the resistor.
2 Calculate the work done in accelerating electrons with a charge of
−1.6 × 10 −19 C through a p.d. of 4800 V.
3 Calculate the current flowing through a 1.1 kΩ resistor with a p.d. of
7.7 V across it.
4 Calculate the electrical energy supplied to a 39 Ω heating element if a
current of 1.2 A flows through it for 3 minutes.
5 A 12 V car battery supplies potential differences across a fixed 5.7 V
GPS unit and a mobile phone charger connected in series. Calculate
the p.d. across the charger.
●● Capacitors
Capacitors are components of electrical circuits that temporarily store
electric charge. The addition of a capacitor into a circuit has two
possible effects: either introducing a time delay into the circuit; or
storing electrical energy for a short period of time. Capacitors are used
extensively in electrical and electronic timing circuits, in power circuits,
for smoothing electrical signals and as part of the signal-receiving
circuits found in radios.
381
Modern capacitors consist of two parallel conducting plates (usually made
of metal foils, films or coatings) separated by a thin insulating layer known
as a dielectric (generally made from thin plastic films, electrolytes, ceramics
469884_20_AQA_A-Level_Physics_381-402.indd 381
17/04/19 10:53 AM
or metal oxides). Most capacitors are then encased in a metal or plastic
housing. Figure 20.1 shows a selection of different capacitors.
There are several different circuit symbols for capacitors depending on
their type, although they are all based on the same simple pattern shown
in Figure 20.2.
electron flow
+Q
++++
+–+–+–+–
+ –+ –+ –+ –
+––+––+––+––
– – – –
–Q
+
Figure 20.1 Different types of capacitor
fixed non-electrolytic
fixed electrolytic
Figure 20.2 The main circuit symbols
for capacitors.
electron flow
Figure 20.3 Capacitor plates.
A potential difference from a battery or a power supply connected across
the metal plates causes electrons to flow off one plate, back through the
battery and onto the second plate (Figure 20.3).
One plate becomes positively charged (where electrons are removed),
while the plate with the excess of electrons becomes negatively charged.
If the capacitor is then disconnected from the source of potential
difference, the charge will stay on the plates until a conducting pathway
allows the excess electrons to flow off the negatively charged plate and
back onto the positive plate, until the two plates have equal charge again
(Figure 20.4). The conducting pathway could be a different part of the
circuit (controlled by a switch) or the charge could gradually leak away
to the surroundings.
20 CAPACITANCE
Capacitance The capacitance of a capacitor
is the ability of the capacitor to store charge
per unit potential difference.
382
Farad The unit of capacitance is the farad
(F), where 1 F is equal to 1 C V−1 (one
coulomb per volt).
electron flow
++++
+–+–+–+–
+ –+ –+ –+ –
+––+––+––+––
– – – –
electron flow
Figure 20.4 Capacitor plates
discharging.
469884_20_AQA_A-Level_Physics_381-402.indd 382
R
The ability of any object to store charge is called capacitance.
Capacitance is given the symbol C, and the SI unit is the farad (F). The
capacitance of a capacitor depends on the area of the metal plates, the
distance between the plates and the electrical properties of the material
separating the plates.
The amount of charge, Q, that can be stored on a capacitor depends on
the size of the capacitance, C, and the potential difference, V, across the
capacitor causing the separation of the charge:
Q = VC
The capacitance of a capacitor can then defined by
Q
V
So one farad is equal to one coulomb per volt. Actually 1 F is quite a large
capacitance, and useful ‘real-life’ capacitors have capacitances measured in
microfarads (µF), nanofarads (nF) or picofarads (pF).
C=
17/04/19 10:53 AM
Capacitors
TIP
A capacitance of 1 F will store a charge of 1 C with a potential difference of 1 V across it.
ACTIVITY
Measuring the capacitance of a capacitor
A fully discharged capacitor is connected to a
variable d.c. power supply and is then gradually
charged to different potential differences. A digital
coulombmeter is then used to measure the charge
stored on the capacitor at each potential difference
(Figure 20.5).
+
V
–
coulomb
meter
C
The data sheet supplied with the coulombmeter states
that the tolerance of the charge readings is ±10%.
Plot a graph of Q against V.
Plot suitable error bars on your data points.
Use your graph to show that Q is proportional to V.
Calculate a value for C, the capacitance of the
capacitor, where Q = CV, and use the error bars to
determine an uncertainty in this value.
5 Another capacitor, with a capacitance of 2 C, is
connected into the same circuit in place of the first
capacitor. Sketch a line on your graph illustrating the
variation of Q with V for this new capacitor.
1
2
3
4
Figure 20.5 Digital coulombmeter measuring charge.
The data from this experiment is shown in Table 20.1.
Table 20.1
Potential difference, V/V
0.0
1.0
2.0
3.0
4.0
5.0
6.0
Charge stored, Q /nC
0
331
664
1023
1328
1670
1996
TEST YOURSELF
1 Define what is meant by the ‘capacitance’ of a
capacitor.
2 Copy this table and correctly match the units in the
first row to the quantities in the second row.
C
Units
Quantities p.d.
A
capacitance
469884_20_AQA_A-Level_Physics_381-402.indd 383
F
V
current charge
3 List three factors that dictate the ability of a
capacitor to store charge.
4 A 4200 µF capacitor is connected to a 6.0 V battery.
Calculate the charge stored on the capacitor.
5 A capacitor stores a charge of 3.2 mC at a p.d. of
6.0 V. Calculate the value of the capacitance.
383
17/04/19 10:53 AM
7 Copy and complete the table.
Q
0.03 C
30 µC
250 nC
V
6.0 V
12.0 V
C
440 µF
0.25
0.20
0.15
0.10
0.05
0.00
0
0.005
0.01
0.015
0.02
0.025
capacitance of each capacitor C/F
10 000 µF
5.0 V
9.0 V
0.30
charge stored on capacitor, Q/C
6 Five different capacitors
are connected one at a time
to the same battery, and a
coulombmeter is used to
measure the charge stored
on each capacitor. The
measurements are shown
graphically in Figure 20.6.
Use the graph to calculate the
output p.d. from the battery.
Figure 20.6
120 nF
Parallel-plate capacitors
Permittivity The permittivity of a material
is the resistance of the material to an
electric field passing through it.
20 CAPACITANCE
Dielectric constant Another term for the
relative permittivity of an insulating material. It
describes the relative resistance of the material
to the propagation of electric field through
it and describes the absolute permittivity
of the material in terms of multiples of the
permittivity of free space, εr ε 0.
384
469884_20_AQA_A-Level_Physics_381-402.indd 384
The capacitance of a parallel-plate capacitor depends on the area of the
plates, their distance apart and the ability of the insulating material between
the plates to separate the charge, a property known as permittivity.
The permittivity of a material is the resistance of the material to an electric
field passing through it. If the permittivity is high, then a larger charge can
be stored on the plates for any given potential difference across them. The
permittivity of capacitor insulating materials is always measured relative
to the permittivity of free space (vacuum), ε0, using a relative permittivity,
εr, sometimes called the dielectric constant of the material. The total
absolute permittivity of an insulator is therefore given by the product εr ε0.
The permittivity of free space ε0 = 8.854 × 10−12 F m−1. Table 20.2 gives
the relative permittivity of a selection of materials commonly used in the
construction of capacitors.
TIP
The permittivity of a material to an electric field is analogous to the
resistance of a material to electric current flowing through it. In the case of
permittivity, current is replaced by electric field.
17/04/19 10:53 AM
Table 20.2
Relative permittivity (dielectric constant)
32
2.8–4.5
2.5–2.7
9.8
Capacitors
Material
Ceramic (ZnMg)TiO2
Polyester
Polystyrene
Aluminium oxide (electrolyte)
The capacitance of a parallel-plate capacitor is given by
εε A
C= r 0
d
where A is the area of the plates and d is their separation. Since the
capacitance is also given by C = Q /V, it follows that
Q ε rε0 A
Q ε r ε 0V
=
or
=
V
d
A
d
V
so the charge density on each plate is proportional to
– which is the
d
electric field E.
TEST YOURSELF
8 A capacitor is constructed from two sheets of
aluminium foil, 45 cm × 95 cm, separated by a thin
layer of polythene cling-film, 12.5 µm thick. The
relative permittivity of polythene is 2.25 and the
permittivity of free space ε0 = 8.854 × 10 −12 F m−1.
Calculate the capacitance of the capacitor.
9 A 6.0 V battery is connected to a 20 nF capacitor.
The area of the capacitor plates is 0.0016 m2 and
they are separated by a ceramic dielectric layer
5 µm thick.
a) Calculate the charge stored on the capacitor.
b) Calculate the relative permittivity of the ceramic
dielectric.
10 A 29–520 pF air-filled variable capacitor is shown in
Figure 20.7.
The five moving capacitor plates behave as five
independent capacitors arranged in parallel,
effectively multiplying the capacitance of one set of
plates by 5. Each moving plate is separated from its
static plate by an air gap 0.5 mm wide. The relative
permittivity of air at room temperature is 1.00.
469884_20_AQA_A-Level_Physics_381-402.indd 385
a) Calculate the maximum area of overlap of the
plates.
b) Calculate the minimum area of overlap between
the plates.
moving
plates
fixed plates
increase
capacitance
Figure 20.7 An air-filled variable capacitor.
385
17/04/19 10:53 AM
ACTIVITY
Measuring the relative permittivity of a
dielectric material
Figure 20.8 shows a digital multimeter being
used to measure the capacitance of a pair of
square capacitor plates separated by a thin
layer of material from a supermarket shopping
bag. Some digital multimeters can measure
capacitance directly by charging and discharging
the capacitor under test with a known current and
then measuring the rate of rise of the subsequent
potential difference. The faster the rate of rise, the
smaller is the capacitance.
The X and Y dimensions of the plates are measured
using a standard ruler to be 30.0 ± 0.1 cm × 30.0
± 0.1 cm. The thickness of the shopping bag
dielectric between the plates is measured randomly
across the dielectric 10 times using a micrometer.
The results are shown (in mm) in Table 20.3.
Figure 20.8 Digital multimeter measuring
capacitance.
Table 20.3
0.13 0.13 0.14 0.14 0.13 0.13 0.13 0.12 0.13 0.13
The area of overlap of the plates is varied by moving
the top plate diagonally relative to the bottom plate.
The plates can be positioned with an uncertainty
of 2 mm (X dimension) × 2 mm (Y dimension) and the
digital multimeter can measure a capacitance to ±5%.
20 CAPACITANCE
1 Use the data given and the data in Table 20.4 to
determine the relative permittivity of the shopping
bag material with a suitable value of uncertainty.
386
δ–
δ+
δ–
X dimension of
plates in
overlap/cm
(±0.2 cm)
Y dimension of
plates in
overlap/cm
(±0.2 cm)
Capacitance, C/
nF (±5%)
30.0
30.0
14.1
28.3
28.3
12.5
26.5
26.5
11.0
24.5
24.5
9.4
22.4
22.4
7.9
20.0
20.0
6.3
17.3
17.3
4.7
14.1
14.1
3.1
10.0
10.0
1.6
Dielectric heating
δ+
δ+
Table 20.4
δ+
Figure 20.9 Water molecules are polar
molecules.
469884_20_AQA_A-Level_Physics_381-402.indd 386
Some molecules, such as water, are called polar molecules because the
opposite ends of the molecule have opposite charges (Figure 20.9).
When water molecules form, the hydrogen atoms become slightly
positively charged, and the oxygen atom becomes slightly negatively
charged. (There is a covalent bond between the hydrogen and oxygen
atoms in a water molecule, but the shared electrons in the bond are
attracted more towards the oxygen atom than they are towards the
hydrogen atoms.) In a polar molecule, the overall charge of the molecule
is zero, but different ‘ends’ of the molecule may have opposite charges.
17/04/19 10:53 AM
Dielectric material An insulating material
where the molecules that make up the
material can be polarised inside an electric
field. Electric charges do not move through
the material, but the polar molecules align
themselves with the field.
charge
+Q
electric
field E
–
–
–
–
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
–
–
–
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
+
+
+
–Q
dielectric
plate
area A
plate
separation d
If the electric field between the plates is suddenly reversed, the polar
molecule will rotate and align itself with the direction of the electric field
again (Figure 20.11). Alternating the electric field between the two plates
will cause a polar molecule, such as water, to continuously rotate between
them (called dipole rotation). This increases its kinetic energy, and causes
it to collide with other adjacent molecules and atoms. These then acquire
more kinetic energy and move in random directions, increasing their
temperature and so dissipating the energy as heat.
+
–
–
+
+
–
+
–
+
–
+
–
–
+
–
–
–
Figure 20.10 The polar molecules in
a dielectric material align themselves
with the electric field.
Most of the dielectric materials used to construct capacitors are solids,
and the atoms and molecules are fixed within the structure. But what
happens when the solid is replaced by a liquid (such as water) or a
gas, where the particles are free to move between the metal plates? The
electric field produced between the two plates of a capacitor will cause
the charged particles in a liquid or a gas to align themselves in the
direction of the field (Figure 20.10). The separated charge in a polar
molecule is particularly able to align itself with the field between two
capacitor plates.
Capacitors
Polar molecules easily stick to metal surfaces, where one end of the polar
molecule induces an opposite charge in the metal, causing the molecule
to be attracted to the metal surface.
+
–
molecules and ions
+
+
–
+
+
–
–
–
+
+
–
+
+
effects of the RF & MW fields
Figure 20.11 Polar molecules rotating in an alternating
microwave field.
TIP
There is a common misconception
that microwave ovens cook food
from the ‘inside out’. This is not
the case. The microwaves cause
water molecules near the surface
(a millimetre or so) to rotate.
These hit nearby molecules,
forcing them into faster, more
energetic motion, and the
molecular vibration is dispersed
into the centre of the food. This
process is quicker than
conventional conduction.
469884_20_AQA_A-Level_Physics_381-402.indd 387
The alternating field between the plates can be produced by
a microwave emitter, such as the magnetron inside a microwave oven.
The frequency of the microwaves is tuned so that it rotates water
molecules within food – causing the food to heat up rapidly. The
optimum frequency for the rotation of water molecules in food is about
10 GHz, but if the frequency of the microwaves was set to this value
then the water molecules in the outer layers of the food would absorb
all the microwave energy, leaving a cool uncooked inner region and an
outer superheated layer. As a result, domestic microwave ovens have a
frequency of 2.45 GHz, which allows the outer layers to heat up more
slowly and then conduct heat deeper into the food.
387
17/04/19 10:53 AM
potential difference
The energy stored by a capacitor
V
∆Q
charge Q
potential difference
Figure 20.12 Graph of Q versus V for a
capacitor.
V
Area = E
= ½ QV
charge Q
Figure 20.13 The energy stored on a
capacitor is equivalent to the area under
a Q against V graph.
When a capacitor is charged up, the p.d. from the electricity supply
(or the energy per unit charge) causes electrons to flow off one plate,
through the external circuit and onto the other plate. This separation
of charge is kept steady provided that the p.d. is continuously applied,
and that there is no leakage of charge. Once the p.d. is removed, and a
complete discharging circuit is connected to the capacitor, the electrical
energy stored by the separated charge can be released as the electrons flow
back off the negatively charged plate and back onto the positively charged
plate – this was shown in Figure 20.4. If the p.d. applied to the plates
is increased, more charge and therefore energy is stored on the plates.
A graph of potential difference against charge for a capacitor is shown in
Figure 20.12.
W
You will remember from the definition of potential difference that V = ,
Q
where W is the amount of work done per unit charge, Q. In the context
of a capacitor, the potential difference V is the amount of work done in
moving unit charge off one plate and onto another plate. At any potential
difference V, the work done moving an amount of charge ΔQ is therefore
W = VΔQ. This is represented by the shaded area in the graph in
Figure 20.12. The total energy E stored on the capacitor, charging the
capacitor from empty up to a charge of Q at a potential difference V, is
calculated by adding up all the similar shaped areas from Q = 0 up to a
charge Q. In other words, this is the whole area under the graph up to Q.
Because the shape is a triangle (Figure 20.13), E = 12 QV . But Q = CV,
1 Q2
.
so E = 12 CV 2 and E =
2 C
20 CAPACITANCE
TIP
388
Use the correct version of the capacitor–energy equation depending on
the question. Use the version containing the data given unless you are
told to do otherwise. If you use calculated values, you are more likely to
make an error.
TEST YOURSELF
11 Explain what is meant by a ‘dielectric material’.
12 Explain how a water molecule can be heated by
an alternating electric field.
13 A 3300 µF capacitor is charged by a 9.0 V battery.
Calculate the energy stored in the capacitor.
469884_20_AQA_A-Level_Physics_381-402.indd 388
14 Calculate the capacitance of a capacitor that
stores 0.25 J of electrical energy when a p.d. of
24 V is connected across it.
15 Calculate the charge on a 220 µF capacitor
storing an energy of 0.12 J.
17/04/19 10:53 AM
V
Y
0.025
charge, Q/C
X
I
Z
V
Capacitor charge and discharge
0.030
16 The graphs X, Y and Z in Figure 20.14 show
possible relationships between electrical
quantities associated with a capacitor
discharging through a resistor.
0.020
0.015
0.010
t
Q
t
Figure 20.14
1
a) Which graph shows the gradient equal to ?
C
b) Which graph could be used to determine the
total charge stored in the capacitor?
c) In which graph does the area under the line
represent the energy stored in the capacitor?
17 The graph in Figure 20.15 shows how the charge
stored on a capacitor varies with the p.d. applied
across it.
