Physics
for
CAPE®
Unit
1
Physics
for
Terry
Joyce
CAPE®
David
Crichlow
Dwight
Carlos
de
Freitas
Hunte
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1
3
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Acknowledgements
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all
at
of
Contents
Introduction
Module
1
Chapter 1
1.
1
1
Chapter 6
Work,
6.
1
Work
and
6.2
Energy
SI quantities
Dimension
1.3
Scalar
and
units
42
and
unit
analysis
Measurements
1
2.2
Measurements
2
2.3
Errors
2.4
Uncertainties
Circular
Chapter 3
Kinematics
3.2
Equations of
3.3
Projectile
motion
7
.
1
Motion
in
a
circle
7
.2
Examples of
circular
motion
1
50
7
.3
Examples of
circular
motion
2
52
Gravitation
8.
1
Gravitational field
8.2
Gravitational
54
12
measurements
potential
and
satellites
56
14
Revision questions
3
58
Module
exam questions
60
1
Practice
16
motion
20
motion
22
Module
2 Oscillations
Chapter 9
and
waves
Oscillations
9.
1
Free oscillations
9.2
Amplitude,
9.3
A
62
Dynamics
Dynamics
1
26
4.2
Dynamics
2
28
4.3
Collisions
simple
period
and frequency
pendulum
mass–spring
and
64
a
system
66
32
9.4
Revision questions
1
Resonance
68
34
Chapter 10
Refraction
Forces
10.
1
Archimedes’
principle, friction
Refraction
72
and
Revision questions
terminal velocity
Polygon of forces
4
74
36
and
gravity
Torque
48
10
4.
1
Chapter 5
46
Kinematics
3.
1
Chapter 4
2
8
measurements
in
44
6
Chapter 8
in
power
4
Measurements
2.
1
and
2
and vector quantities
Chapter 2
5.3
power
energy
Revision questions
Physical quantities and units
1.2
5.2
and
Mechanics
Chapter 7
5.
1
energy
centre of
Chapter 11
Waves
38
and
moment
11.
1
Waves
76
11.2
Transverse
11.3
Superposition
11.4
Interference
11.5
Interference
40
and
longitudinal
waves
and diffraction
78
82
86
experiments
88
iii
Contents
11.6
Measuring the
wavelength of
11.7
Stationary
11.8
Sound
11.9
Electromagnetic
light
waves
90
92
waves
Chapter 17
The first
law of
thermodynamics
94
waves
Revision questions
17
.
1
The first
17
.2
Molar
law of thermodynamics
5
heat
capacities
and p–V
98
diagrams
144
Revision questions
Chapter 12
12.
1
The
The
physics of
physics of
Chapter 13
The
100
physics of
Chapter 18
Thermal
18.
1
Thermal
18.2
Measuring thermal
102
18.3
Radiation
13.2
The
104
18.4
Global
eye
Practice
energy transfer
conduction
Lenses
2
146
and
convection
148
sight
13.
1
Module
8
hearing
hearing
Revision questions
140
96
6
exam questions
warming
and
thermal
108
Revision questions
applications of
energy transfer
Chapter 19
152
154
106
Module 3 Thermal and mechanical
conductivity
158
9
Phases of
160
matter
properties of matter
19.
1
Chapter 14
Temperature
14.2
Thermometers
kinetic
and
Temperature
14.
1
A
and temperature
19.2
Pressure
19.3
Hooke’s
Thermal
164
law
and the
19.5
energy
modulus
and
specific
Experiment to determine the
Elastic
modulus
and
plastic deformation
Revision questions
Module
methods for
heat
Measuring
by
using
a
capacities
Latent
15.5
Measuring
specific
heat
method of
capacities
mixing
heat
3
specific
latent
heats
iv
and
Analysis
and
interpretation
Analysis
and
interpretation:
7
176
gases
178
physical
constants
180
and the
Glossary
gases
kinetic theory of
exam questions
132
Index
The
Analysis
130
kinetic theory
16.2
174
124
List of
Ideal
exam questions
interpretation
Practice
Ideal
Practice
126
Revision questions
16.
1
172
120
20.
1
15.4
Chapter 16
10
measuring
Chapter 20
15.3
170
118
Electrical
specific
168
heat
capacity
15.2
166
properties
Young
Internal
162
114
matter
15.
1
liquids
scales 110
19.4
of
solids,
gases
Young
Chapter 15
model for
134
gases
136
181
185
Introduction
This
Study
Guide
has
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developed
exclusively
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On
ᔢ
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ourself
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in
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areas.
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work
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est
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our
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Physics.
1
1
Physical
1.
1
SI
quantities
Learning outcomes
On
completion
should
be
able
of
this
quantities
and
recall
the
section,
you
A
physical
can
to:
SI
base
units
units
Physical quantities
be
quantities
quantity
measured
product
ᔢ
and
of
a
is
with
the
an
numerical
property
of
instrument.
magnitude
an
It
and
a
object
is
or
a
typically
unit.
For
phenomenon
expressed
example,
as
that
the
suppose
and
a
student
measures
his
mass
and
records
it
as
55 kg.
The
numerical
units
magnitude
ᔢ
determine
the
units
of
in
this
case
is
55.
The
unit
of
mass
is
the
kilogram
(kg).
derived
quantities
SI
ᔢ
recall
commonly
ᔢ
convert
ᔢ
define
used
base quantities
Scientists
units
the
mole
Avogadro’s
Le
and
and
recall
worldwide
Système
or
SI
In
this
have
agreed
Internationale
units.
All
scientific
on
d’Unites
a
common
(The
measurements
system
of
International
are
made
units
System
using
these
known
of
as
Units)
units.
the
system
there
are
seven
base
units
which
have
been
defined
in
constant.
such
a
way
represent
that
the
they
can
standard
be
size
Derived quantities
Table 1.1.1
units
prefixes
easily
of
a
and
reproduced
particular
(T
able
physical
1.1.1).
These
units
quantity.
units
SI base quantities and units
Physical
Physical
Symbol
quantities
other
than
the
base
quantities
are
known
as
Unit
derived
quantities
(T
able
1.1.2).
A
derived
quantity
is
derived
from
a
quantity
combination
Mass
m
kilogram
(kg)
the
of
base
relationship
It
is
v
=
defined
by
quantities.
between
the
the
The
base
following
corresponding
quantities.
unit
Speed
is
a
is
derived
derived
from
quantity.
equation:
s
Length
l
metre
(m)
where
v
is
speed,
s
is
distance
travelled
and
t
is
the
time
taken.
t
Time
t
second
Temperature
T
kelvin
Electric
I
ampere
(s)
From
the
Time
is
definition,
also
a
base
distance
quantity
a
base
quantity
with
is
unit
s
with
unit
m
(metre).
(second).
(K)
s
m
–1
v
=
=
(A)
t
=
m s
s
current
–1
The
Amount
of
n
mole
SI
unit
for
speed
is
therefore
m s
(mol)
Table 1.1.2
Derived quantities and derived units
substance
Derived quantity
Luminous
I
Relationship
Derived
intensity
unit
Name of
candela
in
v
base
units
(cd)
2
Area
(A)
length
×
length
length
×
length
m
–
3
Volume
(V)
×
length
m
–
–3
Density
(ρ)
mass/volume
kg m
displacement/time
m s
–
–1
Velocity
(v)
–
–2
Acceleration
(a)
velocity/time
m s
–
–2
Force
(F)
mass
×
acceleration
kg m s
Work
(W)
force
×
distance
kg m
2
–2
s
2
Power
(P)
Charge
(Q)
work/time
kg m
current
A s
×
time
2
(V)
power/current
kg m
joule
–3
s
watt
coulomb
2
Voltage
newton
–3
s
–1
A
volt
unit
Chapter
1
Physical
quantities
and
units
Prefixes
In
order
to
measured
avoid
is
too
writing
small
too
or
many
too
zeroes
large,
when
prefixes
are
a
quantity
used.
being
T
able
1.1.3
Table 1.1.3
lists
Prefix
commonly
used
List of commonly used prefixes
the
Multiplying
Symbol
prefixes.
factor
Examples:
–12
pico
10
nano
10
micro
10
p
3
6.2 kilometres
6.2 km
=
6.2
×
10
m
–9
n
–3
2.9 milliamperes
2.9 mA
=
2.9
×
10
A
4.1 μm
=
4.1
×
10
–6
μ
–6
4.1 micrometre
m
–3
milli
10
centi
10
deci
10
kilo
10
m
–12
100 picofarads
100 pF
=
100
×
10
3 MW
=
3
10
F
–2
c
6
3 megawatts
×
W
–1
d
3
Conversion of
k
units
6
Physics
key
to
exists
often
requires
converting
between
that
between
them.
you
convert
units
The
is
to
following
from
first
one
unit
determine
examples
will
to
another
.
the
relationship
illustrate
this
mega
10
giga
10
tera
10
M
The
that
9
G
point.
12
T
Example
–1
a
Convert
–1
80 km h
to
m s
2
b
Convert
2.5 mm
2
to
m
–3
c
Convert
a
80 km h
–3
7.9 g cm
to
kg m
3
80
×
3
10
m
80
×
10
m
–1
–1
=
=
=
1 h
60
×
2
b
22.2 m s
60 s
–3
1 mm
=
1 mm
×
1 mm
=
1
×
=
2.5
10
–3
m
×
2
∴
1
×
–6
2.5 mm
×
1
×
10
10
–6
m
=
2.5
×
1
×
2
10
–6
m
=
10
2
m
2
m
–3
7.9 g
7.9
×
10
kg
–3
c
–3
7.9 g cm
=
=
=
–6
1 cm
The
One
mole
of
the
substance
×
1 cm
×
1 cm
1
and the Avogadro
base
can
quantities
refer
to
the
is
‘the
of
10
7900 kg m
3
m
constant
amount
number
×
of
substance’.
particles,
number
The
of
amount
molecules
of
Key points
or
23
number
This
of
ions.
number
is
The
number
called
the
in
this
Avogadro
case
refers
constant
to
(N
a
value
of
6.02
×
10
.
ᔢ
)
A
physical
quantity
consists
of
A
the
The
of
mole
is
particles
the
as
amount
there
are
of
in
substance
12 grams
( n)
of
that
contains
the
same
product
of
a
numerical
value
number
and
a
unit.
The
base
carbon-12.
ᔢ
quantities
are
mass,
23
∴1 mol
=
6.02
×
10
particles
length,
The
amount
given
by
N
of
=
particles
( N)
present
in
an
amount
of
substance
( n)
time,
current,
is
and
nN
temperature,
amount
luminous
of
substance
intensity.
A
ᔢ
All
other
physical
quantities
are
Example
derived from
Calculate
the
number
of
molecules
present
in
1.2 moles
of
helium
the
base
quantities.
gas.
ᔢ
Prefixes
are
used
as
shorthand,
23
1 mol
=
6.02
×
10
for
23
∴
1.2 mol
=
1.2
×
6.02
×
10
small
23
=
7.224
×
10
writing
molecules
helium
very
large
or
very
quantities.
of
gas
ᔢ
The
mole
is
equivalent
to
23
6.02
×
10
particles.
3
1.2
Dimension
Learning outcomes
On
completion
should
be
able
of
this
and
Dimension of
section,
you
The
dimension
physical
to:
lists
ᔢ
recall
the
dimensions
unit
of
of
quantity
some
of
the
analysis
physical quantities
a
physical
and
the
quantity
base
dimensions
shows
quantities
of
the
base
the
listed
relation
in
T
able
between
1.1.1.
the
T
able
1.2.1
quantities.
base
Table 1.2.1
Dimensions of base q uantities
quantities
Base
ᔢ
determine
derived
ᔢ
the
dimensions
quantities
understand
the
dimensional
physical quantity
Dimension
of
importance
Mass
[M]
Length
[L]
Time
[T]
Temperature
[θ]
Electric
[A]
of
analysis.
The
the
current
dimensions
of
relationship
derived
between
physical
the
quantities
corresponding
can
base
be
determined
quantities
is
once
known.
For
example,
mass
[M]
–3
density
=
=
=
[M L
]
3
volume
[L
]
–3
The
dimension
T
able
1.2.2
Table 1.2.2
Derived
for
lists
density
some
is
therefore
derived
[M L
physical
].
quantities
and
their
dimensions.
Dimensions of derived q uantities
physical quantity
Dimension
2
Area
[L]
Volume
[L]
Density
[M L
Acceleration
[L T
Power
[M L
3
–3
]
–2
]
2
–3
T
]
Importance of dimensions
Dimensions
can
1
T
o
deduce
2
T
o
check
In
any
the
the
scientific
must
equal
units
are
to
not
be
dimensions
equation,
the
of
units
same,
of
the
on
derived
an
units
the
then
a
on
the
equation
s
=
ut
the
right-hand
the
equation
+
quantity
(as
shown
previously).
equation.
1
Consider
units
used
homogeneity
the
and
left-hand
side
is
of
the
side
of
the
equation.
equation
If
the
incorrect.
2
at
.
It
represents
the
displacement
2
s
of
a
body
(where
a
is
after
are
t,
constant).
displacement
there
time
two
is
the
where
On
metre
the
u
is
initial
left-hand
(m).
On
the
velocity
side
of
and
the
right-hand
a
is
acceleration
equation
side
of
the
the
unit
equation
terms.
–1
The
unit
for
The
unit
for
ut
is
1
4
×
2
at
2
m s
s
=
–2
=
m s
m
1
2
×
s
=
m
(The
coefficient
is
2
of
ignored.)
Chapter
The
two
unit
on
terms
Since
the
units
on
the
are
being
unit
the
right-hand
on
side
of
the
equation
is
m,
even
though
1
Physical
quantities
and
units
the
added.
the
left-hand
right-hand
side
side
of
the
of
the
equation
equation,
the
is
equal
equation
to
is
the
said
to
be
homogeneous.
An
equation
that
is
not
homogeneous
An
equation
that
is
homogeneous
is
may
not
not
cor rect.
necessarily
be
1
Suppose
the
same
equation
above
is
rewritten
as
s
cor rect.
2
=
at
.
The
units
on
2
both
sides
incorrect
3
T
o
of
string
is
length
that
x
x,
T
y
it
the
is
is
y
are
of
the
a
same
(i.e.
m).
However
,
units
that
tension
the
in
velocity
the
string
v
T,
of
a
wave
mass
of
on
the
are
of
v
of
T
The
units
of
m
The
units
of
l
=
and
the
dimensions
of
v
=
[L T
]
=
kg
m
the
–2
∝
[M]
x
considering
M
L
term
T
term
the
+
x
]
y
x
+
each
we
in
the
=
x
+
y
1
=
x
+
z
–1
=
–2 x
of
of
l
m
=
suggested
these
T
=
=
[M L T
]
[M]
[L]
relationship
we
get
z
–2x
[T]
the
one
at
a
time
equations
on
the
left-hand
side
and
following
Exam tip
1
Solving
of
[L]
dimension
get
dimensions
dimensions
z
[L]
the
dimensions
y
[M]
0
term
the
dimensions
[M L T
side,
and
and
and
∝
right-hand
–2
kg m s
=
–1
Now
and
–1
m s
–1
]
m
l
constants.
=
units
[L T
stretched
z
The
]
a
string
–2
[L T
is
l
z
Considering
equation
term.
–1
The
the
equations.
suggested
to
the
missing
form
string
m
and
it
the
related
of
∝
equation
because
predict
Suppose
v
the
we
get
x
=
1
,
y
=
–
2
1
,
z
=
2
2
To
T l
½
The
equation
now
becomes
v
∝
T
–½
m
½
l
or
v
check
correct,
left-hand
side
limitation
of
this
method
of
trying
to
predict
the
form
of
an
equation
of
it
cannot
equation
can
determine
only
be
the
verified
value
of
the
if
an
equation
the
units
is
on
the
constant
of
proportionality.
side
the
and
the
equation.
If
right-hand
they
are
the
is
same,
that
see
∝
m
A
to
determine
the
equation
is
homogeneous.
This
experimentally.
Key points
ᔢ
The
dimension
of
mass,
length
ᔢ
The
dimension
of
derived
and
time
quantities
can
are
be
[M],
[L]
and
[T]
respectively.
determined from
base
quantities.
ᔢ
Dimensional
ᔢ
An
equation
analysis
is
can
incorrect
be
if
it
used
is
to
not
check
the
homogeneity
of
equations.
homogenous.
5
1.3
Scalar
and vector
Learning outcomes
On
completion
should
ᔢ
be
able
of
section,
you
and vector quantities
Quantities
between
scalar
A
scalar
give
A
vector
ᔢ
ᔢ
add
either
being
a
scalar
or
a
vector
.
magnitude
only.
has
magnitude
and
direction.
of
scalar
of
scalar
quantities
are
mass,
length,
work,
speed,
distance,
and
and
power
.
quantities
and
and
resolve
subtract
subtract
of
acceleration,
and
A
be
vector
end.
vectors.
can
The
Scalar
are
weight,
(Figure
momentum,
velocity,
1.3.1).
represented
length
of
the
of
the
arrow
and
quantities
line
as
a
straight
represents
points
in
the
subtracting
are
added
and
line
the
with
an
arrow
magnitude
direction
of
the
of
at
the
one
vector
.
The
vector
.
scalars
subtracted
numerically
.
For
example,
if
you
vector
were
interested
would
perform
Mass
magnitude
of
of
in
finding
the
the
total
calculation
student
=
50 kg,
as
mass
of
a
student
and
his
back
pack
you
follows:
Mass
of
back
pack
=
4 kg
vector
T
otal
Suppose
of
Figure 1.3.1
quantities
displacement
vectors
Adding
of
vector
scalars
direction
direction
has
quantity
Examples
add
as
and
energy
ᔢ
classified
quantities
examples
vector
be
quantity
Examples
ᔢ
can
to:
differentiate
vector
this
Scalar
quantities
representing a vector quantity
the
mass
90 J
of
of
machine
his
is
is
determine
input
Energy
lost
Adding
and
energy
Energy
Adding
student
60 J,
=
90 J,
inside
back
supplied
Energy
the
to
a
the
pack
50
machine.
energy
output
machine
=
=
=
90
If
lost
+
4
the
=
54 kg
useful
inside
the
energy
output
machine.
60 J
–
60
=
30 J
(combining) vectors
vector
quantities
is
not
as
simple
as
adding
scalar
quantities.
⃗
b
V
ector
quantities
Consider
In
order
drawn
⃗
a
⃗a
Figure 1.3.2
⃗
+ b
of
⃗a
⃗
b
⃗
– b
to
Figure 1.3.3
perform
⃗
a
⃗
– b
is
force
a
6
Subtracting two vectors
are
acting
in
of
the
the
⃗
+ b
form
addition,
from
the
drawn
on
effect
the
two
same
is
of
at
⃗
(– b).
point
at
the
is
which
taken
the
first
of
first
the
same
For
to
the
The
the
vector
but
then
ending
⃗
–b
vector
point
1.3.3).
vector
example
The
⃗
as
b
is
vector
(Figure
example
ended.
length
The
vector
previous
drawn.
which
⃗a
account.
⃗
V
ector b
ended.
vector
⃗a
addition.
is
drawn.
resultant
Using
into
1.3.2.
vector
⃗a
point
vector
be
Figure
vector
⃗a
called
+
to
as
is
⃗
–b
pointing
guide
then
is
in
the
1.3.4).
same direction
The
an
of
shown
vector
⃗a
in the
quantity.
has
starting
having
(Figure
acting
(combined
a
as
point
saying
⃗a
acting
vector
magnitude
Figure 1.3.4
vector
direction
forces
from
is
as
that
vectors
the
vector
⃗a
same
arrow
Vectors
two
drawn
The
starting
an
opposite
Force
⃗
⃗
a – b
by
simply
Adding two vectors
the
⃗
and
b
two
from
subtraction
is
drawn
⃗
a
these
starting
then
direction
vectors
⃗a
add
⃗
vector b.
V
ector
⃗
⃗
a + b
to
by
is
two
have
unit
object
both
forces.
direction.
of
in
forces)
In
is
same
can
Figure
The
force
the
be
1.3.5
resultant
the
newton
direction.
found
two
force
by
simply
forces
is
(N).
The
8 N
3 N
to
Suppose
resultant
adding
and
the
5 N
right.
the
are
Chapter
Vectors
Suppose
in
acting
two
Figure
forces
1.3.6.
Physical
acting
on
resultant
of
the
two
forces.
The
of
the
two
forces.
Therefore,
an
force
resultant
object
is
force
in
found
will
opposite
by
act
directions
subtracting
in
quantities
and
units
3 N
in opposite directions
are
The
1
the
the
direction
as
5 N
shown
magnitude
of
the
larger
8 N
the
resultant
force
is
2 N
to
the
right.
Figure 1.3.5
Adding vectors acting in the
same direction
Vectors
Suppose
Figure
The
R.
forces
In
force
The
The
must
two
1.3.7.
5 N
force.
acting
are
order
is
first
be
of
angle to
acting
add
the
force
decided
at
by
4 N
R
force
be
For
each other
angle
two
of
θ
vectors,
starting
can
upon.
an
the
drawn
of
the
an
to
then
addition
value
at
from
and
the
to
5 N
4 N
ending
force
by
1 cm
other
force
the
determined
example,
each
the
shown
drawn
point
results
scale
can
as
is
of
in
represent
the
the
drawing.
A
1 N.
in
first.
5 N
3 N
4 N
force
scale
This
2 N
means
5 cm
that
the
4 N
respectively.
and
5 N
force
Once
the
ruler
.
Suppose
scale
can
be
represented
drawing
is
as
completed,
lengths
the
4 cm
length
and
R
Figure 1.3.6
is
Adding vectors acting in
opposite directions
measured
using
a
θ
=
60°.
The
length
of
R
will
be
7.8 cm.
–1
Therefore,
R
can
also
R
be
2
c
=
7.8 cm
determined
2
=
a
×
1 N cm
by
=
7.8 N.
calculation
using
the
cosine
rule.
2
+
b
–
5N
2ab cos θ
R
5N
2
c
2
2
=
5
+
4
=
7.8 N
–
(2
×
5
×
4
×
cos 120°)
=
61
θ
c
θ
4N
4N
Figure 1.3.7
Finding the resultant of two
Resolving vectors
vectors acting at an angle to each other
A
vector
can
each
other
.
been
drawn
replaced
be
replaced
Consider
using
with
two
the
the
by
x
two
y
origin
vectors
other
plane
as
the
acting
at
vectors
shown
in
starting
right
acting
Figure
point.
angles
to
at
right
1.3.8.
angles
⃗
vector
P
The
each
to
⃗
vector
P
A
can
has
be
other
.
5 sin 30°
y-axis
1
5 ms
⃗
P
30°
5 cos 30°
y-component
Figure 1.3.9
θ
x-axis
O
x-component
Figure 1.3.8
Key points
Resolving a vector into two components
ᔢ
The
horizontal
component
( x-component)
is
found
by
drawing
a
A
scalar
quantity
has
magnitude
vertical
only.
line
from
the
tip
of
⃗
vector
P
parallel
to
the
y-axis
until
it
meets
the
x-axis.
ᔢ
The
vertical
component
( y-component)
is
found
by
drawing
a
A
vector
quantity
magnitude
line
from
the
tip
of
⃗
vector
P
parallel
to
the
x-axis
until
it
has
a
horizontal
meets
the
and
a
direction.
y-axis.
ᔢ
Scalar
quantities
are
added
and
–1
Suppose
30°
to
a
the
ball
is
struck
horizontal
as
such
that
shown
in
it
travels
Figure
with
a
velocity
of
5 m s
at
subtracted
ᔢ
Vectors
account
The
by
horizontal
and
vertical
components
can
be
found
by
scale
drawing
calculation.
component
=
5
×
cos 30°
=
are
added
their
by
taking
into
directions.
or
ᔢ
–1
Horizontal
algebraically.
1.3.9.
Any
vector
two
vectors
can
be
which
resolved
act
at
into
right
4.33 m s
angles
to
each
other.
–1
V
ertical
component
=
5
×
cos (90
–
30)
=
5
×
sin 30°
=
2.5 m s
7
2
Measurements
2.
1
Measurements
Learning outcomes
On
completion
should
ᔢ
be
able
measure
rule,
lengths
vernier
measure
and
this
section,
using
calliper
screw
weights
lever
Measuring
you
to:
micrometre
ᔢ
of
1
a
and
The
metre
lengths
rule,
vernier
instruments
used
length
metre
is
the
to
calliper
measure
(m).
A
and
micrometre
lengths
metre
rule
in
a
screw
laboratory.
would
be
used
gauge
The
to
SI
are
common
unit
measure
of
the
width
metre
of
a
desk
or
to
measure
of
a
the
length
of
a
pendulum.
A
vernier
calliper
would
be
used
a
the
dimensions
of
a
small
block
of
wood
or
the
diameter
gauge
using
a
spring
the
test
depends
balance
ᔢ
measure
angles
ᔢ
measure
temperature
tube.
diameter
on
micrometre
a
the
Measuring
using
A
of
piece
of
magnitude
length
screw
copper
of
using
gauge
wire.
the
would
The
length
a vernier
be
choice
being
used
of
to
measure
measuring
device
measured.
calliper
a
Figure
2.1.1
shows
a
diagram
of
a
vernier
calliper
.
There
is
a
main
scale
thermometer
and
ᔢ
measure
volumes
laboratory
using
standard
equipment.
a
main
line
vernier
scale
up
scale
with
that
Figure
is
scale.
read
the
must
2.1.2
Reading
is
When
first.
main
be
=
On
as
main
to
object
the
scale.
added
read
an
is
placed
vernier
This
the
gives
main
between
scale,
the
one
of
fraction
scale.
The
outside
the
of
markings
the
vernier
jaws,
the
will
millimetre
calliper
in
follows:
scale
+
vernier
scale
=
56
+
0.7
=
56.7 mm
inside
jaws
main
scale
vernier
scale
object
0
outside
Figure 2.1.1
10
jaws
A vernier calliper
Figure 2.1.2
Measuring
Figure
a
2.1.3
main
The
shows
scale
fractional
length
on
scale
has
micrometre
Reading
=
a
the
using
diagram
shaft
50
screw
main
a
micrometre
of
and
a
a
divisions.
gauge
scale
+
in
screw
micrometre
fractional
One
rotating
screw
scale
on
complete
Figure
2.1.4
scale
Reading a vernier calliper
is
=
a
6.50
gauge.
It
rotating
turn
read
gauge
consists
barrel.
represents
as
+
0.50 mm.
follows:
0.23
=
6.73 mm
5
0
5
10
15
30
0
45
0
5
25
20
mm
15
Figure 2.1.3
8
A micrometre screw gauge
Figure 2.1.4
of
The
Reading a micrometre screw gauge
Chapter
Measuring
The
mass
The
weight
therefore
of
mass
an
of
be
often
can
be
an
be
often
the
is
object
weight
measured
is
Measurements
the
using
using
force
a
a
beam
exerted
spring
on
balance
balance
it
by
(Figure
gravity.
(Figure
2.1.5).
Weight
can
2.1.6).
angles
required
that
measured
trigonometric
can
object
measured
Measuring
It
and
2
by
angles
taking
calculations
measured
required
directly
when
equilibrium
of
be
measured
accurate
to
determine
using
in
angles.
protractors.
performing
some
experiments.
measurements
ray
optic
of
lengths
Where
Angles
and
possible,
Measurement
experiments
or
of
Figure 2.1.5
using
A beam balance
angles
angles
is
demonstrating
forces.
lbs
There
are
optics
experiments
that
require
very
precise
kg
measurements
0
0
of
angles.
In
these
experiments
a
spectrometer
is
used.
Figure
1.0
2
2.1.7
4
illustrates
a
spectrometer
.
2.0
6
3.0
8
4.0
10
5.0
12
Measuring temperature
6.0
14
7
.0
16
8.0
T
emperature
is
measured
using
a
thermometer
.
The
SI
unit
of
18
9.0
20
temperature
Celsius
is
(°C).
the
kelvin
(Refer
to
(K).
14.2,
T
emperature
is
also
measured
in
degrees
Thermometers .)
Measuring volume
The
volume
of
an
The
volume
of
regular
object
is
the
objects
amount
can
be
of
space
taken
determined
by
up
by
the
object.
calculation.
Figure 2.1.6
V
olume
of
a
cuboid
V
=
V
olume
of
a
sphere
V
=
l
×
b
×
h
(length
l,
breadth
b,
height
A spring balance
h)
4
3
πr
(radius
of
sphere
r)
3
2
V
olume
of
cylinder
V
=
πr
(radius
h
of
base
of
cylinder
3
V
olume
The
is
volume
commonly
of
displacement
of
a
small
an
irregularly
method.
stone.
measured
shaped
Suppose
Some
in
water
is
you
cm
are
of
cylinder
h)
m
can
be
required
into
height
3
or
object
place
r,
a
measured
to
measure
measuring
using
the
cylinder
a
volume
and
the
Figure 2.1.7
initial
final
volume
volume
volume
of
recorded.
recorded.
the
stone
The
The
(Figure
stone
is
gently
difference
placed
between
the
into
two
the
water
volumes
and
gives
A spectrometer
the
the
3
2.1.8).
measuring
3
cm
500
cylinder
500
Key points
400
ᔢ
Standard
instruments
calliper
and
the
ᔢ
Mass
measured
ᔢ
Weight
ᔢ
Angles
ᔢ
A
ᔢ
Temperatures
ᔢ
The
volume
of
a
ᔢ
The
volume
of
an
used
to
micrometre
measure
screw
lengths
are
the
metre
rule,
400
vernier
gauge.
300
is
is
using
measured
a
using
beam
a
balance
spring
or
an
electronic
balance.
balance.
200
200
water
are
measured
spectrometer
is
are
using
used
a
protractor.
when
measured
regular
measuring
using
object
irregularly
is
a
angles
optical
experiments.
100
object
is
by
method.
100
calculation.
determined
using
a
Figure 2.1.8
displacement
object
thermometer.
determined
shaped
in
Measuring volume using a
displacement method
9
2.2
Measurements
Learning outcomes
On
completion
should
be
able
of
this
Measuring time
section,
you
The
measure
time
using
a
clock,
and
the
time
base
of
unit
or
time
of
time
is
the
stopwatches.
taken
for
a
second
Suppose
pendulum
(s).
an
to
Time
durations
experiment
complete
is
one
are
measured
performed
oscillation.
to
using
measure
In
order
to
get
stopan
watch
SI
clocks
to:
the
ᔢ
2
accurate
value
for
this
time
interval,
the
time
taken
for
10
oscillations
a
T
is
recorded
using
a
stopwatch.
The
experiment
is
repeated
several
10
cathode-ray
oscilloscope
times
ᔢ
measure
electrical
time
quantities
and
for
the
one
mean
(average)
oscillation
T
is
time
for
ten
determined
oscillations
as
follows:
T
is
=
recorded.
T
/
10
using
standard
The
10.
laboratory
There
are
instances
when
the
time
interval
of
an
event
is
so
small
instruments
that
ᔢ
understand
calibration
ᔢ
to
how
into
a
to
stopwatch
oscilloscope
use
metal
curves
understand
equation
how
a
rearrange
cannot
(CRO)
plates
inside
frequency
is
time-base
setting
can
it
be
be
used.
used.
called
attached
to
the
the
An
instrument
The
CRO
X-plates.
X-plates.
called
consists
A
sweep
This
of
a
a
cathode-ray
pair
generator
frequency
is
of
of
parallel
known
adjusted
using
the
an
on
the
front
panel
of
the
CRO
(Figure
2.2.1).
linear form.
electron
gun
cathode
deflection
anode
plates
Y-plates
grid
X-plates
fluorescent
electron
Figure 2.2.1
Figure
A
screen
beam
A simple diagram of a cathode-ray oscilloscope
2.2.2
shows
a
waveform
on
the
screen
of
a
CRO.
The
time-base
B
–1
setting
is
Suppose
The
calibrated
it
is
at
required
distance
AB
is
2 ms cm
that
6 cm.
the
.
Each
time
square
interval
Therefore,
the
on
the
screen
between
time
A
interval
is
and
1 cm
B
be
between
×
1 cm.
found.
A
and
B
–1
6 cm
×
2 ms cm
Measuring
=
12 ms.
electrical quantities
1 cm
T
wo
Figure 2.2.2
A waveform on the screen
important
difference.
An
electrical
electric
quantities
current
is
are
electric
measured
cur rent
using
an
and
potential
instrument
called
of a CRO
an
ammeter
measured
volt
(V).
type.
in
In
and
using
The
the
to
a
is
Calibration
that
if
10
it
had
is
you
no
not
unit
of
is
the
ampere
instrument
ammeter
measurement
Suppose
an
case
reference
its
the
scale
and
the
called
recorded.
recorded
as
a
seen
In
on
A
potential
voltmeter
voltmeter
analogue-type
is
(A).
can
meter
the
the
be
the
case
of
display
and
an
difference
its
unit
analogue
is
or
location
of
a
meter
,
digital
of
the
the
is
the
digital
pointer
the
instrument.
curves
were
provided
markings
calibrated.
on
The
with
the
a
mercury-in-glass
length
of
it.
thermometer
The
is
thermometer
thermometer
placed
in
known
is
of
(A)
no
use
temperatures
is
Chapter
(temperature
of
pure
melting
ice
0 °C
and
the
temperature
of
steam
2
Measurements
100 °C
temperature / °C
above
pure
boiling
water)
and
the
length
of
mercury
is
measured.
These
calibration
curve
100
temperatures
are
easily
reproducible
thermometer
(B)
which
is
already
and
chosen
calibrated
is
for
this
used
for
reason.
Another
comparison.
Both
80
thermometers
and
100 °C.
are
The
placed
in
substances
temperature
reading
that
on
have
the
temperatures
calibrated
between
thermometer
(B)
0 °C
is
60
recorded
A
and
graph
(Figure
A.
of
the
the
the
This
the
the
length
calibration
mercury
against
graph
is
thermometer
temperature,
from
of
temperature
2.2.3).
When
length
of
curve
length
the
A
is
the
of
placed
is
thermometer
mercury
calibration
mercury
to
in
in
a
curve
then
the
substance
recorded.
determine
is
for
the
This
of
(A)
recorded.
40
plotted
thermometer
20
unknown
length
unknown
is
is
read
off
temperature.
0
5
10
15
20
25
length / mm
Plotting
linear
In
work
practical
two
a
quantities.
linear
plotted
graphs from
it
If
is
relationship,
against
often
two
required
quantities
a
straight
x
non-linear
to
establish
and
line
y
are
graph
relationships
related
would
relationships
be
such
that
obtained
Figure 2.2.3
between
they
have
when
y
Exam tip
is
x
Recall
The
equation
A calibration curve
of
a
straight
line
is
of
the
form
y
=
mx
+
c,
where
m
is
the
rules for
logarithms
the
n
gradient
of
the
straight
line
and
c
is
the
intercept
on
the
log
y-axis.
(A)
=
n log
b
It
is
often
form
of
required
the
that
equation
an
of
a
expression
straight
be
line.
re-written
T
able
2.2.1
so
that
shows
it
resembles
some
log
the
(A)
+
log
b
A
b
(B)
=
log
b
(AB)
b
examples.
A
log
(A)
–
log
b
(B)
=
log
b
b
(
B
)
Table 2.2.1
Expression
What to
plot?
y
x
2
y
=
constants
Gradient
y-intercept
a
and
b
a
(0,b)
k
and
l
n
(0,
lg k)
a
and
b
a
(0,
b)
2
ax
+
b
against
Exam tip
n
T
=
kl
2
y
lg T
2
=
ax
against
lg l
2
+
bx
y
against
log
is
usually
written
as
lg.
10
x
x
log
is
usually
written
as
ln,
e
kt
N
=
Ae
ln N
against
t
A
and
k
–k
(0,
ln A)
where
e
=
2.718
n
Suppose
T
and
l
are
related
by
the
following
equation
T
=
kl
n
lg
T
=
lg
(kl
)
T
ake
log
on
both
sides
of
the
equation
Key points
10
n
lg
T
=
lg
k
+
lg
(l
lg
T
=
lg
k
+
n lg
)
ᔢ
A
plot
of
a
gradient
is
Suppose
N
graph
n
of
and
l
←
lg T
the
Linear
against
lg l
y-intercept
is
form
will
t
are
related
by
the
produce
a
straight
line.
The
ᔢ
lg k
equation
N
=
N
=
ln
(Ae
T
ake
log
on
cathode-ray
both
sides
of
the
used
time
ᔢ
)
measured
using
a
clock
to
oscilloscope
measure
very
can
small
intervals.
Ae
kt
ln
A
be
following
is
stopwatch.
kt
and
Time
Electrical
current
is
measured
equation
e
using
an
ammeter.
kt
ln
N
=
ln
A
+
ln
(e
ln
N
=
ln
A
–
kt ln
ln
N
=
ln
A
–
kt
)
ᔢ
e
Electrical
using
←
Linear
form
ᔢ
a
Calibration
calibrate
A
plot
of
gradient
a
is
graph
–k
of
and
ln
the
N
against
t
y-intercept
will
is
ln
produce
A
a
straight
line.
voltage
is
measured
voltmeter.
curves
are
used
to
instruments.
The
ᔢ
Expressions
quantities
such
can
a
be
involving
can
way
be
that
a
two
rearranged
linear
in
graph
plotted.
11
or
2.3
Errors
in
Learning outcomes
On
completion
should
be
able
of
this
measurements
Systematic
section,
you
Whenever
there
to:
be
ᔢ
differentiate
between
will
a
random
into
identify
ways
systematic
of
and
reducing
random
errors
This
differentiate
between
a
of
or
is
measured,
uncertainty
systematic
large
number
around
of
steel
some
error
the
and
in
the
random
there
is
the
likelihood
measurement.
er rors .
If
a
Errors
physical
that
can
of
times,
it
may
be
revealed
that
quantity
the
is
readings
is
value.
known
experimenter
.
ball
to
fall
Some
as
a
readings
random
Suppose
from
rest
a
may
error
student
through
a
be
and
larger
measures
known
or
usually
the
vertical
smaller
.
occurs
time
as
a
taken
distance
h.
t,
The
precision
distance
and
a
type
result
for
ᔢ
quantity
error
errors
errors
fluctuate
ᔢ
an
random
systematic
measured
and
physical
be
divided
and
h
is
varied
and
the
corresponding
time
is
measured.
accuracy.
It
is
known
1
h
that
h
and
t
are
related
by
the
following
equation
2
=
gt
–2
,
where
g
is
the
acceleration
due
to
free
fall
(g
=
9.81 m s
).
2
√
According
h
through
All
the
and
are
relationship,
The
points
the
do
line.
random
repeating
this
this
origin.
data
below
there
by
to
the
graph
not
The
errors
lie
the
plot
on
in
the
the
√h
against
2.3.1
straight
of
the
finding
could
will
line.
a
mean
taken
give
plot
They
from
Random
the
have
t
shows
points
experiment.
and
student
of
Figure
deviation
measurements
experiment,
a
in
are
the
of
the
straight
√h
be
t
above
indicates
can
line
against
scattered
line
errors
several
a
of
that
reduced
measurements.
measurements
of
In
t
t
for
Figure 2.3.1
a
given
random errors
Suppose
a
obtained
All
√
height
h,
and
find
the
average
of
those
times.
Graph showing the effect of
the
different
are
data
student
shown
points
in
performs
Figure
are
the
same
experiment.
The
results
2.3.2.
scattered
about
the
main
line.
This
illustrates
a
h
random
origin.
In
the
Either
are
ᔢ
This
case
all
zero
However
,
indicates
of
a
the
smaller
A
This
than
error
Graph showing the effect of
error
before
a systematic error
ᔢ
An
no
in
longer
can
when
A
the
For
reading
is
eye
12
easy
to
way
the
for
to
a
error
constant
their
true
systematic
used.
the
not
zeroing
error
In
the
in
or
can
this
through
the
error
value
physical
pass
experiment.
in
all
one
the
occur
case,
quantity
the
direction.
readings
because
of
the
is
instrument,
not
if
present.
possible,
long
period
Instruments
are
manufacturer
being
used
experimenter
quantity,
all
suppose
may
on
be
a
by
In
of
to
of
this
time
be
an
makes
will
be
experimenter
recorded
is
as
instrument
readings
as
experimenter
results
scale
the
the
often
are
as
instrument.
consistently
the
and
calibrated
the
the
case,
0.2 m
to
perform
the
off
same
by
the
same
accidentally
when
it
is
the
mistake
assumes
actually
0.1 m.
1.6 m.
accuracy
understand
game
darts.
a
instrument.
a
division
1.7 m
and
consider
with
of
a
is
being
by
does
experiment.
example,
smallest
Precision
An
the
the
the
A
graph
systematic
when
eliminated
used
measuring
amount.
that
If
a
than
value.
reading
techniques
experiment.
larger
the
there
instrument
accurate.
Improper
is
error
,
calibrated
by
that
there
true
a
be
been
recommended
ᔢ
are
the
gives
performing
have
that
their
incorrectly
may
notice
systematic
readings
instrument
t
Figure 2.3.2
error
.
of
the
darts.
Figure
difference
The
2.3.3
between
objective
illustrates
of
the
the
precision
game
various
is
to
and
hit
scenarios.
accuracy
the
bull’s
Chapter
a
Precise and accurate
Figure 2.3.3
Notice
b
Precise but not accurate
c
Accurate but not precise
d
2
Measurements
Not precise and not accurate
A game of darts
that
accuracy
has
to
do
with
how
close
the
darts
are
to
the
measures
the
bull’s
eye.
Precision
Suppose
has
a
to
do
with
quantity
has
how
a
close
true
the
value
of
darts
x
.
A
are
as
a
student
group.
0
quantity
a
large
experiment
plotted
Figure
to
is
known
show
2.3.4
difference
number
the
to
times
have
number
illustrates
between
of
times
various
n
using
systematic
of
precision
n
n,
appropriate
and
a
scenarios.
and
an
random
particular
The
errors.
reading
graphs
help
x
x
0
is
obtained.
illustrate
the
x
x
n
0
x
0
Precise and accurate
Figure 2.3.4
graph
is
n
0
a
A
x,
The
accuracy.
n
0
instrument.
a
x
0
x
0
a
Precise but not accurate
x
0
Accurate but not precise
a
Not precise and not accurate
Experiments having systematic and random errors
Key points
ᔢ
Measurements
are
accurate
ᔢ
Measurements
are
precise
if
if
the
the
systematic
random
errors
errors
are
are
small.
ᔢ
Random
of
small.
the
result
Example
Suppose
below
two
students
perform
an
experiment
to
determine
the
value
for
ᔢ
errors
occur
experimenter
in
an
the
Random
error
true
errors
as
a
and
that
is
result
can
above
or
value.
can
be
reduced
–2
the
acceleration
due
to
gravity
g.
g
is
known
to
be
9.81 m s
.
The
students
by
repeat
the
experiment
several
times
and
the
value
of
g
is
repeating
A
–
9.80,
9.82,
9.83,
9.81,
and
recorded.
finding
Student
measurements
the
mean.
9.82
ᔢ
Random
errors
cannot
totally
be
eliminated.
Student
B
–
8.45,
8.41,
8.42,
8.45,
8.43
ᔢ
The
results
of
the
experiments
performed
by
student
A
are
both
Systematic
errors
and
errors
are
constant
accurate
in
one
direction.
precise.
ᔢ
The
results
of
the
experiments
performed
by
student
B
are
precise
Systematic
errors
can
be
but
eliminated.
not
accurate.
ᔢ
Precision
is
a
measure
reproducibility
ᔢ
Accuracy
closeness
to
the
is
a
of
true
of
a
measure
the
of
the
result.
of
the
measured
value
value.
13
2.4
Uncertainties
Learning outcomes
On
completion
should
be
able
of
this
understand
section,
you
When
with
to:
the
terms
a
fractional error
quantity
the
is
in derived quantities
measured,
measurement.
of
a
metal
rod.
there
Suppose
The
a
metre
is
an
metre
rule
is
error
rule
able
or
is
to
uncertainty
used
give
to
associated
measure
readings
to
the
nearest
absolute
0.1 cm.
error,
measurements
Uncertainties
length
ᔢ
in
This
means
that
when
a
reading
is
taken,
the
experimenter
will
and
either
record
to
the
nearest
marking
above
or
below
the
actual
length
of
percentage error
the
ᔢ
calculate
derived
the
uncertainties
quantities.
in
rod.
actual
the
of
The
value.
smallest
the
rod
12.40
reading
±
This
error
reading
as
will
on
12.40 cm,
be
in
either
the
the
too
measurement
metre
then
0.05 cm
the
rule).
If
a
is
high
or
written
student
information
too
is
as
low
±0.05 cm
measures
recorded
from
as
the
the
(half
length
follows:
0.05 cm.
The
actual
or
absolute
The
fractional
uncertainty
(error)
is
0.05 cm.
0.05
or
relative
uncertainty
(error)
is
=
0.004.
12.40
0.05
The
percentage
uncertainty
×
is
100%
=
0.4%.
12.40
Adding
Consider
P
=
Q
=
and
two
subtracting
quantities
25.10
±
62.50
The
absolute
The
fractional
±
P
and
Q
0.05
0.05
uncertainty
in
P
is
0.05
0.05
uncertainty
in
P
is
=
0.002
25.10
0.05
The
percentage
uncertainty
in
P
is
×
100%
=
0.2%
25.10
The
absolute
uncertainty
The
fractional
in
Q
is
0.05
0.05
uncertainty
in
Q
is
=
0.0008
62.50
0.05
The
percentage
uncertainty
in
Q
is
×
100%
=
0.08%
62.50
Suppose
P
+
Q
it
and
is
Q
required
–
P
The
each
absolute
+
the
absolute
P
+
Q
absolute
the
absolute
uncertainty
in
the
quantity
=
25.1
is
+
found
62.5
by
=
87.6
adding
the
absolute
uncertainties
–
uncertainty
Q
=
87.6
±
P
=
62.5
–
uncertainty
is
in
of
+
Q
is
0.05
+
0.05
=
0.1.
0.1
25.1
found
P
by
=
37.4
adding
the
absolute
uncertainties
quantity.
Therefore,
∴
14
Q
uncertainty
∴
each
find
quantity.
Therefore,
The
to
P
the
absolute
Q
–
P
uncertainty
=
37.4
±
in
0.1
Q
–
P
is
0.05
+
0.05
=
0.1.
of
Chapter
Multiplication, division,
Whenever
is
found
quantities
by
adding
are
the
powers
multiplied
percentage
or
and
divided
Measurements
roots
the
uncertainties
2
percentage
of
each
of
the
uncertainty
quantities
involved.
Whenever
is
found
a
by
quantity
the
uncertainty
is
quantity
Consider
P
=
Q
is
raised
the
to
a
power
,
power
by
the
the
percentage
percentage
uncertainty
uncertainty
of
the
involved.
Whenever
the
quantity
multiplying
the
by
following
±
6.24
Determine
root
of
a
quantity
multiplying
is
being
1/ n
by
found,
the
the
percentage
percentage
uncertainty
of
involved.
18.2
=
nth
found
0.1
±
the
quantities:
0.02
percentage
uncertainties
in:
P
1
P
×
2
Q
Q
3
2
3
4
P
Q
√
Exam tip
0.1
Percentage
uncertainty
in
P
is
×
100%
=
0.5%
18.2
Ensure
0.02
Percentage
uncertainty
in
Q
that
worked
is
×
100%
=
you
understand
the
examples. Uncertainty
0.3%
questions
6.24
pose
a
challenge
to
students.
1
The
percentage
uncertainty
in
P
×
Q
is
0.5
+
0.3
=
0.8%
P
2
The
percentage
uncertainty
in
is
0.5
+
0.3
is
2
0.5
=
0.8%
Q
2
3
The
percentage
uncertainty
in
P
4
The
percentage
uncertainty
in
×
=
1%
1
3
Q
is
√
×
0.3
=
0.1%
3
Key points
Example
ᔢ
1
A
student
wishes
to
measure
the
volume
V
of
a
wire
of
length
l
Measured
error
obtains
the
following
associated
of
the
wire
d
=
(0.94
±
l
=
(839
±
an
them
as
the
absolute
error.
0.04) mm
ᔢ
Length
have
with
measurements.
known
Diameter
quantities
and
The
absolute
error
is
usually
3) mm
taken
as
half
the
smallest
reading
1
2
The
volume
V
of
the
wire
is
given
by
πd
on
l
the
instrument
scale.
4
Calculate
the
percentage
uncertainty
in
the
measurement
ᔢ
of:
The fractional
absolute
i
the
diameter
error
the
length
iii
the
volume
When
quantities
subtracted,
uncertainties
uncertainty
in
the
measurement
of
d
=
=
×
ᔢ
When
or
4%
uncertainty
in
the
measurement
of
l
=
=
in
the
measurement
of
V
=
2
are
ᔢ
are
multiplied
the fractional
of
the
quantities
+
8.4%
=
l
added.
When
a
power,
Δl
d
=
added.
100%
0.4%
Δd
uncertainty
×
839
l
=
are
quantities
divided,
uncertainties
%
added
absolute
3
Δl
iii
are
the
100%
0.94
d
=
%
the
0.04
Δd
ii
by
value.
V
or
%
the
l
ᔢ
i
is
divided
d
measured
ii
error
2(4)
+
quantity
is
raised
the fractional
to
a
uncertainty
0.4
of
the
the
quantity
is
multiplied
by
power.
15
3
Kinematics
3.
1
Kinematics
Kinematics
Learning outcomes
is
considering
On
completion
should
be
able
of
this
section,
define
displacement,
and
represent
velocity
displacement,
and
a
the
speed,
acceleration
Figure
the
displacement from
under
a
velocity–time
use
the
time
ᔢ
use
slope
graph
the
graph
motion
of
objects
without
motion.
playing
of
a
in
slope
to find
of
a
shows
travelled
displacement
only.
ends
a
the
up
park.
path
at
He
the
starts
taken
point
by
the
by
the
dog.
The
of
the
dog.
Distance
Displacement
is,
arrow
B.
at
the
The
point
dog.
This
joining
is
a
however
,
dotted
A,
moves
A
scalar
a
line
path
and
B
the
represents
quantity
vector
in
gives
and
quantity.
has
This
a
means
the
displacement–
to find
and
it
has
a
magnitude
as
well
as
a
direction.
It
is
possible
for
the
dog
graph
to
ᔢ
is
park
3.1.1
distance
that
area
the
the
and displacement
dog
the
magnitude
find
describe
acceleration
graphically
ᔢ
to
causing
speed,
around
ᔢ
used
actually
you
Suppose
velocity
term
is
to:
Distance
ᔢ
the
what
velocity
have
zero
returned
but
his
to
displacement
the
point
distance
A
after
his
(starting
travelled
would
walk
point)
have
velocity–time
a
around
his
the
park.
displacement
non-zero
distance
If
the
would
dog
be
had
zero,
magnitude.
travelled
acceleration.
Definition
A
Displacement
from
a fixed
is
the
point
in
distance
a
moved
stated
direction.
displacement
B
Figure 3.1.1
Speed
Speed
and velocity
is
defined
quantity.
you
If
might
total
Differentiating between distance and displacement
an
be
as
the
object
rate
interested
distance
of
changes
travelled
in
its
change
its
average
divided
of
speed
by
the
distance.
several
speed.
total
Speed
times
The
time
is
a
during
average
taken
to
scalar
its
journey,
speed
is
the
complete
the
–1
journey.
The
SI
unit
of
speed
is
metres
per
second
Definition
Definition
Speed is the rate of change of distance.
Average
initial
Equation
(m/s
speed
speed
is
plus
or
m s
equal
to
the final
Equation
s
v
u
=
Average
speed
+
v
=
t
2
–1
v
–
speed/m s
–1
s
–
distance
t
–
time/s
u
–
initial
v
–
final
speed/m s
travelled/m
–1
16
speed/m s
).
half
of
speed.
the
Chapter
V
elocity
is
defined
as
the
rate
of
change
of
displacement.
V
elocity
is
3
Kinematics
a
–1
vector
quantity.
The
SI
unit
of
velocity
Definition
Velocity
is
is
metres
per
second
(m s
).
1
Equation
the
rate
of
change
25 m s
Δs
of
v
=
Δt
displacement.
B
–1
v
–
velocity/m s
Δs
–
change
in
displacement/m
Δt
–
change
in
time/s
P
1
25 m s
A
Consider
an
object
P
,
moving
in
a
circular
path
at
a
constant
speed
of
–1
25 m s
(Figure
3.1.2).
At
any
point
on
the
circle,
the
speed
of
the
object
–1
is
25 m s
.
direction
The
of
velocity
motion
has
at
the
points
A
and
B
are
different
since
Figure 3.1.2
An object moving in a
circular path
the
changed.
Acceleration
Exam tip
Acceleration
is
defined
as
the
rate
of
change
of
velocity.
An
object
When the velocity of
accelerates
therefore
when
the
its
velocity
direction
of
changes.
motion
of
Acceleration
the
object
is
is
a
taken
vector
into
quantity
consideration.
–2
The
SI
unit
of
acceleration
is
metres
per
second
squared
(m s
increases, the
acceleration
When the velocity of
).
decreases, the
Definition
an object
and
is
positive.
an object
acceleration
is
negative
Equation
and
Acceleration
is
the
rate
of
change
v
of
a
–
is
referred to
as
a deceleration.
u
=
t
velocity.
–2
a
–
acceleration/m s
u
–
initial
–1
velocity/m s
–1
v
–
final
t
–
time/s
velocity/m s
displacement/m
Δ s
v
=
—
Δ t
Graphical
The
a
motion
graph.
with
of
an
These
object
graphs
non-uniform
equations
not
representation of
of
can
graphical
in
a
straight
particularly
acceleration.
motion
uniform,
moving
are
be
If
used
methods
the
to
motion
useful
line
used
to
a
can
when
acceleration
analyse
are
in
straight
be
the
is
motion.
analyse
represented
object
is
uniform,
If
line
the
by
moving
Δ s
the
acceleration
is
motion.
Δ t
Displacement–time
time/s
graphs
Figure 3.1.3
Suppose
a
an
constant
object
or
P
uniform
displacement–time
The
by
graph
finding
Suppose
Figure
is
a
the
an
stationary
velocity
graph
straight
gradient
object
3.1.4
is
is
shows
in
line
of
a
a
3.1.3
through
the
the
straight
from
point
straight
Figure
dropped
the
in
at
a
O.
It
line
shows
origin.
then
for
moves
some
the
time.
motion
The
of
velocity
Constant velocity
with
is
The
the
object.
displacement/m
determined
line.
height
displacement–time
of
2 m
graph
above
for
the
the
ground.
motion.
The
tangent
graph
for
a
of
this
motion
is
a
curve.
The
object
is
accelerating
uniformly
at
at
–2
rate
9.81 m s
.
The
velocity
at
any
point
on
the
curve
is
point
P
determined
P
by
finding
The
the
velocity
tangent
at
gradient
at
that
the
of
point
point.
A
a
P
tangent
on
the
tangent
is
drawn
curve
a
is
at
that
point
determined
straight
line
on
by
drawn
the
first
such
curve.
drawing
that
it
a
t
just
time/s
touches
the
the
velocity
curve
at
the
at
the
point
point
P
.
P
.
The
gradient
of
the
tangent
is
equal
to
Figure 3.1.4
Non-uniform velocity
17
Chapter
3
Kinematics
Velocity–time
The
motion
The
area
object.
the
of
an
under
The
a
graphs
object
can
be
velocity–time
gradient
of
a
represented
graph
using
measures
velocity–time
graph
a
velocity–time
graph.
the
displacement
of
measures
the
the
acceleration
of
object.
Consider
The
Figure
various
3.1.5.
sections
It
of
represents
the
graph
OA
–
a
velocity–time
can
The
be
described
object
graph.
as
accelerates
follows.
uniformly
from
rest.
–1
v /m s
AB
–
BC
–
CD
area
object
is
travelling
The
object
decelerates
at
constant
velocity.
–
The
object
is
uniformly.
stationary
.
Its
velocity
is
zero.
1
DE
D
O
The
B
A
–
The
object
reverses
direction
and
F
C
accelerates
uniformly
from
The
decelerates
rest.
t /s
area
2
EF
–
object
comes
to
uniformly
and
then
rest.
E
Figure 3.1.5
The
acceleration
gradient
The
of
area
the
at
any
curve
under
a
at
point
that
in
the
journey
is
found
by
finding
the
point.
velocity–time
graph
measures
displacement.
Area
1
gives
the
displacement
of
the
object
from
O
to
C.
Area
2
gives
the
displacement
of
the
object
from
D
to
F
.
Example
Figure
3.1.6
shows
the
velocity–time
graph
for
a
journey
lasting
1
v /m s
R
S
20
Q
P
10
B
A
C
E
D
t /s
0
-10
T
Figure 3.1.6
Use
18
the
information
from
the
graph
to
find:
i
the
velocity
ii
the
acceleration
during
iii
the
acceleration
between
iv
the
distance
travelled
between
10
and
20
seconds
v
the
distance
travelled
between
20
and
40
seconds.
10
seconds
after
the
the
first
40
start
10
and
of
the
journey
seconds
45
seconds
50
seconds.
Chapter
3
Kinematics
–1
i
v
ii
acceleration
=
10 m s
10
–
0
10
–
0
20
–
(–10)
–2
=
gradient
of
line
OP
=
=
1 m s
–2
iii
acceleration
=
gradient
of
line
ST
=
=
40
iv
Distance
travelled
=
Area
B
v
Distance
travelled
=
Area
C
=10
+
×
10
Area
–
=
–3 m s
50
100
m
D
1
=
(10
+
20)10
+
(20
×
10)
=
150
+
200
=
350 m
2
Example
Suppose
surface
hence
a
ball
and
is
dropped
bounces
deduce
the
from
several
a
fixed
times.
height
Sketch
acceleration–time
graph
above
the
for
the
ground
velocity–time
the
motion
on
a
graph
of
the
metal
and
ball.
velocity
The
initial
increases
velocity
of
uniformly
the
ball
because
is
zero.
gravity
As
the
makes
it
ball
falls,
its
accelerate
velocity
at
a
constant
–1
rate
of
in
short
a
9.81 m s
highest
again.
Figure
time
point.
The
.
When
interval.
The
ball
it
Its
the
graph
Acceleration
metal
velocity
changes
velocity–time
3.1.7.
hits
is
surface,
decreases
direction
for
the
defined
until
again
motion
as
the
it
and
of
rate
changes
it
its
the
of
is
direction
zero
at
velocity
ball
is
change
the
increases
shown
of
in
velocity.
The
time
gradient
of
a
velocity–time
graph
gives
acceleration.
The
straight
line
Figure 3.1.8
portions
of
the
velocity–time
graph
represent
constant
acceleration.
displacement
1
v /m s
t /s
time
Figure 3.1.9
2
a /m s
Key points
ᔢ
Displacement
is
the
distance
t /s
moved
in
ᔢ
Speed
is
ᔢ
Velocity
a
a
particular
scalar
and
direction.
quantity.
acceleration
are
Figure 3.1.7
vector
ᔢ
Example
Figure
Velocity
3.1.8
motion
shows
velocity–time
graph
for
the
motion
of
a
car
.
of
of
the
car
and
sketch
the
displacement–time
graph
for
car
velocity
the
of
change
of
Acceleration
accelerates
for
is
the
rate
of
change
velocity.
car
.
some
uniformly
time.
The
from
car
rest.
then
The
car
decelerates
then
travels
uniformly
a
constant
until
it
The
slope
graph
of
a
displacement–time
measures
velocity.
comes
ᔢ
to
rate
the
ᔢ
The
the
Describe
of
motion
is
displacement.
ᔢ
the
quantities.
The
slope
of
a
velocity–time
rest.
graph
Figure
3.1.9
shows
the
displacement–time
measures
acceleration.
graph.
ᔢ
The
area
graph
under
a
measures
velocity–time
displacement.
19
3.2
Equations
Learning outcomes
of
motion
Derivation of the
equations of
motion
–1
On
completion
of
this
section,
you
Consider
an
object
P
,
initially
travelling
at
u m s
.
It
then
accelerates
–2
should
be
able
uniformly
to:
t
ᔢ
derive
ᔢ
use
the
the
solve
equations
equations
of
of
seconds
at
to
a
rate
do
so
of
a m s
and
P
–1
to
travels
achieve
a
final
velocity
through
a
distance
of
s
of
v m s
metres
.
It
takes
(Figure
3.2.1).
motion
motion
to
u
a
v
P
problems.
s
Figure 3.2.1
When
An object P moving with uniform acceleration
deriving
the
equations
of
motion,
the
following
assumptions
are
made.
1
The
acceleration
2
The
motion
The
is
in
velocity–time
Acceleration
is
is
a
uniform
straight
graph
defined
for
as
(constant).
line.
the
the
motion
rate
of
of
P
change
is
shown
of
in
Figure
3.2.2.
velocity.
1
velocity /m s
v
Therefore,
a
–
u
=
t
Rearranging
this
equation
we
get
the
following:
v
∴
The
v
=
u
+
displacement
at
of
Displacement
P
(1)
is
given
by
=
area
under
=
area
of
s
and
is
the
area
under
the
v– t
graph.
graph
u
rectangle
+
area
of
triangle
1
0
t
s
=
(u
×
t)
+
(
×
t(v
–
×
t(at)
u)
2
)
time /s
But
Figure 3.2.2
from
equation
1,
v
–
u
=
at
Velocity–time graph for the
motion of P
1
s
=
(u
×
t)
+
(
2
)
1
2
∴
s
=
ut
+
at
(2)
2
v
From
Equation
1,
t
–
u
=
a
Substituting
this
v
∴
s
=
u
(
–
into
+
)
v
a
2
(2)
we
get
the
(
–
u
)
a
2
2u(v
Exam tip
∴
s
–
u)
+
(v
=
–
2
u)
2uv
under
that
you
which
the
know
the
conditions
equations
Rearranging,
of
2
v
motions
are
2u
2
+
v
we
get
the
2
–
2uv
+
2
u
v
2
–
u
=
2a
2a
following:
2
–
u
=
2as
=
u
applicable.
2
v
20
–
=
2a
Ensure
following:
2
1
u
a
Equation
2
+
2as
(3)
Chapter
Using the
equations of
3
Kinematics
motion
Example
–2
A
car
starts
maintains
comes
to
from
a
rest
and
constant
rest
in
accelerates
speed
for
90 s.
at
a
The
rate
of
brake
2.5 m s
is
then
for
5.2 s.
applied
and
It
the
car
6 s.
Calculate:
i
the
maximum
ii
the
total
i
v
ii
Distance
velocity
distance
of
the
car
travelled.
–1
=
u
+
at
=
0
+
travelled
(2.5
×
during
5.2)
the
=
13 m s
acceleration
stage
is
determined
as
follows:
–1
u
=
0 m s
–2
,
a
=
2.5 m s
,
t
=
1
5.2 s
1
2
s
=
ut
+
2
at
=
(0
×
5.2)
+
Distance
=
travelled
during
the
stage
where
13 m s
the
,
t
=
90 s,
=
(13
a
=
ut
2
at
×
90)
+
(0)(90)
=
travelled
13 m s
v
a
–
1170 m
during
the
deceleration
stage:
–1
,
t
=
0
u
6 s,
–
v
=
0 m s
13
–2
=
=
=
– 2.17 m s
6
t
2
=
2
–1
u
constant:
1
+
2
Distance
is
0 m s
2
=
velocity
–2
1
s
33.8 m
2
–1
u
=
(2.5)(5.2)
2
2
v
=
u
+
2
2as
2
(0)
s
=
(13)
=
38.9 m
T
otal
+
distance
2(– 2.17) s
travelled
=
33.8+
1170
+
38.9
=
1242.7 m
Example
Exam tip
–1
A
ball
is
thrown
Neglecting
air
vertically
resistance,
upwards
with
an
initial
velocity
of
12 m s
.
determine:
When
an
upwards
i
the
distance
travelled
by
the
ball
after
the
velocity
iii
the
time
iv
the
maximum
of
the
ball
after
and
air
is
thrown
vertically
resistance
is
ignored,
0.8 seconds
the
ii
object
acceleration
is
equal
to
–g
0.8 seconds
–2
(– 9.81 m s
taken
for
the
ball
to
reach
its
maximum
When
height
reached
by
the
).
height
an
object
is
released
and
ball.
falls
vertically
downwards,
its
–2
[g
=
9.81 m s
]
–2
acceleration
resistance
Since
the
ball
is
thrown
upwards,
it
decelerates.
The
acceleration
of
is
is g
(9.81 m s
)
if
air
ignored.
the
–2
ball
is
– 9.81 m s
1
1
2
i
s
=
ut
+
at
2
=
(12
×
0.8)
+
2
(– 9.81)(0.8)
=
6.46 m
2
Key point
–1
ii
v
iii
At
=
u
the
+
at
=
12
maximum
v
=
u
+
0
=
12
t
=
+
(– 9.81
height
v
=
×
0
0.8)
=
4.15 m s
ᔢ
The
for
at
line
+
(– 9.81
×
equations
objects
at
a
of
motion
travelling
constant
in
a
are
used
straight
acceleration.
t)
12
=
1.22 s
9.81
2
iv
v
2
=
u
+
2
(0)
2as
2
=
(12)
+
2(– 9.81) s
2
(12)
s
=
=
2
×
7.33 m
9.81
21
3.3
Projectile
Learning outcomes
On
completion
should
ᔢ
be
show
able
that
of
this
Projectile
section,
you
Suppose
a
motion
ball
is
projected
with
a
velocity
V
at
an
angle
of
θ
to
the
horizontal.
to:
the
motion
path
taken
by
a
The
horizontal
component
is
given
by
V
=
V cos θ
H
projectile
is
parabolic
The
vertical
component
is
given
by
V
=
V sin θ
V
ᔢ
perform
calculations
involving
Figure
projectile
3.3.1
shows
the
variation
with
time
of
the
horizontal
and
vertical
motion.
components.
vertical
by
The
force
component
gravity
and
of
of
gravity
the
remains
ball.
acts
The
constant,
vertically
horizontal
provided
and
only
affects
component
that
air
is
resistance
the
unaffected
is
ignored.
–1
v /m s
If
air
resistance
longer
v sin θ
The
In
is
taken
into
account,
the
horizontal
component
is
no
constant.
path
order
taken
to
by
the
analyse
ball
is
parabolic
projectile
as
motion,
shown
the
in
Figure
horizontal
3.3.2.
and
vertical
v
v cos θ
H
motion
are
treated
separately.
v
0
=
v sin θ
V
v
t /s
v
V
θ
Figure 3.3.1
Variation of the horizontal
v
=
v cos θ
H
and vertical components with time
v sin θ
v
v
H
h
max
v
V
θ
v cos θ
Figure 3.3.2
A projectile
Showing that the path taken by a projectile is parabolic
Object projected horizontally (Figure 3.3.3)
Suppose
above
a
the
ball
is
projected
horizontally
with
a
velocity
v
at
a
height
h
ground.
1
velocity /m s
The
horizontal
component
v
=
v
H
v
If
air
the
resistance
is
acceleration
The
horizontal
is
ignored,
this
component
remains
constant.
Therefore,
zero.
displacement
x
at
time
t
is
given
by
h
1
2
x
=
(v)t
+
(0)t
=
(v)t
2
x
∴
t
=
(1)
v
Figure 3.3.3
horizontally
22
An object projected
Initial
vertical
component
of
velocity
v
=
V
0
Chapter
The
vertical
displacement
y
at
time
t
is
given
3
Kinematics
by
1
2
y
=
(0)t
+
(– g)t
(2)
2
[Assuming
the
that
positive
acceleration
Substituting
due
to
Equation
velocity
gravity
(1)
x
y
=
(0)
(
v
into
Equation
(2):
x
(– g)
2
that
– g.]
1
+
)
means
becomes
(
v
the
ball
is
moving
upwards,
2
)
g
2
∴
y
=
–
(
)
2
2v
x
2
This
equation
is
a
parabola
of
the
form
y
=
– ax
Exam tip
Object projected at an angle (Figure 3.3.2)
Suppose
a
ball
is
projected
with
a
velocity
v
at
an
angle
of
θ
to
the
When
asked
to find
the
velocity
of
horizontal.
the
The
horizontal
component
of
velocity
v
=
object
parabolic
v cos θ
at
any
path,
point
along
remember
the
to find
the
H
horizontal
If
air
resistance
is
ignored,
this
component
remains
constant.
acceleration
The
horizontal
is
vertical
components
Therefore,
at
the
and
that
point first.
zero.
displacement
x
at
time
t
is
given
by
1
2
x
=
(v cos θ)t
+
(0) t
=
(v cos θ)t
2
Exam tip
x
∴
t
=
(3)
v cos θ
Do
Initial
vertical
component
of
velocity
v
=
v sin θ
not
learn
of flight,
the formula for
horizontal
range
time
and
V
maximum
The
vertical
displacement
y
at
time
t
is
given
height. Just
resolve
the
by
initial
velocity
into
a
horizontal
and
1
2
y
=
(v sin θ)t
(– g) t
+
(4)
vertical
component
and
apply
the
2
equations
1
[Using
s
=
ut
+
of
motion.
2
at
,
a
is
taken
as
–g
because
the
vertical
component
of
2
the
velocity
decreases
Substituting
Equation
as
the
(3)
ball
into
moves
Equation
upwards]
(4):
2
x
y
=
(v sin θ)
(
1
v cos θ
)
+
x
(– g)
2
(
v cos θ
)
g
sin θ
2
∴
y
=
x tan θ
–
(
2
2v
2
cos
θ
)
x
[
tan θ
=
]
cos θ
2
This
equation
is
a
parabola
of
the
form
y
=
ax
–
bx
Example
A
boy
kicks
(Figure
a
football
such
that
the
ball
follows
the
path
shown
below
3.3.4).
P
1
10 m s
h
v
v
H
60°
7m
R
Figure 3.3.4
23
Chapter
3
Kinematics
Ignoring
air
resistance,
calculate:
i
the
initial
horizontal
ii
the
initial
vertical
iii
the
maximum
iv
the
time
taken
to
reach
v
the
time
taken
to
cover
component
component
height
h
of
of
the
achieved
the
the
the
by
velocity
velocity
the
maximum
distance
of
the
the
ball
ball
ball
height
R,
of
h
where
R
is
the
range
of
the
projectile
vi
the
distance
vii
the
velocity
starting
Sketch
of
air
R
of
the
ball
at
the
point
P
,
where
P
is
7 m
from
the
initial
point.
Figure
3.3.4
resistance
on
and
the
label
ball
it
is
A.
not
Sketch
the
path
B,
such
that
the
effect
ignored.
–1
i
Initial
horizontal
component
V
=
10 cos 60°
=
5 m s
H
–1
ii
Initial
vertical
component
V
=
10 sin 60°
=
8.66 m s
V
–1
iii
u
=
8.66 m s
–1
,
v
=
0 m s
–2
,
a
=
– 9.81 m s
2
v
=
u
+
2as
0
=
(8.66)
2
+
2(– 9.81)s
2
2(– 9.81) s
=
s
=
(8.66)
2
(8.66)
=
3.82 m
2(9.81)
Maximum
height
h
=
3.82 m
–1
iv
u
=
8.66 m s
–1
,
v
=
0 m s
–2
,
v
=
u
+
0
=
8.66
a
=
– 9.81 m s
at
+
(– 9.81) t
8.66
t
=
=
0.883 s
9.81
v
Time
taken
to
reach
Time
taken
to
cover
R
=
2
(0.883)
=
maximum
the
u
=
5 m s
,
horizontal
=
0.883 seconds
distance:
1.766 seconds
–1
vi
height
–2
a
=
0 m s
,
t
=
1.766 s
1
2
s
=
ut
+
at
2
1
2
=
(5
×
1.766)
+
(0)(1.766)
2
=
vii
In
order
velocity
to
in
find
the
the
8.83 m
velocity
vertical
and
of
the
projectile
horizontal
at
P
,
direction
we
at
P
.
need
The
to
find
velocity
the
in
1
v
=
5 m s
H
the
θ
horizontal
acceleration
in
direction
that
is
constant,
since
there
is
no
component
direction.
–1
v
=
5 m s
H
The
time
taken
to
reach
the
point
P
1
2
s
=
ut
+
at
2
1
v
2
7
=
5t
+
(0)t
1
v
=
5.07 m s
2
V
7
Figure 3.3.5
t
=
=
5
24
1.4 s
is
determined
as
follows:
of
Chapter
The
vertical
determined
component
as
follows
v
V
of
velocity
(Figure
+
at
time
t
=
1.4 seconds
3
Kinematics
is
3.3.5):
=
u
at
=
8.66
=
– 5.07 m s
+
(– 9.81)(1.4)
V
–1
V
V
The
resultant
velocity
is
determined
2
v
=
√
as
follows:
2
v
+
v
H
V
2
2
5
+
5.07
v
=
v
=
7.12 m s
θ
=
tan
√
–1
V
5.07
V
–1
(
–1
)
V
=
tan
(
)
5
=
45.3°
H
Figure
3.3.6
shows
the
effect
of
air
resistance
on
the
motion
of
the
ball.
A
1
10 m s
B
60°
Figure 3.3.6
Effect of air resistance on the motion of the ball
Key points
ᔢ
When
ᔢ
The
object
ᔢ
The
horizontal
and
ᔢ
The
horizontal
component
is
ᔢ
an
object
is
travels
projected
both
at
an
angle
horizontally
vertical
motions
of
the
it follows
and
are
vertically
treated
velocity
a
parabolic
path.
simultaneously.
separately.
remains
constant
if
air
resistance
ignored.
Gravity
acts
vertically
and
affects
the
vertical
component
of
the
velocity.
25
4
Dynamics
4.
1
Dynamics
Learning outcomes
On
completion
should
be
able
of
this
1
Linear
section,
you
The
linear
and
to:
momentum
momentum
velocity
( v).
momentum
ᔢ
define
linear
state
of
in
the
same
an
is
object
a
is
vector
direction
as
the
Newton’s first
and
product
quantity.
the
momentum
of
The
velocity
its
mass
direction
(Figure
( m)
of
4.1.1).
the
The
–1
unit
ᔢ
is
(p)
Momentum
of
momentum,
by
definition
is
kg m s
second
v
laws
ᔢ
p = mv
define
the
m
newton.
Figure 4.1.1
Defining linear momentum
Definition
Newton’s first
Linear
the
momentum
product
velocity
in
of
a
a
is
defined
body’s
given
mass
as
and
direction.
If
a
it
to
to
book
move
space
can
is
make
placed
it
law of
on
move.
indefinitely
affects
move
in
its
a
a
table
a
rock
If
in
a
motion.
straight
motion
it
is
will
straight
It
is
line
stay
thrown
line
difficult
forever
.
there
in
until
to
On
until
outer
the
grasp
Earth
a
force
space,
it
gravity
the
of
applied
to
continue
some
concept
frictional
is
will
that
forces
object
an
are
in
object
always
Equation
present.
slows
p
=
If
a
ball
down
the
is
kicked
motion
it
of
will
the
eventually
come
to
rest
because
friction
ball.
mv
Newton
–1
p
–
momentum/kg m s
m
–
mass
stated
continue
of
to
that
move
a
body
with
a
will
stay
constant
at
rest
velocity
or
if
it
unless
is
moving,
acted
will
upon
by
an
body/kg
external
force.
This
law
is
stating
that
a
force
is
required
to
produce
a
–1
v
–
velocity
of
body/m s
change
in
velocity
(acceleration)
(Figure
4.1.2).
a
F
Newton’s first law
v = 0
A force
Newton’s first
that
a
body
continues
law
stays
to
of
at
move
motion
rest
with
or
if
is
required
to
cause
an
object
to
accelerate from
rest
states
moving
a
F
uniform
v = constant
velocity
unless
acted
upon
by
an
A force
external force.
is
required
to
make
an
object
accelerate
when
moving
with
a
constant
velocity
Figure 4.1.2
Newton’s
Newton’s
second
proportional
the
Newton’s second law
force
second
to
law
the
law
states
applied
that
force
the
and
rate
of
takes
change
place
in
of
momentum
the
direction
is
in
which
acts.
Mathematically
this
law
can
be
expressed
as
follows:
Δp
Newton’s
second
law
states
F
that
∝
(1)
Δt
the
rate
of
change
of
momentum
is
Where
proportional
to
the
applied force
in
takes
place
the force
in
the
acts.
direction
in
is
force,
Δp
is
the
change
in
momentum
and
Δt
is
the
change
time.
which
Consider
applied
The
26
F
and
an
to
object
the
initial
of
object
mass
for
momentum
t
is
m
travelling
seconds
mu.
and
with
its
a
velocity
velocity
u.
changes
A
to
force
v.
F
is
Chapter
The
final
momentum
The
change
in
is
Δp
=
Dynamics
mv
momentum
∴
4
mv
is
–
mv
–
mu
mu
(2)
Equation
Substituting
Equation
(2)
into
mv
∴
F
–
Equation
(1):
mu
F
=
ma
∝
t
The
proportionality
proportionality
sign
is
constant
k
now
is
=
k
–
(
=
km
v
But
acceleration
a
–
an
equal
sign
and
F
–
m
–
force/N
mass/kg
a
–
acceleration/m s
a
)
–
(
–2
mu
t
v
F
with
included.
mv
F
replaced
u
)
t
u
=
t
∴
F
1 newton
is
the
=
force
kma
required
to
give
a
mass
of
1 kg
an
acceleration
of
–2
1 m s
the
.
The
reason
equation
why
equal
to
the
newton
is
defined
in
this
way
is
to
make
k
in
1.
Example
A
wooden
block
of
mass
0.50 kg
rests
on
a
rough
horizontal
surface.
A
wooden
force
of
15 N
is
applied
to
the
block.
The
frictional
force
acting
on
a
the
block
block
is
6 N
(Figure
4.1.3).
Calculate
F
15
–
the
acceleration
of
the
block.
15 N
6 N
6
–2
Acceleration
a
=
=
=
m
18 m s
0.5
Figure 4.1.3
Example
A
box
of
Figure
mass
4.1.4.
60 kg
is
being
pulled
along
a
rough
surface
as
shown
in
Calculate:
Exam tip
a
the
component
of
the
80.2 N
b
the
component
of
the
50 N
c
the
acceleration
force
in
the
OX
direction
Always
force
in
the
OX
remember that
equation
acting
on
it
is
of
the
box
in
the
direction
Component
in
the
OX
direction
80.2
b
Component
in
the
OX
direction
50
c
T
otal
horizontal
+
of
OX
if
the
frictional
F
=
ma,
F
is
resultant force.
force
25 N.
a
65.7
in the
direction
19.5
force
=
acting
in
the
×
×
cos 35°
cos 67°
direction
=
=
65.7 N
80.2 N
19.5 N
OX
to
the
right
35°
85.2 N
box
67°
Resultant
force
acting
on
the
F
box
=
85.2
–
25 N
=
60.2 N
60.2
–2
a
=
=
=
m
1.00 m s
60
50 N
Key points
Figure 4.1.4
ᔢ
Linear
ᔢ
Newton’s first
move
ᔢ
momentum
with
Newton’s
law
is
states
uniform
second
proportional
to
the
that
velocity
law
the
product
states
a
acts.
1
is the force
a
body
unless
that
applied force
the force
of
body’s
stays
acted
the
rate
and
mass
at
rest
upon
of
takes
by
and
or
an
change
place
in
if
velocity.
moving
continues
to
external force.
of
momentum
the
direction
in
is
which
–2
ᔢ
newton
required to
give
a
mass of
1 kg
an
acceleration of
1 m s
27
4.2
Dynamics
Learning outcomes
On
completion
should
be
able
of
this
2
Impulse
section,
you
Consider
applied
to:
velocity
ᔢ
understand
ᔢ
draw
the
term
an
to
object
the
to
of
object
increase
mass
for
to
a
v.
ᔢ
mv
understand
–
mu
represents
the
concepts
of
as
Δp,
the
where
and
is
third
give
law
of
examples.
The
quantity
concept
of
example,
when
ball
he
for
period
If
a
ball
to
a
longer
of
time,
while
a
velocity
seconds.
Newton’s
u.
A
This
second
force
force
law
F
is
causes
we
can
the
write
mv
–
mu
in
momentum
of
the
object
and
can
be
momentum.
time.
the
travelling
to
zero.
his
If
ball
to
the
into
will
in.
It
This
force
the
time
exerted
a
he
reduces
the
the
effect
get
bat
the
let
hand
time
for
in
the
in
exerted
a
N s.
force.
ball
to
For
go
further
with
exerted
for
ball
fall
same
the
a
longer
into
his
his
direction
momentum
on
The
momentum.
the
the
the
contact
is
in
is
of
the
bat
change
would
force
impulse
can
by
his
longer
of
time
greater
ball
a
the
keeping
moving
takes
unit
batsman
by
have
catch
same
the
bat,
the
The
account
cricket,
with
had
at
force–time
force
of
F
second
vary
in
a
to
force
applied
the
In
ball.
object
can
acts
to
in
in
a
used
on
an
a
how
ball
cannot
the
it
hand
of
by
the
the
the
ball
ball.
a
player
force
is
it
a
force
graph
varies
shows
a
over
a
constant
t
exerted
as
how
first
strikes
difficult
of
graph
by
for
is
the
ball
the
the
with
racket
player
struck
represents
example
the
force/N
by
on
to
the
his
racket.
the
apply
racket
ball
a
The
would
constant
and
the
the
tennis
the
area
change
under
the
in
momentum
graph
represents
ball.
force/N
F
F
time/s
Δ t
Figure 4.2.1
The
constant.
For
momentum
illustrate
interval
deforms
be
force–time
question.
to
object.
time
tennis,
practice
The
it
be
over
of
shows
under
change
it
game
time.
force
area
as
applied
graph
with
the
graph
time
being
Suppose
the
it
Δp
impulse.
takes
of
=
graphs
period
of
called
game
strikes
reduce
The
is
impulse
in
a
was
F–t
A
Ft
fieldsman
hand,
28
to
t
mass
Ft
Newton’s
motion
=
change
p
weight
state
with
time
interpret force–time
expressed
and
of
impulse
graphs
ᔢ
moving
According
Ft
and
m
period
Force–time graphs
time/s
Chapter
Mass
The
and
mass
of
body
is
Dynamics
Definition
weight
a
4
the
amount
of
matter
contained
in
it.
Bodies
have
a
Mass is the measure of a body’s inertia.
property
to
start
moving
moving,
Mass
associated
is
once
a
with
when
it
is
scalar
in
it
them,
is
at
rest.
motion.
quantity
called
It
Mass
and
the
inertia.
is
is
SI
also
a
It
the
is
is
reluctance
reluctance
measure
unit
the
the
of
a
of
body ’s
kilogram
a
of
body
a
body
to
stop
Definition
inertia.
(kg).
Suppose
The
your
mass
is
60 kg
on
Earth.
If
you
were
to
take
a
trip
to
the
Moon,
weight
exerted
mass
The
will
still
weight
dependent
travel
to
be
of
on
the
of
an
object
is
the force
your
by
gravity
on
it.
60 kg.
an
object
the
is
the
force
gravitational
Moon,
where
the
field
exerted
by
strength.
gravitational
gravity
on
Therefore,
field
it.
if
strength
W
eight
you
is
were
less
W
=
mg
W
–
weight/N
m
–
mass/kg
g
–
gravitational field strength/N kg
is
to
than
that
–1
of
the
Earth,
W
eight
is
a
you
would
vector
quantity
Newton’s third
For
on
in
an
it.
a
object
The
of
on
the
direction.
upwards
a
much
and
the
smaller
SI
unit
is
weight
the
than
newton
on
the
Earth.
(N).
law
resting
weight
vertical
vertically
have
to
the
surface
object
acts
Therefore,
balance
the
of
a
table,
there
downwards.
there
must
weight
of
be
the
are
The
an
two
object
equal
object.
forces
is
not
force
This
acting
acting
force
Newton’s third law
moving
is
If
B,
the
nor mal
reaction
to
the
surface
(Figure
a
body A
exerts
a force
on
body
called
then
body
B
exerts
an
equal
and
4.2.2).
opposite force
R – normal
on
body A.
reaction
W – weight
Figure 4.2.2
Forces acting on an object on a table
Examples of
1.
Force
When
a
each
F
between
two
force
is
parallel
parallel
wire
B.
a
Consider
current
Wire
B
I
in
exerts
the
a
conductors
two
wires
same
force
A
motion
F
A
are
and
direction.
flows
in
the
current
them
is
but
act
same
in
opposite
direction,
the
repulsive
in
on
wire
wires
(Figure
in
force
parallel
Wire
A.
F
A
When
between
opposite
and
them
each
to
F
each
a
are
other
,
other
,
force
of
equal
B
current
directions,
to
exerts
A
directions.
the
flows
conductors
adjacent
B
B
magnitude
the
law of
current-carrying
current-carring
experienced.
carrying
on
Newton’s third
is
in
the
wires
attractive.
the
force
When
between
4.2.3).
I
I
F
F
B
I
F
B
A
F
A
I
A
Figure 4.2.3
B
A
B
Force between current -carrying conductors
29
Chapter
4
Dynamics
2.
Force
When
two
between
a
between
force
charges
them.
F
on
charged objects
are
placed
Consider
object
B.
close
two
to
each
charged
Object
B
other
,
objects
exerts
an
A
equal
a
force
and
B.
and
is
experienced
Object
opposite
A
exerts
force
F
A
object
the
A.
force
(Figure
on
B
If
the
is
charges
repulsive.
are
If
the
the
same
charges
(both
are
positive
different,
or
the
both
force
negative),
is
attractive
4.2.4).
F
F
B
A
+
+
A
B
F
F
B
A
+
A
Figure 4.2.4
Force between charged objects
3. Gravitational force
The
Moon
centripetal
exerts
a
B
orbits
force
force
F
the
Earth.
required
on
between two
the
The
for
the
Moon.
Earth’s
Moon
At
the
masses
gravitational
to
orbit
same
the
time,
field
Earth.
the
provides
The
Moon
the
Earth
exerts
a
force
E
F
on
the
Earth
(Figure
4.2.5).
M
F
F
M
E
Moon
Earth
Figure 4.2.5
4. A
rocket
When
the
Force between two masses
a
rocket
exhaust.
momentum,
law,
the
gas
causing
5. A
A
it
of
air
to
a
force
the
on
accelerates
When
30
is
a
a
volume
out,
downward
equal
and
it
of
force.
opposite
gas
is
expelled
experiences
According
force
on
the
a
from
change
to
in
Newton’s
rocket,
third
therefore
helicopter
able
to
change
If
fly
If
vertical
the
because
exert
in
Newton’s
blades.
resultant
stationary.
in
large
pushed
accelerate.
to
the
a
being
an
helicopter
has
According
is
resulting
hovering
blades
launched,
gas
exerts
helicopter
The
is
As
a
third
this
is
Newton’s
downward
momentum
law,
force
motion
force
of
is
greater
air
equal
zero
on
the
downwards,
the
is
force
third
to
and
than
exerts
the
the
the
law.
air
weight
weight
rise
equal
of
helicopter
of
the
rotating
around
giving
an
The
and
the
to
them.
a
force.
opposite
helicopter
,
remains
helicopter
,
it
upwards.
applying
Newton’s
third
law
ᔢ
The
two
forces
are
at
all
times
ᔢ
The
two
forces
act
in
opposite
ᔢ
The
two
forces
are
of
the
ᔢ
The
two
forces
act
on
same
different
it
is
important
equal
in
to
note
the
following:
magnitude.
directions.
type
(gravitational,
bodies.
electric,
etc.)
Chapter
4
Dynamics
Example
–1
A
ball
of
mass
0.35 kg
hits
a
wall
with
a
speed
of
12 m s
and
rebounds
–1
from
with
wall
the
the
on
wall
wall
the
lasts
its
for
initial
0.2 s.
path
with
Calculate
a
speed
the
of
average
7.2 m s
force
.
The
exerted
impact
by
the
ball.
mv
F
along
–
mu
(0.35
=
×
– 7.2)
–
(0.35
×
12)
=
=
t
33.6 N
0.2
Example
Figure
time.
4.2.6
The
below
mass
shows
of
the
how
object
the
is
force
2 kg
acting
and
is
on
an
initially
object
at
varies
with
rest.
force/N
40
time/s
0
20
Figure 4.2.6
Calculate:
Exam tip
a
the
change
b
the
velocity
c
the
acceleration
in
momentum
of
the
object
of
the
after
mass
10
during
the
first
10
seconds
Always
seconds
take
velocity
of
the
object
during
the
first
10
the
a
Change
b
change
in
in
momentum
momentum
Δp
=
mv
200
=
2(v)
–
v
=
100 m s
of
the
=
area
=
½
mass
under
×
10
×
during
into
direction
account
of
the
when finding
seconds
the
d
the
the
20-second
change
in
momentum
of
a
body.
interval.
graph
40
=
200 N s
mu
–
2(0)
–1
v
–
u
100
–
0
–2
c
Acceleration
=
=
=
t
d
Change
in
10 m s
10
momentum
=
total
area
=
½
20
×
under
×
40
graph
=
400 N s
Key points
ᔢ
Impulse
is
ᔢ
The
ᔢ
Mass
ᔢ
Weight
ᔢ
The
mass
ᔢ
The
weight
ᔢ
Newton’s
area
is
exerts
the
under
the
is
an
time
effect
a force–time
measure
the force
of
a
of
third
equal
of
the
exerted
body
a
of
and
graph
inertia
by
can
states
vary
that
is
of
equal
a
gravity
remains fixed
body
law
a force.
if
if
and
the
a
opposite force
to
the
change
momentum.
body.
on
is
it.
not
affected
by
gravitational field
body A
on
in
exerts
location.
strength
a force
on
varies.
body
B,
body
B
body A.
31
4.3
Collisions
Learning outcomes
On
completion
should
be
able
of
this
The
section,
you
to:
Figure
4.3.1
B
rest
is
at
each
ᔢ
state
the
principle
conservation
of
principle of
ball
shows
and
is
m.
conservation of
two
ball
A
Ball
A
identical
is
moving
balls
A
and
towards
eventually
momentum
B
collides
B
on
with
with
a
a
horizontal
velocity
ball
of
surface.
v.
The
Ball
mass
of
B.
of
During
momentum
exerts
the
on
collision
ball
B
is
of
F
the
and
balls,
the
the
magnitude
magnitude
of
of
the
the
force
force
that
that
ball
B
ball
A
exerts
on
AB
ᔢ
apply
the
principle
of
ball
A
is
F
(Figure
4.3.2).
BA
conservation
of
momentum
The
ᔢ
distinguish
between
elastic
balls
are
in
contact
for
a
length
of
time
Δt.
After
speed
of
ball
A
is
v
,
and
the
speed
of
ball
B
is
v
A
inelastic
the
collision,
the
and
in
the
directions
shown
B
collisions.
in
Figure
4.3.3.
Analysis
A
B
Ball B
v
Figure 4.3.1
The
initial
The
final
momentum
momentum
of
of
ball
ball
B
B
is
is
zero
because
it
is
initially
at
rest.
mv
B
Before collision
The
change
in
momentum
of
the
ball
B
is
therefore
mv
B
A
B
According
F
to
Newton’s
law
F
BA
AB
F
Δt
=
mv
AB
Figure 4.3.2
second
During collision
B
Ball A
v
The
initial
momentum
The
final
The
change
of
ball
A
is
mv
v
A
B
momentum
of
ball
A
is
mv
A
in
momentum
of
the
ball
A
is
therefore
–
mv
mv
A
According
Figure 4.3.3
to
Newton’s
second
law
After collision
F
Δt
=
mv
BA
–
mv
A
Total momentum
T
otal
momentum
before
T
otal
momentum
after
the
the
collision
collision
=
=
mv
mv
+
mv
A
According
ball
B
by
to
Newton’s
ball
A
(F
)
is
third
equal
law ,
to
the
the
B
magnitude
magnitude
of
of
the
the
force
force
exerted
exerted
by
on
ball
AB
A
by
ball
B
(F
).
They
act
in
opposite
directions.
BA
∴
F
=
–F
AB
(Multiplying
Equation
(1)
AB
(1)
by
Δt)
F
Δt
=
–F
AB
Δt
AB
Definition
mv
=
–(mv
B
The
principle
of
conservation
–
of
mv
=
mv
–
mv
B
momentum
states
that for
before
the
collision
momentum
that
total
no
is
after
A
any
∴
system,
mv
=
mv
momentum
equal
to
collision
external forces
act
the
total
provided
on
+
A
the
Equation
is
equal
(2)
to
illustrates
shows
the
the
total
that
the
total
momentum
principle
of
momentum
just
after
conservation
of
the
Momentum
is
conser ved
in
all
just
mv
(2)
B
before
collision.
momentum.
system.
32
mv)
A
collisions
the
This
collision
example
Chapter
Elastic
In
all
and
inelastic
collisions,
is
conserved.
This
is
not
the
case
with
An
kinetic
energy.
Kinetic
energy
is
the
energy
possessed
by
a
body
by
elastic
its
motion.
If
kinetic
energy
is
conserved
in
a
collision,
it
is
said
to
elastic
said
to
T
able
collision .
be
If
kinetic
energy
is
not
conserved,
the
collision
energy
inelastic
kinetic
compares
elastic
and
inelastic
one
in
which
is
conserved.
collision
is
one
in
which
is
inelastic
4.3.1
is
be
An
an
collision
virtue
kinetic
of
Dynamics
Definitions
collisions
momentum
4
energy
is
not
conserved.
collisions.
Example
Table 4.3.1
–1
An
object
of
mass
3 kg
travelling
at
6 m s
strikes
another
object
of
mass
Elastic
Inelastic
collision
collision
Yes
Yes
Yes
No
Yes
Yes
–1
5 kg
and
travelling
move
off
at
1 m s
with
a
in
the
velocity
same
v.
direction.
Calculate
v
The
(Figure
objects
stick
together
4.3.4).
Momentum
1
1
6 ms
1 ms
conserved
v
Kinetic
energy
3 kg
8 kg
5 kg
before
conserved
collision
after
collision
Total
energy
Figure 4.3.4
conserved
Assuming
T
otal
no
external
momentum
forces
before
(3
×
6)
act
on
the
collision
+
(5
system
(the
two
objects):
=
total
momentum
1)
=
(3
5) v
23
=
8v
v
=
×
+
after
collision
23
–1
=
2.88 m s
8
–1
Both
both
objects
objects
Always
move
were
assume
off
with
a
originally
a
sign
velocity
moving
of
2.88 m s
in
the
same
direction
in.
convention.
ᔢ
Motion
to
the
right
ᔢ
Motion
to
the
left
is
is
taken
taken
as
as
positive
negative
velocities.
velocities.
Example
–25
A
stationary
particles
A
nucleus
and
B.
of
The
mass
3.65
particles
A
×
10
and
kg
B
decays
move
off
to
in
produce
opposite
two
directions.
–27
Mass
of
A
Mass
of
B
=
6.64
×
10
Key points
kg
–25
=
3.59
×
10
kg
ᔢ
7
The
initial
(Figure
speed
of
A
is
1.7
×
10
The
principle
.
Calculate
the
initial
speed
of
of
B
4.3.5).
momentum
system,
7
1.7
×
10
before
1
the
ᔢ
momentum
the
In
an
=
total
momentum
×
10
7
×
– 1.7
×
10
ᔢ
–25
)
+
(3.59
×
10
×
–19
=
– 1.13
equal
after
to
collision
external forces
act
system.
×
10
is
collision,
kinetic
conserved.
In
an
inelastic
collision,
kinetic
v)
energy
0
momentum
after
–27
(6.64
no
elastic
energy
=
a
after
Figure 4.3.5
0
is
that for
B
on
before
before
total
collision
provided
A
momentum
conservation
states
ms
the
T
otal
of
–1
m s
is
not
conserved.
–25
+
3.59
×
10
v
ᔢ
Total
energy
is
conserved
in
all
–19
1.13
×
10
5
v
=
=
3.14
×
10
–1
m s
collisions.
–25
3.59
×
10
33
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
11
accompanying CD.
1
An
object
two forces
Q
and
1
When
a
substance
is
heated,
the
energy
required
its
Energy
where
a
temperature
required
c
is
State
a
=
is
given
mass
×
c
by
×
the
=
acted
act
at
8 N,
upon
an
by
angle
determine
two forces
of
the
θ. Given
P
and
that
Q. The
P
resultant force
=
6 N
acting
to
on
raise
is
the
object
when:
expression:
change
in
a
θ
=
0°
b
θ
=
180°
c
θ
=
90°
d
θ
=
120°.
temperature
[8]
constant.
the
quantities
that
are
SI
base
quantities. [2]
12
Distinguish
between
a
systematic
and
a
random
error.
b
Determine
SI
base
the
units
of
the following,
in
terms
units:
13
i
energy
ii
the
required
A
student
constant
c.
base
three
base
wire
using
quantities
and
their
How
corresponding
units.
[3]
b
can
a
A
small
metal
sphere
viscous fluid. As
drag force
SI
base
F,
units
of
of
it falls
where
F
=
radius
at
a
krv.
r
is
speed
k
is
a
dropped
v,
it
into
experiences
constant. State
r
is
the
of
a
error
in
How
can
the
14
Distinguish
15
A
acting
the
is
on
a
glass
given
marble,
η
by
is
F
marble falling
=
the
6πrηv,
P
where
quantity
viscosity
of
and
v
is
the
velocity
of
the
marble.
=
=
P
–
6.
12
the
units for viscosity.
With
the
aid
of
The
statement
written
as
an
example,
“The
the
Explain
why
respect
to
an
the
explain
what
is
meant
Write
an
magnitude
of
a
physical
a
product
equation
units
equation
of
a
number
and
a
unit”.
of
b
the
State
the
if
it
must
is
which
principle
balance
one
the
[2]
random
error
in
diameter?
[2]
between
precision
and
accuracy.
[3]
S
is
determined from
the
equation
of
0.02 m
and
Q
=
1.84
uncertainty
in
±
0.02 m. Calculate
S.
[2]
of
wishes
piece
to
of
wire
determine
is
given
the
by
R
resistance
=
V/I. A
of
wire.
V
to
be
be
is
homogenous
with
correct.
equations
limitation
of
underlies
using
using
base
I
are
a
piece
measured.
=
12
±
0.4 V
=
1.0
±
0.2 A
absolute
the
resistance
uncertainty
of
the
wire
and
include
the
R.
[3]
[1]
homogenous,
which
and
[2]
but
the
base
mean
±
diameter
0.02 mm.
uncertainty
of
a
piece
Calculate
the
of
wire
is
percentage
in:
a
the
diameter
[2]
b
the
cross-sectional
checking
units.
units
The
0.6
[2]
Explain
systematic
quantity
incorrect.
8
any
by
17
7
gauge.
Q.
±
Calculate
6
screw
eliminate
[3]
I
is
reduce
the
resistance
V
the
student
Determine
of
5
piece
measurement?
he
percentage
student
SI
a
the
16
fluid
of
[3]
a viscous fluid
radius
diameter
the
k.
The frictional force
through
the
a
S
4
measure
micrometre
the
measuring
3
to
[2]
a
State
wishes
[2]
of
2
[3]
of
area
of
the
wire.
[3]
[2]
to
18
The volume
of
a
cylinder
is
given
by
the
expression
2
balance
equations.
[1]
V
=
πr
h. The volume
measured
9
a
State
the
difference
between vector
and
and
height
of
a
cylinder
is
as:
scalar
3
V
quantities.
h
b
Give
two
examples
=
12.0
±
examples
of
a
of
scalar
a
vector
quantity
and
=
21.0
±
0.
1 cm
the
radius
two
quantity.
[4]
Calculate
uncertainty.
10
An
at
object
an
is
angle
horizontal
being
of
30°
pulled
with force
to
horizontal. Calculate
the
and vertical
components
of
of
5 N
the
acting
the
5 N force.
[2]
34
0.5 cm
[2]
of
the
cylinder,
with
its
[5]
Revision questions
19
a
Differentiate
between
displacement
and
distance
24
A
metal
ball
is
thrown vertically
upwards
with
1
an
–1
travelled.
b
State
[2]
how
you
would
displacement–time
c
State
and
20
Define
how you
determine
velocity from
initial velocity
a
graph.
a
Ignoring
[1]
i
the
would determine the displacement
acceleration from
the following
a velocity–time
graph.
air
[2]
distance
determine:
travelled
by
the
ball
after
[2]
the velocity
iii
the
time
of
the
taken for
ball
the
after
ball
0.9 seconds
to
reach
its
[2]
maxi-
height
[2]
[2]
acceleration.
[2]
b
the
The
variation
with
time
t
of
the
velocity
v
of
a
maximum
ball falls
Sketch
The
resistance,
ii
iv
21
.
terms:
velocity
b
15 m s
0.9 seconds
mum
a
of
to
height
the
reached
ground
the velocity–time
and
by
the
bounces
graph for
the
ball.
[2]
twice.
metal
ball.
cyclist
[3]
travelling
down
a
slope
is
illustrated
below.
c
Show
how
travelled
it
by
is
possible
the
ball
to
determine
between
the
the first
distance
bounce
and
–1
v /m s
the
25
An
second
object
is
bounce.
projected
[2]
with
an
initial velocity
of
–1
15
5 m s
at
Ignoring
an
air
angle
of
30°
to
the
horizontal.
resistance:
10
i
Calculate
the
the
velocity
initial
of
the
horizontal
component
of
object.
[1]
5
ii
Calculate
velocity
0
t /s
10
20
30
40
iii
Sketch
the
of
a
On
the
a
constant velocity
after
the
graph
the
velocity
of
the
cyclist
initial
c
the
total
acceleration
of
the
cyclist
car
the
graph,
with
show
with
time.
the
[1]
variation
of
the
the
maximum
time.
[2]
height
h
achieved
by
distance
travelled
before
approaches
a
reaching
vii
light
with
a
speed
height
Calculate
[3]
traffic
[2]
the
time
taken
to
reach
the
maxi-
taken
to
cover
the
distance
[3]
constant velocity.
A
of
object.
Calculate
mum
22
variation
[1]
vi
the
the
estimate:
maximum
b
the
[1]
show
component
Calculate
the
a
of
40 s.
v
Using
to
component
same
vertical
reaches
component
object.
graph
horizontal
cyclist
the
vertical
50
iv
The
initial
R,
where
h.
the
R
is
[2]
time
the
range
viii
Calculate
the
range
ix
Calculate
the
velocity
of
the
projectile.
[2]
R.
[2]
of
of
the
ball
at
the
point
–1
20 m s
. The
light
changes
to
red. The
driver
applies
P,
the
brake
when
at
a
distance
of
40 m from
the
where
P
is
2 m from
the
initial
starting
lights.
point.
Calculate
rest
at
the
the
deceleration
of
the
car
if
it
comes
lights.
[3]
26
A
box
plane
23
At
a
sports
day
at
school, Akil
runs
a
100 m
[3]
to
race.
of
at
mass
40°
25 kg
to
the
is
pulled
up
horizontal
a
by
smooth
a
rope
inclined
which
is
He
parallel
to
the
plane.
–2
accelerates from
the
blocks
at
a
rate
of
2 m s
for
a
4.5 seconds.
He
runs
the
remainder
of
the
race
at
Sketch
a
diagram
to
show
the forces
acting
on
box.
constant
b
Calculate
Calculate
his
the
speed
after
distance
the first
travelled
5 s.
during
acting
component
parallel
The
tension
to
the
of
the
weight
inclined
of
the
plane.
[2]
in
the
rope
is
250 N.
[2]
Determine
the
time
taken for
the
race.
the
velocity–time
graph for
the
Calculate
the
acceleration
Calculate
the
reaction force
of
the
box.
[3]
[3]
d
Sketch
the
box
the first
c
d
Calculate
[2]
5 s.
c
[3]
speed.
b
a
the
a
race.
between
the
box
and
[3]
the
plane.
[2]
35
5
Forces
5.
1
Archimedes’
and
terminal
Learning outcomes
On
completion
should
ᔢ
be
able
understand
upthrust
of
this
principle, friction
velocity
Archimedes’
section,
you
When
a
beach
principle
ball
is
placed
in
water
there
are
two
forces
acting
on
it.
to:
the
acting
origin
of
on
object
an
the
1
The
weight
2
The
upthr ust,
of
the
ball,
which
which
is
a
acts
force
vertically
acting
downwards.
vertically
upwards.
in
When
the
ball
is
placed
in
water
,
it
displaces
some
water
.
The
weight
of
a fluid
this
ᔢ
explain
effects
ᔢ
the
of
cause
terminal
and
use
the
water
is
equal
to
the
upthrust
(Figure
5.1.1).
and
resistive forces
understand
of
nature,
displaced
Origin of the
concept
velocity.
Consider
h
below
a
upthrust
cylinder
the
of
surface
height
of
a
h
fluid
and
of
cross
density
sectional
ρ
(Figure
area
A
at
a
distance
of
5.1.2).
1
(See
19.1
and
19.2
for
coverage
of
density
and
pressure.)
upthrust
V
olume
of
fluid
displaced
=
volume
Mass
of
fluid
displaced
=
density
Weight
of
fluid
displaced
=
mass
=
ρAhg
=
ρgh
of
cylinder
=
Ah
=
ρAh
ball
×
×
volume
gravitational
field
strength
water
(1)
weight
Pressure
p
1
Figure 5.1.1
Force
The forces acting on an
exerted
on
top
of
1
surface
F
=
pA
=
ρgh
1
A
1
object immersed in a fluid
Pressure
p
=
ρgh
2
Force
p
exerted
on
bottom
surface
F
2
=
pA
=
ρgh
2
1
Upthrust
h
A
2
=
F
1
–
F
2
=
1
ρgh
A
–
ρgh
2
A
1
A
=
h
ρgA(h
2
–
h
2
But
h
=
h
–
)
1
h
2
1
h
Upthrust
From
=
Equations
(1)
and
(2),
upthrust
=
ρgAh
(2)
weight
of
fluid
displaced.
Resistive forces
When
p
a
ball
is
rolled
on
the
floor
,
it
eventually
comes
to
rest.
The
reason
2
why
Figure 5.1.2
Archimedes’
a
body
an
upthrust
principle
is
totally
states
or
the
1
slow
down
2
prevent
that
each
of
in
a fluid,
which
is
it
rest
is
because
of
friction.
Friction
is
a
force
the fluid
motion
the
of
an
motion
movement
object.
of
Frictional
that
moving
between
two
forces:
objects
and
stationary
objects
in
contact
with
partially
is
produced
when
work
is
done
against
friction.
experiences
equal
to
the
In
machines,
there
forces
are
cause
frictional
thermal
forces
energy
acting
to
be
on
moving
produced.
parts.
This
The
makes
displaced.
machines
36
to
other
.
frictional
weight
comes
opposes
Heat
submerged
ball
Origin of an upthrust
Definition
when
the
less
efficient.
Lubricating
moving
parts
with
oil
or
grease,
Chapter
inside
wear
the
and
Friction
they
tear
Frictional
that
forces
can
using
lubricants
ᔢ
using
rollers
ᔢ
polishing
force
through
acts
on
a
reduce
friction.
machine
surfaces
way,
at
and
are
the
This
makes
not
reduces
it
more
completely
microscopic
the
amount
of
efficient.
smooth.
level
Forces
they
Even
are
though
actually
rough.
ᔢ
Drag
the
because
appear
and
helps
inside
arises
may
jagged
machine
5
be
such
and
ball
surfaces
is
the
fluid .
objects
resistance
is
velocities,
the
the
(F
to
force
Air
as
reduced
as
oil
or
grease
bearings
ensure
that
opposes
travel
proportional
between
they
resistance
they
by:
to
is
a
are
smooth
the
motion
special
through
the
surfaces
air
.
velocity
type
At
of
as
of
of
low
the
in
contact
possible.
an
object
frictional
velocities,
object
( F
=
as
it
moves
force
which
the
air
kv).
At
higher
velocity
air
resistance
is
proportional
to
the
square
of
the
velocity
of
2
object
=
kv
).
terminal
velocity
Terminal velocity
Consider
jumps
a
parachutist
from
an
jumping
aircraft,
his
from
initial
an
aircraft.
velocity
is
When
zero.
The
the
parachutist
initial
acceleration
–2
is
9.81 m s
.
The
initial
force
acting
on
the
parachutist
is
his
weight,
0
time
which
acts
because
downwards.
the
initial
There
velocity
is
is
no
zero.
drag
As
force
he
acting
falls,
his
on
him
velocity
at
the
start
increases
and
drag force
his
acceleration
decreases.
his
velocity
acts
The
and
resultant
force
The
upwards.
acting
on
drag
The
him
force
drag
is
F
acting
force
=
W
on
him
therefore
–
D,
is
proportional
increases
where
W
is
as
the
he
D
to
falls.
weight
R
of
the
the
parachutist
fall,
his
weight
force
acting
on
he
falling
at
is
(Figure
and
is
the
becomes
him
a
D
is
zero.
constant
drag
equal
His
force
to
acting
drag
force.
acceleration
velocity.
He
has
is
now
on
At
also
him.
this
At
some
point,
zero,
reached
point
the
which
means
terminal
in
resultant
that
velocity
5.1.3).
weight
W
Example
Figure 5.1.3
An
object
has
a
mass
of
2.2 kg.
When
the
object
falls
in
air
,
the
Terminal velocity
air
2
resistance
F
is
given
2
k
=
0.039 N s
by
F
=
kv
,
where
v
is
the
velocity
of
the
object
and
–2
m
Key points
Calculate:
ᔢ
a
the
weight
b
the
terminal
c
the
a
W
b
At
of
the
Archimedes’ principle states that
object
when a body is totally or partially
velocity
of
the
object
submerged in a fluid, it experiences
–1
acceleration
of
the
object
when
it
is
falling
with
a
velocity
of
10 m s
an upthrust which is equal to the
weight of the fluid displaced.
=
mg
=
terminal
∴
2.2
×
9.81
velocity,
the
W
=
F
F
=
kv
21.6
=
(0.039) v
=21.6 N
resultant
force
acting
on
the
object
is
ᔢ
Friction
ᔢ
When
opposes
motion.
zero.
work
friction,
is
heat
done
is
against
produced.
2
ᔢ
An
object
reaches
terminal
2
velocity
when
it falls
through
a
21.6
fluid.
–1
v
=
=
√
23.5 m s
0.039
ᔢ
–1
c
When
v
=
10 m s
,
At
resistance
F
=
(0.039)(10)
=
force
velocity,
the
weight
3.9 N
of
Resultant
terminal
2
air
acting
on
object
=
W
–
F
=
21.6
–
3.9
=
the
object
equal
to
the
drag
17.7 N
force. The
F
is
17.7
and
the
resultant force
acceleration
is
is
zero
zero.
–2
a
=
=
m
=
8.04 m s
2.2
37
5.2
Polygon
Learning outcomes
of forces
completion
of
this
section,
be
able
use
a
three
body
vector
triangle
to
ᔢ
in
ᔢ
the
concept
of
find
the
object
is
in
(Figure
magnitude
of
drawn
be
vector
an
experiment
centre
irregular
Q
free
and
R
body
act
on
an
diagram
object.
shows
Figure
all
the
5.2.1(a)
forces
shows
acting
on
the
it.
equilibrium.
The
forces
can
be
used
to
draw
a
vector
5.2.1(b)).
of
shaped
the
forces.
The
sides
Since
the
of
the
object
triangle
is
in
represent
equilibrium,
the
the
triangle
centre
gravity
describe
P,
A
equilibrium
understand
of
forces
diagram.
represent
triangle
forces
gravity
to:
The
ᔢ
of
you
free
should
centre
Polygon of forces
Suppose
On
and
will
a
triangle
closed
will
triangle.
not
be
If
the
closed.
object
The
was
vector
not
in
triangle
equilibrium,
is
drawn
as
the
follows:
to
gravity
of
1
One
2
Moving
of
the
forces
acting
on
the
object
is
selected
and
drawn
first
(e.g.
P).
an
in
an
anticlockwise
direction,
the
next
force
is
drawn.
The
lamina.
force
3
The
R
is
drawn
force
Q
is
by
starting
then
drawn
from
by
the
arrowhead
starting
at
the
of
P
arrowhead
of
R
Q
The
as
directions
in
the
free
of
all
body
the
forces
in
the
vector
triangle
are
exactly
the
same
diagram.
P
Suppose
and
P
S.
used
an
In
for
object
this
the
Suppose
is
case
a
vector
two
in
equilibrium
vector
polygon
triangle,
forces
P
and
the
Q
when
can
polygon
act
on
an
acted
be
upon
drawn.
can
be
object.
by
four
Using
drawn
The
the
forces,
same
(Figure
object
P,
Q,
R
principles
5.2.2).
is
not
in
R
equilibrium.
Q
object
to
be
direction
The
in
resultant
force
equilibrium,
opposite
to
R
a
must
acting
force
be
of
on
the
equal
exerted
on
object
is
magnitude
the
body
R.
to
In
R,
(Figure
order
acting
for
in
the
a
5.2.3).
R
(a)
(b)
P
P
Figure 5.2.1
(a)
Three forces acting on a body
(b)
A vector triangle
R
S
Q
P
Q
P
Q
(a)
(b)
P
Q
R
S
Q
R
 R
(a)
Figure 5.2.2
(a)
The free body diagram
(b)
The vector polygon
(c)
(b)
Figure 5.2.3
(a) Free body diagram
(b)
Resultant force
(c)
The vector triangle
Example
A
particle
point.
38
It
of
is
mass
being
a
Sketch
b
Draw
c
Calculate
a
a
0.51 kg
pulled
diagram
vector
the
to
by
a
supported
the
and
that
by
horizontal
show
triangle
angle
is
forces
hence
the
a
string
force
of
acting
calculate
string
makes
attached
to
a
fixed
3.2 N.
on
the
with
the
particle.
tension
the
in
the
vertical.
string.
Chapter
5
Forces
θ
T
θ
T
W = mg
3.2 N
3.2 N
W = mg
Figure 5.2.4
W
=
mg
Using
=
0.51
×
Pythagoras’
2
9.81
=
T
2
5
=
√
5.0 N
theorem
2
T
=
+
3.2
35.24
=
5.9 N
3.2
tan θ
=
5.0
3.2
–1
θ
=
Centre of
tan
(
5.0
=
)
32.6°
(Figure
5.2.4)
gravity
Finding the
Definition
centre of
gravity of
an
irregularThe
shaped
is
A
lamina
gravity
is
of
1
T
wo
2
A
a
an
thin
nail
is
holes
placed
freely
3
A
string
4
A
pencil
sheet
of
stiff
irregular-shaped
small
hang
centre of
a
The
of
near
one
of
In
the
the
the
order
to
following
edge
of
holes
the
and
locate
steps
the
are
centre
of
the
point
weight
of
a
through
body
which
appears
A
mass
nail
7
The
8
A
the
act.
the
lamina
is
made
to
to
attached
mark
to
several
it
is
then
points
on
attached
the
to
the
lamina
nail.
where
the
string
lamina
and
points
hang
all
to
lamina.
string
made
in
is
removed
step
and
a
straight
line
is
drawn
through
is
placed
attached
4.
to
6
body
taken:
mass
the
a
it.
mass
used
lamina,
made
through
from
with
is
are
material.
hangs.
5
gravity
lamina
through
the
second
hole
and
the
lamina
is
made
a
string
to
freely.
string
with
a
mass
attached
to
it
is
then
attached
to
the
nail.
centre
pencil
is
used
to
mark
several
points
on
the
lamina
where
the
string
of
gravity
hangs.
9
The
the
10
mass
and
points
The
point
gravity
of
string
made
of
in
is
removed
step
intersection
the
lamina
and
a
straight
line
is
drawn
through
8.
of
the
(Figure
two
lines
drawn
is
the
centre
of
5.2.5).
Key points
Figure 5.2.5
Finding the centre of gravity
of an irregular shaped lamina
ᔢ
A free
ᔢ
For
an
body
object
polygon
ᔢ
The
a
diagram
can
centre
body
in
be
of
seems
shows
equilibrium
all
the forces
when
acting
several forces
on
a
act
body.
on
it,
a
closed
vector
drawn.
gravity
to
of
a
body
is
the
point
through
which
all
the
weight
of
act.
39
5.3
Torque
and
Learning outcomes
On
completion
should
be
able
of
this
moment
The
section,
moment of
There
you
is
to:
not
is
a
required
ᔢ
understand
the
concept
of
open
a
the
the
door
knob
middle
door
.
It
is
of
is
positioned
the
door
positioned
at
where
at
the
the
a
edge
larger
edge
the
of
a
force
door
door
.
would
in
It
be
to
be
a
large
moment
about
the
hinges
of
the
door
.
The
order
for
moment
of
a
couple
force
ᔢ
to
why
at
torque
there
and
reason
positioned
a force
understand
moment
of
the
concept
about
defined
of
of
a force
ᔢ
state the conditions for equilibrium
ᔢ
state
ᔢ
apply
as
action
a
the
of
Consider
pivot
is
the
product
the
force
Figure
turning
of
the
from
effect
force
the
of
and
pivot.
a
the
The
force.
The
moment
perpendicular
SI
unit
is
the
of
distance
newton
a
force
of
the
metre
is
line
(N m).
5.3.1.
F
the
principle
of
moments
d
the
principle
of
moments.
O
d
Definition
θ
F
The
moment
of
a force
is
defined
Figure 5.3.1
as
the
product
perpendicular
action
of
of
the force
distance
of
the force from
O
and
the
the
Defining moment of a force
Figure 5.3.2
the
line
of
The
moment
of
force
F
about
O
=
F
×
d
pivot.
Always
used
is
Figure
The
remember
,
at
right
5.3.2,
the
moment
The
when
angles
of
a
the
moment
force
F
principle of
Consider
calculating
to
plank
P
line
of
the
about
of
the
force
O
=
moment
action
F
F
×
of
the
of
a
force,
force
determined
from
as
The
as
shown
force
F
in
force
balancing
Figure
produces
on
a
pivot
O.
Forces
F
,
F
an
an
anticlockwise
moment
F
anticlockwise
moment
The
force
F
produces
a
O
d
d
Therefore,
3
2
clockwise
moment
F
Sum
of
d
3
for
the
plank
clockwise
F
act
on
the
3
P
to
be
moments
in
about
O.
about
O.
1
d
2
3
1
d
F
2
d
and
2
5.3.3.
1
produces
F
In
follows:
moments
1
The
distance
pivot.
d cos θ
1
plank
the
the
2
about
O.
3
equilibrium
=
Sum
=
F
of
anticlockwise
moments
F
2
F
d
3
3
d
1
+
F
1
d
2
2
F
1
F
3
Figure 5.3.3
Example
Applying the principle of
Consider
a
wheelbarrow
filled
with
some
sand.
A
construction
worker
is
moments
about
to
lift
the
wheelbarrow
with
a
force
P
(Figure
5.3.4).
Calculate:
Definition
i
The
principle of
that for
a
body
moments
to
be
in
states
sum
of
the
minimum
the
wheelbarrow
value
the
magnitude
off
of
the
the
vertical
force
P,
needed
to
raise
the
legs
clockwise
of
R
when
the
legs
of
the
wheelbarrow
just
leave
moments
ground.
must
be
equal
anticlockwise
same
to
the
sum
moments
of
the
about
the
i
T
aking
moments
about
the
centre
of
the
wheel.
pivot.
Sum
40
of
ground
equilibrium,
ii
the
the
of
clockwise
moments
=
Sum
of
anticlockwise
moments
the
Chapter
480
×
0.6
=
P
×
Forces
1.4
0.8 m
0.6 m
480
P
5
×
0.6
=
1.4
P
=
ii
Just
as
the
wheelbarrow
is
206 N
about
to
leave
the
ground,
it
will
be
in
equilibrium.
R
Sum
of
upward
forces
R
+
=
Sum
of
206
=
480
R
=
480
R
=
274 N
downward
forces
480 N
The torque of
A
couple
do
not
forces
consists
coincide.
F
and
F
1
The
of
A
206
Figure 5.3.4
couple
two
equal
couple
acting
on
and
tends
a
to
opposite
produce
steering
wheel
forces
whose
rotation
of
a
car
only.
lines
of
Consider
(Figure
F
action
1
two
5.3.5).
2
forces
F
and
F
1
a
a
–
are
equal
and
have
a
turning
effect
or
moment
called
2
torque
d
T
orque
of
couple
=F
d
or
F
1
The
It
SI
unit
should
steering
of
be
equilibrium
2
torque
noted
wheel
is
that
rotates
when
d
the
the
newton
metre
resultant
turning
anticlockwise.
subjected
to
(N m).
The
forces
F
effect
steering
and
F
1
is
not
wheel
zero.
is
not
The
in
F
only.
2
2
Figure 5.3.5
The torque of a couple
Example
A
ruler
the
of
length
torque
of
0.5 m
the
is
couple
pivoted
when
at
its
equal
centre.
and
Calculate
opposite
forces
the
of
magnitude
magnitude
of
3 N
3 N
60°
are
applied
T
orque
of
as
shown
couple
=
Conditions for
In
order
for
a
in
F
Figure
×
d
=
5.3.6.
3
×
0.5 sin 60°
=
1.3 N m
equilibrium
system
to
be
in
equilibrium,
the
following
conditions
must
apply.
ᔢ
The
resultant
force
acting
ᔢ
The
resultant
torque
ᔢ
The
resultant
moment
on
the
system
is
zero.
60°
is
3 N
zero.
Figure 5.3.6
is
zero.
Key points
ᔢ
The
moment
distance
ᔢ
The
of
the
principle
sum
of
the
of
a force
line
of
of
is
the
action
moments
clockwise
product
of
the force
the force from
states
moments
of
that for
is
equal
a
to
the
body
the
and
the
perpendicular
pivot.
to
sum
be
of
in
equilibrium,
the
the
anticlockwise
moments.
ᔢ
A
couple
not
ᔢ
A
ᔢ
The
consists
of
two
equal
and
opposite forces
whose
lines
of
action
do
coincide.
couple
tends
torque
of
perpendicular
a
to
produce
couple
distance
is
rotation
the
only.
product
between
of
one
of
the forces
and
the
the forces.
41
6
Work,
6.
1
Work
energy
and
Learning outcomes
On
completion
should
be
able
of
this
and
energy
Energy
section,
Energy
you
is
ᔢ
identify
state
various
the
describe
forms
principle
conservation
ᔢ
the
capacity
or
ability
to
do
work.
to:
V
arious
ᔢ
power
of
of
forms
of
energy
include:
energy
of
ᔢ
mechanical
ᔢ
ᔢ
(kinetic
and
potential)
ᔢ
electrical
thermal
ᔢ
nuclear
chemical
ᔢ
solar
energy
examples
of
(T
able
6.1.1).
energy
conversion
Energy
ᔢ
define
conversion
work.
The
Sun
reaching
Energy
Table 6.1.1
is
the
the
primary
Earth
cannot
be
is
in
source
the
created
example,
gasoline
chemical
energy
has
of
but
to
of
can
chemical
converts
energy
form
for
light
be
the
and
converted
energy
thermal
locked
and
Earth.
infrared
from
up
Most
of
radiation
one
inside
mechanical
form
it.
the
energy
(heat
to
energy).
another
.
When
burnt,
For
the
energy
.
Examples of energy conversion
Example
Energy
1
An
incandescent
bulb
2
A
book falling from
3
A
photovoltaic
4
A
battery
5
A
hydroelectric
6
A
lighted
7
A
catapult
a
being
shelf
switched
to
on
conversion
Electrical
the floor
Gravitational
cell
Light
energy
Chemical
plant
Chemical
released
Elastic
Energy
to
light
potential
into
energy
Gravitational
candle
being
energy
into
potential
into
heat
energy
electrical
to
energy
light
energy
kinetic
energy
and
sound
energy
energy
electrical
potential
energy
and
into
and
into
energy
kinetic
energy
and then
into
electrical
energy
heat
kinetic
energy
conservation
Renewable
sources
of
energy
are
derived
from
natural
sources
(sunlight,
Definition
waves,
The
principle
energy
be
states
created
of
conservation
that
nor
can
neither
destroyed,
but
can
one form
to
be
another.
Most
of
fossil
fuels.
the
energy
Fossil
replenished.
gas.
As
T
o
The
global
Therefore,
42
geothermal,
hydroelectric)
and
are
replenished
over
time.
of
energy
converted from
wind,
in
conserve
used
fuels
the
factories
oil
depends
and
gas
there
and
heavily
on
diminish,
is
a
transportation
This
greater
means
fossil
prices
need
that
fuels,
of
for
lights
ᔢ
Use
fluorescent
ᔢ
Use
natural
ᔢ
Do
ᔢ
Switch
ᔢ
Car
ᔢ
W
alk
bulbs
instead
refrigerator
electrical
with
use
leaving
a
of
incandescent
a
other
doors
bulbs.
open.
appliances
when
people.
bicycle
instead
of
a
car
.
not
in
use.
such
energy
room.
lighting.
leave
off
when
comes
they
these
energy:
off
or
of
Caribbean,
Switch
pool
homes,
non-renewable.
Caribbean
reserves
ᔢ
not
in
are
from
cannot
as
fuels
oil
be
and
increase.
conservation.
Chapter
Alternative
sources of
energy
6
Work,
energy
and
power
in the Caribbean
Table 6.1.2
Alternative
source of
1
Main feature/use
Advantages
Disadvantages
Region for
energy
potential
Solar
Solar
from
energy
the
Solar
water
homes
harnessed
Abundance
Sun.
in
heaters
and
–
Can
hotels.
of
sunlight
the Caribbean.
the
be
attached
roof
of
to
Sunlight varies throughout the day so
All Caribbean
insulated storage tanks are required.
territories
For
existing
use
large
many
amounts
solar
panels
of
electricity
are
required.
buildings.
Batteries
Solar
panels
sunlight
Solar
that
into
driers
convert
Direct
electricity.
–
used
to
dry
Very
Wind
into
required to
store
energy.
of
electricity.
effective
at
drying
crops.
crops.
2
conversion
sunlight
are
Kinetic
energy
converted
of
into
wind
Efficient
electrical
method
converting
wind
of
Large
into
capital
cost. Affects
Cuba
environment.
Jamaica
energy
using
wind
turbines.
electricity.
Wind
is
seasonal
and
variable.
Barbados
Batteries
3
Hydroelectric
Gravitational
energy
of
potential
water
stored
Efficient,
in
method
reliable
of
Huge
through
turbines
required to
capital
store
energy.
cost.
Dominica
producing
Affects
dams flow
are
ecology.
Guyana
electricity.
(Amaila falls)
Problems
to
4
Geothermal
produce
Thermal
the
energy
Earth
is
inside
used
Small
to
land
area
Very
site-specific
steam
to
and
expensive.
Guadelope
required.
Harmful
produce
with flooding.
electricity.
gases
may
come
up from
St
Lucia
generate
the
ground.
Dominica
electricity.
5
Biofuels
All Caribbean
territories
Biogas
ᔢ
Produced
breaking
by
bacteria
down
Gas
plant
can
be
cooking
used for
and
Greenhouse
Agricultural
and
animal
waste.
–
Gasohol
Mixture
Biodiesel
ᔢ
of
way
of
getting
rid
land
crops for fuel
is
used
instead
to
plant
of food for
methane.
of
ᔢ
produced.
Main
One
constituent
gases
heating.
gasoline
and
waste
Used
material.
consumption.
as
a fuel
in
some
as
a fuel
in
diesel
alcohol.
cars.
Made by chemically reacting
Used
vegetable oil with an alcohol.
engines.
Some
of
the feedstock
biodiesel
is
also
used for
used for food.
Work
The
work
moved
in
done
the
by
a
force
direction
of
is
the
the
product
of
the
force
and
the
distance
force.
Key points
W
=
Fs
(Figure
6.1.1)
W
=
Fs cos θ
(Figure
6.1.2)
F
ᔢ
Energy
to
do
is
the
capacity
or
ability
work.
θ
The
SI
unit
of
work
is
the
joule
(J).
ᔢ
F
1 joule
is
the
work
done
by
a
The
principle
energy
of
1 N
when
it
moves
through
of
1 m
in
the
direction
neither
s
energy
of
can
be
created
nor
destroyed,
=
W
Fs
=
but
Fs cos θ
can
be
converted from
one
force.
form
Figure 6.1.1
1 J
=
hand.
done
to
another.
Figure 6.1.2
1 N m
ᔢ
When
the
conservation
that
of
W
the
states
a
s
distance
of
force
someone
lifts
If
the
object
by
the
upward
muscles
in
the
an
is
object,
held
force
arm
work
stationary
of
get
the
tired
is
in
hand
even
done
its
by
the
final
because
though
upward
position,
it
no
is
force
no
work
stationary.
work
is
of
the
is
being
However
,
being
The
work
product
distance
of
done
of
by
a force
the force
moved
in
and
the
is
the
the
direction
the force.
done.
43
6.2
Energy
and
Learning outcomes
On
completion
should
be
able
of
this
power
Kinetic
section,
you
A
cricket
ball
to:
distinguish
potential
between
kinetic
ball
strikes
break
ᔢ
energy
the
glass
An
energy
object
of
1
the formula for
through
window
comes
it
from
the
will
the
air
possesses
break
kinetic
the
glass.
energy
kinetic
The
energy.
energy
possessed
by
If
used
the
the
to
ball.
mass
m
moving
with
a
velocity
v
has
a
kinetic
energy
of
2
=
mv
K
derive
travelling
glass
and
E
ᔢ
a
2
kinetic
energy
Deriving the
ᔢ
derive
equation for the
kinetic
energy of
a
body
the formula for
Consider
gravitational
potential
force
ᔢ
define
ᔢ
appreciate
F
acts
distance
power
the
an
object
of
mass
m,
travelling
with
a
velocity
of
v.
A
constant
energy
concept
Work
of
s.
on
the
The
done
by
object
object
the
and
brings
decelerates
force
F
is
W
=
it
at
a
to
rest
rate
of
while
a
travelling
(Figure
through
a
6.2.1).
Fs
efficiency.
Using
Newton’s
∴
Definition
W
Considering
Initial
The
kinetic
energy
energy
possessed
of
by
a
body
virtue
of
is
second
the
=
law
=
ma)
mas
motion
velocity
(F
of
=
v
=
u
the
object:
Final
velocity
=
0
Acceleration
=
– a
the
2
its
2
v
+
2as
motion.
2
2
0
=
v
–
=
2as
2as
2
v
1
F
2
∴
as
=
v
2
v
1
2
Work
The
s
done
loss
by
in
the
kinetic
force
F
energy
is
of
W
=
the
m
(
v
)
2
object
is
equal
to
the
work
done
by
the
force.
1
2
Figure 6.2.1
Deriving the formula for
Therefore,
the
kinetic
energy
of
the
object
is
E
=
mv
K
2
kinetic energy
Potential
energy
Definition
Potential
The
potential
energy
of
a
body
energy
possessed
by
it
by
its
state
or
be
classified
as
follows:
Gravitation
potential
energy
–
The
energy
of
a
body
by
virtue
of
its
virtue
position
of
can
is
ᔢ
the
energy
in
a
gravitational
field.
position.
ᔢ
Electrical
due
ᔢ
to
Elastic
its
potential
position
potential
deformed.
an
energy
(Example
Deriving the
potential
energy
in
–
a
–
The
electric
–
The
energy
stretched
equation for the
energy of
a
energy
possessed
by
a
charged
body
field.
possessed
spring
has
change
in
by
a
elastic
body
when
potential
energy.)
gravitational
body
h
Consider
an
vertically
upwards
of
h.
the
In
object
order
object
to
must
of
mass
with
a
maintain
be
equal
Upward
W = mg
Figure 6.2.2
44
Downward
∴
force
m
at
a
constant
a
to
the
above
velocity
constant
force
(Weight)
height
v
velocity,
weight
=
F
W
=
mg
F
=
mg
of
the
the
and
the
ground.
travels
upward
object
It
moves
through
force
(Figure
a
distance
acting
6.2.2).
on
Chapter
Work
The
work
energy
of
done
done
the
by
by
the
force
force
=
is
F
×
equal
d
to
=
6
Work,
energy
and
power
mgh
the
gain
in
gravitational
potential
object.
Equation
∴
E
=
mgh
P
P
o
Efficiency
(η)
=
×
100%
P
Power
and
efficiency
Power
defined
P
–
useful
–
input
power
output/W
o
power
The
is
is
the
work
distance
as
watt
done
rate
at
which
work
is
being
done.
The
SI
unit
of
P
power/W
(W).
by
moved
the
a
in
force
the
is
defined
direction
of
as
the
the
product
force.
W
=
of
F
the
×
force
and
the
d
Equation
W
Power
is
defined
as
the
rate
at
which
work
is
being
done.
P
=
W
t
P
=
t
F
Therefore,
we
can
write
P
×
d
=
d
.
But
recall
that
v
=
t
∴
The
P
efficiency
output
to
the
Machines
and
a
not
100%
Friction
greasing
input.
is
It
defined
is
parts
as
the
expressed
efficient.
between
moving
–
power/W
W
–
energy/J
t
–
time/s
Fv
machine
power
are
machines.
of
=
t
P
There
moving
reduces
as
are
parts
ratio
a
of
useful
power
percentage.
energy
generate
friction
the
and
losses
present
unwanted
increases
the
Equation
in
heat.
Oiling
efficiency
P
of
=
Fv
machines.
P
–
power/W
F
–
force/N
v
–
velocity/m s
Example
A
cyclist
–1
pedalling
along
a
horizontal
road
provides
a
power
of
210 W
and
–1
reaches
bicycle
a
steady
is
speed
of
6.2 m s
.
The
combined
mass
of
the
cyclist
and
112 kg.
1
v
a
b
=
6.2 m s
F
Calculate:
i
the
kinetic
ii
the
total
The
cyclist
energy
resistive
stops
of
the
force
cyclist
acting
pedalling
and
and
on
(549
+
33.9) N
bicycle
the
allows
=
forward
the
motion.
bicycle
to
(112
come
to
×
9.81
×
sin 30°) N
rest.
33.9 N
Assuming
distance
c
The
that
the
travelled
cyclist
resistive
by
decides
the
to
force
cyclist
go
up
a
remains
before
slope.
constant,
coming
The
to
angle
calculate
the
30°
rest.
of
the
weight
slope
is
30°
=
(112
×
9.81) N
to
Figure 6.2.3
–1
the
horizontal.
the
slope,
bicycle.
the
In
order
cyclist
Calculate
maintain
pedals
this
1
to
harder
i
E
=
constant
and
supplies
speed
more
of
6.2 m s
power
to
up
the
power
.
1
2
a
a
2
mv
=
Key points
3
(112)(6.2)
=
2.15
×
10
J
K
2
ii
P
=
2
ᔢ
Fv
P
F
=
=
=
v
b
Loss
in
its
energy
kinetic
energy
energy
of
possessed
a
body
by
is
virtue
of
33.9 N
6.2
kinetic
The
the
210
=
work
=
33.9
done
by
resistive
motion.
force
ᔢ
The
gravitational
potential
3
2.15
×
10
×
d
energy
of
a
body
is
the
energy
3
2.15
d
×
10
possessed
=
=
Component
of
the
weight
of
the
Frictional
force
=
112
×
=
33.9 N
cyclist
9.81
×
and
bicycle
sin 30°
=
down
the
slope
the
cyclist
must
supply
its
is:
549 N
power
to
provide
a
Power
which
ᔢ
Therefore,
of
position.
ᔢ
mg sin θ
virtue
63.5 m
33.9
c
by
force
is
defined
work
Efficiency
is
is
as
the
being
defined
rate
at
done.
as
the
ratio
of
of
the
useful
power
output
to
the
–1
549
+
33.9
=
582.9 N
up
the
incline
to
maintain
a
speed
of
6.2 m s
power
input.
3
∴
P
=
Fv
=
582.9
×
6.2
=
3.61
×
10
W
(Figure
6.2.3)
45
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
2
d
be
Calculate
of
accompanying CD.
e
1
A
small toy of
mass
50 g
attached to
a
string
the
roof of the
horizontally
inside of
and the
string
a
car. The
car
a
a
steady
Sketch
toy
of
position
a
and
the
at
diagram
indicate
an
to
the
Sketch
acting
c
on
the force–time
by
ball
on
graph
the
to
player’s
show
the force
hand.
[2]
show
the forces
direction
a
Explain
b
Distinguish
what
c
State
d
Use
is
meant
of
the
acting
on
between
by
the
mass
term
and
‘impulse’.
[1]
weight.
[2]
the
Newton’s
third
law
of
motion.
[2]
acceleration
Newton’s
laws
to
explain
how
a
rocket
is
able
[2]
the
the
Calculate
[2]
25° to the vertical.
car.
Calculate
acceleration
attached to the toy takes
angle of
to
b
horizontal
accelerates
5
up
maximum
hangs
exerted
from the
the
ball.
magnitude
of
the
toy.
the
[3]
acceleration
of
leave
the
Earth’s
surface.
[3]
resultant force
the
toy.
6
[2]
a
Define
b
State
linear
the
momentum.
law
of
[1]
conservation
of
linear
momentum.
2
a
State
Newton’s first
and
second
laws
of
[3]
motion.
–1
c
An
object
of
mass
4 kg
travelling
at
5 m s
strikes
[4]
–1
another
b
Using
Newton’s
laws
of
motion,
explain
how
is
able
to
hover
above
the
ground.
opposite
a
Define
b
State
c
An
linear
momentum.
object
of
unit
of
mass
linear
0.6 kg
momentum.
is
2 kg
travelling
at
1 m s
direction. The
objects
stick
in
together
move
off
with
a velocity
v. Calculate
v.
[3]
[2]
7
the SI
mass
[4]
and
3
of
a
the
helicopter
object
travelling
[1]
with
The
north
poles
together. When
a
in
opposite
of
two
bar
released,
directions.
magnets
the
are
magnets
Explain
how
the
held
move
off
principle
–1
velocity
of
Calculate
of
the
2.5 m s
the
of
kinetic
energy
and
the
why
a
direction
is
required for
and
not
the
other.
A
cricketer
State
throws
shows
to
this
[3]
a
ball
of
two
how
mass
the force
on
physical
elastic
quantities
that
are
conserved
in
collision.
[2]
A
skydiver
has
a
mass
of
80 kg.
He
jumps from
an
0.
15 kg. The figure
aircraft
below
applied
[2]
9
4
is
one
an
quantity
momentum
[3]
8
Explain
of
situation.
momentum
object.
conservation
the
ball from
and free falls.
He
reaches
a
terminal velocity
the
–1
of
cricketer’s
hand varies
rest
thrown
and
is
with
time. The
horizontally
to
ball
80 m s
before
opening
his
parachute. Calculate:
starts from
another
a
the
weight
b
the
air
of
the
skydiver
[1]
player.
resistance
F
acting
on
the
skydiver
when
force / N
travelling
at
terminal velocity
[1]
2
20
c
the
magnitude
d
the
acceleration
of
k
of
if
F
=
the
kv
.
[2]
skydiver
when
his velocity
–1
15
is
10
42 m s
[3]
.
a
Explain
what
b
Explain
why
is
meant
by
terminal velocity.
[2]
10
a viscous
oil
a
small
metal
eventually
sphere falling
reaches
a
through
terminal
5
velocity.
c
An
[3]
object
has
a
mass
of
1.9 kg. When
the
t /s
0
0.
1
0.2
0.3
object falls
0.4
in
air,
the
air
resistance
F
is
given
by
2
F
=
kv
,
where
k
=
0.028 N s
v
is
2
a
Estimate
b
What
under
c
is
area
change
the
Calculate
it
46
the
in
graph
the
released.
under
physical
the
graph.
quantity
[2]
does
the
velocity
velocity
of
the
object
and
–2
m
.
Calculate:
area
represent?
horizontal
the
i
the
weight
ii
the
terminal
of
the
iii
the
acceleration
object
[2]
[1]
of
the
ball
when
[2]
velocity
of
of
the
the
object
object
when
[3]
it
is falling
–1
with
a velocity
of
5 m s
.
[3]
Revision questions
11
A
cuboid
and
1
×
a
with
mass
of
3
–3
10
kg m
dimensions
3.8 kg
so
that
30 cm
is floating
its
×
in
25 cm
water
largest faces
×
of
are
15 cm
15
density
A
uniform
rests
horizontal.
B
Calculate:
are
the
b
the fraction
upthrust
of
on
of
the
the
the
cuboid
cuboid
that
is
beneath
a
the
water.
State
two
acting
a
body
to
be
weight
worker
a
(Side
of
end
weight
of
to
the
and
B. The
each
closer
diagram
on
90 N
and
0.2 m from
end.
Sketch
[2]
conditions for
one
of
supports A
located
from
[3]
b
a
plank
two
construction
a
surface
12
on
length
the
800 N
support
plank
2.00 m
supports A
of
to
2
and
plank. A
stands
0.45 m
B)
show
the forces
it.
Calculate
[2]
the force
acting
on
the
plank
at
support
the force
acting
on
the
plank
at
support
in
B.
equilibrium.
c
b
Three
[3]
[2]
co-planar force A,
B
and C
act
on
a
Calculate
body
A.
that
is
in
16
i
[2]
equilibrium.
Explain
how
a vector
triangle
can
be
used
State
two
conditions
necessary for
a
body
to
be
in
to
equilibrium.
represent
ii
Explain
the forces A,
how
the
B
triangle
and C.
illustrates
that
the
17
forces A,
B
and C
are
in
equilibrium.
Three forces
A
toy
of
mass
0.75 kg
hangs from
two
strings
below.
on
Find
an
the
object O
as
shown
resultant
of
these forces
in
the
and
as
its
shown
act
[1]
figure
c
[2]
[3]
direction
with
respect
to
the
horizontal.
below.
38 N
T
1
30°
28°
45°
T
2
O
34°
40°
20 N
22 N
W
18
The
toy
is
in
equilibrium.
Draw
a vector
A
car
of
mass
500 kg
is
travelling
along
a
horizontal
triangle
–1
to
determine
the
magnitudes
of
T
and
T
1
13
Explain
what
is
meant
by
the
centre
of
.
road
with
then
descends
gravity
of
through
a
of
a
between
couple.
the
moment
of
a force
and
a
hill
a vertical
frictional force
down
torque
of
10 m s
. The
car
[4]
[2]
Distinguish
constant velocity
2
body.
14
a
the
of
of
length
distance
200 N
of
acts
300 m
while
20 m. A
on
the
travelling
constant
car
as
it
moves
hill. Calculate:
the
a
the
initial
b
the
total
kinetic
energy
of
the
car
[2]
of
the
energy
possessed
by
the
car
at
the
top
hill
c
the
work
done
d
the
velocity
of
by
the frictional force
the
car
at
the
bottom
of
the
hill.
47
7
Circular
7
.
1
Motion
motion
in
Learning outcomes
On
completion
should
ᔢ
be
able
express
of
this
circle
Angular velocity
section,
Consider
you
to:
angular
a
displacement
an
object
object
is
made
to
object
is
initially
attached
travel
at
the
in
to
a
a
string
circular
point
A.
As
of
length
path
it
at
a
travels
r
(Figure
constant
in
an
7.1.1).
speed
v.
The
The
anticlockwise
direction
in
an
angle
is
swept
out.
When
the
object
reaches
the
point
B,
an
angle
of
radians
θ
ᔢ
understand
angular
the
concept
is
swept
angle
of
of
out.
one
The
distance
radian
is
travelled
defined
such
along
that
s
the
=
arc
of
the
circle
arc
length
is
s.
An
r
velocity
s
Therefore
if
the
radius
of
the
circle
is
r
and
the
is
s,
then
θ
=
r
ᔢ
understand
the
concepts
centripetal force
and
of
When
centripetal
the
object
circumference
acceleration.
Therefore,
the
returns
of
the
arc
to
the
length
in
one
s
∴
point
θ
one
the
distance
travelled
will
be
the
complete
revolution
s
=
2πr
2πr
=
=
=
r
Therefore,
A,
circle.
2π
r
revolution
is
equivalent
to
2 π
radians.
v
B
π
radians
=
180°
1
radian
=
57.3 °
O
The
θ
angular
velocity
ω
is
the
rate
of
change
of
angular
displacement.
θ
r
ω
=
t
B
s
v
The
A
time
taken
to
complete
on
revolution
is
called
the
period
T
A
θ
t
Figure 7.1.1
=
Defining the radian
ω
The
angular
displacement
Definition
during
one
revolution
is
2 π
2π
∴
T
=
ω
The
radian
is
defined
as
the
angle
s
The
subtended
arc
equal
at
in
centre
length
of
to
a
circle
the
by
radius
linear
of
the
object
at
any
point
in
the
circle
is
given
by
v
=
t
of
Since
the
velocity
an
the
angular
velocity
is
ω,
the
angle
swept
out
in
time
t
is
ωt
circle.
s
But,
ωt
=
r
s
Therefore,
Definition
v
=
ωtr
=
=
t
Angular velocity
ω,
is
defined
Centripetal
as
the
rate
of
change
rω
t
of
acceleration
angular
–1
displacement. The
SI
unit
is
rad s
An
object
moving
accelerating.
speed
is
in
a
However
,
accelerating.
straight
an
line
object
Consider
at
a
constant
moving
an
in
object
a
speed
circular
travelling
at
is
not
path
a
at
a
constant
constant
speed
v
Equation
in
a
circular
point
v
=
on
path
the
as
shown
circular
path
in
Figure
remains
7.1.2.
The
unchanged.
speed
The
of
the
object
direction
of
at
the
any
object
rω
is
continuously
changing.
Since
velocity
is
a
vector
quantity,
the
velocity
–1
v
–
linear
r
–
radius/m
ω
–
angular
at
velocity/m s
the
point
A,
v
is
different
from
the
velocity
at
the
point
A
change
in
velocity
B,
v
.
The
B
is
found
using
vector
subtraction.
–1
velocity/rad s
Change
in
velocity
Δv
=
v
–
B
V
ector
48
subtraction
can
be
v
A
thought
of
as
a
vector
addition
as
follows:
Chapter
Δv
=
v
–
v
B
In
order
to
perform
the
=
v
A
+
(– v
B
addition,
7
Circular
motion
)
A
the
vector
v
is
first
drawn.
The
vector
B
–v
is
then
drawn.
The
starting
point
for
this
vector
is
the
ending
point
of
A
vector
v
B
–v
v
A
B
O
v
B
Δv
Δθ
B
Δθ
v
A
A
Figure 7.1.2
Uniform circular motion
Δv
=
v
–
v
B
=
v
A
+
(– v
B
)
A
Equation
Acceleration
is
the
rate
of
change
of
velocity.
2
Δv
a
a
=
ω
r
=
Δt
–2
If
Δt
is
small,
Δθ
is
small
and
Δv
=
=
=
–
acceleration/m s
ω
–
angular
r
–
radius/m
–1
vΔθ
a
a
vΔθ
velocity/rad s
vω
Δt
2
But
v
This
=
rω
∴
acceleration
a
=
(rω)ω
is
always
=
ω
r
directed
towards
the
centre
of
the
circle.
Equation
2
mv
F
Centripetal force
=
r
It
in
has
a
already
circular
been
path
established
with
that
constant
the
speed
acceleration
is
directed
of
an
towards
object
the
travelling
centre
F
–
centripetal force/N
m
–
mass/kg
v
–
velocity/m s
r
–
radius/m
–1
of
the
circle.
required
force.
to
produce
This
According
According
force
to
to
this
must
Newton’s
acceleration.
also
Newton’s
act
second
law
F
of
force
the
=
law
motion,
is
centre
called
of
the
a
force
the
ma
∴
F
Key points
=
r
2
a
=
ω
centripetal
circle.
mv
=
r
Since
is
2
v
a
This
towards
2
But
second
2
r,
the
centripetal
force
can
be
written
as
F
=
mω
r
ᔢ
Angular
change
velocity
of
is
angular
the
rate
of
displacement.
Example
ᔢ
An
object
of
mass
1.2 kg
is
travelling
in
a
circular
path
of
radius
An
object
travelling
in
a
circular
0.8 m,
path
at
a
constant
speed
is
–1
with
a
constant
speed
of
0.5 m s
.
Calculate:
accelerating.
a
the
angular
b
the
time
c
the
centripetal
acceleration
d
the
centripetal
force
velocity
of
the
object
ᔢ
taken
for
the
object
acting
to
of
on
complete
the
the
one
revolution
direction
changing
ᔢ
The
object
directed
v
ω
=
=
r
=
=
=
the
ω
accelerating
the
and
centre
is
of
circle.
An
r
unbalanced force
centripetal force
10.1 s
to
=
is
towards
is
called
the
required
0.625
2
a
is
velocity
2π
=
ω
c
its
0.625 rad s
ᔢ
T
object
0.8
2π
b
the
hence
0.5
–1
a
of
and
changing.
object
object.
The
2
=
(0.625)
(0.8)
produce
a
centripetal
–2
=
0.313 m s
acceleration.
d
F
=
ma
=
1.2
×
0.313
=
0.375 N
49
is
7
.2
Examples
Learning outcomes
On
completion
should
ᔢ
be
able
analyse
of
this
of
An object
section,
you
to:
circular
If
motion
in
a
circular
horizontal
a
stone
tension
the
motion
attached to
a
1
string
being
whirled
in
a
path
is
in
attached
the
circular
string
to
a
string
provides
and
the
whirled
in
centripetal
a
circular
force
path,
necessary
the
to
maintain
motion.
circle
ᔢ
analyse
motion
in
a
vertical
circle
Horizontal
ᔢ
analyse
the
motion
of
a
circle
conical
Consider
an
object
of
mass
m,
attached
to
a
string
of
length
r
being
pendulum.
whirled
As
v
in
a
horizontal
mentioned
force
needed
earlier
,
for
the
circle
the
with
tension
object
to
a
T
move
constant
in
in
the
a
speed
string
circular
v
as
in
provides
path.
Figure
the
The
7.2.1.
centripetal
acceleration
a
of
O
the
object
is
directed
towards
the
centre
of
the
circle.
According
to
T
Newton’s
second
law
F
=
ma
T
=
2
mv
∴
W = mg
r
Figure 7.2.1
An object moving in a
If
the
string
breaks,
the
tension
T
will
not
be
present.
As
a
result,
the
horizontal circle
centripetal
Newton’s
circle),
force
first
in
the
acting
law,
the
direction
on
the
object
of
the
object
will
fly
will
off
no
in
longer
a
instantaneous
exist.
straight
velocity
line
at
According
(tangent
the
time
to
to
the
when
the
Exam tip
string
from
Make
the
sure
that
diagrams
you
and
breaks.
the
centre
when
an
vertical
the
method
tension
object
is
of
object
the
does
not
move
off
in
a
direction
radially
away
circle.
understand
used
to
Vertical
calculate
The
in
the
whirled
in
a
Consider
whirled
circle.
circle
string,
the
an
in
object
a
at
determined.
tension
is
object
of
vertical
different
The
at
a
mass
circle
m,
attached
with
positions
tension
is
maximum
at
at
a
the
a
to
constant
and
how
the
minimum
bottom
a
of
string
speed
v.
of
tension
at
the
the
length
Figure
top
in
the
of
r
being
7.2.2
the
shows
string
circle.
is
The
circle.
mg
T
a
O
O
a
a
O
T
T
W = mg
F
=
ma
F
=
ma
F
=
ma
2
2
mv
2
T
mv
W = mg
mv
∴
=
T
T
+
mg
–
mg
=
=
r
r
r
2
mv
2
mv
∴
∴
T
=
–
T
=
+
mg
r
r
Figure 7.2.2
An object moving in a vertical circle
Example
An
object
of
mass
0.80 kg
is
attached
to
a
string
and
spun
in
a
vertical
–1
circle
50
of
radius
0.90 m
with
a
constant
speed
of
9 m s
.
Calculate:
mg
Chapter
7
Circular
motion
P
a
the
minimum
b
the
maximum
a
Minimum
circular
tension
in
tension
tension
the
in
string
the
occurs
θ
string.
when
the
mass
is
at
the
highest
point
in
the
path.
2
2
mv
Minimum
tension
T
0.80(9)
=
–
mg
T
=
–
r
(0.80)(9.81)
0.90
v
=
Q
64.2 N
m
O
b
Maximum
circular
tension
occurs
when
the
mass
is
at
the
point
in
the
r
path.
2
2
mv
Maximum
tension
T
0.80(9)
=
+
mg
=
+
r
=
A
lowest
conical
(0.80)(9.81)
0.90
W = mg
79.8 N
Figure 7.2.3
A conical pendulum
pendulum
T cos θ
T
Consider
The
an
point
P
horizontal
the
object
object
is
fixed
circle
is
of
of
mass
to
a
attached
support
radius
constant,
m
r
the
and
shown
string
to
the
in
a
string
mass
Figure
makes
an
is
PQ
made
7.2.3.
angle
of
length
to
When
of
θ
to
l.
rotate
the
the
in
a
velocity
θ
of
vertical.
m
T sin θ
Figure
Since
force
of
7.2.4
the
mass
acting
the
shows
is
forces
moving
toward
tension,
the
the
T sin θ
F
in
acting
a
centre
circular
of
provides
=
on
the
the
the
object.
path,
circle
there
O.
centripetal
The
must
be
a
centripetal
horizontal
component
force.
W = mg
ma
Figure 7.2.4
Analysing the forces acting
2
mv
∴
T sin θ
on the object
=
(1)
r
The
mass
does
not
move
in
a
vertical
direction.
60°
T
∴
T cos θ
Equation
(1)
=
divided
mg
by
(2)
(2)
F
2
T sin θ
mv
=
÷
mg
r
T cos θ
2
v
tan θ
(0.06 × 9.81) N
=
rg
Figure 7.2.5
Example
A
to
small
a
mass
rigid
of
60 g
support.
is
The
attached
mass
is
to
a
made
string.
to
One
travel
in
end
a
of
the
string
horizontal
is
circle
fixed
Key points
of
ᔢ
radius
0.18 m.
The
string
makes
an
angle
of
60°
to
the
vertical.
The
For a mass attached to a string the
mass
tension in the string provides the
takes
0.65 s
to
complete
one
revolution.
Calculate:
centripetal force required to keep
a
the
angular
b
the
centripetal
acceleration
c
the
centripetal
force
velocity
of
the
mass
an object moving in a circular path.
of
the
mass
ᔢ
acting
on
the
The tension
acting on
being whirled
d
the
a
T
tension
in
the
string.
with
ᔢ
2π
a
mass
mass
The
a
in
constant
tension
a
horizontal
speed
acting
on
is
a
circle
constant.
mass
2π
–1
=
0.65s,
ω
=
=
T
=
9.67 rad s
being
2
a
=
ω
c
F
=
ma
=
0.06
×
d
T cos 60°
=
0.06
×9.81
=
(9.67)
a
vertical
a
constant
speed
(0.18)
=
16.8 m s
ᔢ
16.8
=
When
analysing
a
conical
1.01 N
pendulum
it
is
necessary to
resolve the tension
0.06
T
circle
varies.
–2
b
r
in
0.65
with
2
whirled
×9.81
=
and
=
1.18 N
(Figure
horizontal
into
its vertical
components.
7.2.5)
cos 60°
51
7
.3
Examples
Learning outcomes
On
completion
should
be
able
of
this
of
circular
Vehicles
section,
you
In
order
the
to:
roads
ᔢ
analyse
the
motion
of
a
around
a
analyse
the
a
car
to
and
the
road
are
road
being
go
a
around
provide
designed,
care
a
bend
bend
the
is
(arc
of
a
necessary
taken
to
circle),
the
centripetal
ensure
that
friction
force.
cars
do
between
When
while
going
around
bends.
Instead
of
making
not
curved
skid
roads
off
flat,
bend
they
ᔢ
around
2
vehicle
the
going
for
tyres
going
motion
motion
of
an
are
banked.
Figure
7.3.1
shows
a
vehicle
on
a
banked
road.
aircraft
R
banking.
R cos θ
R sin θ
θ
W = mg
Figure 7.3.1
Using
A vehicle travelling along a banked road
Newton’s
second
law
F
=
ma
2
mv
∴
R sin θ
=
(1)
r
Assuming
the
vehicle
R cos θ
Equation
(1)
does
=
divided
not
move
in
a
vertical
direction.
mg
by
(2)
Equation
(2)
2
R sin θ
mv
=
÷
mg
r
R cos θ
2
v
tan θ
=
rg
An
aircraft
An
aircraft
the
weight
wings
at
flying
W
an
component
make
the
banking
of
horizontally
the
angle
of
the
aircraft
θ
aircraft.
to
lift,
the
experiences
In
order
vertical.
L sin θ
for
a
lift
the
As
the
aircraft
provides
the
necessary
turn.
L
L cos θ
θ
L sin θ
W = mg
Figure 7.3.2
52
An aircraft banking
force
aircraft
L
to
which
turn,
banks,
the
balances
it
tilts
its
horizontal
centripetal
force
to
Chapter
Using
Newton’s
second
law
F
=
7
Circular
motion
ma
2
mv
∴
L sin θ
=
(3)
r
Assuming
the
aircraft
L cos θ
Equation
(3)
=
does
not
move
in
a
vertical
direction.
mg
divided
(4)
by
Equation
(4)
2
L sin θ
mv
=
÷
mg
r
L cos θ
2
v
tan θ
=
rg
Example
4
An
in
aircraft
a
of
mass
horizontal
3.5
×
direction
10
of
kg
flies
radius
r.
with
It
is
its
wings
travelling
tilted
at
a
in
order
constant
to
fly
speed
of
–1
200 m s
Calculate:
a
the
vertical
b
the
lift
c
the
horizontal
d
the
acceleration
e
the
value
component
force
of
of
L
L
component
of
the
of
L
aircraft
towards
the
centre
of
the
circle
r
L
40°
r
O
W
Figure 7.3.3
a
V
ertical
component
L
=
weight
of
aircraft
W
=
mg
3.5
4
=
×
10
5
×
9.81
=
3.43
×
10
N
Key points
5
b
L cos 40°
=
L
=
3.43
×
10
5
3.43
×
10
ᔢ
5
=
4.48
×
10
The friction
between
the
tyres
N
cos 40°
and
c
Horizontal
component
of
L
=
L cos (90
–
the
road
provides
40)°
centripetal force
5
=
4.48
×
10
the
required
to
keep
5
×
cos 50°=
2.88
×
10
N
a
car
moving
in
a
circular
path.
5
F
2.88
×
10
–2
d
a
=
=
=
8.23 m s
ᔢ
The
horizontal
component
4
m
3.5
×
10
of
an
aircraft’s
lift,
while
2
v
e
a
banking,
=
provides
the
necessary
r
centripetal force
2
(200)
travel
3
=
=
a
it
2
v
r
required for
=
4.86
×
10
m
(Figure
along
an
arc.
7.3.3)
8.23
53
to
8
Gravitation
8.
1
Gravitational field
Learning outcomes
On
completion
should
be
able
of
this
Gravitational field
section,
you
A
gravitational
placed
to:
inside
field
this
and field
exists
field,
it
around
lines
bodies
experiences
a
that
force.
have
mass.
This
force
If
is
an
object
attractive
is
in
nature.
ᔢ
understand
the
concept
of
a
Figure
gravitational field
as
ᔢ
define
gravitational field
ᔢ
state
ᔢ
describe
an
determine
to
law
of
gravitation
experiment
the
illustrates
Earth.
If
an
the
object
gravitational
P
of
mass
m,
field
is
around
placed
spherical
inside
the
object
such
gravitational
strength
field
Newton’s
8.1.1
the
to
The
acceleration
due
gravity.
of
the
towards
Earth,
the
gravitational
lines.
The
field.
The
spaced
The
it
centre
out
the
direction
of
the
field
of
the
field
spacing
closer
experiences
of
around
the
the
The
are,
Earth
lines
lines
lines
force.
the
field
field
a
direction
of
this
force
is
Earth.
are,
the
is
gives
the
field
idea
stronger
weaker
gravitational
represented
an
is
of
by
the
the
field.
the
field.
the
direction
using
field
strength
of
The
the
of
the
more
force
on
a
Definition
test
A
a
gravitational field
body where
force when
a
is
mass
placed
a
region
mass
placed
in
the
field.
around
experiences
a
Newton’s
law of
gravitation
in the field.
All
an
bodies
that
attractive
force
on
keeps
the
the
Consider
have
force
Earth
Moon
two
mass
on
the
exert
according
in
orbit
bodies
a
Moon.
to
on
the
masses
each
Moon
Newton’s
around
having
force
The
third
law.
of
m
and
m
1
by
a
distance
r.
Newton
stated
that
The
an
Earth
equal
This
and
force
exerts
opposite
is
what
Earth.
m
M
other
.
exerts
there
respectively
and
separated
2
exists
a
force
of
attraction
between
P
these
two
product
square
bodies.
of
of
the
Newton’s
Figure 8.1.1
the
The
magnitude
masses
distance
law
can
be
of
the
of
the
bodies.
between
the
expressed
It
is
two
as
force
is
also
directly
inversely
bodies
(Figure
proportional
proportional
to
to
the
the
8.1.2).
follows.
Diagram showing the
gravitational field around the Earth
Definition
Equation
Definition
Gm
Newton’s law of gravitation states
m
1
F
=
2
–
2
that the force of attraction between
The
direction
of
a
r
gravitational field
any two bodies is directly proportional
is
the
direction
of
the force
on
a
G
–
placed
in
constant
–11
to the product of their masses and
mass
gravitational
test
(6.67
×
10
2
N m
–2
kg
)
the field.
inversely proportional to the square of
m
–
mass
of
one
body/kg
–
mass
of
other
–
distance
1
the distance between them.
m
body/kg
2
m
1
r
m
between
the
centres
2
F
F
of
mass
of
the
two
bodies/m
r
Figure 8.1.2
Newton’s law of gravitation
The
constant
of
proportionality
G
is
called
the
gravitational
–11
has
been
experimentally
determined
as
6.67
×
10
2
N m
constant.
It
–2
kg
.
The
minus
Did you know?
sign
to
If
there
is
no
negative
sign,
it
in
stated
that
the force
is
that
indicates
Newton’s
that
third
the
law
A,
is
exerting
a
force
F
on
force
is
applies.
another
attractive.
equal
54
equation
attractive.
This
It
means
is
important
that
if
one
should
body
be
the
remember
and
opposite
force
on
body
A.
body
B,
then
body
B
will
exert
an
Chapter
Gravitational field
The
gravitational
field
8
Gravitation
strength
strength
is
the
force
acting
per
unit
mass.
This
Definition
means
mass
that
the
placed
in
gravitational
the
field.
field
On
the
strength
Earth’s
is
the
surface
force
the
exerted
on
gravitational
a
1 kg
field
The
gravitational field
strength
is
the
–1
strength
is
g
=
9.81 N kg
.
Gravitational
field
strength
is
a
vector
quantity
.
force
acting
per
unit
mass.
GMm
In
Figure
8.1.1,
the
force
exerted
on
object
P
is
F
=
–
,
where
r
is
2
r
the
distance
between
the
centre
of
mass
of
M
and
m
Equation
The
gravitational
field
strength
at
P
due
to
the
mass
M
is
g
F
g
The
force
exerted
on
P
in
terms
of
the
gravitational
field
strength
is
F
=
GMm
mg
=
=
mg
m
g
–
gravitational field
–
2
–1
r
strength/N kg
GM
∴
g
=
F
–
force/N
m
–
mass/kg
–
2
r
From
the
this
point
distance
The
is
it
can
dependent
from
its
centre
be
on
of
seen
the
an
the
object
object.
is
released
The
that
mass
the
of
gravitational
the
object
field
creating
strength
the
field
at
and
mass.
acceleration due to
When
on
equation,
P
it
gravity
falls
gravitational
to
the
force
ground.
produces
The
an
force
of
gravity
acceleration
acts
which
electromagnet
is
–2
equal
to
9.81 m s
.
This
magnitude
can
be
determined
experimentally
as
metal
follows.
the
An
time
t
iron
bearing
taken
is
is
made
to
fall
through
a
known
distance
h
ball
and
recorded.
light
gate
light
gate
1
2
s
=
ut
+
at
2
s
=
h,
u
=
0
a
=
g
h
1
1
2
h
=
(0)t
+
2
gt
=
gt
2
2
2h
∴
g
=
2
t
timer
T
wo
light
distance
gates
are
between
set
up
them
as
is
shown
in
measured.
Figure
The
8.1.3
metal
and
ball
is
the
held
vertical
in
place
using
Figure 8.1.3
Measuring the acceleration
due to gravity (free fall method)
an
electromagnet.
begins
starts.
stops.
falling.
When
The
As
the
due
h
to
the
ball
ball
height
acceleration
When
the
passes
and
electromagnet
passes
through
the
gravity
through
the
measured
is
the
switched
first
second
time
t
light
light
are
off,
gate,
used
the
gate,
to
metal
the
the
ball
timer
timer
calculate
the
Did you know
g
g
varies
around
the
Earth. The
Earth
Key points
is
at
ᔢ
A
gravitational field
is
the
region
around
a
body
where
a
mass
not
the
a
perfect
sphere.
It
is
squashed
poles.
experiences
a force.
ᔢ
The
direction
mass
ᔢ
placed
Newton’s
two
law
bodies
is
proportional
ᔢ
of
in
a
gravitational field
is
the
direction
of
the force
on
a
test
the field.
of
gravitation
proportional
to
the
Gravitational field
square
states
the force
to
the
product
of
the
distance
strength
is
the force
of
of
attraction
their
masses
between
per
unit
between
and
inversely
them.
mass.
–2
ᔢ
The
acceleration
measuring
the
due
time
to
gravity
taken for
a
is
9.81 m s
mass
to
and
travel
can
be
through
determined
a
known
by
vertical
distance.
55
8.2
Gravitational
Learning outcomes
On
completion
should
be
able
of
this
potential
Gravitational
section,
When
you
an
object
gravitational
to:
floor
ᔢ
define
gravitational
ᔢ
understand the term equipotential
of
a
done
of
h,
against
gravitational
discuss
the
motion
state
the
applications
given
to
in
a
energy.
a
higher
gravitational
If
an
one,
object
while
field,
of
it
mass
possesses
m
travelling
is
moved
through
a
from
one
vertical
mass
force
gains
of
potential
gravitational
gravity.
energy.
The
The
potential
work
gain
in
done
is
energy.
equal
gravitational
to
Work
the
is
gain
potential
in
energy
by
ΔE
=
mgΔh
P
was
assumed
that
the
gravitational
potential
g
remains
constant
as
the
of
mass
geostationary
the
the
satellites
It
ᔢ
present
of
is
geostationary
is
potential
distance
ᔢ
satellites
potential
potential
building
and
is
being
moved
vertically
upwards.
For
distances
close
to
the
Earth’s
satellites.
surface,
away
this
from
remains
M
longer
can
the
be
assumed
Earth’s
constant
applies.
and
Recall
to
surface,
the
be
true.
the
equation
that
the
However
,
gravitational
for
as
gravitational
gravitational
we
field
field
move
strength
potential
strength
due
further
no
longer
energy
to
a
no
mass
M
1 kg
varies
with
distance
r
as
follows:
∞
GM
force
of
g
=
–
2
E
gravity
is
P
r
increasing
Consider
Figure 8.2.1
a
1 kg
mass
moving
away
from
the
Earth’s
surface
to
some
Moving a 1 kg mass to a
point
where
the
gravitational
field
strength
due
to
the
Earth
is
negligible.
point far away from the Earth
Assume
Earth,
gains
that
work
is
point
done
gravitational
maximum
M
this
is
infinity.
against
the
potential
gravitational
As
force
energy.
potential
the
of
At
mass
moves
gravity.
The
infinity,
away
mass
from
the
therefore
the
mass
would
some
point
P
have
its
energy.
∞
1 kg
P
Suppose
Figure
r
the
work
done
same
The
φ
=
0
in
a
1 kg
8.2.1.
mass
The
direction.
gravitational
moving
unit
is
moved
movement
from
of
Therefore,
potential
mass
(1 kg)
the
infinity
mass
negative
at
the
from
to
and
work
point
infinity
P
to
is
the
is
force
being
defined
that
of
done
as
point.
as
shown
gravity
on
the
act
the
work
in
in
mass.
done
Gravitational
GM
φ
=