0.005
0.000
0
2
6
4
8
10
12
14
p.d./V
Figure 20.15
a) Use the graph to calculate the capacitance of
the capacitor.
b) Use the graph to calculate the energy stored on
the capacitor, when 8.0 V is applied across it.
●● Capacitor charge and discharge
The circuits in Figure 20.16 show a battery, a switch and a fixed resistor
(circuit A), and then the same battery, switch and resistor in series with a
capacitor (circuit B). The capacitor is initially uncharged.
I
A
I
A
R
R
V
V
circuit A
C
circuit B
I
I
The graphs underneath the circuit diagrams show how
the current varies with time from the moment that the
switches are closed. In the case of the resistor alone
(circuit A), the current immediately jumps to a value I,
where I = V/R, and stays at that value regardless of how
long it has been since the switch was closed. Time is not
a factor in this circuit. In the case of circuit B, where an
initially uncharged capacitor is connected in the circuit,
the current also immediately rises to the same value, I,
determined by I = V/R, but it then starts to decay away
with time, eventually reaching zero. The series capacitor
limits the way that current flows through the resistor.
389
If the capacitor is initially uncharged, the amount of
∆Q
= I , is
charge that can be stored on it per second,
∆t
initially determined by I = V . As the capacitor starts to
R
t
t
Figure 20.16 Circuit diagrams for a battery, resistor and
capacitor network.
store charge, so a p.d. is developed across the capacitor,
Q
VC = . As the emf of the battery, ε, remains constant,
C
then the potential difference, VR, across the fixed resistor,
R, reduces because
ε = VR + VC
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Reducing VR reduces the current, I, flowing. The initial current flowing onto
the capacitor gradually decays away as the capacitor stores more charge,
increasing VC.
capacitor charging
time
t
charge on capacitor Q
charge on capacitor Q
Graphs of charge Q stored on the capacitor with time are shown in
Figure 20.17, one representing the capacitor charging, and one discharging.
(a)
capacitor discharging
time
t
(b)
Figure 20.17 Graph of Q against t for a capacitor (a) charging
and (b) discharging.
The charging graph (Figure 20.17a) shows that, initially, the capacitor is
TIP
Remember that the gradient of a
charge–time graph is the electric
current.
Q or V
Q or V
t
t
capacitor
discharging
capacitor
charging
20 CAPACITANCE
I
I
t
t
Figure 20.18 Graphs of Q or V and I
against t, for charging and discharging
capacitor.
Q
390
current at time t' =
∆Q
∆t
∆Q
∆t
VR
.
I=
R
uncharged, and the gradient of the graph,
(equal to the current, I),
is at a maximum and is determined by
As more charge is stored
on the capacitor, so the gradient (and therefore the current) drops, until
the capacitor is fully charged and the gradient is zero. As the capacitor
discharges (Figure 20.17b), the amount of charge is initially at a maximum,
as is the gradient (or current). The amount of charge then drops, as does the
gradient of the graph. This is described by
∆Q
∝Q
∆t
The shape of the discharging graph is an exponential decay, meaning that
the rate of decay of the charge (or the gradient or the current) depends on
the amount of charge stored at any given time. For a discharging capacitor,
the current is directly proportional to the amount of charge stored on the
capacitor at time t.
Graphs of V (the p.d. across the capacitor) against t follow the same pattern
as the graph of Q against t, because Q ∝ V (from Q = VC). When current–
time graphs are plotted, you should remember that current can change
direction and will flow one way on charging the capacitor and in the other
direction when the capacitor is discharging. The size of the current is always
at a maximum immediately after the switch is closed in the charging or
discharging circuit, because the charging current will be highest when the
capacitor is empty of charge, and the discharging current will be highest
when the capacitor is full of charge. This is shown in the graphs in
Figure 20.18.
∆Q
, is the current, I, as shown in
The gradient of the Q against t graph,
∆t
Figure 20.19.
∆Q
∆t
0
t'
t
Figure 20.19 Graph illustrating how the
gradient of a Q against t graph is the
current.
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When the capacitor is discharging, during a small time interval Δt, the element
of charge transferred is ΔQ, and this is calculated by ΔQ = I'Δt. The total charge
transferred, Q, during the discharge from t = 0 to t = t' is the sum of all the
charge elements during this period – this is the area under the graph.
I'
Discharging a capacitor
∆Q
charge stored
t'
t
Figure 20.20 Relationship between
charge and the area under a graph of I
against t.
Applying Kirchhoff’s second circuit law around the capacitor–resistor loop:
V
S
Consider the circuit shown in Figure 20.21. When switch S is closed, the
capacitor C immediately charges to a maximum value given by Q = CV. As
switch S is opened, the capacitor starts to discharge through the resistor R
and the ammeter. At any time t, the p.d. V across the capacitor, the charge
stored on it and the current, I, flowing through the circuit and the ammeter
are all related to each other by two equations.
Capacitor charge and discharge
During discharge the area under an I against t graph, up to time, t, is the
charge transferred off the capacitor, as shown in Figure 20.20.
I
VC + VR = 0
C
R
A
I
Figure 20.21 A capacitor discharging
circuit.
where VC and VR are the p.d.s across the capacitor and the resistor,
respectively. Substituting for both using Q = VC and V = IR gives
Q
+ IR = 0
C
But
∆Q
I=
∆t
so Q
∆Q
=−
R
∆t
C
or
Q
∆t
RC
∆Q = −
This is a differential equation and requires calculus to provide the solution
(for those interested, see the Maths box):
Q = Q0 e
t
− RC
As Q = VC and V = IR, at any time during the discharge, Q ∝ V and V ∝ I,
so there are corresponding equations for the p.d. across the capacitor and
the current flowing in the circuit during discharge:
V = V0 e
t
− RC
and
I = I0 e
−
t
RC
391
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The time constant
MATHS BOX
This shows how to use calculus
to solve the differential equation
∆Q = −
Q
∆t
RC
When integrating a differential
equation, such as the one here,
we let Δt → 0, and then change
to calculus notation (Δ to d) and
group like terms on each side.
Here this gives
dQ
1
=−
dt
Q
RC
(showing a constant ratio), and
Q
t
dQ
1
=−
dt
RC
Q
0
Q0
Integrating gives
 t t
[ln Q]QQ 0 = −  
 RC  0
and so
t
ln Q − ln Q0 = −
RC
Then using the rules of logs
Q
t
ln
=−
RC
Q0
so finally,
− t
Q = Q0 e RC
∫
∫
The quantity RC is called the time constant of a capacitor circuit. The time
constant is related to the half-life of the decay of charge off the capacitor,
and is analogous to the half-life of radioactive decay. We define the halflife of capacitor discharge as the time taken for the charge stored on the
capacitor (or the current or the voltage) to halve.
Q
When t = t 12 , Q = 0 , so
2
Q0
−
= Q0 e RC
2
Cancelling Q0 from each side gives
t1/ 2
1 − t1/RC2
=e
2
and taking the natural logarithm of both sides of the equation leads to
t1
ln 0.5 = − 2
RC
t 1 = 0.693RC
2
Also when t = RC, we have
Q = Q0 e
− RC
RC
= Q0 e−1
= 0.37Q0
This means that one time constant, RC, is the time for the charge stored on
the capacitor to drop to 37% of the initial value, Q0.
Graphical analysis
20 CAPACITANCE
The equations of exponential decay can be rewritten in the form of a straight
line, so that a graph can be drawn and the gradient and y-intercept measured.
This allows you to calculate the time constant of the circuit, if it is unknown.
Consider the current–time equation:
TIP
Remember that t 1 = 0.693RC .
I = I0 e
2
t
− RC
Rearranging and taking logarithms of both sides gives
So
I 
−t
ln   = ln(e RC )
 I 0 
ln I − ln I0 = −
392
or
ln I = ln I0 −
469884_20_AQA_A-Level_Physics_381-402.indd 392
t
RC
1
1
t or ln I = −
t + ln I A
RC
RC
17/04/19 10:55 AM
This is the equation of a straight line of the form
y-intercept = In I0
y = mx + c
where m is the gradient and c is the y-axis intercept of the line. Plotting
ln I on the y-axis against t on the x-axis produces a gradient of −
gradient = –1/RC
y-axis intercept equal to ln I0, as illustrated in Figure 20.22.
1
and a
RC
t
Figure 20.22
TEST YOURSELF
18 The flash unit on a small disposable camera consists
of a 2.0 nF capacitor charged from a 1.5 V battery,
discharging through a 3.9 MΩ resistance flash bulb.
a) Calculate the energy stored in the capacitor
before discharge.
b) Calculate the time constant of the circuit.
The capacitor needs to discharge by 75% (two
half-lives) before it can be recharged from the
battery circuit.
c) Calculate the minimum recharge time for the
capacitor.
19 A 2800 µF capacitor and a variable resistor are
used as part of the timing controls in a traffic light.
Part of the timing circuit is shown in Figure 20.23.
12 V
S
C = 2800 μF
Capacitor charge and discharge
In I
f) Switch S is closed again and the variable
resistor is adjusted so that the resistance is
now halved to a value of 4.0 kΩ. Use the words
halves doubles stays the same
reduces to zero
t o determine what happens to the quantities below when switch S is opened:
i) the initial charge stored on the capacitor
ii) the total initial energy stored in the capacitor
iii) the initial current flowing through the
variable resistor
iv) the time constant of the discharge circuit.
20 A capacitor of capacitance C is fully charged
by connecting it directly to a 3.0 V battery. The
capacitor is then disconnected from the battery
and connected to a 12 kΩ resistor in series with an
ammeter. The graph in Figure 20.24 shows how the
discharge current from the capacitor varies with
time following its connection to the resistor.
0.30
0.25
A
I
Figure 20.23
The initial value of the variable resistor R is set
to 8.0 kΩ and the timing circuit is controlled by
closing and opening switch S, which is initially
closed, completing the charging circuit. Calculate:
a) the initial charge stored in the capacitor
b) the total energy stored by the capacitor
c) the initial current flowing through variable
resistor R.
Switch S is then opened.
d) Explain why the current flowing through
the variable resistor reverses and starts to
decrease.
e) Calculate the time constant of the discharge
circuit.
469884_20_AQA_A-Level_Physics_381-402.indd 393
current, I/mA
R
0.20
0.15
0.10
0.05
0.00
0
5
10
15
time, t/s
20
25
393
Figure 20.24
a) Explain why the quantity represented by the
area under the graph is the initial charge stored
on the capacitor.
b) Use the graph to estimate the initial charge
stored on the capacitor.
c) Calculate the capacitance of the capacitor.
d) Calculate the initial energy stored on the capacitor.
17/04/19 10:55 AM
Charging a capacitor
VS
R
C
In many cases, the charging of a capacitor is designed to be as quick as
possible, and so the resistance of the charging part of a capacitor circuit is
kept as low as possible (Figure 20.25). However, if the charging process is
part of a circuit that requires a higher resistance, the charging time must
be taken into account. There is no need for you to know how to derive
the formula for charging a capacitor. The Maths box is included here for
interested mathematicians.
Summarising the equations for charging a capacitor (from the Maths box):
VR
VC
Figure 20.25 Circuit for charging a
capacitor.
Q = Q0 (1 − e
t
− RC
)
and therefore
V = V0 (1 − e
t
− RC
)
and
I = I0 e
t
− RC
You do not need to be able to derive these formulae, but you do need to
understand how to use them.
MATHS BOX
Consider the circuit shown in Figure 20.25. Kirchhoff’s
second circuit law tells us that the sum of the emfs in
the circuit equals the sum of the p.d.s in the circuit, so
20 CAPACITANCE
VS = VR + VC
394
where VS is the emf of the source, VR is the p.d. across
the resistor and VC is the p.d. across the capacitor.
Using Ohm’s law and the basic capacitor equation,
this becomes
Q
VS = IR +
C
During a small time interval Δt when the capacitor
is charging, VS and C do not change, ∆ VS = 0, unlike I
∆t
and Q, which do change. So in the small time interval
we have
∆I
1 ∆Q
0=
R+
C ∆t
∆t
∆Q
But
= I, so
∆t
0=
or
I
∆I
R+
∆t
C
∆I
I
=–
∆t
RC
Rearranging and replacing with calculus notation
gives
dI
1
=−
dt
I
RC
469884_20_AQA_A-Level_Physics_381-402.indd 394
Integrating in a similar way to before gives
I = I0 e
t
− RC
At t = 0, VC = 0 and I = I0 = VS , and so at any time, t,
R
we can write
I=
or
VS − RCt
e
R
VR = VS e
t
− RC
Going back to Kirchhoff’s second circuit law,
VC = VS − VR
and substituting for VR gives
VC = VS − VS e
t
− RC
Factorising this gives
VC = VS (1 − e
t
− RC
)
and then using the capacitor equation we obtain
Q = CVS (1 − e
t
− RC
)
But CVS is the maximum charge that can be stored on
the capacitor when it is fully charged. That is equal
to Q0, the initial charge on the capacitor when it is
about to discharge. So finally we have
Q = Q 0(1 − e
t
− RC
)
17/04/19 10:56 AM
Capacitor charge and discharge
EXAMPLE
Charging a capacitor
Answer
A 3.0 V battery charges an initially uncharged 10 000 µF
capacitor through a 1000 Ω resistor, as shown by the
circuit diagram in Figure 20.26.
The charging voltage is given by
−
V = V0 (1− e
t
RC
)
so
25 s
× 0.01F
V = 3.0 V × (1− e− 1000 Ω
C = 10 000 μF
)
= 2.75 V = 2.8 V (2 s.f.)
2 Calculate the charging current after 15 s.
Answer
V = 3.0 V
The current is given by
R = 1000 Ω
A
Figure 20.26
1 Calculate the voltage across the capacitor 25 s after
the switch is closed to charge the capacitor, which
was initially uncharged.
−
I = I0 e
t
RC
where I0 is the initial charging current given by
, so
I=
V
I0 = 0
R
V0 − RCt
3.0 V − 1000 Ω15×s0.01F
e
e =
R
1000 Ω
= 6.7 × 10−4 A
REQUIRED PRACTICAL 9
Investigating the charging and discharging of capacitors
Note: This is just one example of how you might tackle this required
practical.
A student carries out an experiment to determine the capacitance of an
unknown capacitor that she has recovered from a large stage amplifier.
She connects up the circuit in Figure 20.27 using a battery pack, a
variable resistor, a two-way switch and a data logger set to measure the
current in mA as a function of time.
She sets the battery pack to 6.0 V and the variable resistor to 880 Ω.
She ensures that the capacitor is completely discharged by earthing
its connections before reconnecting into the circuit and moving the
switch to position X. She then records the charging current (in mA)
every 10 s for 100 s from the data logger, before moving the switch
immediately to position Y and recording the discharging current every
469884_20_AQA_A-Level_Physics_381-402.indd 395
X
Y
C
Vs
R
A
Figure 20.27
395
17/04/19 10:56 AM
10 s for a further 100 s. Her results are shown in
Table 20.5. The data logger acting as an ammeter is
able to measure the charging current as a positive
current and the discharging current as a negative
current.
Table 20.5
Status
Time, t /s
Current, I/mA
Charge
0
6.81
10
3.97
20
2.31
30
1.34
40
0.78
50
0.46
60
0.27
70
0.15
80
0.09
90
0.05
100
0.03
100
−6.81
110
−3.50
120
−2.20
130
−1.40
140
−0.65
150
−0.46
160
−0.30
170
−0.15
180
−0.08
190
−0.06
200
−0.03
20 CAPACITANCE
Discharge
1 Plot a graph of the current (y-axis) against time
(x-axis) – show the charging and discharging
phases on the same graph.
2 Use your graph to estimate the total charge
stored on the capacitor when it is fully charged.
(Remember, the currents are in mA.)
3 Use your graph to estimate a value for the time
constant, RC, of the circuit. Hence make an
estimate of the capacitance of the capacitor.
4 Make a copy of the part of Table 20.5 showing the
capacitor discharging. Add a further column to
your table showing the natural logarithm of the
modulus (magnitude) of the discharging current –
for example, the first value, at 100 s, is the natural
logarithm of 6.81 (= 1.92), followed by ln(3.50) =
1.25.
5 Plot a second graph of ln I (y-axis) against t (x-axis).
6 Use your log graph to calculate a value for the
time constant, RC, of the circuit and hence another
value for C, the unknown capacitance. including an
estimate of the uncertainty of the value.
7 Compare your values from parts 3 and 6.
396
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1 What is the charge stored on a 220 µF capacitor with a 6 V p.d.
across it?
A 36.6 mC
C 1.32 mC
B 2.70 mC
D 3.96 mC
Practice questions
Practice questions
2 What is the energy stored in a 4200 µF capacitor with a 3 V p.d.
across it?
A 13 mJ
C 1.3 mJ
B 0.93 mJ
D 19 mJ
3 A constant current of 208 µA is used to charge an initially uncharged
capacitor or with a capacitance of 24 µF. How long will it take for the
p.d. across the capacitor to rise to 2600 V?
A 54 s
C 166 s
B 108 s
D 300 s
4 Two capacitors, P and Q, are both charged by the same 12 V battery. The
capacitance of P is 4200 µF, and the capacitance of Q is 420 µF. Which
row of the table gives the correct ratios of energy and charge stored by
each capacitor?
energy stored in P
charge stored in P
energy stored in Q
charge stored in Q
A
1
1
B
10
1
C
10
10
D
1
10
5 A 2000 µF capacitor has been fully charged by a 6.0 V battery, before
it is discharged through a 10 kΩ resistor. What is the charge stored by
the capacitor 20 s after the discharge begins?