—
potential
is
a
scalar
quantity.
The
gravitational
potential
at
infinity
is
r
defined
Figure 8.2.2
as
being
equal
to
zero.
W
e
have
already
indicated
that
at
infinity,
Defining gravitational
the
gravitational
potential
energy
is
at
a
maximum.
Therefore,
at
any
point
potential
closer
to
the
Therefore,
The
Definition
Earth,
the
gravitational
gravitational
gravitational
potentials
potential
φ,
at
a
potential
have
energy
negative
distance
r
will
values
from
a
be
less
(Figure
point
than
zero.
8.2.2).
mass
M
is
given
GM
by:
φ
=
–
r
The
gravitational
potential
φ,
at
a
Figure
point
is
the
work
done
in
mass from
infinity
shows
the
variation
of
gravitational
potential
with
distance
moving
from
unit
8.2.3
to
that
the
Earth.
point.
gravitational
1
potential / J kg
Equation
GM
φ
=
–
r
distance/m
–1
φ
–
gravitational
potential/J kg
0
G
–
gravitational
constant
–11
(6.67
×
10
2
N m
M
–
mass/kg
r
–
distance from
–2
kg
the
)
centre
of
mass/m
Figure 8.2.3
56
Variation of gravitational potential with distance from the centre of the Earth
Chapter
Field
A
lines
and
gravitational
point
is
There
are
the
are
the
field
same.
potential
is
is
A
of
line
called
orbit
the
the
a
force
using
acting
gravitational
drawn
an
Geostationary
Satellites
represented
within
called
through
equipotential
field
on
a
field
point
where
points
line
lines.
The
mass
the
having
(Figures
direction
placed
of
at
that
gravitational
the
same
8.2.4
and
field
at
a
point.
potentials
gravitational
8.2.5).
satellites
Earth
geostationary
in
distinct
satellites .
paths.
They
There
orbit
the
are
particular
Earth
above
satellites
the
equator
.
equipotential
4
They
have
above
(1
the
day),
all
the
a
period
Earth’s
it
time.
satellites
of
to
These
have
24 hours
surface.
appears
be
Since
orbit
the
stationary
satellites
many
and
at
satellite
above
orbits
a
has
the
from
distance
a
to
3.6
period
same
west
of
point
east.
of
on
×
10
field
24 hours
the
ᔢ
weather
ᔢ
television
lines
km
equator
Figure 8.2.4
lines
Diagram showing field lines
around a spherical body
Geostationary
uses.
field
Geostationary
Gravitation
equipotentials
direction
points
8
satellites
are
used
lines
in:
monitoring
ᔢ
telephone
communication
equipotential
transmission.
lines
Global
positioning
satellites
(GPS)
are
not
geostationary.
They
have
a
4
period
of
12 hours.
They
orbit
at
a
height
of
approximately
2.02
×
10
km.
surface
Global
positioning
satellites
are
Figure 8.2.5
ᔢ
for
time
ᔢ
in
cellular
ᔢ
to
determine
synchronisation
telephony
precise
of
the
Earth
used:
location
on
ᔢ
to
track
ᔢ
to
guide
the
vehicles
Diagram showing field lines
close to the surface of the Earth
missiles
Earth.
Example
A
satellite
of
mass
2500 kg
is
placed
in
a
geostationary
orbit
at
a
distance
7
of
4.23
×
10
m
from
the
centre
of
the
Earth.
Calculate:
i
the
ii
the
speed
iii
the
acceleration
iv
the
force
v
the
mass
i
The
angular
velocity
of
the
period
the
of
a
the
satellite
of
exerted
of
of
the
by
satellite
in
its
orbit
satellite
the
Earth
on
the
satellite
Earth.
geostationary
satellite
2π
=
Key points
24 hours
2π
–5
Angular
velocity
ω
=
=
T
ii
=
24
×
7.27
×
10
–1
rad s
ᔢ
3600
The
gravitational
point
Speed
of
satellite
v
=
the
work
done
in
at
a
moving
rω
7
=
is
potential
4.23
×
10
5
×
7.27
×
10
3
=
3.08
×
10
unit
–1
mass from
infinity
to
that
m s
point.
2
iii
Acceleration
of
satellite
a
=
–ω
r
ᔢ
–5
=
(7.27
×
10
2
)
7
×
4.23
×
10
An
equipotential
Force
exerted
by
the
Earth
on
the
satellite
F
=
is
a
line
0.224 m s
drawn
iv
line
–2
=
through
points
having
the
ma
same
=
2500
=
560 N
×
gravitational
potential.
0.224
ᔢ
A
geostationary
period
of
satellite
24 hours
and
has
a
appears
GMm
v
Using
F
=
–
be
2
at
the
same
point
above
the
r
Earth
2
7
Fr
560
×
(4.23
all
the
time.
2
×
10
×
2500
)
24
Mass
of
Earth
=
=
=
6.00
×
10
kg
–11
Gm
6.67
×
10
57
to
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
7
accompanying CD.
3
An
object
circular
of
mass
path
of
0.85 kg
radius
is
travelling
0.5 m,
with
a
in
a
horizontal
constant
speed
–1
of
1
Explain
what
is
a
energy
b
the
meant
1.2 m s
by:
a
the
angular
b
the
time
2
a
principal
of
conservation
of
energy
the
object
[2]
taken for
the
object
to
complete
revolution
[2]
[2]
what
is
meant
by
the
terms
‘work’
c
the
centripetal
acceleration
d
the
centripetal force
of
the
object
[2]
object.
[2]
and
‘power’.
b
of
[3]
work.
Explain
velocity
[1]
one
c
. Calculate:
acting
of
the
[2]
At an amusement park in Trinidad, a ride consists of
8
Calculate
the force
revolving
a carriage being pulled up a ramp by a steel cable.
period
The carriage and the passengers have a combined
in
of
a
required
horizontal
to
circle
keep
of
a
mass
radius
of
0.6 m
1.2 kg
with
0.8 seconds.
a
[3]
mass of 480 kg. Initially, the carriage is being pulled
9
–1
such that it is travelling at 8 m s
An
object
of
mass
200 g
is
attached
to
a
string
towards the
and
spun
in
a vertical
circle
of
radius
0.50 m
with
a
ramp which is inclined at 30° to the horizontal. The
–1
constant
speed
of
6 m s
. Calculate:
carriage is brought to rest after travelling for some
a
the
minimum
tension
b
the
maximum
in
the
string
[3]
time up the slope. During the process, the carriage
travels a vertical distance of 3.0 m.
tension
in
the
string.
[3]
Calculate:
10
i
the
initial
kinetic
energy
of
the
carriage
A
small
of
the
passengers
the
gain
the
the
iii
in
gravitational
carriage
and
the work done
string
of
80 g
is fixed
is
to
attached
a
rigid
to
a
string. One
support. The
mass
end
is
[3]
made
ii
mass
and
the
potential
energy
passengers
against the
to
travel
in
a
horizontal
circle
of
radius
0.60 m.
of
The
string
The
mass
makes
an
angle
of
40° to
the
vertical.
[2]
resistive force
takes
0.
15 s
to
complete
one
revolution.
F
Calculate:
acting on the
carriage
as
it
moves
up the
ramp.
a
the
angular
velocity
of
b
the
centripetal
c
the
centripetal force
d
the
tension
a
A
the
mass
[2]
[1]
iv
3
the
Explain
your
in
magnitude
what
answer
is
to
gravitational
mass
the
m
is
of
meant
derive
F.
by
an
the
Earth’s
concept
of
expression for
potential
energy
raised vertically
when
through
a
A force
F
is
an
object
distance
h
of
11
near
mass
of
whirled
[4]
on
a
string
body
that
is
moving
with
v
in
the
direction
the
mass
[2]
the
mass
[2]
of
the force.
in
the
string.
[2]
0.50 kg
in
will
Derive
a
is
attached
horizontal
break
the
when
circle
the
to
of
a
string
radius
tension
and
1.
10 m. The
exceeds
maximum frequency
of
50 N.
rotation.
Describe
the
motion
of
the
mass
if
the
string
an
breaks.
expression
F
5
and
a
relating
power
P
dissipated
by
the force
v.
Define
radian.
12
a
State
b
The
Newton’s
Earth
Convert
can
law
be
of
gravitation.
considered
to
be
[2]
a
uniform
[1]
sphere
b
[3]
to
[2]
the
[5]
a
b
velocity
of
on
increase
surface.
acting
acting
work. Use
the
Calculate
4
acceleration
[2]
the following
to
of
radius
R.
R
is
assumed
to
be
radians
6
i
c
30°
ii
Convert
the following
140°
to
6.4
×
the
Earth.
3.
14 radians
ii
m. A
geostationary
satellite
is
orbiting
degrees
i
i
10
[2]
1.57 radians
Explain
what
is
meant
by
a
geostationary
[2]
orbit.
6
a
Explain
what
b
Describe
is
meant
qualitatively
by
angular velocity.
how
it
is
that
a
[2]
body
which
ii
Show
is
[3]
that
given
by
the
radius
of
a
the
expression
geostationary
orbit
2
is
travelling
in
a
circular
path
with
uniform
speed
3
r
has
acceleration.
gR
=
,
2
3]
ω
2
c
Derive
the
equation for
circular
motion
a
=
ω
r,
where
where
a
is
the
centripetal
acceleration,
ω
is
at
angular velocity
58
and
r
is
the
radius
of
the
g
is
the
acceleration
due
to
gravity
the
circle. [5]
the
Earth’s
surface
and
ω
is
the
angular
Revision questions
velocity
of
the
satellite
about
the
centre
of
16
3
The Earth may be assumed to be a sphere of mass
24
the
Earth.
[3]
6.0 × 10
kg. The Moon may also be considered a
22
Determine
the
radius
of
a
geostationary
orbit.
[3]
sphere of mass 7
.35 × 10
centre
of
3.84
10
the
Earth
to
kg. The
the
distance from
centre
of
the
Moon
the
is
8
13
a
Explain
what
is
meant
×
gravitational
ii
equipotential.
potential
why
gravitational
potentials
are
A
at
speed
in
a
circular
orbit
around
the
a
Earth.
Calculate
on
the
the
gravitational force
exerted
by
the
Moon.
ii
Calculate
iii
Sketch
the
acceleration
of
the
Moon.
[3]
communications
385 km
travels
always
negative.
14
Moon
[1]
Earth
Explain
the
[1]
i
b
that
by:
constant
i
m. Assume
above
the
satellite
is
Earth. The
located
mass
of
at
a
the
height
of
satellite
is
a
diagram
showing
the
direction
of
this
acceleration.
iv
Explain
why
this
acceleration
does
not
increase
3
4.2
×
10
kg. The
6370 km. The
radius
Earth
is
of
the
Earth
assumed
to
is
be
assumed
a
point
to
be
mass
the
speed
of
the
Moon.
of
v
Determine
the
gravitational field
strength
of
the
24
6.0
×10
kg. Calculate:
Earth
15
a
the force
b
the
centripetal
c
the
speed
A
global
acting
of
on
that
satellite
the
orbit
the
[2]
(GPS)
Earth
Moon.
[2]
satellite.
system
the
[2]
acceleration
positioning
satellites
the
at
in
uses
circular
a
number
orbits
at
of
a
4
distance
of
Calculate
2.22
the
×
10
angular
km
above
speed
of
the
one
Earth’s
such
surface.
satellite.
24
Mass
of
Earth
=
5.99
×
10
kg
3
Radius
of
Earth
=
6.38
×
10
km
[3]
59
Module
Answers
to
the
selected
structured
1
Practice
multiple-choice
questions
questions
can
and
be found
exam
to
on
questions
8
the
Which
of
the following
is
not
true
about
inelastic
collisions?
accompanying CD.
a
Momentum
is
b
Total
c
Kinetic
energy
is
conserved
d
Kinetic
energy
is
not
energy
conserved
is
conserved
Multiple-choice questions
1
Which of the following
a
ampere,
degree
b
coulomb,
kelvin
c
kilogram,
kelvin
d
metre,
pairs of
units
are SI
base
units?
celsius
9
A
compact
starts
A
student
degree Celsius
wishes
to
measure
the
density
of
He
has
cube
of
material X.
He
measures
the
Mass
average
of
Length
cube
of
=
one
length
±
16.5
side
of
=
one
side
of
the
the
any
point
is
the
percentage
the
the
density
is
time,
b
3%
b
10
A
satellite
in
has
units
unit for
outer
the
the
player
edge
distance
axis
of
of
as
and
the
the
rotation. The
linear velocity
of X
is
b.
ratio
a/b
c
is
4
d
¼
a
W
before
height
h
=
launch.
5R
above
It
is
the
then
Earth’s
when
the
R
is
the
radius
of
the
Earth. What
is
the
student
acting
on
the
satellite
when
it
is
material X?
2%
c
specific
the
weight
at
9%
d
the
Earth?
5%
W
The SI
of
2
a
orbit
W
a
3
the
½
a CD
cube.
orbiting
a
at
is
a. The
inside
0.
1 cm
error
of
is
disc
centre
gravitational force
determines
placed
point Y
the
of Y
in
½
surface.
What
on
is from
velocity
placed
±
rotate. A
is
mass
0.5 g
4.2
(CD)
material
a
and
disc
point X
point Y
At
X.
to
disc. A
linear
2
conserved
heat
capacity
in
terms
of
base
W
b
W
c
5
d
6
25
36
is:
2
a
kg m
c
m
–2
s
–1
2
K
b
m
d
m
–2
s
–1
K
Structured questions
2
2
s
–1
–2
K
–2
–1
s
K
11
(Refer
4
What
to
is
a
Distinguish
b
Explain
between
precision
and
accuracy.
[2]
15.
1.)
the
number
of
atoms
present
in
0.090 kg
of
what
random
carbon-12?
is
error
meant
and
by
give
a
an
systematic
example
of
and
each
a
type
error.
22
a
4.5
×
10
c
4.5
×
10
of
[4]
21
b
4.5
×
10
d
4.5
×
10
23
c
A
student
is
given five
marbles
and
asked
to
24
determine
make
the
the
density
of
marbles. The
the
material
student
used
decides
to
to
line
the
1 μg
5
What
is
the
ratio
?
marbles
up
in
a
straight
line
against
the
edge
of
1 kg
a
–2
a
–12
10
b
–9
10
c
metre
d
A
small
object
is
projected
horizontally from
a
wall
–1
of
height
velocity
7 m
of
with
the
a
speed
object
just
of
30 m s
before
. What
striking
–1
a
32.2 m s
c
11.8 m s
notes
the
beginning
and
ending
10
point
6
rule. She
–3
10
is
the
the
ground?
–1
b
30.0 m s
d
34.2 m s
–1
measure
are
X
Y
–1
along
the
the
ruler. She
mass
of
then
the five
uses
a
balance
marbles.
Her
to
results
as follows:
=
=
12.5
20.0
Mass
±
0.
1 cm
±
of five
0.
1 cm
marbles
=
20.5
±
0.5 g
Calculate:
7
A
uniform
plank
of
length
l
is
supported
by
two
i
the
diameter
of
one
marble,
including
the
T
1
straps
as
shown
below. What
is
the
ratio
absolute
?
uncertainty
[2]
T
ii
2
the
mass
absolute
T
2
iii
the
60
b
marble,
including
the
uncertainty
density
marbles,
l /4
½
one
[2]
T
1
a
of
3l /4
2
c
¼
d
4
of
the
material
including
the
used
to
uncertainty.
make
the
[3]
Module
12
A
cyclist
total
is
training
mass
of
the
in
a
hilly
cyclist
region
and
his
is
Practice
exam
questions
Calculate:
in Jamaica. The
bicycle
1
85 kg.
i
the
time
taken for
the
cannon
ball
to
hit
the
–1
Initially
on
a
on
to
he
level
is
travelling
dirt
road.
another
vertical
level
distance
of
at
He
a
constant
then
road
travels
while
speed
down
travelling
of
a
12 m s
surface
slope
ii
through
the
a
iii
b
kinetic
the
level
the
loss
energy
of
the
cyclist
and
his
bicycle
on
road
in
15
a
[2]
potential
energy
while
body
a
vertical
distance
of
to
the
speed
bottom
Given
of
of
the
the
that
cyclist
and
5.0 m
his
at
the
how
[3]
cyclist
was
by
the
the
be
in
cannon
ball
just
before
sea.
[3]
that
must
be
satisfied for
a
equilibrium.
P,
Q
and
R
[2]
act
on
an
object O. The
[2]
bicycle
slope.
the
travelled
[2]
of
conditions
Three forces
object O
c
the
two
[2]
distance
travelling
b
through
State
sea
ball
the velocity
hitting
the
the
horizontal
cannon
5.0 m.
Calculate:
a
of
providing
a
a
is
in
equilibrium.
vector
triangle
is
Explain
drawn
to
using
a
sketch
represent
these
forces.
power
c
How
[3]
does
the
triangle
show
that
the
object O
is
–1
of
320 W
calculate
when
the
he
was
total
travelling
at
resistive force
12 m s
acting
,
on
in
a
[2]
Explain
what
is
meant
by
the
terms
‘work’
16
a
State
b
The
Newton’s
mass
and
and
‘energy’.
5.98
×
car
of
of
radius
10
gravitation.
of
the
Earth
[2]
are
assumed
to
6
kg
and
6.40
×
10
m
respectively.
[2]
Determine
A
law
24
be
b
[1]
the
cyclist.
13
equilibrium?
mass
900 kg
is
travelling
at
a
constant
strength
g
a value for
at
the
the
Earth’s
gravitational field
surface.
[3]
–1
speed
of
18 m s
down
a
sloped
road. The
angle
c
of
the
road
to
the
horizontal
is
12°. The
A
geostationary
driver
another vehicle
in front
of
her. She
brakes
to
bring
the
car
to
a
complete
of
3000 N
opposes
the
a
distance
of
×
10
m from
above
the
the
centre
of
the
Earth
and
is
equator.
stop. A
i
constant force
at
applies
orbiting
the
is
7
4.23
notices
satellite
motion
Explain
what
is
meant
by
a
geostationary
of
satellite.
the
ii
i
Sketch
a
diagram
to
show forces
acting
on
Calculate
when
it
Calculate
car
down
is
the
the
at
rest
on
the
component
slope.
of
the
Calculate
the
weight
of
slope.
brakes
v
are
Calculate
from
the
normal
reaction
acting
on
the
deceleration
of
the
car
when
the
where
where
Calculate
the
loss
travelled
the
the
of
satellite
Calculate
the
speed
Calculate
the
acceleration
by
brakes
the
car
are
Explain
State
is
located.
[1]
of
the
satellite.
[3]
of
[2]
the
satellite.
law
Distinguish
inelastic
car
is
meant
of
by
linear
conservation
momentum.
of
[2]
linear
[3]
between
an
elastic
collision
and
an
collision.
[2]
applied
stops.
kinetic
what
the
momentum.
a
Explain
how
is
it
that
an
object
travelling
[2]
in
vi
the
iv
b
c
[2]
distance
point
point
where
iii
a
the
18
to
point
the
applied.
the
at
[2]
[2]
Calculate
strength
the
car.
iv
gravitational field
[2]
17
iii
the
the
the
car
ii
[1]
car.
energy
of
the
a
circular
path
with
uniform
speed
has
car.
acceleration.
State
the
direction
of
the force
[2]
producing
vii
Calculate
the
work
done
by
the
this
acceleration.
[4]
3000 N force.
2
[1]
b
Derive
where
14
a
Distinguish
between
scalar
and vector
the
a
is
equation for
the
circular
centripetal
motion
acceleration,
a
ω
=
is
ω
r,
the
quantities.
angular velocity
and
r
is
the
radius
of
the
circle.
[2]
[4]
b
Give
c
A
an
example
of
each.
[2]
c
cannon
in Tobago
horizontally
cannon
ball
on
the
is fired
is
positioned
edge
of
a
cliff
horizontally
such
that
20 m
with
it
lies
high. A
a velocity
A
mass
rotated
of
in
minimum
0.85 kg
is
a vertical
tension
in
attached
circle
the
of
to
a
string
radius
string
is
and
1.50 m. The
2.5 N.
of
i
Determine
ii
Determine the
speed
of
rotation.
[3]
–1
45 m s
.
maximum tension
in the
string.
[3]
61
9
Oscillations
9.
1
Free
oscillations
Learning outcomes
On
completion
should
be
able
of
this
Examples of free oscillations
section,
you
Examples
of
free
oscillations
are
shown
in
Figures
9.1.1–3.
to:
glider
spring
ᔢ
describe
examples
of free
oscillations
ᔢ
understand
the
term
simple
harmonic motion
air
ᔢ
state
the
conditions
Figure 9.1.1
simple
ᔢ
harmonic
illustrate
motion
simple
track
necessary for
motion
A glider oscillating back and
Figure 9.1.2
forth on an air track using two springs
A small marble oscillating
in a dish
harmonic
Simple
graphically.
Simple
of
a
harmonic
har monic
mass
directed
is
towards
following
sign
(–)
the
motion
proportional
The
to
motion
the
fixed
equation
indicates
direction
that
of
is
to
periodic
the
motion
displacement
in
which
from
a
the
fixed
acceleration
point
and
point.
is
used
the
the
a
(SHM)
to
define
direction
of
SHM.
the
Note
that
acceleration
is
the
negative
always
opposite
displacement.
spring
Definition
Equation
2
oscillation
SHM
is
a
periodic
motion
in
which:
a
=
–ω
x
mass
of
mass
–2
1
Figure 9.1.3
the
acceleration
is
proportional
to
the
a
–
acceleration/m s
ω
–
angular frequency/rad s
x
–
displacement/m
A mass oscillating on a spring
–1
displacement from
2
directed
towards
a fixed
the fixed
point
and
point.
2
acceleration /m s
Figure
9.1.4
shows
displacement
displacement /m
Consider
or
O.
fixed
The
the
a
position).
points
9.1.5.
respectively.
Graph showing the
A
can
be
seen
At
the
an
particle
equilibrium
Figure
Figure 9.1.4
for
relationship
object
M
It
initially
begins
and
B
point
The
that
to
oscillate
P
O.
the
A
and
a
velocity
are
at
in
of
with
of
and
M
the
point
SHM
maximum
snapshot
displacement
v
the
acceleration
and
SHM.
starting
represent
position
between
undergoing
the
(equilibrium
about
the
point
displacement
motion
acceleration
from
opposite
its
O
is
of
from
illustrated
M
equilibrium
are
v
in
and
position
a
is
x.
It
directions.
relationship between acceleration and
displacement
The
conditions
ᔢ
A
mass
that
ᔢ
A
fixed
ᔢ
A
restoring
it
is
necessary
for
SHM
are
as
follows:
oscillates.
point
at
which
the
mass
is
in
equilibrium.
a
force
which
returns
the
mass
to
its
equilibrium
position
if
M
displaced.
v
x
Displacement, velocity
A
Figure 9.1.5
62
O
P
Simple harmonic motion
B
Consider
one
side
the
motion
gently
and
of
a
then
and
simple
acceleration
pendulum.
released,
gravity
When
pulls
on
the
it.
bob
This
is
displaced
force
causes
to
it
Chapter
to
return
and
left
causes
and
SHM
can
process
of
the
to
at
the
O
time
point
→
oscillation
is
position.
be
t
A.
A
T.
0.
One
→
The
→
time
So
(Figure
bob
the
9.1.6).
pendulum
oscillation
O
This
again.
the
passes
bob
this
oscillates
Oscillations
point
to
the
position.
graphically
=
However
,
repeated
equilibrium
illustrated
O
follows
one
equilibrium
the
be
at
reaches
as
its
right
initially
it
to
9
is
B
is
the
→
called
is
motion
O.
the
The
pendulum
displaced
The
of
time
period
of
to
the
the
bob
right
is
until
pendulum
taken
to
bob
complete
oscillation.
B
Figure
9.1.7
motion
If
the
of
a
(Graph
simple
amplitude
displacement
from
is
A,
the
of
displacement–time
The
graph
is
x
is
graph
sinusoidal
in
for
A
the
shape.
0
Equation
distance
Figure 9.1.6
position)
displacement
time
a
the
(maximum
the
of
illustrates
pendulum.
equilibrium
then
function
I)
=
as
x
=
The motion of a simple
pendulum
A sin ωt
a
x
–
displacement/m
A
–
amplitude/m
A sin ωt
displacement /m
–1
T
o
determine
point
on
graph
at
a
the
that
time
t
velocity
gradient
point
Therefore,
At
the
at
any
displacement–time
at
=
of
must
time
T/4,
the
be
t
–
angular frequency/rad s
–
time/s
Graph I
time /s
tangent
determined.
=
the
ω
t
0,
the
gradient
gradient
is
zero.
is
a
At
maximum
time
t
=
and
T/2,
is
the
velocity /m s
positive.
gradient
is
1
a
Graph II
maximum
t
=
T,
the
and
negative.
gradient
is
a
At
time
t
maximum
=
3T/4,
and
the
gradient
is
zero.
At
time
positive.
time /s
The
velocity–time
graph
(Figure
Equation
9.1.7
Graph
II)
can
mathematically
in
be
the
expressed
acceleration /m s
2
equation
v
=
v
Graph III
cos ωt
0
opposite.
–1
v
–
velocity/m s
–
maximum
ω
–
angular frequency/rad s
t
–
time/s
time /s
Similarly,
the
velocity–time
–1
v
velocity/m s
0
graph
can
be
used
acceleration–time
to
obtain
graph
the
for
–1
Figure 9.1.7
the
Graphs showing the variation
of displacement, velocity and acceleration
motion
of
the
pendulum.
The
with time
acceleration–time
by
finding
(Figure
the
9.1.7
graph
gradient
Graph
is
at
found
various
points
on
the
velocity–time
graph
1
III).
velocity
v /m s
Equation
The
equation
gives
mathematical
the
expression
for
the
2
v
velocity
of
an
object
=
±ω
(x
√
undergoing
2
– x
displacement
)
 x
–1
SHM
related
to
displacement
x /m
0
0
v
–
velocity/m s
ω
–
angular frequency/rad s
x
0
–1
(Figure
9.1.8).
x
–
maximum
–
displacement/m
displacement/m
0
x
Figure 9.1.8
Graph of velocity against
displacement
Key points
ᔢ
ᔢ
Free
oscillations
to
SHM
periodic
is
a
a
include
attached
motion
displacement from
The
conditions for
ᔢ
The
displacement–time
ᔢ
The
velocity–time
SHM
in
a fixed
ᔢ
undergoing
the
motion
of
a
simple
pendulum
and
a
mass
spring.
SHM
are
an
also
and
the
and
acceleration
directed
oscillating
graph for
graphs
are
which
point
the
an
mass
object
is
proportional
towards
and
a
the fixed
the
restoring force.
undergoing
acceleration–time
to
point.
SHM
is
graphs for
sinusoidal.
an
object
sinusoidal.
63
9.2
Amplitude,
Learning outcomes
On
completion
should
be
able
of
this
period
Displacement
section,
you
Displacement
equilibrium
to:
graph
ᔢ
understand the terms
for
frequency,
a
and
express
period
frequency
and
simple
amplitude
distance
Figure
moved
9.2.1
pendulum.
in
a
stated
shows
The
part
dashed
direction
of
line
a
from
the
displacement–time
point
P
represents
the
position
represents
of
the
the
equilibrium
pendulum
at
a
angular
instant
in
time.
The
distance
x
represents
the
displacement.
phase difference
The
ᔢ
the
position.
The
particular
frequency
is
and
amplitude,
position.
period,
and frequency
in
terms
maximum
amplitude.
of
The
displacement
SI
unit
for
from
the
equilibrium
displacement
and
is
called
amplitude
is
the
the
metre.
angular frequency.
P
a
x
Definitions
Displacement
in
a
stated
equilibrium
Amplitude
is
the
distance
direction from
moved
the
position.
is
the
Figure 9.2.1
Diagram illustrating displacement and amplitude
maximum
displacement from
the
equilibrium
Period, frequency
and
angular frequency
position.
Suppose
T
is
the
a
simple
time
Figure
9.2.2
period
T
is
pendulum
taken
for
illustrates
one
a
is
displaced
oscillation.
and
The
left
SI
displacement–time
to
unit
graph
oscillate.
is
the
and
The
second
shows
period
(s).
how
the
determined.
Definitions
displacement/m
The
period
is
the
time
taken for
one
r
oscillation.
The frequency
oscillations
is
per
the
unit
number
of
time.
r
0
time/s
Equation
1
f
=
Figure 9.2.2
Determining T from a displacement–time graph
T
f
–
frequency/Hz
T
–
period/s
Frequency
hertz
1/T
(Hz).
f
is
If
the
one
oscillations
Angular
number
of
oscillation
are
frequency
oscillations
is
produced
(angular
produced
in
1
per
in
unit
a
time.
time
T,
The
then
SI
it
unit
follows
is
the
that
second.
velocity)
ω
is
defined
as
the
rate
of
change
of
–1
angular
displacement
Angular
frequency
(see
and
7.1,
Motion
frequency
Equation
are
in
a
circle ).
related
by
The
the
SI
unit
following
is
rad s
equations.
Equation
2π
ω
=
2πf
ω
=
T
–1
ω
–
angular frequency/rad s
–1
f
64
–
ω
–
angular frequency/rad s
T
–
period/s
frequency/Hz
Chapter
Linking
Simple
simple
harmonic
Consider
a
turntable.
beam
t
=
of
0,
The
The
B.
shadow
The
is
moves
Distance
of
parallel
of
a
produces
at
O.
from
At
O
a
to
=
r sin (ωt)
peg
moves
this
t,
back
the
and
movement
(Figures
disc
of
are
closely
radius
angular
of
circular
the
disc
r,
peg
motion
related.
attached
velocity
on
rotates
a
Oscillations
ω.
to
A
screen.
through
a
horizontal
an
At
time
angle
θ.
P
.
OP
of
a
shadow
time
ωt
the
to
and
motion
constant
=
motion
beam
attached
at
motion
circular
θ
amplitude
harmonic
and
rotates
light
shadow
shadow
peg
disc
parallel
the
motion
vertical
The
harmonic
9
9.2.3
is
and
forth
r.
The
between
shadow
the
points
moves
A
with
and
simple
9.2.4).
light
parallel
beam
of
light
A
wooden
shadow
peg
P
r
θ
0
ω
B
turntable
turntable
Figure 9.2.3
screen
Experiment showing the relationship between SHM
Figure 9.2.4
and circular motion (side view)
Experiment to show the relationship between
SHM and circular motion (top view)
Example
The
graph
in
Figure
9.2.5
shows
how
the
acceleration
of
an
2
a /m s
object
undergoing
simple
harmonic
motion
varies
with
time.
Determine:
45
a
the
period
b
the
frequency
of
oscillation
t /ms
0
c
the
angular
d
the
amplitude
frequency,
x
of
ω
the
oscillation.
45
0
a
Period
b
frequency
T
=
50 ms
1
f
1
=
Figure 9.2.5
=
=
20 Hz
–3
T
50
×
10
–1
c
angular
d
a
frequency
ω
=
2πf
=
2π(20)
=
126 rad s
2
=
–ω
x
Maximum
acceleration
a
occurs
at
maximum
displacement
(amplitude)
45
–3
x
=
–
=
–
=
2
0
– 2.83
×
10
m
2
ω
(126)
Key points
ᔢ
Displacement
equilibrium
is
the
distance
ᔢ
Amplitude
is
the
maximum
ᔢ
The
is
the
time
ᔢ
The frequency
ᔢ
Circular
period
moved
in
a
stated
direction from
the
position.
is
motion
the
and
displacement from
taken for
number
simple
of
one
the
equilibrium
position.
oscillation.
oscillations
harmonic
per
motion
unit
are
time.
closely
related.
65
9.3
A simple pendulum and a mass–spring system
Learning outcomes
On
completion
should
ᔢ
be
derive
of
a
able
the
of
this
A
section,
The
mass
pendulum
m
is
displaced
to
the
right
through
an
arc
x
(Figure
9.3.1).
to:
equation for
simple
you
simple
the
Restoring
force
=
For
angles
mg sin θ
period
x
small
(in
radians)
sin θ
≈
θ
≈
pendulum
l
x
ᔢ
derive
of
a
the
mass
equation for
on
a
the
period
Therefore,
Using
ᔢ
describe the
kinetic
and
the
interchange
potential
simple
force
=
mg
(
)
l
Newton’s
second
law
F
=
ma
between
energy
F
Acceleration
during
restoring
spring
harmonic
towards
O
mgx
=
gx
=
motion.
÷
m
m
=
l
l
gx
Acceleration
in
the
direction
of
x,
a
=
–
l
2
Comparing
with
the
equation
for
simple
harmonic
motion
a
=
–ω
x
gx
θ
2
=
ω
x
l
2π
l
But
ω
=
T
2
g
2
2π
∴
=
l
(
4π
=
)
T
2
T
2
x
4π
l
2
T
=
g
mg sin θ
l
T
=
2π
mg
Figure 9.3.1
√
g
A simple pendulum
A
mass
Consider
obeys
attached to
a
spring
Hooke’s
applied
and
e
is
small
by
k(e
mass
=
the
mass
is
oscillations
+
x).
The
–
=
pulled
m
in
a
+
second
k
is
=
to
the
produced
it.
Assume
spring
(Figure
that
constant,
the
F
spring
is
the
force
9.3.2).
ke
vertical
x)
attached
where
downwards
resultant
k(e
Newton’s
ke,
spring
extension
mg
ke
Using
F
the
∴
When
with
law
a
a
plane.
The
downward
ke
–
law
ke
–
distance
F
=
ma
– kx
=
ma
a
=
–
e
tension
force
kx
x
=
and
T
released,
in
the
it
spring
makes
is
given
is
– kx
kx
∴
(downwards)
in
the
direction
m
2
Comparing
with
the
equation
for
simple
harmonic
x
k
2
=
ω
m
2π
But
ω
=
T
mg
2
k
Figure 9.3.2
66
A mass-spring system
m
2
2π
=
(
T
4π
)
=
2
T
motion
a
=
–ω
x
of
x
Chapter
9
Oscillations
energy /J
2
4π
m
2
T
=
k
total
energy
m
T
=
2π
k
e
Rewriting
in
terms
of
e
and
g:
T
=
2π
since
√
mg
=
ke
=
2π
g
2e
The
period
of
oscillation
for
two
similar
springs
in
series
is
T
T
√
g
time /s
e/2
The
period
of
oscillation
for
two
similar
springs
in
parallel
is
T
=
potential
energy
2π
g
kinetic
Figure 9.3.3
Energy
When
a
in
simple
system
is
oscillating
interchange
between
total
of
not
is
energy
damped.
displaced
harmonic
the
motion.
energy
(velocity
=
at
At
a
0).
Figure
at
and
equilibrium
9.3.3
a
When
maximum
the
When
is
the
bob
is
maximum.
the
bob
at
its
its
kinetic
variation
of
energy
energy
kinetic
that
is
is
the
is
with
at
this
also
a
at
a
energy /J
an
The
system
When
the
is
total
energy
bob
simple
its
point
position,
at
energy,
there
system.
displacement,
energy
equilibrium
potential
the
oscillates
maximum
kinetic
the
of
pendulum.
system
its
The
motion,
energy
provided
simple
the
passes
therefore
the
a
its
is
zero
velocity
maximum.
–A
A
displacement /m
minimum.
potential
energy
and
potential
total
energy
with
time
for
the
motion
of
a
simple
energy
of
a
system
oscillating
with
simple
energy
pendulum.
kinetic
The
Relationships between
energy and time
harmonic
kinetic
constant,
of
released,
position,
shows
and
motion
and
motion
simple
remains
the
side
with
potential
system
one
potential
is
the
Consider
to
harmonic
harmonic
motion
energy
can
Figure 9.3.4
be
represented
graphically
as
a
function
of
energy
displacement.
At
Relationship between
maximum
energy and displacement
displacement,
energy
is
at
maximum
a
the
potential
minimum.
and
the
energy
At
zero
potential
is
at
a
maximum
displacement
energy
is
at
a
the
and
the
kinetic
minimum
kinetic
energy
(Figure
is
at
E
a
/ mJ
K
9.3.4).
3.00
Example
A
spring
end
of
a
is
through
a
harmonic
from
hung
spring.
the
from
The
small
a
fixed
mass
distance
motion.
y
Figure
equilibrium
is
point.
pulled
and
9.3.5
position
is
A
mass
released.
shows
of
of
150 g
downwards
the
the
The
energy
hung
its
mass
variation
kinetic
is
from
undergoes
with
of
from
the
equilibrium
free
position
simple
displacement
the
x /cm
x
mass.
– 1.2
1.2
Figure 9.3.5
Using
the
figure:
a
Determine
b
Determine
the
angular
c
Determine
the
frequency
the
distance
y
through
which
the
mass
was
initially
displaced.
Key points
frequency.
of
oscillation.
ᔢ
To
derive
period
in
a
y
=
an
of
equation for
oscillation
simple
harmonic
a
the
system
motion,
the
1.2 cm
restoring force
b
of
Maximum
kinetic
energy
=
energy
=
determined. The
acceleration
the
mass
compared
the
defining
of
1
2
kinetic
be
3.00 mJ
1
Maximum
must first
2
mv
=
2
mω
2
(a
2
–
x
)
is
then
with
2
equation for
simple
1
–3
×
150
×
10
2
×
ω
–2
{(1.2
×
10
2
)
2
–
0
}
–3
=
3.00
×
10
harmonic
motion.
There
constant
2
–3
2
×
3.00
×
ᔢ
10
is
a
interchange
–1
ω
=
=
√
150
–3
×
10
–2
×
(1.2
×
10
16.7 rad s
2
between
)
potential
energy for
ω
c
Frequency
of
oscillation
f
=
system
kinetic
oscillating
16.7
=
2π
a
and
=
2.66 Hz
with
simple
harmonic
motion.
2π
67
9.4
Resonance
Learning outcomes
On
completion
should
be
able
of
this
Damped oscillations
section,
you
When
a
oscillate
to:
simple
with
gradually
ᔢ
describe
practical
examples
describe
forced
ᔢ
displaced
over
time
at
its
equilibrium
practical
examples
of
oscillations
understand
the
concept
resistance
pendulum.
causes
That
is
and
position.
decreases
pendulum
is
said
energy
to
amplitude
the
Since
say,
in
to
be
work
an
be
to
and
released,
amplitude
pendulum
the
of
it
begins
the
eventually
pendulum
transferred
is
done
exponential
damped
to
oscillation
is
comes
oscillating
to
in
against
from
the
air
The
the
air
,
oscillating
resistance.
oscillation
The
of
the
9.4.1).
displacement/m
situations
is
is
higher,
larger
air
resisitance;
where
energy
resonance
away
manner
.
(Figure
Velocity
identify
The
of
resonance
ᔢ
slightly
motion.
oscillations
air
ᔢ
decreases
is
harmonic
of
rest
damped
pendulum
simple
useful
and
when
lost
rapidly,
amplitude
decreases
it
rapidly
should
be
avoided.
time/s
displacement/m
I
Velocity
Light damping
air
is
lower,
smaller
resistance;
energy
lost
more
slowly,
exponential
amplitude
decay
of
decreases
less
the
rapidly
time/s
amplitude
Note:
Figure 9.4.1
Figure 9.4.2
the
period
of
oscillation
remains
constant
Graph showing damped oscillations
Lightly damped oscillations
Initially,
This
into
is
the
in
pendulum
the
kinetic
form
energy
of
as
has
its
maximum
potential
the
energy.
pendulum
energy
This
bob
when
energy
begins
is
it
was
displaced.
converted
moving
back
to
the
displacement/m
equilibrium
II Critical damping
dependent
The
on
the
of
velocity
resistance
amplitude
decreases
the
Figure 9.4.3
Critically damped oscillations
motion
decreases
There
displacement/m
III
an
of
but
are
the
now
bob
the
lightly
ᔢ
critically
ᔢ
heavily
is
therefore
more
of
degrees
is
the
It
to
which
(e.g.
oscillations
a
start
energy
initial
lost
This
at
the
car
at
As
the
a
rapid
some
slower
be
The
air
the
amplitude
even
the
The
passes,
and
the
that
of
rate.
time
rate
why
noted
is
oscillations.
oscillations.
a
though
oscillations
constant.
can
pendulum
a
of
amplitude
system
(e.g.
resistance
explains
remains
a
air
rapidly.
should
damped,
oscillation
the
the
lose
be
slowly.
manner
.
oscillations
damped
will
of
at
reduces
than
Energy
pendulum
damped
greatest
smaller
decrease
period
be
will
exponential
the
magnitude
therefore
smaller
.
different
ᔢ
the
will
pendulum
would
in
it
oscillation
of
is
Since
velocity,
oscillating
amplitude
time/s
position.
be
damped.
oscillating
suspension
in
They
are:
air)
system)
Heavy damping
For
a
damped
system
eventually
that
oscillations.
is
decreases
lightly
to
damped
zero
as
the
the
amplitude
system
comes
of
to
the
rest
oscillation
(Figure
9.4.2).
time/s
For
a
system
oscillation
For
Figure 9.4.4
68
Heavily damped oscillations
a
system
(Figure
that
(Figure
that
9.4.4).
is
critically
damped,
the
system
comes
to
rest
after
9.4.3).
is
heavily
damped,
the
system
fails
to
oscillate
one
Chapter
9
Oscillations
Resonance
Systems
that
frequency.
f
of
the
oscillate
This
with
particular
system.
In
the
simple
harmonic
frequency
case
of
a
is
motion
known
simple
as
do
the
pendulum
so
at
natural
of
length
a
particular
frequency
l,
the
natural
0
frequency
is
given
by
g
1
f
=
0
√
2π
In
the
the
case
of
natural
mass
m
frequency
l
attached
is
given
to
1
f
periodic
force
the
amplitude
the
periodic
the
referred
Figure
as
the
to
9.4.5
equal
the
the
is
the
at
a
spring
constant
k,
m
a
system
increases
to
as
√
to
the
such
force.
that
significantly
natural
resonance.
periodic
forces
when
frequency
At
The
it
of
resonance,
periodic
it
the
the
is
oscillate,
frequency
system.
energy
force
to
is
of
This
transferred
sometimes
driver
.
illustrates
frequency
to
is
known
by
as
oscillations
equal
is
system
applied
vibration
force
phenomenon
to
of
is
having
k
2π
a
spring,
=
0
If
a
by
of
a
how
the
the
periodic
maximum
natural
amplitude
force
when
frequency
of
of
an
changes.
the
the
oscillating
The
frequency
system
amplitude
of
the
of
varies
the
periodic
force
is
system.
A
0
f
driver
0
0
frequency
f /Hz
Figure 9.4.5
A
simple
Resonance
laboratory
phenomenon
pendulums.
are
fixed
varying
of
A
A
driver
paper
and
OB,
the
string
cones
Q,
at
AOB
B.
The
R
attached
driver
S
set
The
to
pendulum,
up
OB.
as
Each
set
OD
them.
has
The
all
shown
up
a
to
is
a
mass
P
,
in
Q,
R
Figure
of
will
and
S
pendulum
larger
have
to
is
the
as
9.4.6.
have
pendulum
are
to
several
attached
pendulums
The
demonstrate
referred
consists
driver
the
oscillating.
builds
up
pendulum
pendulums
Since
begin
be
arrangement
arrangement
pendulum
oscillating.
and
is
to
can
The
The
attached
pendulum.
begins
P
,
experiment
resonance.
and
lengths
oscillation.
the
to
at
of
The
ends
pendulums
a
it
Barton’s
different
and
small
is
called
inverted
displaced
slightly
attached
to
having
length
amplitude
than
a
the
of
period
the
rest
string
similar
of
them.
O
S
R
Q
D
driver
pendulum
Figure 9.4.6
P
Investigating resonance
69
Chapter
9
Oscillations
Experiment to
A
mass
is
attached
investigate the
between
two
springs
effect of damping
as
fixed
shown
in
Figure
9.4.7.
support
ruler
spring
equilibrium
mass
position
spring
signal
oscillator
generator
Figure 9.4.7
A
ruler
is
placed
measured.
oscillator
slightly.
rule.
The
A
Experiment to demonstrate resonance
The
is
connected
The
The
adjacent
to
amplitude
period
frequency
to
equilibrium
of
f
the
a
of
the
mass
position
signal
the
1
g
2π
l
is
that
first
oscillation
is
then
displacements
noted
generator
.
oscillation
=
so
is
is
The
on
the
mass
measured
determined
determined.
a
be
An
displaced
from
using
The
is
can
ruler
.
the
stop
signal
metre
watch.
generator
0
is
then
signal.
mass
turned
This
to
oscillate.
amplitude
signal
Figure
f
0
The
on.
of
same
signal
causes
The
is
shows
the
of
is
oscillator
to
the
measured
the
experiment
generator
frequency
oscillation
generator
9.4.8
The
signal
be
the
mass
using
results
can
of
of
is
a
to
produce
move
signal
the
The
ray
a
sinusoidal
spring
generator
recorded.
cathode
the
used
set
is
and
forces
varied
frequency
and
of
the
the
the
oscilloscope.
experiment.
to
show
the
effect
of
damping
on
f
0
0
the
Figure 9.4.8
Results of experiment
resonance
Figure
The
card
shows
A
small
card
is
attached
to
the
mass
the
damps
the
is
the
motion
repeated
effect
of
and
of
the
the
damping
mass
by
resonance
on
the
increasing
curve
resonant
is
are
important
resonant
to
note.
frequency
is
The
also
peak
of
the
curve.
curve
is
There
flatter
lower
.
spring
A
peak
is
no
damping
flatter
equilibrium
damping
position
card
spring
signal
oscillator
generator
0
f
0
lower
resonant
frequency
when
damped
70
the
plotted.
ruler
oscillations
shown
in
drag.
The
Figure
are
9.4.10
two
things
support
that
Figure 9.4.9
as
9.4.9.
experiment
fixed
curve.
Use of a card to damp the
Figure 9.4.10
Effect of damping on the resonant curve
and
wider
and
Chapter
Unwanted
In
cities,
tall
common,
by
buildings
large
to
seconds.
to
buildings.
2010,
an
in
Haiti.
In
reduce
Situations
Microwave
Even
though
examples
is
the
of
the
As
the
of
The
at
a
temperature
force
of
the
the
resonance
resonance
body.
are
hydrogen
Electric
T
uning
is
The
Large
earthquakes
waves
amounts
buildings
7.0
buildings
energy
are
destroyed
caused
are
produced
of
are
are
major
designed
with
useful
The
patient
emit
produce
radio
an
of
the
thermal
region
this
food
many
W
ater
microwave
exploit
are
appliance
resonance.
the
inside
and
there
household
cookers
to
energy
fact.
oscillate.
spreads
it.
(MRI)
non–invasive
structures
body
placed
in
molecules
a
disastrous,
use
lies
increases
is
the
be
common
make
warming
nuclei
is
to
can
A
Microwave
internal
human
transmitted
to
where
occur
.
magnitude
that
imaging
imaging
view
the
ovens
water
thereby
Hydrogen
nuclei
processed
there
are
in
are
used
a
person
frequency
medical
the
signals
occurring
amounts
basis
magnetic
cause
diagnostic
processes
huge
as
large
and
and
for
field.
are
the
imaging
Radio
to
in
hydrogen
this
resonance
which
of
frequency
occur
.
detected
The
and
image.
circuits
circuits
resonance.
capacitors
two
In
present.
pulses
food,
to
useful.
water
Magnetic
used
is
region.
the
technique.
resonance
earthquakes.
frequency
throughout
nuclei
of
countries,
resonance
Microwave
oscillate
microwaves
human
of
can
areas
occurs,
resonance
effects
The
technique
resonance
effects
In
oscillate.
earthquake
resonance
electromagnetic
Magnetic
to
with
Oscillations
cookers
microwave.
the
destruction
wealthy
where
where
molecules
of
naturally.
buildings
In
destruction
dampers
forces
associated
oscillate
amounts
earthquakes
transferred
in
problems
9
in
These
and
electrical
types
of
inductors.
components
at
the
devices
circuits
An
such
have
electric
circuit’s
as
radios,
reactive
current
resonant
make
use
elements
will
oscillate
of
such
as
between
the
frequency.
Key points
ᔢ
Air
resistance
ᔢ
The
ᔢ
Oscillations
ᔢ
Resonance
amplitude
frequency
ᔢ
Damping
ᔢ
There
the
and friction
are
of
can
oscillations
be
occurs
of
the
affects
lightly,
when
a
can
be
oscillations
decreases
critically
or
the frequency
oscillating
where
to
when
the
damped.
damped.
heavily
of
be
damped.
driver
is
equal
to
the
natural
system.
resonant frequency
situations
effects
cause
resonance
curve.
can
be
useful
and
situations
where
catastrophic.
71
10
Refraction
10.
1
Refraction
Learning outcomes
On
completion
should
be
able
of
this
Refraction
section,
you
Consider
side
to:
of
water
ᔢ
state
the
laws
of
refraction
a
large
the
container
container
section
to
(Figure
of
water
.
create
10.1.1).
a
A
deep
A
piece
water
straight
of
wood
section
bar
is
is
placed
and
then
a
used
on
one
shallow
to
produce
of
straight
wavefronts
travelling
from
the
deep
water
to
the
shallow
water
light
(Figure
ᔢ
understand
the
term
10.1.2).
boundary
refractive
direction
index
at
understand
the
terms
ᔢ
and
discuss
The
angle,
as
practical
internal
wavefronts
of
a
of
change
the
change
in
wave
deep
direction.
speed
also
water–shallow
of
the
changes.
water
This
wave
The
change
and
is
in
called
as
the
wave
travels
from
deep
water
to
frequency
shallow
water
.
applications
behaves
when
a
ray
densities.
as
shown
the
water
shallow
as
of
light
a
wave
light
angle
Figure
where
of
and
travels
Consider
in
surface
The
a
ray
of
10.1.3.
the
ray
incidence
i
as
such
between
light
The
the
be
travelling
the
refracted.
media
nor mal
strikes
is
can
two
is
of
from
a
medium
line
boundary
angle
Refraction
different
drawn
the
1
at
between
between
occurs
optical
to
medium
right
the
normal
angles
two
and
2
to
media.
the
water
incident
Figure 10.1.1
the
reflection.
Light
deep
the
result
strike
total internal reflection
the
total
a
wavelength
constant
Refraction of
of
wavefronts
critical
remains
angle
the
some
occurs
refraction.
ᔢ
If
A container of water
and
the
ray.
The
angle
refracted
of
refraction
r
is
the
angle
between
the
normal
ray.
sin i
The
ratio
is
a
constant,
and
is
called
the
refractive
index
sin r
deep
water
The
refractive
index
of
medium
2
with
respect
to
medium
1
is
n
1
2
sin i
Therefore,
n
1
=
.
This
is
known
as
Snell’s
law
2
sin r
c
1
It
can
also
be
shown
that
n
1
=
,
where
c
2
is
the
speed
of
light
in
1
c
2
medium
1
and
c
is
the
speed
of
light
in
medium
2.
2
If
the
ray
of
light
was
travelling
from
medium
2
to
medium
1,
the
Note:
refractive
index
of
medium
1
with
respect
to
medium
2
is
n
2

speed
and
wavelength
1
decreases
1

frequency
is
unchanged
Therefore,
n
2

direction
of
wavefronts
=
1
n
change
1
2
c
v
The
Figure 10.1.2
absolute
refractive
index,
n,
of
a
material
is
defined
as
n
=
,
Refraction of water waves
c
m
where
c
is
the
speed
of
light
in
a
vacuum
and
v
the
c
is
the
speed
of
light
in
m
material.
normal
Suppose
n
c
2
and
1
the
n
absolute
refractive
index
of
medium
1
and
medium
2
were
respectively.
2
c
v
For
r
medium
medium
1,
n
2
=
Equation
(1)
Equation
(2)
1
c
1
medium
c
1
i
v
For
medium
2,
n
=
2
c
2
n
c
1
c
1
Equation
(2)
÷
Equation
(1)
72
c
v
÷
c
2
Figure 10.1.3
c
v
=
n
2
c
v
=
c
1
c
1
×
c
2
1
=
c
v
c
2
Chapter
Snell’s
law
can
be
rewritten
Refraction
n
sin i
Therefore,
10
1
as
1
=
or
n
sin i
=
n
1
sin r
sin r
2
n
2
Laws of
1
The
refraction
incident
ray,
the
refracted
ray
and
the
normal
at
the
point
of
i
incidence
lie
in
the
same
plane.
sin i
2
The
ratio
is
constant,
where
i
is
the
angle
of
incidence
and
r
is
sin r
i <
the
When
a
angle
a
ray
medium
normal
When
of
of
(e.g.
a
of
ray
light
travels
higher
air
of
C
refraction.
to
from
optical
a
medium
density,
the
of
lower
refracted
optical
ray
density,
bends
to
towards
2
the
glass).
light
travels
from
a
medium
of
higher
optical
density,
to
a
C
medium
normal
of
lower
(e.g.
optical
glass
to
density,
the
refracted
ray
bends
away
from
the
air).
i =
Critical
angle
and total
internal
C
reflection
3
Consider
a
incidence
90°.
ray
of
called
There
are
light
the
travelling
critical
three
from
angle,
scenarios
to
glass
C,
for
angle
of
incidence
<
C.
The
ray
2
The
angle
of
incidence
=
C.
The
angle
along
the
surface
of
the
is
There
the
(Figure
The
refracted
air
.
which
consider
1
is
to
angle
an
of
angle
of
refraction
is
10.1.4).
refracted
of
is
out
refraction
of
is
the
i
glass.
90° and
the
ray
glass.
i >
3
The
n
angle
sin 90°
of
incidence
=
n
1
sin C,
>
but
C.
The
sin 90°
ray
is
reflected
back
inside
the
C
glass.
Figure 10.1.4
=1
2
n
1
g
Therefore,
sin C
=
=
n
n
a
g
a
cladding
In
scenario
3,
the
ray
of
light
is
said
to
be
totally
internally
reflected.
The
θ
angle
of
incidence
and
the
angle
of
reflection
are
3
equal.
θ
4
core
θ
θ
2
Conditions
for
total
inter nal
5
θ
6
reflection
θ
1
cladding
1
Light
must
dense
2
The
be
travelling
medium
angle
on
(e.g.
from
glass
incidence
internal
surrounded
density
fibre
at
critical
occurs.
Fibre
than
one
a
The
is
material
glass.
end.
angle
optic
reflection
by
A
ray
The
for
the
light
cables
is
used
of
be
in
light,
of
fibre
guided
used
dense
to
an
optically
less
cables,
The
the
a
fibre
in
critical
laser
,
is
the
and
until
the
very
cladding
inside
boundary
extensively
the
angle.
Figure 10.1.5
glass
Key points
A fibre optic cable
reflection
typically
incidence
along
than
optic
cladding.
glass-cladding
are
greater
internal
called
angle
optically
air).
must
Applications of total
T
otal
to
an
it
a
is
of
into
greater
internal
reaches
fibres
lower
projected
fibre
total
field
thin
has
the
optical
the
than
glass
ᔢ
reflection
other
a
end.
communications
Refraction
direction
the
result
the
is
of
of
the
a
a
change
wave
that
change
in
in
occurs
speed
as
of
wave.
to
sin i
ᔢ
transmit
data
in
the
form
of
light
pulses
(Figure
The
is
ratio
a
constant,
and
10.1.5).
sin r
Expensive
Cheaper
engagement
rings
made
rings
from
contain
glass
diamond
sparkle
much
crystals
less,
which
because
is
‘sparkle’.
glass
refractive
of
2.42.
index
The
respectively.
from
the
many
total
1.50,
critical
Each
flat
angle
internal
of
are
while
angles
ray
surfaces
reflections
critical
of
of
for
light
inside
reflections,
glass
that
the
perceived
diamond
diamond
by
our
that
more
a
larger
diamond
diamond
diamond
means
hence
and
enters
has
before
eyes
there
as
is
are
refractive
41.8°
reflects
finally
‘sparkle’.
a
‘sparkle’
greater
than
in
and
index.
The
critical
angle,
of
incidence for
C,
is
of
refraction
which
the
the
angle
angle
24°
is
90°.
times
emerging.
The
refractive
index
many
The
smaller
chance
the
has
ᔢ
a
called
for
many
ᔢ
Total
when
internal
the
greater
reflection
angle
than
of
the
occurs
incidence
critical
is
angle.
glass.
73
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
7
4
a
accompanying CD.
mass
b
1
Explain
what
is
meant
Calculate
A
of
the
120 g
simple
is
in
potential
raised
pendulum
energy
through
consists
of
when
a
1.
1 mm.
a
[2]
light
by:
inextensible
a
oscillations
[1]
b
free
[1]
c
simple
The
oscillations
harmonic
gain
motion.
variation
horizontal
[2]
string
of
and
the
a
bob
of
potential
displacement
of
mass
of
energy
the
bob
120 g.
with
is
x,
the
shown
below.
4
potential
2
Describe
3
The
an
example
of
a free
oscillation.
energy / 10
J
[2]
15
centre
with
of
simple
a
cone
of
harmonic
a
loudspeaker
motion
is
oscillating
of frequency
1200 Hz
10
and
amplitude
0.07 mm. Calculate
a
the
angular frequency
b
the
maximum
of
the
oscillations
[2]
5
acceleration
of
the
centre
of
the
cone.
[2]
x /mm
Sketch
a
graph
displacement
the
to
x
of
show
the
the variation
acceleration
with
of
the
30
centre
A
cone.
10
20
30
[3]
pendulum
bob
oscillates
with
simple
bob
Its
displacement varies
with
of
the
pendulum
is
displaced
sideways
harmonic
until
motion.
10
of
The
4
20
time
as
its
centre
of
mass
is
raised
through
a vertical
shown
distance
of
1.
1 mm
and
then
released.
below.
c
displacement /m
Copy
the figure
variation,
as
and
the
i
the
total
ii
the
kinetic
sketch
pendulum
graphs
to
show
oscillates,
of
x
the
with:
energy
[2]
energy.
[2]
0.
18
d
Determine
the
amplitude
of
oscillation
of
the
pendulum.
t /s
[2]
0
8
0.
18
a
Explain
what
b
A
of
a
piece
meant
Plasticine®
spring. The
is
is
spring
is
is
able
to vibrate
in
begins
to vibrate
and
forced
to
by
the
term
attached
attached
a vertical
the
to
‘resonance’.
to
a
one
end
support
plane. The
mass-spring
[2]
of
that
support
system
is
Determine:
a
the
amplitude
b
the
period
c
the frequency
of
of
the
the
oscillation
oscillate.
[1]
oscillation
i
Explain
ii
Sketch
what
is
meant
by forced
oscillations.
[1]
[2]
d
the
angular frequency
[1]
e
the
acceleration
[1]
f
5
[1]
i
when
the
displacement
is
zero
ii
when
the
displacement
is
at
the
Derive
maximum velocity
an
expression for
of
the
the
a
of
iii
[1]
maximum
pendulum
period
tude
of
bob.
oscillation
[2]
pendulum.
the
to
mass
show
the variation
with frequency
of
of
ampli-
vibration
Plasticine
the
same
axes
show
the
®
[3]
is
now flattened
oscillations
as
in
effect
b
(ii)
of
to
be
sketch
so
that
it
damped. On
another
the
graph
damping.
to
[3]
a
c
simple
graph
support.
causes
[2]
of
of
the
The
a
State
one
situation
in
which
resonance
is
useful.
[6]
[1]
d
6
Derive
an
expression for
the
period
of
oscillation
State
one
hazard.
two
74
identical
springs
in
parallel.
situation
in
which
resonance
can
be
a
of
[6]
[1]
Revision questions
9
a
State
the
laws
of
refraction.
[2]
8
b
The
speed
of
of
light
light
in
in
air
glass
is
is
3.00
1.99
×
10
×10
Discuss
11
The
one
application
of
total
m s
. The
refractive
index
of
glass for
incident
on
a face
of
a
red
light
is
1.510.
Red
. Consider
light
ray
reflection. [2]
–1
m s
8
a
internal
–1
8
speed
10
4
prism
as
travelling
at
3.00
×
10
–1
m s
is
incident
at
an
shown
angle
of
32°
on
an
air–glass
boundary. Calculate:
below.
a
the
angle
b
the
speed
of
c
the
critical
of
refraction for
red
light
in
red
light
glass
60°
angle for
the
air–glass
interface.
30°
i
Calculate
used
ii
to
the
make
Calculate
the
refractive
the
index
of
the
glass
prism.
critical
[2]
angle for
a
glass/air
interface.
iii
Sketch
the
ray
a
[2]
diagram
inside
to
the
show
prism
what
and
as
happens
it
leaves
to
the
prism.
iv
Calculate
[3]
the
emerges from
angle
the
of
refraction
prism.
when
the
ray
[4]
75
11
Waves
11.
1
Waves
Learning outcomes
On
completion
should
be
able
of
this
Describing
section,
you
When
to:
point
ᔢ
understand
a
stone
moving
that
a
wave
away
of
a
is
wave
dropped
from
impact
P
.
on
into
The
the
a
lake
ripples
surface
at
or
of
a
point
waves
the
lake.
P
,
circular
transmit
The
ripples
energy
point
Q
is
are
away
a
seen
from
short
the
distance
transmits
away
from
P
.
A
side
profile
from
the
point
P
to
Q
is
shown
in
Figure
11.1.1.
energy
It
ᔢ
define
the
terms
is
the
amplitude,
period,
frequency
when
applied
to
a
understand
the
terms
P
phase
and
Q.
The
is
of
the
either
energy
the
stone
of
the
surface
of
move
is
which
wave
transfers
actually
energy
travelling
from
from
along
the
line
PQ
oscillate
about
a
fixed
P
point.
is
to
transmitted
A
propagate.
or
will
be
make
up
it
this
is
an
type
of
wave.
The
is
In
that
can
order
wave
example
wave
There
one
waves
waves .
considered.
into
the
wave
Mechanical
longitudinal
dropped
by
mechanical
are
be
to
a
in
at
right
a
the
ways
substance
classified
transverse
oscillate
a
describe
produced
of
various
requires
as
wave,
lake
wave.
angles
a
when
The
propagation
way
to
free
produce
end
up
of
a
and
the
to
the
wave.
transverse
down
with
wave
your
is
to
fix
hand
one
until
end
of
waves
a
are
rope
and
produced.
the
the
wave
from
the
side
will
produce
a
picture
similar
to
the
wave
lake.
energy
Figure
at
water
particles
lake
in
of
of
the
Viewing
flow
that
wave
was
that
Another
Q
P
stone
the
water
wave
direction
point
that
any
particles
waves.
which
particles
travel
understand
without
transverse
transverse
of
to
Q,
water
classifying
through
phase difference
direction
to
wave.
It
ᔢ
point
and
to
speed
important
displacement,
the
The
11.1.2
dashed
illustrates
line
the
represents
main
the
characteristics
rope
in
its
of
the
undisturbed
wave.
or
rest
position.
The
dropped
up
Figure 11.1.1
A
Profile of a wave
and
down
motion
progressive
energy
along
wave
with
of
is
it.
the
one
hand
that
The
generates
moves
wave
in
in
a
Figure
a
transverse
particular
11.1.2
is
progressive
direction,
made
up
of
wave.
carrying
a
series
of
crest
crests
λ
point
P
The
and
P
troughs.
on
the
The
wave
wavelength
λ
distance
from
of
a
the
wave
x,
rest
is
represents
position.
the
the
The
distance
displacement
SI
unit
between
is
two
the
of
the
metre.
successive
crests
a
x
or
two
The
successive
amplitude
from
the
Suppose
a
period
rest
a
of
a
the
time.
The
The
SI
movement
The
SI
progressive
position.
that
of
troughs.
point
a
is
wave
unit
of
P
unit
is
the
is
the
single
oscillates
metre.
the
maximum
displacement
metre.
particle
about
(point
the
rest
P)
or
is
observed
over
equilibrium
trough
position
Figure 11.1.2
of
the
rope.
Figure
11.1.3
shows
how
the
displacement
of
the
Snapshot of a wave
particle
varies
with
time.
The
particle
varies
sinusoidally
graph
with
shows
that
the
displacement
of
the
time.
displacement /m
One
complete
cycle
or
oscillation
is
made
up
of
one
crest
and
one
trough.
The
T
time
the
The
T
0
taken
wave
for
and
number
one
is
cycle
shown
of
or
by
oscillation
T
in
oscillations
Figure
per
unit
is
the
time
the
The
frequency
frequency
of
and
the
wave
period
of
and
a
is
T
wave
are
a
the
SI
period
unit
point
=
Variation of displacement
T
–
period/s;
f
–
in
related
f
76
of
measured
1
of P with time
The
on
is
a
of
the
oscillation
of
second.
wave
is
known
time/s
as
Figure 11.1.3
called
11.1.3.
frequency/Hz
hertz
by
the
(Hz).
following
equation.
Chapter
The
speed
case
refers
of
a
wave
is
the
distance
travelled
per
unit
time.
Speed
in
11
Waves
this
Exam tip
and
the
to
the
rate
wavelength
at
can
which
be
energy
used
to
is
being
determine
transferred.
the
velocity
The
of
a
frequency
wave.
The
When
derivation
is
shown
reading
period from
Time
taken
During
this
for
one
time
the
wavelength
or
below.
oscillation
the
wave
=
would
T
on
travel
a
distance
=
the
a
graph
check
the
label
x-axis.
λ
distance
speed
=
time
1
v
=
λ
×
T
displacement /m
λ
direction
of
the
wave
=
T
T
Q
P
1
But
R
=
f
∴
v
=
fλ
distance
along
λ
–1
v
–
speed/m s
Phase
;
and
f
–
frequency/Hz;
λ
–
the
wavelength/m
phase difference
Figure 11.1.4
All
the
Not
particles
all
the
particular
some
For
are
the
other
.
The
wave
measure
Phase
be
of
relationships
P
and
the
with
the
are
the
vibrate
wave
particles
about
move
may
be
their
mean
together
.
moving
At
positions.
at
Q
on
the
same
while
Key points
the
point
particle
fraction
Q.
of
sometimes
transverse
time.
R
The
the
P
is
and
are
moving
phase
of
oscillation
measured
wave
Q
in
in
said
Figure
to
be
downward
a
particular
that
has
degrees
or
been
11.1.4
in
ᔢ
Waves
are
in
phase
with
each
other
have
zero
phase
and
is
point
on
ᔢ
Mechanical
material
a
to
completed.
radians.
difference.
transmit
completely
out
of
phase
have
a
phase
difference
of
π
two
waves
P
and
Q
shown
in
Figure
two
exists
waves
between
are
the
not
two
in
phase
waves.
with
The
each
phase
Mechanical
by
the
waves
or
can
either
be
longitudinal.
or180°.
Displacement
of
a
point
on
a
11.1.5
other
.
A
phase
difference
φ
is
the
distance from
equilibrium
difference
between
them
the
position.
is
ᔢ
determined
which
travel.
wave
The
a
that
ᔢ
Consider
require
through
Particles
Particles
radians
waves
medium
transverse
are
energy.
phase
ᔢ
that
The phase of a point on a
wave
any
upwards,
downwards.
at
particle
phase
wave
up
some
moving
particles
out
a
make
time,
upwards
therefore
is
of
mechanical
that
in
may
moving
each
a
particles
example,
with
in
instant
particles
both
wave/m
The
wavelength
of
a
wave
is
the
following:
distance
between
two
successive
y
points
Phase
difference
φ
=
×
in
phase.
2π
λ
ᔢ
The
graph
in
Figure
11.1.6
shows
two
waves
P
and
Q
that
completely
The
amplitude
maximum
of
phase.
The
phase
difference
between
them
is
π
radians
or
wave
is
the
the
180°.
ᔢ
The
period
position.
of
a
wave
is
the
time
displacement /m
taken for
P
Q
P
a
displacement from
equilibrium
displacement /m
of
out
ᔢ
Q
one
oscillation.
The frequency
of
the
oscillations
per
wave
rate
number
of
a
wave
is
second.
y
ᔢ
distance
λ
distance
speed
which
the wave/m
the
The
a
energy
is
is
being
the
at
transferred.
wave/m
ᔢ
The
on
phase
a
wave
fraction
Figure 11.1.5
of
along
along
Phase difference between
Figure 11.1.6
of
of
is
a
a
the
particular
measure
point
of
oscillation
the
that
Waves that are out of phase
has
been
completed.
two waves
77
11.2
Transverse
Learning outcomes
On
completion
should
be
able
of
this
and
longitudinal
Transverse
section,
Mechanical
you
particles
to:
ᔢ
understand
between
a
the
and
In
a
appreciate
that
a
transmits
to
progressive
the
energy from
that
the
wave
is
amplitude
ᔢ
explain
ᔢ
understand
be
waves
wave.
based
T
wo
longitudinal
on
types
the
of
movement
mechanical
of
the
waves
are
slinky
spring
waves.
differentiate
and
In
order
the
between
to
other
the
produce
end
is
a
two
types
transverse
moved
up
and
of
waves,
wave ,
down
a
one
end
repeatedly.
of
the
In
slinky
this
the
particles
oscillate
at
right
angles
to
the
direction
of
type
travel
of
one
wave.
intensity
of
Light
is
an
example
of
a
transverse
wave.
order
the
term
that
polarised
to
produce
a
longitudinal
wave ,
one
end
of
the
slinky
is
fixed
a
and
the
other
this
type
end
is
moved
back
and
forth
as
shown
in
Figure
11.2.1.
In
proportional
of
wave,
the
particles
oscillate
in
the
same
direction
of
travel
of
squared
the
can
and
classified
the
another
understand
to
to
used.
wave,
In
ᔢ
be
up
waves
progressive
the
point
be
fixed
of
wave
order
can
wave
is
ᔢ
can
make
waves
longitudinal
difference
transverse
longitudinal
waves
that
transverse
and
waves
wave.
Sound
is
an
example
of
a
longitudinal
wave.
polarisation
transverse
and
waves
a
Transverse wave
direction
of
travel
of
the
wave
direction
of
travel
of
the
wave
longitudinal
cannot.
b
Longitudinal wave
Figure 11.2.1
How a slinky spring can be used to illustrate transverse and longitudinal
waves
A
transverse
case
of
When
of
a
in
a
a
is
easily
longitudinal
longitudinal
series
which
region
wave
of
illustrated
wave,
wave
the
particles
where
the
are
particles
visualising
is
compressions
using
travelling
and
can
towards
moving
sinusoidal
be
through
rarefactions .
moving
are
it
a
each
away
In
the
difficult.
a
A
curve.
medium,
it
compression
other
.
from
each
A
consists
is
a
region
rarefaction
is
a
other
.
Definitions
A
longitudinal
However,
In
a
transverse
wave,
the
particles
medium
vibrate
at
right
the
direction
of
energy
troughs
a
in
the
longitudinal
medium
wave,
vibrate
the
in
particles
the
same
of
energy
each
a
sinusoidal
wave
produced
other
variation
At
and
molecules
low
a
and
illustrated
rarefactions
wavefor m.
by
pressure.
of
a
as
a
transverse
cor respond
Figure
loudspeaker.
pressure
compression,
this
are
transfer.
a
78
on
the
loudspeaker.
air
direction
sometimes
to
wave.
peaks
11.2.2
The
illustrates
sinusoidal
a
wave
transfer.
illustrates
In
is
angles
longitudinal
to
wave
compressions
in
and
the
the
the
cor responds
moving
away
to
of
air
the
air
molecules
molecules
are
a
high
pressure.
from
each
other
and
in
front
moving
At
a
this
of
the
towards
rarefaction,
the
cor responds
to
Chapter
loudspeaker
T
C
T
C
T
C
T
C
11
Waves
T
direction
the
of
travel
longitudinal
of
wave
pressure
distance
the
Figure 11.2.2
Visualising a longitudinal wave
Progressive
A
progressive
and
transfers
wave
transverse
is
example
a
wave
is
is
and
of
longitudinal
direction,
waves
energy
Both
an
along
wave
a
a
wave
from
energy
is
to
produced
wave.
to
waves
progressive
transmitted
its
is
location
longitudinal
traverse
progressive
related
that
one
As
in
can
be
wave.
the
a
result
of
progressive
Sound
wave
that
as
vibrations
another
.
is
moves
same
an
in
a
direction.
waves.
Light
example
of
particular
The
intensity
of
a
amplitude.
2
I
The
of
intensity
the
more
the
is
wave.
low,
the
wave
a
is
is
a
with
a
a
smaller
by
small
be
by
a
the
the
a
of
ear
by
the
to
human
feel
as
energy.
the
T
ake
a
amplitude
low
as
for
a
example,
volume
with
decibel
the
waves
high
vibrations
transmits
the
waves
When
sound
ear
the
When
progressive
human
of
amplitude
amplitude.
longitudinal
able
square
large
loudspeaker
.
amount
produces
will
to
with
longitudinal
perceived
you
wave
produced
perceived
speaker
This
proportional
that
produces
transmit
instances,
is
wave
being
This
the
amplitudes.
A
means
speaker
waves
increased,
some
than
waves
amplitudes.
These
a
This
energy
sound
of
∝
small
level.
volume
with
decibel
from
the
is
larger
level.
In
speaker
.
Example
–2
A
sound
wave
Calculate
an
the
amplitude
of
amplitude
intensity
of
of
a
0.15 mm
sound
has
wave
an
of
intensity
the
same
of
3.2 W m
frequency,
.
which
has
0.45 mm.
2
Recall
that
I
∝
I
=
A
2
∴
kA
–2
Initially,
A
=
0.15 mm
and
I
=
3.2 W m
–3
3.2
=
k
=
k
×
(0.15
×
10
2
)
3.2
8
=
–3
(0.15
Now
the
changed
frequency
to
of
the
sound
×
10
wave
1.422
×
10
2
)
is
the
same,
but
the
amplitude
has
0.45 mm.
2
I
=
kA
=
(1.422
=
28.8 W m
8
×
10
)
–3
×
(0.45
×
10
2
)
–2
79
Chapter
oscillations
several
11
Waves
in
Polarisation
planes
direction
In
of
travel
of
wave
transverse
of
propagation
of
oscillation
illustrated
plane
at
right
direction
the
Figure 11.2.3
of
angles
travel
to
of
In
a
in
Figure
An
of
easy
a
the
be
wave.
in
Figure
oscillations
In
any
an
one
are
perpendicular
unpolarised
of
an
infinite
to
transverse
number
the
direction
wave,
of
the
planes.
plane
This
is
11.2.3.
wave,
the
oscillations
are
restricted
to
one
plane
as
shown
way
to
and
understand
move
the
the
free
concept
end
up
and
of
polarisation
down
with
is
your
to
fix
hand
one
end
until
waves
wave
An unpolarised wave
are
produced.
are
said
to
side
be
to
are
The
waves
vertically
side,
said
the
to
be
produced
polarised.
waves
are
If
produced
horizontally
in
you
a
vertical
move
lie
in
a
polarised.
It
plane.
your
hand
These
horizontal
plane.
is
to
possible
waves
horizontally
These
produce
waves
in
in
plane
in
the
11.2.4.
rope
waves
one
of
can
polarised
from
oscillations
waves,
different
planes
of
polarisation
by
simply
adjusting
the
angle
at
which
only
direction
your
of
travel
of
wave
hand
Consider
moves.
a
situation
where
the
light
from
a
filament
lamp
is
viewed
with
®
the
is
naked
placed
appears
eye.
in
to
An
front
have
oscillations
of
intensity,
of
the
lamp
reduced.
the
I,
light
will
and
The
be
then
reason
waves
to
observed.
viewed,
for
one
this
plane.
the
is
It
When
a
sheet
intensity
that
the
of
sheet
therefore
of
Polaroid
the
light
restricts
prevents
the
most
®
of
plane
at
right
angles
the
light
rotated
direction
of
travel
in
Figure 11.2.4
from
reaching
the
eye.
If
the
sheet
of
Polaroid
a
plane
perpendicular
to
the
direction
of
travel
of
light,
is
now
the
of
intensity
the
waves
to
is
unchanged
(Figure
11.2.5).
wave
A polarised wave
I
light
waves
are
not
polarised
Polaroid®
sheet
I
intensity
of
light
is
light
reduced
Figure 11.2.5
are
waves
polarised
The effect of a sheet of Polaroid®
®
Suppose
rotated
and
to
plane
rotated
second
through
reduced
one
a
the
T
ransverse
direction
in
of
90°,
a
it
have
planes.
propagation.
longitudinal
is
first
the
is
to
not
In
a
test
waves.
only
used
have
these
zero
Since
light
are
restricts
the
the
the
first
is
pass
by
transverse
only
sheet
the
oscillations
oscillation
to
the
polarised.
between
to
the
can
one
direction
Therefore,
transverse
is
through
waves.
to
but
now
sheet,
second
to
first,
light
11.2.6).
parallel
be
of
the
perpendicular
the
oscillations
cannot
of
waves
(Figure
wave,
distinguish
front
through
exhibited
which
waves
to
pass
transverse
Polarisation
waves
Longitudinal
a
is
to
in
intensity
through.
allow
oscillations
placed
The
waves
pass
reduced
that
now
sheet.
light
will
property
Longitudinal
Polaroid
allowed
propagation.
polarisation
80
is
of
the
intensity
waves
different
plane.
of
is
to
When
light
through
hence
Polarisation
be
90°
zero.
of
sheet
and
Chapter
11
Waves
®
There
are
many
sunglasses
Liquid
to
reduce
cr ystal
engineers
glare
displays
often
measure
under
examples
use
the
of
by
the
limiting
found
the
in
effect
amount
polarising
of
the
digital
of
effect
amount
of
watches
polarisation
stress
distributed
to
of
light.
light
are
Polaroid
entering
polarised.
per for m
the
eye.
Str uctural
experiments
throughout
a
component
test.
I
intensity
Polaroid®
sheet
is
Polaroid®
reduced
sheet
to
rotated
zero
through
90°
Figure 11.2.6
T
able
11.2.1
transverse
Table 11.2.1
summarises
wave
and
a
the
similarities
longitudinal
and
differences
between
a
wave.
Comparing transverse and longitudinal waves
Transverse
Similarities
Transfers
direction
the
energy
of
reflection,
Particles
wave
and
that
the
Transfers
of
direction
be
Light,
refraction,
at
of
up
right
energy
energy
of
the
Particles
wave
transfer
polarised
and
that
the
of
direction
of
be
refraction,
interference
make
oscillate
Cannot
microwaves
in
propagation
reflection,
diffraction
angles
wave
wave
Shows
interference
make
oscillate
direction
Can
Examples
in
propagation
the
diffraction
to
Longitudinal
wave
Shows
Differences
wave
in
up
the
energy
the
same
transfer
polarised
Sound
Key points
ᔢ
In
a
the
ᔢ
In
a
transverse
direction
A
ᔢ
The
of
progressive
direction
ᔢ
wave,
as
energy
wave,
energy
progressive
intensity
wave
of
the
a
particles
in
the
medium
vibrate
at
right
angles
to
transfer.
the
particles
in
the
medium
vibrate
in
the
same
transfer.
transfers
energy from
progressive
wave
is
one
point
proportional
to
to
another.
the
amplitude
squared.
ᔢ
In
an
unpolarised
ᔢ
In
a
ᔢ
Transverse
ᔢ
Longitudinal
plane
wave,
polarised
waves
wave,
can
waves
the
be
oscillations
the
are
oscillations
in
are
various
planes.
restricted
to
one
plane.
polarised.
cannot
be
polarised.
81
11.3
Superposition
Learning outcomes
On
completion
should
be
able
of
this
The
section,
move
two
ᔢ
state
the
principle
two
off
steel
The
ᔢ
draw
the
marbles
term
and
wide
gap
to
recall
show
narrow
wavelength
using
towards
to
a
very
a
a
towards
It
point
different
point
something
amplitude
is
diffraction
the formula
projected
directions.
is
P
each
easy
on
a
to
other
,
they
visualise
this
collide
if
you
and
project
table.
P
in
with
two
space,
waves.
they
pass
When
two
through
waves
each
are
other
.
diffraction
4A.
each
ᔢ
is
towards
However
,
diagrams
are
different
situation
directed
understand
particles
in
superposition
of
superposition
ᔢ
diffraction
principle of
When
you
to:
and
of
one
Assuming
other
,
the
of
interesting
the
that
waves
the
two
amplitude
of
happens
is
A
and
waves
the
at
the
arrive
resultant
the
point
P
.
amplitude
at
the
wave
point
at
Suppose
of
the
P
the
in
the
second
phase
point
P
is
wave
with
5 A.
If
determine
the
waves
arrive
the
amplitude
at
the
point
P
completely
out
of
phase
with
each
other
,
diffraction
of
the
resultant
wave
at
P
is
3 A.
This
phenomenon
can
be
grating.
explained
The
at
wave
a
1
the
of
they
algebraic
Consider
P
source
principle
of
superposition
superimpose
superposition
states
on
that
each
when
other
and
(Figure
two
the
or
11.3.1).
move
resultant
waves
arrive
displacement
two
sum
of
waves,
the
individual
wave
1
and
displacements
wave
2
as
shown
of
in
each
wave.
Figure
11.3.2.
Both
2
w
waves
two
is
source
principle
point,
the
2
is
e
v
a
using
are
in
waves
phase
with
combine,
produced.
Notice
by
each
other
.
applying
that
the
(Phase
the
difference
principle
amplitude
of
the
of
is
zero.)
When
superposition,
resultant
wave
is
these
wave
twice
3
as
1
large
as
wave
1
and
wave
2.
Figure
11.3.3
shows
what
happens
when
Figure 11.3.1
wave
Definition
1
and
or
superposition
more waves
2
arrive
out
of
phase.
displacement /m
a
Principle of
wave
arrive
at
a
displacement /m
wave
wave
1
1
a
– when two
point, the
time/s
resultant displacement
is the
algebraic
at that
sum of the
displacements of
point
individual
time/s
–a
each wave.
–a
a
wave
2
a
wave
2
time/s
–a
time/s
resultant
wave
–a
2a
wave
3
resultant
wave
a
wave
3
time/s
time/s
–a
2a
Figure 11.3.2
Applying the principle of
superposition (waves in phase)
82
Figure 11.3.3
Applying the principle of
superposition (waves out of phase)
Chapter
11
Waves
Diffraction
Diffraction
edge
of
an
observed
is
the
object
or
through
of
spreading
through
a
to
the
width
the
in
the
wavelength
out
gap,
gap.
of
the
It
the
a
of
gap.
wavefronts,
In
order
wavelength
is
important
wave
when
of
to
for
the
when
a
wave
significant
wave
note
that
diffraction
occurs.
the
diffraction
must
there
passes
be
is
Definition
be
comparable
no
This
to
change
Diffraction
wavefronts
be
illustrated
when
sketching
diffraction
diagrams.
the
of
a
spreading
wave
when
out
of
they
pass
information
through
should
is
Light
has
a
gap
or
pass
the
edge
of
an
a
obstacle.
very
to
small
occur
,
For
this
This
player
generated
to
be
can
and
experiment
Sound
They
a
CD
The
be
a
of
be
player
light
easily
be
light
easily
can
be
doesn’t
in
order
for
comparable
is
on
not
the
heard
bend
diffraction
to
normally
other
in
the
have
through
another
around
a
light
in
our
much
doorway.
room
corners
of
wavelength.
observed
hand,
diffracted
demonstrated
are
wavelength
bar
.
As
produced
what
performed
Figure 11.3.4 (a)
that
should
waves
can
straight
shows
is
means
gap
but
to
not
allow
the
seen.
waves
(c)
This
the
diffraction
seen.
using
plane
(b)
why
be
Diffraction
(a),
the
wavelengths.
explains
tank,
of
experiences.
necessarily
CD
width
reason
everyday
larger
wavelength.
the
before
=
is
the
and
seen
using
in
bar
move
from
gaps
a
ripple
vibrates
towards
above
and
tank.
on
the
the
edge
λ
a
Plane
surface
gap.
Figure
ripple
of
waves
the
an
tank
of
are
the
11.3.4
when
the
object.
wavelength
after
=
λ
wavelength
after
=
λ
wavelength
after
=
λ
Narrow gap diffraction
wavelength
Figure 11.3.4 (b)
before
=
λ
Wide gap diffraction
wavelength
before
=
λ
object
Figure 11.3.4 (c)
If
the
gap
is
diffraction
same
the
made
will
principle
door
doesn’t
Diffraction at the edge of an object
that
allows
allow
much
occur
.
waves
applies
light
the
smaller
The
to
to
pass
a
than
will
wavelength
pass
microwave
through
microwaves
the
not
to
so
escape
cooker
.
that
from
of
through
you
The
can
inside
the
the
metal
view
the
wave,
gap.
the
It
no
is
grid
this
on
food,
but
microwave.
83
Chapter
11
Waves
Single
When
a
occurs.
slit diffraction
single
The
slit
is
placed
intensity
in
front
distribution
of
is
a
parallel
shown
in
beam
Figure
of
light,
diffraction
11.3.5.
intensity
I
0
θ
0
single
slit
Figure 11.3.5 (a)
screen
0
Single slit diffraction
distance
Figure 11.3.5 (b)
along
screen
Variation of intensity
with distance along screen
Diffraction
A
diffraction
ruled
on
a
e.g.
interference
the
bright
zeroth
order
grating
piece
wavelength
The
gratings
of
red
fringes
maximum
zeroth
order
produces
a
or
light)
pattern
order
consists
glass
of
are
is
=
called
1).
maximum.
fringe
a
large
and
at
dark
maxima.
(n
The
=
0).
fringe
Figure
number
When
incident
bright
maximum
(n
of
plastic.
right
The
are
fringe
next
pattern
11.3.6
equally
angles
fringes
The
of
monochromatic
to
at
the
is
how
on
centre
called
symmetrical
illustrates
a
(a
lines
single
grating,
observed
fringe
is
the
spaced
light
a
an
screen.
is
the
called
first
about
diffraction
the
grating
pattern.
n = 2
n = 1
monochromatic
n = 0
θ
light
screen
n = 1
diffraction
grating
n = 2
Figure 11.3.6
Figure
Producing a fringe pattern with a diffraction grating
11.3.7
shows
the
variation
of
intensity
along
the
intensity
n=3
Figure 11.3.7
84
n=2
n=1
n=0
n=1
n=2
n=3
Variation of intensity along the fringe pattern
fringe
pattern.
Chapter
The
light
coherent
the
path
The
At
incident
sources
light
slits
waves
is
an
wavelength
of
of
in
phase
regions
and
is
and
of
a
ever y
where
obser ved,
half
bright
fringes
of
path
The
are
half
diffraction
obser ved,
wavelengths.
occurs.
difference
light
inter ference
is
from
waves
occurs.
grating
Waves
producing
inter ference
the
wavelengths.
the
slit,
number
constr uctive
are
on
at
whole
destr uctive
incident
as
diffracts
slits
fringes
number
light
At
the
phase
deviation
grating
waves.
dark
odd
out
the
from
are
where
completely
angular
of
difference
regions
the
on
11
are
The
related
to
the
follows:
Equation
d sin θ
=
nλ
d
–
separation
θ
–
angle
of
λ
–
wavelength
n
–
nth
slits
between
of
in
the
grating/m
zeroth
order fringe
and
the
nth
order fringe
light/m
order fringe
Example
When
plane
at
monochromatic
diffraction
angles
500
of
lines
30°
per
light
grating,
to
the
of
the
normal
millimetre.
wavelength
second
to
order
the
Calculate
λ
is
grating.
the
incident
diffraction
The
value
of
normally
lines
are
diffraction
on
a
Exam tip
formed
grating
has
When
λ
1
×
white
diffraction
–3
light
passes
grating,
a
through
central
a
bright
10
–6
Separation
of
slits
d
=
=
2
×
10
m
fringe
is
observed. On
either
side
of
500
the
d sin θ
=
nλ
)(sin 30°)
=
2
λ
=
central
spectra
bright fringe,
(red
→
violet)
a
are
series
of
seen.
–6
(2
×
10
×
λ
–6
(2
×
10
)(sin 30°)
2
–7
λ
=
5
×
10
m
Example
Determine
the
highest
order
of
diffracted
beam
that
can
be
produced
–6
when
light
a
grating
of
with
wavelength
a
spacing
of
2
×
10
m
is
illuminated
normally
with
640 nm.
d sin θ
=
nλ
Key points
Recall,
sin θ
≤
1
(i.e.
the
sine
of
an
angle
cannot
exceed
1)
nλ
∴
ᔢ
≤
Principle
of
superposition
–
when
1
d
two
or
more
waves
arrive
at
a
–9
n
×
(640
×
10
point,
)
∴
≤
the
resultant
displacement
1
–6
2.5
×
10
–6
2.5
n
×
10
that
of
the
of
each
point
–9
n
has
to
be
an
≤
×
10
3.91
ᔢ
three
corresponds
on
either
to
integer
,
7
algebraic
sum
displacements
wave.
Diffraction
the
highest
order
possible
is
n
=
is
principal
maxima
(the
zero
order
the
of
a
spreading
wave
when
out
maxima,
plus
the
edge
through
of
an
a
gap
or
pass
of
they
3.
pass
This
the
individual
wavefronts
Since
is
≤
640
n
at
the
obstacle.
side).
85
11.4
Interference
Learning outcomes
On
completion
should
be
able
of
this
Young’s double
section,
Thomas
you
and
to:
on
ᔢ
explain
what
is
meant
Y
oung
produces
a
card
performed
an
with
slit
experiment
an
interference
two
slits
in
experiment
pattern.
it.
The
that
showed
that
Monochromatic
slits
are
a
fraction
light
light
of
a
is
was
made
millimetre
a
wave
to
in
shine
width
by
and
are
about
one
millimetre
apart.
A
screen
is
set
up
about
one
metre
or
interference
more
ᔢ
understand
and
ᔢ
the
terms
coherence
dark
shows
path difference
explain Young’s
and
away
double
from
fringes
the
the
is
slits
(Figure
observed
variation
of
the
on
11.4.1).
the
screen
intensity
of
using
the
series
of
(Figure
light
equally
11.4.2).
along
the
spaced
Figure
fringe
bright
11.4.3
pattern.
screen
slit
double
experiment
A
principle
slit
of
single
slit
superposition
S
A
1
ᔢ
state
to
the
conditions
produce
an
interference
necessary
monochromatic
interference
light
region
observable
source
pattern.
S
B
2
Figure 11.4.1
Young’s double slit experiment (Top view)
0
intensity
x
bright fringe
central
bright fringe
distance
0
Figure 11.4.2
x
Fringe pattern produced on
the screen
Figure 11.4.3
Variation of intensity of light along the fringe pattern
Coherent
waves
Coherent
waves
difference
between
have
the
same
frequency
and
hence
a
constant
phase
Definition
phase
Two
waves
are
said
to
be
coherent
is
a
constant
phase
are
T
o
said
each
At
Constructive
produce
interference
the
waves
arrive
coincide). The
displacement
two
waves
is
in
between
Q.
If
P
them
and
is
Q
zero.
are
P
in
and
the
waves.
phase
If
P
and
difference
Q
are
between
completely
them
is
out
180°.
P
of
phase
and
Q
are
waves.
double
where
light
slit,
the
sources,
a
diffraction
single
takes
experiment
crests
of
waves
slit
place
(light
is
as
coincide,
used.
and
the
a
As
the
light
light
waves
passes
overlap.
wave)
(Figure
11.4.4).
These
points
constr uctive
correspond
to
bright
interference
fringes
on
phase
resultant
greater
than
interference
arrive
screen.
At
interference
either
fringes
out
of
occurs
phase
on
points
occurs
the
where
(Figure
screen.
In
crests
and
11.4.5).
both
cases,
troughs
These
the
coincide,
points
waves
are
destr uctive
correspond
to
dark
superposed.
when
Path difference
(crest
Figure
trough
coincide). The
displacement
is
11.4.6
illustrates
two
slits,
resultant
S
and
1
is
a.
The
central
zero.
the
86
difference
and
waves.
Destructive
and
coherent
coherent
the
points
the
the
phase
P
occurs
occurs
the
be
Explaining the
Definitions
of
to
other
,
coherent
through
(crests
the
waves
them.
still
when
other
,
two
difference
with
between
each
Consider
if
Q
there
with
them.
point
P
.
bright
fringe
is
at
O
S
.
The
distance
between
them
2
and
the
first
bright
fringe
occurs
at
Chapter
displacement /m
wave
11
Waves
1
P
a
time/s
x
displacement /m
wave
1
S
a
–a
1
θ
θ
time/s
a
wave
0
a
2
–a
time/s
S
2
λ
a
–a
wave
2
D
resultant
wave
time/s
2a
wave
Figure 11.4.6
3
Understanding path
difference
–a
resultant
a
time/s
wave
wave
S
3
P
1
time/s
S
2a
Figure 11.4.4
Constructive interference
P
2
–a
Figure 11.4.5
Destructive interference
λ
Suppose
S
and
S
1
bright
emit
wave
crests
at
the
same
time.
The
point
P
is
a
2
fringe
and
distance
therefore
constructive
interference
must
occur
at
this
Figure 11.4.7
point.
The
distance
S
P
is
greater
than
S
P
.
2
from
S
has
to
travel
Therefore,
the
wave
crest
interference
1
a
longer
distance
to
arrive
at
P
when
compared
Condition for constructive
to
2
the
wave
the
two
crest
from
S
.
In
order
for
constructive
interference
to
occur
at
P
,
1
crests
travelled
by
must
the
coincide.
wave
from
S
This
is
can
equal
only
to
happen
exactly
if
one
the
extra
distance
wavelength
λ.
S
P
1
This
2
extra
distance
is
called
the
path
difference
(Figure
11.4.7).
S
P
2
For
constructive
zero
or
a
whole
interference
number
of
to
occur
at
P
,
the
path
difference
must
be
wavelengths.
λ/2
S
P
–
S
2
P
=
nλ
n
=
0,
1,
2,
…
distance
1
Figure 11.4.8
So
the
path
difference
must
be
0,
λ,
2λ,
Condition for destructive
…
interference
Where
For
P
is
the
first
destructive
dark
fringe,
interference
to
destructive
occur
,
a
interference
crest
and
a
must
trough
occur
must
at
P
.
coincide.
Key points
This
can
only
happen
if
the
extra
distance
travelled
by
the
wave
from
S
2
is
equal
to
exactly
½λ
(Figure
11.4.8).
ᔢ
For
destructive
odd
number
of
interference
to
occur
at
P
,
the
path
difference
must
be
an
Young’s
shows
half-wavelengths.
double
that
slit
light
experiment
behaves
as
a
wave.
1
S
P
–
S
2
P
(
=
1
n
)
+
ᔢ
λ
n
=
0,
1,
2,
The
principle
of
superposition
…
2
can
1
So
the
path
difference
must
be
3
⁄2 λ,
⁄2 λ,
…
There
normal
are
interference to take
conditions,
conditions
interference
that
the
interference
must
be
Constructive
when
place
satisfied
of
light
to
is
not
produce
easily
ᔢ
must
be
ᔢ
The
waves
must
meet
at
a
the
waves
are
arrive
in
occurs
phase
with
interference
waves
each
arrive
out
occurs
of
phase
other.
Coherent
polarised,
waves
are
waves
that
point.
have
If
interference
coherent.
ᔢ
ᔢ
waves
Destructive
with
waves
the
observed.
observable
patterns:
The
explain
other.
when
ᔢ
to
experiment.
each
Under
used
5
⁄2 λ,
ᔢ
Conditions for
be
they
must
be
in
the
same
plane
a
constant
phase
difference
of
between
them.
polarisation.
ᔢ
ᔢ
ᔢ
The
The
waves
must
amplitudes
be
of
of
the
the
same
waves
Observable
interference
can
observed
must
be
only
The
waves
must
have
the
same
be
only
when
similar
.
certain
ᔢ
patterns
type.
conditions
are
met.
frequency.
87
11.5
Interference
Learning outcomes
On
completion
should
be
able
of
this
experiments
Demonstrating
section,
A
you
straight
travel
to:
each
ᔢ
describe
experiments
of
interference
waves,
sound
wood
gaps.
gaps
This
can
as
be
used
shown
causes
the
in
to
a
ripple tank
produce
Figure
waves
to
plane
11.5.1.
overlap
wave
fronts
Diffraction
in
the
that
occurs
region
at
beyond
openings.
As
a
result,
an
interference
pattern
is
produced.
There
are
of
points
water
the
of
two
in
to
the
demonstrate
piece
towards
interference
waves
on
the
surface
of
the
water
where
it
is
stationary,
and
points
where
and
it
is
disturbed.
If
we
assume
that
the
wavefronts
represent
crests,
then
microwaves.
points
of
occurs.
curved
intersection
This
line
CD
The
the
vibrating
waves
At
occurs.
with
experiment
same
of
corresponds
wavefronts.
interference
are
would
to
This
can
of
where
at
which
constructive
intersection
crests
corresponds
to
and
the
of
the
troughs
points
of
interference
line
AB
meet,
with
the
destructive
intersection
of
the
to
the
wavefronts.
also
be
(Figure
coherent.
points
points
points
curved
strip
be
The
performed
11.5.2).
using
This
interference
two
would
pattern
dippers
ensure
would
attached
that
be
the
sources
similar
to
that
B
produced
in
Figure
11.5.1.
A
X
0
C
vertical
vibrations
D
dipper
dipper
Figure 11.5.1
Demonstrating interference
Figure 11.5.2
E
D
Using dippers to produce circular wavefronts
in a ripple tank
Demonstrating the
In
order
to
demonstrate
interference of
the
interference
of
sound
sound
waves
waves
the
following
are
required:
ᔢ
two
ᔢ
a
The
identical
signal
two
generator
.
loudspeakers
loudspeakers
facing
so
A
the
that
it
line
(large
zero
produces
and
moved
the
in
in
is
a
front
of
on
O
amplitude
waves
the
of
are
frequency
to
an
the
out
Y
,
line
the
XY
.
The
in
At
loudness
phase
and
At
An
O.
will
500 Hz
observer
point
will
O,
When
reach
destructive
of
apart
a
the
is
to
2 kHz.
notice
in
and
adjusted
walks
sound
difference
point
The
1.0 m
generator
observer
the
a
to
range
path
at
0.5
signal
the
The
occurs
generator
.
about
The
oscilloscope).
of
signal
oscilloscope.
interference
the
a
are
11.5.3).
oscilloscope).
towards
on
to
they
loudspeakers.
along
the
that
(Figure
signal
constructive
sound
connected
so
connected
loudness
from
are
adjusted
direction
amplitude
(small
88
same
XY
variation
is
are
microphone
the
loudspeakers
is
this
along
a
heard
case
is
microphone
minimum
minimum
interference
value
loudness,
occurs.
Chapter
11
Waves
Y
loudspeaker
P
oscilloscope
signal
a
O
generator
loudspeaker
microphone
X
Figure 11.5.3
Demonstrating interference of sound waves
Demonstrating the
In
order
to
demonstrate
interference of
the
interference
of
microwaves
microwaves
the
following
are
required:
ᔢ
a
microwave
ᔢ
a
metal
ᔢ
a
microwave
transmitter
plate
Microwaves
with
two
slits
detector
.
from
the
transmitter
T
are
incident
on
the
two
slits
S
1
and
S
which
are
equidistant
from
T
.
2
of
region
which
is
is
at
O
is
moved
from
a
are
S
plane
of
maximum.
O
towards
to
occurs.
approximately
A,
another
the
In
slits
The
path
occurs
the
at
as
is
coherent
slits
at
sources
in
the
decreases
the
out
point
of
the
superpose
along
11.5.4).
When
demonstration,
and
moved
difference
O.
are
the
(Figure
intensity
maximum
microwaves
this
by
detector
the
act
2
diffracted
microwave
the
intensity,
interference
A
interference
increasing
minimum
waves
it.
to
constructive
before
is
The
beyond
parallel
measured
and
and
1
microwaves.
the
S
the
The
this
to
phase
a
AB,
intensity
case
is
microwave
X.
in
line
zero
detector
minimum
At
and
distance
a
point
of
destructive
between
the
slits
1–8 cm.
metal
with
A
plate
two
slits
X
S
1
O
T
S
2
microwave
microwave
transmitter
detector
B
Figure 11.5.4
Demonstrating interference of microwaves
Key points
ᔢ
An
interference
ᔢ
Interference
connected
ᔢ
Interference
a
metal
of
to
sound
the
of
sheet
pattern
of
water
waves
same
microwaves
with
two
can
signal
slits
waves
be
can
be
produced
produced
using
two
in
a
ripple
tank.
loudspeakers
generator.
can
be
and
a
produced
by
microwave
using
a
microwave
source,
detector.
89
11.6
Measuring
the
wavelength
of
Learning outcomes
light
ax
Derivation of the formula
λ
=
D
On
completion
of
this
section,
you
Consider
should
be
able
two
coherent
light
sources
is
the
midpoint
of
S
and
S
1
derive
the formula for
separation
in Young’s
the fringe
double
screen
slit
the
experiment
ᔢ
describe
measure
light
an
experiment
the
and
S
1
to:
M
ᔢ
S
to
wavelength
using Young’s
of
double
is
path
A
central
difference
The
first
The
formula
central
slit
D.
bright
difference
S
be
P
equal
–
S
2
P
fringe
to
occurs
derived
fringe
and
is
separated
by
a
distance
a.
2
distance
between
the
slits
and
the
2
bright
fringe
can
bright
is
.The
at
by
the
equal
is
zero.
the
point
P
considering
first
to
located
adjacent
one
at
the
Constructive
(Figure
the
point
wavelength
In
this
case,
occurs
at
O.
11.6.1).
distance
bright
O.
interference
fringe.
between
The
the
path
λ
1
experiment
Consider
ᔢ
describe
an
experiment
the
triangle
PMO.
to
x
measure
the
wavelength
of
tan θ
light
=
D
using
a
diffraction
grating.
Consider
the
triangle
S
S
2
Q
1
λ
sin θ
=
a
P
If
D
For
>>
a,
small
the
angle
angles
θ
θ
is
≈
very
sin θ
small.
≈
tan θ
x
S
λ
1
x
∴
=
θ
a
θ
a
D
O
M
ax
∴
Q
λ
=
D
S
2
λ
The
equation
is
only
applicable
when
D
>
>
a
and
that
S
and
S
1
act
as
2
D
coherent
(diagram
not
drawn
to
sources
of
scale)
Measuring the
Figure 11.6.1
light.
wavelength of
light
using the Young’s
Deriving the formula for
wavelength
double
slit
arrangement
D
light
of
wavelength
S
λ
1
a
S
2
travelling
microscope
double
slit
with
(diagram
Figure 11.6.2
The
double
1
and
S
.
The
to
scale)
a
monochromatic
arrangement.
light
width
of
of
In
this
wavelength
the
slits
is
light
λ,
is
source
can
experiment
incident
approximately
a
be
determined
parallel
normally
0.5 mm
beam
on
and
a
using
of
pair
the
of
slits
distance
2
between
the
microscope
produced
slits
is
is
is
used
(Figure
microscope
90
of
slit
monochromatic
S
drawn
Measuring the wavelength of light using Young’s double slit experiment
wavelength
Y
oung’s
not
scale
approximately
to
measure
11.6.2).
The
approximately
1 mm.
the
distance
2 m.
In
distance
In
this
between
order
experiment,
between
to
the
the
slits
measure
a
bright
and
the
travelling
fringes
the
travelling
distance
between
Chapter
the
fringes
accurately
,
the
distances
between
several
fringes
are
11
Waves
measured
Equation
and
the
average
measurement.
separation
The
taken.
disadvantage
This
of
minimises
this
method
the
is
random
the
error
intensity
of
in
the
the
ax
light
λ
decreases
difficult
the
as
to
you
move
know
away
exactly
monochromatic
where
light
Measuring the
from
a
source
the
central
bright
is
bright
fringe
is
calculated
wavelength of
fringe.
located.
as
shown
This
The
makes
=
it
wavelength
D
of
x
–
fringe
λ
–
wavelength
separation/m
D
–
perpendicular
(right).
monochromatic
between
light
double
using
a diffraction
laser
A
laser
means
can
is
be
used
used
that
it
as
a
because
diffraction
grating
of
diffraction
grating
is
monochromatic
it
produces
produces
a
the
light/m
distance
screen
and
the
between
the
slits/m
grating
a
A
of
very
known
a
thin
light
highly
beam
number
of
source
for
collimated
of
light
lines
per
with
this
light
little
metre
N
experiment.
source.
used.
the
separation
slits/m
This
spreading.
is
–
A
The
Equation
is
projected
on
the
set
up
on
diffraction
a
spectrometer
.
grating
such
The
that
it
light
is
at
from
right
the
laser
angles
to
it
d sin θ
λ
(Figure
11.6.3).
spectrometer
.
The
The
angular
deviation
wavelength
of
the
can
light
be
is
measured
calculated
accurately
as
shown
using
n
(right).
λ
The
experiment
about
1.5 m
zero - order
using
is
a
One
from
by
the
rule.
be
done
diffraction
light
The
and
using
grating.
the
angular
a
screen.
The
nth- order
deviation
The
screen
separation
diffracted
of
the
nth
is
between
light
order
is
light
of
is
measuring
that
it
the
allows
wavelength
for
a
larger
using
the
distance
to
–
the
measured
diffracted
wavelength
of
monochromatic
light/m
placed
d
–
distance
θ
–
angular
n
–
light
calculation.
advantage
diffracted
also
diffracted
metre
found
can
=
the
between
slits/m
deviation for
the
nth
order
nth
order
diffracted
light
second- order
be
measured
accurately.
One
disadvantage
diffracted
light
intensity
of
diffracted
is
the
of
measuring
that
light
it
is
the
difficult
decreases
as
wavelength
to
pinpoint
you
move
using
its
the
actual
away
from
second- order
position,
the
because
zero - order
light.
top view
n
= 2
spectrometer
n
= 1
angular
deviation
directly from
n
laser
diffraction
(known
Figure 11.6.3
measured
spectrometer
= 0
grating
spacing)
Measuring the wavelength of monochromatic light using a diffraction grating
Example
Key points
A
laser
produces
monochromatic
light
which
is
incident
at
right
angles
5
to
a
diffraction
grating.
The
diffraction
grating
has
5.6
×
10
lines
per
ᔢ
metre.
A
screen
is
located
1.50 m
from
the
diffraction
grating.
The
wavelength
monochromatic
spots
are
observed
at
0.54 m
on
either
side
of
the
central
bright
the
wavelength
of
the
monochromatic
determined
slits
d
=
can
double
slit
1
experiment.
–6
between
source
experimentally
light.
using Young’s
1
Distance
a
light
spot.
be
Calculate
of
Bright
=
=
1.79
×
10
m
5
N
5.6
×
10
ᔢ
The
wavelength
of
a
0.54
–1
Angle
between
zero - order
and
first- order
light
θ
=
tan
(
)
=
19.8°
monochromatic
light
source
can
1.5
be
determined
experimentally
–6
d sin θ
λ
=
1.79
×
10
×
=
n
sin 19.8°
=
606 nm
using
a
diffraction
grating.
1
91
11.7
Stationary
Learning outcomes
On
completion
should
be
able
of
this
waves
Stationary
section,
Stationary
you
two
to:
understand
what
is
meant
by
describe
experiments
stationary
transverse
to
produce
transverse
transverse
travelling
waves
produce
ᔢ
compare
progressive
stationary
waves
same
speed
can
progressive
progressive
from
a
right
in
form
wave
wave,
to
stationary
A
waves.
stationary
In
kept
taut
produces
2
reflected
leaving
left.
wave
are
formed
type,
of
opposite
stationary
travelling
of
same
These
by
equal
the
superposition
amplitude
directions.
waves.
from
speed,
two
(dashed
The
can
means
Both
and
of
frequency
Figure
longitudinal
11.7.1
left
to
right.
shows
frequency
progressive
Another
and
waves
amplitude
superpose
is
to
line).
the
they
of
the
the
a
which
superpose
string
of
passes
small
pulley
Figure
on
length
string
waves
oscillator
See
one
several
reach
wave.
velocity
produced
The
of
progressive
as
be
experiment,
oscillator
.
by
stationary
wave
wave
this
mechanical
a
over
weights.
travel
and
with
a
causing
string
is
it
to
right.
right
reflected
to
waves
oscillate
pulley
mechanical
left
from
to
attached
frictionless
The
from
travel
the
by
long
and
a
is
oscillator
The
left.
to
and
waves
The
are
waves
produce
a
11.7.2.
incident
wave
and
the
reflected
wave
along
the
string
1
is
Figure 11.7.1
same
waves)
the
and
rapidly.
wave
the
of
wave
a
ᔢ
with
(standing
waves
a
and
stationary
waves
progressive
travelling
ᔢ
waves
given
by:
Producing a stationary wave
Equation
T
v
string
=
pulley
√
m
–1
v
–
velocity
T
–
tension
m
–
mass
of
in
wave/m s
the
string/N
–1
oscillator
per
unit
length
of
the
string/kg m
weight
Figure 11.7.2
Producing a stationary
wave on a string
Suppose
Only
a
stretched
specific
illustrates
string
modes
the
three
of
is
fixed
vibrations
simplest
at
its
are
modes
end
and
possible
of
then
on
vibrations
the
made
to
string.
(i.e.
first,
vibrate.
Figure
second
11.7.3
and
A
third
λ
harmonics).
frequency
at
The
which
it
first
harmonic
vibrates
is
is
called
called
the
the
fundamental
fundamental
and
frequency .
the
The
2
higher
first
frequencies
are
called
overtones.
harmonic
The
fundamental
frequency,
(fundamental)
f
,
of
a
vibrating,
stretched
string
is
given
by:
o
Equation
A
A
1
f
T
=
0
N
2l
N
√
m
N
f
–
fundamental frequency/Hz
0
second
(first
harmonic
l
–
length
T
–
tension
of
m
–
mass
string/m
overtone)
in
the
string/N
–1
A
A
per
unit
length/kg m
A
N
N
There
N
third
are
points
harmonic
(second
at
rest.
These
is
zero.
At
92
points
points
a
are
stationary
called
wave
midway
nodes.
where
between
The
particles
nodes
displacement
are
on
a
stationary
at
permanently
these
wave
points
the
overtone)
displacement
Figure 11.7.3
along
N
points
are
is
twice
called
the
amplitude
antinodes
(Figure
of
either
11.7.4).
progressive
wave.
These
Chapter
The
wavelength
of
the
wave
is
given
11
Waves
by:
A
A
A
Equation
λ
=
2d
N
λ
–
wavelength
d
–
distance
Figure
From

node
shows
diagram
two
the
it
adjacent
position
can
be
nodes/m
of
seen
the
that
Figure 11.7.4
string
all
the
at
various
particles
points
between
in
time.
positions of
at various
adjacent
are
in
phase
with
each
other
.
All
the
particles
between
N
and
Nodes and antinodes
string
string
string
2
moving
moving
in
phase
with
each
other
.
All
the
particles
between
N
and
N
2
also
in
phase
with
each
other
.
However
,
the
particles
are
3
between
N
out
of
phase
with
the
particles
between
N
and
N
2
between
N
and
N
1
N
and
N
2
are
moving
downwards,
.
All
the
N
2
particles
3
while
the
particles
between
N
N
2
are
upwards
upwards
and
1
are
antinode
N
1
are