A 4.4 mC
C
8.8 mC
B 2.2 mC
D
1.1 mC
6 A capacitor of capacitance C is charged through a resistor of resistance R
to a p.d. of 10 V. The capacitor is then allowed to discharge back through
the same resistor, R, as shown in Figure 20.28, until the p.d. across the
capacitor is 5 V, before being recharged again back up to 10 V.
397
C
10 V
V
R
Figure 20.28
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17/04/19 10:56 AM
Which of the graphs in Figure 20.29 best shows how the p.d. across
the capacitor varies with time during the discharge and the subsequent
recharging?
10
p.d.
10
p.d.
5
5
0
0
0
0
t
A
10
p.d.
t
B
10
p.d.
5
5
0
0
0
t
0
C
t
D
Figure 20.29
7 An initially fully charged capacitor, C, discharges through
a resistor, R, and loses half its charge in 20 s. The time constant,
RC, of the circuit is
C 39 s
B 9s
D 29 s
20 CAPACITANCE
8 Figure 20.30 shows how the p.d. across a capacitor varies with
the charge stored on it.
398
The table shows possible values of the capacitance and the
energy stored by the capacitor when the p.d. across it is 12 V.
Which row gives the correct values?
Capacitance, C/mF
Energy stored, E /mJ
A
2.5
90
B
2.5
180
C
0.4
90
D
0.4
180
potential difference, V/V
A 19 s
12
10
8
6
4
2
0
0
5
10
15
20
25
30
charge, Q/mC
Figure 20.30
9 A 12 V car battery is used to fully charge an 18 mF capacitor. The
capacitor then fully discharges through a small electric
motor, which lifts a 200 g mass stack. If the motor lifts the
mass stack with a 10% efficiency, through what height will
the mass stack be lifted?
A 3 cm
C 30 cm
B 6 cm
D 60 cm
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17/04/19 10:56 AM
10 Figure 20.31 shows how the charge stored on a capacitor, of
capacitance C, varies with time as it discharges through a resistor, R.
0.20
C 4.4 s
B 4.2 s
D 3.0 s
charge/μC
What is the time constant RC of this circuit?
A 6.3 s
11 Figure 20.32 shows how the charge, Q, stored on a capacitor
varies with the p.d. across it, V.
0.15
0.10
0.05
0.00
Which of the following statements is not correct?
A The energy stored on the capacitor can be calculated by
measuring the area under the graph.
Practice questions
0.25
0
2
4
6
8
10
50
60
time, t/s
Figure 20.31
C If the charge stored on the capacitor was doubled, the energy
stored would quadruple.
D Doubling the capacitance would halve the gradient of the
graph.
12 A capacitor, C, is charged with a potential difference of 12.0 V, and
then discharges through a resistor, R, where R = 50 kΩ. A student
measured the p.d. across the capacitor every 5 s using a data
logger with graph-plotting software. The graph in Figure 20.33
shows his results.
charge/Q, C
B The gradient of the graph is numerically equivalent to the
capacitance of the capacitor.
p.d./V
Figure 20.32
14
a) Use the graph to calculate:
ii) the time constant of the circuit, including the
correct unit(4)
iii) the capacitance, C, of the capacitor(2)
iv) the charge stored on the capacitor after 30 s.(3)
b) A garden automatic sprinkler system contains a time
delay circuit using an identical capacitor to that used
in part (a). The capacitor is charged using a 6 V battery
and discharges through a similar 50 kΩ resistor.
i)
12
the initial discharge current flowing through
the resistor(1)
10
p.d./V, V
i)
8
6
4
2
0
0
10
Figure 20.33
The smaller 6 V p.d. from the battery changes the
energy stored by the capacitor. State and explain
how the energy stored on the capacitor changes
compared to the value calculated in a)iv).(2)
20
30
40
time, t/s
399
ii) State and explain the effect of this change in p.d.
on the time constant of the circuit.(1)
13 A resistor-capacitor circuit is used as a timing mechanism
for an experiment to measure the acceleration due to
gravity. The experimental set-up is shown in Figure 20.34.
469884_20_AQA_A-Level_Physics_381-402.indd 399
17/04/19 10:56 AM
Switch 1 is initially closed, keeping the capacitor C charged
at 6.0 V. Switch 2 is also initially closed. The steel ball is
dropped, opening switch 1, disconnecting the capacitor
from the battery. The capacitor then starts to discharge
through resistor R. The ball falls, opening switch 2,
stopping the discharge through the resistor.
6.0 V
steel ball
Switch 1
C
V
a) Describe the measurements that need to be made in this
experiment, and explain how these measurements could
be used to calculate the acceleration due to gravity. The
quality of your written communication will also be
assessed in this question.(6)
R
b) In one such experiment, the value of C was 440 µF and
R was 10 kΩ. The p.d. of the battery was 6.0 V and the
distance between the switches was 1.0 m. The voltmeter
Figure 20.34
reading was initially 6.0 V and dropped to 5.4 V. Using
this information, calculate the time for the ball to drop between
the two switches.(3)
Switch 2
c) Use your answer to part b) to calculate the acceleration
due to gravity.(2)
14 Many tablet computers use a capacitor as well as a rechargeable
battery to store electrical energy. The rechargeable battery provides
most of this electrical energy, but the capacitor is used as an
emergency back-up, if the battery is suddenly removed or fails
during operation. The capacitor stores just enough electrical energy
to shut down the tablet safely.
400
b) The 3.6 V lithium-ion battery can deliver a steady current
of 0.84 mA for 3 hours. Show that the battery can store
about 400 times more electrical energy than the capacitor.(2)
3.5
c) State two reasons why a capacitor would be unsuitable as
the main energy store for a tablet computer.(2)
2.5
15 a) Explain what is meant by the ‘capacitance’ of a capacitor. (2)
b) A capacitor of ‘capacitance’ C is charged from a 9 V battery
through a fixed resistor of resistance R. The graph in
Figure 20.35 shows how the charge Q varies with time after
the capacitor and resistor are connected in series with the
battery.
Using the maximum charge stored on the capacitor, as
determined from the graph, calculate the capacitance of
the capacitor.(3)
469884_20_AQA_A-Level_Physics_381-402.indd 400
3.0
charge stored, Q/mC
20 CAPACITANCE
a) Calculate the electrical energy stored in a capacitor of capacitance
12 800 µF found in a tablet computer
operating with a lithium-ion 3.6 V battery.(2)
2.0
1.5
1.0
0.5
0.0
0
10
20
30
40
50
60
time, t/s
Figure 20.35
17/04/19 10:57 AM
Practice questions
c) The time constant for a capacitor-charging circuit is the time
taken for the charge to rise to 63% of its maximum value.
Use the graph to determine the time constant of this circuit.(2)
d) Calculate the resistance of the resistor.(2)
e) State what value is represented by the gradient of the graph.(1)
f) Calculate the initial current flowing through the resistor
during charging.(1)
g) Sketch a graph to show how the current flowing though the resistor
varies with time for the 50 s following connection to the battery.(2)
16 A 640 µF capacitor is initially fully charged from a 12 V battery.
The capacitor is then discharged through a 48 kΩ resistor.
a) Use this data to calculate:
i)
the time constant of this circuit(1)
ii) the initial discharge current through the resistor(1)
iii) the initial charge stored by the capacitor(1)
iv) the initial energy stored by the capacitor.(1)
b) The capacitor is disconnected from the resistor after 40 s.
Without losing any of its charge, it is connected to a second
resistor of resistance 24 kΩ. Calculate:
i)
3.0 V
the charge stored by the capacitor at the start of the
discharge through the 24 kΩ resistor(2)
C
ii) the initial p.d. across the 24 kΩ resistor(1)
A
iii) the total energy transferred to the 24 kΩ resistor.(1)
17 A physics technician finds an unlabelled capacitor in a drawer
and sets up the circuit shown in Figure 20.36 with a
data-logging ammeter probe together with graph plotting
software, to measure the capacitance of the capacitor.
a) Use the graph to show that the resistance of
R is 120 kΩ.(2)
b) Use the graph to determine the ‘half-time’ of the decay,
and hence calculate a value for the time constant of
the circuit.(2)
c) Calculate the capacitance, C, of the capacitor.
469884_20_AQA_A-Level_Physics_381-402.indd 401
(1)
Figure 20.36
30
25
20
current, I/μA
The technician closes the switch, which charges the capacitor.
She then opens the switch, allowing the capacitor to
discharge through the data-logging ammeter and the resistor.
The data logger records the current every 5 s after she opens
the switch, and then draws a graph of the results, which is
shown in Figure 20.37.
R
15
10
401
5
0
0
50
100
150
200
250
time, t/s
Figure 20.37
17/04/19 10:57 AM
Stretch and challenge
The questions that follow are British Physics Olympiad questions.
18 A thundercloud has a horizontal lower surface area of 25.0 km2, 750 m
above the surface of the Earth.
a) Using a capacitor as a model, calculate the electrical energy, E stored
when its potential is 1.00 × 105 V above the earth potential (0 V).
b) The cloud rises to 1250 m.
i)
Explain whether the energy stored, has increased or decreased.
ii) What is the change in electrical energy, ΔE?
(BPhO R1-2012 Q1(h))
19 Two uncharged capacitors C1 and C2, with capacitances C1 and C2, are
connected in series with a battery and a switch S. When the switch is
closed there is a charge Q1 on C1 and Q2 on C2.
a) What is the relation between Q1 and Q2?
b) Give an expression for the potential difference across each capacitor.
c) Derive an expression for the capacitance C of a single capacitor
equivalent to C1 in series with C2.
d) Calculate the total energy stored in the capacitors.
(BPhO R1-2007 Q5)
20 Two capacitors, of capacitance 2.0 µF and 4.0 µF, are each given a charge
of 120 µC. The positive plates are now connected together as are the
negative plates. Calculate:
a) the new p.d. between the plates of the capacitors
b) the change in energy. Explain this change.
20 CAPACITANCE
(BPhO R1-2010 Q5)
402
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21
Magnetic fields
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
●
●
●
●
Magnets have two poles: a north-seeking pole (N, or north pole) and a
south-seeking pole (S, or south pole).
A magnetic field is a region where the magnet exerts a force on other
magnets or magnetic materials, even if they are not in contact.
A magnet attracts magnetic materials (e.g. iron) placed in its
magnetic field provided the field is non-uniform. The magnet attracts
another magnet if unlike poles are facing (north–south), and repels
another magnet if like poles are facing (north–north or south–south).
A current-carrying wire creates a magnetic field, which encircles the
wire carrying the field.
An electromagnet can be made by making a coil of wire and passing a
current through it. The strength of an electromagnet increases if the
current in the wire increases, if more turns are added to the coil, and
if a magnetic core is placed inside the coil.
Electromagnets made using a soft magnetic core (e.g. iron) lose their
magnetism if the current is turned off. If a hard magnetic material is
used (e.g. steel), the electromagnet retains its magnetism when the
current is turned off.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 Magnetism causes a non-contact force. Give examples of two other
non-contact forces, and what these forces affect.
2 Describe how to make a strong, temporary magnet.
3 A compass needle is magnetic. Explain why the north pole of a
compass needle is attracted to the Earth’s geographic North Pole.
4 Iron can be magnetised if it is stroked repeatedly in one direction
using a magnet. Iron atoms behave like mini-magnets. Magnetic
domains are regions where groups of atoms line up in one direction.
Explain how iron can be magnetised in terms of magnetic domains.
403
Experiments using radiation up to 10 billion times brighter than sunlight are
carried out at the Diamond Light Source facility in Oxfordshire (Figure 21.1).
Diamond is a synchrotron which accelerates beams of electrons to the speed
of light. Very strong magnets direct the electrons through a pipe 56 Zm
circumsference and under ultra-high vacuum. The electrons lose energy,
emitted as synchrotron light, as they change direction in the magnetic field.
Synchrotron light ranges from infrared radiation to x-rays.
Electrons also travel through magnets set out in arrays called insertion devices.
These are even stronger magnets, some of which are super-conducting. Since
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17/04/19 11:04 AM
the direction of the magnetic field repeatedly changes, the electrons are forced
to wiggle through the device. The electrons release energy when they change
directions, either as extremely intense electromagnetic radiation turned into
specific frequencies, or a broad spectrum of radiation, depending on the
arrangements of the magnets.
Figure 21.1 Diamond in Oxfordshire.
●● Magnetic flux lines
21 MAGNETIC FIELDS
S
N
field lines around a bar magnet
Figure 21.2 Flux lines for a bar magnet.
The flux density is highest at the poles,
where the field is strongest.
Magnets have a north pole and a south pole, so we call magnets dipoles.
A magnetic field is the region where a magnet exerts a force on objects made
from magnetic materials, or on other magnets. We represent the magnetic
field using arrows, or flux lines, to indicate the direction and strength of the
field in the region surrounding the magnet. The rules for drawing magnetic
flux lines, electric field lines and gravitational field lines are similar. For
magnets, these rules are as follows:
●
●
●
404
N
X
N
●
●
●
Figure 21.3 The combined field from
two bar magnets. Neutral points are
marked X.
469884_21_AQA_A-Level_Physics_403-419.indd 404
Flux lines represent the direction of the force experienced by the north
pole of a magnet at any point in the magnetic field. They run from the
north pole to the south pole.
A magnetic field is strongest where its flux density is highest, and this is
shown as flux lines closest together (Figure 21.2).
A magnetic field with twice the strength is drawn with twice the number
of flux lines per unit area in the same region.
The magnetic field of more than one magnet is the combined field of the
individual magnets.
Flux lines never cross.
If there is more than one magnet, the magnetic fields cancel out in some
places and there is a neutral point (Figure 21.3).
A magnet freely suspended in a magnetic field will align itself with the field.
17/04/19 11:04 AM
Magnetic flux density The number of
magnetic flux lines that pass through
an area of 1 m2
Tesla The flux density that causes a force of
1 N on a 1 m wire carrying a current of 1 A at
right angles to the flux.
The strength, or intensity, of a magnetic field is its magnetic flux density,
B, also known as a B-field. Magnetic flux density is measured in teslas (T).
Magnetic field or flux lines are a model that helps us to visualise the field.
You will learn more about magnetic flux, ϕ, in Chapter 22.
If you look at a diagram of a magnetic field, you can see that the flux lines
are closer together in places (for example, at the poles of a bar magnet as
shown in Figure 21.2). You can tell that the magnetic field and the force
produced by the magnet are stronger at the poles because this is where the
flux lines are closer together.
Magnetic flux lines
The strength of a magnetic field
Magnetic fields from current-carrying wires
(a)
(b)
Figure 21.4 The magnetic flux of a
current-carrying wire: (a) with the
current flowing into the page and (b)
with the current coming out of the page.
Moving charges cause a magnetic field, which we describe using flux lines.
The magnetic flux around a current-carrying wire is shown as concentric
circles, indicating the magnitude and direction of the flux pattern. Moving
away from the wire, flux lines are further apart because the field gets weaker.
If you look at the wire with the conventional current flowing away from you,
the flux lines circle the wire in a clockwise direction. Symbols inside the wire
indicate the current direction: ⊗ indicates a current flowing away from you
(Figure 21.4a) and  indicates a current flowing towards you (Figure 21.4b).
The combination of flux lines for a loop of wire is shown in Figure 21.5.
B
S
N
A
Figure 21.5 A magnetic field circles a current-carrying
wire.
I
I
Figure 21.6 The magnetic field for a
current-carrying loop of wire.
–
I
+
I
A solenoid is a current-carrying coil of wire that produces magnetic flux
(Figure 21.7) – this is also described as an electromagnet formed from a coil
of wire. The magnetic flux outside a solenoid is similar to the magnetic flux
for a bar magnet, with the north pole at one end of the coil and the south
pole at the other end, depending on the current direction.
A current-carrying wire in a magnetic field moves because a force acts on
it. The magnetic field making the wire move is called a catapult field. The
catapult field is due to the combined effect of the current-carrying wire’s
flux and the static flux. Figure 21.8 shows the separate fluxes, and how they
combine to form a catapult field.
405
Figure 21.7 The magnetic field for a
solenoid.
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Measuring magnetic field strength
(a)
N
S
+
A magnetic field runs through a coil of wire as well as outside it, as shown
in Figures 21.6 and 21.7. You can measure the flux density of the field
using a magnetic field sensor called a Hall probe. The probe contains a slice
of semiconducting material. If a current flows in the semiconductor when
it is perpendicular to the magnetic flux, a potential difference is generated
across the sides of the semiconductor. This potential difference is directly
proportional to the flux density.
●● Forces on current-carrying wires
(b)
You need to know how to calculate the size and direction of a force on a
current-carrying wire in a magnetic field.
Calculating the direction of the force
Magnetic flux density is a vector quantity. When the directions of the
magnetic flux, the current in the conductor and the force are all at
right angles to each other, Fleming’s left-hand motor rule, shown in
Figure 21.9, helps you see the three-dimensional arrangement of these
vectors.
= FORCE
(c)
N
S
21 MAGNETIC FIELDS
Figure 21.8 (a) A uniform magnetic field.
(b) The field around a current-carrying
wire. (c) The catapult field. The force is
from the strong to the weak field.
force, F
Hold your thumb, first finger and second finger of your left hand
at right angles to each other. The thuMb represents the direction of
the force (Motion), the First finger represents the direction of the
magnetic Field, and the seCond finger represents the direction of the
Current.
EXAMPLE
Fleming’s left-hand rule
A current-carrying wire is held so that the current is into the page,
and the magnetic field direction is from the bottom to the top of the
page. Use Fleming’s left-hand rule to find the direction of the force on
the wire.