points
in time
nodes
A
wave/m
between
11.7.5
this
of
N
2
N
3
N
4
5
1
moving
upwards.
distance
3
along
Consider
within
the
one
phase
cycle
of
of
the
the
particles
wave
is
at
in
a
a
progressive
different
phase
wave.
from
Each
each
point
the
string
Therefore,
Points
A
points
and
D
A,
are
B,
C
and
moving
D
are
at
upwards,
a
different
while
phase
points
B
in
and
their
C
are
cycle.
A
Points
E
and
are
F
in
are
phase
in
with
phase
each
with
other
.
each
They
other
.
are
They
both
are
moving
The
The
one
(Figure
amplitude
wave.
It
varies
speed
end
lined
of
both
a
of
the
from
and
fine
a
B
moving
progressive
wave
E
11.7.6).
sound
open
with
downwards
upwards.
A
downwards
moving
downwards
moving
Figure 11.7.5
and
C
string
moving
downwards.
Points
string
other
.
stationary
zero
can
the
at
be
a
to
A
closed
depends
a
measured
other
powder
.
wave
node
using
as
on
the
maximum
a
shown
loudspeaker
,
at
long
in
position
the
cylindrical
Figure
connected
to
tube
11.7.7.
a
along
the
C
F
antinode.
signal
D
with
The
tube
Figure 11.7.6
A progressive wave
is
generator
,
loudspeaker
is
placed
at
adjusted
the
so
tube
nodes.
open
that
create
The
wavelength
speed
the
of
the
a
the
The
powder
stationary
distance
of
end.
the
wave
eventually
wave.
between
sound
can
frequency
the
waves
be
The
the
settles.
powder
nodes
and
of
is
since
The
to
generator
sounds
settles
used
the
signal
at
long
is
waves
tube
inside
displacement
determine
frequency
is
the
known,
the
heaps
of
powder
signal
determined.
generator
Comparing
stationary
and
progressive
Figure 11.7.7
waves
Producing a stationary
sound wave
Progressive
wave
Standing
wave
Key points
Transfers
point
to
energy from
one
another.
Even
though
transfer
the
wave
energy from
has
one
energy,
point
to
it
does
not
another.
ᔢ
All
up
the
the
particles
wave
that
have
make
the
same
amplitude.
The
particles
different
zero
that
make
up
amplitudes. The
(nodes)
to
a
the
wave
amplitude
maximum
have
Stationary
waves)
ranges from
superposition
(antinodes).
waves
equal
All
the
particles
that
make
There
are
particles
that
make
up
the
wave
of
the
the
wave
are
in
motion.
are
stationary
phase
difference
exists
particles
neighbouring
of
the
by
two
same
the
progressive
type,
of
with
and frequency
the
same
speed
in
(antinodes).
Between two
nodes the
neighbouring
particles
phase difference
is
zero.
(A
and
directions.
between
ᔢ
between
of
amplitude
opposite
A
(standing
that
travelling
up
waves
are formed
A
node
is
permanently
is
a
point
on
a
wave
that
B)
at
rest.
wave.
The
phase
difference
between
particles A
and C
ᔢ
is
π
An
antinode
is
the
point
midway
radians.
between
nodes.
93
11.8
Sound
waves
Learning outcomes
On
completion
should
be
able
of
this
Sound
section,
Sound
you
describe
practical
waves
in
applications
discuss
the
in
to
liver
.
application
musical
understand
of
that
reflected
widely
in
industry.
frequency
pregnant
women,
sound
and
In
the
waves)
internal
field
is
of
used
organs
medicine,
to
produce
such
as
the
images
kidney
is
also
used
in
physiotherapy.
same
principle
used
components
to
determine
used
in
if
there
are
hairline
fractures
in
industries.
instruments
sound
and
is
sound
waves
use
refracted.
are
way
to
is
to
earth.
In
Sound
The
from
use
determine
the
the
depths
submarines
waves
such
as
and
frequency
the
fundamental
If
note
the
distances
location
waves
lowest
determine
The
sound
countries
determining
harmonic
bats
determined.
Geologists
L
how
transmitted
wave
first
sonar
of
objects.
It
operates
in
a
can
similar
be
used
high
Ultrasound
Submarines
ᔢ
industry
of
mechanical
waves
in
industry
The
ᔢ
are
(very
foetuses
and
sound
waves
ultrasound
to:
of
ᔢ
waves
to
oil
and
that
a
this
of
objects.
time
be
the
taken
Sound
for
the
waves
reflected
calculated.
structure
technique
is
of
the
crucial
underlying
in
gas.
instruments
vibrating
This
the
then
determine
musical
frequency.
and
can
T
rinidad
of
location
string
note
is
or
called
pipe
the
can
produce
is
called
fundamental
(fundamental)
a
is
n
times
the
fundamental,
it
is
called
the
nth
har monic
λ
L
=
2
When
a
string
together
A
second
musical
of
a
the
note
guitar
is
plucked,
fundamental
is
notes
note.
characterised
by
of
These
high
notes
loudness,
frequencies
are
pitch
called
and
are
produced
overtones
timbre .
Loudness
harmonic
of
(first
with
a
note
depends
on
the
amplitude
of
sound.
Pitch
of
a
note
is
dependent
overtone)
on
L
=
the
frequency
the
relative
sound.
strengths
Percussion
third
of
Timbre
or
quality
of
a
note
is
dependent
on
λ
of
the
instrument
overtones
–
steel
produced
with
the
note.
pan
harmonic
(second
The
steel
by
stick
pan
is
a
percussion
instrument.
It
produces
sound
when
struck
overtone)
a
with
a
piece
of
rubber
attached
to
its
end.
When
a
particular
note
3 λ
L
is
=
struck,
stationary
waves
are
set
up
on
that
particular
section
of
the
steel
2
pan.
Figure 11.8.1
Only
certain
modes
of
vibration
are
possible.
The
largest
amplitude
of
Stationary waves in an
vibration
is
that
of
the
fundamental
frequency.
Overtones
are
also
present
open-ended tube
and
give
the
Stringed
characteristic
instrument
‘steel
–
pan’
sound.
guitar
tuning fork
antinode
end-correction,
c
A
guitar
has
transverse
resonant
tube
the
ends.
certain
several
waves
strings
travel
Stationary
modes
of
attached
along
waves
the
are
vibrations
to
it.
string
produced
are
allowed.
When
and
are
(refer
The
a
string
reflected
to
Figure
string
is
plucked,
on
reaching
11.7.3).
vibrates
Only
with
the
L
fundamental
frequency
as
well
as
the
overtones.
node
Wind
A
flute
across
water
instrument
is
one
produces
and
is
end,
of
a
the
hollow
air
inside
longitudinal
wave
at
the
tube.
wave
far
superpose
the
tube
ends
The
a
are
begins
(pressure
end.
and
Both
wave)
incident
stationary
open.
to
When
vibrate.
which
wave
is
The
travels
longitudinal
air
is
vibration
along
wave
produced.
blown
the
and
As
in
tube
the
the
case
A resonance tube
of
94
a
reflected
reflected
Figure 11.8.2
made
– flute
the
steel
pan
and
guitar
,
only
certain
modes
of
vibrations
are
allowed.
Chapter
11
Waves
Resonance tubes
A
resonance
Figure
the
tube
11.8.2
can
shows
experiment.
A
be
used
the
to
determine
arrangement
tuning
fork
of
of
the
the
known
speed
of
apparatus
frequency
is
sound.
used
to
placed
perform
over
the
λ
air
column.
The
air
above
the
tube
vibrates.
A
longitudinal
wave
is
L
L
=
4
transmitted
the
water
.
Depending
vibrations
to
the
down
The
on
are
the
two
the
tube
waves
length
possible.
frequency
of
of
If
the
and
is
reflected
superpose
the
the
air
and
column,
fundamental
tuning
when
fork,
it
produce
only
the
surface
stationary
certain
mode
resonance
hits
a
of
modes
vibration
of
wave.
of
corresponds
occurs.
first
At
resonance,
the
sound
heard
inside
the
tube
is
enhanced
(a
loud
harmonic
sound
(fundamental)
is
heard).
Resonance
of
the
air
to
the
different
surface
the
column
of
the
modes
tube.
to
will
occur
slowly
water
.
resonance
calculations
is
of
An
An
cater
at
specific
increased.
vibrations
antinode
occur
this
(Figure
of
when
lengths
possible.
will
end-correction
for
lengths
These
A
node
just
c
is
length
correspond
will
above
length
the
will
exist
the
at
open
the
end
sometimes
of
used
in
11.8.3).
3 λ
L
=
4
Example
A
a
small
short
loudspeaker
distance
allowed
to
increases
is
run
at
emitting
length
=
emitting
above
slowly
specific
sound
21 cm.
a
long
out
of
lengths
at
It
a
the
of
next
State
b
Calculate
the
wavelength
c
Calculate
the
speed
a
Resonance
phenomenon
is
air
occurs
of
taking
at
of
place
in
of
taking
the
of
frequency
containing
the
the
water
.
intensity
tube.
400 Hz,
length
When
the
=
of
the
the
effect
is
is
positioned
When
sound
water
is
heard
loudspeaker
first
noticed
at
64 cm.
third
harmonic
(first
overtone)
place.
the
sound
sound
in
constant
tube
tube,
frequency
a
the
sound
glass
the
waves
waves
in
in
the
the
air
air
column.
column.
tube.
5 λ
L
=
4
λ
b
When
resonance
first
occurs,
=
21,
∴
λ
=
=
64,
∴
λ
=
4
×
21
=
84 cm
=
85.3 cm
4
3λ
When
resonance
occurs
again,
4
×
4
84
c
Average
wavelength
+
64
3
85.3
=
=
84.7 cm
fifth
harmonic
2
(second
–2
Speed
of
sound
=
f
×
λ
=
400
×
84.7
×
overtone)
–1
10
=
339 m s
Figure 11.8.3
Modes of vibration in a
resonance tube
Reflection
Sound
waves
concert
Sound
halls
can
so
material
be
that
absorbing
Refraction
a
and
refraction of
reflected
the
when
medium
in
echoes.
amount
materials
occurs
as
are
the
of
to
be
to
of
a
waves
Acoustical
reflection
used
speed
order
sound
line
of
the
wave
engineers
sound
design
is
limited.
walls.
changes.
transmitted.
waves
The
Sound
speed
of
waves
require
sound
Key points
waves
ᔢ
in
air
depends
through
sunny
upper
on
layers
day,
the
layers
of
the
air
temperature
of
ground
are
cooler
different
heats
than
up
of
the
air
.
A
temperatures
the
the
lower
lower
sound
will
layers
layers.
of
be
wave
refracted.
the
When
Sound
waves
On
atmosphere.
someone
applications
a
The
ᔢ
Sound
waves
are
refracted
upwards
away
from
the
ground.
At
the
temperature
of
the
lower
layers
of
the
atmosphere
is
waves
the
upper
downwards
layers.
towards
When
the
someone
shouts,
sound
waves
are
produced
by
instruments.
Sound
waves
can
be
reflected
lower
and
than
many
industry.
night
ᔢ
time,
in
shouts,
various
sound
have
travelling
are
refracted.
refracted
ground.
95
11.9
Electromagnetic
Learning outcomes
On
completion
should
ᔢ
be
state
able
the
of
this
Properties of
section,
you
to:
properties
waves
An
electromagnetic
electromagnetic
at
wave
magnetic
field
right
graphical
representation
consists
angles
of
to
of
each
such
a
an
waves
oscillating
other
.
Figure
electric
11.9.1
field
and
shows
a
wave.
of
electromagnetic
waves
list
magnitude
direction
of
travel
of
electric field
ᔢ
of
the
the
orders
of
wavelengths
electromagnetic
of
electromagnetic
wave
the
spectrum.
magnetic field
Figure 11.9.1
Properties
An electromagnetic wave
of
electromagnetic
waves:
8
ᔢ
They
travel
ᔢ
They
consist
to
each
at
the
of
They
are
ᔢ
They
can
travel
ᔢ
They
can
be
oscillating
transverse
in
a
wavelengths.
waves
Radio
shortest
electromagnetic
Table 11.9.1
(3
electric
×
10
and
–1
m s
)
in
magnetic
a
vacuum.
fields
at
right
angles
vacuum.
diffracted,
electromagnetic
the
light
waves.
refracted,
Electromagnetic
have
of
other
.
ᔢ
The
speed
can
waves
and
polarised.
spectrum
be
arranged
have
wavelength.
spectr um
reflected
the
They
(T
able
based
largest
form
on
the
magnitude
wavelength,
part
of
what
while
is
of
their
gamma
referred
to
rays
as
the
11.9.1).
The electromagnetic spectrum
Electromagnetic
wave
Frequency/Hz
Wavelength/m
–1
Radio
waves
Lowest frequency
>10
–2
Microwaves
10
–4
to
10
–6
Infrared
~10
–7
Visible
7
×
–7
10
to
4
×
10
(ROYBGIV)
(Red
Violet)
–8
Ultraviolet
~10
Exam tip
–9
X-rays
It
is
expected
that
you
are
10
–11
to
10
able
–12
Gamma
to
recall
the
the
magnitude
wavelength
of
each
of
waves
listed
in
10
–16
to
10
the
visible
part
of
the
electromagnetic
spectrum,
red
light
has
the
the
longest
wavelength
table.
wavelength.
96
Highest frequency
the
In
electromagnetic
rays
of
(640 nm)
and
violet
light
(470 nm)
has
the
shortest
Chapter
Table 11.9.2
Type of
Radio
11
Waves
Sources of em waves and their uses
electromagnetic
wave
waves
Source
ᔢ
Microwaves
Infrared
Use
Transmitters
ᔢ
Klystron
ᔢ
Microwave
ᔢ
All
hot
tubes
ovens
objects
ᔢ
Telecommunications
ᔢ
Navigation
ᔢ
Telecommunications
ᔢ
Heating
ᔢ
Radar
ᔢ
Enables
systems
of food
pictures
to
be
taken
in
the
dark.
ᔢ
Used
in fibre
optic
cables
–
telecommunications
Visible
Ultraviolet
X-rays
ᔢ
Hot
ᔢ
The
objects
ᔢ
Incandescent
ᔢ
Mercury
ᔢ
Electric
ᔢ
X-ray
ᔢ
Remote
control for
ᔢ
To
be
ᔢ
To
kill
ᔢ
Produces Vitamin
ᔢ
Medical
ᔢ
Cancer
ᔢ
Used
able
and fluorescent
objects.
vapour
arcs
lamps
lamp
(sparks)
tubes
bacteria
in
imaging
ᔢ
Cosmic
ᔢ
Nuclear
radiation
ᔢ
Radioactive
power
plants
D
–
in
the
skin
Radiography
treatment
industry
welded
fractures
rays
see
Sun
of
Gamma
to
your TV
joints
in
to
check
and
the
detect
quality
‘hair
line’
metals.
ᔢ
To
determine
ᔢ
Cancer
ᔢ
Sterilisation
the
structure
of
crystals
treatment
of
surgical
instruments
elements
Key points
ᔢ
Electromagnetic
right
ᔢ
angles
The
waves
each
Electromagnetic
vacuum,
ᔢ
to
waves
transverse,
in
order
of
have
can
electromagnetic
arranged
consist
of
oscillating
electric
and
magnetic fields
at
other.
be
similar
properties
diffracted,
spectrum
increasing
can
consists
of
be
a
(e.g.
same
speed
in
a
polarised)
list
of
electromagnetic
waves
wavelengths.
97
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
5
8
be
Determine
that
accompanying CD.
can
the
be
highest
produced
order
when
of
a
diffracted
grating
beams
with
a
spacing
–6
of
1
Explain,
using
appropriate,
clearly
the
labelled
meaning
of
diagrams
where
the following
amplitude
phase
In
order
d
the
e
coherent
be
particle
in
difference
a
wave
a
Explain
when
between
two
waves
i
[2]
waves
what
is
applied
meant
to
i
wavelength
ii
frequency
iii
speed
that
3
a
by
the following
these
is
three
State
one
c
State
two
conditions
a
an
[1]
Describe
a
deduce
an
Using
wave.
between
[3]
transverse
and
12
a
the
of
Describe
and
example
of
each
similarities
describe
type
between
with
phenomenon
of
wave.
the
two
types
the
that
aid
of
applies
a
to
diagram,
one
experiment
of
principle
this
an
why
is
meant
by
plane
polarisation
of
Draw
waves
cannot
be
An
wave
how
a
the
[4]
ripple
to
demonstrate
tank.
[4]
waves
are
produced
in
the
[2]
diagram
on
letter C,
to
the
show
water.
where
Label
the
interference
Label
two
two
is
meant
the
passing
diagram
wide
of
is
by
the
term
diffraction.
width
of
affects
to
the
the
illustrate
gap
through
points
pattern
with
blue
what
light from
at
light
grating
is
D,
where
occurs.
[4]
diffraction
both
a
of
narrow
pattern
red
is
a
right
seen
at
a
and
produced on
Coherent
a
passes through
a
reason, the
screen
a double-slit
effect on the
spacing of the fringes observed when,
blue
light
is
changes
used
as
the
are
made:
red
source
instead
light.
[2]
b
The
distance
between
c
The
width
each
the
slits
is
decreased.
[2]
wave.
of
slit
in
the
double-slit
gap
monochromatic
angles
to
a
is
gradually
increased.
[3]
One
of
by
the
to
angles. Use
explain
principle
the
slits
is
covered
by
an
opaque
card.
[1]
light
diffraction
emerging from
certain
superposition
meant
is
light
[3]
projected
of
letter
[2]
Young’s
double
this
slit
experiment
can
be
used
to
the
measure
principle
interference
the
which
gap.)
Bright
diffraction
with
independently, the following
14
grating.
constructive
points
interference
coherent
appearance
d
the
wavelength
of
monochromatic
light.
the
a
Describe Young’s
b
State
double
slit
experiment.
[6]
effect. [2]
how
of
superposition.
the
[2]
c
7
a
arrangement. State with
plane-
[2]
how
is
beam
Explain
explain
tank.
arrangement
6
the
[5]
superposition,
experiment
interference
when
[1]
sound
what
Describe
source
of
coherent
of
A
demonstrate
experiment.
Explain
destructive
a
c
to
microwaves.
[3]
what
Explain
a
[3]
in
occurs.
transverse
polarised.
and
the
of
waves.
a
observe
[2]
only.
Explain
(Use
to
[2]
Explain
a
[4]
longitudinal
13
b
interference.
necessary
interference
the
a
by:
equation for
waves.
5
sources
[2]
produced
light
two
coherent.
light.
interference
c
b
be
[2]
b
a
must
ii
of
ripple
4
meant
interference
b
waves
waves from
waves:
definitions,
of
Distinguish
wave
of
waves
[1]
speed
State
the
terms
waves.
d
of
[3]
interference
what
results
the
light
400 nm.
coherence
State
b
Using
with
[2]
11
b
normally
[1]
10
2
illuminated
[1]
a
phase
is
observed,
Explain
of
m
[1]
may
c
10
wavelength
displacement
b
×
terms:
9
a
2.0
what
these
measurements
measurements
are
are
taken
used
and
to
explain
calculate
wavelength.
State
[4]
approximate values for:
When monochromatic light of wavelength λ is incident
i
the
width
ii
the
distance
of
each
between
slit
the
two
[1]
iii
the
distance
between
the
double
normally on a plane diffraction grating, the second
slits
[1]
order diffraction lines are formed at angles of 28° to the
slit
and
the
normal to the grating. The diffraction grating has 600
screen.
lines per millimetre. Calculate the value of λ.
98
[4]
[1]
Revision questions
d
Explain
the
roles
interference
in
played
the
by
diffraction
production
of
the
and
18
observed
fringes.
One
The
[4]
taut
of
15
A
laser
produces
monochromatic
red
light
end
by
the
620 nm
which
is
incident
at
right
a
diffraction
grating. The
diffraction
long
means
of
oscillator
string
over
a
is
is
attached
a frictionless
weight
of
adjusted
to
an
oscillator.
pulley
and
is
kept
3.50 N. The frequency
is
produced. Three
so
that
a
antinodes
stationary
are
present
on
angles
the
to
a
passes
of
wave
wavelength
of
string
5
wave. The
distance
between
the
two
adjacent
grating
antinodes
is
16.2 cm
and
the frequency
is
118 Hz.
5
has
5.5
×
10
lines
per
metre. A
screen
is
located
Calculate
1.50 m from
observed
the
at
the
diffraction
different
distance
points
between
second-order
grating.
the
diffracted
on
the
central
Bright
spots
screen. Calculate
bright
spot
and
19
the
light.
Distinguish
between
progressive
wave
a
the
by
amplitude
a
wave
reference
of vibration
of
and
a
to:
particles
in
the
particles
difference
in
each
three
they
between
the
energy
A
string
fixed
of
length
points.
It
is
80 cm
along
is
plucked
each
stretched
at
to form
a
its
stationary
of
the
sensations
to
string.
sound
and
state
wave.
and
wave. A
the
single
sound of
constant
containing water. When water
run
a
short distance
slowly out of the tube, the
heard
increases
=
at
specific
loudspeaker
400 Hz, the
19.7 cm.
It
next occurs
is
a
long
allowed
intensity of the
lengths of
is
effect
above
emitting
is first
at
air
in
sound
noticed
length
=
a
at
the
centre
of
the
two
Calculate
the
59.
1 cm.
air
the
wavelength
of
the
sound
waves
column.
[3]
Calculate
the
speed
of
the
sound
waves
in
the
is
meant
by
a
what
is
meant
by
an
iii
why
nodes
stationary
wave
a
State
are formed
at
the
c
State
the
wavelength
The frequency
Calculate
the
of
of
the
vibration
of
the
speed
of
the
wave.
of
the
b
[2]
ends
string
Explain
in
c.
what
is
mean t
by
wave.
the
sp eed
three features
regions
of
is
State
a
typical
i
red
[1]
ii
ultraviolet
[1]
iii
infrared
iv
X-rays.
340 Hz.
the
of
waves
which
are
electromagnetic
common
spectrum. [3]
wavelength
of:
light
[1]
radiation
[1]
radiation
[1]
[1]
[2]
22
d
all
[2]
antinode
string.
b
[3]
string.
to
what
air
antinode
21
ii
in
string
Explain:
i
at
at
[2]
between
centre
emitting
positioned
column.
is formed
[3]
[6]
glass tube
b
vibrates
length
is
a
17
unit
on.
loudspeaker
a frequency of
[3]
transferred
subjective
depend
the tube. When the
neighbouring
wave
small
length
c
per
frequency
sound
[2]
phase
A
to
each
wave
b
mass
[4]
stationary
making
State
what
20
16
the
are
calc ulat ed
[2]
Arrange
the following
wavelength
and
infrared
: X-rays,
waves.
in
increasing
microwaves,
magnitude
ultraviolet
of
waves
[4]
99
12
The
12.
1
The
physics
physics
Learning outcomes
On
completion
should
be
able
of
this
of
you
to:
The
explain
how
the
ear
responds
12.1.1
outer
end
incoming
sound
understand
the
the
terms
diagram
collects
the
waves
and
auditory
of
the
directs
canal
is
human
sound
the
ear
.
waves
tympanic
into
the
auditory
membrane
(ear
canal.
drum).
At
The
of
cause
three
the
tiny
tympanic
bones
membrane
called
the
to
vibrate.
hammer
,
the
The
anvil
middle
and
the
ear
stirrup.
significance
When
of
a
wave
consists
ᔢ
shows
ear
of
ear
to
sound
an
hearing
human
Figure
the
ᔢ
of
The
section,
hearing
sensitivity
the
tympanic
membrane
vibrates,
the
bones
in
the
middle
ear
and
vibrate
as
well.
The
three
bones
act
like
a
lever
system.
They
reduce
the
frequency response
amplitude
ᔢ
state
the
magnitude
threshold
of
of
time,
the
hearing
The
state
the
intensity
at
is
use
the
is
produced
pressure
connected
nor mal
on
on
to
the
the
oval
back
conditions,
tympanic
the
of
membrane.
window
the
is
throat
pressure
on
At
the
same
increased.
via
the
both
eustachian
sides
of
the
which
membrane
is
the
same.
When
someone
ascends
in
an
aircraft,
experienced
pressure
ᔢ
ear
Under
tympanic
discomfort
vibration
vibrational
middle
tube.
ᔢ
the
of
equation for
difference
exists
on
either
side
of
the
tympanic
membrane
intensity
resulting
in
the
ear
“popping”.
level
The
ᔢ
understand
the
terms
noise
and
loudness
inner
attached
that
run
signals
is
filled
auditory
its
stiffness.
vibrates
ear
to
entire
When
which
length.
the
causing
oval
the
are
with
nerve.
Inside
These
hairs
It
contains
the
vary
vibrates,
to
via
in
are
length,
liquid
The
the
cochlea,
there
the
resonate.
transmitted
the
cochlea
hairs
window
tiny
then
liquid.
the
produce
nerve
to
is
tiny
thickness
inside
hairs
auditory
which
many
hairs
and
inner
ear
electrical
the
brain,
anvil
where
they
are
then
interpreted.
stirrup
hammer
Frequency
auditory canal
response
and
intensity
auditory
nerve
The
tympanic
cochlea
membrane
eustachian
outer
human
This
tube
range
ear
is
the
upper
the
human
is
able
called
limit
of
to
the
detect
frequencies
frequency
20 kHz
in
the
range
of
the
ear
.
response
decreases.
In
the
frequency
20 Hz
As
a
range
to
20 kHz.
person
60 Hz
gets
to
older
,
1 kHz,
inner
ear
ear
can
detect
changes
of
2 Hz
to
3 Hz.
At
frequencies
above
ear
1 kHz,
it
is
difficult
for
the
human
ear
to
detect
small
changes
in
frequencies.
middle
ear
P
Intensity
Figure 12.1.1
is
the
sound
power
per
unit
area
(
I
=
A
at
)
a
stated
frequency.
The human ear
The
smallest
sound
intensity
that
can
be
detected
by
the
human
ear
is
–12
the
threshold
frequency
Figure
ear
.
of
The
The
shows
person.
human
hearing.
3 kHz.
12.1.2
certain
the
of
ear
.
The
variation
Above
the
of
of
curve
Intensities
sensitivity
threshold
threshold
the
of
the
of
below
the
is
of
1.0
with
cannot
the
be
ability
to
×
10
W m
at
a
frequency
.
hearing
intensities
curve
ear
is
varies
threshold
represents
human
hearing
hearing
called
–2
with
that
frequency
can
detected
detect
be
by
the
of
detected
the
a
by
human
smallest
ΔI
Equation
fractional
change
ΔI
of
intensity
I.
Sensitivity
depends
on
the
ratio
.
I
Sensitivity
Intensity
is
the
power
per
unit
increases
with
frequency
to
a
maximum
and
then
decreases
with
area
increasing
frequency
.
The
maximum
sensitivity
is
at
1–3 kHz.
P
I
=
at
a
stated frequency.
The
human
ear
can
detect
a
wide
range
of
intensities.
The
minimum
A
–12
intensity
that
can
be
detected
is
1.0
×
10
–2
W m
at
a
frequency
of
3 kHz.
–2
I
–
intensity/W m
–2
The
P
–
upper
limit
of
the
range
is
100 W m
in
the
frequency
range
1 kHz
6 kHz.
The
upper
limit
is
called
the
threshold
of
pain .
Persons
exposed
2
A
–
to
power/W
area/m
100
–2
intensities
of
100 W m
can
experience
pain
and
temporary
deafness.
to
Chapter
The
logarithmic
response of the
12
The
physics
of
hearing
sound
ear
intensity
intensity
level /dB
2
Equal
changes
in
intensity
are
not
perceived
as
equal
changes
in
loudness.
/W m
120
Loudness
level
the
is
may
the
be
subjective
used
following
as
a
response
measure
equation.
of
Changes
of
a
person
loudness.
in
to
a
given
Intensity
loudness
depend
intensity
.
level
on
is
the
1.0
Intensity
defined
fractional
2
100
10
80
10
60
10
40
10
20
10
0
10
using
4
6
change
8
ΔI
in
intensity
(
)
.
I
Loudness
is
a
logarithmic
response
to
intensity
.
10
12
Example
2
1
A
person
and
an
with
normal
intensity
level
hearing
at
the
is
ear
exposed
of
to
20 dB.
a
sound
Calculate
of
frequency
the
intensity
10
at
the
of
4
10
5
10
10
f /Hz
3 kHz
this
Figure 12.1.2
sound
3
10
ear
.
Variation of threshold of
hearing with frequency
I
Intensity
level
=
(
10 log
)
10
I
0
Equation
I
20
=
(
10 log
–12
10
1
×
10
)
I
Intensity
level
=
10 log
10
I
log
10
(
–12
1
×
10
)
=
(
)
I
2
0
Intensity
I
level
is
measured
in
dB
(decibels)
=
100
–12
1
×
10
I
–
intensity
of
sound
incident
on
–2
–10
I
=
1
×
ear/W m
–2
W m
10
I
–
threshold
of
hearing
0
–12
(1.0
Loudness
Intensity
level
threshold
of
Loudness
is
depends
T
o
a
on
define
a
standard.
being
the
as
is
the
defined
at
The
as
the
standard
then
of
the
source
2
W m
)
of
a
of
person
frequency
a
to
the
of
be
test
of
to
a
the
frequency
is
adjusted
evaluated.
found
loudness
intensity
sound
and
the
3 kHz.
source
being
is
the
of
loudness,
standard
source
the
of
response
and
unit
1 kHz
terms
frequency
individual
consistent
source,
in
a
subjective
the
10
noise
hearing
loud
1 kHz
test
and
×
given
of
when
source
is
1 kHz
until
Suppose
90 dB
intensity
and
sound.
it
the
it
is
is
chosen
perceived
intensity
is
as
loud
as
as
level
as
of
the
90 phons.
Key points
Example
ᔢ
The
intensity
of
sound
measured
at
a
distance
2
loudspeaker
is
measured
as
1.2
×
10
of
0.8 m
from
.
Given
that
the
sound
varies
ear
is
able
to
detect
in
the
range
20 Hz
to
intensity
20 kHz.
1
of
human
frequencies
–2
W m
The
a
as
,
where
d
is
the
distance
measured
from
the
2
d
ᔢ
loudspeaker
,
calculate
the
intensity
of
the
sound
at
a
distance
of
The
threshold
of
hearing
is
the
3.0 m.
minimum
detected
intensity
by
the
that
can
be
ear.
1
I
∝
2
ᔢ
d
ear
k
I
The
sensitivity
is
the
of
ability
the
to
human
detect
the
=
2
smallest fractional
d
intensity
change
ΔI
of
I.
k
2
1.2
×
10
=
2
(0.8)
k
=
ᔢ
Loudness
response
76.8
is
the
subjective
of
a
person
to
a
has
a
logarithmic
given
intensity.
Therefore
the
intensity
at
a
distance
of
3.0
is
given
by:
ᔢ
k
76.8
–2
I
=
=
ear
response
to
sound.
=8.53 W m
2
d
The
2
(3.0)
101
13
The
13.
1
Lenses
physics
Learning outcomes
On
completion
should
be
able
of
this
of
sight
Lenses
section,
you
A
converging
When
to:
a
refraction
ᔢ
differentiate
between
diverging
understand
length
and
the
terms
the
focal
f
principal axis
relate focal
lens
in
recall
length
to
power
of
ᔢ
draw
optical
the
The
diagrams
are
the
or
lens
pole
centre
of
beam
is
(Figure
P
of
the
optical
length
of
light
point
to
show
at
the
the
middle
surface
focused
of
through
than
a
a
at
the
edges.
converging
point
lens,
known
13.1.1).
The
point
P
is
known
as
as
the
the
the
lens
centre
of
the
lens.
is
of
The
known
the
horizontal
as
lens
P
the
and
line
drawn
principal
the
focal
axis .
point
through
The
F
is
distance
known
lens.
F
,
actually
an
image
pass
will
through
be
seen.
the
The
point
F
.
image
If
a
screen
produced
in
is
placed
this
case
is
produced
in
a
a
and
a
magnifying
real
diverging
or
concave
lens
is
thinner
at
the
centre
than
at
the
edges.
simple
a
parallel
beam
of
light
strikes
the
surface
of
a
diverging
rays
spread
out
as
leave
lens,
glass.
refraction
rays
occurs
appear
point
conver
image
how
When
camera
the
thicker
strikes
lens formula
A
images
of
the
focal
rays
the
called
ray
is
light
a
dioptres
the
F
centre
between
at
ᔢ
and
lens
of
lenses
as
ᔢ
beam
occurs
point
optical
ᔢ
convex
converging
focal
and
or
parallel
F
,
an
to
and
have
image
the
diverged
will
not
from
be
the
seen,
point
because
they
F
.
If
the
a
the
screen
rays
do
lens.
is
not
The
placed
at
actually
the
pass
ng
through
the
point
F
.
The
image
in
this
case
is
called
a
virtual
image .
The
(convex)
distance
between
the
optical
centre
the
diverging
of
the
lens
and
the
point
F
is
called
len
principal
axis
the
focal
The
point
power
of
of
a
lens
is
defined
lens.
Equation
by
parallel
F
P
the
beam
equation
here.
1
of
light
P
The
unit
dioptre.
for
Its
power
symbol
of
is
a
lens
is
=
the
f
D.
f
Figure 13.1.1
A converging (convex) lens
A
more
shorter
diverging
principal
powerful
focal
lens
length
will
have
(Figure
a
P
–
power
f
–
focal
of
a
lens/dioptres
length
of
lens/m
13.1.3).
axis
Converging
lenses
have
positive
values.
(concave)
Diverging
lenses
have
negative
values.
len
For
example:
parallel
ᔢ
A
converging
lens
of
focal
length
0.4 m
will
have
a
power
of
F
beam
1
P
of
li
ht
+
=
+2.5 D.
0.4
ᔢ
A
f
diverging
lens
of
focal
length
– 0.3 m
will
have
a
power
of
1
–
=
– 3.33 D.
0.3
Figure 13.1.2
A diverging (concave) lens
F
F
f
Figure 13.1.3
102
f
less
converging
less
powerful

lens
Relating power and focal length of a lens
more
converging
more
powerful

lens
Chapter
The
13
The
physics
object
between
T
o
sight
lens formula
u
An
of
is
placed
the
in
object
determine
the
front
and
(to
the
optical
position
of
left)
centre
the
of
of
image
a
converging
the
lens
several
is
lens.
the
rays
The
object
are
v
distance
distance
u.
constructed.
image
object
Rules
ᔢ
A
for
constructing
ray
of
light
point
of
the
ray
parallel
diagrams:
to
the
principal
axis
passes
through
the
focal
f
lens.
Figure 13.1.4
ᔢ
A
ray
Using
of
these
the
lens.
the
image
The
light
rules,
Note
lens
passing
and
an
that
the
through
image
the
optical
formula
can
image
is
centre
(shown
right)
the
be
optical
centre
constructed
inverted
of
the
can
and
lens
be
is
is
used
of
on
the
the
real.
the
for
lens
is
undeviated.
right-hand
The
image
side
distance
distance
converging
and
of
between
v
diverging
Equation
lenses.
A
sign
convention
is
used
(real
is
positive).
1
Quantity
Positive
sign
(+)
(Real)
Negative sign (–) (Virtual)
1
=
f
1
Object distance,
2
Image
3
Focal
distance,
length,
u
Object
v
f
is
in front
of
lens
Object
is
Image is at the back of lens
Image
Converging
Diverging
(convex)
ᔢ
The
focal
length
of
a
converging
ᔢ
The
focal
length
of
a
diverging
lens
lens
lens
is
is
is
at the
back of
in front
of
v
lens
lens
(concave)
1
+
u
f
–
focal
u
–
object
length
v
–
image
of
the
lens/m
distance/m
distance/m
lens
positive.
camera
negative.
object
ᔢ
Real
object
ᔢ
Virtual
and
real
image
distances
are
positive.
lens
object
and
virtual
image
distances
are
film
negative.
Example
An
object
20 cm.
is
placed
Calculate
12 cm
the
in
front
position
of
of
the
a
converging
image
and
lens
state
of
its
focal
length
nature.
Figure 13.1.5
u
=+12 cm
(real
object)
1
f
1
=
=
+20 cm
(converging
lens)
1
+
f
u
1
v
1
1
=
+
20
12
1
v
1
1
=
1
–
v
=
20
–
12
30
image
v
Since
the
image
=
object
– 30 cm
distance
is
negative,
it
is
virtual
and
is
in
front
of
the
Figure 13.1.6
converging
A
A
lens
simple
simple
move
(same
side
of
the
lens
as
the
and
is
Key points
shown
forth
from
in
Figure
the
13.1.5.
photograph
The
film.
lens
This
of
the
camera
allows
for
can
ᔢ
A
converging
objects
at
different
distances
to
be
focused
on
the
film.
The
parallel
is
real,
diminished
and
A
called
diverging
of
magnifying
an
upright
object
and
magnifying
is
placed
enlarged
glass
a
a
beam
single
the focal
point.
parallel
lens
rays
spreads
so
that
a
beam
they
glass
appear
When
through
inverted.
ᔢ
A
rays
image
point
produced
lens focuses
light
of
from
A magnifying glass
object).
camera
camera
back
A simple camera
between
image
is
the
focal
produced.
point
This
is
and
the
the
lens,
principle
a
by
virtual,
which
a
a
to
single
have
point
diverged from
called
the focal
point.
works.
103
13.2
The
eye
Learning outcomes
On
completion
should
be
able
of
this
The
section,
you
Figure
eye
to:
human
13.2.1
must
vitreous
ᔢ
explain
how
the
at
explain
different
and
the
humour)
terms
depth of field
the
various
defects
of
and
human
before
eye.
media
reaching
The
rays
(cornea,
the
retina.
of
light
aqueous
Each
entering
humour
,
medium
has
the
lens,
methods
Light
first
strikes
the
air-cornea
boundary.
a
different
Most
of
used
to
occurs
=
at
indices
1.38).
the
this
The
pupil
boundary
(refractive
light
then
towards
the
because
index
of
travels
lens.
of
the
air
=
1.0,
through
The
large
main
difference
the
refractive
the
aqueous
function
of
between
index
of
humour
,
the
lens
is
then
to
the
fine
eye
index.
refractive
cornea
depth of focus
describe
the
several
distances
through
ᔢ
through
eye forms
bending
ᔢ
illustrates
pass
refractive
images
eye
tune
the
focusing
of
light
so
that
an
image
is
formed
on
the
retina.
correct
The
retina
consists
of
nerve
endings
that
generate
electrical
impulses
that
them.
are
sent
The
of
to
lens
the
is
muscles
lens
is
brain
via
suspended
called
long
and
the
the
by
optic
ligaments
ciliary
thin.
nerve.
which
muscles.
When
the
are
When
muscles
attached
the
to
muscles
contract,
the
a
circular
are
lens
ring
relaxed,
becomes
the
short
retina
and
fat
(more
powerful
or
shorter
focal
length).
cornea
Accommodation
lens
so
as
to
The
closest
is
focus
the
ability
images
of
the
formed
eye
from
to
change
objects
at
the
focal
different
length
of
the
distances.
lens
pupil
image
vitreous
aqueous
The
on
far
point
the
to
retina
point
of
the
is
the
eye
at
25 cm
normal
which
for
a
eye
the
eye
normal
is
taken
can
eye
to
still
and
be
at
is
produce
called
a
the
focused
near
point.
infinity.
humour
humour
Depth of field
optic
O
O
O
Figure 13.2.1
and depth of focus
nerve
2
1
The human eye
I
I
I
1
depth
2
of field
depth
Figure 13.2.2
Consider
of
field
Depth of field and depth of focus
an
of
of focus
object
the
eye
O
is
being
the
viewed
distance
at
some
moved
by
distance
the
from
object
eye.
The
(between
O
depth
and
1
O
),
while
the
image
remains
in
focus
(Figure
13.2.2).
2
For
a
given
slightly
parallel
accommodation,
nearer
(O
)
and
the
slightly
eye
is
able
further
(O
2
light
variation
rays
the
in
depth
The
depth
see
from
clearly
some
an
object
fixed
point.
This
1
distance
of
to
)
through
which
the
eye
can
still
see
clearly
is
called
focus.
of
field
and
the
depth
of
focus
are
affected
by
the
size
of
the
iris.
image
Figure 13.2.3
Short-sightedness
Short-sightedness
parallel
The
light
are
rays
occurs
(less
diverging
lens
image
person
blurred.
is
when
sightedness
104
the
powerful).
diverging
lens
sightedness,
Figure 13.2.4
able
The
to
(myopia)
only
image
lens
focus
forms
is
not
in
able
objects
front
to
Short-sightedness
is
the
used
far
to
correct
point
is
close
of
relax
also
the
in
to
the
retina.
order
occurs
if
short-sightedness.
closer
than
infinity
closer
than
25 cm
(Figures
13.2.3
and
13.2.4).
Distant
become
eyeball
In
and
Correction for short-
be
to
the
eye.
objects
Short-sightedness
the
the
long
is
too
case
near
of
and
thin
long.
A
short-
point
may
Chapter
Long-sightedness
13
The
physics
of
sight
(hypermetropia)
near
The
person
close
to
is
the
eye
sightedness
is
not
also
able
long
sight.
near
point
to
are
the
In
case
greater
is
of
objects
The
the
short
eyeball
the
focus
because
become
if
is
only
blurred.
occurs
to
occurs
able
image
ciliary
and
too
fat
that
forms
A
25 cm
(Figures
from
powerful).
the
far
13.2.5
the
the
become
converging
long-sightedness,
than
far
behind
muscles
(more
short.
are
eye.
retina.
weak
Objects
Long-
and
the
250 mm
is
used
to
point
is
infinity
image
lens
Long-sightedness
lens
and
point
Figure 13.2.5
Long-sightedness
correct
and
the
13.2.6).
near
point
Astigmatism
image
The
person
has
difficulty
focusing
light
rays
from
objects
in
different
Figure 13.2.6
planes
at
the
same
time.
The
problem
is
caused
because
the
surface
Correction for long-
of
sightedness
the
cornea
adjusted
is
uneven.
such
cornea-lens
that
Astigmatism
its
system
axis
is
is
is
corrected
perpendicular
cylindrical
(Figure
to
using
the
a
axis
cylindrical
in
which
lens,
the
eye
13.2.7).
Cataracts
Persons
light
with
enters
removed.
cataract
the
eye.
Surgeons
converging
lenses
have
In
can
are
lenses
order
to
implant
used
to
which
correct
a
new
correct
have
this
lens
the
become
defect,
inside
opaque.
the
the
lens
eyes
of
or
V
ery
the
little
eye
glasses
is
with
defect.
Figure 13.2.7
Example
Correction lens for
astigmatism
A
person
retina
has
a
distance
near
is
point
1.7 cm.
of
30 cm
and
Calculate
a
the
far
point
power
of
90 cm.
a
the
eye
b
the
corrective
lens
needed
to
view
an
object
at
a
c
the
corrective
lens
needed
to
view
an
object
at
infinity.
This
The
when
person
person
is
is
p
also
1
=
=
f
b
Long
p
object
+
correction
1
f
Power
u
of
+
1
+
1
lens
near
the
point
point
far
of
30 cm.
distance
is
point
of
greater
is
less
25 cm.
than
than
25 cm.
infinity.
=
62.2 D
=
62.8 D
1
+
0.25
corrective
near
0.017
=
v
the
because
0.30
sight
the
cornea–
1
=
v
at
because
1
1
=
viewed
long-sighted
1
u
=
is
short-sighted
1
a
an
The
of:
Key points
0.017
=
62.8
–
62.2
=
+0.6 D
(Converging
lens)
ᔢ
Alternatively,
since
the
image
at
30 cm
is
on
the
same
side
of
the
Accommodation
of
as
the
object,
it
is
virtual.
Therefore,
the
image
distance
is
the
eye
to focus
p
=
1
=
f
1
+
u
1
Short
sight
=
v
–
0.25
=
0.30
Depth
of field
by
an
is
the
distance
object,
=58.8 D
while
the
image
remains
in focus.
0.017
1
=
1
=
f
Power
different
1
=
f
p
at
correction
1
=
objects
distances.
0.67 D
moved
p
ability
1
ᔢ
c
the
images
negative.
formed from
1
is
lens
1
+
u
of
1
1
=
v
corrective
ᔢ
+
0.90
lens
=
=
58.8
–
59.9
Common
defects
of
the
eye
59.9 D
0.017
include
=
short-sightedness,
sightedness,
–1.1 D
astigmatism
long-
and
cataracts.
Alternatively
,
the
image
distance
is
negative,
because
the
image
is
virtual.
ᔢ
1
p
=
1
=
1
+
1
=
There
=
u
v
∞
various
techniques
–1.1 D
used
f
are
1
–
to
correct
these
defects.
0.90
105
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
6
6
accompanying CD.
A
lens
20 cm
a
1
a
Explain
what
is
meant
by
loudness
and
b
c
what
state
Describe
is
meant
by
the
threshold
of
how
the
[3]
sensitivity
across
the
of
the
range
of
ear
Explain
of
the
the focal
length
of
is
lens
to
is
meant
sound
by
the
waves
of
it
+2.5 D. An
lens.
of
the
object
is
placed
Determine:
lens
and
state
what
type
[2]
average
position
A
diverging
is
optical
‘frequency
an
the
object
to
nature
[2]
what
response’
7
audible
frequencies.
a
in front
of
and
nature
of
the
image
has
a focal
length
produced. [3]
hearing
its value.
loudness varies
2
power
[4]
Explain
and
a
intensity
level.
b
has
8
person.
State
lens
placed
centre
of
the
three
simple
at
of
a
distance
the
image
of
of
20 cm. An
30 cm from
lens. Calculate
the
the
position
and
produced.
properties
of
the
[3]
image
produced
by
a
camera.
[3]
[3]
9
b
Describe
and
explain
how
this frequency
Draw
glass
varies
with
age.
A
person
listening
has
to
produces
an
eardrum
music
0.
14 μW
ray
diagram
to
explain
how
a
magnifying
works.
[3]
[3]
10
2
3
a
response
using
of
of
area
53 mm
headphones,
sound
power
to
Explain
what
is
meant
by:
. When
the
the
a
accommodation
b
depth
of
the
eye
[2]
headphone
eardrum.
of focus.
[2]
Calculate:
11
a
the
intensity
of
the
sound
incident
on
The
range
3.8 m.
eardrum
In
the
intensity
level
at
the
eardrum.
a
Describe
how
the
ear
responds
to
an
wave.
Sketch
a
The
to
show
the
variation
of
power
of
the
intensity
of
sound
is
the
threshold
old
boy.
Indicate,
of
no r ma l
using
range
Indicate,
at
the
ear
h earing
in
a
State
two
of
values
on
the
x-axis,
State
the
a
the
aid
of
a
the following
i
focal
ii
principal
iii
focal
State
two
pairs
of
spectacles. Calculate
lens
in
each
pair
of
spectacle.
[3]
the
lens
required
to
correct
an
eye
type
of
lens
calculate
being
used
the focal
to
correct
length
in
the
cm.
[2]
b
State
the
c
Calculate
eye
defect
being
corrected.
[1]
a value
on
in
the
y-axis,
the
13
[1]
which
the
70-year-old
diagram,
graph
may
man.
near
point
and far
point
of
the
[3]
explain
a
Explain
what
b
Draw
diagram
be
a
is
meant
to
by
show
short-sightedness.
how
this
eye
defect
[2]
is
corrected.
[3]
[2]
what
is
a
Explain
b
State
what
is
meant
by
astigmatism.
[2]
meant
the
type
of
lens
used
to
correct
this
eye
terms:
axis
point.
the
the
eye.
[2]
length
[3]
lens formula
expression for
106
the
and
defect.
b
to
for
14
by
1.
1 m
the
hearing.
ways
different for
With
is
optometrist
15-yea r-
of frequencies.
using
threshold
a
an
[3]
audible
5
with
of
unaided
iii
gentleman
his vision,
+1.5 D.
defect
ii
elderly
with
a
i
an
[4]
graph
frequency
him
power
defect
b
of
correct
incoming
12
sound
to
[3]
the
4
order
[2]
provides
b
of vision
the
the
and
power
write
of
a
down
lens.
an
[3]
[1]
Revision questions
15
a
Explain
what
is
meant
by
the following
b
terms:
An
are
i
accommodation
elderly
person
situated
ii
depth
of field
[2]
iii
depth
of focus.
[2]
him
A
student
clearly
complains
unless
they
see
clearly
1.2 m
and
objects
9 m from
which
the
[2]
eyes. Calculate
b
can
between
6
that
are
she
more
cannot
than
see
objects
60 cm from
her
to
see
the
objects
i
located
ii
very
at
the
power
which
near
distant from
of
lens
required for
are:
point
the
the
of
the
eye
[3]
eye.
[3]
eyes.
17
i
State
the
student’s
eye
defect.
a
Explain
sound
ii
State
the
iii
Calculate
near
point
of
a
normal
power
of
the
person.
lens
the
her
defect.
used
Draw
ray
diagram
a
A
conversation
[3]
to
show
how
a
point
object
placed
two
at
point
enters
the
rays
Draw
another
the
to
taking
place
between
×
10
intensity
of
the
W m
Explain
lens fixes
her
is
meant
by
the
threshold
a
Explain
the
eye
what
and
is
meant
state
how
by
show
how
and
Calculate
state
the
is value.
intensity
level
[2]
of
the
the
problem.
the following
they
of
[3]
are
[3]
[3]
c
16
is
.
what
conversation.
correcting
two
sound
normal
eye.
diagram
incoming
of
ii
v
an
–2
hearing
near
is
average
–6
i
light from
to
to
6.2
iv
responds
[4]
students. The
correct
ear
wave.
[1]
b
the
how
[1]
defects
of
Explain
why
response
to
the
ear
is
intensity.
said
to
have
a
logarithmic
[3]
corrected.
i
astigmatism
[3]
ii
cataract.
[2]
107
Module
Answers
to
the
selected
structured
2
Practice
multiple-choice
questions
questions
can
and
be found
exam
to
on
questions
7
the
Light of wavelength 640 nm falls on a pair of slits,
forming fringes 2.80 mm apart on the screen. What is
accompanying CD.
the new fringe spacing if the wavelength were 400 nm?
a
2.28 mm
b
1.75 mm
c
4.48 mm
d
2.80 mm
Multiple-choice questions
1
A
mass
m
is
attached to
a vertical
helical
spring
having
8
a
spring
constant
k. The
mass
is displaced
slightly
A
ray
of
light
happens
undergoes
simple
harmonic
motion with
amplitude
maximum velocity
v of the
mass
is
given
m
a
v
=
b
v
=
=
√
A
particle
d
The
by
P
=
the frequency,
speed
and
wavelength
Frequency
speed
wavelength
a
increases
increases
decreases
b
decreases
increases
increases
c
unchanged
increases
increases
d
unchanged
increases
decreases
m
v
=
a
is
undergoing
x
in
simple
mm
of
the
0.58cos(0.
12π)t. What
is
harmonic
particle
motion.
is
given
the frequency
9
of
What
one
is
is
the
ratio
6.0 dB
of
louder
the
intensity
than
the
of
two
sounds
if
other?
oscillation?
6
a
a
0.06 Hz
b
0.58 Hz
c
0.
12 Hz
d
A
sound
wave
of
amplitude
0.25 mm
has
0.40
b
4.0
c
60
d
10
0.22 Hz
10
3
of
by:
m
displacement
x
to
light?
k
a
k
2
air. What
a
m
v
into
k
a
k
c
water
a.
the
The
travels from
and
A
person
has
a far
point
which
is
9.5 cm from
his
and
eyes. What
is
the
power
of
the
corrective
lens
needed
–2
intensity
sound
of
3.5 W m
wave
amplitude
of
of
the
. What
is
the
intensity
same frequency
which
0.43 W m
c
7
.0 W m
has
a
stationary
to view
a
–1.
1 D
an
object
at
infinity?
an
14 W m
d
1.75 W m
b
+1.
1 D
c
–2.9 D
d
+2.9 D
Structured questions
–2
11
A
him
–2
b
–2
4
for
0.50 mm?
–2
a
of
wave
is
produced
when
two
a
Explain
what
is
meant
by
simple
harmonic
similar
motion.
progressive
waves
of frequency
400 Hz
b
in
opposite
directions
[2]
travelling
superimpose. The
Show
that
two
adjacent
nodes
is
period
of
oscillation
T
of
a
mass
m
distance
attached
between
the
1.2 m. The
speed
to
a
spring,
having
a
spring
constant
k
is
of
m
the
progressive
waves
is:
given
–1
a
480 m s
c
960 m s
by
T
=
2π
b
900 m s
d
167 m s
c
–1
.
√
–1
A
small
particle
P
[6]
k
undergoes
simple
harmonic
–1
motion. The displacement
x of
P
in
metres
is
given
–3
5
A
diffraction
grating
Monochromatic
right
the
is
angles
second
to
has
light
the
order
of
N
lines
per
wavelength
grating. The
maxima
is
by
metre.
λ
is
angular
θ. Which
incident
deviation
of
at
of
the following
x
=
1.4
×
(5π)
t., where
What
ii
Determine
the
iii
Sketch
the
velocity–time
the
acceleration
of
iv
the
sin
i
true?
is
10
angular frequency
period
of
t
is
of
in
seconds.
P
?
[2]
oscillation.
graph for
[2]
the
motion
P.
[3]
Sketch
time
graph for
the
λ
a
sin θ
=
2Nλ
b
sin θ
=
2
motion
of
P.
[3]
N
v
λ
c
sin θ
=
Ultraviolet
d
ultraviolet
rays
sin θ
=
have
a
differ from
microwaves,
in
that
cannot
c
have
a
108
have
a
1
the
displacement
2
the
velocity
3
the
acceleration
State
be
of
P
[2]
of
P
[2]
of
P.
[2]
a
the
conditions
stationary
necessary for
the formation
wave.
[2]
polarised
higher
Explain
what
is
meant
by
a
node
and
an
antinode
wavelength
on
d
determine:
lower frequency
b
a
1.0 s
rays:
of
b
=
2Nλ
12
a
t
1
N
2
6
When
higher frequency.
a
stationary
wave.
[2]
Module
c
A
stationary
wave
string. There
between
is formed
are four
the four
nodes
nodes
is
on
a
stretched
present. The
1.40 m. The
b
distance
speed
In
the
experiment
2
Practice
you
exam
described,
questions
state
a
typical
value for:
of
i
the
width
ii
the
distance
of
the
between
slits.
the
slits
[1]
iii
the
distance
between
the
slits
–1
the
i
waves
State
on
the
the
string
is
wavelength
220 m s
of
the
.
waves
on
the
string.
ii
A
Calculate
laser
the
the frequency
produces
light
the first
is
light
incident
order
the
of
vibration.
[2]
Explain
the
of
wavelength
role
normally
maximum
is
on
a
diffraction
produced
at
grating,
17
a
14°.
State
between
the
lines
on
the
number
of
positions
of
the
grating
b
[3]
maximum
the
a
parallel
normally
the
on
grating
directions.
and
b
has
on
the
this
order
a
is
third
of
maxima
Explain
wavelength
of
beam
red
diffraction
consisting
normally
51°,
a
interference
Light
observable fringes.
[4]
conditions
necessary
patterns
in
to
obtain
the Young’s
double
an
[3]
expression
the
to
show
wavelength
λ
of
the
relationship
the
light
source,
the
separation
a,
the
distance
between
the
slits
[3]
and
When
and
experiment.
Derive
slit
a
diffraction
light
intensity.
14
by
producing
observable fringe
slit
spacing
in
played
640 nm. When
between
b
screen.
[1]
c
Calculate:
a
the
[2]
interference
13
[1]
and
the
in
of
of
incident
light
in
by
these
is
coincides
produced
produced
with
by
the
light
of
[4]
by
of
d
light
and
the
spacing
between
the
[6]
Young’s fringes
separation
of
screen
the
light
were formed
and
is
light
and
0.50 mm. The
slits
is
using
two
slits
with
distance
1.2 m. The
12 mm. Calculate
a
between
distance
the
the
between
wavelength
used.
of
[2]
Describe the
effect,
if
any, on the
separation and
maximum brightness of the fringes when the
second
the
D
x.
ten fringes
shorter
angle
screen
monochromatic
incident
diffraction
maximum
c
particular
diffraction
grating. The
a
fringes
leaving
maxima.
wavelengths
470 nm. At
maximum
played
producing
diffraction
wavelength
is
the
intensity
role
two
order
light
grating,
the
following
changes
are
made
in the
experiment
in
c.
other
i
The
distance
between
the
slit
and
the
screen
wavelength. Calculate
is
i
the spacing between the lines on the grating
[2]
ii
the number of lines per metre on the grating
[1]
iii
the
other
iv
the
highest
wavelength
observed
order
with
present
maximum
470 nm
in
the
that
light
can
the
ii
[2]
The
spectrum
of
be
[2]
waves
is
a
number
of
The
State
electromagnetic
of
waves
common
to
c
State
waves.
the following
light, X-rays,
a
typical
red
ii
X-rays
in
increasing
magnitude
of
microwaves,
infrared
light
18
a
Explain
b
Describe
diagrams
wavelength for
light
microwaves.
wavelength
a
an
wavelength
red
and
of
monochromatic
calculate
double
experiment
of
a
light. State
explain
and
the
and
the
screen
of
the
light
incident
on
the
is
the
increased,
slits
and
keeping
the
the
screen
distance
and
the
light
constant.
[3]
an
meant
by
experiment
of
to
is
water
the
to
waves
illustrate
term
diffraction.
demonstrate
in
both
a
ripple
narrow
the
tank.
gap
[2]
Draw
and
wide
[6]
Explain
how
diffraction
is
affected
by
the
width
gap.
of
the
lamp
what
how
[2]
[1]
using Young’s
Describe
slits
diffraction.
light
can
Explain
to
slit
measure
red
of
the
sound
waves
diffraction
of
is
more
light.
[3]
the
are
measurements
of
than
monochromatic
measurements
wavelength
diffraction
observed
experiment.
producing
these
why
be
easily
measured
is
[1]
d
The
the
source
separation
[1]
the
16
light
slit
[2]
slit
what
diffraction
[4]
c
iii
between
the
the
[3]
gap
i
of
[2]
all
wavelength:
visible
and
regions.
three features
Arrange
separation
divided
source
b
slit
constant.
keeping
intensity
between
a
source
the
constant.
double
into
keeping
wavelength
distance
wavelength.
electromagnetic
The
light
increased,
iii
15
increased,
light.
taken
are
used
to
[5]
109
14
Temperature
14.
1
Temperature
Learning outcomes
On
completion
should
be
able
of
this
understand
section,
you
T
emperature
two
to:
the
temperature
scales
Temperature
bodies
other
,
ᔢ
and
concept
is
A
the
and
thermal
measure
B,
of
energy
of
the
different
flows
degree
of
hotness
temperatures,
between
them.
If
are
the
in
of
a
body.
contact
When
with
temperature
of
each
body
A
is
of
higher
than
the
temperature
of
body
B,
thermal
energy
flows
from
A
to
B
temperature
(Figure
ᔢ
define
a
thermometric
property
are
at
14.1.1).
the
ther mal
ᔢ
state
examples
of
same
Thermal
energy
temperature.
equilibrium
with
At
each
will
continue
this
stage,
other
.
to
both
There
is
flow
until
bodies
zero
are
net
both
said
flow
of
bodies
to
be
in
thermal
thermometric
energy
between
the
bodies.
However
,
thermal
energy
is
continuously
properties
moving
ᔢ
explain
is
ᔢ
how
a
temperature
back
and
forth
between
the
what
flow
is
meant
thermodynamic
two
bodies.
scale
defined
explain
the
of
thermal
energy
by
scale
of
A
B
temperature.
T
T
1
2
T
>T
1
Figure 14.1.1
Thermal energy flows as a result of a temperature difference
Defining
In
order
to
property
property
a temperature
define
and
is
volume
The
length
temperature.
mercury
capillary
ᔢ
mercury
it
the
a
scale,
to
be
in
a
capillary
the
volume
therefore
of
of
A
varies
of
a
ther mometric
thermometric
continuously
thermometric
with
properties.
tube
capillary
temperature
its
concepts
that
examples
the
up
the
understood.
material
are
in
moves
decreases,
tube
of
mercury
As
and
need
following
of
of
scale
temperature
points
property
The
The
increases
a
fixed
some
temperature.
ᔢ
1
tube
the
capillary
will
vary
mercury
tube.
decreases.
If
The
with
increases,
the
its
volume
temperature
length
of
mercury
of
the
in
the
decreases.
The
electrical
resistance
of
a
coil
of
The
electrical
resistance
of
a
metal
wire
increases
with
increasing
temperature.
ᔢ
The
pressure
of
gas
contained
in
a
fixed
container
(volume
kept
constant)
If
some
varies
air
increases.
ᔢ
The
is
with
As
e.m.f.
placed
inside
temperature.
the
a
As
container
,
the
temperature
generated
when
the
pressure
temperature
decreases,
two
the
dissimilar
inside
increases,
pressure
pieces
of
the
the
container
pressure
decreases.
metals
are
connected
When
two
generated,
which
110
the
metals
which
two
such
is
as
copper
dependent
metals
are
on
and
the
connected.
iron
are
connected,
temperature
of
the
an
e.m.f.
junction
is
at
Chapter
A
thermometric
property
14
Temperature
should:
Definition
ᔢ
vary
smoothly
ᔢ
be
ᔢ
respond
over
the
range
of
temperatures
changes
in
temperature
being
measured
A
sensitive
to
small
thermometric
property
rapidly
to
changing
of
a
Once
a
thermometric
A
fixed
reproduced.
points
ᔢ
as
The
ice
For
property
is
a
example,
is
selected,
standard
the
that
a
degree
Celsius
fixed
of
scale
points
hotness
is
defined
need
that
by
to
can
varies
temperature.
be
easily
using
with
two
be
fixed
follows.
lower
at
point
material
is
temperatures.
continuously
defined.
property
fixed
standard
point
(ice
point)
atmospheric
is
the
pressure.
temperature
This
of
pure
temperature
is
melting
defined
as
0 °C.
ᔢ
The
upper
above
pure
fixed
temperature
The
chosen
point
boiling
is
(steam
water
defined
as
thermometric
conditions
stated
on
as
above.
at
point)
the
temperature
atmospheric
two
X,
is
points
measured
(0,
X
)
under
and
(100,
0
a
graph
linear
fixed
shown
relationship
points.
intervals.
The
In
relationship
and
Figure
to
14.1.2.
therefore
range
order
is
in
is
of
steam
pressure.
This
100 °C.
property
The
is
standard
a
divided
determine
straight
up
an
The
into
line
unknown
X
two
)
are
plotted
100
Celsius
one
the
is
scale
drawn
hundred
assumes
between
equally
temperature
θ,
a
the
two
spaced
the
following
used.
Equation
X
–
X
θ
θ
0
=
×
X
–
100 °C
X
100
0
θ
–
unknown
X
–
value
–
value
of
temperature
thermometric
θ
property
X
of
at
temperature
θ
thermometric
100
property
X
–
value
of
at
100 °C
thermometric
0
property
at
0 °C
thermometric
property
X
100
X
θ
X
0
0
θ
100
temperature/°C
Figure 14.1.2
Defining a temperature scale
111
Chapter
14
Temperature
Thermometers
thermometer
uses
thermometers
ice
at
are
standard
reading.
are
This
used
a
would
they
fixed
may
all
thermometric
are
give
used
to
measure
different
relationship
temperature
may
not
order
to
avoid
called
the
of
ther modynamic
temperatures
are
thermodynamic
ᔢ
Absolute
zero
ᔢ
The
triple
which
The
scale
water
the
the
or
the
melting
the
well.
reason
same
same
However
,
for
way
thermometric
of
a
this
if
glass
is
that
between
property
the
and
thermometers.
defined
particular
the
different
pure
give
as
of
temperature
The
vary
the
was
any
kelvin
which
of
is
the
that
(K).
The
is
totally
substance.
absolute
lowest
energy
water
and
thermodynamic
of
in
internal
point
ice,
all
point
If
of
would
type
scale.
fixed
This
scale
this
scale,
On
points
used
in
is
the
are:
(0 K),
the
for
scale
measured
scale
temperature,
a
properties
not
between
linear
confusion,
the
steam
readings.
may
property.
them
the
The
independent
of
at
points.
In
all
Each
temperature
true
properties
be
the
point),
be
temperatures.
thermometric
measure
(ice
also
of
the
to
pressure
thermometers
water
,
measure
particular
used
different
all
to
of
(273.16 K).
water
vapour
temperature
T
This
are
is
temperature
substances
in
is
is
the
at
At
this
minimum.
temperature
thermal
given
possible.
a
at
equilibrium.
by:
Equation
P
T
T
=
×
273.
16 K
p
tr
T
–
thermodynamic temperature/K
P
–
pressure
of
an
ideal
gas
at
a
T
temperature
p
–
pressure
of
T/Pa
the
same
volume
of
tr
an
of
ideal
Although
pressure
The
SI
gas
at
the
triple
point
water/Pa
the
scale
variation
unit
of
is
of
theoretical,
an
ideal
temperature
is
gas
the
it
at
is
identical
constant
kelvin
(K).
to
the
scale
based
on
the
volume.
An
interval
of
1
kelvin
is
1
defined
as
measured
Another
being
on
the
⁄273.16
the
temperature
thermodynamic
commonly
following
of
used
unit,
scale
the
of
of
degree
the
triple
point
of
water
temperature.
Celsius
( °C)
is
defined
by
equation.
Equation
θ
θ
=
–
T – 273.
15
temperature
in
degrees
in
kelvin/K
Celsius/ °C
T
It
–
temperature
follows
that
temperature
112
a
temperature
change
of
1 °C.
change
of
as
1 K
is
exactly
equal
to
a
the
Chapter
14
Temperature
Example
A
resistance
resistance
thermometer
is
found
thermometer
resistance
used
i
to
is
is
to
placed
found
measure
Calculate
be
to
the
the
is
placed
3825 Ω.
in
be
a
liquid
976 Ω.
unknown
value
of
the
in
At
of
A
pure
water
100 °C,
its
unknown
at
unknown
and
is
temperature
constant-volume
temperature
0 °C
resistance
and
it
gas
is
and
The
the
thermometer
found
temperature
its
185 Ω.
using
to
be
the
is
65 °C.
resistance
thermometer
.
ii
State
iii
Suggest
your
a
answer
reason
resistance
R
and
–
θ
for
kelvin.
the
difference
constant-volume
between
gas
the
readings
on
the
thermometer
.
R
θ
i
in
0
=
×
R
–
100 °C
R
100
0
976
–
3825
185
–
3825
=
×
100
– 2849
=
×
100
– 3640
θ
ii
=
In
θ
T
78.3 °C
order
=
T
–
θ
=
78.3
The
convert
+
degrees
Celsius
to
kelvin
we
use:
273.15
+
273.15
351.45 K
difference
properties
in
reading
respond
temperature
the
property
assumption
is
due
differently
between
thermometric
This
from
273.15
=
=
iii
to
may
and
fixed
varies
not
be
to
the
fact
that
uniquely
points.
linearly
true
for
The
to
different
the
changes
assumption
between
different
thermometric
the
two
is
in
that
fixed
the
points.
thermometric
properties.
Key points
ᔢ
Temperature
is
ᔢ
A
thermometric
property
ᔢ
A
thermometric
property
Thermometric
change
gas
ᔢ
A
and
in
are
ᔢ
There
are
the
The
triple
a
is
scale
can
the
one
can
hotness
that
be
used
include
wire,
be
defining fixed
–
the
to
the
two
defined
by
points
on
the
temperature.
volume
pressure
of
a
liquid,
a fixed
metals
a
of
are
suitable
the
volume
of
connected.
thermometric
scale
(ice
thermodynamic
point
scale
and
steam
(absolute
point).
zero
and
water).
scale
is
called
thermometric
one
in
in
choosing
two fixed
of
temperature.
points.
the Celsius
any
with
dissimilar
on
of
body.
measure
change
when
a
change
points
point
of
varies
two fixed
on
change
of
of
generated
thermodynamic
depend
A
e.m.f.
and
There
ᔢ
properties
temperature
ᔢ
measure
resistance
the
property
ᔢ
a
an
absolute
property
degree Celsius
is
and
exactly
is
scale
because
it
does
not
theoretical.
equal
to
a
change
of
one
kelvin.
113
14.2
Thermometers
Learning outcomes
On
completion
should
ᔢ
be
able
describe
of
a
a
of
this
Types of thermometers
section,
you
Liquid-in-glass thermometer
to:
the
principal features
liquid-in-glass,
thermistor,
a
a
resistance,
thermocouple
narrow
and
a
constant-volume
bore
gas
thermometer
ᔢ
state
the
advantages
disadvantages
of
and
various
types
of
thermometers
ᔢ
select an appropriate thermometer
mercury
to measure temperature.
large
bulb
(thin-walled
Figure 14.2.1
In
a
liquid
mercury.
it
to
rise
A liquid-in-glass thermometer
glass-in
When
up
transparent
it
Mercury
while
is
it
is
a
up
freezes
measure
very
respond
or
at
low
large
quickly
and
is
will
be
a
234 K,
which
rapidly
affect
lower
easy
that
than
it
is
good
means
the
the
it
that
it
This
Mercury
stick
of
is
the
to
using
it
is
used
glass.
the
glass
mercury
be
used
to
thermometers
that
It
they
also
order
to
have
cannot
means
measured.
In
These
because
heat.
means
value.
with
causes
through
cannot
being
used
not
advantage
temperatures.
actual
is
seen
does
of
filled
14.2.1).
mercury.
easily
that
is
mercury
Glass
Mercury-in-glass
capacities.
bulb
the
(Figure
conductor
temperature
than
glass
of
use.
Another
changing
the
to
means
bore.
very
heat
walled
volume
more
This
the
in
thermometer
temperatures.
to
will
therefore
down
thin
and
means
glass.
thermal
thermometer
the
expand
This
a
change
of
not
wet
metal
Mercury
relatively
bore
does
not
the
inexpensive
opaque.
does
moving
that
fine
are
and
is
thermometer
,
heated,
the
thermometers
because
glass)
The
make
that
the
reading
this
circuit for
thermometer
sensitive,
the
bore
is
made
very
thin.
The
drawback
to
this
measuring
is
that
it
limits
the
range
of
the
thermometer
of
a
given
length.
resisitance
Resistance thermometers
In
a
platinum-resistance
insulating
material
like
thermometer
,
mica
(Figure
a
platinum
14.2.2).
It
wire
relies
is
on
coiled
the
fact
on
an
that
oil
the
is
platinum
coil
resistance
used
means
that
melting
range.
114
is
a
a
high
large
makes
2046 K.
Wheatstone
wire
change
the
in
with
is
means
used
temperature.
coefficient
resistance
thermometer
This
bridge
varies
temperature
that
to
for
very
the
of
a
small
sensitive.
the
change
This
change
in
Platinum
thermometer
measure
Platinum
resistance.
has
in
a
has
very
a
large
resistance
A platinum-resistance
of
thermometer
of
platinum
has
This
point
A
the
it
there
temperature.
Figure 14.2.2
of
because
the
platinum.
Since
a
Wheatstone
bridge
is
used,
very
slight
changes
in
Chapter
resistance
type
of
can
be
measured
thermometer
means
that
it
is
is
slow
very
that
to
it
accurately.
has
respond
a
to
The
relatively
rapidly
biggest
large
drawback
heat
changing
capacity.
of
14
Temperature
this
This
temperatures.
Thermistors
A
ther mistor
The
device
temperature.
which
respond
of
A
type
accurate
small.
that
its
have
of
the
to
the
that
being
is
less
periodic
also
measured.
stable
than
over
a
temperature.
that
they
means
The
a
calibration.
temperatures
with
coefficient,
means
It
resistor
.
varies
increasing
This
temperatures.
it
sensitive
thermistor
with
capacities.
needs
measure
thermally
temperature
decreases
heat
is
and
a
of
negative
changing
thermometer
used
a
temperatures
thermometer
when
with
resistance
has
small
rapidly
on
device
The
resistance
very
to
effect
resistance
non-linear
thermistor
quickly
little
this
a
very
means
Thermistors
have
is
is
can
that
they
disadvantage
platinum-
This
long
makes
period
of
it
less
time.
Thermocouples
A
ther mocouple
effect.
an
When
e.m.f.
value
the
of
is
makes
two
produced
the
e.m.f.
temperature
use
different
and
of
the
metals
an
electric
generated
difference
ther moelectric
are
is
joined
current
dependent
between
the
effect
together
flows
on
the
in
as
the
metals
junctions
J
the
e.m.f.
with
temperature
of
J
is
almost
the
being
and
Seebeck
Figure
circuit.
J
1
of
or
in
.
14.2.3,
The
used
The
and
variation
2
always
parabolic,
when
2
kept
J
is
at
0 °C.
placed
In
in
order
to
melting
measure
ice,
which
J
is
1
an
unknown
provides
the
temperature,
reference
one
junction
temperature,
1
while
the
other
junction
J
is
placed
in
contact
with
the
object
of
2
unknown
therefore
respond
Figure
temperature.
have
quickly
14.2.4
the
and
that
for
being
is
a
very
the
the
changing
variation
neutral
small
The
in
e.m.f.
is
in
region.
two
This
e.m.f.
less
temperature
are
very
small
means
junctions.
that
they
They
can
temperatures.
of
change
this
there
have
capacities.
thermocouple
e.m.f.).
given
heat
rapidly
The
above
maximum
change
to
small
shows
thermocouple.
below
very
Thermocouples
with
in
the
(temperature
e.m.f.
Also,
temperature
sensitive
it
as
a
can
possible
result
be
that
of
seen
values
for
for
region
a
produces
temperature
from
the
a
just
the
graph,
temperature
measured.
G
e.m.f. / mV
thermocouple
is
not
in
sensitive
this
region
J
1
parabolic
curve
J
2
θ
θ
ice
water
neutral
Figure 14.2.4
Figure 14.2.3
A thermocouple
temperature
Variation of e.m.f. with
temperature for a thermocouple
115
Chapter
14
Thermometers
Constant-volume
In
this
type
measured
holds
a
of
gas
remains
the
mercury
tube.
the
of
a
left
the
is
always
the
is
of
at
The
of
the
then
the
gas
tube
=
of
fixed
is
expands
+
of
to
in
the
that
contact
fixed
of
gas
bulb
of
adjusted
up
the
so
that
measuring
the
that
volume
is
glass
volume
When
so
gas
The
is
with
the
adjusted
the
tube
marker
.
pushes
is
mass
14.2.5).
movable
volume
and
fixed
ensure
the
tube
a
(Figure
to
brought
returns
p
pressure
order
movable
measured
p
In
height
bulb
the
the
manometer
gas.
the
glass
height
hand
gas
of
constant,
measured.
The
mercury
volume
temperature,
being
thermometer
,
using
fixed
gas thermometers
substance
the
movable
the
mark.
mercury
The
on
pressure
using
ρhg
θ
where
p
is
the
pressure
of
the
gas
at
a
temperature
θ,
p
is
atmospheric
θ
–1
pressure,
the
g
is
density
the
of
Earth’s
mercury
gravitational
and
h
is
the
field
height
strength
of
the
(9.81 N kg
mercury
in
),
the
ρ
is
U-tube.
h
fixed
volume
mark
gas
Table 14.2.1
Figure 14.2.5
A constant-volume gas thermometer
T
able
compares
14.2.1
the
various
Range/K
thermometer
Mercury-in-glass
have
been
discussed.
Resistance
Thermometric
Advantages
Disadvantages
property
234–710
thermometer
Volume
a fine
25–1750
thermometer
of
Electrical
of
a
mercury
column
resistance
platinum
coil
in
ᔢ
Easy
ᔢ
Limited
ᔢ
Portable
to
use
ᔢ
Fragile
ᔢ
Accurate
ᔢ
Slow
ᔢ
Wide
ᔢ
Can
range
Thermocouple
80–1400
e.m.f.
across
junction
of
dissimilar
Constant-volume
thermometer
3–1750
Pressure
mass
of
two
metals
of
a fixed
gas
constant
the
at
volume
range
response
suitable for
measure
temperature
116
that
Comparison of different types of thermometers
Type of
gas
thermometers
small
differences
range
changing
ᔢ
Not
a
ᔢ
Not
as
temperatures)
direct
ᔢ
Wide
ᔢ
Small
resistance
ᔢ
Fast
volume
ᔢ
Can
take
and
send
response
ᔢ
Wide
ᔢ
Accurate
remote
to
range
a
reading
accurate
or
Bulky
Slow
ᔢ
Fragile
a
constant-
readings
ᔢ
as
thermometer
computer
ᔢ
(not
rapidly
response
Chapter
In
order
to
measure
thermometer
the
would
of
use
depend
ᔢ
the
ease
ᔢ
the
range
ᔢ
the
response
time
ᔢ
the
accuracy
of
of
of
the
temperature
the
of
a
substance,
the
choice
14
Temperature
of
on:
thermometer
thermometer
of
the
the
thermometer
thermometer
.
Example
Select
a
suitable
choice
of
thermometer
to
measure
the
following
temperatures.
a
The
temperature
b
The
boiling
c
A
small
d
A
rapidly
a
For
a
heat
point
change
very
small
capacity
the
its
A
in
small
very
will
have
of
the
resistance
to
lower
of
ethanol.
(90 K).
ethanol,
be
the
since
used.
thermometer
reason
temperature
The
it
a
The
most
has
of
the
suitable
small
for
with
this
is
ethanol
a
and
thermometer
junctions
and
very
that
small
the
will
not
would
therefore
has
be
a
capacity.
platinum-resistance
Boiling
of
temperature.
temperature.
heat
quantity
temperature.
quantity
will
small
oxygen
body
thermocouple,
very
b
true
a
of
changing
thermometer
give
of
oxygen
thermometer
occurs
thermometers
at
are
a
would
constant
very
be
suitable
in
temperature.
accurate
and
can
this
case.
Platinum-
measure
steady
temperatures.
c
A
mercury-in-glass
temperature
d
The
have
best
thermometer
around
heat
would
be
suitable
in
this
case.
Body
37 °C.
thermometer
small
changing
is
would
capacities
be
and
the
thermocouple.
therefore
respond
Thermocouples
quickly
to
rapidly
temperatures.
Key points
ᔢ
There
are
advantages
and
disadvantages
to
using
the following
thermometers:
–
liquid-in-glass
–
resistance
–
thermistor
–
thermocouple
–
constant-volume
gas
thermometer.
117
15
Thermal
properties
15.
1
Internal
energy
Learning outcomes
On
completion
should
ᔢ
be
define
able
the
of
this
you
to:
term
relate
body
a
rise
to
an
in
are
heated
in
Bunsen
energy
and
temperature
increase
in
of
a
internal
from
a
is
and
terms
more
other
.
supplied
increases.
the
made
vibrate
If
each
random
define
specific
heat
up
of
flame.
about
many
The
their
particles.
particles
mean
to
Their
the
The
energy
is
positions.
supplied,
potential
iron,
the
inter nal
distribution
Internal
of
the
the
energy
kinetic
energy
and
of
the
kinetic
energy
In
a
in
motion.
The
the
gas
there
U
is
motion
Definition
the
sum
of
the
of
the
kinetic
of
the
particles
of
a
random
and
system
a
piece
iron
Their
of
gain
iron
being
thermal
particles
kinetic
increases.
potential
piece
and
of
move
energy
So
as
energy
iron
potential
further
is
of
the
energy
apart
thermal
the
particles
sum
of
energy
of
the
the
particles.
=
E
+
E
also
sum
causes
forces
of
that
are
The
internal
molecules
that
the
the
that
the
random
present.
The
molecules
molecules
exist
potential
P
to
between
energy
kinetic
as
and
molecules
possess
move
the
further
The
potential
constantly
energy.
apart
molecules).
well.
are
kinetic
(against
Therefore,
internal
energies
energy
of
the
the
of
the
gas
molecules
present.
energy
energy
of
is
a
system
supplied
to
is
a
determined
system,
the
by
the
state
particles
of
the
within
it
system.
begin
is
vibrating
more
rapidly.
potential
energy
energy
the
The
kinetic
energy
of
the
particles
increases.
The
distribution
potential
that
many
means
possess
Whenever
energy
are
This
attractive
molecules
the
the
heat capacity
specific heat capacity
internal
Consider
inside
K
The
capacity
internal energy
energy
ᔢ
matter
energy
Substances
increases.
ᔢ
and
Internal
section,
of
make
of
the
particles
also
increases.
Consequently,
the
internal
energies
up
of
system
increases.
the
In
an
ideal
gas,
the
force
of
attraction
between
the
molecules
is
system.
negligible.
energies
There
of
are
the
two
ᔢ
supply
ᔢ
compress
Suppose
the
of
kinetic
result
of
the
the
they
energy
Heat
gas
a
of
off
amount
on
of
the
material
When
a
up
is
greater
the
and
gas
of
gas
Some
with
the
to
the
a
make
the
a
kinetic
gas:
gas
molecules
greater
up
the
using
speed.
gas
a
The
the
the
is
the
The
increases.
temperature
piston,
energy
that
be
up
average
gas
moving
kinetic
result
strike
The
the
can
make
increases.
of
piston,
strike
average
heat
increase
are
gas
Energy
that
the
of
all
the
internal
increase.
substance,
materials
particles
temperature
The
increases.
specific
the
of
container
.
The
molecules
speed.
required
of
that
metal
When
off
and
a
compressed
the
temperature
itself.
rapidly.
bounce
energy
the
it).
container
.
particles
on
it)
inside
the
mass
energy
the
vibrate
dependent
gas.
gas
they
energy
the
internal
(heating
on
only
of
cylinder
capacity
the
up
is
heating
to
with
energy
make
work
internal
make
the
gas
(do
by
all
a
that
quantity
begin
the
that
internal
increase
the
increases.
and
depends
the
to
gas
inside
bounce
particles
The
that
gas
to
container
energy
gas
is
the
the
is
ways
the
container
walls
If
to
the
molecules
energy
there
supplied
118
Therefore,
the
capacity
the
temperature
required
easier
to
of
a
temperature
heat
up
than
substance
rise
others.
and
Chapter
The
amount
of
energy
needed
to
raise
the
temperature
of
1 kg
of
15
Thermal
properties
of
matter
a
Definition
substance
by
1 °C
(or
1 K)
is
called
the
specific
heat
capacity
( c)
of
the
substance.
The
The
amount
of
energy
E
required
to
produce
a
temperature
change
of
Δθ
specific
substance,
heat
c,
is
capacity
the
of
amount
a
of
heat
H
in
a
substance
of
mass
m kg
is
given
energy
by
required
temperature
E
=
c
is
the
specific
heat
substance
capacity
of
the
–1
unit
of
specific
heat
capacity
is
c
=
=
=
mΔθ
–1
The
unit
can
also
be
expressed
as
J kg
by
one
kilogram
of
a
degree.
J
H
SI
one
the
substance.
E
The
increase
mcΔθ
H
where
of
to
J kg
–1
K
Equation
kg K
–1
°C
.
This
is
so
because
a
change
E
=
mcΔθ
H
in
1 °C
is
exactly
equal
to
a
change
in
1 K.
E
–
amount
m
–
mass/kg
of
c
–
specific
energy
absorbed
H
The
larger
that
is
the
specific
required
to
heat
increase
capacity
its
of
the
substance,
temperature.
The
the
specific
more
heat
energy
capacity
–1
–1
of
aluminium
and
copper
are
–1
880 J kg
–1
K
and
380 J kg
heat
2 kg
of
aluminium
and
copper
are
heated
to
produce
The
amount
E
=
change
of
of
=
2
absorbed
×
880
×
by
the
10
=
aluminium
is
given
of
energy
absorbed
by
the
copper
is
given
=
mcΔθ
=
2
×
380
×
10
=
The
°C
simple
calculation
energy
that
is
shows
that
required
to
heat
capacity
of
a
substance,
7 600 J
C,
more
the
or
by
H
of
temperature/K
17 600 J
The
E
the
°C
in
Definition
amount
This
or J kg
change
by
H
The
–
10 °C.
energy
mcΔθ
–1
a
Δθ
temperature
–1
K
respectively.
–1
Suppose
capacity/J kg
–1
K
the
higher
raise
the
the
specific
temperature
heat
of
capacity,
one
kilogram
is
the
required
of
a
amount
to
of
heat
increase
substance
by
the
one
energy
temperature
degree.
substance.
same
released
10 °C.
equation
when
The
a
can
be
used
substance
amount
of
to
cools.
energy
determine
Suppose
released
is
the
2 kg
of
amount
of
energy
aluminium
cools
by
Equation
17 600 J.
E
Another
useful
substance
is
quantity
the
energy
is
heat
capacity
required
to
raise
(C) .
the
The
heat
capacity
temperature
of
a
of
a
=
CΔθ
H
substance
E
–
amount
C
–
heat
Δθ
–
change
of
energy
absorbed
H
by
one
–1
degree.
–1
The
The
unit
of
heat
amount
of
capacity
energy
E
is
the
–1
J K
or
required
to
capacity/JK
in
–1
or J °C
temperature/K
or
°C
J °C
produce
a
temperature
change
of
Δθ
H
in
a
substance
is
given
by
E
=
CΔθ
H
Equation
Key points
C
=
mc
–1
ᔢ
The
the
internal
kinetic
energy
and
of
a
system
potential
is
energies
the
of
sum
the
of
the
random
particles
that
distribution
make
up
the
C
–
heat
capacity/J K
m
–
mass/kg
c
–
specific
of
system.
–1
ᔢ
The
the
ᔢ
The
gas
ᔢ
The
to
internal
energy
molecules
internal
or
doing
specific
raise
the
of
an
of
a
ideal
gas
is
dependent
The
ᔢ
The
the
SI
unit
energy
work
heat
on
gas
the
can
gas
capacity
temperature
heat
of
the
kinetic
energy
heat
capacity/J kg
–1
K
of
only.
of
of
be
increased
(compressing
a
substance
1 kg
of
a
is
by
the
the
substance
specific
capacity
temperature
of
of
a
heat
a
capacity
substance
substance
is
by
is
the J kg
the
supplying
energy
to
the
gas).
amount
by
–1
ᔢ
on
1 °C
of
or
energy
required
1 K.
–1
K
amount
of
energy
required
to
raise
1 °C.
119
15.2
Electrical
heat
completion
should
be
able
of
this
describe
section,
you
Specific
by
to:
experiments
capacities
specific
of
heat
using
an
with
liquids
the
and
of
substances
method
electrical
or
by
methods
can
a
of
be
experimentally
method
of
mixtures.
determining
the
determined
This
specific
section
heat
capacity
solids.
heat
substances
by
Specific
electrical
capacities
electrical
to
of
determine
specific
Electrical methods for measuring specific heat capacities
deals
ᔢ
measuring
capacities
Learning outcomes
On
methods for
heat
capacity of
a
liquid
methods.
The
is
apparatus
shown
in
calorimeter
initial
used
Figure
(the
to
measure
15.2.1.
vessel
temperature
of
in
A
the
which
the
specific
known
the
liquid
θ
mass
heat
is
heat
of
capacity
liquid
m
is
measurements
measured
using
of
liquid
placed
are
the
in
a
made).
The
thermometer
.
1
An
for
electric
a
current
length
recorded.
heating
of
time
During
of
is
the
then
t.
the
liquid.
The
made
flow
ammeter
heating
The
to
period
final
through
and
the
the
voltmeter
stirrer
temperature
θ
heating
reading
is
used
to
is
recorded.
I
element
and
provide
V
are
uniform
2
voltmeter
stirrer
thermometer
variable
ammeter
resistor
insulation
calorimeter
heating
element
liquid
Figure 15.2.1
An electrical method to determine the specific heat capacity of a liq uid
Experimental
Mass
of
Initial
results
liquid
=
m kg
temperature
of
liquid
=
θ
°C
1
Final
temperature
of
liquid
=
θ
°C
2
Ammeter
reading
V
oltmeter
Time
reading
heating
Electrical
Energy
=
I amps
=
V volts
element
energy
was
supplied
absorbed
by
the
switched
to
liquid
liquid
=
on
=
=
IVt
mc(θ
–
θ
1
where
If
it
is
liquid
120
c
is
the
assumed
can
be
specific
that
found
no
by
heat
heat
using
capacity
losses
the
of
occur
,
t seconds
the
the
following:
)
2
liquid.
specific
heat
capacity
of
the
Chapter
Electrical
energy
supplied
by
heating
IVt
=
element
mc(θ
–
θ
1
=
energy
absorbed
by
of
is
15
Thermal
properties
of
matter
liquid
)
2
IVt
∴
c
=
m(θ
–
θ
1
Specific
The
apparatus
similar
the
A
in
heat
to
solid.
the
used
one
The
second
to
is
of
in
A
the
solid
placed
hole.
in
small
the
thermometer
to
the
thermometer
and
a
measure
used
mass
thermometer
the
capacity of
ensure
the
solid
the
specific
previous
m
is
one
hole
that
heat
and
of
there
The
capacity
experiment.
determined
amount
solid.
)
2
a
oil
is
using
heating
is
placed
good
initial
T
wo
an
a
are
electronic
element
inside
thermal
solid
holes
is
the
of
in
balance.
placed
hole
contact
temperature
drilled
the
with
between
solid
θ
1
is
measured
flow
using
through
voltmeter
the
the
thermometer
.
heating
reading
I
and
element
V
are
An
for
a
electric
length
recorded.
The
current
of
time
final
t.
is
then
The
made
to
ammeter
temperature
θ
is
and
recorded
2
(Figure
15.2.2).
voltmeter
thermometer
variable
ammeter
resistor
insulation
electric
heater
solid
Figure 15.2.2
An electrical method to determine the specific heat capacity of a solid
Experimental
Mass
of
Initial
results
solid
=
m kg
temperature
of
solid
=
θ
°C
1
Final
temperature
of
solid
=
θ
°C
2
Ammeter
reading
V
oltmeter
Time
reading
heating
Electrical
Energy
=
I amps
=
V volts
element
energy
absorbed
was
supplied
by
the
switched
to
solid
solid
=
on
=
=
IVt
mc(θ
–
θ
1
where
If
it
is
solid
c
is
the
assumed
can
Electrical
be
specific
that
found
energy
no
by
heat
heat
using
supplied
capacity
losses
the
by
t seconds
of
occur
,
)
2
the
the
solid.
specific
heat
capacity
of
the
following:
heating
element
IVt
=
=
energy
mc(θ
absorbed
–
θ
1
by
solid
)
2
IVt
∴
c
=
m(θ
–
1
θ
)
2
121
Chapter
15
Thermal
properties
of
matter
A
continuous flow
capacity of
In
1899
heat
a
The
liquid
tube
of
is
specific
heat
liquid
Callendar
capacity
method to determine the
a
and
allowed
containing
Barnes
liquid
a
by
to
a
flow
heating
devised
a
method
continuous
from
a
element
flow
to
constant-head
as
shown
measure
the
specific
method.
in
tank
Figure
through
a
glass
15.2.3.
inflow
outflow
vacuum
θ
°C
θ
1
+
Figure 15.2.3
The
glass
heat
a
steady
this
tube
is
At
glass
experiment
accurate
insulated
escape
state.
the

heater
Continuous flow method
cannot
leaving
°C
2
and
is
from
this
a
of
a
glass
liquid.
the
remains
jacket,
The
at
temperature
constant.
measuring
liquid
which
entering
A
steady
glass
is
experiment
platinum-resistance
good
T
emperature
stage,
tube
is
by
the
of
the
suitable
evacuated,
is
allowed
liquid
It
is
that
reach
entering
thermometer
thermometer
.
so
to
and
for
extremely
temperatures.
tube
=
θ
°C
1
T
emperature
of
liquid
leaving
glass
tube
=
θ
°C
2
Mass
of
water
collected
as
it
leaves
the
tube
=
m
kg
1
Time
taken
Ammeter
to
collect
reading
liquid
=
I
=
t
seconds
amps
1
V
oltmeter
reading
=
V
volts
1
Under
steady
element
is
state
carried
conditions,
away
I
V
1
where
h
specific
is
the
heat
The
rate
that
the
at
heat
t
which
=
m
of
to
c(θ
energy
θ
supplied
by
the
heating
liquid.
)
+
h
(1)
2
surroundings
in
a
time
t
and
c
is
the
liquid.
liquid
collected
–
1
the
the
the
flowing
1
lost
the
all
the
1
capacity
amount
by
is
in
a
fl owing
time
t
into
is
m
the
.
glass
The
tube
cur rent
is
altered
and
so
voltage
2
are
adjusted
to
bring
the
temperature
θ
back
to
its
original
value.
The
2
temperature
of
of
leaving
the
liquid
the
liquid
the
entering
the
glass
constant-head
tube
tank.
is
This
the
temperature
temperature
θ
is
1
constant.
time
t
is
Since
again
all
the
temperatures
are
the
same,
the
heat
h
T
emperature
of
liquid
entering
glass
tube
=
θ
°C
1
T
emperature
of
liquid
leaving
glass
tube
=
θ
°C
2
Mass
of
water
collected
as
it
leaves
the
tube
=
m
kg
2
Time
122
taken
to
collected
liquid
=
t
seconds
lost
in
a
Chapter
Ammeter
reading
=
I
15
Thermal
properties
of
matter
amps
2
V
oltmeter
reading
=
V
volts
2
Under
steady
element
is
state
carried
∴
I
V
2
where
h
is
the
Subtracting
(I
V
2
conditions,
away
t
=
–
V
1
the
c(θ
2
heat
I
m
2
lost
Equation
2
by
)t
to
(1)
=
all
–
the
–
(I
The
inlet
using
ᔢ
The
in
a
m
V
–
lost
is
main
is
not
heating
(2)
–
V
1
in
a
time
t
(2):
θ
2
–
m
)
1
)t
1
)(θ
1
the
–
θ
2
continuous
outlet
calculations.
apparatus
The
the
in
)(θ
I
2
)
1
flow
temperatures
platinum-resistance
heat
the
and
the
h
Equation
2
ᔢ
by
=
(m
of
+
1
2
advantages
)
supplied
liquid.
surroundings
from
c(m
energy
1
2
c
The
θ
2
1
the
flowing
method
can
be
are:
measured
very
accurately
thermometer
.
experiment
Therefore,
is
the
not
required
heat
since
capacity
of
it
is
eliminated
various
parts
of
the
required.
disadvantage
of
this
method
is
that
a
large
volume
of
the
liquid
required.
Key points
ᔢ
The
specific
electrical
heat
capacity
of
a
liquid
and
a
solid
can
be
determined
by
an
method.
123
15.3
Measuring
by
using
Learning outcomes
On
completion
should
ᔢ
be
able
describe
specific
and
of
this
section,
liquids
to
capacities
by
method
you
The
a
experiments
using
a
heat
of
Measuring the
to:
heat
a
specifi c
measure
of
solids
method
of
specific
liquid
of
Suppose
a
small
you
are
piece
electronic
metal
heat
is
of
specific
specific
required
metal.
balance.
then
mixing
capacity
known
capacities
a
solid
heat
to
The
The
placed
of
a
capacity of
can
be
a
determined
solid
by
mixing
it
with
capacity.
determine
mass
metal
into
heat
is
of
the
the
attached
container
specific
metal
of
to
is
a
heat
piece
boiling
capacity
measured
of
water
.
using
thread.
It
is
left
c
of
an
The
there
for
mixtures.
approximately
the
same
using
In
a
the
mass
temperature.
to
The
ensure
that
temperature
the
of
metal
the
and
boiling
the
water
water
is
are
at
measured
thermometer
.
meantime,
of
the
measured.
The
is
used
by
water
Mass
some
water
stirrer
the
10 minutes
of
hot
to
is
is
cold
water
measured.
piece
stir
the
of
metal
water
is
The
placed
initial
is
quickly
gently.
inside
a
polystyrene
temperature
The
added
highest
to
of
the
the
cold
cup.
water
water
.
temperature
The
is
A
achieved
recorded.
piece
of
metal
=
m
kg
1
Initial
temperature
of
the
metal
=
θ
°C
1
Mass
of
cold
water
=
m
kg
2
Initial
temperature
of
cold
water
=
θ
°C
2
Final
temperature
of
cold
water
=
θ
°C
3
–1
Specific
heat
capacity
of
water
=
c
J kg
–1
K
w
Heat
lost
by
m
piece
×
c
of
×
metal
(θ
1
–
θ
1
)
=
heat
=
m
3
gained
×
c
2
×
c
2
The
with
specific
a
heat
liquid
similar
to
or
that
capacity
solid
used
of
of
a
–
θ
3
×
(θ
w
×
)
2
–
θ
3
(θ
1
specific
water
)
2
=
m
Measuring the
(θ
w
m
c
×
by
heat
liquid
known
–
1
θ
)
3
capacity of
can
specific
be
heat
a
determined
capacity.
liquid
by
The
mixing
it
calculation
is
above.
Example
Calculate
the
energy
required
to:
a
Increase
the
temperature
of
0.75 kg
b
Increase
the
temperature
of
1.2 kg
of
of
aluminium
copper
from
from
10 °C
25 °C
to
heat
–1
380 J kg
a
E
b
E
capacity
of
aluminium
and
copper
are
880 J kg
–1
K
respectively]
=
mcΔθ
=
0.75
=
mcΔθ
=
1.2
×
880
×
(60
–
25)
=
23 100 J
H
H
124
×
380
×
(45
–
10)
=
15 960 J
60 °C.
45 °C.
–1
[Specific
to
–1
K
and
Chapter
15
Thermal
properties
of
matter
Example
Calculate
the
energy
released
when
1.5 kg
of
water
cools
from
90 °C
to
25 °C.
–1
[Specific
E
heat
=
capacity
mcΔθ
=
of
1.5
water
×
=
4200
4200 J kg
×
(90
–
–1
K
25)
]
=
409 500 J
H
Example
A
kettle
from
rated
30 °C
to
at
2 kW
is
used
to
raise
the
temperature
of
1.5 kg
of
water
100 °C.
Calculate:
a
the
energy
required
to
raise
the
temperature
of
the
water
from
30 °C
to
100 °C
b
the
time
taken
to
raise
the
temperature
of
the
water
from
30 °C
to
100 °C.
–1
[Specific
heat
capacity
of
water
=4200 J kg
–1
K
]
5
a
Energy
required
E
=
mcΔθ
=
1.5
×
4200
×
(100
–
30)
=
4.41
×
10
J
H
b
Energy
=
power
×time
5
E
t
=
4.41
×
10
=
=
220.5 s
3
P
2
×
10
Key points
ᔢ
The
specific
with
heat
another
capacity
substance
of
a
(solid
liquid
or
or
liquid)
solid
of
can
be
known
determined
specific
heat
by
mixing
capacity.
125
15.4
Latent
heat
Learning outcomes
On
completion
should
be
able
of
this
Change of
section,
Matter
you
energy
to:
gas.
ᔢ
explain
the
concepts
of
ᔢ
explain
a
of
to
solid,
a
three
solid,
there
are
states
it
–
solids,
changes
many
liquids
into
particles
a
and
liquid
held
gases.
and
When
eventually
together
by
into
strong
a
bonds.
melting
a
solid
the
term
specific latent
to
vibrate
of
the
solid.
is
temperature–time
graphs
particles
heated,
melting
and
rapidly
these
about
particles
increases,
particles
increases.
also
their
gain
mean
energy.
The
resulting
begin
Eventually,
in
an
moving
enough
positions.
particles
increase
further
energy
The
in
apart
is
kinetic
begin
energy
temperature
and
their
absorbed
by
of
the
potential
the
solid,
to
causing
determine
more
The
energy
use
one
boiling
heat
ᔢ
in
supplied
Inside
When
and
exists
is
state
the
bonds
between
the
particles
to
break.
The
motion
of
the
boiling
particles
becomes
more
disordered.
The
particles
move
freely
within
the
points
structure.
ᔢ
explain
the
cooling
accompanies
has
that
At
become
this
a
point
in
the
heating
the
solid
has
melted
and
liquid.
evaporation.
As
more
rapidly.
energy
The
increase
further
energy
to
in
is
supplied
and
At
and
this
their
by
the
of
of
the
the
the
liquid
the
and
liquid
temperature
liquid,
a
The
energies
to
cause
particles
in
the
particles
liquid.
potential
the
stage,
randomly
process,
to
energy
temperature
absorbed
break.
heating
is
kinetic
apart
rapidly
are
become
particles
increases,
particles
increase.
the
a
also
At
even
begin
between
further
state.
gas
vibrate
resulting
Eventually
bonds
much
disordered
has
apart,
this
(Figures
more
an
moving
enough
the
particles
moving
point
15.4.1
in
in
the
and
15.4.2).
temperature
gas
boiling
process,
gas
point
boiling
point
liquid
melting
point
liquid
liquid
melting
solid
solid
and
liquid
and
point
gas
solid
mixture
mixture
time
Figure 15.4.1
A heating curve (solid to liq uid to gas)
At
a
Figure 15.4.2
change
between
the
increases
not
the
energy
is
change
in
Boiling
is
without
a
all
the
the
that
potential
the
in
Melting
state,
particles
kinetic
A cooling curve
energy
make
energy
temperature
of
change
of
the
increase
Since
126
time
of
energy
of
particles,
it
a
supplied
up
the
the
of
particles.
the
used
The
particles
substance
follows
is
substance.
is
that
to
a
the
energy
during
dependent
at
break
All
change
a
bonds
supplied
supplied
change
on
of
the
energy
the
of
does
state.
kinetic
state,
there
is
no
temperature.
the
process
whereby
a
solid
changes
into
a
liquid
without
temperature.
the
process
change
in
whereby
a
liquid
temperature.
changes
into
a
gas
(vapour)
a
Chapter
Latent
The
Thermal
properties
of
matter
heat
energy
latent
15
required
to
change
the
state
of
a
substance
is
known
as
the
heat
When
a
solid
changes
into
a
liquid,
latent
heat
is
required.
The
specific
Definition
latent
a
heat
of
substance
fusion
from
a
is
the
solid
energy
into
required
liquid
to
without
a
convert
change
unit
in
mass
(1 kg)
of
temperature.
The
specific
latent
heat of fusion
l
f
–1
The
SI
latent
unit
heat
of
of
specific
fusion
latent
is
heat
of
calculated
fusion
using
is
the
the
J kg
following
.
The
specific
is
equation.
the
of
energy
required
substance from
without
a
change
a
in
to
convert
solid
to
a
1 kg
liquid
temperature.
Equation
E
=
ml
H
E
f
–
energy/J
m
–
mass/kg
l
–
specific
H
latent
heat
of
f
–1
fusion/J kg
Example
5
The
specific
amount
of
latent
energy
heat
of
required
fusion
to
of
ice
convert
is
3.3
80 g
of
×
10
ice
at
5
E
=
ml
H
=
0.08
×
3.3
×
10
–1
Jkg
0 °C
.
Calculate
into
water
the
at
0 °C.
4
=
2.64
×
10
J
f
Definition
The
specific
unit
mass
latent
(1 kg)
of
heat
a
of
vaporisation
substance
from
a
is
the
liquid
energy
into
a
required
vapour
to
convert
without
a
The
change
in
specific
latent
heat of
temperature.
vaporisation
l
,
is
the
energy
v
–1
The
SI
latent
unit
heat
of
of
specific
latent
vaporisation
is
heat
of
vaporisation
calculated
using
is
the
the
J kg
following
.
The
specific
equation.
required
from
a
change
to
convert
liquid
in
to
a
1 kg
of
vapour
substance
without
a
temperature.
Equation
E
=
ml
H
E
v
–
energy/J
H
m
–
mass/kg
l
–
specific
latent
heat
of
v
–1
vaporisation/J kg
Example
6
The
specific
Calculate
100 °C
to
latent
the
heat
amount
steam
at
of
of
vaporisation
energy
of
required
water
to
is
convert
=
ml
H
=
×
10
250 g
of
–1
J kg
.
water
at
100 °C.
6
E
2.3
0.25
×
2.3
×
10
amount
of
energy
5
=
5.75
×
10
J
v
Example
Calculate
to
steam
the
at
required
to
convert
1.2 kg
of
ice
at
–10 °C
100 °C.
3
Specific
heat
capacity
Specific
latent
Specific
heat
Specific
latent
of
ice
=
2.1
×
10
–1
J kg
–1
K
5
heat
of
fusion
of
ice
=
3.3
×
10
3
capacity
of
water
=
4.2
×
10
–1
J kg
–1
J kg
6
heat
of
vaporisation
of
water
=
2.3
×
10
–1
J kg
127
Chapter
15
Thermal
properties
of
matter
Energy
required
to
raise
the
temperature
of
ice
to
0 °C
Exam tip
3
E
=
mcΔθ
=
1.2
×
2.1
×
10
×
(0
–
–
10)
=
25 200 J
H
If
it
is
required
to
determine
the
Energy
energy
needed
to
change
required
to
convert
ice
at
0 °C
to
water
at
0 °C
the
5
E
temperature
a
change
E
=
of
of
a
substance
state,
use
the
without
equation
=
ml
H
=
1.2
×
3.3
×
10
=
396 000 J
f
Energy
required
to
raise
the
temperature
of
water
from
0 °C
to
100 °C
mcΔθ
H
3
E
=
mcΔθ
=
1.2
×
4.2
×
10
×
(100
–
0)
=
504 000 J
H
If it is required to determine the
Energy
required
to
convert
water
at
100 °C
to
steam
at
100 °C
energy needed to change the state of
a substance, use the equation
E
6
= ml
E
H
=
ml
H
=
1.2
×
2.3
×
10
=
2 760 000 J
v
Energy
required
to
convert
1.2 kg
of
ice
at
–10 °C
to
steam
at
100 °C
6
=
25 200
+
396 000
+
504 000
+
2 760 000
=
3.69
×
5
The
specific
latent
heat
of
fusion
of
ice
is
3.3
×
the
heat
of
specific
latent
can
heat
be
gases.
vaporisation
latent
of
fusion
attributed
In
ice
In
In
a
solid.
steam
separate
greater
separation
heat
of
when
is
only
When
is
energy
supplied,
are
seen
in
converted
Boiling
a
gas
–
into
A
is
in
J kg
reaching
Factors
its
its
that
is
affect
area
area
air
of
move
changing
from
from
a
solid
slightly.
that
to
Hence
the
so
the
far
about
to
a
of
of
the
why
to
is
by
that
liquid.
the
to
the
strong
mean
tightly
that
gas
where
and
by
their
required
reason
heat
quantities
similar
body
that
specific
less
is
apart
liquid
the
the
together
them
liquid,
latent
specific
seen
liquids
together
energy
a
the
be
two
solids,
vibrate
within
are
than
tightly
held
can
these
of
between
molecules
a
liquid,
remains
The
T
able
in
it
its
eventually
force
of
completely
much
mean
the
latent
fusion.
temperature
start
to
constant
energy
15.4.1
boil.
(i.e.
energy
and
its
supplied
compares
rises.
At
this
If
boiling
causes
state
energy
the
point).
the
evaporation
changes
more
point,
Bubbles
liquid
and
from
a
to
be
boiling.
liquid
into
temperature.
of
a
room
eventually
water
process
point.
the
rate
is
25 °C.
If
evaporates.
changes
liquid
the
by
which
the
a
Evaporation
of
state
some
The
from
with
exposed
to
the
surface
surface).
liquid
water
is
boiling
liquid
to
changes
occurs
evaporation
increases
exposed
above
above
are
The
absorbs
the
(rate
is
to
and
It
spilled
point
into
a
of
gas
on
pure
without
point.
boiling
surface
of
yet
the
ᔢ
blown
vapour
.
boiling
held
free
.
larger
between
negligible.
liquid.
temperature
temperature
flow
to
will
liquid
room,
to
J kg
boiling
change
ᔢ
larger
128
a
free
greater
boiling
a
are
molecules
the
10
much
are
separation
are
×
–1
structures
the
increased
the
in
particles
when
and
is
the
changing
is
2.3
difference
particles
are
supplied
the
Evaporation
ᔢ
them
liquid
100 °C,
reaching
and
substance
the
floor
water
of
the
without
Suppose
the
is
the
temperature
the
water),
vaporisation
Evaporation
The
The
molecules
than
is
differences
molecules
between
the
ice.
water
,
forces
(vaporised
attraction
of
water
vaporisation
water)
liquid
The
of
the
forces.
intermolecular
of
to
(solid
intermolecular
positions.
heat
of
J
–1
10
6
latent
10
at
a
gas
without
temperature.
are:
increasing
the
any
into
temperature)
atmosphere
(rate
increases
if
a
atmosphere)
of
the
liquid
(rate
increases
when
air
is
Chapter
Evaporation
an
athlete
runs
increases.
sweat
into
the
plays
In
skin
body
to
as
the
for
race.
the
body.
This
the
important
100 m
order
covers
vapour
.
a
an
to
Thermal
cooling
the
cool
energy
energy
evaporates.
is
(latent
The
race,
down,
the
the
human
body
sweat
required
heat)
is
is
to
of
matter
temperature
convert
of
properties
Suppose
produced.
taken
evaporation
body.
Thermal
from
the
The
the
the
sweat
sweat
surface
causes
of
the
cool.
Table 15.4.1
Comparing evaporation and boiling
Evaporation
Change
Occurs
in
During
body
thermal
sweat
role
15
of
at
Boiling
state
the
occurs
surface
Change
of
the
Occurs
of
state
within
occurs
the
body
of
the
liquid
liquid
No
bubbles
are
seen
Bubbles
are
seen
rising
molecules
having
escape from
Occurs
at
any
Using the
in
a
The
of
temperature
kinetic
container
molecules
kinetic
in
from
the
surface
from
the
surface,
decreases.
kinetic
can
as
be
The
energy
a
of
of
it
as
liquid
Some
the
the
it
the
have
average
the
specific
explain
a
range
temperature
why the
Since
of
of
have
kinetic
particles
that
a
(boiling
remaining
sufficient
surface
of
energy
the
liquid
point)
liquid
evaporates
liquid.
evaporates
at
molecules
temperature
of
concluded
some
model to
cools
energies.
Occurs
the
the
most
energy
substance
that
is
make
temperature
(Figure
speeds
and
sufficient
energetic
of
the
the
the
to
a
range
break
molecules
remaining
dependent
up
of
therefore
energy
on
the
substance.
remaining
free
average
escape
molecules
of
the
kinetic
energy
remaining
molecules
decreases
average
Therefore,
liquid
liquid
it
decreases
Figure 15.4.3
molecules
Cooling effect associated
with evaporation
15.4.3).
Key points
ᔢ
Melting
change
ᔢ
Boiling
in
ᔢ
Latent
At
All
ᔢ
is
the
a
or
process
process
The
is
the
liquid
change
the
The
heat
of
energy
specific
change
ᔢ
the
by
which
a
solid
changes
into
a
liquid
liquid
changes
into
a
gas
without
a
temperature.
by
which
a
without
a
change
temperature.
liquid
ᔢ
is
in
unit
specific
required
to
energy
into
gas
state,
mass
is
heat
of
latent
without
there
supplied
latent
a
is
unit
a
no
used
of
into
to
convert
change
change
to
of fusion
solid
heat
convert
required
break
of
a
a
of
a
in
substance from
bonds
substance
without
of
liquid
a
a
solid
into
temperature.
temperature
the
liquid
vaporisation
mass
in
a
is
a
in
the
a
vapour
the
substance.
substance.
energy
change
substance
into
of
the
is
in
the
required
to
temperature.
energy
without
a
change
in
temperature.
ᔢ
Evaporation
reaching
its
is
the
process
boiling
whereby
a
liquid
changes
into
a
vapour
without
point.
129
15.5
Measuring
Learning outcomes
On
completion
should
ᔢ
be
able
describe
an
of
this
specific
Measuring the
section,
you
to:
solid
The
experiment
the
specific
specific
mixtures.
to
latent
of
a
describe
an
experiment
the
heat
mass
Some
inside
of
water
a
of
a
fusion
of
ice
calorimeter
which
is
calorimeter
.
specific
The
a
about
can
is
be
found
measured
15 °C
above
by
using
room
a
method
an
of
electronic
The
mass
of
the
water
temperature
and
the
is
calorimeter
vaporisation
of
a
specific
heat
capacities
of
the
water
and
the
is
calorimeter
to
latent
known.
The
liquid.
initial
temperature
of
the
water
is
measured
using
a
heat
thermometer
.
of
heat of fusion of
solid
are
measure
latent
The
measured.
ᔢ
latent
heat
placed
of fusion
specific
heats
(ice)
balance.
measure
latent
placed
the
gently
ice
mass
used
The
Some
into
melts.
of
to
the
The
water
and
T
emperature
of
are
blotted
stirrer
is
calorimeter
mass
heat
ice
A
temperature
the
the
latent
of
water
.
lowest
determine
specific
pieces
the
of
water
of
the
fusion
=
θ
ice
of
of
the
is
to
they
stir
water
is
was
are
the
is
measured
that
ice
until
used
dry
and
mixture
then
again.
then
until
recorded.
This
will
all
The
be
added.
determined
as
follows:
°C
1
Mass
of
calorimeter
=
m
kg
1
Mass
of
calorimeter
and
water
=
m
kg
2
Mass
of
water
=
m
=
(m
3
Mass
of
water
and
–
m
2
) kg
1
calorimeter
at
end
of
experiment
=
m
kg
4
Mass
of
ice
used
=
m
=
(m
5
Final
temperature
of
–
m
4
water
=
) kg
2
θ
°C
2
–1
Specific
heat
capacity
of
water
=
4200 J kg
Specific
heat
capacity
of
calorimeter
–1
K
–1
Energy
lost
Energy
lost
by
Energy
+
by
∴
×
4200
×
(θ
–
θ
1
specific
latent
)
+
to
m
2
heat
used
×
c
melt
×
(θ
1
of
–1
K
Energy
used
to
increase
+
calorimeter
3
The
c J kg
=
water
m
=
ice
–
θ
1
fusion
can
be
temperature
)
=
m
2
l
5
calculated
+
m
f
×
of
melted
4200
×
θ
5
from
ice
2
the
expression
above.
Measuring the
a
apparatus
liquid
until
the
is
latent
it
is
inner
Mass
to
vessel
of
of
boil
over
in
to
a
measure
Figure
boiling.
walls
allowed
collected
used
shown
begins
receiving
the
The
is
all
then
period
liquid
vapour
jacket
until
and
parts
used
of
the
15.5.1.
to
time
collected
=
Current
=
I
amps
1
V
oltage
=
V
volts
1
Time
=
t
seconds
specific
The
the
collect
t
is
m
latent
liquid
produced
passes
of
is
through
apparatus
the
heat
heated
passes
the
of
at
vaporisation
a
through
constant
the
condenser
.
become
liquid.
determined.
kg
1
130
heat of vaporisation of
liquid
The
a
specific
The
mass
of
holes
The
steady.
of
rate
in
liquid
The
the
liquid
Chapter
15
Thermal
properties
of
matter
hole
jacket
heater
liquid
water
out
condenser
water
receiving
Figure 15.5.1
Electrical
energy
supplied
V
1
the
is
t
=
is
=
m
1
experiment
heater
vessel
Measuring specific latent heat of vaporisation
I
The
in
l
1
energy
+
and
to
vaporise
liquid
+
heat
losses
h
(1)
v
repeated,
changed
required
but
the
this
new
time
mass
the
m
potential
of
vapour
difference
which
across
condenses
2
in
the
same
before.
liquid,
time
This
is
which
Mass
of
t
is
measured.
because
remains
liquid
heat
The
loss
fixed
at
collected
is
its
=
heat
dependent
boiling
m
loss
h
in
on
time
the
t
is
the
same
temperature
of
as
the
point.
kg
2
Current
=
I
amps
2
V
oltage
=
V
volts
2
Time
=
t
seconds
I
V
2
Equation
(1)
–
Equation
I
V
1
t (I
t
–
V
I
–
1
I
m
t
=
m
2
V
2
l
2
gives
2
V
=
(2)
1
1
t
2
l
1
)
=
+
–
m
v
(m
2
h
(I
–
m
v
) l
2
V
1
l
2
1
l
(2)
v
–
1
I
v
V
2
)t
2
=
v
(m
–
1
m
)
2
131
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
7
9
a
Describe
accompanying CD.
a
Outline
how
a
physical
of
a
mercury-in-glass
thermometer.
b
1
the features
property
Give
one
[3]
reason
why
glass
is
used
in
this
type
of
which varies
thermometer.
with
temperature
may
be
used
to
temperature.
b
Suggest
why
[1]
measure
c
State
two
d
State
one
reasons
the
thermodynamic
scale
advantage
is
called
an
absolute
scale.
a
Explain
is
used.
[2]
what
is
meant
by
a
and
one
disadvantage
thermometer
over
a
of
a
platinum-
[2]
resistance
2
mercury
of
mercury-in-glass
temperature
why
[3]
thermometer
over
the
same
range.
[2]
thermometric
10
property.
a
Explain
what
is
meant
by
the
internal
energy
of
substance.
b
State
c
Suggest
three
thermometric
properties.
why,
on
the Celsius
scale,
A
piece
of
thermometers
agree
at
the fixed
may
not
between
necessarily
the fixed
agree
at
piece
A
resistance
points.
thermometer
is
[3]
placed
11
a
Explain
in
pure
0 °C
and
its
resistance
is found
to
of
resistance
is
160 Ω. The
be
has
in
a
liquid
of
A
is found
unknown
to
constant-volume
measure
the
be
gas
unknown
Bunsen flame.
energy
of
[2]
is
meant
by
specific heat capacity
a
[3]
specific
heat
capacity
of
temperature
and
–1
K
. Calculate
the
amount
heat
energy
is
to
raise
temperature
of
65 g
of
copper
the
from
resistance
a
internal
3625 Ω. At
thermometer
required
placed
in
the
heat capacity.
Copper
–1
its
to
copper.
what
385 J kg
100 °C,
heated
water
b
at
is
happens
temperatures
and
3
what
points
the
but
copper
different
Explain
of
[2]
[3]
b
types
a
[1]
25 °C
to
65 °C.
[2]
850 Ω.
thermometer
temperature
is
used
and
12
to
is found
A
bicycle
The
to
and
bicycle
is
a
rider
have
travelling
a
total
along
a
mass
of
90 kg.
horizontal
road
brakes
applied
at
a
–1
be
constant
65 °C.
a
Calculate
the
using
the
State
your
value
of
resistance
the
unknown
until
temperature
thermometer.
the
braking
[3]
bicycle
b
answer
in
kelvin.
Suggest
a
reason for
the
difference
between
on
the
resistance
bicycle
process,
and
rider
6.2 m s
and
and
brake
is
thermometer.
5
State
the
principal features
of
a
thermocouple.
advantages
0.
15 kg. The
is
Briefly
gas
describe
of
a
thermocouple
over
thermometer.
a
suitable
the features
of
a
the
initial
the
the
choice
of
thermometer
a
the
temperature
of
a
b
the
temperature
of
various
to
measure
blast furnace
temperature
of
boiling
d
the
temperature
of
a
measuring
period
132
of
the
one
positions
in
the flame
[1]
the
whether
a
water
[1]
room.
thermistor
temperature
year.
energy
brake
in
blocks
heat
capacity
of
the
brake
. Calculate:
energy
of
the
bicycle
and
rider
kinetic
in
the
energy
brake
converted
into
blocks
[1]
temperature
rise
of
the
brake
[3]
Describe
specific
an
is
of
a
[1]
a
electrical
heat
and
capacity
explain
are
used
method
of
water.
how
to
to
the
measure
Include
a
the
circuit
measurements
determine
the
in
specific
the
heat
[1]
burner
c
Discuss
the
[4]
capacity
8
of
energy
maximum
diagram
Bunsen
of
–1
K
kinetic
amount
experiment
a
mass
blocks.
constant-volume
the following:
of
the
the
[2]
thermometer.
State
thermal
of
a
13
7
into
During
[2]
c
6
total
specific
1180 J kg
thermal
platinum-resistance
rest.
energy
[4]
b
two
to
kinetic
[2]
a
Describe
comes
the
converted
blocks. The
–1
4
of
are
constant-volume
blocks
gas
rider
65%
. The
the
is
readings
of
[2]
the
c
speed
suitable
choice for
laboratory
over
a
[3]
of
water.
[8]
Revision questions
14
a
Define
the
terms
specific latent of fusion
and
16
specific latent heat of vaporisation.
b
25 g
of
ice
at
containing
–12 °C
180 g
of
is
added
water
at
to
a
[3]
polystyrene
cup
26 °C.
Explain
7
the following:
a
Sweating
b
The
helps
an
temperature
takes
athlete
of
a
cool
liquid
down.
drops
as
[3]
evaporation
place.
[3]
Calculate:
17
i
the
amount
of
thermal
energy
required
Describe
heat
convert
ii
the
ice
to
water
the final
temperature
[specific
heat
3
4.2
×
10
of
capacity
at
0 °C
the
of
[3]
=
specific
latent
5
15
a
Explain
and
b
10
one
c
State
capacity
heat
two
specific
latent
[6]
Describe
of
an
experiment
vaporisation
of
to find
water.
the
specific
latent
[6]
of
ice
=
of fusion
2.
1
of
×
ice
10
–1
J kg K
,
=
is
.]
meant
by
the
terms
evaporation
[2]
similarity
and
between
and
the
processes
of
boiling.
differences
evaporation
the
–1
J kg
what
evaporation
to find
ice.
,
boiling.
State
18
heat
3
heat
×
of
–1
J kg K
specific
3.3
experiment
of fusion
[3]
water.
water
an
to
between
boiling.
[1]
the
processes
of
[2]
133
16
Ideal
16.
1
Ideal
gases
and
Charles’
A
completion
of
this
section,
be
able
fixed
state
the
gas
mass
use
the
a
of
gas
graph
is
of
heated
at
volume
constant
against
pressure.
temperature
For
is
each
temperature
plotted.
the
result
of
such
an
experiment.
When
the
graph
is
Figure
16.1.1
extrapolated,
it
laws
cuts
ᔢ
law
to:
shows
ᔢ
theory
you
measured,
should
kinetic
gases
Learning outcomes
On
the
equation
of
state for
the
temperature
axis
at
– 273.15 °C.
an
3
V/m
ideal
ᔢ
gas
understand
absolute
the
concept
of
zero.
Definition
The
volume
directly
of
a
fixed
proportional
mass
to
its
of
gas
is
absolute
 273.
15
temperature,
pressure
is
provided
kept
that
temperature /°C
the
constant.
Figure 16.1.1
Charles’ law
V
Equation
V
1
Charles’
law
can
be
used
in
the
form
2
=
T
T
1
V
∝
V
=
2
T
Boyle’s
law
kT
A
fixed
mass
of
gas
is
compressed
at
a
constant
temperature.
For
3
V
–
volume
of
gas/m
each
T
–
absolute
temperature/K
k
–
proportionality
pressure,
experiment
the
are
volume
shown
in
of
the
Figure
gas
is
recorded.
The
results
of
this
16.1.2.
constant
p/Pa
p/Pa
Definition
The
pressure
of
a fixed
inversely
proportional
provided
that
kept
the
mass
to
its
of
gas
is
volume,
temperature
is
constant.
1
3
3
/m
V/m
V
Figure 16.1.2
Equation
Boyle’s law
1
p
∝
Boyle’s
law
can
be
used
in
the
form
pV
=
p
V
1
V
=
p
1
V
2
2
k
3
V
–
volume
of
p
–
pressure
k
–
proportionality
Pressure
law
gas/m
of
the
gas/Pa
constant
A
fixed
the
are
mass
of
gas
temperature
shown
in
temperature
of
is
Figure
axis
heated
the
at
gas
is
16.1.3.
at
constant
recorded.
When
the
volume.
The
graph
is
p
1
pressure
law
can
be
used
in
the
form
T
1
134
2
=
T
of
each
this
pressure,
experiment
extrapolated,
– 273.15 °C.
p
The
For
results
2
it
cuts
the
Chapter
p/Pa
16
Ideal
gases
and
the
kinetic
theory
Definition
The
pressure
directly
of
proportional
temperature,
volume
a fixed
is
mass
to
provided
kept
its
of
gas
is
absolute
that
the
constant.
Equation
 273.
15
Figure 16.1.3
zero
The
experimental
that
when
the
temperature
data
graphs
temperature
is
The
SI
from
are
It
is
unit
equation of
Charles’
law
extrapolated,
called
possible.
temperature.
The
p
∝
p
=
T
The pressure law
Absolute
This
temperature /°C
absolute
the
of
zero
and
they
zero .
of
the
temperature
the
pass
It
is
Pressure
through
believed
absolute
on
this
or
law
is
T
–
absolute
p
–
pressure
temperature/K
k
–
proportionality
of
the
gas/Pa
constant
show
– 273.15 °C.
to
be
Kelvin
scale
kT
the
the
lowest
scale
of
kelvin
(K).
state
Equation
All
the
gas
equation
laws
of
can
state.
be
The
combined
constant
into
in
one
this
equation
equation
–1
molar
obeys
gas
the
constant.
gas
It
is
equal
to
8.31 J mol
is
known
R
and
is
as
the
called
the
pV
=
nRT
–1
K
.
An
ideal
gas
is
one
that
n
–
number
p
–
pressure/Pa
of
moles
V
–
volume/m
T
–
absolute
R
–
molar
of
gas
laws.
3
Example
An
to
ideal
gas
prevent
is
contained
heat
losses.
A
in
a
cylinder
piston
is
which
attached
is
to
surrounded
one
end
of
by
the
temperature/K
insulation
gas
constant
cylinder
.
–1
–4
The
initial
volume
of
the
gas
is
1.3
×
10
(8.31 J mol
3
m
.
The
pressure
–1
K
)
and
5
temperature
of
i
Calculate
ii
The
the
the
gas
are
number
1.2
of
×
10
moles
Pa
of
and
gas
310 K
in
the
respectively.
cylinder
.
Key points
piston
is
–5
1.9
×
10
moved
to
compress
the
gas.
The
volume
decreases
to
3
m
.
The
temperature
increases
to
650 K.
Calculate
the
new
ᔢ
Charles’
law
states that the
pressure.
volume of
is directly
i
Using
the
equation
of
state
pV
=
a fixed
mass of
proportional to
pV
number
of
moles
of
gas
in
the
cylinder
n
its
nRT
absolute temperature,
The
gas
that the
pressure
is
provided
kept
constant.
=
RT
ᔢ
5
1.2
×
Boyle’s
law
states
that
the
–4
10
×
1.3
×
10
pressure
=
8.31
×
of
a fixed
mass
of
gas
310
is
inversely
proportional
to
–3
=
6.06
×
10
mols
its
ii
Using
the
equation
of
state
pV
=
nRT,
n
is
unchanged.
ᔢ
nRT
The
new
pressure
of
the
gas
is
p
volume,
provided
temperature
is
The
law
Pressure
kept
that
the
constant.
states
that
the
=
V
pressure
–3
6.06
×
10
×
8.31
×
is
650
of
directly
a fixed
mass
proportional
of
to
gas
its
=
–5
1.9
×
absolute
10
that
the
temperature,
volume
is
provided
kept
constant.
6
=
1.72
×
10
Pa
ᔢ
An
gas
ideal
gas
is
one
that
obeys
the
laws.
135
16.2
The
kinetic
Learning outcomes
On
completion
should
ᔢ
be
state
able
the
kinetic
ᔢ
use
ᔢ
a
speed
section,
of
kinetic
the
and
The
you
terms
derive
the
Assumptions
the
to
explain
molecules
a
ᔢ
use
pV
=
by
equation for
deduce
a
the
high
that
a
gas
gases
is
made
up
of
many
small
speeds.
theory
molecules
that
The
collisions
ᔢ
The
intermolecular
are
make
elastic
up
the
(kinetic
gas
are
energy
is
similar
(same
masses).
conserved).
The
ᔢ
the
volume
The
ᔢ
A
of
time
large
of
forces
during
the
occupied
between
the
except
gas
by
molecules
the
collisions
between
the
molecules
of
the
gas
are
collisions.
is
negligible
when
compared
to
the
gas.
is
negligible
when
compared
with
the
time
collisions.
number
of
molecules
analysis
can
be
are
present;
therefore
the
rules
of
applied.
–2
translational
ᔢ
the
statistical
Nmc
monatomic
assumes
at
kinetic
ᔢ
ᔢ
to
deduce
3
the
the
All
negligible,
gas
1
gases
randomly
ᔢ
mean square
equation for
of
of
of
kinetic theory of
pressure
root mean square
exerted
theory
moving
volume
pressure
gases
gases
theory
exerts
of
speed
ᔢ
kinetic
particles
assumptions
gas
define
this
of
Assumptions of the
to:
theory
the
why
of
theory
the
kinetic
gas
energy
The
ᔢ
Newtonian
motion
of
the
molecules
kinetic
mechanics
is
random.
apply.
of
Using the
molecules
total
monatomic
ᔢ
average
energy
of
exerts
kinetic theory to
explain
why
a
gas
pressure
gas.
Suppose
walls
or
a
of
a
container
the
bourdon
exerts
a
gauge.
pressure
gas
are
the
container
.
they
is
container
.
moving
undergo
The
on
the
around
When
a
filled
This
kinetic
inner
in
the
change
with
a
a
of
random
The
can
theory
walls
molecules
in
gas.
pressure
be
can
the
exerts
be
used
manner
because
the
to
Newton’s
second
law
of
motion
(
F
why
the
of
direction
–
over
the
exert
surface
a
force
area
on
of
the
the
walls
inner
of
the
walls
of
the
of
of
changes.
Δp
)
=
,
the
Δt
This
container
.
gas
the
container
,
mu
=
container
.
the
the
walls
the
Δt
molecules
on
manometer
molecules
with
walls
their
a
explain
mv
According
pressure
The
colliding
with
a
using
to
container
.
collide
momentum,
gas
measured
force
As
a
is
exerted
result
a
F
pressure
is
exerted
on
it
(
p
)
=
A
If
the
of
temperature
the
container
the
temperature
the
average
with
of
of
a
change
greater
pressure
the
gas
the
energy
walls
on
exerted
the
on
is
walls
the
increased,
kinetic
When
of
of
momentum
force
is
The
increases.
kinetic
collisions
of
increases.
the
the
container
inner
the
can
occur
greater
of
the
be
of
and
exerted
used
the
increases.
container
.
walls
pressure
temperature
molecules
much
of
the
the
theory
more
on
the
force
the
explain
is
walls
why
increased,
means
frequently.
greater
container
gas
This
therefore
A
to
that
The
molecules
means
the
rate
exert
that
the
increases.
Mean square speed and root mean square (r.m.s.) speed
Suppose
c
,
1
If
you
value,
136
c
a
,
gas
c
2
,
…,
made
up
of
N
atoms
having
speeds
as
follows:
c
3
were
you
is
N
to
determine
would
have
the
square
calculated
of
the
each
mean
speed
and
square
find
speed
the
of
mean
the
atoms
Chapter
16
Ideal
gases
and
the
kinetic
theory
2
that
make
up
the
gas.
Mean
square
speed
can
be
represented
as
either
c
Equation
2
or
⟨c
⟩
Mean
The
root
mean
square
(r
.m.s.)
speed
is
the
square
root
of
the
square
speed
2
2
c
square
=
mean
speed.
+ c
1
2
⟨c
2
+ c
2
2
+ … + c
3
N
=
⟩
N
Example
–1
Five
gas
molecules
have
–1
speeds
of
100 m s
–1
,
– 250 m s
–1
,
300 m s
,
Equation
–1
275 m s
and
– 200 m s
Calculate
Root
mean
square
2
c
a
The
mean
square
speed
of
the
gas
2
molecules.
⟨c
√
⟩
speed
2
+ c
1
=
2
2
+ c
2
+ … + c
3
N
=
N
b
The
root
c
The
mean
a
⟨c
mean
square
speed
of
speed
the
2
of
+
gas
2
(– 250)
molecules.
molecules.
2
100
the
+
2
300
+
2
275
+
(– 200)
2
4
⟩
=
=
5.56
×
10
–1
m s
5
z
2
2
100
+
2
(– 250)
+
300
2
+
275
2
+
(– 200)
2
⟨c
b
–1
⟩
√
=
=
236 m s
5
100
+
(– 250)
+
300
+
B
275
+
(– 200)
A
–1
c
c
=
=
45 m s
5
c
l
Derivation of the
pressure
exerted
by
a
gas
m
c
x
Consider
a
cubic
container
of
side
l,
containing
N
gas
molecules,
each
of
y
mass
m
(Figure
16.2.1).
C
If
you
were
randomly.
to
consider
Suppose
at
a
single
one
molecule,
instant
in
it
time
would
its
be
velocity
moving
is
c
about
(Figure
x
16.2.2).
O
D
This
velocity
can
be
resolved
into
three
components.
(O
,
O
x
,
O
y
).
The
z
Figure 16.2.1
molecule
has
velocity
components
as
Deriving the pressure
follows:
exerted by a gas
In
the
Ox
direction
–
c
x
In
the
Oy
direction
–
c
c
z
y
c
In
the
Oz
direction
–
c
z
2
Also,
c
2
=
2
c
+
c
x
2
+
c
y
2
c
z
2
=
c
2
+
x
c
2
+
c
y
z
c
y
Assuming
strikes
that
the
the
face
molecule
that
was
chosen
moves
back
and
forth
and
ABCD.
c
x
The
distance
travelled
by
the
molecule
=
2l
Figure 16.2.2
Resolving the velocity into
components along the three axes
distance
The
time
between
collisions
with
the
face
ABCD
2l
=
=
speed
c
x
Momentum
of
molecule
just
before
hitting
face
ABCD
=
mc
x
Momentum
of
molecule
just
after
hitting
face
ABCD
=
– mc
x
Change
in
momentum
=
(– mc
)
–
mc
x
=
– 2mc
x
x
2
2mc
Δp
mc
x
Rate
of
change
of
momentum
=
=
x
=
Δt
l
2l
(
)
c
x
(The
the
–
sign
face
is
removed
because
we
are
interested
in
the
force
acting
on
ABCD).
2
mc
x
Therefore,
the
force
exerted
on
the
face
ABCD
is
l
137
Chapter
16
Ideal
gases
and
the
kinetic
theory
2
mc
x
(
force
The
pressure
exerted
on
the
face
ABCD
=
)
l
2
mc
x
=
=
3
2
area
l
l
3
But
the
volume
of
the
cube
V
=
l
2
mc
x
(
force
The
pressure
exerted
on
the
face
ABCD
=
)
l
2
mc
x
=
=
2
area
This
is
the
pressure
exerted
by
only
one
molecule,
V
l
acting
on
the
face
ABCD.
Since
there
are
N
molecules,
all
of
them
must
be
considered.
2
mc
x
1
Pressure
(exerted
by
first
molecule)
p
=
1
V
2
mc
x
2
Pressure
(exerted
by
second
molecule)
p
=
2
V
2
mc
x
3
Pressure
(exerted
by
third
molecule)
p
=
3
V
2
mc
x
N
Pressure
(exerted
by
Nth
molecule)
p
=
N
V
T
otal
pressure
acting
of
face
ABCD
=
p
+
p
1
+p
2
2
+
…
mc
2
mc
x
2
mc
x
1
mc
x
2
x
3
+
N
+
V
p
N
2
=
+
3
+
V
…
+
V
V
m
(
=
V
This
can
be
written
in
terms
of
mean
2
2
c
+
2
c
x
+
square
c
mean
square
speed
be
⟨c
c
x
N
)
speeds.
2
+
c
2
+
c
x
1
⟩
+
3
x
2
the
2
…
x
2
2
Let
c
x
1
2
+
…
+
c
x
2
x
3
N
=
N
2
Nm⟨c
⟩
x
Therefore
p
=
V
Since
the
motion
of
the
molecules
is
2
equation
for
p
in
terms
of
⟨c
random,
we
or
⟨c
y
2
2
⟩
Nm⟨c
=
2
⟨c
=
⟨c
But,
c
=
+
c
x
2
Therefore,
⟨c
⟨c
⟩
⟨c
2
⟩
+
⟨c
⟩
⟩
y
2
⟨c
c
z
x
Therefore,
2
+
y
2
=
⟩
2
c
Equation
V
z
2
=
⟩
z
2
⟩
y
2
Nm⟨c
=
V
2
⟩
x
2
⟩
y
=
V
Therefore,
obtained
z
x
p
have
⟩
Nm⟨c
So
could
2
⟩
2
+
⟨c
⟩
z
2
=
3⟨c
⟩
x
1
2
pV
Nm
=
⟨
c
⟩
1
2
3
Or
⟨c
⟩
2
=
⟨c
⟩
x
3
p
–
pressure
V
–
volume
of
the
gas/Pa
1 Nm
2
Therefore,
3
of
the
pressure
of
the
gas
now
becomes
p
=
⟨c
gas/m
3
N
–
number
of
m
–
mass
one
molecules
–
mean
mass
of
V
molecule/kg
But,
density
ρ
=
Nm
=
volume
V
2
⟨
c
⟩
square
speed
of
the
1
2
molecules
Therefore,
the
pressure
is
p
=
ρ ⟨c
3
138
⟩
⟩
the
Chapter
Total
kinetic
energy of
a
monatomic
16
Ideal
gases
and
the
kinetic
theory
gas
1 Nm
2
The
equation
p
=
⟨c
to
several
microscopic
molecules
equation
and
with
the
relates
properties
speed
the
⟩
a
macroscopic
property
(pressure)
V
3
of
the
equation
of
(number
of
molecules).
state
for
an
molecules,
Let
us
ideal
now
gas
mass
of
compare
( pV
=
the
this
nRT).
1
2
pV
=
Nm ⟨c
⟩
(1)
3
pV
=
nRT
(2)
Definition
By
comparing
equations
(1)
and
(2)
we
obtain
the
following.
Kinetic
1
energy
of
all
the
gas
2
Nm ⟨c
⟩
=
nRT
molecules
3
3
1
3
2
Multiply
both
sides
of
the
equation
by
=
Nm
2
3
1
2
Nm ⟨c
⟩
)
=
c
=
⟩
nRT
2
3
2
(
⟨
2
m
–
mass
of
N
–
total
n
–
number
–
mean
gas
molecule/kg
nRT
3
2
1
number
of
of
gas
moles
of
molecules
gas
3
(
2
Nm ⟨c
⟩
)
2
=
nRT
2
⟨
c
⟩
square
speed
of
the
2
molecules
3
Therefore,
the
total
kinetic
energy
of
all
the
gas
molecules
is
nRT
R
–
molar
gas
constant
2
–1
(8.31 J mol
Therefore,
the
kinetic
energy
of
the
gas
molecules
is
proportional
to
T
absolute
temperature
of
the
gas
(E
∝
–1
K
)
the
–
absolute
temperature/K
T).
k
In
order
you
to
divide
determine
the
the
average
expression
for
the
kinetic
total
energy
kinetic
of
a
single
energy
by
molecule,
the
number
of
Definition
molecules
present.
1
T
otal
kinetic
energy
of
all
the
molecules
3
(
=
2
Nm ⟨c
⟩
)
=
2
1
Kinetic
energy
of
a
single
molecule
(
=
2
m ⟨c
⟩
)
N
=
nN
,
where
N
is
the
total
number
of
molecules,
n
kinetic
energy
of
a
gas
molecule
3
nRT
2
N
1
3
2
=
m
2
But
Translational
nRT
2
is
the
⟨
c
⟩
number
A
kT
=
2
2
m
–
mass
–
mean
of
gas
molecule/kg
23
of
moles
and
N
is
the
Avogadro’s
constant
(6.02
×
10
).
2
⟨
A
c
⟩
k
speed
of
the
molecules
R
And
square
=
,
where
k
is
known
as
the
Boltzmann
constant.
k
N
–
Boltzmann
constant
A
–23
(1.38
3
Therefore,
the
kinetic
energy
of
a
single
molecule
=
×
10
–1
J K
)
kT
T
2
–
absolute
temperature/K
Example
A
balloon
contains
a
the
number
b
the
average
c
the
total
of
0.60 mol
helium
of
helium
atoms
in
the
at
310 K.
Key points
Calculate:
balloon
ᔢ
kinetic
kinetic
energy
energy
of
of
a
the
helium
helium
atom
atoms
in
the
in
the
balloon
[Boltzmann
constant
=
1.38
×
10
kinetic
theory
has
several
can
be
assumptions.
balloon.
ᔢ
–23
The
The
kinetic
theory
used
to
–1
J K
explain
why
a
gas
exerts
pressure.
23
Avogadro
constant
=
6.02
×
10
]
ᔢ
The
average
energy
23
a
Number
of
atoms
=
N
=
nN
=
0.60
×
6.02
×
10
of
a
translational
monatomic
kinetic
gas
23
=
3.612
×
10
A
3
3
molecule
3
is
given
by
–23
b
Average
kinetic
energy
=
kT
=
×
2
1.38
×
10
×
kT
2
310 J
2
ᔢ
The
total
kinetic
energy
of
all
given
by
the
–21
=
6.42
×
10
J
molecules
23
c
T
otal
kinetic
energy
=
3.612
×
10
–21
×
6.42
×
10
in
a
gas
is
3
=
2.32
×
10
J
3
nRT
2
139
17
The first
17
.
1
The first
law
Learning outcomes
On
completion
should
ᔢ
be
state
able
of
this
law
law
thermodynamics
of
thermodynamics
The first
section,
you
to:
the first
of
The
first
law of thermodynamics
law
of
energy.
to
the
It
of
ther modynamics
states
energy
that
supplied
the
to
is
change
the
simply
in
system
the
internal
plus
the
principle
energy
work
of
a
done
of
conservation
system
on
the
is
equal
system.
of
The
thermodynamics
first
law
of
thermodynamics
can
be
expressed
mathematically
as
follows:
ᔢ
derive
done
an
by
expression for
a
the
work
gas.
Equation
The first
Definition
The first
law
of
ΔU
=
ΔU
–
law
ΔQ
that
the
change
energy
of
energy
supplied
the
work
a
system
done
to
on
is
the
the
in
ΔW
change
in
internal
energy
of
system/J
internal
equal
to
system
ΔQ
–
energy supplied to the system/J
ΔW
–
work
the
done
on
the
system/J
plus
system.
It
is
important
equation
to
understand
the
sign
convention
used
when
applying
the
above.
ΔU
is
positive
ΔU
is
negative
ΔQ
is
positive
ΔQ
is
negative
(+)
(–)
is
positive
ΔW
is
negative
when
(–)
when
by
a
is
energy
when
energy
internal
energy
when
(+)
internal
the
when
(–)
work done
the
when
(+)
ΔW
The
gas
thermodynamics
thermodynamics
the
states
+
of
work
is
is
work
is
is
energy
being
being
being
being
is
increasing.
decreasing.
supplied
to
removed
done
on
done
the
from
the
by
the
system.
the
system.
system.
system.
gas
inlet
Consider
(Figure
movable
piston
traps
a
gas
cylinder
17.1.1).
the
gas
The
inside
that
has
a
movable
cross-sectional
the
cylinder
.
area
As
long
of
as
piston
the
attached
piston
there
is
a
is
to
A.
one
The
pressure
end
piston
difference
valve
between
the
pressure,
gas
the
contained
piston
will
in
the
begin
cylinder
moving
and
freely.
the
external
Now,
atmospheric
suppose
a
gas
of
atmospheric
volume
V
is
inside
the
cylinder
.
gas
is
The
piston
will
move
to
the
right
until
pressure
gas
at
pressure
p
the
and
volume
pressure
of
the
equal
to
atmospheric
pressure
p
V
If
the
valve
is
opened
to
allow
more
gas
to
enter
the
cylinder
,
the
pressure
Figure 17.1.1
inside
the
because
more
Δ x
cylinder
more
collisions
piston
will
F
external
per
begin
atmosphere
will
begin
molecules
second
to
with
a
pressures
of
move
force
are
the
increase
are
acting
to
F.
to
gas
the
The
same.
being
on
the
right.
piston
The
(Figure
added
inner
The
gas
stops
piston
17.1.2).
and
walls
of
slowly
moving
moves
It
there
increases
are
the
going
pushes
when
to
cylinder
.
back
the
through
a
be
The
the
internal
distance
and
x
atmospheric
pressure
gas
and
at
pressure
volume
The
work
done
by
the
gas
against
ΔW
=
force
=
F
×
Figure 17.1.2
140
atmospheric
pressure
is
given
by
p
V + ΔV
×
Δx
distance
moved
in
the
direction
of
the
force
Chapter
17
The first
law
of
thermodynamics
force
But
pressure
=
area
∴
F
Work
=
done
p
by
×
A
the
gas
against
But
the
atmospheric
change
in
pressure
volume
=
ΔV
p
=
×
A
A
×
×
Δx
Equation
Δx
ΔW
Work
When
done
work
is
by
the
being
gas
against
done
by
atmospheric
the
gas,
ΔW
is
pressure
ΔW
=
pΔV
=
pΔV
ΔW
–
work
p
–
pressure/Pa
done
/J
ΔV
–
change
negative.
3
The
In
same
this
expression
case,
ΔW
is
can
be
used
when
the
gas
is
being
in
volume/m
compressed.
positive.
Applying the first
law of thermodynamics
Example
A
fixed
mass
of
an
ideal
gas
absorbs
1500 J
of
heat
energy
and
expands
4
under
a
constant
pressure
of
–2
to
a
the
volume
of
4.2
×
2.5
×
10
–2
Pa
from
a
of
2.1
×
10
3
m
3
10
m
.
Calculate
the
change
in
internal
energy
of
gas.
Heat
Since
supplied
the
Work
to
pressure
done
by
the
gas
remains
gas
ΔW
is
ΔQ
=
1500 J
constant,
=
pΔV
=
2.5
=
420 J
4
Using
the
first
Change
Note
of
volume
that
in
law
of
work
–2
(4.2
10
×
–2
10
–
2.1
×
10
)
thermodynamics
internal
since
×
energy
was
of
done
gas
by
ΔU
the
=
ΔQ
+
=
1500
=
1080 J
gas
a
ΔW
+
minus
(– 420)
sign
is
needed
in
front
ΔW
Example
An
ideal
gas
is
contained
in
a
cylinder
which
is
surrounded
by
insulation
–4
to
prevent
heat
losses.
The
initial
volume
of
the
gas
is
2.5
×
10
3
m
.
The
5
pressure
and
i
Calculate
ii
The
temperature
the
piston
×
new
10
number
is
–5
2.1
of
moved
of
to
the
gas
moles
is
of
compress
1.1
gas
the
×
in
10
the
gas.
Pa
and
320 K
respectively.
cylinder
.
The
volume
decreases
to
3
m
pressure
.
The
of
temperature
the
gas
and
increases
explain
why
to
750 K.
the
Calculate
temperature
of
the
the
gas
increases.
ii
The
work
internal
i
Using
done
energy
the
on
of
the
the
equation
of
gas
is
86 J.
Determine
the
increase
in
the
gas.
state
pV
=
nRT
pV
The
number
of
moles
of
gas
in
the
cylinder
n
=
RT
5
1.1
×
10
–4
×
2.5
×
10
=
8.31
×
320
–2
=
1.03×
10
mol
141
Chapter
17
The first
law
of
thermodynamics
ii
Using
the
equation
of
state,
pV
=
nRT
is
unchanged.
nRT
The
new
pressure
of
the
gas
is
p
=
V
–2
1.03
×
10
×
8.31
×
750
=
–5
2.1
×
10
6
=
The
molecules
When
a
the
greater
the
to
molecules
mean
ΔW
=
gas
the
moving
with
means
increase.
of
are
collide
This
kinetic
temperature
iii
the
velocity.
molecules
the
of
Since
energy
the
of
all
compressed
about
the
that
3.06×
piston,
average
temperature
its
Pa
randomly
moving
the
10
at
they
kinetic
of
molecules,
a
velocities.
rebound
energy
gas
there
high
is
is
of
with
all
proportional
an
increase
in
gas.
+86 J
Energy
supplied
to
the
system
ΔQ
=
0 J
since
the
cylinder
is
insulated.
Using
The
the
first
change
in
law
of
thermodynamics
internal
energy
of
the
gas
ΔU
=
ΔQ
+
=
0
+
=
+86 J
ΔW
86
Example
A
gas
movable
cylinder
is
fitted
with
a
piston
which
can
move
without
friction
piston
contains
0.055 mols
of
an
ideal
gas
at
a
temperature
of
300 K
and
a
5
pressure
of
1
×
10
Pa
(Figure
17.1.3).
Calculate:
insulated
i
the
volume
ii
the
internal
of
the
gas
energy
of
the
gas.
cylinder
Suppose
the
temperature
of
the
gas
is
increased
to
360 K
and
the
pressure
Figure 17.1.3
is
kept
constant.
Calculate:
iii
the
change
iv
the
external
v
the
total
in
internal
work
amount
energy
done
of
by
of
the
energy
the
gas
supplied
–1
[molar
i
gas
Using
constant
the
=
equation
8.31 J mol
of
gas
state
pV
nRT
to
the
gas.
–1
K
=
]
nRT
0.055
×
8.31
×
300
–3
V
olume
of
gas
V
=
=
=
1.37
×
10
3
m
5
p
ii
The
the
of
an
internal
gas.
all
The
the
ideal
energy
an
temperature
molecules
gas
of
1
is
given
that
ideal
of
gas
the
make
is
gas
up
10
dependent
is
the
×
on
dependent
gas.
The
the
on
total
temperature
the
kinetic
kinetic
energy
of
by
3
E
=
nRT
K
2
3
∴
U
=
nRT
2
3
Internal
energy
U
=
142
3
nRT
2
=
×
2
0.055
×
8.31
×
300
=
of
energy
206 J
Chapter
iii
Change
in
temperature
Change
in
internal
of
the
gas
=
360
–
300
=
17
The first
law
of
thermodynamics
60 K
3
energy
ΔU
=
nRΔT
2
3
=
×
0.055
×
8.31
×
60
=
41 J
2
iv
External
The
new
work
done
volume
by
of
the
the
gas
gas
is
given
must
nRT
first
0.055
×
by
be
ΔW
=
pΔV
determined.
8.31
×
360
–3
V
=
=
=
1.65
×
10
3
m
5
p
External
work
done
by
1
the
gas
ΔW
×
10
=
pΔV
5
=
v
Using
Since
T
otal
the
first
work
law
was
energy
of
done
supplied
1
×
10
–3
(1.65
thermodynamics
by
the
gas
to
the
gas
ΔQ
=
×
–3
10
–
1.37
ΔU
=
ΔQ
ΔW
=
– 28 J
ΔU
–
ΔW
+
=
×
10
)
=
28 J
(– 28)
=
69 J
ΔW
41
–
Example
5
At
a
temperature
of
100 °C
and
pressure
of
1.01×
10
Pa,
1 kg
of
steam
3
occupies
1.67 m
.
At
occupies
1.04
10
If
water
the
–3
1 kg
of
×
at
same
temperature
and
pressure,
1 kg
of
water
3
m
100 °C
is
converted
to
1 kg
of
steam
at100 °C.
Calculate:
i
the
energy
ii
the
work
iii
the
increase
in
internal
iv
the
increase
in
potential
supplied
done
against
6
[l
=
2.26
×
10
to
produce
the
this
change
atmosphere
energy
energy
of
the
molecules
of
water
.
–1
J kg
]
v
i
Energy
supplied
to
produce
change
Q
=
ml
v
6
=
1
×
2.26
=
2.26
×
10
6
ii
Work
done
against
the
×
10
J
atmosphere
ΔW
=
pΔV
=
1.01
×
10
=
1.69
×
10
Key points
5
3
(1.67
–
1.04
×
10
5
)
ᔢ
J
The first
states
iii
Increase
in
internal
energy
ΔU
=
ΔQ
+
6
2.26
of
thermodynamics
change
in
internal
ΔW
energy
=
law
the
×
10
of
a
system
is
equal
to
5
+
(–1.69
×
10
)
the
energy
supplied
to
it
plus
the
6
=
2.09
×
10
J
work
iv
At
a
change
of
state,
there
is
no
change
in
temperature.
All
ᔢ
energy
supplied
is
used
to
increase
the
potential
energy
of
done
and
not
their
kinetic
The
the
increase
in
system.
work
done
by
pressure
a
gas
is
at
equal
to
energies.
the
Therefore,
the
the
constant
molecules
on
the
potential
energy
of
the
pressure
multiplied
by
the
molecules
change
in
volume
of
the
gas.
6
=
2.09
×
10
J.
143
17
.2
Molar
heat
Learning outcomes
On
completion
should
be
able
of
this
capacities
Molar
section,
you
The
define
the
molar
required
to:
In
ᔢ
heat
term
the
case
p–V
diagrams
capacities
heat
to
and
capacity
increase
the
of
large
a
gas,
of
a
substance
temperature
changes
of
in
is
the
one
amount
mole
pressure
of
and
it
of
by
thermal
one
volume
energy
degree.
occur
when
molar heat
supplied
with
thermal
energy.
The
change
in
volume
and
pressure
when
capacity
a
ᔢ
explain
why
capacity
is
at
greater
capacity
a
the
constant
than
at
molar
the
or
defined
heat
liquid
under
is
heated
two
is
negligible.
particular
The
heat
capacity
of
a
gas
can
be
conditions.
pressure
molar
constant
solid
ᔢ
Molar
heat
capacity
at
constant
pressure.
ᔢ
Molar
heat
capacity
at
constant
volume.
heat
volume for
gas
The
molar
heat
capacity
of
a
gas
at
constant
pressure
C
,
is
the
amount
p
of
ᔢ
perform
calculations
using
energy
required
to
raise
the
temperature
of
one
mole
of
a
gas
by
one
p –V
degree,
when
the
pressure
remains
constant.
diagrams.
The
molar
heat
capacity
at
constant
volume
C
,
is
the
amount
of
v
energy
required
degree
The
movable
when
molar
to
the
heat
raise
the
volume
temperature
remains
capacity
at
of
one
mole
of
a
gas
by
one
constant.
constant
pressure
is
greater
than
the
molar
piston
heat
capacity
at
constant
volume.
heat
Consider
then
gas
atmospheric
pressure
gas
must
external
gas
contained
to
the
expand
work
potential
Figure 17.2.1
a
supplied
and
gas.
and
by
a
cylinder
order
move
moving
kinetic
in
In
the
the
energy
for
piston.
piston.
of
at
the
a
the
In
this
Also,
gas
pressure
pressure
p.
to
Heat
process,
the
heat
molecules
energy
remain
the
gas
supplied
(Figure
is
constant,
has
to
the
do
increases
the
17.2.1).
Heating a gas at constant
pressure
Now
consider
When
fixed
volume
the
the
gas
is
its
temperature.
no
external
same
gas
heated,
Since,
contained
all
the
the
in
energy
volume
of
a
cylinder
supplied
the
to
having
it
container
is
is
a
used
volume
to
fixed,
V.
increase
the
gas
does
V
heat
work.
temperature
of
a
This
gas
means
when
its
that
less
volume
energy
is
fixed.
is
It
required
follows
to
raise
that
C
the
>
C
p
Molar
heat
capacity
at
constant
pressure
and
constant
volume
v
are
related
gas
as
Figure 17.2.2
follows
(Figure
17.2.2):
Heating a gas at constant
–1
C
Equations
=
nC
H
=
R,
where
R
is
the
molar
gas
constant
(R
=
–1
8.31 J mol
K
).
p–V
indicator
diagrams
diagram
is
a
graph
showing
how
the
pressure
p
of
a
gas
varies
Δθ
nC
H
E
=
v
p
with
E
C
Using
An
E
–
p
volume
volume
V
during
a
change.
The
work
done
in
each
stage
can
be
Δθ
determined.
v
–
its
is
energy
Suppose
represented
by
the
a
gas
point
has
A.
a
pressure
The
gas
and
then
volume,
expands
such
while
that
its
its
state
pressure
H
n
–
number
of
remains
moles
by
C
–
molar
heat
capacity
the
fixed.
gas
is
Its
new
given
by
state
ΔW
is
=
represented
pΔV.
If
the
by
the
volume
point
of
the
B.
gas
The
is
work
kept
done
fixed
and
at
p
constant
C
–
molar
its
pressure
C.
No
is
increased,
the
new
state
would
be
represented
by
the
point
pressure
heat
capacity
work
is
done
by
the
gas
during
this
change.
Suppose
the
gas
is
now
at
v
compressed,
constant
represented
Δθ
–
change
while
keeping
the
pressure
fixed.
The
new
state
would
be
volume
in
temperature
of
by
the
point
D.
The
work
done
on
the
gas
is
given
by
ΔW
=
the
pΔV.
If
the
volume
of
the
gas
is
kept
fixed
and
its
pressure
is
decreased,
gas
the
new
the
gas
energy
state.
that
144
state
of
the
The
it
would
during
gas
gas
hasn’t
this
is
be
represented
change.
zero.
starts
off
changed
During
The
at
the
state.
by
internal
point
The
the
the
A
net
point
entire
energy
and
effect
A.
cycle,
of
a
all
work
the
gas
returns
of
No
to
the
is
is
change
done
in
dependent
the
point
changes
is
A,
by
internal
on
its
meaning
zero.
Chapter
17
The first
of
fixed
of
thermodynamics
pressure/Pa
Example
A
law
mass
pressure,
of
an
ideal
volume
gas
and
in
a
heat
pump
temperature
as
undergoes
shown
in
a
cycle
Figure
of
17.2.3
changes
B
(not
919 K
5
20 × 10
drawn
to
scale).
C
No
heat
is
internal
supplied
energy
of
to
the
the
gas
gas
is
from
as
A → B
and
C → D.
The
increase
in
5
11 × 10
505 K
follows
A
5
492 K
2.5 × 10
A → B
Using
the
1300 J
the
first
B → C
law
of
–1200 J
thermodynamics
C → D
and
the
data
– 450 J
supplied,
5
determine
1 × 10
190 K
D
following:
i
the
work
done
on
the
gas
from
A
ii
the
work
done
on
the
gas
from
B
iii
the
heat
iv
the
increase
v
the
heat
vi
the
number
to
B
and
C
C
and
D
to
D
-3
-3
0.42 × 10
to
to
1.8 × 10
A
3
volume/m
supplied
to
the
gas
from
B
to
C
Figure 17.2.3
First
i
ii
law
of
in
internal
supplied
of
to
the
moles
of
gas
gas
thermodynamics
of
the
from
D
being
ΔU
=
gas
to
from
D
to
A
A
used.
ΔQ
+
ΔW
From
A
to
B,
ΔW
=
ΔU
+
ΔQ
=
1300
–
0
=
1300 J
From
C
to
D,
ΔW
=
ΔU
+
ΔQ
=
– 450
–
0
=
– 450 J
From
B
ΔV
0.
=
to
C
there
This
is
no
means
∴
From
D
to
ΔV
0.
This
=
A
there
is
means
supplied
to
the
no
=
no
volume
is
done
of
the
on
or
gas,
by
therefore
the
gas.
0
no
=
gas
in
work
change
that
ΔW
Heat
change
that
ΔW
∴
iii
energy
in
work
volume
is
done
of
the
on
or
gas,
by
therefore
the
gas.
0
for
B
to
C,
ΔQ
=
ΔU
=
–
1200
ΔW
–
0
=
1200 J
Key points
iv
The
internal
energy
is
is
fixed,
energy
of
dependent
the
total
a
on
gas
is
dependent
temperature.
change
in
internal
on
Since
its
the
energy
of
state.
The
internal
temperature
the
gas
at
during
A
ᔢ
the
The
molar
heat
capacity
of
constant
pressure
C
gas
at
the
amount
a
,
is
p
complete
From
A
cycle
to
is
zero.
B,
ΔU
=
to
+1300 J
raise
mole
From
B
to
From
C
C,
ΔU
=
–1200 J
ΔU
=
– 450 J
of
when
to
D,
the
a
the
of
energy
required
temperature
gas
by
one
pressure
of
one
degree,
remains
constant.
From
D
to
A,
ΔU
=
?
ᔢ
All
these
internal
energy
values
should
add
up
to
zero,
because
The
molar
heat
capacity
at
ΔU
constant
volume
C
,
is
the
v
from
A
to
A
is
zero.
amount
From
D
to
A,
ΔU
=
0
–
(1300
+
–1200
+
– 450)
=
Heat
ΔQ
supplied
=
ΔU
–
to
ΔW
the
=
gas
350
for
–
D
0
to
=
energy
required
to
+350 J
raise
v
of
A,
the
mole
of
when
350 J
temperature
a
the
gas
by
one
volume
of
one
degree
remains
constant.
vi
Using
the
equation
of
state
pV
=
nRT
pV
The
number
of
moles
of
gas
in
the
cylinder
n
ᔢ
=
The
molar
heat
capacity
at
RT
constant
5
2.5
×
10
is
greater
–3
×
1.8
×
10
than
=
=
8.31
pressure
×
the
molar
heat
capacity
at
0.11 mols
492
constant
ᔢ
p–V
diagrams
illustrate
and
volume.
the
volume
can
be
used
changes
of
a
in
to
pressure
gas.
145
Revision
Answers
found
to
on
questions
the
that
require
questions
calculation
can
be
8
accompanying CD.
8
A
cylinder
without friction
gas
1
Explain
what
is
meant
an
ideal
b
absolute
at
a
with
and
a
piston
contains
temperature
by:
of
which
can
0.022 mols
280 K
and
a
move
of
an
pressure
ideal
of
5
1.6
a
is fitted
gas
×
10
Pa.
[1]
Calculate:
2
A
bubble
the
of
zero.
gas
[1]
rises from
surface. The
pressure
the
at
bottom
the
of
bottom
a
pond
of
the
to
pond
i
the
volume
ii
the
internal
Suppose
the
of
the
gas
energy
temperature
[2]
of
of
the
the
gas.
gas
[3]
is
increased
to
5
is
2.5
×
10
Pa
and
the
pressure
at
the
surface
is
330 K
and
the
pressure
is
kept
constant.
5
1.0
×
10
Pa. The
volume
of
the
bubble
at
the
bottom
Calculate:
3
of
the
pond
bubble
3
A
at
is
2.5 cm
the
cylinder
the
volume
of
the
surface.
contains
temperature,
. Calculate
1.8 mol
25 °C
iii
the
change
in
iv
the
external
v
the
total
internal
energy
of
the
gas
and
of
the
a
gas
at
pressure
room
inside
work
amount
done
of
by
the
energy
gas
[2]
supplied
is
3.6
when
×
an
10
Pa. The
additional
temperature
4.5 mol
of
gas
increases
is
[Molar
to
the
4
the
cylinder. Calculate
the
new
pressure
a
gas
is
Explain
what
is
of
a
gas
heated
the
is
to
V
at
330 K
a
temperature
at
constant
is
meant
by:
the
molar
heat
capacity
at
constant volume
the
molar
heat
capacity
at
constant
of
percentage
increase
in
pressure.
[2]
220 K.
Explain
why
the
molar
heat
capacity
at
constant
the volume
of
is
greater
than
the
molar
heat
capacity
the
constant volume for
a
gas.
[3]
[3]
c
a
Explain
b
Write
what
down
is
meant
the
by
an
equation
of
ideal
gas.
state for
[1]
an
ideal
the
symbols
ideal
gas
has
a volume
of
the
a
temperature
of
a
State four
1.9
×
3
10
m
gas
is
1.0
×10
300 K. The
b
The
two
[1]
assumptions
kinetic
pressure
of
the
kinetic
theory.
[4]
gas
is
p
theory
and
of
gases
the volume
can
V
of
show
an
that
ideal
the
gas
is
of
given
Pa. The
the
and
5
the
between
[3]
pressure
at
relationship
gas,
used.
–3
An
State
quantities.
10
explaining
is
]
pressure.
gas?
c
–1
K
ii
at
5
8.31 J mol
i
pressure
What
=
[5]
b
The
constant
inside
cylinder.
The volume
gas
pumped
9
into
the
[2]
–1
60 °C
to
gas.
the
5
cylinder
[2]
[3]
by
the
expression
compressed
1
–3
until
its volume
and
pressure
is
1.
1
×10
2
3
m
pV=
and
Nm
⟨
c
⟩
3
5
1.6
×
10
Pa
i
the
ii
the final
respectively. Calculate:
where
number
of
moles
of
gas
present
i
temperature
of
the
3
6
A
cylinder
of volume
6.5
×
10
gas.
at
a
pressure
of
1.8
×
10
m
contains
and
a
the
mass
of
a
gas
molecule.
ii
helium
temperature
what
the following
symbols
mean:
N
2
3
Pa
is
State
[3]
7
gas
m
[2]
c
and
⟨
Use
the
kinetic
of
[2]
⟩
expression
energy
temperature
300 K.
c
i
State
what
is
of
a
to
deduce
gas
the
molecule
mean
at
a
T.
[3]
meant
by
the
internal
energy
of
Calculate:
a
a
the
amount
b
the
mean
of
gas
in
the
container
ii
kinetic
energy
of
the
substance.
helium
atoms
Using
the
total
the
container.
kinetic
energy
of
the
helium
atoms
the
expression
the
change
mass
energy
and
of
an
ideal
expands
gas
under
absorbs
a
1100 J
constant
2.
1
×
10
–2
Pa from
a volume
–2
volume
of
Calculate
146
of
3.6
the
×
10
of
1.9
×
heat
pressure
4
of
internal
in
b
ii
energy
to
of
explain
an
gas
is
proportional
to
the
change
in
[3]
temperature
A fixed
in
in
ideal
7
derived
[2]
why
c
[2]
[2]
10
3
m
to
a
3
m
change
in
internal
energy
of
the
gas.
[5]
of
the
gas.
[2]
Revision questions
11
A fixed
as
mass
shown
of
an
ideal
gas
undergoes
a
8
cycle ABCA
below.