Answer
With your second finger (current) pointing into the page, and your first
finger (field) pointing from bottom to top of the page, you should find that
the direction of the force is from left to right.
magnetic
fields, B
406
current, I
left hand
Figure 21.9 Fleming’s left-hand rule.
469884_21_AQA_A-Level_Physics_403-419.indd 406
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If the wire is at an angle θ to
the flux, the force on the wire
is calculated using F = BIl sin θ,
where θ is the angle between the
wire, carrying the current, and
the flux lines (Figure 21.10).
When the wire lies parallel to
the flux lines (θ = 0), there is no
force on the wire.
I
We can measure the flux density at any point by measuring the force
exerted on a current-carrying wire at that point. The tesla is defined as the
flux density that causes a force of 1 N on 1 m of a wire carrying a current
of 1 A at right angles to the magnetic field. Written as an equation, this
becomes
F = BIl
where F is force (N), B is magnetic flux density (T), I is current (A) and l is
the length of the conductor in the field (m)
For the interested student, the Maths box gives more information about the
size of the force.
Forces on current-carrying wires
Calculating the size of the force
MATHS BOX
B
θ
ℓ
Figure 21.10
EXAMPLE
Horseshoe magnet
Figure 21.11 shows a current-carrying wire held
perpendicular to the field between the two poles of a
horseshoe magnet. Assume there is 2 cm of the wire in
the field.
force
B=
F
0.03 N
=
= 0.71T
Il 2.1A × 0.02 m
2 What current is required for the wire to experience
an upward force of 0.09 N?
Answer
t
n
rre
cu
Rearrange the same equation to give I on the lefthand side, and substitute for the flux density, B,
from the answer to part (a):
I=
Figure 21.11
F
0.09 N
=
= 6.3 A
Bl 0.71T × 0.02 m
1 Calculate the magnetic flux density (B-field) for the
wire if the current is 2.1 A and it experiences an
upward force of 0.03 N.
3 The current flows into the page. State whether the
direction of the magnetic field is from left to right, or
right to left.
Answer
Answer
The length of the wire in the field is 0.02 m.
Substitute known values into the equation:
469884_21_AQA_A-Level_Physics_403-419.indd 407
Using Fleming’s left-hand rule, the field is from the
right side to the left side of the page.
407
17/04/19 11:04 AM
TEST YOURSELF
1 A 50 cm wire carries a current of 0.2 A. Calculate
the force due to the magnetic field on the wire if
it is placed:
a) perpendicular to the Earth’s magnetic field of
flux density 50 µT
b) perpendicular to the magnetic field of an
electromagnet of flux density 1.5 T.
2 A wire is placed at right angles to a magnetic
field of flux density 6 mT.
a) If the force per metre of wire is 0.03 N,
calculate the current in the wire.
b) State the value of the angle, θ, if the
magnitude of the force is zero.
3 Copper has a density of 8960 kg m −3.
a) Calculate the mass of an insulated piece of
copper wire, 10 cm long with a cross-sectional
area of 2 × 10−6 m2. Ignore the mass of the
insulation. Two such pieces of wire are used as
weights in Figure 21.12.
Figure 21.12 shows a current balance formed
from a rectangular frame balanced on two pivots
halfway along the sides AD and BC. Part of side
AB lies at right angles to a uniform magnetic
field. Sides AD and BC are 240 mm long. There is
a small gap in side CD a small weight hangs on
the wire hear the gap.
The frame balances when there is no current
in the circuit. When a current of 3.2 A flows
through the circuit, the weight must be moved
25 mm closer to the pivot to balance the frame.
b) Calculate the change in the moment of the
copper wire when the current is off, and
when the current is on.
c) Use your answer to part (b) to calculate the
force on side AB when the current is on.
d) The length of the wire frame in the field is
10 cm. Calculate the flux density.
4 Use Fleming’s left-hand rule to find the direction
of the force, magnetic field or current in the four
situations shown in Figure 21.13.
magnadur
magnets on yoke
A
B
current
B
N
current
pivot
A
(a)
pivot
C
S
D
(b)
D
gap
C
21 MAGNETIC FIELDS
small weight
408
stiff copper
wire frame
bush
N
Figure 21.12
–
+
S
Figure 21.13
w
I
v
(c)
S
N
brush contact
(d)
5 Part of a circuit, VWXY, lies in the same
horizontal plane as a uniform magnetic field
of flux density B. Two sides of the circuit are
parallel to the flux, as shown in Figure 21.14.
magnetic field
I
x
thin copper disc
(free to rotate)
I
y
Figure 21.14
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c) An extra twist of wire is made so that the wire
loops and now leaves the field at X, as shown
in Figure 21.15. Explain how the magnitude
of the force varies in different sections of the
circuit, VW, WX, XY and YV.
Calculate the flux density of the field, B, giving
your answer to an appropriate number of
significant figures.
w
I
x
v
I
I
magnetic field
I
I
Forces on current-carrying wires
a) When the current I flows through the wire, the
side WX is forced upwards. State the direction
of the magnetic field.
b) The length of wire, WX, in the field is 8.0 cm
and this stem, experiences a force of 0.241 N
when the current I is 4.6 A.
y
Figure 21.15
REQUIRED PRACTICAL 10
The force on a current-carrying conductor
Note: this is just one example of
how you might tackle this required
practical.
A sensitive electronic balance can be
used to investigate how magnetic flux
density, current and length of wire
are related.
supported
terminals
T1
T2
thick
copper
wire
A
U-magnet
A current-carrying conductor
experiences a force when it is
perpendicular to a magnetic field.
When magnets and a conductor are
set up as shown in Figure 21.16, the
electronic balance reading changes
Circuit for varying the current
electronic
when current flows in the wire.
balance
Newton’s third law means that the
Figure 21.16 Experimental set-up for the activity, including the circuit.
magnetic force on the wire is equal in
size but acts in the opposite direction
to the magnetic force on the magnet. The change of reading of the
TIP
balance shows the magnetic force on the magnet.
If this experiment is actually
T1
T2
409
performed, remember that the
Two ceramic magnets are fixed inside a steel yoke to create a uniform
wire will heat up, so turn the
field. The yoke is placed on the electronic balance. The circuit is set up
circuit off between readings.
as shown in Figure 21.16, and the copper wire is supported horizontally
A heavy-duty rheostat is also
between the magnets, perpendicular to the field. The balance is zeroed,
required.
and then the circuit is switched on. Readings are taken of the current
(in amps) and electronic balance reading (in grams). At least six sets of
readings are taken for different current values. The effect of changing the length of wire in the field can be
measured by adding an extra magnet and then re-zeroing the balance before taking readings.
469884_21_AQA_A-Level_Physics_403-419.indd 409
17/04/19 11:04 AM
You are investigating the relationship
F = BIl
Force F is calculated by converting readings in grams from the electronic
balance to force. The magnetic flux density B (T) is constant if you use the same
magnets and same arrangement. The length of wire between the magnets
is l, measured in metres. Some sample results are shown in Table 21.1. The
length of the magnets is 4.5 cm.
Table 21.1
Current/A
1.0
2.0
3.0
4.0
5.0
Reading from
balance/g
0.46
0.91
1.35
1.83
2.30
Force/N
TIP
Tare (zero) the balance between
readings as necessary.
1 Copy and complete the table by filling in the missing values in the force column.
2 Explain why it is acceptable to have a different number of decimal places
for the calculations of force.
3 Explain why it is acceptable to have a different number of significant
figures for the balance readings.
4 Using a graph, or otherwise, show that the magnetic flux density B is 100 mT.
●● Forces on a charged particle moving in
21 MAGNETIC FIELDS
a magnetic field
410
Charged particles moving in a magnetic field also experience a force. Old-style
televisions and computer monitors use electron guns to produce beams of
rapidly moving electrons in evacuated tubes, and their direction is controlled
using a varying magnetic field. You can calculate the force on a single charge,
Q, travelling perpendicular to a magnetic field, with flux density B.
If charge Q travels a distance l in t seconds, then the charge has a velocity
v = tl . But I = Qt and we can substitute for I in
F = BIl
This gives
BQI
F= t
Since the velocity of the charge is v = tl , this gives
F = BQ v
As before, you can use your left hand to predict the direction of the force.
The thumb represents force, the first finger represents the magnetic field
and the second finger represents the direction of a moving positive charge.
The sign of the charge is important – a positively charged particle and a
negatively charged particle will move in opposite directions in the same field.
This is because if a negative charge moves to the left (for example), the
conventional current flows to the right.
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Conventional current – in many circuits the charge carriers are electrons
but this is not always the case. In an electrolyte, for example, positive and
negative ions move in opposite directions when a current flows.
By convention, we always who the direction of movement of positive
charge carries and so, in a simple wire circuit, this is in the opposite
direction to electron flow.
Work done
When a charged particle moves at right angles to a uniform flux, the
charged particle moves in a circle (Figure 21.17), because the force is always
perpendicular to the motion, provided no energy is lost (for example, when
the charge is in a vacuum). Work is force multiplied by distance moved in
the direction of the force. Since the force is perpendicular to the motion, no
work is done by the magnetic field on the charge, so the kinetic energy of
the charge does not change.
(a)
(b)
electron
path of electron
e
v
e
Forces on a charged particle moving in a magnetic field
TIP
F
v
F
F
e
v
magnetic flux density B into the page
B into the page
Figure 21.17 An electron travels in a circuit path when it moves perpendicular to a
magnetic field because the force is perpendicular to its motion.
Applying ideas about circular motion
You can link the ideas of circular motion covered in Chapter 15 with the
motion of a charged particle in a magnetic field.
Since the charged particle follows a circular path in a magnetic field
(Figure 21.17), we know it experiences a centripetal force. For circular
motion, the centripetal force must equal the force exerted by the magnetic
field. If we link the equations for centripetal force and the movement of a
particle in the magnetic field, we find that
411
2
F = mv
r = BQ v
Dividing both sides by v gives
mv = BQ
r
This equation has many applications. For situations where B and Q are
constant, the radius is proportional to the momentum of the particle:
mv = BQr
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EXAMPLE
Paths of ionising radiation
Answer
Figure 21.18 shows ionising radiation travelling
through a uniform magnetic field in a vacuum, at right
angles to the lines of magnetic flux.
magnetic field
alpha particles
beta particles
Figure 21.18
1 Compare the directions of motion of an alpha
particle, a beta particle and a gamma ray if they
move through the magnetic field as shown.
Answer
Using Fleming’s left-hand rule, the magnetic field
is into the page. The alpha particle is positively
charged, so the force on it is upwards, and the alpha
particle travels in a circular path and clockwise.
The gamma ray has no charge, so it continues to
move in a straight line.
The beta particle has a negative charge, so the
force on it is initially downwards, and it follows a
clockwise circular path.
2 The diagram in Figure 21.18 is not to scale.
Assuming the particles travel at the same speed,
calculate the ratio of the radius of the paths for
alpha particles and beta particles.
21 MAGNETIC FIELDS
mv = BQ
r
But v and B are constant, so r is proportional to
m/Q. The alpha particle has a charge of 2e, and a
mass of 8000me. The beta particle is an electron,
with charge e and mass me. So the ratio is
gamma rays
412
From the text,
radius(alpha) m(alpha) Q (beta)
=
radius(beta) Q (alpha) m(beta)
8000me e
=
= 4000
2e me
In reality, beta particles travel much faster than
alpha particles, and relativistic effects increase
the mass of beta particles. This means that the
observed ratio is likely to be smaller. However, it
shows that it is extremely difficult to deflect alpha
particles using a typical school magnet.
3 Describe how the path will be different if the
radiation travels in air, rather than in a vacuum.
Answer
In a vacuum, the charged particles travel in a
circle. In air, charged particles in a magnetic field
travel in a spiral, because they lose energy and
slow down. Because r is proportional to v, as v
decreases so does r.
You can also use the equation to find the radius r of the circle in which
the charged particle travels, since r = BmvQ . This idea is used in mass
spectrometers, since charged particles with different mass/charge ratios
travelling at the same speed will follow different paths.
TEST YOURSELF
6 D
escribe two situations in which a charged
particle experiences no magnetic force when it is
in a magnetic field.
7 An alpha particle travelling at 1% of the speed
of light enters a field of flux density 1 × 10 −3 T.
Calculate the force experienced by the particle
if it travels
a) parallel to the direction of the magnetic flux
b) perpendicular to the direction of the magnetic
flux.
469884_21_AQA_A-Level_Physics_403-419.indd 412
8 C
alculate the flux density when an electron,
travelling at right angles to the field direction,
at a velocity of 1 × 107 m s−1, experiences a force
of 1 × 10 −15 N.
9 A n electron enters a uniform magnetic field of
flux density 0.036 T, travelling at right angles
to the lines of flux. Calculate the speed of the
electron if the radius of its path is 20 mm.
10 Figure 21.19 shows the paths of three particles in
a magnetic field coming out of the page.
17/04/19 11:05 AM
α source
fast
(higher
energy)
A
slow
(lower
energy)
B
β source
γ undeflected
γ source
Figure 21.19
Cyclotrons
a) Write down an expression for the centripetal
force acting on the electron.
b) Show that the electron’s speed is 5.6 × 106 m s−1.
c) How does the radius of its path change if
i) the kinetic energy of the electron doubles
ii) allowance is made for the relativistic mass
of the electron?
12 The mass of positively charged particles can be
measured using a mass spectrometer. Charged
particles are accelerated in a vacuum, and a
velocity selector is used to select particles
travelling at a specific velocity. These particles
pass into a magnetic field applied perpendicular
to their path and their position on a detector is
recorded.
a) Explain why particles of the same charge but
different masses (e.g. M, 2M and 3M) have
different paths in a magnetic field.
b) E xplain how the paths would be different for
the following three particles travelling in the
same field: mass M and charge Q; mass 2M
and charge 2Q; mass 3M and charge 2Q.
ompare the relative charges and speeds of the
C
different particles A and B.
11 A
n electron travels in a circular path of radius
74 mm at right angles to a uniform magnetic field
of flux density 0.43 mT.
●● Cyclotrons
Cyclotron A particle accelerator that
accelerates charged particles through a
spiral path using a fixed magnetic field and
an alternating potential difference.
A cyclotron is used to force charged particles into a circular path that
accelerates them to very high speeds. Cyclotrons are often used with heavier
particles like alpha particles and protons. Experiments using particle
accelerators investigate the structures of complex molecules like proteins, as
well as sub-nuclear structures.
D1
proton
beam
The cyclotron is formed from two semi-circular ‘dees’,
separated by a small gap and connected to a highfrequency alternating potential difference (Figure 21.20).
A strong magnetic field is applied perpendicular to the
dees. The perpendicular magnetic field forces charged
particles to move in a circular path inside the dees.
The particles experience a potential difference when
they travel across the gap, and gain energy equal to QV
(where Q is the particle’s charge in coulombs, and V is
the potential difference in volts). Since the particles have
more kinetic energy, they move faster and accelerate to the
next dee.
proton
source
D2
high-frequency voltage supply
Figure 21.20 Structure of a cyclotron, a proton accelerator.
469884_21_AQA_A-Level_Physics_403-419.indd 413
413
The ac voltage is timed to change direction every time the
particles reach the gap between the dees. It must alternate
to accelerate the particles each time they reach a gap. If
the voltage did not alternate, the particles would follow a
cycle of accelerate–decelerate–accelerate.
Particles spend the same time inside each dee, but the
radius of their path increases after each gap and they
travel further in the same time.
17/04/19 11:05 AM
Remember that the centripetal force acting on the charged particle equals
the magnetic force on the charged particle, so
mv2
= BQ v
r
or
BQr
v = m (i)
Because the radius is proportional to the speed of the charged particle, the
particles spiral outwards as they accelerate through the cyclotron.
The time, t, spent in each dee is given by
t = πvr
(ii)
because πr is half the circumference of the circle. Substituting for v in
equation (ii) using equation (i) gives t, the time spent in one dee:
πr
BQr /m
πm
=
BQ
t=
which does not depend on either radius or speed.
21 MAGNETIC FIELDS
Synchrotron A particle accelerator that
accelerates charged particles through a
circular path using a varying magnetic field.
414
The effect of special relativity limits a particle’s speed in a cyclotron.
Particles get more massive as they travel close to the speed of light. As
particles move faster and their mass increases, the time spent in each dee
increases and the more massive particles get out of step with the alternating
potential difference. A synchrotron overcomes this problem by increasing
the magnetic field as the speed of the particles increases. The radius of their
path remains constant even though the particles travel faster.
EXAMPLE
Alpha particles in a cyclotron
Alpha particles are accelerating in a cyclotron. The
magnetic flux density is 0.8 T and the voltage across
the gaps between the dees is 20 kV. The mass of an
alpha particle is 6.64 × 10−27 kg and its charge is
3.2 × 10−19 C. Ignore relativistic effects.
So
1 Show that the frequency needed for the voltage
supply to synchronise with the arrival of protons at
the gaps is 6.14 MHz.
2 How many circles should the alpha particles make
to reach an energy of 10 MeV?
Answer
πm
The time spent in one dee is t = BQ ; the period T for
one complete circle (circling through two dees) is 2t.