4
volume / 10
10
3
m
A
8
6
4
2
C
B
0
5
10
15
20
25
30
5
pressure / 1
0
a
State
when
b
the
it
Copy
and
energy
in
undergoes
Calculate
change A
c
change
the
to
the
work
energy
of
the
on
the
gas
[1]
during
the
B.
[2]
the
during
table
the
below
relating
the
cycle ABCA.
[6]
Work done
Energy
Change
on
supplied
internal
to
energy of
gas/J
gas/J
the
B
gas
cycle ABCA.
done
complete
changes
Change
internal
Pa
A
→
B
→ C
+650
C
→ A
+315
in
gas/J
-520
147
18
Thermal
18.
1
Thermal
energy
conduction
Thermal
Learning outcomes
one
On
completion
should
ᔢ
be
able
explain
of
this
section,
transfer
or
energy
more
of
can
the
and
be
convection
transferred
following
three
from
one
point
to
another
by
either
methods:
you
ᔢ
conduction
–
a
ᔢ
convection
–
ᔢ
radiation
no
medium
is
required.
to:
what
is
meant
by
a
medium
is
required.
thermal
–
medium
is
required.
Energy
can
be
transferred
in
a
conduction
vacuum.
ᔢ
describe
the
vibrations
process
and
of
electron
lattice
diffusion
Thermal
ᔢ
explain
and
use
the
concept
Thermal
thermal
explain
conduction
what
is
meant
by
thermal
region
any
net
of
high
of
one
vibrations
explain
a
process
how
land
and
sea
temperature
movement
because
convection
ᔢ
is
by
which
thermal
energy
flows
from
conductivity
a
ᔢ
conduction
of
of
and
of
the
two
to
a
region
material
itself.
mechanisms.
electron
of
low
temperature
Thermal
These
without
conduction
mechanisms
are
occurs
called
lattice
diffusion
breezes
are formed
Lattice vibrations
ᔢ
explain
how
ocean
currents
are
In
a
non-metal
such
as
ceramic
there
are
no
free
electrons.
If
some
hot
produced.
coffee
is
poured
ceramic
the
mug,
coffee
atoms
in
adjacent
their
walls
is
it
into
in
thermal
atoms
energy
wall
with
atoms
to
the
the
flows
in
mug,
are
the
as
is
inner
as
process
outside
called
the
well.
walls
These
of
lattice
of
until
the
heat
flows
mug,
outer
wall
room
temperature
vibrate,
collide
from
the
When
the
to
with
the
process
(Figure
In
bonds.
begin
then
This
hot.
by
inner
which
18.1.1).
at
gain
kinetic
energy
net flow
kinetic
of
thermal
energy
energy
transferred
Figure 18.1.1
strong
they
atoms
mug.
becomes
ceramic
As
vibrations
coffee
it
by
the
energy.
repeats
of
together
walls
gain
contact
hot
outside
bonded
they
vibrating
and
to
with
vibrate
begin
mug
inner
atoms
contact
begin
the
ceramic
adjacent
neighbours
of
a
to
Conduction by lattice vibrations
Electron diffusion
Metals
of
hot
spoon.
the
in
the
their
148
In
a
the
conductors
begin
energy
there
When
collide
the
energy.
fast-moving
process,
is
are
a
free
The
collide
this
energy
If
free
with
moving
in
the
Since
other
thermal
process
transfers
metal
electrons
atom.
This
a
transmitted
placed
process,
electrons.
thermal
heat.
electrons
is
vibrating
they
In
of
quickly
spoon
vibrating.
with
structure,
kinetic
the
good
thermal
metal,
metal
they
within
this
very
structure.
time
by
are
soup,
at
a
is
spoon
up
around
pot
of
gain
the
hot
is
called
is
the
energy
are
electron
rate
a
pot
the
within
atoms
every
moving
transfer
being
faster
in
of
randomly
soup,
kinetic
and
energy
placed
handle
electrons
atoms
much
the
some
of
transferred
diffusion.
than
In
lattice
Chapter
vibrations.
In
a
vibrations,
but
metal,
some
of
the
thermal
energy
is
transferred
by
18
Thermal
energy
transfer
lattice
Definition
are
good
this
occurs
conductors
of
to
heat
a
lesser
and
degree
electricity
than
electron
(Figure
diffusion.
Metals
18.1.2).
The
coefficient
conductivity
kinetic
energy
transferred
metal
atoms
in
hot
thermal
heat
per
unit
rate
area
of flow
per
unit
electrons
temperature
electrons
metal
of
the
fast-moving
of
to fast-moving
is
end
gradient,
heat flow
is
faces
thin
of
of
the
a
at
right
when
angles
parallel-sided
material,
under
the
to
the
slab
steady
state
soup
conditions.
metal
atoms
vibrate
gain
Figure 18.1.2
as
kinetic
net flow
of
thermal
energy
l
they
θ
energy
Thermal
2
conductivity
sample
Conduction
at
2
θ
1
rate
θ
Conduction by electron diffusion
θ
The
>
1
occurs
which
at
different
thermal
rates
energy
in
is
different
thermal
transferred
by
lagging
conductors.
thermal
conduction
Figure 18.1.3
Coefficient of thermal
conductivity
depends
on:
ᔢ
the
cross-sectional
ᔢ
the
material
ᔢ
the
temperature
from
area
which
the
conductor
is
made
Equation
Consider
a
slab
of
gradient
material
across
of
the
conductor
.
cross-sectional
area
A
and
length
l.
The
θ
Q
–
θ
1
temperatures
at
the
ends
of
the
conductor
are
maintained
at
θ
and
1
(θ
>
θ
1
).
This
is
referred
to
as
steady
state
=
θ
2
– kA
t
(
2
)
l
conditions.
2
Q
–1
The
sides
of
the
conductor
are
completely
lagged
θ
so
–
that
The
temperature
gradient
is
defined
are
no
heat
–
rate
of flow
–
coefficient
of
heat/W
or J s
t
θ
1
losses.
there
k
2
of
thermal
as
l
–1
conductivity/W m
–1
K
2
Q
The
rate
of
flow
of
heat
through
the
conductor
(
t
A
)
is
expressed
as
–
cross-sectional
area/m
follows:
θ
–
1
θ
2
–1
–
θ
Q
–
1
∝
equation
can
be
temperature
gradient/K m
l
2
A
t
The
θ
l
re-written
as
follows:
θ
Q
–
– kA
(
temperature
θ
1
=
t
2
)
l
θ
1
The
constant
of
conductivity .
transferred
(Figure
Good
proportionality
The
in
the
negative
direction
k,
sign
in
is
called
indicates
which
the
the
that
coefficient
thermal
temperature
is
of
ther mal
energy
is
being
decreasing
θ
18.1.3).
thermal
2
conductors
have
a
high
coefficient
of
thermal
distance
along
sample
conductivities.
–1
(e.g.
Poor
copper
=
thermal
390 W m
conductors
–1
(e.g.
glass
The
graph
conductor
=
0.8 W m
in
Figure
varies
–1
K
–1
silver
have
a
low
=
420 W m
coefficient
–1
K
,
18.1.4
with
,
of
–1
wood
shows
distance
=
0.15 W m
how
along
it
the
–1
K
)
thermal
K
)
temperature
when
conductivity.
–1
the
sides
across
are
sample
a
insulated.
Figure 18.1.4
temperature
gradient
is
constant.
with
lagging
The
Temperature difference
across a thermal conductor (lagging)
149
Chapter
18
Thermal
energy
transfer
The
temperature
graph
varies
θ
in
with
Figure
distance
18.1.5
along
shows
it
not
how
when
temperature
gradient
is
temperature
gradient
decreases
the
constant.
the
temperature
sides
Heat
are
not
escapes
across
insulated.
from
the
a
conductor
The
sides
and
the
1
as
the
distance
from
the
hotter
side
increases.
Example
θ
2
–5
distance
along
A
copper
rod
of
is
arranged
is
maintained
length
such
that
0.6 m
it
is
and
cross-sectional
completely
lagged
on
area
the
7.85
sides.
×
One
at
18.1.6).
100 °C
and
Calculate
the
the
other
rate
of
end
flow
is
of
maintained
thermal
at
rod
(thermal
conductivity
copper
Figure 18.1.5
end
of
copper
rod
=
0 °C
energy
–1
copper
without
2
m
sample
(Figure
sample
10
through
the
–1
390 W m
K
).
lagging
lagging
Temperature difference
across a thermal conductor (without
0 °C
100 °C
lagging)
0.6 m
Figure 18.1.6
Under
steady
state
conditions
θ
Q
–
θ
1
=
– kA
t
2
(
)
l
Q
100
–5
=
– 390
×
7.85
×
10
×
t
(
–
0
)
0.6
W
=
5.10 W
Example
An
to
ideally
an
The
lagged
free
end
aluminium
at
compound
aluminium
the
of
is
point
bar
the
of
copper
maintained
at
which
bar
length
the
bar
at
consists
20 cm
is
of
long
a
maintained
0 °C
copper
(Figure
and
copper
and
of
at
100 °C
18.1.7).
aluminium
conductivity
of
copper
–1
of
aluminium
=
220 W m
copper
=
390 W m
12 cm
bars
long
joined
cross-sectional
and
Calculate
–1
(Thermal
bar
equal
are
the
free
the
area.
end
of
the
temperature
joined.
–1
K
,
thermal
conductivity
–1
K
)
rod
lagging
aluminium
rod
0 °C
100 °C
12 cm
20 cm
Figure 18.1.7
Under
and
Let
steady
state
aluminium
the
conditions,
are
the
temperature
at
same.
the
the
point
Q
where
100
=
390
×
A
×
t
(
×
–
(
–
0.12
+
flow
of
the
two
heat
heat
bars
through
loss
meet
θ
)
=
210
θ
×
A
×
(
from
be
–
=
(
0.20
=
7800
78 θ
=
7800
θ
=
)
–
78 θ
7800
=
103.2
150
θ
0
0.20
210 θ
)
25.2 θ
25.2 θ
of
prevents
θ
0.12
100
390
rate
(Lagging
75.6 °C
)
the
the
copper
sides.)
Chapter
18
Thermal
energy
transfer
Convection
In
is
fluids
(liquids
convection.
temperature
of
in
the
a
fluid
therefore
liquid
to
a
result
liquid
above
is
of
low
a
to
sets
main
is
transferred
of
transferred
change.
bottom
than
the
up
form
temperature
density
the
dense
sinks
process
heat
of
at
less
the
energy
region
becomes
The
process
a
the
from
rise.
gases)
Thermal
to
as
beaker
,
and
from
due
liquid
bottom
and
convection
of
to
When
becomes
the
because
thermal
region
the
bulk
some
warmer
.
above
causes
currents
the
a
energy
it.
the
in
of
high
movement
water
It
is
heated
expands
The
less
(Figure
change
transfer
and
cooler
dense
18.1.8).
density
of
denser
liquid
In
the
this
liquid.
liquid
Land
and
sea
convection
breezes
currents
The
specific
During
above
the
the
replace
it.
heat
day,
land
capacity
the
land
rises
This
and
creates
of
heats
the
a
water
up
is
much
cooler
sea
much
air
breeze
larger
faster
from
(Figure
than
than
above
that
the
the
of
sea.
sea
soil.
The
rushes
hot
in
air
to
18.1.9).
heat
Figure 18.1.8
warm
air
Convection currents in a
rises
liquid
cooler
above
rushes
land
air from
the
in
replace
land
heats
the
air
up
above
faster
sea
to
than
the
the
land
sea
sea
during
the
Key points
day
ᔢ
Figure 18.1.9
Thermal
energy
is
transferred
The formation of sea breezes (day time)
by
conduction,
convection
and
radiation.
During
than
the
cooler
,
a
the
land
night,
sea.
The
denser
breeze
air
the
air
reverse
above
from
(Figure
process
the
above
sea
the
occurs.
is
land
The
warmer
rushes
land
and
in
to
cools
therefore
replace
more
rises.
it.
This
quickly
The
ᔢ
Thermal
by
creates
conduction
electron
ᔢ
Electron
diffusion
mechanism for
warm
air
transfer
rises
ᔢ
air from
above
The
in
rushes
in
to
air
above
the
coefficient
faster
heat
per
is
thin
the
of
Colder
of flow
per
unit
at
right
when
angles
the
a
to
parallel-sided
the
slab
sea
of
the
material,
under
steady
ᔢ
The formation of land breezes (night time)
conditions.
Thermal
region
convection
of
thermal
of
high
is
the
energy from
temperature
to
a
currents
region
currents
sun
rate
area
night
a
the
thermal
gradient,
faces
transfer
Ocean
unit
heat flow
state
Ocean
of
the
sea
sea
the
main
energy
down
than
during
Figure 18.1.10
is
temperature
cools
the
replace
land
land
is
thermal
the
of
the
lattice
metals.
conductivity
land
or
vibrations.
18.1.10).
cooler
occurs
diffusion
heats
denser
occur
up
the
water
as
a
result
ocean,
rushes
of
warm
in
and
the
change
water
rises
convection
in
density
because
its
currents
of
sea
water
.
density
are
set
As
decreases.
the
as
a
of
bulk
low
temperature
movement
result
of
a
of
density
due
to
the fluid
change.
up.
151
18.2
Measuring
Learning outcomes
On
completion
should
ᔢ
be
able
describe
of
this
thermal
conductivity
Experiment to find the thermal
section,
you
good
conductivity of
a
conductor
to:
an
experiment
to find
θ
3
θ
the
thermal
good
conductivity
of
4
a
conductor
θ
water flowing
1
ᔢ
describe
an
experiment
to find
2
the
thermal
poor
conductivity
of
in
θ
at
a
constant
rate
a
conductor.
l
A
lagging
electrical
Figure 18.2.1
In
the
long
there
is
case
to
a
lagged
used
One
the
to
bar
rate.
of
good
a
conductors,
measurable
measurable
to
prevent
determine
end
coil
Experiment to measure the thermal conductivity of a good conductor
give
is
heating
of
is
the
cooled
When
heat
is
by
steady
thermometers
temperature
the
bar
losses
state
length
the
an
sides
are
steady),
θ
θ
of
flow
of
heat
through
the
θ
– kA
t
Where
A
between
is
θ
the
and
cross-sectional
θ
1
The
rate
and
k
is
–
the
(
area
bar
.
shown
(i.e
and
in
The
that
other
coils
the
at
bar
18.2.1.
end
a
reading
are
the
experiment
Figure
The
θ
3
is
sufficiently
of
constant
on
recorded.
all
the
The
4
given
by:
θ
2
)
l
of
the
heater
.
θ
2
1
=
,
be
ensures
experiment,
conducting
conductor
Q
should
This
this
reached
,
1
rate
bar
of
is
electric
through
conditions
the
In
conductivity
with
flowing
become
of
difference.
gradient.
from
thermal
heated
water
have
the
temperature
the
thermal
(1)
conductor
conductivity
and
of
l
the
is
the
distance
conductor
.
2
at
which
thermal
energy
is
being
removed
by
the
water
is
given
by:
Q
m
=
t
where
m/t
is
the
rate
of
×
c
flow
of
×
(θ
w
t
the
–
3
water
θ
)
(2)
4
and
c
is
the
specific
heat
w
capacity
of
Equation
water
.
(1)
and
equation
θ
–
1
– kA
(2)
θ
l
equated
to
)
=
c
(θ
w
t
–
3
θ
In
a
the
large
value
of
k
)
conductivity of
a
conductor
case
of
a
poor
conductor
,
cross-sectional
through
152
the
4
Experiment to find the thermal
poor
determine
m
2
(
are
the
sample.
area.
the
This
sample
allows
for
must
a
be
large
thin
rate
of
and
should
flow
of
have
heat
Chapter
steam
18
Thermal
energy
transfer
in
steam
chest
steam
brass
θ
out
base
2
sample
θ
brass
1
Figure 18.2.2
In
this
Experiment to measure the thermal conductivity of a poor conductor
experiment,
Thermometers
heated
it
is
a
using
poor
disk
a
are
the
steam
negligible.
are
constant
Under
in
chest
conductor
,
are
sample
inserted
it
is
placed
holes
(Figure
can
steady
be
in
between
the
18.2.2).
assumed
state
brass
two
Since
that
the
heat
conditions,
brass
slabs.
the
slabs.
The
upper
sample
losses
is
from
the
temperature
slab
thin
θ
sides
and
θ
1
the
of
rate
of
loss
thermal
The
rate
and
the
of
heat
energy
of
flow
rate
loss
of
of
flow
from
is
heat
by
the
of
heat
bottom
through
of
the
the
brass
sample
disc.
is
The
is
and
2
equal
main
to
form
convection.
through
the
sample
θ
Q
–
– kA
given
insulation
by:
θ
1
=
t
is
2
(
)
l
(1)
brass
disk
θ
1
where
2
A
=
πr
and
r
is
the
radius
of
the
disc
and
l
is
the
thickness
of
the
sample.
thermal
The
rate
of
loss
of
heat
from
the
bottom
brass
slab
is
given
energy
loss
by:
Figure 18.2.3
Q
Δθ
=
mc
(
×
t
where
m
is
the
mass
of
the
brass
disc
)
Δt
(2)
and
c
is
the
specific
heat
capacity
of
temperature / °C
brass.
In
is
order
to
heated
determine
directly
the
with
rate
the
at
which
steam
the
chamber
.
brass
disc
When
the
cools,
the
brass
temperature
of
disc
the
θ
1
brass
disc
is
steady,
brass
disc
is
covered
θ
is
recorded
over
the
a
sample
with
an
period
and
steam
insulator
of
time
chamber
(Figure
until
it
are
18.2.3).
removed
The
approximately
and
the
temperature
5 °C
less
than
1
the
of
previous
the
graph
steady
at
θ
state
is
value.
determined
A
cooling
(Figure
curve
is
then
plotted.
The
slope
18.2.4).
time /s
1.
Equation
(1)
and
equation
(2)
are
equated
to
determine
the
value
of
k
Figure 18.2.4
Measuring the rate of heat
loss by convection from the brass disc
θ
–
θ
1
– kA
(
Δθ
2
)
l
=
mc
Δt
Key points
ᔢ
When
determining
conductor
ᔢ
When
should
determining
sample
should
be
the
be
thermal
long
the
thin
and
thermal
and
conductivity
of
a
good
of
a
poor
conductor,
the
thin.
have
conductivity
a
large
conductor,
cross-sectional
the
area.
153
18.3
Radiation
Learning outcomes
On
completion
should
be
able
of
this
Thermal
section,
you
Hot
objects
higher
to:
away
ᔢ
explain
what
is
meant
by
radiation
the
per
emit
energy
temperature
second
by
by
of
a
process
the
thermal
called
object,
the
radiation.
thermal
more
radiation.
energy
Thermal
is
The
transmitted
radiation
is
made
up
thermal
of
electromagnetic
waves
with
a
continuous
range
of
wavelengths.
radiation
These
ᔢ
explain
what
is
meant
by
thermal
surface
equilibrium
ᔢ
understand
are
also
wavelengths
electromagnetic
that
good
good
absorbers
emitters
of
radiation
of
Suppose
a
a
body
small
it
can
from
be
object
X,
either
at
a
infrared
thermal
chamber
T
18.3.1).
(Figure
whose
reflected
or
temperature
Since
walls
the
concept
of
by
the
convection
are
the
visible
is
regions
incident
of
the
on
the
absorbed.
T
is
placed
maintained
chamber
or
is
at
a
inside
an
evacuated,
temperature
conduction.
All
the
heat
cannot
energy
is
be
transmitted
by
a
thermal
radiation.
If
T
until
temperature
>
T
1
,
the
object
loses
energy
by
thermal
radiation
2
body.
its
becomes
T
.
If
T
2
ᔢ
to
radiation
2
understand
black
the
When
1
evacuated
transmitted
ᔢ
span
spectrum.
sketch
the
spectrum
of
a
black
thermal
radiation
until
its
<
T
1
,
the
object
gains
energy
by
2
temperature
becomes
T
.
In
either
case,
the
2
body
radiator
and
the
effect
of
state
Stefan’s
Prevost’s
law.
will
become
T
eventually.
It
should
be
noted
that
energy
2
absorption
temperature
ᔢ
temperature
and
emission
theory
temperature
as
surroundings
of
its
is
does
exchanges
not
states
surroundings
equal
to
the
cease
its
rate
that
rate
of
at
this
point.
when
of
a
body
emission
absorption
of
is
of
at
the
same
radiation
radiation
to
from
the
the
surroundings.
non-conducting
Therefore,
is
thread
at
and
the
in
the
same
emitting
equilibrium
previous
demonstration
temperature
energy
with
by
its
as
its
thermal
even
though
surroundings,
radiation.
It
is
it
is
said
the
still
to
object
X
absorbing
be
in
ther mal
surroundings.
T
1
The
rate
at
depends
which
on
the
energy
flows
temperature
from
of
the
an
object
by
ther mal
radiation
object.
X
It
follows
also
evacuated
chamber
T
2
a
would
Good
Figure 18.3.1
rise
the
water
the
heat
equal
of
that
the
the
in
water
better
in
can
perfect
one
and
is
can
that
B
cools
the
of
is
of
thermal
otherwise
of
at
outer
painted
over
faster
a
the
radiation
is
temperature
A
Bunsen
water
water
black.
are
of
If
of
that
water
and
placed
can
the
time,
in
at
B.
burner
.
over
in
100 °C
surface
radiation
can
A
into
is
two
painted
temperature
it
A.
can
The
be
silver
of
the
shown
black
surface
surface.
using
cans
ther mal
period
than
silvered
in
of
the
water
The
repeated
the
drying
surfaces.
of
absorbs
out
B.
lighted
becomes
absorber
that
boiling
placed
a
absorber
emitters
measured
B
good
a
can
room
The
A
cans
are
thermometer
period
B
temperature.
two
of
time.
increases
It
more
is
can
Equal
placed
used
be
at
to
shown
quickly
than
A.
example
is
of
temperature
bitumen
A
good
of
a
radiation;
surroundings.
and
be
are
from
dried
on
its
than
can
Another
pointed
A
can
water
the
of
is
is
thermal
also
temperature
radiation
154
cans
distance
measure
are
cans,
experiment
volumes
that
surface
water
radiates
The
both
which
of
volumes
metal
outer
in
body
above
equal
identical
that
a
emitter
absorbers
Suppose
and
that
good
the
hot.
the
crops.
black
This
thermal
all
of
The
is
concept
an
radiation
radiation
of
a
In
is
called
is
body
countries,
absorbs
effective
which
black
some
surface
means
a
black
incident
is
a
crops
energy
of
it.
theoretical
It
A
are
thermal
drying
body.
on
by
crops.
black
should
one.
body
be
Chapter
Energy distribution
body
A
spectrum of
a
Thermal
energy
transfer
black
radiator
black
body
radiator
characteristic
constant
Figure
in the
18
its
is
at
a
one
that
temperature.
temperature
18.3.2
radiator
of
shows
consists
a
constant
typical
emits
Black
of
a
thermal
body
continuous
spectrum
radiation
radiation
range
obtained
which
from
of
from
a
a
is
source
at
wavelengths.
black
body
temperature.
power
radiated
2
per
unit
area / W m
visible
infrared
region
region
T
0
500
1000
1500
2000
wavelength /μm
Figure 18.3.2
Since
the
Spectrum of a black body radiator
graph
is
curved
it
means
that
the
energy
carried
away
by
Exam tip
radiation
the
area
is
not
under
evenly
the
distributed
graph
gives
the
across
total
the
range
energy
of
wavelengths.
radiated
per
unit
Also,
time
per
Notice
unit
surface
area
at
a
temperature
of
Figure
18.3.3
shows
what
that,
happens
to
the
distribution
curves
as
a
black
of
a
black
body
radiator
is
the
curve
shifts
to
the
left
increased.
the
visible
emitted
region
of
radiation falls
the
in
the
electromagnetic
radiated
2
per
temperature
increases,
the
and
power
the
body
distribution
temperature
as
T
unit
spectrum. The
area / W m
λ
changes from
max
colour
red
to
of
the
yellow
body
to
white.
red
violet
5500 K
5000 K
4500 K
4000 K
3500 K
0
500
1000
1500
2000
wavelength /μm
Figure 18.3.3
From
the
the
Effect on spectrum as temperature increases
graphs
temperature
car ried
This
away
can
cur ve
be
peaks
emitting
by
it
can
of
the
the
said,
at
a
be
shorter
because
lower
radiation
seen
black
in
that
body
distribution
wavelengths
at
higher
wavelength.
the
the
changes.
visible
The
increases
temperatures,
If
a
region
body
of
is
the
cur ve
changes
proportion
with
the
hot
of
the
as
energy
temperature.
distribution
enough,
it
electromagnetic
will
begin
spectr um.
155
Chapter
18
Thermal
energy
transfer
At
around
the
red
body
1200 K,
end
of
changes
There
is
a
the
the
visible
spectr um.
from
red
relationship
to
wavelength
At
higher
yellow
between
to
that
is
emitted,
temperatures,
the
lies
mostly
colour
of
in
the
white.
λ
of
each
cur ve
and
its
temperature.
max
It
can
be
expressed
as
follows
λ
T
=
constant
max
This
relationship
Stefan’s
The
area
is
known
as
Wien’s
displacement
law
.
law
under
each
cur ve
gives
the
total
energy
E,
radiated
per
unit
time
Definition
per
unit
under
Stefan’s
law
states
that
the
rate
energy
is
radiated from
a
the
is
power
proportional
of
the
to
graph
is
at
a
temperature
proportional
the
T.
Stefan’s
fourth
power
law
of
states
the
that
the
temperature
area
of
body
expressed
in
kelvin.
black
Stefan’s
body
area
at
the
which
sur face
law
can
be
expressed
mathematically
as
follows.
the fourth
temperature
of
the
Equation
body
expressed
in
kelvin.
Stefan’s
Law
4
P
=
σAeT
P
–
energy
σ
–
Stefan’s
radiated
A
–
surface
e
–
emissivity
T
–
surface
per
second/W
–8
constant
(5.67
×
10
–2
W m
–4
K
)
2
When
an
area/m
of
object
sur roundings.
difference
the
body
(e
=
1 for
a
black
body)
temperature/K
radiates
energy,
it
also
absorbs
The
net
energy
emitted
by
an
between
the
energy
emitted
by
the
absorbed
by
equation
below.
the
object.
The
net
energy
energy
object
is
object
radiated
per
from
the
equal
and
to
the
second
the
energy
is
given
by
the
Equation
4
P
=
σAe(T
4
–
T
net
P
)
s
–
net
–
Stefan’s
energy
radiated
per
second/W
net
–8
σ
constant
(5.67
×
10
–2
W m
–4
K
)
2
A
–
surface
area/m
e
–
emissivity
T
–
surface
T
–
temperature
of
the
body
(e
=
1 for
a
black
body)
temperature/K
of
surroundings/K
s
Example
A
of
blackened
40 °C
is
sphere
temperature
156
(e
suspended
of
=
0.25)
inside
a
of
radius
box
12 cm,
which
is
initially
held
at
a
at
a
temperature
constant
110 °C.
a
Calculate
the
rate
b
Calculate
the
initial
of
absorption
net
rate
of
of
energy
by
absorption
of
the
sphere
energy
by
inside
the
the
box.
sphere.
Chapter
a
T
=
273.15
+
110
=
Rate
area
of
of
sphere
energy
Thermal
energy
transfer
383.15 K
2
Surface
18
A
=
–2
4πr
=
4π(12
×
10
2
–1
)
=
1.81
×
10
2
m
absorption
4
P
=
σAeT
=
(5.67
=
273.15
–8
b
T
Net
rate
×
of
10
+
40
=
=
×
10
4
)(0.25)(383.15)
=
55.25 W
313.15 K
absorption
4
P
–1
)(1.81
of
energy
4
σAe(T
–
T
)
s
–8
=
(5.67
×
10
–1
)(1.81
×
10
4
)(0.25)(383.15
4
–
313.15
)
=
30.5 W
Example
The
emissivity
of
a
person’s
skin
is
0.74.
Normal
body
temperature
2
is
310 K.
0.37 m
temperature
the
exposed
of
of
skin
308K.
is
exposed
Calculate
to
the
the
net
atmosphere
energy
which
radiated
per
is
at
a
second
from
skin.
4
P
=
4
σAe(T
–
T
net
)
s
–8
=
5.67
×
=
3.66 W
10
4
×
0.37
×
0.74
×
(310
4
–
308
)
Key points
ᔢ
Thermal
from
a
radiation
region
of
The
rate
at
the
high
electromagnetic
ᔢ
is
process
by
temperature
which
to
a
thermal
region
of
energy
low
is
transferred
temperature
by
waves.
which
thermal
energy flows
by
radiation
depends
on
temperature.
ᔢ
When
a
body
thermal
ᔢ
Good
ᔢ
A
black
body
Stefan’s
body
in
is
is
thermal
back
absorbers
theoretical
ᔢ
is
energy
are
a
also
perfect
good
with
between
emitters
absorber
of
another
body,
there
is
a flow
of
them.
of
thermal
thermal
energy.
energy. The
concept
is
a
one.
law
states
that
proportional
expressed
equilibrium
and forth
in
to
the
rate
at
the fourth
which
power
energy
of
the
is
radiated from
temperature
of
a
the
black
body
kelvin.
157
18.4
Global
warming
thermal
Learning outcomes
On
completion
should
be
able
of
this
energy
The
section,
you
greenhouse
Energy
from
explain
the
greenhouse
discuss
applications
transfer
of
thermal
conduction,
radiation.
of
This
Sun
and
effect
reaches
the
radiation.
heats
causes
wavelength
the
energy
convection
of
up
the
Earth
The
in
the
radiation
Earth.
This
form
of
passes
light
easily
temperature
of
and
short
through
the
the
Earth
effect
rises.
ᔢ
the
infrared
atmosphere
ᔢ
applications
transfer
wavelength
to:
and
by
and
the
Sun.
Sun.)
of
infrared
the
fuels
warming
ᔢ
is
produces
significantly
infrared
dioxide
of
Earth
Earth
from
the
known
of
the
to
as
emitting
the
is
the
rise
is
the
The
further
.
carbon
war ming .
The
The
term
of
the
wavelength
of
causes
the
heating
burning
This
from
that
energy
process
gas.
The
coming
than
longer
trapped
effect .
long
that
the
This
dioxide
radiation.
lower
prevents
greenhouse
of
than
much
Earth.
even
infrared
longer
Earth
atmosphere
volumes
global
start
leaving
Earth
large
to
in
to
radiation
temperature
radiation
temperature
up
this
(The
Carbon
the
of
fossil
contributes
consequences
of
global
are:
melting
of
Caribbean
ᔢ
climate
ᔢ
extinction
the
is
polar
very
ice
caps
prone
to
resulting
rising
sea
in
increasing
sea
levels.
The
levels
changes
of
plant
and
animal
Applications of thermal
life
forms.
energy transfer
The vacuum flask
The
is
vacuum
designed
out
of
the
between
flask.
the
convection.
radiation
the
heat
flask
such
The
walls.
The
by
it
hot
is
flask
The
glass
entering
loss
keeps
that
is
a
the
conduction
hot
for
and
cold
thermal
double-walled
vacuum
walls
from
liquids
difficult
are
eliminates
silvered.
outside
(Figure
of
the
glass
heat
The
liquids
energy
loss
The
18.4.1).
cork
stopper
silvered
walls
contents
vacuum
plastic
Figure 18.4.1
158
A vacuum flask
by
container
cold.
travel
vessel
silvered
flask.
to
with
The
flask
into
a
conduction
surfaces
cork
or
vacuum
and
reduce
stopper
reduces
Chapter
The
solar
Solar
in
water
homes
box
and
copper
The
an
the
which
insulated
box
the
copper
later
is
Solar
pipes
heats
tank.
with
either
effect).
The
warms
conductor)
warm
glass
are
water
the
in
is
or
infrared
trapped,
up
in
an
energy
transfer
box,
plastic
is
a
surface
an
it,
the
good
the
18.4.2).
up
escape
inside
of
absorber
because
water
insulated
are
heats
cannot
and
use
surface.
and
up
for
there
(Figure
glass
radiation
with
water
insulated
blackened
surface
contact
heat
of
warming
black
stored
a
through
blackened
to
insulated
with
enters
gets
used
consists
the
contact
wavelength
and
are
heater
Inside
radiation
plastic
The
suggests,
water
thermal
covered
radiation
(good
up.
storage
Longer
or
name
solar
infrared
box.
glass
the
The
make
(greenhouse
radiation.
them
box
wavelength
through
as
hotels.
insulated
pipes,
Thermal
heater
heaters,
and
insulated
Short
the
water
18
of
the
inside
storage
tank
for
use.
sunlight
pressure
storage
resistant
outlet
tank
inlet
absorber
glass
or
plate
plastic
(black
coated)
electric
sheet
thermally
heat
support
(roof )
Figure 18.4.2
exchanger
contact
with
(pipe
the
in
insulated
booster
box
thermal
absorber
plate)
A solar water heater
Key points
ᔢ
Short
wavelength
ᔢ
The
Earth
absorbs
ᔢ
The
Earth
then
cannot
dioxide
the
heat
ᔢ
The
temperature
ᔢ
The
more
warming
some
carbon
takes
to
of
radiation
of
re-radiates
penetrate
cause
infrared
this
easily
enters
the
atmosphere.
energy.
longer
wavelength
infrared
atmosphere. Greenhouse
be
the
gases
radiation,
such
as
which
carbon
trapped.
Earth
dioxide
is
rises
and
pumped
this
into
is
the
called
global
atmosphere,
warming.
the
more
global
place.
159
Revision
Answers
found
1
to
on
questions
the
Copper
that
require
questions
calculation
can
be
9
7
accompanying CD.
and
glass
conductivities.
mechanisms
have
used
to
this
thermal
difference
transfer
in
thermal
8
terms
energy
of
in
the
State
Explain
Write
the
thermal
glass.
what
what
rod
of
length
0.8 m
and
cross-sectional
the
[6]
linear
heat flow
symbols
by
conduction
mean.
[3]
is
meant
by
thermal radiation.
[2]
an
equation for Stefan’s
–5
law
of
black
body
area
radiation
×
to find
[4]
copper
6.0
of
the formula for
explaining
9
A
experiment
each
material.
2
an
conductivity
different
Explain
Describe
and
explain
what
each
term
means
in
the
2
10
m
is
arranged
such
that
it
is
completely
equation.
lagged
on
100 °C
and
Calculate
through
the
sides. One
the
the
the
other
rate
end
3
An
=
copper
rod.
390 W m
ideally
bar
12 cm
lagged
long
26 cm
and
of
copper
the
end
the
and
of
the
of
K
aluminium
thermal
(Thermal
at
an
bar
consists
aluminium
is
the
at
a
of
its
the
sphere
of
radius
6 cm
surroundings. Calculate
sphere
if
it
absorbs
is
the
40 kW
of
in
equilibrium
temperature
power from
the
conductivity
copper
120 °C
at
which
and
=
conductivity
of
aluminium
a
Explain
b
What
c
0 °C. Calculate
the
are
is
the
meant
by
global
consequences
of
warming.
global
[3]
warming?
Suggest
three
ways
we
can
help
reduce
global
warming.
[3]
copper
390 W m
=
what
[3]
the free
[3]
copper
11
[3]
end
joined.
of
surroundings.
length
area. The free
maintained
point
are
at
of
bar
12
–1
K
220 W m
An
electric
iron
is
coils. The
electric
electrical
power
opened
iron
is
to
expose
switched
the
on
bare
and
a
heating
constant
,
–1
thermal
blackened
of
of
–1
(Thermal
A
with
[2]
maintained
bars
10
10 °C.
energy
conductivity
cross-sectional
is
at
[4]
at
)
to
aluminium
temperature
maintained
maintained
of
compound
equal
is
–1
joined
bar
is
of flow
–1
copper
end
–1
K
is
supplied
to
it. The
coils
glow
red
)
hot.
Sketch
a
graph
to
show
the
variation
of
temperature
a
along
the
compound
bar.
One
of
from
4
Explain
5
When
how
land
and
the
ways
by
which
thermal
energy
is
lost
[3]
sea
breezes
are formed.
the
other
[6]
heating
two
coil
methods
is
by
by
convection.
which
thermal
Name
energy
the
is
lost.
determining
the
thermal
conductivities
b
copper
and
glass,
the
choice
of
dimension
used
Explain
different. State
what
dimensions
are
used
[4]
13
Describe
heater
Describe
an
experiment
to find
the
160
of
copper
in
the form
the
air
heating
is
blown
coil
over
glows
less
brightly
it.
[2]
the
and
principal features
explain
the
design
of
in
a
solar
terms
water
of
thermal
thermal
energy
conductivity
some
and
why.
6
why
is
when
very
[2]
of
of
a
cylinder.
[6]
transfer.
[5]
Revision questions
9
161
19
Phases
19.
1
A
of
kinetic
Learning outcomes
On
completion
should
be
able
of
this
matter
model for
solids,
liquids
and
gases
Density
section,
you
The
density
Density
to:
is
of
a
substance
defined
as
the
gives
mass
an
per
idea
unit
of
how
concentrated
volume.
The
symbol
matter
used
is.
to
–3
represent
ᔢ
define
ᔢ
relate
the
term
density
is
ρ
(rho).
The
SI
unit
Definition
the
difference
structures
and
of
density
is
kg m
density
in
Equation
the
densities
of
solids,
m
Density
is
defined
as
the
mass
per
ρ
liquids
and
gases
to
simple
=
ideas
V
unit
of
the
spacing,
ordering
volume.
and
–3
ρ
motion
of
–
density/kg m
molecules
m
–
mass/kg
V
–
volume/m
3
ᔢ
describe
for
ᔢ
a
solids,
describe
simple
liquids
an
the
appreciate
by
the
of
such
model
gases
Brownian
movement
provided
and
experiment
demonstrate
and
kinetic
to
motion
evidence for
The
and
distinguish
an
between
crystalline
and
therefore
of
by
a
the
substance
density
of
is
defined
water
.
as
the
Relative
density
density
of
has
the
no
units
dimensionless.
experiment
the
kinetic
model for
solids,
liquids
and
gases
structure
A
of
is
density
divided
molecules
A
ᔢ
relative
substance
kinetic
model
may
be
used
to
explain
the
properties
of
solids,
liquids
non-crystalline
and
gases.
Matter
is
thought
to
be
made
up
of
many
tiny
particles.
These
solids.
particles
particle
tend
can
to
attract
refer
to
solids,
each
atoms,
matter
are
liquids
typical
arrangement
of
other
.
ions
and
the
or
They
also
molecules.
gases.
particles
Figures
that
tend
to
The
three
19.1.1
make
up
move
(a),
a
about.
common
(b)
solid,
and
(c)
liquid
The
term
states
show
and
a
of
the
gas.
Equation
Solids
Relative
density
density
of
substance
In
a
solid,
the
particles
are
held
closely
together
by
strong
forces
of
=
attraction.
density
of
The
spacing
between
the
particles
is
therefore
very
small.
The
water
particles
to
move
are
arranged
about
positions.
As
a
within
result
in
an
the
of
orderly
solid.
this,
manner
.
Instead,
solids
have
The
they
particles
vibrate
fixed
are
about
shapes
and
not
their
free
fixed
volumes.
Liquids
In
a
that
in
spacing

ordering
of
very

small
well
arrangement
about
their fixed
Figure 19.1.1 (a)
162
a
 free
to
vibrate
positions
Structure of a solid
the
solid.
solid.

ordering
of
particles
of
spacing
ordered
particles
motion of
a
liquid,
force
The
There
very

is
of
attraction
separation
less
ordering
small
well
about
Figure 19.1.1 (b)
a
liquid
particles
particles
than
arrangement

ordering
in
 free
to
vibrate
positions
Structure of a liquid
motion of
at
high
is
a
very

is
weaker
slightly
solid.
than
more
The
than
particles
large
particles
distributed;
particles
their fixed
in
the
the
spacing
ordered
particles
motion of
between
between
there
is
randomly
no
particles