The frequency for a complete circle is f
=
BQ
1
=
2t 2πm
f=
0.8 T × 3.2 × 10−19 C
2π × 6.64 × 10−27 kg
= 6.14 × 106 Hz
Answer
The energy gained when an alpha particle (charge
2e) crosses the gap is 2 × 20 keV = 40 keV. The alpha
particle crosses two gaps per cycle, so it gains
80 keV (80 × 103 eV) per cycle. To reach an energy of
10 MeV (10 × 106 eV), the alpha particle must make
10 × 106
80 × 103
469884_21_AQA_A-Level_Physics_403-419.indd 414
circles, which is 125 circles.
17/04/19 11:05 AM
13 In a cyclotron, explain the role of
a) alternating electric fields
b) fixed flux density.
14 Charged particles move at constant speed as they travel around a
single dee. These particles are also accelerating. Explain how these
two statements can both be true.
15 Protons are accelerated in a cyclotron. If the voltage supply
alternates at a frequency of 4 MHz, calculate the magnetic flux
density required.
The mass of a proton is 1.67 × 10 −27 kg.
16 E xplain why identical particles with different energies can be
accelerated in a cyclotron together.
17 Find an expression for the maximum kinetic energy for a proton in a
cyclotron of radius R.
Cyclotrons
TEST YOURSELF
415
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Practice questions
1 Charged particles enter a magnetic field of flux density T at right angles
to the field. Which one of these would decrease the radius of the circular
path of the charged particles?
A decrease in charge Q
B decrease in mass m
C increase in velocity v
D decrease in flux density B
2 A positively charged particle travels north at a steady speed, v. A
magnetic field is applied in the horizontal plane in which the particle is
travelling. The flux is directed from east to west. Which of the following
describes the motion of the particle after it enters the field?
A The particle accelerates upwards in the vertical plane.
B The particle’s motion is unchanged.
C The particle accelerates downwards in the vertical plane.
D The particle accelerates in the westerly direction.
3 An electron enters a uniform magnetic field, travelling at a steady speed
at right angles to the field. The shape of the electron’s path in the field is
A a circle
B a straight line
C a parabola
21 MAGNETIC FIELDS
D an ellipse
416
4 The speed of an electron of charge e and mass m travelling in a circular
path of radius r in a magnetic field B is given by
A Bemr
B Bem
2πr
C mer
B
D Ber
m
5 A 50 cm wire carries a current of 1.2 A. The force the wire experiences if
it is placed in a flux density 0.3 mT is
A 0.18 N
B 1.8 × 10−4 N
C 18 N
D 0.018 N
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Practice questions
6 A wire carrying a current is placed parallel to a magnetic field. When
the wire is gradually turned so it is perpendicular to the field, the force
experienced by the wire
A stays at a constant value
B decreases to zero from a maximum value
C increases from zero to a maximum value
D remains zero
7 A magnetic field is from south to north. A current-carrying wire in the
field experiences an upwards force. The direction of the current is
A into the page
B from east to west
C from west to east
D downwards
8 When a proton in a cyclotron travels through a dee, which of these
statements is true?
A The proton is constantly accelerating and its speed is increasing.
B The proton travels at a constant speed and is not accelerating.
C The proton travels at a constant velocity and is constantly
accelerating.
D The proton travels at a constant speed but is constantly accelerating.
9 The magnetic field of a cyclotron is
A constant in magnitude and applied perpendicular to the plane of the
dees
B varying in magnitude and applied perpendicular to the plane of the
dees
C constant in magnitude and applied parallel to the plane of the dees
D varying in magnitude and applied parallel to the plane of the dees
10 A beta particle travels from east to west across a magnetic field of strength
0.6 mT, which is directed northwards. The beta particle travels at
4 × 105 m s−1. The force the beta particle experiences in the field is
A 3.8 × 10−14 N downwards
B 3.8 × 10−17 N downwards
C 3.8 × 10−14 N upwards
D 3.8 × 10−17 N upwards
417
N
I
S
Figure 21.21
11 A current-carrying wire of length 10 cm is placed in a magnetic
field as shown in Figure 21.21.
a) Predict the direction of the force acting on the wire.
(1)
b) Calculate the flux density if the force on the wire is 2 mN when
the current is 0.4 A.
(3)
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c) A student wants to use the magnetic field to lift the wire up. The
wire’s mass per unit length is 0.5 g cm−1. Calculate the minimum
current required to allow this piece of wire to lift.
(4)
d) Explain why a larger current would be needed in reality.
(1)
e) Explain what happens if an alternating current is used instead.
(4)
12 A hospital cyclotron uses magnetic fields to accelerate protons around a
circular path. The diameter of the cyclotron is 2.0 m.
a) If the field is in a suitable direction, the protons move in a circular
path of constant radius. By referring to the force acting on the
protons, explain how this happens.
(4)
b) Calculate the centripetal force acting on a proton when travelling
(3)
around the cyclotron at a speed of 6 × 107 m s−1.
c) Calculate the flux density of the magnetic field needed to produce
this force.
(3)
d) A cyclotron accelerates the protons to their final speed by applying
a varying potential difference at the gaps between the two dees in
which the protons travel. Explain how the potential difference is used
to accelerate the protons.
(4)
13 Electrons travel around a tube placed in a magnetic field of flux density
0.3 mT (see Figure 21.22).
path of
electrons
a) State and explain the direction of the magnetic field that
forces the electrons to travel in this path.(2)
b) If the radius of the orbit is 15 cm, calculate the flux density
producing this motion.(4)
centre of
circle for
path of the
electrons
R
evacuated tube
in the plane
of the paper
21 MAGNETIC FIELDS
c) Explain why the electrons are accelerating without
speeding up.(2) Figure 21.22
418
d) Predict the approximate speed required for protons to travel in the
same orbit in the same flux density.
(3)
14 An experiment is set up in which particles travel from left to right
across a uniform magnetic field, directed into the plane of the page. The
particles are travelling at the same speed. Describe in detail how their
path through the magnetic field could be used to identify
(8)
a) the sign of the charge of electrons, protons, alpha particles and
neutrons
b) the relative masses of electrons, protons and alpha particles.
Stretch and challenge
15 A stream of charged particles is originally moving at velocity v and
directed perpendicular to a uniform magnetic field. The particles follow
a circular path in a plane perpendicular to the field and the original
motion.
a) Describe how this path changes if the magnetic field varies
continuously, becoming weaker then stronger.
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Practice questions
The Aurora Borealis is the appearance of coloured lights in the sky
usually seen near the poles of the Earth. This is caused by charged
particles that spiral along the Earth’s magnetic field lines, and are
channelled to the poles (Figure 21.23). As the particles interact with
gases in the atmosphere, they cause coloured light with the emission
of electron transitions.
North
particle
paths
South
magnetic
field lines
Figure 21.23
b) Describe how spiral motion is caused when a charged particle travels
close to, but not quite parallel to a magnetic field line, by explaining:
i) what is meant by ‘spiral motion’
ii) how the field affects components of velocity that are parallel and
perpendicular to it
iii) how these components of velocity combine.
Use a diagram if this is helpful.
c) A charged particle is travelling at speed v at an angle θ to a magnetic
field B with flux density B.
i) State the components of velocity perpendicular and parallel to
the field B .
ii) Calculate the radius of the spiral for a particle of charge Q, and
the forward distance travelled while the particle circles the field
once.
iii) Prove that these expressions are consistent with the expression
tan θ = 2πr .
d
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419
17/04/19 11:05 AM
22
Magnetic flux
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
●
●
●
Magnetic flux density B is measured in teslas (T).
The force on a current-carrying wire perpendicular to a magnetic
field is given by F = BIl, where I is the current (A) and l is the length of
wire in the field (m).
The force on a charged particle moving perpendicular to a magnetic
field is given by F = BQv, where Q is the charge (C) and v is its velocity
(m s −1).
Fleming’s left-hand rule is used to determine the directions of force,
magnetic field and current or velocity.
The angular speed ω (also called angular frequency) is calculated as
ω = 2πf, where f is the frequency (Hz).
TEST YOURSELF ON PRIOR KNOWLEDGE
1 a) Work out the direction of the force acting on the current-carrying
wire shown in Figure 22.1.
b) If 3 cm of the wire is in the field
and experiences a force of
2 mN when the current flowing
in the wire is 2 A, calculate the
magnetic flux density in teslas.
+
2 a) A proton travels at 1% of the
–
speed of light perpendicular to
N
S
a magnetic field of flux density
1 mT. Calculate the force acting
on the proton.
b) How does the force change if
Figure 22.1
it is travelling parallel to the
field?
420
469884_22_AQA_A-Level_Physics_420-434.indd 420
In this chapter, you will learn how we create an emf using electromagnetic
induction. We use electromagnetic induction frequently, for example when
generating mains electricity in power stations. Electromagnetic induction
is also used in induction loops, for example to detect vehicles approaching
traffic lights. A conducting loop buried under the road surface carries
an alternating current, creating an alternating magnetic field, which is
monitored constantly. When ferromagnetic material, such as a car, passes or
remains over the loop, the change of flux causes a change in current, so the
presence of a car is detected.
17/04/19 11:06 AM
area
A
Magnetic flux
●● Magnetic flux
magnetic
field B
In Chapter 21, we came across the idea of magnetic flux density, which
measures the strength of a magnetic field B, or B-field, in teslas (T). A
diagram of a magnetic field indicates the density of magnetic flux by
showing the number of flux lines per square metre (Figure 22.2).
Magnetic flux is defined as magnetic flux density B (in teslas, T) multiplied
by the area of the surface, A (in m2), where the area A is perpendicular to
the lines of flux (Figure 22.2). Written as an equation, this becomes ϕ = BA.
Magnetic flux is measured in webers (Wb), where 1 Wb equals 1 T m2.
When an area is not perpendicular to the lines of magnetic flux, as shown
in Figure 22.3, the flux through the area A is now the component
Figure 22.2
ϕ = BA cos θ
Magnetic flux ϕ is magnetic flux density ×
cross-sectional area perpendicular to field
direction (measured in webers).
Figure 22.3 shows the area of a loop perpendicular to a field if the loop
itself is at angle θ to the flux lines.
Weber (Wb) is the unit of magnetic flux,
equal to 1 tesla metre² (1 T m2)
Cutting flux lines
When an object passes through a magnetic field, we can say that it ‘cuts’
the magnetic flux lines. Figure 22.4 shows a wire of length l moving
downwards in a magnetic field with horizontal field lines. You can see that
the wire cuts across the flux lines as long as it moves perpendicular to them.
The wire cuts through more flux lines each second if
q
B
length l is longer
the wire moves faster
●the magnetic flux density is stronger.
●
q
area A
●
area A cos q
Figure 22.3 Acos θ is the area
perpendicular to the flux lines.
Q
l
v
P
flux density (B)
If a conductor moves perpendicular to field lines, it ‘cuts’ the flux lines.
But if the conductor moves parallel to field lines, they are not cut.
Magnetic flux linkage
Magnetic flux linkage is defined as Nϕ, where ϕ is the number of flux lines
that pass through , or link, with each of the turns of a coil of N turns. Since
flux ϕ = BA for a single loop of wire, then the flux linkage is Nϕ = BAN if
the coil of wire has N turns that are perpendicular to the lines of flux. Flux
linkage is measured in weber-turns.
Flux linkage depends on several factors, as shown in Figure 22.5:
Figure 22.4 A wire ‘cuts’ field lines
when it moves perpendicular to them.
(a) the flux density
(b) the orientation of the coil and flux lines
(c) the coil’s cross-sectional area
Magnetic flux linkage Magnetic flux linked
by a coil, calculated as magnetic flux ϕ ×
number of turns N of the coil (measured in
weber-turns).
Weber-turns The unit of magnetic flux
linkage.
421
(d) the number of turns on the coil.
Flux linkage is important because an emf is induced in a coil, in which
the flux linkage changes. You will learn more about this in the next
section.
Figure 22.6 shows a coil being turned in a magnetic field. As the coil turns
in the magnetic field, the area of the coil perpendicular to the field is given
by A cos θ, and the magnetic flux linkage is given as
Nϕ = BAN cos θ
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17/04/19 11:06 AM
(a) The flux through an area is proportional
to the flux density.
where N is the number of turns on the coil, ϕ is the magnetic flux
(in Wb), B is the magnetic flux density (in T), A is the cross-sectional
area of the coil (in m2) and θ is the angle between the axis of the coil
and the flux lines. The flux linkage changes as shown in Figure 22.7.
Nf
A
B
(b) Flux varies depending on the relationship
between the boundary faces and the
direction of flux.
0
90º
Figure 22.6 The magnetic flux linkage
changes as a coil of cross-sectional
area A and with N turns rotates in a
flux density B.
180º
270º
360º q
Figure 22.7 How flux linkage changes
when a coil of N turns rotates in a field.
(c) Flux is proportional to the area within the
boundary.
Figure 22.5
EXAMPLE
Magnetic flux linkage through a coil
Figure 22.8 shows a coil of wire formed as a 60° triangle with sides of
length 30 cm. The coil has 50 turns. Calculate the magnetic flux linkage
with the coil when it is placed with the axis at 40° to a vertical in a
uniform horizontal flux of 0.02 T.
40°
A
Answer
The area of the coil is
( 12 × base × height) =
B
1
2
× 0.30 m × 0.30 m × sin 60° = 0.039 m2
22 MAGNETIC FLUX
The flux linkage is
422
BAN cos θ = 0.02 T × 0.039 m2 × 50 × cos 40°
Figure 22.8
= 0.03 Wb turns
TEST YOURSELF
1 Calculate the magnetic flux through the face of a
magnet, if the face measures 2.0 cm by 6.0 cm and
the magnetic flux density of the magnet is 0.03 T.
2 Calculate the flux through the horizontal surface
of the British Isles. The average flux density in the
region is 53 µT at 20° to the vertical and the area of
the British Isles is 3.0 × 1011 m2.
3 a) Calculate the magnetic flux passing through a
copper sphere of radius 3.0 m placed in a region
of uniform magnetic flux density 2.0 T.
b) Explain whether or not the amount of magnetic
flux changes if the sphere is sliced in half, along
the axis perpendicular to the field.
469884_22_AQA_A-Level_Physics_420-434.indd 422
4 A square coil of wire has sides 12 cm long, and 250
turns. Calculate the magnetic flux linkage when:
a) the cross-sectional area of the coil is
perpendicular to a field of flux density 0.08 T.
b) the face of the coil makes an angle of 60° to the
magnetic flux lines.
c) Sketch a diagram showing how the flux linkage
changes if the coil is initially perpendicular to
the field and is turned until it is parallel and
then perpendicular again.
17/04/19 11:06 AM
Our lives would be completely different without Michael Faraday’s
discovery of electromagnetic induction in 1831 using insulated coils of
wire and changing magnetic fields.
microammeter
coil of
many turns
N
Figure 22.9 Using a moving magnet to
induce an emf
magnet
microammeter
wire
Figure 22.10 Moving a wire into a
magnetic field induces an emf
You can easily demonstrate electromagnetic induction using a coil of wire
connected to a microammeter, as shown in Figure 22.9. The microammeter
flicks one way when a bar magnet is moved into the coil, and the other way
when the magnet is pulled out. It is zero when the magnet is stationary inside
the coil. An emf is induced if there is relative movement between the coil and
a magnetic field (either the magnet or the coil moves) or the magnetic flux
linkage changes (for example, the strength of an electromagnet changes).
Electromagnetic induction
●● Electromagnetic induction
Figure 22.10 shows electromagnetic induction caused by a length of wire
moving between two magnets. The wire is connected to the microammeter,
which flicks one way when the wire moves down, and flicks in the opposite
direction when the wire moves up.
An emf is induced in the wire because an electric charge moving
perpendicular to a magnetic field experiences a force, BQv (see Chapter 21).
Using Fleming’s left-hand rule, you can see that electrons in a wire move
towards one end of the wire when the wire moves perpendicular to the
magnetic field. This leaves one end of the wire negatively charged overall
and the other end positively charged, creating a potential difference across
the wire. A current can flow if the wire is part of a complete circuit – for
example, when the wire is connected to a microammeter.
Faraday’s law
Faraday’s law The induced emf equals the
rate of change of (magnetic) flux.
S
S
We can calculate the magnitude of the induced emf in a coil using
Faraday’s law. This states that the magnitude of the induced emf
equals the rate of change of magnetic flux linkage, and is written as
∆ ( Nφ )
ε=
∆t
where Δ(Nϕ) is the change in flux linkage and Δt is the time over which
that changes takes place. Since a coil of wire has a fixed number of turns,
this becomes
∆φ
ε =N
x
∆t
We can use this law to help us to understand some earlier observations:
●
N
N
●
I
●
downward flux increases ●
Figure 22.11 Flux linkage in the coil
increases as the coil moves closer to
the magnet.
Relative movement between a magnet and a coil changes the flux linkage
in the coil (Figure 22.11). This generates an emf.
Rotating a coil in the plane perpendicular to the field changes the crosssectional area through which the flux passes. This changes the flux linkage,
and generates an emf.
Increasing the relative motion, or the speed at which the coil rotates, increases
the rate of change of the flux linkage, which increases the induced emf.
If there is no relative movement or rotation, the flux linkage does not
change, so no emf is generated.
423
Lenz’s law
Faraday’s law calculates the magnitude of the induced emf and is often combined
with Lenz’s law, which indicates the direction of the induced emf. Lenz’s law states
that the direction of the induced emf opposes the changes causing it.
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Lenz’s law The direction of the induced
emf causes effects that oppose the change
producing it.
Pushing a magnet’s south pole into the coil
induces a south pole – this repels the magnet
N
S
S
0
Pulling a south pole out of the coil induces a
north pole – this attracts the magnet
N
S
N
0
Figure 22.12 A magnet being pushed
into, or pulled out of, a coil of wire.
Lenz’s law determines the direction
of the induced emf
Figure 22.12 shows the south pole of a magnet moving into a coil. This
induces an emf when there is a complete circuit, a current flows and the
coil behaves as an electromagnet, with its south pole facing the magnet’s
south pole, repelling the magnet.
Pulling the magnet out of the coil induces an emf such that the same end of
the coil becomes a north pole, which attracts the magnet.
We can combine Lenz’s law with Faraday’s law and write
∆ ( Nφ )
ε =−
∆t
where Δ(Nϕ) is the change in flux linkage and Δt is the time over which that
changes takes place. Since a coil of wire has a fixed number of turns, this
becomes
∆φ
ε =− N
∆t
Lenz’s law is the result of conservation of energy. When the south pole of
a magnet is pushed into the coil, a current is induced in the wire, which
becomes an electromagnet. If the south pole of the electromagnet faces the
moving magnet, the poles repel and work must be done to keep pushing
the magnet into the coil of wire. If you try this with a very strong magnet in
a large coil, you may feel the force you are working against.
22 MAGNETIC FLUX
If Lenz’s law did not apply and, instead, the north pole of the coil faced the
magnet’s south pole, the magnet would be attracted. This would make the
magnet accelerate into the coil, increasing the induced emf. This would
start a process in which increasing the emf increased the acceleration, which
increased the emf and so on. That would imply that energy can be created
without doing any work. This, of course, cannot happen.
EXAMPLE
Magnitude of an induced emf
= 3 × 10−3 T × 2π × (0.02)2 m2 × 200
= 7.5 × 10−4 Wb turns
Calculate the magnitude of the induced emf when
a flat coil of radius 2.0 cm, with 200 turns, is placed
at right angles to a varying magnetic field. The field
strength is increased from 0 to 3.0 mT in 0.30 s.
Answer
Maximum flux linkage is
Minimum flux linkage = 0 Wb turns
Magnitude of induced emf is
ε=
=
Nϕ = BAN
∆ (Nφ )
∆t
7.5 × 10-4 Wb turns
0.30 s
-3
= 2. 5 × 10 V
424
Eddy currents
A metal sheet moving into (or out of) a magnetic field can become very hot.
This happens if very large currents, called eddy currents, are set up in the
metal sheet. Eddy currents are circulating electric currents flowing in the
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17/04/19 11:06 AM
pivot
v
N
(a)
(b)
Figure 22.13 (a) Eddy currents form
in an aluminium pendulum, slowing it
down as it moves in to and out of a fixed
magnetic field unless (b) slits are cut in
the pendulum.
Eddy currents can become very large because metals have a low resistance.
Eddy currents are put to good use in induction cookers. Here a highfrequency alternating current in the cooker produces a rapidly changing
magnetic field, which induces a large alternating current in the base of a
saucepan causing it to heat up.
Eddy currents can also cause magnetic braking. A pendulum swinging
between two magnets slows down quickly because eddy currents are set up
in the metal when it enters and leaves the field (Figure 22.13a). Magnetic
fields created by these eddy currents interact with the fixed magnetic field,
opposing the motion and stopping the pendulum. Cutting slits in the
pendulum prevents eddy currents forming and the pendulum continues to
swing (Figure 22.13b).
REQUIRED PRACTICAL 11
Using a search coil to investigate
changes in magnetic flux density
Electromagnetic induction
pivot
plane of the metal. They are caused by the change of flux linkage when the
metal moves perpendicular to the field, and the currents flow in a direction
to oppose the motion creating them.
ac supply
oscilloscope
Note: This is just one example of how you might tackle
this required practical.
A search coil is a small flat coil made from 500–2000
turns of insulated wire mounted on a handle
(Figure 22.14). An emf is induced in the coil when
the coil is placed in a magnetic field that varies. The
amplitude of induced emf is directly proportional to the
amplitude of the varying flux density of the field.
search coil
leads
plastic handle
Figure 22.14 A search coil.
You can calibrate the search coil by connecting it
to an oscilloscope. When the search coil is placed
perpendicular to a known flux density that varies, the
trace on the oscilloscope is used to find the amplitude
of the emf induced in the coil.
A calibrated search coil can be used with the
oscilloscope to measure the strength of an unknown
flux density, or the effect of changing the angle of a
coil in a known flux density (Figure 22.15).
469884_22_AQA_A-Level_Physics_420-434.indd 425
slinky spring
search coil
Figure 22.15
To investigate the effect of changing the angle of a coil
in a flux density, the calibrated search coil is placed
in a known magnetic field that varies. The amplitude
of the induced emf is measured by connecting the
search coil to an oscilloscope with the timebase
turned off, so the emf is displayed as a vertical line
on the screen. Measurements are taken when the
search coil is held at different angles in the magnetic
field. The area of the coil is given by A cos θ, where θ
is the angle between the axis of the coil and the flux
lines and A is the cross-sectional area of the coil. We
find that the amplitude of the induced emf also varies
with cos θ, which is consistent with flux linkage given
by BAN cos θ.
425
Note: The slinky spring will heat up – don’t leave the
current on for only longer than necessary.
17/04/19 11:07 AM
EXAMPLE
An experiment with a search coil
A student places a search coil of radius 8 mm and
500 turns in a varying magnetic field of maximum flux
density 0.4 T. The search coil is then connected to an
oscilloscope and placed in the magnetic field with
its face perpendicular to the flux lines. It is turned
through 360° taking 0.2 s.
Nf
0
90º
180º
270º
360º q
1 Calculate the maximum flux linkage.
Answer
Maximum flux linkage is
BAN = 0.4 T × π × (0.008 m)2 × 500
= 0.040 Wb turns
2 Sketch a graph showing how the maximum flux
linkage varies as the coil is rotated through 360°.
Include as much detail as possible.
Answer
The graph is shown in Figure 22.16. Make sure you
sketch a cosine curve with the peak values and the
x-axis labelled.
3 Explain at least two additional steps the student
could take to reduce errors.
Figure 22.16
Answer
Use a data logger to record changes of emf with
time precisely and permanently (i.e. so the data can
be processed).
Use a calibrated motor to rotate the search coil at a
steady rate to improve repeatability and accuracy.
Repeat measurements of data to prepare mean
values for different angles to reduce the impact of
random errors.
Reduce systematic errors by calibrating the search
coil using a known magnetic field and oscilloscope.
22 MAGNETIC FLUX
TEST YOURSELF
426
5 The north pole of a magnet is pushed into a coil of
wire.
a) Describe what happens to the coil as the
magnet moves into the coil, rests inside the coil
and is pulled back out again.
b) Sketch a diagram showing how the induced emf
changes with time.
6 The magnetic flux density between the poles of
an electromagnet is 0.20 T. A coil, with 500 turns
and cross-sectional area 2.0 × 10 −4 m2, is placed in
the field perpendicular to the flux lines. The field
increases steadily to 0.60 T in 10 ms.
a) Calculate the initial magnetic flux linkage, and
the flux linkage at 10 ms.
b) Calculate the emf induced in the coil.
c) Calculate the emf induced in the coil if it was
held with its axis at 30° to the field while the
field changed.
7 A search coil with a cross-sectional area of 1.0 cm2
and 2500 turns is placed between the poles of a
magnet. If the coil is pulled out of the magnetic
field in 5 ms, and the average induced emf is 0.9 V,
calculate the strength of the magnetic field.
8 a) Explain why a copper ring heats up if it is placed
in a region with an alternating magnetic field.
469884_22_AQA_A-Level_Physics_420-434.indd 426
b) You have two coils of insulated wire, both have
the same circumference. One coil is made up
of just one loop, but the wire in the other coil is
three times as long and has been twisted into
three loops. Explain why the current induced by
passing a magnet through each of the two coils
of insulated wire is the same.
c) Explain why a magnet dropped through a
vertical copper pipe falls more slowly than the
same magnet falling through a vertical plastic
pipe.
9 Figure 22.17 shows a seismometer made from
a bar magnet suspended on a spring, which is
attached to a metal rod that transmits vibrations
from the Earth. Use the diagram to explain how the
seismometer detects waves from an earthquake.
+1
metal rod
voltage / V
magnet
A
B
0
–1
time /s
Figure 22.17
17/04/19 11:07 AM
Changing magnetic field
lines of
magnetic
flux B
steel
string
coil wound
on a pole of a
permanent
magnet
alternating
emf
4 mm
Figure 22.18 A guitar pick-up.
Some microphones uses electromagnetic induction to change a sound wave
into an electrical signal. The microphone has a lightweight coil suspended
in a circular groove between the poles of a permanent magnet (Figure 22.19).
The coil is attached to a diaphragm that vibrates when a sound wave reaches
it. Since the coil and magnetic field are perpendicular, an emf is induced in
the vibrating coil, which depends on the frequency and amplitude of sound
waves. The induced emf is amplified, and a loudspeaker changes these
signals back to audible sound.
Conductor moving in a straight line
flexible
diaphragm
An induced emf can be caused by a conductor moving in a magnetic field. For
example, a straight wire may be dropped through a uniform magnetic field, or
a plane may fly at a constant height and speed in the Earth’s magnetic field.
diaphragm
support
S
sound
waves
Electric guitar strings are made of magnetised steel. When the strings
are plucked, they vibrate directly above pick-ups, which are fixed on the
guitar’s neck. The pick-ups are bar magnets wrapped in up to 7000 coils of
very fine wire (Figure 22.18). The vibrating string causes vibrations in the
magnetic field surrounding the coil of the pick-up. This change is converted
into an emf and amplified. The distinctive electric guitar sounds come from
deliberate distortion when the note is amplified.
Applications of electromagnetic induction
●● Applications of electromagnetic induction
magnet
N
moving coil
wound onto
a former
S
electrical
leads
electrical
signal
output
Figure 22.19 A microphone uses
electromagnetic induction to change
sound into electrical signals.
A credit card includes information stored on a magnetic strip. The credit
card reader has a small coil in it, and when the credit card is swiped
through the reader, an emf is induced in the coil. It is important to swipe
the card quickly enough so that the induced emf is large enough to be
interpreted.
When a conductor moves at a velocity v perpendicular to the flux lines,
Faraday’s law applies and an emf is generated. For a conductor of length
l travelling in a flux density B, the area swept out per second is length ×
velocity. The induced emf equals the rate of change of flux linkage, so
dA
dt
Because the area swept out per second is lv, this becomes
ε=B
ε = Blv
where B is the magnetic flux density (in T), l is the length of the conductor (in
m) and v is the velocity of the conductor perpendicular to the field (in m s−1).
Electrical power
Q
Power is the rate of doing work, and we can show that the expression for
electrical power when a wire cuts flux lines (Figure 22.20) is consistent with this.
l
v
flux density (B)
427
P
You already know that P = VI in a circuit, where P is power (W), V is
potential difference (V) and I is current (A). When power is generated by
electromagnetic induction, we write
P = εI
Figure 22.20 A wire moving at right
angles to flux lines.
469884_22_AQA_A-Level_Physics_420-434.indd 427
where ε is the emf generated (V).
17/04/19 11:07 AM
You also know that work done is F × d, where d is the distance travelled in
the direction of the force F, so the rate of doing work is F × v, where F is
force (N) and v is velocity in the direction of the force (m s−1).
Each second, the change of flux linkage for a wire moving through a
magnetic field is BA, or Blv, where l is the length of the wire perpendicular
to the field. Substituting in P = EI gives
P = (Blv)I = (BIl)v
Since BIl is the force on a conductor in a field, BIl × v is consistent with the
rate of doing work, or power generated, F × v.
EXAMPLE
Induced emf between wing tips
Calculate the magnitude of the induced emf generated
between the wing tips of an aircraft flying at 220 m s−1
at a constant height. Assume that the average vertical
component of the Earth’s magnetic field is
4.1 × 10−5 T. The wing tips measure 30 m from tip to tip.
Answer
The area swept out by the wing tips each second is
lv = 30 m × 220 m s−1 = 6600 m2 s−1
∆φ
= Blv
∆t
-5
2 -1
= 4.1 × 10 T × 6600 m s
-1
= 0.27 Wb s
Since
ε=
∆φ
∆t
the induced emf is 0.27 V.
Then we obtain
●● Calculating an induced emf for a
rotating coil
w
q
B
22 MAGNETIC FLUX
coil
428
When a coil rotates in a magnetic field (Figure 22.21), an a.c. voltage is
induced in the coil.
To calculate the value of the induced emf at time t, you can use the
following equation for a plane coil in a uniform magnetic field so long as
the axis of rotation is at right angles to the field:
ε = BANω sin ωt
where ε is the induced emf (in V), B is the magnetic flux density (in T), A is the
E0
B
cross-sectional are a of the coil (in m2), N is the number of turns on the coil, ω
is the angular speed of the rotating coil (which can also be expressed as ω = 2πf,
where f is the frequency of rotation current) and t is the time (in s).
Since the maximum value of sin ωt is 1, the maximum induced emf is
εmax = BAN
Figure 22.21 Remember to specify the
axis of rotation for a coil in a magnetic
field, which is shown here as the black
dot.
469884_22_AQA_A-Level_Physics_420-434.indd 428
Figure 22.22 shows how the magnetic flux linkage and the induced emf
are linked:
d(Nϕ)
● At A, Nϕ is a maximum, gradient
is 0, so ε = 0.
dt
d(Nϕ)
● At B, Nϕ is 0, gradient
is a maximum and negative, so ε is a
dt
maximum positive.
d(Nϕ)
● At C, Nϕ is a minimum, gradient
is 0, so ε = 0.
dt
d(Nϕ)
● At D, Nϕ is 0, gradient
is a maximum and positive, so ε is a
dt
minimum negative.
17/04/19 11:07 AM
θ = 2πft
Nϕ = BAN cos 2πft
However,
ω = 2πf
C
a
b
0
−e max
−Nf
time
f
em
to give
a
A
age
link
Nϕ = BAN cos θ
b
flux
into the equation
emf or flux linkage
We can deduce an expression for
the induced emf by substituting
B
a
b
D
a
Figure 22.22
so
Nϕ = BAN cos ωt
But since
ε = −N
dφ
dt
we find
ε = BANω sin ωt
and the magnitude of the emf is
given by
ε = BANω sin ωt
You have come across a similar
situation in Chapter 16 (simple
harmonic motion).
EXAMPLE
Induced emf in a rotating
coil
A coil of 500 turns and crosssectional area 0.18 m2 rotates at a
frequency of 5 Hz in a uniform flux
density of strength 0.04 T.
1 Calculate the angular frequency
of the coil.
Answer
Angular frequency = 2πf
= 2π × 5 Hz = 10π rad s−1.
2 Calculate the maximum value of
induced emf for the coil.
Answer
Since the maximum value of
sin ωt is 1,
ℰ = BANω sin ωt = BANω
= 0.04 T × 0.18 m2 × 500 × 10π
= 36π V = 113 V
3 Calculate the value of the emf
when t = 0.20 s and 0.21 s if the
emf is zero when t = 0.
Answer
When t = 0.20 s,
ℰ = BANω sin ωt
= 0.04 T × 0.18 m2 × 500 ×
10π sin(10π × 0.20)
= 36π sin 2π
= 0 V
When t = 0.21 s,
ℰ = BANω sin ωt
= 0.04 T × 0.18 m2 × 500 ×
10π sin(10π × 0.21)
= 36π sin(2.1π)
= 11.2π V = 35 V
Calculating an induced emf for a rotating coil
+Nf
+e max
b
MATHS BOX
TEST YOURSELF
10 A
wire of length 15 cm is moved perpendicular to
a magnetic field of flux density 1.2 T. If the wire
moves at a speed of 5 cm s−1, calculate the induced
emf in the wire.
11 A wire of length 8.0 cm, and negligible crosssectional area, is dropped through a uniform
magnetic field of strength 5.0 mT so that it cuts
the flux lines.
a) The wire is dropped horizontally. Explain why the
emf induced in the wire increases as it falls.
469884_22_AQA_A-Level_Physics_420-434.indd 429
b) Calculate the induced emf when the speed of
the wire is 3.2 m s−1.
12 A dynamo has a coil of wire of 800 turns. When it
is used, the coil spins three times a second in a
region of uniform flux density 2.4 T.
a) Calculate the angular frequency of the coil.
b) The radius of the coil is 5 mm. Calculate the
maximum value of the induced emf
c) The emf is given by ε = BANω sin ωt. Calculate
the emf at time 0.36 s.
429
17/04/19 11:07 AM
Practice questions
1 A coil rotates in a plane perpendicular to flux lines in a magnetic field.
The flux linkage and induced emf vary during the cycle. Which one of
the following is always true?
A When the flux linkage is a maximum, the induced emf has a
maximum value.
B When the flux linkage is zero, the induced emf is zero.
C When the flux linkage is a maximum, the induced emf is zero.
D When the flux linkage is increasing, the induced emf is increasing.
2 The unit of magnetic flux is
A weber
C volt metre2
B weber-turns
D tesla metre
3 A metal sheet is pulled through a magnetic field, with its plane
perpendicular to the flux lines. Once the sheet is moving at a
steady speed, the force needed to pull the sheet at a constant speed
A increases
C is zero
B decreases
D is constant
22 MAGNETIC FLUX
4 A coil of wire is moved at right angles into,
through and out of a uniform magnetic field at
a steady speed. Which diagram in Figure 22.23
shows how the induced emf varies in the coil as it
enters, moves through and leaves the field?
430
5 A large square coil of insulated copper placed in a
storeroom has 50 turns. Each of its sides measures
80 cm. The coil is leaning at 45° to the vertical
against a wall. The Earth’s vertical magnetic
flux density, B, at that point is 50 µT. Calculate the
magnetic flux linkage of the coil.
A 1.41 mWb
C 1.13 kWb
B 1.13 mWb
D 1.13 Wb
(a)
t
(b)
t
(c)
t
(d)
enters field
t
leaves field
Figure 22.23
Nf
6 Figure 22.24 shows how the flux linkage, Nϕ, changes when a
coil moves into a magnetic field.
The induced emf in the coil
A increases until t1 and then is constant between t1 and t2
time
0
t1
t2
t3
Figure 22.24
B is constant between t1 and t2 and then decreases to zero at t3
C decreases and then is zero between t1 and t2
D is zero between t1 and t2 and is constant between t2 and t3
469884_22_AQA_A-Level_Physics_420-434.indd 430
17/04/19 11:07 AM
Practice questions
7 A small magnet is dropped through a narrow copper tube and then
through a plastic tube of the same diameter and length. Which one of
the following statements is true?
A The magnet falls at the same speed in both tubes.
B The magnet falls slower in the copper tube because copper
is magnetic.
C The magnet falls slower in the copper tube because of eddy currents
in the copper tube.
D The magnet falls faster in the copper tube because of eddy currents in
the magnet.
8 A coil of 100 turns has a cross-sectional area of 3.5 × 10−3 m2. It is
placed in a uniform magnetic field of flux density 4.9 mT, making
an angle of 40° to the flux lines (Figure 22.25).
The change in flux linkage when the coil is rotated
anticlockwise until θ = 90° is:
A an increase of 0.4 mWb turns
coil
axle
q
magnetic
field B
Figure 22.25
B a decrease of 0.4 mWb turns
C an increase of 0.6 mWb turns
D a decrease of 0.6 mWb turns
9 A dynamo spins with its axis perpendicular to the flux lines in a
magnetic field. The period of rotation is 0.01 s. If the period doubles,
which of the following changes will occur?
A The maximum emf and number of cycles per second
will double.
B The maximum emf and number of cycles per second
will halve.
C The maximum emf will double and the number of cycles
per second will halve.
D The maximum emf will halve and the number of cycles
per second will double.
10 Calculate the time taken for a search coil to be pulled out of a magnetic
field if the maximum emf generated is 0.6 V. The search coil has an area
0.001 m2 and 2000 turns and is perpendicular to the magnetic flux. The
magnetic flux density is 400 mT.
A 3.0 s
C 1.3 s
B 0.48 s
D 800 s
11 a) Describe the function of a simple ac generator.
431
(2)
A generator with 600 turns and a cross-sectional area of
3.0 × 10−3 m2 is placed so it can spin in a horizontal magnetic
field of flux density 0.049 T. The coil spins about a vertical axis.
b) Calculate the maximum magnetic flux linkage for the coil.(3)
469884_22_AQA_A-Level_Physics_420-434.indd 431
17/04/19 11:07 AM
Figure 22.26 shows how the magnitude of the
flux linkage varies as the coil turns.
d) Calculate the maximum emf generated when
the coil spins in the field.(4)
e) Use the graph to state when the emf has its
maximum value.(1)
f) Explain how the maximum emf generated
changes when the coil spins at half the speed
in the field.(2)
12 a) State the SI unit of magnetic flux.
8
6
f/10–2 Wb turns
c) Explain why the flux linkage changes in this
way as the coil turns.(3)
10
4
2
0
–2
0
0.01
0.02
0.03
t/s
–4
–6
–8
–10
Figure 22.26
(1)
Two wire coils A and B are placed next to each
other (Figure 22.27). Coil A is connected to a
switch and a battery. Coil B forms a circuit with
a millivoltmeter.
soft iron core
A
b) Describe and explain what is seen on the
millivoltmeter when circuit A is switched on
and off.(5)
c) Explain how the readings would change if
circuit A contained a second cell.(2)
B
S
mV
Figure 22.27
13 a) Explain the difference between magnetic
flux and magnetic flux linkage.
(1)
22 MAGNETIC FLUX
b) A straight wire of length 12.4 cm is held horizontally, then
released so that it falls through magnetic flux density of 0.3 mT. If the
emf generated across the wire at time t is 14.0 µV, calculate the speed
of the wire at this time.
(3)
432
14 A metal rod of length 2.3 m is pivoted at one end. It is moved in a
circle making a complete circuit in 4 s.
a) Calculate the area swept out by the rod in 1 s.(1)
b) If the rod is orientated so that it is always perpendicular to a magnetic
field of strength 1.2 T, calculate the maximum emf generated by this
(2)
movement.
15 A coil of 600 turns rotates at a frequency of 4 Hz perpendicular
to a field of flux density 30 mT. The area of the coil is 15 cm2.
a) Calculate the magnitude of the maximum flux linkage.
(4)
b) Calculate the maximum induced emf
(2)
c) If the flux linkage has its maximum value at time t = 0, calculate
when the induced emf first has its maximum value.
(2)
469884_22_AQA_A-Level_Physics_420-434.indd 432
17/04/19 11:07 AM
Practice questions
16 Figure 22.28 shows a way to measure the flow of oil through a pipeline.
A small turbine is placed in the pipe so that the oil flow turns the blades
around. Some magnets have been placed in the rim of the turbine so that
they move past a solenoid.
leads to an oscilloscope
solenoid
oil
magnets in the
rim of the wheel
Figure 22.28
turbine blades
These moving magnets induce a voltage in the solenoid, which can
be measured using an oscilloscope. Figure 22.29 shows the trace
obtained. The faster the turbine rotates, the larger is the voltage
induced in the solenoid. By measuring this voltage, an engineer can
tell at what rate the oil is flowing.
a) The poles on the magnet rim are arranged alternately with a
north then a south pole facing outwards. Use this fact to explain
the oscilloscope trace.(1)
b) Sketch the trace on the oscilloscope for the following
(separate) changes.
i)
The number of turns on the solenoid is made
Figure 22.29
Hodder Science: Physics Matters 3rd Edition
1.5 times larger.(1)
Fig J07.04
Mac/eps/Illustrator 8.0/4-col & B/W s/s
ii) The flow of oil is increased so that the turbine
Text: Helvetica 8/9
HODDER & STOUGHTON EDUCATIONAL
rotates twice as quickly.(1)
c) i)
Studio: Peters & Zabransky
Use the trace in Figure 22.29 to show that there is a time
of 0.08 s between each magnet passing the solenoid.(2)
ii) Calculate how long it takes for the turbine to rotate once.
(1)
iii) Calculate how many times the turbine rotates each second.
(1)
Stretch and challenge
17 An electrodynamic tether is a cable used to control the motion of a
satellite as it is taken into orbit. A tether of 20 km in length is used
to connect a satellite to a space shuttle. The strength of the Earth’s
magnetic field at this altitude has a flux density of 50 µT.
433
a) If the satellite is travelling at an orbital speed of 8.0 km s−1
perpendicular to the Earth’s field, calculate the emf generated
across the tether.
b) Explain why this figure is likely to be inaccurate, and suggest a
more accurate value.
469884_22_AQA_A-Level_Physics_420-434.indd 433
17/04/19 11:07 AM
18 A copper ring falls through a region of horizontal magnetic field
of flux density B (Figure 22.30). Describe how the flux linkage,
the induced emf and the current in the ring change as it enters the
field, passes through and leaves it.
falling
A
B
C
magnetic field going
into page
D
E
Figure 22.30 A ring falling through a magnetic field.
19 Figure 22.31 shows three coils connected in series to a data logger.
A magnet is dropped through the three coils.
The graph in Figure 22.32 shows the voltage measured by the data
logger as the magnet falls.
S
N
Explain the shape of the graph by commenting on the height of
the peaks, the width of the peaks, the gaps between the peaks and
the direction of the peaks.
434
voltage/mv
22 MAGNETIC FLUX
Comment on the area under the peaks.
Figure 22.32
469884_22_AQA_A-Level_Physics_420-434.indd 434
data logger
Figure 22.31
time/ms
17/04/19 11:07 AM
23
Alternating currents
and transformers
PRIOR KNOWLEDGE
Before you start, make sure that you are confident in your knowledge and
understanding of the following points:
●
●
Electric current, I, is the rate of flow of charge, ∆Q , measured in
∆t
amperes (A).
Potential difference (voltage), V, is the amount of electrical work done
per unit charge, W . Potential difference is measured in volts (V).
Q
●
●
●
●
Electric current, I, potential difference, V, and resistance, R, in a circuit
are related to each other through the equation V = IR.
V2
Electrical power is the rate of doing electrical work, P = VI = I 2R =
R
The frequency, f, of a waveform is the number of complete waves per
second and is measured in hertz (Hz).
The time period, T, of a waveform, measured in seconds (s), is related to
the frequency of the waveform, f, by f
●
●
=
1
.
T
The emf induced in a coil is ε = −N d ϕ /dt, where ε is the induced emf
(V), N is the number of turns on the coil, ϕ is the magnetic flux (Wb)
and t is time (s).
Eddy currents are generated in metal sheets by changes in magnetic
flux. Eddy currents transfer electrical energy to heat energy.
TEST YOURSELF ON PRIOR KNOWLEDGE
1 An ac voltage has a time period of 0.004 s. What is the frequency of the
voltage supply?
2 A current of 13 A flows through an electric fire element, which has a
resistance of 14 Ω. Calculate the power dissipated by the fire in kW.
3 A bar magnet, which has poles measuring 1.5 cm × 1.5 cm, is
pulled out of a coil in a time of 0.2 s. The coil has 10 000 turns and a
resistance of 50 Ω. The average current flowing in the coil while the
magnet is moving is about 35 mA. Estimate the flux density near the
pole of the magnet.
435
●● Mains electricity
Electricity is generated and transmitted around the country in the form of
alternating currents (ac) and voltages. These are used because they can be
transformed to high voltages and very low currents in order to minimise the
thermal energy lost as the current travels through the wires of the National
Grid (Figure 23.1). Only about 2–3% of the electrical energy from the
generators is lost as heat, saving energy, carbon emissions and money.
469884_23_AQA_A-Level_Physics_435-448.indd 435
17/04/19 11:10 AM
Alternating current is delivered by the National Grid to consumers as a
sinusoidally varying supply with a frequency of 50 Hz, and a range of different
voltages, depending on the customer. Household mains has a nominal voltage
of 230 V, although this value varies throughout the day depending on the
demand and supply of electricity. The maximum current that can be drawn
by a single domestic supply is about 65 A. The electrical socket ring main in
your house has a maximum current of 13 A protected by a fuse or a circuit
breaker. However, it is only lamps, heaters, cookers and devices such as
vacuum cleaners and mowers, with large electric motors, that use ac directly
off the mains. Most other devices work at much lower voltages and as direct
currents (dc). This means that devices such as televisions, computers and
games consoles all require a separate (or built-in) step-down transformer that
converts 230 V ac into (for example) 12 V dc.
●● Alternating current and voltage
23 ALTERNATING CURRENTS AND TRANSFORMERS
Figure 23.1 Electricity supply pylons –
part of the National Grid.
Peak voltage is half the peak-to-peak
voltage, and is equivalent to the amplitude
of the waveform.
Alternating currents and voltages move in one direction for half of their
cycle and in the opposite direction for the other half. Mains electricity
comes in a sinusoidally changing pattern, with the magnitude of the
current or the voltage continuously varying between maximum positive and
negative values. The peak value of the voltage (or potential difference) is
the maximum value in either the positive or negative direction, with respect
to zero. The peak-to-peak value of the voltage is measured from one peak
in the positive direction to the other peak (called a trough) in the negative
direction (see Figure 23.2).
The peak voltage, V0, of the alternating waveform is half the peak-to-peak
voltage, and is equivalent to the amplitude of the waveform. For a given
component such as a resistor, the peak current I0 and peak voltage V0 are
related to each other through the equation
V0 = I0R
peak-topeak
voltage, 2V0
Comparing ac and dc equivalents
As alternating currents and voltages vary continuously, what value is used in
calculations that gives the same effect as the equivalent direct current or voltage?
peak
voltage, V0
The average values cannot be used, because the average values are both
zero – there is the same amount of signal above zero as there is below zero.
The values chosen are the root mean square (r.m.s.) voltage and current.
When multiplied together, these quantities produce the same power in a
resistor as would be produced by the same dc values. This can be expressed
more easily in the form of the equation:
Figure 23.2 An alternating electrical
waveform.
P = VdcIdc = VrmsIrms
436
A sinusoidal alternating voltage, V, varying with time, t, can be represented
by the equation
sin(2πft)
V = V0 sin(2πft)
+1
0
3T/2
T/2
T
2T
–1
Figure 23.3 Alternating voltage.
469884_23_AQA_A-Level_Physics_435-448.indd 436
t
where V0 is the peak voltage, and f is the frequency of the supply. This is
shown on the graph in Figure 23.3.
If this voltage is applied across a fixed resistor, R, then the power dissipated
by the resistor is equal to
17/04/19 11:10 AM
sin2(2w ft)
1
0.5
0
V 2 V02 sin 2 (2π ft )
=
R
R
The average power is the power we need to compare to an equivalent
T/2
T
3T/2
Figure 23.4 The sin2 graph.
2T
t
2
constant dc value, but as VR0 is constant in this equation, we only need to
find the average value of sin2(2πft). This can be done by analysing the graph
of the function in Figure 23.4.
It can be seen from Figure 23.4 that the average value of sin2(2πft) = 0.5, so
P=
V02
R
1
2
Alternating current and voltage
P=
This will be the same power as that for an equivalent constant dc value of
voltage, Vdc:
P=
V02 Vdc2
=
R
R
1
2
Hence we obtain
V02
= Vdc2
2
V02
2
Vdc =
and
Vdc = Vrms =
V0
2
As the alternating current varies in phase with the voltage, using a similar
reasoning yields
I rms =
I0
2
As a result, the mean alternating power, Pmean, which is equivalent to the dc
power, is given by
Pmean = VrmsIrms
and the peak alternating power, Ppeak, is given by
Ppeak = V0I0
Hence, finally we have
Pmean
V
I
= 0 × 0
2
2
VI
= 00
2
Ppeak
=
2
437
In other words, the mean power dissipated through a fixed resistor by an
alternating current and voltage is equal to half the peak power dissipated.
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EXAMPLE
Mains power
Answer
The given UK mains voltage is 230 V ac – this is an r.m.s.
value. The largest power that can usually be drawn from
a local grid is 15 kW – this is the mean value, equivalent
to a dc supply.
= 325.3 V = 330 V (2 s.f.)
Answer
Use the equation for Pmean from the text:
Pmean = VrmsIrms
P
= mean
Vrms
15 kW
=
230 V
I0 = Irms × 2
= 65.2 A × 2
= 92.2 A = 92 A (2 s.f.)
Ppeak = V0I0
= 325.3 V × 92.2 A
= 29993 W = 30 kW (2 s.f.)
= 65.2 A = 65 A (2 s.f.)
2 What are the peak power, peak voltage and peak
current?
23 ALTERNATING CURRENTS AND TRANSFORMERS
V0 = Vrms × 2
= 230 V × 2
1 What is the r.m.s. current?
Irms
Combine the equations for V0 and I0 from the text to
give the equation for Ppeak:
We can then use these values to calculate the peak-topeak values of p.d. and current:
Vpeak-to-peak = 2V0 = 650 V (2 s.f.)
Ipeak-to-peak = 2I0 = 180 A (2 s.f.)
TEST YOURSELF
1 What is meant by the 'root mean square' (r.m.s.)
voltage?
2 An ac power supply delivers Vrms = 6.0 V to a fixed
resistor of resistance R = 2.5 Ω. Calculate:
a) the r.m.s. current through the resistor
b) the mean power delivered to the resistor
c) the peak power delivered to the resistor.
3 In the USA, the nominal r.m.s. voltage is 120 V.
If the mean power delivered to a US domestic
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469884_23_AQA_A-Level_Physics_435-448.indd 438
house is the same as that to a UK house (15 kW),
calculate:
a) the r.m.s. current delivered
b) the peak voltage delivered
c) the peak power delivered.
4 The ac motor for a (UK) mains washing machine
works with a peak power of 400 W. Calculate:
a) the mean power drawn by the motor
b) the r.m.s. current through the motor.
●● Using an oscilloscope to display waveforms
Analysing alternating waveforms is best done by displaying the waveform
using an oscilloscope. Oscilloscopes are a form of visual, calibrated voltmeter,
where the operator is able to alter how the waveform is displayed. First, it is
possible to control the time taken for the signal to move across the screen by
adjusting the timebase – often labelled time/div. Secondly, the amplitude of
the signal displayed on the calibrated screen can be controlled by adjusting
the y-sensitivity – this is also known as the vertical sensitivity, y-gain or simply
volts/div. The timebase (in seconds) provides a scale for the x-axis of the
screen and indicates the time taken for the signal to move horizontally across
one square on the screen. Using the square grid on the screen to measure
the number of horizontal squares between two successive peaks (or troughs)
allows the period of the waveform to be determined and hence the frequency.
The oscilloscope also makes it very easy to measure the peak-to-peak value of
the wave by counting vertical squares and then using the y-sensitivity (usually
calibrated in volts, millivolts or microvolts) to apply a scale.
17/04/19 11:10 AM
calibrated screen
timebase
Oscilloscope with a dc signal
The oscilloscope in Figure 23.5 is displaying a dc signal (from a
battery for example). Describe the signal.
Answer
The timebase is set to 20 ms/div, so, because there are 10 horizontal
divisions on the screen grid, the signal takes 200 ms (0.2 s) to travel
from one side of the screen to the other. The y-sensitivity is set to
1 V/div, and the signal is 2.4 divisions vertically up from the centre
line. This makes the voltage of the signal 2.4 V.
y -sensitivitiy
Figure 23.5 Oscilloscope displaying a dc signal.
EXAMPLE
Oscilloscope with an ac waveform
Using an oscilloscope to display waveforms
EXAMPLE
The oscilloscope in Figure 23.6 illustrates an ac waveform
from a signal generator. Describe the waveform.
Answer
In this case the timebase and y-sensitivities have not changed.
There are five horizontal divisions between the two successive
peaks or troughs, and this corresponds to a time period of
100 ms. The frequency of the signal is therefore
1
1
frequency =
=
= 10 Hz
timeperiod 100 × 10−3 s
The peak-to-peak voltage measured from the bottom of a trough
to the top of a peak on the screen is six divisions, corresponding
to 6 V. This corresponds to a peak voltage of 3 V and
Vrms =
ACTIVITY
Virtual oscilloscopes and signal generators
There are many excellent virtual oscilloscope and
signal generator simulations and apps available online. Using the keywords in italics as search terms in
a search engine will take you to a range of different
versions, although they all operate using the same
principles as the real thing illustrated in Figures 23.5
and 23.6.
Some simulations are just oscilloscopes, and these
rely on an external signal being generated and fed
through the computer’s sound card or microphone.
Be careful when doing this – use an external device
that does not exceed the sound card’s input voltage
(a tablet is ideal for this).
Other simulations have a built-in signal generator
469884_23_AQA_A-Level_Physics_435-448.indd 439
3V
2
Figure 23.6 Oscilloscope displaying an ac waveform.
= 2.1V.
that allows you to generate an alternating signal
directly for display on the oscilloscope screen.
Use one of these simulations to familiarise yourself
with the controls of the oscilloscope, so that when you
come to use the real thing you will be able to analyse
alternating waveforms and extract the relevant key
information, such as the frequency and the peakto-peak values. You could also use your simulation
to analyse the voltage signals coming off different
music tracks, although the rapidly varying voltages
may be tricky to measure unless the simulation has
a ‘Hold’ or ‘Freeze’ function. Alternatively you could
speak or sing directly into the sound card and use
the oscilloscope and your voice to analyse some
alternating signals.
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●● Transformers
Many devices use transformers to increase or reduce the
voltage of an alternating voltage supply. Transformers
are used by the National Grid to increase the voltage
generated in power stations up to 400 000 V, so that
energy can be saved as electricity is transmitted around
the country. Then transformers are used to reduce
this high voltage for safe use in the home. Figure 23.7
shows the transformers used to step up the voltage in a
power station.
Structure of a transformer
The structure of a transformer is simple. It consists of
two coils of wire linked by a soft iron core as shown in
Figure 23.8.
23 ALTERNATING CURRENTS AND TRANSFORMERS
Figure 23.7 Step-up transformers increase the
generated voltage from 25 000 V to 400 000 V.
An alternating current in the primary coil creates a
changing magnetic field in the core, which is made of a
soft magnetic material such as iron. The secondary coil is also wound round
the core. As the magnetic flux in the core changes, the magnetic flux linkage
to the secondary coil changes and an emf is induced in the secondary coil.
Because transformers use electromagnetic induction, they only work with
an ac supply.
The turns rule
For an ideal transformer, with no power losses, the ratio of the turns on
each coil equals the ratio of the primary and secondary voltages. That is
ac input
ac output
Vs N s
=
Vp N p
where Vs = secondary voltage (V), Vp = primary voltage (V), Ns = turns on
the secondary coil and Np = turns on the primary coil.
magnetic field in core
Figure 23.8 The structure of a
transformer.
A step-up transformer is a transformer that increases voltage, so Ns/Np
is more than 1. A step-down transformer is a transformer that decreases
voltage, so Ns/Np is less than 1. Figure 23.9 shows a simple circuit diagram
for a transformer, with the symbols for an ac supply, a step-up transformer
and a bulb.
You already know that ℰ depends on the number of turns on the coil. The
induced emf is given by
ε = −N
440
dφ
dt
The rate of flux change
from AQA paper Jan 2011
Figure 23.9
469884_23_AQA_A-Level_Physics_435-448.indd 440
dϕ
dt
in the core of the transformer is the
same for both coils, but the number of turns N is different, so
the induced emf is different in the secondary coil, and depends
on the ratio of Ns to Np.
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VpIp = VsIs
where Vs = secondary voltage (V), Vp = primary voltage (V), Is = current in
the secondary coil (A) and Ip = current in the primary coil (A).
Transformers
Transformers cannot increase the power output of the supply. In an ideal
transformer, with no power losses, the power input to the transformer must
be equal to the power output. Therefore we can write the following equation:
This means that a transformer that reduces the output voltage compared to
the input voltage has a larger current in the secondary coil compared to the
primary coil.
EXAMPLE
Step-down transformer
A step-down transformer has 2500 turns on the primary
coil. It transforms mains voltage, 230 V ac, into a 12 V ac
supply.
2 When the current in the secondary coil is 1.5 A, what
is the current in the primary coil? Assume that the
transformer is 100% efficient.
1 Calculate the number of turns on the secondary
coil.
Answer
Rearranging the equation VpIp = Vs Is gives
Answer
V
Rearranging the equation s
Vp
Vs
N s = Np ×
Vp
=
Ns
Np
Ip =
gives
Vs
× Is
Vp
12 V
× 1.5 A
230 V
= 0.078 A
=
12 V
230 V
= 130 turns
= 2500 ×
Transformer efficiency
Transformers can be very efficient, but they are never 100% efficient. The
efficiency of a transformer is calculated using this equation:
efficiency =
VsIs
VpIp
where Vs = secondary voltage (V), Vp = primary voltage (V), Is = current in
the secondary coil (A) and Ip = current in the primary coil (A).
EXAMPLE
Efficiency of a transformer
The efficiency of a mains transformer is 90%. The mains
supply is 230 V ac and the output of the transformer
is 12 V ac. Calculate the current in the secondary coil
when the current in the primary coil is 0.5 A.
Answer
Use the equation for efficiency and substitute the
values known:
469884_23_AQA_A-Level_Physics_435-448.indd 441
efficiency =
VsIs
VpIp
441
12 V × Is
230 V × 0.5 A
Rearranging to make Is the subject gives
0.9 =
0.9 × 230 V × 0.5 A
12 V
= 8.6 A
Is =
17/04/19 11:11 AM
Energy losses in transformers
Energy losses in transformers occur because of the following effects:
●
●
●
23 ALTERNATING CURRENTS AND TRANSFORMERS
●
Heat is produced in the copper wires of the primary coil and secondary
coil when a current flows. Using low-resistance wire reduces these losses.
This is particularly important for the secondary coil of a step-down
transformer, because the current is larger in the secondary coil compared
to the primary coil. A thicker wire is often used in the secondary coil of
a step-down transformer.
Some magnetic flux produced by the primary coil does not pass through
the iron core, which means the flux linkage to the secondary coil is not
100%. This can be reduced by designing the transformer with coils close to
each other or wound on top of each other, which improves the flux linkage.
There is an effect called hysteresis. Some energy is lost as heat every time
the direction of the magnetic field changes because energy is needed
to realign the magnetic domains in the core. This is reduced by using a
soft magnetic material such as iron, rather than steel which needs more
energy to demagnetise and magnetise.
Eddy currents form in the iron core due to the continuously changing flux.
These currents heat the core up, increasing energy losses. Eddy currents
are reduced by making the core using laminated sheets separated by thin
layers of insulation. Eddy currents are discussed in more detail below.
Eddy currents in transformers
solid core
laminated core
with no laminations
high eddy currents
with laminations
low eddy currents
Figure 23.10 Eddy currents are reduced
by building the core from thin, insulated
layers of iron.
442
In Chapter 22, you learnt that eddy currents are created in metal sheets when
there is a change in magnetic flux. In the core of a transformer, the alternating
supply creates alternating magnetic flux changes, and these create eddy currents.
Eddy currents flow in loops, in a direction that opposes the magnetic flux
changes that cause them. The result is that eddy currents in the iron core will
reduce the emf induced in the secondary coil. In a core made from solid iron,
eddy currents could become large enough to melt the core, because the resistance
of the iron core is very low. To prevent these problems, the core is built from very
thin laminations, or layers, of metal (Figure 23.10). The eddy currents are smaller
when there are thin laminations, because the induced voltage drives the current
round longer paths – so the resistance to flow increases. The laminations are
insulated from each other, for example using layers of insulating varnish.
●● Transmission of electrical power
Energy losses due to the heating of transmission lines in the National
Grid can be very significant because electrical energy can be transmitted
very long distances from the power stations to the end users. Electricity is
transmitted throughout the UK (Figure 23.11), and also between European
countries – for example, between the UK, France and the Netherlands.
Transformers are used to step up the voltage generated in power stations.
Since power transmitted is equal to the product V × I, stepping up the
voltage in transmission lines reduces the current. Smaller currents have
a smaller heating effect on the power lines, so reducing the current in
transmission lines reduces energy losses to the surroundings.
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Calculating power losses in transmission lines
Power losses in the National Grid total about 3% of demand, and mainly
occur in the generator transformers, overhead lines, underground cables and
grid supply transformers. Two-thirds of the losses in the National Grid occur
in the overhead lines of the transmission system. However, the percentage
losses in power lines in the distribution system are bigger than in transmission
lines because the voltage is stepped down, so currents in the power lines are
larger. Losses in the distribution system can reach as much as 15%.
Transmission of electrical power
Power stations generate electrical energy at a potential of about 25 kV. This
voltage is stepped up using transformers shortly after it leaves the power
station and is transmitted using transmission lines operating at 275 kV
and 400 kV. Overhead transmission lines are supported using the familiar
large steel pylons. Transformers in substations step down the voltage for
distribution of electricity to the end user. Distribution lines operate at
132 kV, with cables supported on smaller steel pylons. Wooden poles are
used to support power lines operating at 11 kV and 33 kV.
Power losses are calculated using P = I2R. Because the power losses are
proportional to the square of the current, doubling the current quadruples
the power losses. Power cables are made from aluminium supported by
steel cores, and the low resistance of these cables reduces losses in power
lines, since losses are proportional to R.
Step-down transformers in distribution systems are made more efficient
by using thicker wire in the secondary coil. The current is higher in the
secondary coil of step-down transformers, so I2R losses due to the heating
of the secondary coil can be significant. Reducing the resistance of the
secondary coil reduces I2R losses.
EXAMPLE
Transmission line
A power transmission line in a factory operates at 25 kV. The power input
to the cable is 750 kW.
1 Calculate the current in the transmission line.
Answer
Rearranging the equation P = VI for power gives
P 750 × 103 W
=
V
25 × 103 V
= 30 A
I=
2 The resistance of the cable is 40 Ω. Calculate the power supplied by the
cable.
443
Answer
Use the equation power supplied = input power − power losses
power losses = I2R = (30 A)2 × 40 Ω = 36 kW
power supplied = 750 kW − 36 kW = 714 kW (710 kW 2 s.f.)
3 Calculate the efficiency of the transmission line.
Use the equation for efficiency and substitute the values known:
efficiency =
469884_23_AQA_A-Level_Physics_435-448.indd 443
Pout 714 kW
=
× 100 = 95.2% (95% 2s.f.)
Pin 750 kW
17/04/19 11:11 AM
TEST YOURSELF
5 Explain why:
a) transformers do not work using dc
b) the iron core of a transformer is laminated
c) a thicker wire is used in the secondary coil of a
step-down transformer.
8 A transformer is 95% efficient. The transformer
uses mains voltage, 230 V ac, and the output
voltage is 6 V ac. Calculate the current in the
primary coil if the current in the secondary coil
is 4.8 A.
23 ALTERNATING CURRENTS AND TRANSFORMERS
6 a) C
alculate the turns ratio for a transformer that
steps up a 25 kV input to an output of 132 kV.
b) The transformer in part (a) is 90% efficient,
and it has a current of 40 A flowing in the
primary coil. Calculate the power output in the
secondary coil.
c) Calculate the current in the secondary coil.
7 A step-up transformer transforms the input
voltage, 12 V ac, into a 48 V ac supply.
a) If the primary coil has 200 turns, calculate the
number of turns on the secondary coil.
b) When the current in the primary coil is 2.4 A,
what is the current in the secondary coil?
Assume that the transformer is 100% efficient.
444
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17/04/19 11:11 AM
Practice questions
Practice questions
1 The r.m.s. voltage from a power supply with a peak voltage
of 6 V is:
A 3.0 V
C 4.2 V
B 0.12 V
D 0.85 V
se the information in Figure 23.12 about the voltage waveform
U
from an ac power supply to answer questions 2, 3 and 4.
2 The peak-to-peak voltage shown in
Figure 23.12 is:
A 128 V
80
60
C 90.5 V
40
B 64 V
A 128 V
C 90.5 V
B 64 V
D 45.3 V
voltage, V
D 45.3 V
3 The r.m.s. voltage of the signal shown
in Figure 23.12 is:
C 0.1 Hz
B 200 Hz
D 100 Hz
0
–20 0
–40
2
4
6
8
10
12
time/ms
–60
–80
4 The frequency of the ac power supply shown
in Figure 23.12 is:
A 0.2 Hz
20
Figure 23.12
5 Which of the waveforms in Figure 23.13 shows
a 4.24 V r.m.s. voltage?
y-gain: 2 V/div
y-gain: 1 V/div
A
B
y-gain: 2 V/div
y-gain: 1 V/div
445
C
D
Figure 23.13
469884_23_AQA_A-Level_Physics_435-448.indd 445
17/04/19 11:11 AM
6 A transformer has 1500 turns on the primary coil and 600 turns on the
secondary coil. The transformer uses 230 V ac mains supply, and draws
a current of 0.4 A in normal use. If the efficiency of the transformer is
85%, what is the current in the secondary coil, when the secondary
voltage is 92 V?
A 0.65 A
C 0.85 A
B 1.00 A
D 0.14 A
7 The primary coil of a step-down transformer uses an ac mains supply.
The secondary coil is connected to a phone charger. Which line A–D in
the table correctly describes the potential difference and current in the
secondary coil in relation to the primary coil?
Secondary current/primary current
Secondary p.d./primary p.d.
A
>1
>1
B
>1
<1
C
<1
>1
D
<1
<1
23 ALTERNATING CURRENTS AND TRANSFORMERS
8 Which of these does not reduce the efficiency of a transformer?
446
A heating of the primary and secondary coils
B eddy currents in the iron core
C leakage of magnetic flux from the primary coil
D insulation between the primary and secondary coils
9 The National Grid transmits electrical power from power stations using
transmission lines. Substations link transmission lines to distribution
systems that distribute electrical power to the final users. Which line
A–D in the table correctly describes the arrangement of step-up and
step-down transformers in the National Grid?
Transformers in power stations
Transformers in substations
A
Step-up
Step-down
B
Step-up
Step-up
C
Step-down
Step-down
D
Step-down
Step-up
10 A cable, 4 cm2 in cross-section and of resistivity 5 × 10−8 Ω m,
carries a current of 2500 A. The power loss per km is:
A 391 W
C 391 kW
B 781 W
D 781 kW
11 An alternating voltage from a signal generator is displayed
on an oscilloscope screen with the following settings: timebase,
25 ms per division; and y-sensitivity, 3 V per division. The
waveform of the voltage signal is shown in Figure 23.14.
Figure 23.14
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17/04/19 11:11 AM
a) the peak-to-peak voltage of the signal
(1)
b) the r.m.s. voltage
(1)
c) the time period of the signal
(1)
d) the frequency of the signal.
(2)
Practice questions
Calculate:
Make a sketch copy of the trace on the oscilloscope screen.
e) On your copy, sketch the dc voltage signal that would produce
the same power dissipation in a resistor, of resistance R,
equivalent to that produced by the signal generator.
(2)
12 Domestic electricity in the USA is delivered with a peak value of
170 V and a frequency of 60 Hz.
a) State what is meant by the ‘peak value’ and show how this
value is related to the root mean square (r.m.s.) value.
(2)
b) Calculate the r.m.s. voltage.
(2)
c) A light bulb is connected to the mains supply in the USA and
draws an r.m.s. current of 0.50 A. Calculate the mean power
of the bulb.
(1)
d) Using a suitable set of axes, sketch the voltage waveform
of mains electricity in the USA. Include suitable numerical
scales on your sketch graph.
(4)
13 A student is using an oscilloscope to measure the voltage from
a range of different voltage sources. She connects the voltage
sources to the y-input of the oscilloscope. The y-gain of the
oscilloscope is set to 0.5 V/div, and the timebase is set to
4 ms/div. The screen of the oscilloscope is divided into
a 10 × 10 grid as shown in Figure 23.15.
a) Copy the diagram twice and draw sketches of the
oscilloscope screen illustrating the voltage waveforms
of the following sources:
i) 1.5 V cell (battery)
(1)
ii) UK mains low-voltage ac power supply, 2 V (peak).
(2)
b) Calculate the r.m.s. voltage of the ac power supply.
14 a) Describe how you would use an oscilloscope to compare the
output from an ac, 12 Vrms, 15 Hz wind turbine and a 12 V dc
car battery. You need to consider the quality of your written
communication in your answer.
Figure 23.15
(2)
447
(6)
b) The car battery is connected to a car headlight bulb and the
current is measured to be 2.5 A. Calculate the power of