order
moving
randomly
speeds
Figure 19.1.1 (c)
Structure of a gas
Chapter
are
able
take
to
the
move
shape
throughout
of
the
the
container
body
they
of
are
the
in
liquid.
and
As
their
a
result,
volumes
19
Phases
of
liquids
are
fixed.
Gases
In
a
microscope
gas,
solid
with
matter
or
the
particles
liquid.
each
particles
The
other
are
and
far
are
far
apart
particles
the
apart
walls
from
from
move
of
each
each
about
the
other
container
other
,
when
randomly
the
at
holding
force
of
compared
high
the
speeds,
gas.
attraction
with
a
colliding
Since
the
between
them
light
is
negligible.
As
a
result,
a
gas
fills
up
any
space
that
is
available.
smoke
cell
Brownian
The
motion
apparatus
used
to
demonstrate
Brownian
motion
is
shown
in
Figure
Figure 19.1.2
19.1.2.
The
It
cell
using
a
is
called
of
a
small
illuminated
microscope.
particles
around
consists
are
seen
as
the
side
The
specks.
is
containing
the
smoke
adjusted
The
haphazard
cell
and
microscope
bright
haphazardly.
Brownian
from
The
transparent
bright
motion
so
particles
that
specks
of
the
some
the
are
are
observed
a
smoke
seen
smoke
Brownian motion
smoke.
regular
atoms
moving
particles
array
throughout
is
of
extending
the
lattice
motion
atoms
Explanation
–
The
movement
of
the
smoke
particles
is
due
to
air
Figure 19.1.3
molecules
bombarding
them
from
all
sides.
The
haphazard
motion
The structure of a
of
crystalline solid
the
smoke
rapidly
in
particles
all
suggests
that
the
air
molecules
are
moving
around
directions.
a
If
larger
smoke
haphazardly
particles
because
are
the
used,
the
bright
randomness
of
the
specks
are
collisions
seen
moving
averages
monomer
out.
Figure 19.1.4
Brownian
motion
provides
evidence
for
the
kinetic
model
of
polymer
chain
less
units
A polymer chain
matter
.
Key points
Structure
of
crystalline
and
non-crystalline
solids
ᔢ
In
a
crystalline
regular
pattern
crystalline
shows
a
solid ,
that
solid
simple
is
the
atoms,
repeats
quartz.
structure
ions
itself
within
Metals
of
a
or
are
molecules
the
also
crystalline
are
crystal.
An
crystalline
arranged
example
solids.
in
of
Figure
Density
is
per
volume.
unit
an
amorphous
throughout
an
the
amorphous
substances.
solid,
there
structure.
material
The
atoms
is
in
is
There
glass.
this
a
ᔢ
solid.
random
Relative
the
is
no
definite
Glass
case
arrangement
are
is
a
pattern.
mixture
arranged
to
of
of
the
An
form
an
the
mass
density
density
of
of
a
is
the
ratio
substance
water.
It
has
no
of
to
the
units.
atoms
example
silica
as
a
19.1.3
density
In
defined
a
and
of
ᔢ
other
A
kinetic
explain
amorphous
liquids
model
the
and
can
be
properties
used
of
to
solids,
gases.
structure.
ᔢ
Polymers
units
consist
that
(Figure
repeat
of
long
chains
themselves.
of
Each
molecules.
repeating
Each
unit
chain
is
called
consists
a
that
of
random
monomer
ᔢ
the
be
can
be
coiled
polymer
19.1.6
motion
molecules
demonstrates
are
in
continuous
motion.
19.1.4).
Polymers
may
Brownian
up
is
below
semi-crystalline
or
tangled.
The
semi-crystalline
show
the
or
amount
or
difference
amorphous.
of
tangling
amorphous.
between
The
the
Figure
two
long
polymer
determines
19.1.5
and
in
whether
Figure
structures.
In
a
ions
chain
ᔢ
a
crystalline
or
regular
within
In
amorphous
an
random
atoms
the
the
are
pattern
itself
a
ᔢ
solid,
molecules
atoms,
arranged
that
repeats
crystal
solid,
there
arrangement
throughout
the
of
is
the
structure.
Polymers consist of long chains of
molecules. Each chain consists of
units that repeat themselves. Each
repeating unit is called a monomer.
ᔢ
Polymers
or
Figure 19.1.5
A semi-crystalline polymer
Figure 19.1.6
can
be
semi-crystalline
amorphous.
An amorphous polymer
163
19.2
Pressure
Learning outcomes
On
completion
should
be
able
of
this
Pressure
section,
you
The
per
to:
pressure
unit
area
acting
on
on
the
a
surface
surface.
is
The
defined
SI
unit
of
as
the
force
pressure
is
exerted
the
normally
pascal
(Pa).
–2
From
ᔢ
define
pressure
ᔢ
derive
an
A
expression for
the
block
definition
of
in
of
follows
mass
2 kg
that
1 Pa
rests
on
=
a
1 N m
table
as
shown
in
Figure
19.2.1.
the
Calculate
pressure
wood
it
the
pressure
exerted
on
the
surface
of
the
table.
a fluid
–1
[Use
ᔢ
recall
the
various
g
=
to
measure
]
instruments
Area
used
9.81 N kg
in
contact
with
the
surface
of
the
table
=
l
×
b
pressure.
–2
=
5
×
=
0.004 m
–2
10
×
8
×
10
2
Definition
Force
Pressure
is
defined
as
exerted
on
table
=
mg
the force
=
acting
normally
per
unit
2
×
9.81
=
19.62 N
area.
force
19.62
3
∴
pressure
=
=
=
area
4.91
×
10
Pa
0.004
Equation
Pressure
F
p
in
a fluid
=
A
A
fluid
p
–
pressure/Pa
fluids.
F
–
force/N
called
A
–
area/m
is
a
substance
Above
the
that
Earth’s
has
the
surface,
ability
there
to
is
a
flow
.
thick
Liquids
layer
of
and
gas.
gases
This
are
layer
is
5
the
atmosphere.
The
atmosphere
exerts
a
pressure
of
1.01
×
10
Pa
on
2
the
Earth’s
felt
by
This
us.
surface.
This
prevents
atmosphere.
of
a
lake.
In
is
though
because
our
The
this
Even
bodies
the
pressure
insides
from
pressure
case
the
this
becomes
diver
of
our
collapsing
is
seems
body
under
more
large,
exert
the
on
a
is
usually
outward
pressure
significant
experiencing
an
it
atmospheric
pressure.
exerted
diver
at
not
by
the
pressure
the
bottom
as
well
5.5 cm
as
the
In
5 cm
a
acts
pressure
fluid,
exerted
pressure
equally
in
all
by
the
water
increases
with
in
the
lake
depth.
At
above
a
him.
particular
depth,
pressure
directions.
8 cm
In
Figure
19.2.2,
a
container
is
filled
with
a
liquid.
The
pressure
in
the
Figure 19.2.1
liquid
is
at
same
the
point
D
greatest
is
Consider
A
B
C
depth
deeper
a
the
in
than
area
bottom
the
container
cross-sectional
h
at
liquid
A,
of
of
of
B
and
and
height
the
the
C,
h,
base
of
container
.
are
therefore
and
is
filled
the
The
at
therefore
with
a
the
at
fluid
container
is
points
same
a
of
A
A,
B
C
pressure.
higher
are
The
pressure.
density
(Figure
and
ρ.
The
19.2.3).
pressure
increases
V
olume
Mass
D
of
of
W
eight
fluid
fluid
of
fluid
in
the
container
=
density
=
mass
of
×
=
cross-sectional
volume
fluid
×
=
area
×
height
=
Ah
ρAh
gravitational
field
strength
=
ρAhg
liquid
Therefore,
the
force
exerted
on
A
=
ρAhg
Figure 19.2.2
force
Pressure
exerted
on
the
surface
A
=
ρAhg
=
=
area
It
follows
by
a
that
vertical
the
difference
distance
Δh
in
a
in
pressure
fluid
of
Δp,
density
between
ρ
ρgh
A
is
two
given
by
points
ρgΔh.
separated
The
h
A
expression
indicates
area
base.
of
the
derivation
Figure 19.2.3
164
the
that
This
same
the
simply
pressure
means
expression
at
that
would
any
if
have
a
depth
is
cylinder
been
independent
was
used
obtained.
in
of
the
the
Chapter
Example
A
diver
the
is
10 m
diver
at
below
this
of
matter
the
surface
of
a
lake.
Calculate
the
pressure
exerted
of
water
=
ρgΔh
depth.
3
[Density
Phases
Equation
Δp
on
19
=
1
×
–3
10
kg m
Δp
5
,
atmospheric
pressure
=
1
×
10
–
difference
–
density
in
pressure/Pa
g
–
gravitational field strength/N kg
Δh
–
difference
Pa,
–3
–1
g
=
9.81 N kg
ρ
]
of fluid/kg m
–1
Pressure
due
to
water
=
ρgh
3
=
T
otal
pressure
acting
on
1
×
in
height/m
4
10
×
9.81
×
10
=
9.81
×
10
Pa
to
diver
gas
at
a
open
particular
=
pressure
due
to
atmosphere
+
pressure
due
to
=
1
×
5
10
×
9.81
×
the
water
pressure
5
to
atmosphere
p
5
10
=
1.981
×
10
Pa
A
Instruments
used to
measure
pressure
h
The U-tube
This
device
density
ρ.
manometer
consists
Mercury
of
is
a
U-shaped
commonly
tube
used.
containing
The
U-tube
a
liquid
of
known
manometer
can
B
be
used
to
measure
connected
to
the
the
pressure
container
of
a
holding
gas.
the
One
gas.
end
The
of
the
other
tube
end
is
is
open
to
the
liquid
atmosphere.
The
pressure
at
the
point
A
is
atmospheric
pressure
P
.
at
The
A
density
pressure
at
height
plus
gas
is
h
the
point
B
is
the
atmospheric
actually
the
same
pressure
pressure
as
the
exerted
(Figure
pressure
at
by
the
19.2.4).
the
liquid
The
point
column
pressure
of
of
the
Figure 19.2.4
A U-tube manometer
B.
curved
The
pressure
of
the
gas
is
given
by
p
=
p
+
metal
tube
ρgh
A
The
Bourdon
The
at
Bourdon
one
end,
the
gauge,
the
pointer
(Figure
The
gauge
gauge
which
the
to
consists
is
of
a
connected
pressure
move.
causes
The
curved
to
a
the
scale
is
metal
pointer
.
tube
to
tube.
When
The
a
straighten
calibrated
to
gas
out
measure
tube
is
is
closed
connected
slightly
pressure
to
causing
directly
19.2.5).
barometer
A
barometer
is
inverted
measure
consists
in
a
tray,
of
a
also
atmospheric
long
glass
containing
pressure.
tube
partially
mercury.
Depending
on
The
the
filled
with
barometer
mercury.
is
atmospheric
used
It
to
pressure
Figure 19.2.5
the
height
(Figure
of
the
19.2.6).
mercury
The
in
the
glass
atmospheric
tube
pressure
will
is
either
given
increase
being
measured
pressure,
or
A Bourdon gauge
decrease
by
vacuum
p
=
ρgh
A
where
ρ
is
the
density
h
is
the
height
of
of
mercury,
mercury
in
g
is
the
the
gravitational
field
strength,
column
long
glass
tube
Key points
h
mercury
ᔢ
Pressure
is
defined
as
the force
acting
normally
per
unit
p
area.
A
ᔢ
The
pressure
in
a fluid
depends
on
depth,
gravitational field
strength
and
density.
ᔢ
Pressure
Bourdon
can
be
measured
using U-tube
manometer,
barometer
or
a
gauge.
Figure 19.2.6
A barometer
165
ρ
19.3
Hooke’s
law
Learning outcomes
On
completion
should
be
able
of
this
Hooke’s
section,
you
A
spring
the
to:
by
ᔢ
understand
the
and
terms
is
shape
and
state
stretching,
law
and
define
the
tensile
length
define
by
by
making
spring,
the
a
coil
from
a
piece
a
force
is
required.
applied
force
is
called
pressing,
the
terms
stress,
strain
force
is
19.3.1).
constant
using
ᔢ
the
a
If
of
the
metal.
spring
tensile
force.
In
is
If
order
to
change
lengthened
the
applied
force
is
called
a
the
spring
compressive
is
force.
When
compressive force
Hooke’s
spring
made
tensile
a
ᔢ
modulus
law
of
shortened
force
the Young
a
is
called
A
an
mass
metre
applied
to
a
spring
extension.
is
attached
rule.
A
to
Suppose
its
spring
it
and
length
is
attached
the
additional
increases.
to
extension
masses
are
a
The
rigid
increase
support
produced
attached
is
to
in
(Figure
measured
the
spring
and
and
the
extension
produced
in
each
case
is
measured.
Young modulus
The
data
Force
is
obtained
the
Extension
and
rigid
is
can
be
manipulated
is
the
therefore
variable
plotted
used
to
variable
being
on
plot
a
and
is
graph
measured
the
y-axis.
of
extension
therefore
as
a
Figure
plotted
result
19.3.2
of
a
against
on
the
change
shows
a
force.
x-axis.
in
force
typical
graph
support
that
is
obtained
by
stretching
a
spring.
extension
spring
metre
rule
A
mass
O
Figure 19.3.2
From
O
Figure 19.3.1
Demonstrating Hooke’s law
the
and
can
be
Graphical representation of Hooke’s law
graph
A.
force
In
you
this
expressed
can
region,
seen
the
that
mathematically
F
there
extension
=
is
is
a
straight
line
proportional
to
portion
the
between
force.
This
as
ke
Exam tip
The
term
k,
in
the
expression
is
called
the
spring
constant.
It
is
defined
–1
Sometimes
pay
on
close
a
F
against
attention
question
e
to
is
plotted.
the
axes
as
the
force
per
In
the
region
unit
extension.
Its
unit
is
N m
.
So
given
OA,
the
spring
is
said
to
obey
Hooke’s
law.
paper.
Definition
Hooke’s
law
extension
to
elastic
Stress,
F
Consider
states
that
produced
the force
the
Definition
limit
is
The
proportional
provided
not
strain
a
is
applied,
the
the force
constant
acting
per
k
is
defined
unit
as
extension.
that
exceeded.
and the Young
cylindrical
spring
piece
of
material
modulus
of
length
l,
having
a
cross-
A
sectional
(Figure
l
area
of
19.3.3).
material.
The
A.
Suppose
This
stress
force
is
exerted
a
force
called
on
F
a
the
is
applied
tensile
on
force
material
is
one
end
because
the
force
of
it
is
the
material
stretching
acting
normally
the
per
–2
Figure 19.3.3
166
Defining tensile stress
unit
cross-sectional
area.
The
SI
unit
for
stress
is
the
pascal
(Pa)
or
N m
Chapter
Definition
19
Phases
of
matter
Equation
F
Stress
is
defined
as
the force
acting
stress
=
A
per
unit
cross-sectional
area.
F
–
force/N
A
–
cross-sectional
2
area/m
–2
Unit
Suppose
(Figure
the
force
19.3.4).
F
The
causes
the
strain
material
exerted
on
to
the
of
stress
produce
/Pa
an
material
is
or
N m
extension
the
ratio
e
of
the
A
extension
units
produced
because
it
is
to
a
the
ratio
original
of
length
of
the
material.
Strain
has
F
no
lengths.
e
l
Definition
Equation
Figure 19.3.4
Defining tensile strain
e
Strain
is
defined
as
the
ratio
of
the
strain
=
l
extension
the
to
the
original
length
of
material.
e
–
extension/m
l
–
original
Strain
no
Another
quantity
of
importance
is
the
is
a
length/m
ratio
of
lengths
and
has
units
Y
oung
modulus .
This
is
the
Equation
ratio
of
the
tensile
stress
to
the
tensile
strain.
The
SI
unit
for
the
Y
oung
–2
modulus
is
also
the
pascal
(Pa)
or
tensile
N m
Young
modulus
(E)
tensile
Example
A
metal
strain
–2
E – Young
wire
of
length
2.0 m
is
clamped
vertically.
A
load
is
attached
to
–7
it
stress
=
extends
0.56 mm.
The
cross-sectional
area
of
the
wire
is
1.1
×
10
modulus
/Pa
or
N m
and
2
m
Calculate
i
the
strain
ii
the
force
in
the
applied
wire
to
the
wire.
11
[Y
oung
modulus
of
wire
=
2.0
×10
Pa]
–3
e
0.56
×
10
–4
i
Strain
=
=
=
l
F/A
ii
E
=
2.8
×
10
2.0
Fl
=
e/l
Ae
EAe
11
F
=
=
EA
×
strain
=
(2
×
10
=
6.16 N
–7
×
1.1
×
10
)
–4
×
2.8
×
10
l
Key points
ᔢ
Hooke’s
law
applied,
provided
spring
states
that
that
constant
k
the
the
is
extension
elastic
ᔢ
The
defined
ᔢ
Stress
is
defined
as
the force
ᔢ
Strain
is
defined
as
the
ratio
limit
as
the
is
not
the force
acting
of
produced
per
unit
is
proportional
to
the force
exceeded.
acting
per
unit
cross-sectional
extension
to
the
original
extension.
area.
length
of
the
material.
ᔢ
The Young
modulus
is
the
ratio
of
the
tensile
stress
to
the
tensile
strain.
167
19.4
Experiment to determine the Young
Learning outcomes
On
completion
should
be
able
of
this
Experiment to
section,
you
The
a
to:
to
ᔢ
describe
an
experiment
Y
oung
length
a
of
piece
the Young
modulus
of
metal
in
the form
of
wire
of
a
wire.
can
material
The
be
like
copper
extension
very
small.
can
produced
In
order
to
be
determined
when
a
produce
load
is
by
using
applied
a
this
experiment,
a
piece
of
wire
having
a
measurable
length
metre
is
of
at
least
used.
wire.
In
a
this
rigid
wire
experiment
support,
and
mass
the
support
the
to
because
supporting
(Figure
adjacent
other
attached
needed,
is
used
the
it
19.4.1),
to
end
each
as
to
the
test
keep
compensates
two
other
.
the
for
pieces
One
wire.
wire
of
wire
The
wire
is
are
used
attached
as
reference
vertical.
temperature
The
a
wire
has
reference
changes
to
reference
and
a
wire
sagging
is
of
beam.
beam
The
diameter
gauge.
test
in
of
one
a
modulus
copper
E
to
extensions
determine
measure
modulus
of
Several
the
test
wire,
measurements
d,
is
are
measured
taken
at
using
different
a
micrometre
positions
on
screw
the
wire
wire
and
the
average
value
is
determined.
This
reduces
the
random
error
in
reference
the
value
of
d
wire
The
original
measuring
vernier
scale
A
load
wire
The
fixed
mass
is
and
length
first
the
This
extension
is
ᔢ
The
test
wire
using
is
a
l,
is
measured
balance.
measured
is
repeated
several
also
repeated
when
can
be
Measurements
load
the
measured
experiment
experiment
wire.
of
using
a
metre
rule
or
tape.
used
to
original
be
to
It
using
is
a
times,
if
using
the
attached
V
ernier
unloading
determine
then
masses
elastic
the
test
scale.
increasing
the
to
limit
loads.
from
was
the
The
test
exceeded.
taken:
length
of
the
wire
l
is
measured
using
a
metre
rule
or
tape.
Figure 19.4.1
Experiment to determine
ᔢ
The
diameter
of
the
wire
d
is
measured
using
a
micrometre
the Young modulus of a metal
gauge.
ᔢ
The
mass
of
the
ᔢ
The
extension
each
of
the
load
wire
is
is
measured
measured
Determination of the Young
The
results
Original
of
the
experiment
length
Diameter
of
of
can
wire
wire
using
using
a
a
balance.
vernier
scale.
modulus
be
represented
=
l m
=
d m
=
π
2
as
follows:
2
d
πd
2
Cross-sectional
The
168
results
can
be
area
of
entered
wire
in
a
table
(
2
)
such
=
m
4
as
the
following.
screw
a
Chapter
19
Phases
of
matter
Table 19.4.1
m/kg
F
=
mg/N
Extension/m
(loading)
Extension/m
(unloading)
1
2
3
4
5
6
A
graph
results
of
in
force
the
against
extension
is
plotted
(Figure
19.4.2)
using
the
table.
force /N
extension /m
Figure 19.4.2
The
Plot a graph of force against extension
gradient
of
the
linear
region
of
stress
Y
oung
modulus
the
F
=
graph
e
=
determined.
Fl
÷
strain
is
=
A
l
eA
F
The
gradient
of
the
graph
is
e
l
Therefore,
the
Y
oung
modulus
=
gradient
×
A
force/N
Example
Figure
19.4.3
length
2.0 m
shows
the
results
of
an
experiment
using
a
metal
wire
with
42
and
diameter
1.1 mm.
2
area
of
wire
=
π
(
1.1
)
=
π
×
)
2
–7
9.5
10
(
2
=
2
–3
d
Cross-sectional
×
10
2
m
42
4
Gradient
of
straight
line
=
=
7.0
×
10
–1
N m
–3
0.6
×
10
0
0.6
extension
/mm
l
Y
oung
modulus
=
gradient
of
straight
line
×
Figure 19.4.3
A
2.0
4
=
7.0
×
10
11
×
=
1.47
×
10
Pa
–7
9.5
×
10
169
19.5
Elastic
and
Learning outcomes
On
completion
should
be
able
of
this
distinguish
brittle
and
section,
you
Ductile
recall
the
between
example
ductile,
polymeric
materials
stress–strain
ductile,
materials
ductile
brittle
and
material
of
a
is
can
ductile
force-extension
obeyed.
easily
material
graph
The
be
for
point
a
is
stretched
copper
.
ductile
A
is
and
Figure
material.
called
formed
the
19.5.1
In
limit
the
of
into
a
wire.
shows
region
a
An
typical
OA,
Hooke’s
proportionality.
Beyond
graphs
this
for
graphs
to:
law
ᔢ
deformation
Force–extension
A
ᔢ
plastic
point
Hooke’s
law
is
not
obeyed
and
the
graph
is
no
longer
linear
.
polymeric
The
point
B
is
called
the
elastic
limit.
This
is
the
maximum
load
which
materials
a
ᔢ
understand the terms
proportionality,
the
limit of
elastic limit
body
and
the
ᔢ
distinguish
between
elastic
a
is
experience
removed.
point
point
C
is
is
material
and
still
When
reached
called
breaks.
regain
the
force
where
the
The
yield
in
there
is
point.
graph
also
its
original
size
increased
a
marked
The
point
illustrates
and
shape,
beyond
the
increase
in
D
the
is
the
elastic
extension.
point
regions
once
at
which
where
there
and
is
plastic
load
limit,
The
yield point
can
elastic
defor mation
(O
to
B)
and
plastic
defor mation
(B
to
D).
In
the
deformation.
elastic
force
region,
is
the
removed.
original
shape
material
In
and
the
size
returns
plastic
when
to
its
region,
the
original
the
force
is
shape
material
and
does
size
not
when
return
to
the
its
removed.
Definition
force /N
Elastic
deformation
returns
to
its
–
original
the
material
shape
and
size
C
B
when
the
external force
is
D
removed.
A
Definition
Plastic
does
deformation
not
return
to
–
its
the
material
original
shape
O
and
size
when
removed from
the
external force
is
extension /m
it.
elastic
plastic
deformation
Figure 19.5.1
material
deformation
Force–extension graphs for ductile materials – metal (copper)
breaks
force /N
at
this
point
Brittle
Glass
materials
and
extension
gradient
is
only
cast
iron
graph
(Figure
an
material
for
good
brittle
19.5.2).
elastic
will
are
There
region.
return
to
examples
materials
This
its
is
no
a
brittle
that
shape
materials.
straight
plastic
means
original
of
is
line
region
once
and
the
with
on
the
force
The
a
force–
steep
graph.
is
There
removed,
the
size.
extension /m
Figure 19.5.2
Force–extension graphs for
Polymeric
materials
An
of
brittle materials – (glass and cast iron)
is
example
usually
Hooke’s
force /N
a
polymeric
non-linear
law.
A
small
Stress–strain
Strain–strain
show
extension /m
Figure 19.5.3
Force–extension graphs for
polymeric materials – (rubber and elastic)
170
the
polymeric
not
return
has
been
its
are
produces
similar
applying
materials.
to
force
is
19.5.3).
rubber
.
This
a
The
means
large
force–extension
that
it
does
not
graph
obey
extension.
graphs
graphs
effects
material
(Figure
In
the
original
permanently
and
to
force–extension
removing
case
length
of
the
when
deformed.
loads
on
ductile
the
load
graphs.
ductile,
material,
is
Figures
brittle
the
removed.
19.5.4–6
and
material
The
does
material
Chapter
In
the
case
of
the
brittle
material,
the
material
will
return
to
its
19
Phases
of
matter
original
Definition
shape
In
and
the
shape
size
case
and
unloading
of
unit
the
the
size.
The
ultimate
polymeric
It
should
process,
hysteresis.
per
if
volume.
pointed
is
lost
between
The
stress
material,
be
energy
area
tensile
as
the
rubber
the
out
exceeded.
gets
returns
during
The
curves
actually
not
material
that
heat.
two
is
effect
the
is
in
its
known
represents
warmer
to
loading
the
the
as
ultimate
maximum
It
maximum
lost
tensile
stress
withstand.
elastic
energy
is
a
stress
shown
point
is
material
on
a
as
the
can
the
stress–strain
curve.
process.
breaking
stress /Pa
The
original
and
point
stress /Pa
stress /Pa
loading
loading
unloading
unloading
strain
Figure 19.5.4
Strain
Figure
that
strain
strain
A ductile material
Figure 19.5.5
A brittle material
Figure 19.5.6
A polymeric material
energy
19.5.7
obeys
shows
Hooke’s
how
law.
the
The
extension
work
varies
done
for
with
an
force
for
extension
of
a
e
material
is
given
force /N
by
1
the
shaded
area.
The
area
of
the
triangle
is
×
base
×
F
height.
2
1
∴
work
done
=
Fe
2
Since
as
work
strain
is
being
done
on
the
material,
energy
is
stored
in
the
material
energy.
1
∴
strain
energy
=
Fe
2
e
extension /m
Example
A
thin
strain
steel
suspended
of
the
wire
from
wire.
the
the
extension
ii
the
energy
an
of
length
ceiling.
A
1.2 m
mass
of
and
diameter
4.0 kg
is
0.6 mm
attached
to
produced
stored
assumption
in
in
the
you
the
steel
modulus
F/A
i
E
=
lower
energy
Strain energy
end
wire.
made
in
your
calculation.
for
steel
=
2.0
×10
–1
Pa,
g
=
9.81 N kg
Key points
]
Fl
ᔢ
Elastic
deformation
–
the
=
e/l
Force
Figure 19.5.7
wire
11
[Y
oung
the
is
Calculate:
i
State
initially
exerted
Ae
on
material
wire
=
4.0
×
shape
9.81 N
force
2
0.6
×
and
is
size
to
its
when
original
the
removed from
external
it.
2
–3
d
returns
10
2
Cross-sectional
area
of
wire
=
π
(
2
)
=
π
(
)
2
m
ᔢ
Plastic
deformation
material
F/A
(4.0
×
does
not
–
the
return
to
its
9.81)(1.2)
–4
E
=
=
=
e/l
×
×
10
original
m
(
and
size
when
the
10
11
π
shape
2
–3
0.6
8.3
)
2
(2.0
×
10
external force
)
ᔢ
1
When
a
is
removed from
material
is
it.
stretched,
1
4
ii
Energy
stored
in
the
wire
=
Fe
=
2
×
(4.0
×
9.81)
×
(8.3
×
10
)
external
work
is
done
on
the
2
material. The
energy
stored
in
–2
=
In
the
calculations,
it
is
assumed
that
1.63
Hooke’s
×
10
law
is
J
obeyed.
the
material
is
stored
as
strain
energy.
171
Revision
Answers
found
1
to
on
a
questions
the
that
require
questions
calculation
can
10
Calculate:
be
accompanying CD.
Distinguish
between
the
structure
of
i
the
spring
ii
the
work
constant
done
in
[2]
compressing
the
spring
by
crystalline
35 mm.
and
non-crystalline
solids.
7
Give
b
an
example
Describe
the
example
2
a
Explain
b
Describe
of
of
each.
structure
a
what
of
commonly
is
meant
you
a
polymer. Give
used
by
would
a
Explain
what
one
polymer.
Brownian
[3]
motion.
by
the
terms:
demonstrate
i
stress
[1]
ii
strain
[1]
iii
Young
modulus.
[1]
Some
bridges
are
designed
such
that
many
cables,
Brownian
made
of
high-tensile
steel,
are
used
to
support
the
[3]
roadway. A
Interpret
the
results
of
such
an
experiment.
[2]
8.0
cable
Distinguish
between
a
solid
and
a
liquid
by
particular
–3
of
3
meant
[2]
motion.
c
is
[2]
8
how
[2]
[2]
×
is
cable
has
a
cross-sectional
area
2
10
m
and
length
40 m. The
tension
in
this
650 kN.
reference
11
to
the
spacing,
ordering
and
motion
of
the
[Young
particles.
[3]
4
a
b
Define
the
Derive
an
term
pressure
expression for
and
the
state
its SI
pressure
unit.
modulus
exerted
by
b
A
the Young
modulus
of
a
i
the
extension
ii
the
strain
9
a
Describe,
material.
can
steel
wire
initially
of
length
1.4 m
mass
0.5 mm
of
3.5 kg
is
is
suspended
attached
to
from
the
the
10
Pa]
the
cable
stored
in
[5]
the
cable.
[2]
the
to
aid
of
a
measure
diagram,
the
the Young
apparatus
modulus
of
a
of
Describe
copper
the
wire.
method
[4]
used
to
determine
the
ceiling.
lower
end
measurements.
[4]
of
c
the
of
energy
with
used
required
A
×
and
b
diameter
2.8
[2]
length
thin
=
Calculate:
that
Define
steel
a
[4]
a
the
[2]
fluid.
5
of
Describe
how
the
measurements
taken
can
be
wire.
used
to
determine
the Young
modulus.
[4]
Calculate
i
the
extension
ii
the
energy
State
an
produced
stored
in
the
assumption you
in
the
steel
wire
[3]
wire.
made
10
[2]
in your
a
Define
b
Distinguish
calculation.
the
terms
stress
between
and
elastic
strain.
and
[2]
plastic
deformation.
[2]
11
[Young
modulus for
steel
=
2.0
×
10
Pa,
11
–1
g
6
=
9.81 N kg
a
State
Hooke’s
b
Explain
what
law.
is
[2]
mean
by
the
term
spring
constan t.
c
A
spring
is
Give
[1]
compressed
variation
example
by
applying
a force
to
for
each
a
a
b
a
polymeric
c
a
ductile
of
this force
F
with
and
sketch
a force–extension
graph
the following:
brittle
compression
x
material
[2]
material
[2]
material.
[2]
each
material,
use
the
terms
plastic
and
elastic
to
is
describe
shown
of
it.
For
The
an
]
their
behaviour
as
indicated
by
the
graphs.
below.
[6]
force /N
12
Give
an
example
of
each
of
these
materials.
120
80
For
40
a
ductile
ii
a
brittle
iii
a
polymeric
each
strain
0
10
20
30
40
compression/mm
172
i
of
material
material
these
graph.
[1]
[1]
material.
materials,
sketch
[1]
a
possible
stress–
[3]
Revision questions
13
Define
the following
a
limit
b
elastic
of
c
yield
terms:
15
proportionality
[1]
limit
A
and
light
steel
rigid
wire
bar
as
is
suspended
shown
horizontally
by
a
10
brass
below.
[1]
point.
[1]
10
14
Aluminium
A
piece
to
A
a
of
rigid
load
of
has
a Young
aluminium
wire
support. The
30 N
is
modulus
of
length
diameter
attached
to
of
of
7
.
1
×
10
1.5 m
the
the free
Pa.
is
wire
attached
is
brass
1.5 mm.
steel
end.
Calculate:
a
the
stress
b
the
extension
on
the
c
the
strain
wire
[2]
produced
energy
stored
[2]
in
the
wire.
[2]
12 kg
The
length
12 kg
is
of
each
attached
wire
to
the
is
2.2 m. When
bar,
it
remains
a
mass
of
horizontal.
11
[Young
modulus
of
brass
=
1.0
×
10
Pa
11
Young
modulus
a
Calculate
b
Given
of
the
that
steel
tension
the
the
is
6.0
brass
×
2.0
in
×
each
10
10
Pa]
wire.
cross-sectional
–7
wire
=
[2]
area
of
the
brass
2
m
,
calculate
the
extension
of
wire.
c
Calculate
d
Determine
the
[3]
energy
the
stored
diameter
of
in
the
the
brass
steel
wire
wire.
[2]
[3]
173
Module
Answers
to
the
selected
structured
3
Practice
multiple-choice
questions
questions
can
and
be found
exam
to
on
questions
6
the
Which
of
evaporation
accompanying CD.
a
the following
and
Boiling
boiling
occurs
of
at
statements
is
true
about
the
water?
any
temperature
evaporation
occurs
100 °C
Evaporation
occurs
at
and
Multiple-choice questions
b
boiling
1
Which
of
the following
thermometers
can
be
measure
the
temperature
at various
positions
of
produced
by
a
Bunsen
A
platinum-resistance
The
A
thermocouple
c
A
mercury-in-glass
external
and
pressure.
rate of
evaporation
changes
in the
and
surface
boiling
are
unaffected
area of the water.
Evaporation
and
boiling
occur
only
at
the
surface
thermometer
of
b
temperature
the
burner?
d
a
any
on
a
by
flame
dependent
used
c
to
is
the
water.
3
7
thermometer
What
is
the
pressure
of
a
gas
of
density
0.
12 kg m
–1
d
A
constant-volume
gas
and
thermometer
root
mean
square velocity
of
1200 m s
?
4
a
2
Which
of
the following
is
not
an
assumption
of
a
theory
The
size
of
compared
b
of
The force
the
molecules
volume
is
negligible
occupied
attraction
by
between
the
the
8
when
144 Pa
The first
ΔU
gas.
molecules
=
is
change
3
The
collisions
d
The
duration
A
car
to
is
has
rest
a
by
is
and
–1
a
10
Pa
2.5
×
10
can
be
Pa
into
thermodynamics
ΔW. When
there
law
is
of
no
a
gas
change
undergoes
in
expressed
an
temperature
thermodynamics for
an
as
isothermal
of
the
gas.
isothermal
is
inelastic.
a
the
+
of
collision
time
of
thermal
mass
of
their
average
60%
energy
the
the
in
brake
specific
car
is
brought
kinetic
the
9
ΔU
c
– ΔQ
=
ΔW
energy
brakes
is
is
ΔW
the
wire from
of
capacity
=
A force–extension
What
components
heat
a
b
ΔU
d
ΔQ
=
ΔQ
when
collisions.
55 kJ. The
brake.
total
negligible
between
energy
the
is
graph
amount
extension
a
0.004 J
b
0.001 J
c
0.009 J
d
0.005 J
of
of
wire
work
2 mm
is
done
to
=
ΔW
shown
in
–
ΔU
below.
stretching
the
3 mm?
–1
510 J kg
the
of
applying
car. The
25 kg
are
kinetic
converted
the
d
law
ΔQ
change,
negligible.
to
×
5
The first
c
compared
5.76
gases?
the
of
b
the
c
kinetic
48 Pa
K
brake
. What
is
the
increase
in
temperature
of
components?
4.3 °C
b
0.004 °C
c
2.6 °C
d
0.23 °C
F /N
4
An
insulated
of two
composite
metals
P
and Q.
rod of
P
length
and Q
80 cm
are of the
consists
same
30
lengths. The thermal
–1
400 W m
of
P
is
at
K
a
–1
and
300 W m
a temperature of
12 ° C. What
and Q
conductivities of
–1
are
P
and Q
are
–1
K
respectively. One
90 °C
is the temperature
and one
at the
end
end of Q
is
point where
20
at
P
e /mm
2
joined?
62 °C
b
57 °C
c
48 °C
d
51 °C
10
5
According
to Stefan’s
law,
the
rate
at
which
The
length
l
of
transferred from
a
black
body
is
given
by
a
material X
is
measured
when
energy
various forces
is
3
F
are
applied
to
it. The
graph
below
which
shows
the
results
of
this
investigation. Which
of
the
equation?
following
is false?
2
a
P
=
σAT
c
P
=
σAT
b
P
=
σAT
d
P
=
σTA
4
a
The
b
The force
material
c
The
ultimate
d
The
work
Hooke’s
law
up
to
30 N.
–1
1.22 m
174
obeys
4
to
constant
of
tensile
done
in
1.24 m
the
material
stress
is
stretching
is
0.65 J
at
is
750 N m
35 N.
the
material from
Module
13
F /N
a
b
Explain
what
is
specific
heat
ii
specific
latent
an
Practice
meant
i
Describe
3
by
questions
the following:
capacity
heat
exam
of
copper
of fusion
experimental
of
method
[2]
ice.
to
[2]
measure
35
the
specific
heat
capacity
of
a
liquid.
[6]
33
30
c
A
kettle
water
is
at
rated
25 °C
switched on
at
is
2.0 kW. A
poured
and
mass of
into
it takes
a
800 g of
kettle. The
kettle
is
2.3 minutes for the water
15
to
start
boiling. The
8.6 minutes for the
half
kettle
is
left on
and
it takes
mass of water to decrease
its original value. Use the data to
by
calculate:
l /m
1.20
1.24
1.22
i
the
specific
heat
capacity
ii
the
specific
latent
heat
of
water
[3]
of vaporisation
of
water.
iii
[2]
State one assumption used in your calculations.
Structured questions
[1]
11
a
Explain
what
b
Explain
how
c
is
a
meant
by
physical
temperature.
property
which varies
with
measure
temperature
Describe
the
the
resistance
temperature
of
a
of
a
may
[1]
substance
be
used
of
a
a
State four
b
Explain
i
to
substance.
principal features
14
ii
platinum-
thermometer.
Discuss
the
advantages
and
disadvantages
of
the
a
[4]
iii
d
the
the following
air
inner
[2]
inside
a
in-glass
same
e
A
a
thermometer
temperature
student
using
a
which
measures
the
be
a
the
used
temperature
of
in
a
b
the
temperatures
do
not
exerts
kinetic
a
has
air
[4]
theory:
pressure
on
the
[4]
been
inside
the
tyre
increases
after
driven
pressure
[3]
decreases
the
when
some
air
tyre.
Three
molecules
have
[2]
speeds
v,
2v
and
4v.
Calculate:
liquid
thermometer
thermometer. The
the
theory.
mercury-
[4]
platinum-resistance
that
may
and
kinetic
a
range.
mercury-in-glass
notices
thermometer
the
using
tyre
pressure
escapes from
platinum-resistance
of
walls
air
car
the
assumptions
and
i
the
mean
ii
the
root
square
speed
mean
square
gas
heated
of
the
speed
molecules
of
the
[2]
molecules.
[2]
student
c
agree. They
2.4 mol
of
a
is
at
a
constant
pressure
5
do,
however,
temperature
Explain
why
agree
of
when
steam
this
is
used
above
to
measure
pure
boiling
of
the
×
10
Pa. The
increases from
water.
so.
4.0
Molar
[3]
heat
temperature
250 K
capacity
to
of
–1
pressure
12
a
b
Explain
what
is
i
Conduction
iii
Radiation.
Explain
what
meant
by
ii
meant
Convection
by
thermal
c
state
Describe
energy
i
a
its SI
and
transfer
metal
unit.
state
in
State
and
the
i
the
amount
ii
the
change
iii
the
work
a
Explain
mechanism for
thermal
i
used
to
at
constant
of
energy
in volume
done
by
the
supplied
of
the
to
the
gas
gas
[2]
[2]
gas.
[2]
what
is
meant
by:
i
tensile
stress
[1]
ii
tensile
strain
[1]
iii
Young
thermal
ii
a
ceramic
cup.
the
choice
of
dimensions
experimentally
conductivity
a good conductor
determine
modulus.
[1]
[6]
Sketch
stress–strain
graphs for
the following
and
the
give
material
gas
.Calculate
the following:
spoon
explain
the
–1
K
[3]
b
d
gas
conductivity
15
and
29 J mol
the
the following:
[6]
is
is
of
300 K.
an
example
of
each
type.
the
i
A
brittle
ii
A
ductile
material
iii
A
polymeric
[2]
of:
ii
a poor conductor.
material
[2]
[4]
material
[2]
175
20
Analysis
20.
1
Analysis
Learning outcomes
On
completion
should
be
able
of
this
and
interpretation
and
interpretation
Plotting
section,
you
In
experimental
One
to:
way
of
unknown
ᔢ
draw
straight
line
graphs
determine
data
data
is
is
often
by
collected.
drawing
a
This
data
suitable
must
graph
to
be
analysed.
determine
constants.
following
guidelines
should
be
used
when
drawing
graphs.
data
ᔢ
ᔢ
work,
analysing
using
The
experimental
graphs
gradients
Choose
a
scale
that
is
suitable
(e.g.
1 cm
to
5 units,
1 cm
to
10 units).
and
Do
not
use
scales
such
as
1 cm
to
3 units.
intercepts
ᔢ
ᔢ
rearrange
equations
to
obtain
The
at
straight
scale
should
be
chosen
such
that
the
points
being
plotted
occupy
a
least
half
the
graph
paper
.
line form.
ᔢ
The
and
x-
and
the
ᔢ
The
graph
ᔢ
A
or
ᔢ
If
×
the
data
either
If
be
should
be
between
T
wo
an
one
usually
the
mx + c
of
are
to
is
called
general
the
line
to
m.
be
It
linear
a
are
quantities
being
plotted
title.
the
points
relationship,
equal
being
a
numbers
non-linear
the
data
when
the
line
of
plotted.
of
data
relationship,
points.
Do
changed
is
other
a
at
usually
y-axis.
of
plotting
not
on
graph
time.
variable.
only
quantities
straight
a
plotting
Since
intercept
a
relationship
responding
the
The
appropriate
represent
a
the
temperature/°C).
best
fit
points
on
a
use
smooth
a
curve
straight
line
points.
the
equation
is
follow
required
all
an
there
through
can
on
with
height/m,
to
follow
that
labelled
line.
variable.
plotted
used
investigate
quantity
experiment,
The
=
the
drawn
quantities
quantity
y
of
be
T/s,
given
be
to
adjacent
manipulated
be
so
appear
experiment
only
y-axis
appear
data
d/cm,
should
drawn
side
the
should
(e.g.
should
a
should
ᔢ
y-axes
units
are
line
the
is
This
on
y
=
y-axis
as
c.
is
is
+
c.
called
The
In
the
second
variable
being
constant
Consider
y-axis).
quantities,
The
responding
mx
is
x-axis.
quantity
known
and
two
quantity
the
The
one
( x-axis
between
is
changed
in
variables.
gradient
Figure
or
slope
20.1.1.
y
2
B(x
,y
2
)
A
and
B
lie
on
the
straight
line.
The
coordinates
of
A
and
B
are
( x
2
,
y
1
and
(x
,
y
2
y
)
respectively.
The
gradient
of
the
line
passing
through
A
)
1
and
B
2
1
A(x
,y
1
is
)
given
by:
1
(0,c)
y
–
y
–
x
2
Gradient
(slope)
1
=
x
2
1
Example
x-axis
x
1
x
2
A
system
is
oscillation
Figure 20.1.1
oscillating
T
is
with
dependent
simple
on
the
harmonic
length
of
motion.
the
The
oscillator
x.
period
It
is
of
known
A straight-line graph
that
T
and
x
are
related
by
the
following
equation:
–b
T
The
of
following
the
case.
176
table
oscillator
It
is
is
required
shows
varied
that
the
and
the
=
Ax
results
the
of
period
constants
A
an
of
experiment
oscillation
and
b
be
where
is
the
measured
determined.
x/m
0.40
0.50
0.60
0.70
0.80
0.90
T/s
2.85
2.32
2.
15
2.01
2.01
1.90
length
in
each
Chapter
The
equation
has
to
be
rearranged
such
that
a
linear
form
is
20
Analysis
and
interpretation
obtained.
b
T
T
aking
log
on
both
sides
of
=
Ax
the
equation:
10
b
lg
T
=
lg (Ax
)
lg
T
=
lg A
+
lg (x
lg
T
=
lg A
–
b
A
straight
line
graph
The
gradient
The
y-intercept
lg T
and
lg x
of
the
is
are
would
line
is
be
)
b lg x
obtained
if
lg
T
is
plotted
against
lg x
– b
lg A
determined
as
shown
in
the
following
table.
x/m
0.40
0.50
0.60
0.70
0.80
T/s
2.85
2.32
2.
15
2.01
2.01
1.90
–0.397
–0.301
–0.097
–0.046
lg (x/m)
lg (T/s)
A
line
The
In
0.455
of
best
graph
order
of
to
fit
lg
find
0.407
is
T
–0.222
–0.
155
0.365
0.332
0.90
0.303
0.279
drawn.
against
the
lg x
gradient
is
of
shown
the
in
line,
Figure
two
20.1.2.
points
A
and
log (T / s)
B
are
chosen
0.5
on
the
line.
These
points
are
not
one
of
the
experimental
data
points
and
A
should
be
sufficiently
far
apart
to
produce
a
large
right-angled
triangle.
0.4
A
B
–
–
(– 0.38,
0.45)
(– 0.065,
0.285)
B
0.285
Gradient
=
–b
–
0.3
0.45
=
=
– 0.52
– 0.065–(– 0.38)
Therefore,
From
the
b
graph,
=
the
0.2
0.52
y-intercept
=
0.25
0.
1
lg A
=
0.25
A
=
10
0.25
Therefore,
=
1.78
0
In
the
event
that
measurement
point
A
is
the
from
chosen.
y-intercept
the
The
graph,
a
cannot
point
y-intercept
lg T
=
lg A
–
b lg x
0.45
=
lg
–
0.52
lg A
=
0.252
A
=
10
A
is
be
on
determined
the
line
calculated
is
as
by
direct
chosen.
follows:
Suppose
0.4
the
0.3
0.2
0. 1
0
log (x / m)
Figure 20.1.2
(– 0.38)
0.252
=
1.79
177
Analysis and interpretation: Practice exam questions
Answers to questions that
require
calculation
can
be found on the
accompanying CD.
1
A
system
is
oscillating
such
that
the
period
T
is
related
by
the following
expression:
n
T
=
,
kx
where
x
is
the
length
of
the
oscillator
in
mm,
and
k
and
n
are
constants.
The
length
shows
2
the
x
is varied
and
the
period
T
is
measured. The
table
below
results.
x/mm
200
250
300
350
400
450
500
T/s
0.622
0.696
0.762
0.823
0.880
0.933
0.984
a
Plot
b
Use
A
a
the
graph
resonant tube
movable
a
suitable
given
is
to
is
length
adjusted
that
determine
loudspeaker
allow
the
so that
f. The
length of
is
placed
resonance occurs.
length of the tube
below
Fundamental frequency
you
value
l, the frequency of the
measured. The table
Length
will
such that the
piston. A
frequency
is
graph
l/m
of
k
air
determine
and
inside
sound
and
n.
[8]
it
[4]
can
be varied
using
end of the tube.
coming from the
Resonance occurs
is varied
k
n.
at the opened
shows the
f/Hz
to
a
For
loudspeaker
at the fundamental
and the fundamental frequency
results of the
experiment.
429
286
215
172
143
0.20
0.30
0.40
0.50
0.60
1
a
Plot
a
graph
of
against
l.
[4]
f
b
3
Use
In
an
of
the
experiment
copper,
at various
results
4
graph
an
to
determine
measure
electric
times
are
to
is
shown
heater
the
is
specific
used
measured. The
in
the
table
speed
to
of
sound
heat
heat
electric
it
in
the
capacity
and
heater
the
is
tube.
of
rated
at
90
180
270
360
420
Temp
26.2
49.2
72.3
95.3
118
134
a
Plot
b
Use
c
Determine
A
long
a
graph
the
of
graph
tube
temperature
to
the
is filled
radii
are
determine
heat
with
the viscous fluid. When
time
The
taken for
results
radius
time
of
them
the
r/mm
t/s
–1
velocity
v/cm s
–1
lg (v/cm s
lg (r/mm)
)
to
against
the
capacity
to
the
of
reach
specific
the
travel
T
150 W. The
[4]
heat
block
of
are falling
through
are
capacity
a
small
with
distance
shown
in
the
of
copper.
[5]
copper.
terminal velocity
spheres
experiment
block
time.
a viscous fluid. Several
allowed
1.5 kg
below.
0
T/°C
a
[5]
temperature
Time/s
different
178
the
of
metal
spheres
when falling
terminal
90.0 cm
table
[1]
below.
1.00
1.50
2.00
2.50
3.00
60.0
26.7
15.0
9.6
6.7
of
through
velocity,
is
the
measured.
Analysis
In
this
experiment,
sphere
by
the
terminal
the following
velocity
v
is
related
to
the
radius
and
r of
interpretation:
Practice
exam
questions
the
equation:
n
v
5
=
kr
,
where
a
Copy
b
Plot
c
Determine
In
is
an
and
a
a
and
to
of
value
to
the
recorded. The
is
moved
closer
is
constants.
the
table.
that
of
k
edge
of
a
[5]
will
allow
and
measure
string. The
is
experiment
are
graph
the
experiment
piece
n
complete
suitable
attached
by
k
of
the
table. The
modulus
300 g
the
corresponding vertical
to
determine
k
and
n.
[4]
[3]
table. A
l
to
n.
the Young
length
you
mass
metre
is
of
repeated for various values
l
and
of
l
wood,
a
attached
rule
distance
values
of
hanging
d
is
d
are
and
metre
to free
over
rule
end
the
recorded. The
table
string
recorded. The
d
l
d
d/m
0.060
0.050
0.032
0.020
0.011
0.005
l/m
0.800
0.750
0.650
0.550
0.450
0.350
3
l
3
/m
Width
of
ruler
Thickness
It
is
of
known
w
=
ruler
that
3 cm
t
d
=
6 mm
and
l
are
related
by
the following
equation:
4mg
d
where
due
in
to
m,
a
m
(
=
is
the
gravity,
and
Copy
3
3
Ewt
t
is
mass
E
is
the
and
)
l
attached
the Young
thickness
complete
of
the
the
metre
modulus
the
ruler
rule
in
of
wood,
in
m.
kg,
w
g
is
is
the
the
acceleration
width
of
the
ruler
table.
[5]
3
6
b
Plot
c
Determine
d
Use
An
graph
your
alloy
length
are
a
in
of
table
d
to
the form
wire
to
is
the
below
against
gradient
graph
the
attached
The
of
the
l
.
of
estimate
of
a
wire
1.80 m
wire
shows
line
obtained.
the Young
is
and
and
the
[5]
the
attached
its
the
[3]
modulus
to
a
diameter
rigid
is
of
wood.
[3]
support. The
0.5 mm. Several
corresponding
extension
x
is
original
weights
measured.
results.
F/N
5
10
15
20
25
30
35
40
x/mm
0.40
0.70
1.
10
1.45
1.80
2.30
3.30
5.
10
a
Plot
a
graph
b
Estimate
the
c
Use
graph
your
of
F
against
work
to
F
done
x.
in
[4]
stretching
determine
the
the Young
wire
by
modulus
3.0 mm.
[3]
of
[4]
the
alloy.
179
List
of
physical
constants
−11
Universal
gravitational
constant
G
=
6.67
×
10
g
=
9.80 m s
=
6380 km
=
5.98
2
N m
–2
kg
–2
Acceleration
Radius
of
due
the
to
gravity
Earth
R
E
24
Mass
of
the
Earth
M
×
10
kg
E
22
Mass
of
the
Moon
M
=
7.35
×
10
kg
M
5
Atmosphere
atm
=
1.00
×
10
–2
N m
–23
Boltzmann’s
constant
k
=
1.38
×
10
–1
J K
9
Coulomb
constant
=
9.00
×
10
2
N m
–2
C
–31
Mass
of
the
electron
m
=
9.11
×
10
kg
e
–19
Electron
charge
e
=
1.60
×
10
C
3
Density
of
water
=
1.00
×
10
Resistivity
of
steel
=
1.98
×
10
Resistivity
of
copper
=
1.80
×
10
=
400 W m
–3
kg m
–7
Ω m
–8
Ω m
–1
Thermal
conductivity
of
copper
–1
K
–1
–1
Specific
heat
capacity
of
aluminium
=
910 J kg
K
Specific
heat
capacity
of
copper
=
387 J kg
Specific
heat
capacity
of
water
=
4200 J kg
Specific
latent
heat
of
fusion
=
3.34
×
10
Specific
latent
heat
of
vaporisation
=
2.26
×
10
–1
–1
K
–1
–1
K
5
of
ice
–1
J kg
6
of
water
–1
J kg
23
Avogadro
constant
N
=
6.02
×
10
per
mole
A
8
Speed
of
light
in
free
space
c
=
3.00
×
10
=
4π
=
8.85
×
10
h
=
6.63
×
10
u
=
1.66
×
10
–1
m s
–7
Permeability
of
free
space
μ
×
10
–1
H m
0
12
Permittivity
of
free
space
ε
–1
F m
0
–34
The
Planck
constant
J s
–27
Unified
atomic
mass
constant
kg
–27
Rest
mass
of
proton
m
=
1.67
×
10
R
=
8.31 J K
σ
=
5.67
×
10
=
1.67
×
10
kg
p
–1
Molar
gas
constant
–1
mol
–8
Stefan–Boltzmann
constant
–2
W m
–27
Mass
of
neutron
m
n
180
kg
–4
K
Glossary
A
Destructive
C
out
Absolute
refractive
index
The
ratio
of
Centre of
gravity
The
point
speed
of
light
in
a
vacuum
to
the
which
all
the
weight
of
a
body
speed
of
light
180°,
in
the
The
medium.
lowest
to
temperature
or
a
number
a
Waves
phase
meet
difference
path
of
difference
half
of
an
wavelengths)
act.
giving
zero
(with
appears
odd
Absolute
interference
phase
through
of
the
of
Centripetal force
The
a
reduced
or
zero
resultant
unbalanced force
displacement.
possible
(0 K
or
–273.
15 °C).
required
to
keep
an
object
moving
in
a
Diffraction
Acceleration
The
rate
of
change
of
circular
The
wavefronts
velocity.
Charles’
law
The
volume
of
a fixed
The
eye’s
ability
to
of
gas
is
directly
proportional
to
of
an
the
shape
of
its
lens
in
order
to
absolute
temperature,
provided
produce
a focused
image
measure
of
on
the
the
retina.
closeness
the
pressure
Coherent
is
kept
waves
object
a
the
measured
value
to
the
true
that
have
and
difference
the
gap.
piece
of
number
glass
of
or
equally
line
drawn.
the
hence
a
between
This
is
the
distance
constant
moved from
phase
a
constant.
Waves
same frequency
value.
of
passes
through
A
large
Displacement
of
or
grating
with
spaced
The
out
wave
that
plastic
Accuracy
a
its
Diffraction
adjust
when
mass
edge
Accommodation
spreading
path.
a fixed
point
in
a
stated
them.
direction.
Air
resistance
A force
which
opposes
Compression
The
region
on
a
Drag force
the
motion
of
an
object
moving
longitudinal
wave
where
the
motion
through
air.
are
moving
towards
each
A
solid
where
there
is
a
Conduction
The
process
by
arrangement
of
the
atoms
thermal
energy flows
in
an
which
object
in
opposes
the
a fluid.
a
material
A
material
that
can
which
be
random
of
other.
Ductile
Amorphous
A force
particles
easily
stretched
and formed
into
a
solid from
wire.
throughout
Amplitude
from
the
the
The
structure.
maximum
equilibrium
a
displacement
position.
region
region
the
of
of
higher
lower
temperature
temperature
movement
of
the
to
a
E
without
material
itself.
Efficiency
Angle of
incidence
The
angle
between
Constructive
interference
Waves
output
the
incident
ray
and
the
normal.
in
phase
(with
a
phase
difference
reflection
The
angle
between
zero,
or
a
path
difference
of
a
reflected
ray
and
the
normal.
number
of
wavelengths)
ratio
the
of
power
collision
A
giving
energy
is
refraction
The
angle
between
greater
resultant
refracted
ray
and
the
normal.
Convection
The
collision
process
by
The
rate
of
change
of
displacement.
thermal
of
energy flows from
higher
temperature
to
a
a
Positions
midway
between
lower
temperature
due
to
its
to
the
on
a
stationary
is
wave. The
maximum
at
this
movement
point.
density
of
the fluid
as
the
a
is
and
size
removed
it.
of
potential
energy
The
energy
bulk
result
of
by
a
body
when
deformed.
a
Electric
amplitude
material
shape
external force
possessed
nodes
The
original
region
region
Elastic
Antinode
which
which
from
angular
in
a
when
Angular velocity
power
displacement.
returns
the
useful
conserved.
Elastic deformation
Angle of
the
input.
whole
kinetic
the
to
of
Elastic
Angle of
The
meet
current
The
rate
of flow
of
change.
charge.
Archimedes’
principle
When
a
body
Couple
Two
equal
and
opposite forces
Electric
is
totally
or
partially
submerged
in
a
whose
lines
of
action
do
not
potential
possessed
fluid,
it
experiences
an
upthrust
which
Critical
angle
The
angle
of
equal
to
the
weight
of
the fluid
which
the
angle
of
by
a
charged
refraction
is
in
an
a
ray
travelling from
a
energy
due
to
its
electric field.
90° for
Electromagnetic
displaced.
The
body
incidence for
position
is
energy
coincide.
dense
to
spectrum
less
Electromagnetic
waves
arranged
in
23
Avogadro constant
6.02
×
10
per
mole
optically
dense
medium.
order
Critical damping
The
system
comes
of
Electromagnetic
rest
after
one
Crystalline
Are
kilogram,
metre,
wave
solid
The
atoms,
ions
of
oscillating
mole
and
that
and
or
magnetic fields
ampere,
wave
electric
second,
molecules
kelvin,
A
oscillation.
consists
units
wavelengths.
to
B
Base
their
are
arranged
in
a
at
right
angles
to
each
regular
candela.
other.
pattern
Black
body
A
perfect
absorber
that
repeats
itself
within
the
of
Electron diffusion
A
mechanism
by
crystal.
thermal
radiation.
which
Boiling
A
substance
absorbs
energy
state from
a
liquid
to
a
in
a
change
in
law
The
temperature.
pressure
of
place
gas
Damping
The
process
whereby
a fixed
mass
amplitude
of
an
oscillating
A
line
drawn
through
the
points
Boyle’s
takes
metals.
Equipotential
without
conduction
and
D
changes
thermal
having
the
same
gravitational
system
potential.
of
gas
is
inversely
proportional
to
its
decreases
over
time.
Energy
volume,
provided that the temperature
Degree Celsius
A
unit
used
to
The
capacity
or
ability
to
do
measure
work.
is
kept
constant.
temperature.
Evaporation
Brownian
motion
The
haphazard
Density
The
mass
per
unit
liquid
motion
of
particles.
Derived quantities
Physical
changes
than
the
base
process
into
a
by
gas
which
a
without
quantities
reaching
other
The
volume.
its
boiling
point.
quantities.
181
Glossary
F
Mechanical
I
a
First
law of thermodynamics
The
Ideal
gas
A
gas
that
obeys
the
gas
wave
substance
A
wave
through
that
which
requires
to
laws.
propagate.
change
in
internal
energy
of
a
system
Image distance
Distance
between
the
Molar
is
equal
to
the
energy
supplied
to
the
image
and
the
optical
centre
of
heat
thermal
system
plus
the
work
done
on
the
capacity
energy
the
Impulse
The
product
of
a force
and
A
standard
degree
to
of
increase
temperature
of
for
which
it
acts
on
an
of
one
mole
of
a
time
substance
point
amount
required
lens.
system.
Fixed
The
the
by
one
degree.
object.
Molar heat capacity of a gas at constant
hotness
that
can
easily
be
reproduced.
Inelastic
collision
A
collision
in
which
pressure
Focal
length
The
distance
between
the
kinetic
energy
is
not
The amount of energy
conserved.
required to raise the temperature of one
optical
centre
of
the
lens
and
the focal
Inertia
The
reluctance
of
a
body
to
mole of a gas by one degree, when the
point
of
the
lens.
start
moving
when
it
is
at
rest
or
the
pressure remains constant.
Free
body diagram
A
diagram
showing
reluctance
of
a
body
to
stop
moving
Molar heat capacity of a gas at constant
the forces
acting
on
a
body.
when
it
is
in
motion.
volume
Frequency
The
number
of
oscillations
Intensity
The
power
per
unit
The amount of energy required
area.
to raise the temperature of one mole of
per
unit
time.
Internal
energy
The
random
a gas by one degree when the volume
Frequency
response
The
range
of
distribution
of
the
kinetic
and
remains constant.
frequencies
that
can
be
detected
by
potential
energies
of
the
particles
that
Mole
the
human
ear.
make
up
the
The
amount
contains
Friction
A force
which
opposes
of
substance
that
substance.
the
same
number
of
particles
motion.
in
12 g
of
carbon-12.
K
Fundamental frequency
The
lowest
Moment
frequency
that
a
vibrating
string
can
product
of
a force
and
the
or
Kelvin
pipe
The
The
SI
unit
of
perpendicular
temperature.
distance
of
the
line
of
produce.
Kinetic
body
energy
by
Energy
virtue
of
its
possessed
by
a
action
of
Monomer
motion.
the force from
The
unit
that
a
pivot.
repeats
itself
in
G
Kinetic theory of
Geostationary
satellite
A
satellite
that
assumes
that
a
gases
gas
is
A
theory
made
up
that
a
polymer.
of
N
has
to
a
period
be
at
the
of
24 hours
same
point
and
appears
above
the
many
at
small
high
particles
moving
randomly
speeds.
Natural frequency
Earth
all
the
The frequency
with
time.
which
a
system
oscillates
without
L
Global
warming
The
increase
in
applying
temperature
of
the
Earth
as
a
result
Latent
the
greenhouse
Gravitational field
heat
where
a
The
The
region
the
periodic
driving
energy
mass
state
of
a
required
to
Newton
substance
The force
required
to
give
a
around
–2
without
body
external
force.
effect.
change
a
an
of
experiences
a
change
in
mass
temperature.
of
1 kg
an
acceleration
of
1 m s
.
a
Lattice vibration
A
mechanism
Newton’s first
by
law
A
body
stays
at
rest
force.
which
Gravitational field
strength
in
acting
per
unit
thermal
solids
without free
potential
The
work
moving
unit
mass from
infinity
if
moving
uniform
continues
velocity
to
unless
move
acted
with
upon
The
amplitude
of
an
the
by
external force.
a
eventually
period
of
decreases
to
zero
Newton’s
of
time.
law of
attraction
gravitation
between
two
The force
bodies
is
point.
Linear
Gravitational
potential
energy
momentum
of
a
body
by
virtue
of
in
a
mass
which
of
directly
a
proportional
to
their
masses
and
velocity.
A
wave
in
which
the
inversely
square
of
the
proportional
distance
to
the
between
them.
gravitational field.
effect
The
heat
gets
trapped
or
in
the
in
the
same
direction
Newton’s
second
law
The
rate
the
energy
of
transfer.
change
of
mome ntum
is
in
Loudness
greenhouse
are
process
as
by
and
wave
oscillations
Greenhouse
product
its
Longitudinal
position
The
The
body’s
energy
a
or
to
over
a
place
done
oscillation
in
takes
electrons.
mass.
Light damping
Gravitational
conduction
The force
The
subjective
sensation
of
proportional
to
the
appl ied
fo rce
and
Earth’s
sound
that
depends
of
sound.
on
the
amplitude
takes
place
in
the
directio n
in
which
atmosphere.
the
Lower fixed
the
point
The
temperature
force
acts.
Newton’s third
law
If
a
body A
exerts
a
H
of
Harmonic
The
nth
harmonic
is
n
times
pure
melting
atmospheric
ice
at
standard
Node
M
and
the
Mass
The
amount
of
matter
substance by one degree.
in
a
body.
It
is
a
measure
then
body
opposite force
on
a
B
on
is
energy
wave
zero.
of
a
replenished
Energy
that
(fossil fuels).
body’s
Normal
to
be
A
line
drawn
at
right
angles
inertia.
a
oscillate.
Mean
Hooke’s
law
The
extension
produced
square
speed
2
c
1
2
+ c
2
2
+ c
3
2
proportional
to force
applied,
provided
surface.
Normal
is
⟨c
⟩
A force
the
182
elastic
limit
is
not
exceeded.
that
acts
N
at
=
reaction
2
+ … + c
right
angles
when
one
N
that
an
body A.
stationary
amplitude
exerts
contained
cannot
system fails
B,
Non-renewable
required to increase the temperature of a
The
body
Positions
where
The amount of heat energy
Heavy damping
on
equal
pressure.
the fundamental frequency.
Heat capacity
force
contact
with
another.
body
is
in
to
Glossary
provided
O
the
Object distance
that
no
external forces
act
on
system.
Specific latent heat of fusion
energy
required to
convert
The
unit
mass
Distance between the
Principle of
moments
For
a
body
(1 kg) of
a
substance from
a
solid to
object and the optical centre of the lens.
that
Oscillation
An
object
is
moving
is
in
equilibrium,
clockwise
and forth
about
a fixed
The
notes
other
than
is
sum
equal
of
to
the
liquid without
the
Specific
latent
a
change
in temperature.
heat of vaporisation
of
the
anticlockwise
moments
The
energy
required
to
convert
unit
the
about
fundamental
moments
point.
sum
Overtone
the
back
the
same
pivot.
mass
(1 kg)
of
a
substance from
a
note.
Progressive
energy
is
wave
A
wave
in
transferred from
which
one
liquid
point
to
to
a
vapour
without
a
change
in
temperature.
P
another. The
Path difference
wave
profile
moves.
Speed
The
Stefan’s
The difference in distance
rate
law
of
The
change
rate
at
of
distance.
which
energy
R
is
travelled by two waves that interfere.
Period
The
time
taken
to
complete
proportional
one
Radian
cycle
Phase
or
A
The
angle
subtended
at
oscillation
that
of
the
the fraction
has
been
of
centre
of
a
length
to
the
circle
by
an
arc
equal
in
of
the
An
error
that
occurs
as
φ
×
result
2π
of
the
experimenter
and
property
of
an
or
phenomenon
that
can
error
that
with
an
The
depends
on
below
sensation
of
the frequency
The
region
on
wave
where
the
moving
away from
each
return
to
its
original
shape
An
image
where
the
pass
the
external force
is
The
change
wave
when
in
travelling
direction
To
restrict
the
oscillations
of
of
between
because
Polymers
A
wave
to
one
material
chains
of
the
of
a
Energy
one
index
The
body
by
virtue
of
its
at
that
of
the
of
error
unit
charge
electrical
in
The
Relative density
by
a
or
position.
work
done
converting from
to
rate
which
at
other forms.
work
is
Renewable
energy
natural
sources
geothermal,
Resonance
Energy
(sunlight,
amplitude
system
the frequency
the
The
degree
of
the
of
a
of
is
waves,
of
vectors
wind,
velocity
falling
a fluid.
driver
The force
acting
normally
per
Root
acting
mean
The
on
a
in
of
square
by
an
object
when
the
of
conductivity
heat
per
vector
unit
The
area
rate
per
of
unit
system.
sum
of
gradient,
when
the
heat
the
is
at
right
angles
to
the faces
of
a
body.
parallel-sided
slab
of
the
material,
speed
steady
state
conditions.
area.
2
c
mass
law
of
gas
The
is
pressure
directly
of
a fixed
⟨c
√
2
+ c
1
2
Pressure
⟩
=
2
+ c
2
2
Thermal
+ … + c
3
its
absolute
The
condition
N
under
√
equilibrium
which
two
objects
in
physical
N
proportional
contact
to
maximum
reached
matches
under
unit
The
constant
thin
Pressure
of
an
enhanced
the
natural frequency
Resultant vector
result.
hotness
body.
flow
reproducibility
of
derived from
temperature
measure
in
etc.).
The
oscillating
being
done.
The
error
The ratio of the density
flow
Precision
wave.
constant
of a substance with the density of water.
state
energy
the
direction.
Thermal
Power
is
individual
each
A
Terminal velocity
per
resultant
point
T
of
possessed
The
more
ratio
a
Potential difference
or
the
sin r
consists
molecules.
energy
two
point,
two
Temperature
Potential
a
change
sin i
plane.
that
per
area.
wave.
a
Refractive
transverse
at
sum
Systematic
speed
length.
of
it.
Polarise
When
arrive
displacements
occurs
unit
normally
it.
removed
and
per
acting
cross-sectional
algebraic
in
long
through
size
media
from
The force
displacement
a
when
Stress
rays
does
and
and
opposite
superimpose.
extension
Superposition
image
Refraction
not
in
other.
the
actually
material
and
The
waves
The
when
particles
sound.
Plastic deformation
wave formed
same frequency
travelling
Strain
unit
Real
A
the
a
sound
of
of
expressed
the
instrument.
subjective
of
are
directions
are
that
or
be
longitudinal
Pitch
above
power
body
value.
Rarefaction
measured
is
an
true
object
wave
waves
amplitude
in
The
the
results
λ
Physical quantity
of
kelvin.
two
=
is
a
π
Phase difference
the fourth
temperature
Stationary
error
body
circle.
completed.
Random
black
in
the
radius
to
a
the
revolution.
measure
radiated from
temperature,
with
each
other
exchange
no
provided
heat
energy. The
two
objects
are
at
the
S
that
the
volume
is
kept
constant.
same
Principal
axis
A
line
that
passes
centre
of
curvature
of
a
lens
light
is
neither
reflected
are
said
to
be
A
quantity
that
has
in
thermal
equilibrium.
such
magnitude
that
and
through
Scalar quantity
the
temperature
only.
Thermal
radiation
The
process
by
nor
Simple harmonic motion
Period motion
which
thermal
energy
is
transferred
refracted.
in which the acceleration is proportional
Principle of
to the displacement from a fixed point
energy
Energy
can
neither
be
destroyed,
but
can
be
one form
to
of
of
high
low
temperature
temperature
by
waves.
The International System of Units.
Thermistor
A
non-linear
device
with
a
another.
Snell’s
law
The
ratio
thermally
is
a
sensitive
resistor.
constant.
sin r
For
any
system,
the
momentum
before
collision
is
after
heat
capacity
The
amount
of
measuring
A
device
used for
temperature.
equal
heat
momentum
Thermocouple
total
Specific
the
region
conservation of
momentum
to
a
region
electromagnetic
sin i
Principle of
a
converted
SI units
from
to
created
and directed to the fixed point.
nor
from
conservation of
energy
required
to
increase
the
Thermodynamic
scale
A
temperature
collision,
temperature
one
degree.
of
1 kg
of
a
substance
by
scale
that
properties
is
independent
of
any
of
particular
the
substance.
183
Glossary
Thermoelectric
of
an
are
e.m.f.
joined
The
two
generation
dissimilar
metals
together.
Thermometric
property
effect
when
varies
A
physical
continuously
Triple
with
wave
oscillations
direction
property
that
Transverse
at
are
of
wave
energy
point of
which
A
ice,
in
which
perpendicular
to
the
the
water
The
and
Virtual
temperature
water
The
rate
of
change
of
displacement.
transfer.
water
Velocity
image
appear
to
An
pass
image
where
the
rays
through.
vapour
W
temperature.
Threshold of
are
hearing
The
minimum
in
thermal
equilibrium
(273.
16 K
or
0.01 °C.
Wavelength
intensity
of
sound
that
can
The
distance
between
two
be
successive
crests
or
troughs.
U
detected
by
the
human
ear.
Weight
Threshold of
pain
The
Upper fixed
intensity
at
which
pain
is
the
point
The
temperature
an
Timbre
steam
The
subjective
sensation
sound
that
depends
on
the
number
Torque of
a
produced
couple
above
pure
boiling
with
The
a
atmospheric
water
at
pressure.
Work
The
distance
A force
acting
vertically
product
and
on
an
object
placed
in
of
of
a force
and
the
between
A
quantity
that
has
and
incidence
within
the
angle,
is
the
greater
ray
is
than
the
internally
A
triangle
showing
denser
three forces
medium
direction.
the
Vector triangle
of
modulus
Tensile
the
magnitude
When
direction
of
the
Y
two forces.
reflection
the
V
one
Vector quantity
internal
in
a fluid.
note.
product
distance
moved
force.
Young
the forces
184
on
of
upwards
overtones
angle
gravity
of
Upthrust
Total
by
ear.
standard
of
exerted
object.
experienced
of
in
The force
minimum
critical
reflected.
acting
on
an
object.
by
tensile
strain.
stress
divided
Index
Headings
in
bold
indicate
glossary
terms.
cast
iron
170
cataracts
destructive
105
cathode-ray
oscilloscopes
(CRO)
10,
70
interference
diffraction
82,
83,
dimensions
4–5
84–5,
86,
87
,
88,
89
91
A
CD
players
centre of
absolute
refractive
absolute
zero
index
displacement
83
gravity
38,
16–17
,
62–4,
graphs
67
,
76
17
72
centripetal
acceleration
48–9
distance
16
135
centripetal force
absorbers
9,
displacement–time
39
49
diverging
lenses
102,
103
154
changes
acceleration
17
,
18–19,
of
charged
centripetal
acceleration
gravitation
55
circuits,
division,
uncertainty
15
objects
law
30
134,
double
electrical
135
slit
experiments
drag force
86,
90–1
71
drivers
37
69
104
circular
accuracy of
126
48–9
Charles’
accommodation
state
62–3
measurements
motion
48–53,
65
ductile
materials
170,
171
12–13
coefficient
of
resistance
coefficient
of
thermal
114
dynamics
26–33
addition
scalars
conductivity
149
6
E
coherent
uncertainty
6,
32–3
7
ears
combining
air
resistance
vectors
Earth
echoes
concave
energy
lenses
amorphous
solids
amperes
10
efficiency
angle of
116
45
collision
33
29
elastic deformation
76,
pendulums
electric
conservation
of
energy
conservation
of
momentum
reflection
angle of
refraction
electric
current
electrical
velocity
constant-volume
gas
thermometers
29
measurement
proportionality
electromagnetic
spectrum
10
electromagnetic
waves
96
116
angles
of
10,
quantity
17
72
constants
71
32
73
constant
angles
circuits
42
72
angle of
170–1
51
93
incidence
critical
57
,
152–3
elastic
conical
64,
56,
95
103
148–50,
163
conductors
amplitude
102,
43
conduction
(A)
54,
78
52–3
alternative
100–1
6
37
compression
aircraft
86
14
collisions
vectors
waves
96–7
,
154
54
73
electron diffusion
148–9
constructive interference 86, 87
, 88, 89
measurement
9
emitters
continuous flow
projectile
motion
converging
angular velocity
92,
of
units
lenses
equations
of
state
cooling
curves
cooling
effects
F
41
angle
fields,
(cathode-ray
focal
conductors
focal
currents,
126,
Bourdon
128,
gauges
ocean
law
134,
129
D
165
materials
170,
motion
102
102
focus
depth
force
36–41
104
centripetal force
charged
time
sea
breezes
objects
49
30
151
171
circular
deformations
Brownian
length
point
68
135
day
brittle
164
151
155–6
damped oscillations
Boyle’s
pressure
94
29
43
154,
motion
51
170–1
163
dynamics
degrees Celsius
(°C)
26–7
112
gravitational force
C
density
depth
of field
depth
of focus
of
38–9
10–11
104
vectors
cameras
40–1
104
polygons
curves
30
162
moment
calibration
57
29
9
radiation
54–5,
111
70
flutes
10,
110,
163
2
body
boiling
10,
points
fluids,
solids
current-carrying
black
oscilloscopes)
4
current
104
gravitation
68
fixed
quantities
balances
biofuels
depths
73
52–3
crystalline
beam
102–5
104
couples
CRO
units
12–13
129
129
critically damped oscillations
base
128,
3
165
physical
measurement
126
field
base
in
evaporation
critical
barometers
154
57
71
eyes
aircraft
135
110,
102
errors
B
of
41,
3
36
cornea
banking
171
20–1
103
105
constant
148–59,
motion
42
104
principle
67
,
of
equipotentials
cookers
Avogadro
42–5,
equations
equilibrium
convex
astigmatism
102,
energy
93
humour
Archimedes’
of
48
conversion
antinodes
lenses
64
conversion
6,
151
7
angular frequency
aqueous
148,
154
122–3
energy
convection
vectors
methods
23
6,
7
103
derived quantities/units
2,
14
185
Index
force–extension
force–time
free
graphs
(F–t)
body diagrams
free fall
170
graphs
methods
horizontal
28
projectile
hydroelectric
38
motion
energy
hypermetropia
specific
22–3
heat
logarithmic
43
capacity
responses
long-sightedness
105
longitudinal
55
120–3,
124–5
101
105
waves
76,
78–81,
95
I
free
oscillations
loudness
62
frequency
64,
frequency
response
69,
76,
92,
95,
94,
lower fixed
101
ice
111,
127
,
128,
101
point
(ice
point)
111
130
100
ideal
gases
118,
134–9
M
friction
36–7
impulse
fuel
28
43
in
fundamental frequencies
92,
94,
phase
waves
incidence
fundamental
notes
92,
specific
angles
82
magnetic
72
resonance
magnifying
glasses
imaging
(MRI)
71
103
94
indices
fusion,
77
,
95
latent
heat of
72
manometers
165
127
industry,
sound
inelastic
waves
collision
94
marbles,
33
mass
oscillations
29,
62
30
G
inertia
gamma
gases
rays
118,
97
of
internal
128
molar
heat
capacity
matter
thermodynamics
geostationary
iron
144
162,
97
oscillations
100,
101
phases
118
of
thermal
170
mean
lamina,
66–7
,
70
9
62
matter
86–7
energy
systems
measurement
94
irregular-shaped
163
centre
of
gravity
39
140–1
satellites
96,
levels
interference
136–9
heat
of
intensity
126
latent
phases
mass-spring
waves
instruments
state
theory
29
infrared
134–9
changes
kinetic
96,
162–71
properties
square
measurements
mechanical
57
speed
melting
118–31
136–7
8–15
waves
76
126
K
geothermal
glass
energy
43
mercury-in-glass
170
metals
kelvin
gliders,
oscillations
(K)
62
metre
kinematics
global
positioning
satellites
(GPS)
global
warming
energy
44,
158–9
models
129,
kinetic theory of
force–time
graphs
17
,
96,
8
97
specific
heat
capacity
molar
heat
capacity
144
149
11
mole
sea
breezes
heat
3
151
126–9,
40–1
130–1
momentum
17
lattice vibrations
graphs
gauges
89,
methods,
moment
motion
velocity–time
71,
124–5
176–7
line
screw
microwaves
136–9
28
relationships
latent
straight
134,
L
land,
plotting
gases
63
lagging
non-linear
micrometre
mixing
graphs
8
64
162–3
170–1
displacement–time
(m)
67
graphs
deformations
rules
metres
kinetic
116
16–25
57
kinetic
thermometers
168
112
26–7
,
32
148
monatomic
18–19
gases
139
laws
gratings, diffraction
84–5,
monochromatic
91
gravitation
gravitation
monomers
motion
gravitational
constant
26–7
,
gravitational field
29,
32,
motion
strength
116
circular
equations
LCDs
(liquid
crystal
potential
gravity,
centres
of
38,
26–7
,
projectile
simple
102,
29,
32,
49
22–5
103
44–5
lenses
65
8
56
energy
48–53,
20–1
81
laws
potential
measurement
lens formulae
gravitational
displays)
30
length
gravitational
motion
140–5
54–5
gravitational force
17
73
thermodynamics
gravitational fields
90–1
163
49
54
refraction
light
54
54–7
103,
harmonic
62,
63,
65,
67
104
39
MRI
(magnetic
resonance
imaging)
light
greenhouse
guitars
effect
158
multiplication,
diffraction
gratings
double
experiments
musical
86,
polarisation
refraction
harmonic
notes
80–1
N
94
capacity
narrow
(C)
118–25,
graph
loss
momentum
Newton,
crystal
displays
(LCDs)
night
liquid-in-glass
83
69
Isaac
26–7
,
29,
32,
81
126
heavily damped oscillations
diffraction
26
153
curves
gap
11
natural frequency
liquid
heating
plotting
144–5
linear
heat
68
100–1
linear
heat
94
104
72–3
lightly damped oscillations
hearing
instruments
90–1
myopia
H
15
84–5
94
slit
uncertainty
thermometers
68
114,
time
sea
breezes
151
115
nodes
92,
93
liquids
helicopters
30
noise
changes
homogeneous
equations
of
law
non-crystalline
circular
motion
heat
128,
solids
129,
relationships
non-renewable
matter
11
130–1
50
of
163
151
non-linear
phases
186
currents
166
latent
horizontal
126
5
convection
Hooke’s
state
101
162–3
energy
42
49,
54
71
Index
non-uniform
normal
velocity
reactions
17
speed
radians
normals
16–17
,
9,
92,
148,
154–7
damping
94
waves
96,
70
97
mass-spring
O
random
errors
12,
currents
rarefactions
78
ray
103
direction
vectors
diagrams
spring
balances
waves
78
images
102
73,
waves
phase
waves
77
,
state
changes
angles
72–3,
state
equations
73
92,
94,
index
relative density
P
renewable
energy
111,
thermometers
114–15,
projectiles
pan
22–5
resistive force
51,
conductors
interference
63,
29
86–7
66
resolution
of
resonance
resonance
instruments
law
vectors
instruments
94
line
motion
48,
63,
64,
166–7
,
170–1
stress
166–7
,
170–1
68–71
tubes
94,
resultant vectors
76
ripple
17
strain
95
tanks
83,
graphs
170–1
6
stringed
periods
94
156
7
stress–strain
percussion
150
128
36–7
straight
pendulums
95
149,
116
Stefan’s
current-carrying
conditions
42
steel
resistance
state
162
steam
100
path difference,
92–3,
72
steady
parallel
waves
95
refractive
pain thresholds
135
95
stationary
parabolic
95
126
82
refraction
overtones
92–3,
95
62–71
reflection
of
9
102
reflection
out
70
7
standing
centres
oscillations
66–7
,
62
151
real
optical
systems
13
oscillations
opposite
136–7
springs
musical
radio
ocean
91
77
,
48
72
radiation
notes,
spectrometers
R
29
instruments
94
88
subtraction
periods of oscillation
63,
76
rockets
30
scalars
phase
77
,
82
root
mean
square
(r.m.s.)
speed
6
136–7
uncertainty
phase difference
77
roots,
uncertainty
vectors
phases
of
matter
162–71
rubber
14
15
6
170
superposition
physical quantities/units
systematic
pitch
errors
plastic deformations
170–1
direction
vectors
7
thermometers
satellites
56,
57
temperature
114–15,
110–31,
6–7
conduction
polarisation
breezes
148,
151
81
second-order
diffracted
light
91
energy
lenses
transfer
148–59
102
seconds
(s)
10
equilibrium
of force
110,
154
38–9
semi-crystalline
polymers
163
heat
polymeric
materials
polymers
163
170,
loss
153
171
SHM
(simple
harmonic
motion)
62,
63,
measurement
65,
gravitational
56,
10–11
154–7
57
short-sightedness
104
temperature
potential difference
9,
67
radiation
potential,
152–3
151
convection
80,
148–50,
80
sea
polygons
135
116
scalar quantities
P,
13
T
same
platinum-resistance
pole
12,
S
94
Polaroid
82
2–7
coefficients
of
resistance
10
SI
units
2,
64
114
potential
energy
44–5,
67
sight
102–5
tensile
power
42,
44,
45,
stresses
166
102
simple
harmonic
motion
(SHM)
62,
terminal velocity
powers,
uncertainty
63,
65,
measurements
148–50,
slit
diffraction
152–3
84
thermal
convection
148,
thermal
energy
loss
thermal
energy
transfer
151
3
Snell’s
pressure
kinetic
law
72
153
164–5
solar
theory
136,
heat
energy
42,
43,
159
148–59
137–8
solar
molar
conduction
12–13
single
prefixes
37
67
thermal
precision of
36,
15
capacity
water
heaters
159
thermal
equilibrium
thermal
radiation
110,
154
144
solids
p
V
diagrams
144–5
changes
of
state
126
thermistors
thermodynamics
115
140
latent
heat
128,
130
thermocouples
Pressure
law
principal
axes
of
matter
162,
specific
moments
progressive
waves
thermodynamic
scales
heat
capacity
thermodynamics
112,
112
121
waves
88,
89,
94–5
thermoelectric
76,
79,
heat
capacity
115
118–25
thermometers
10–11,
112,
114–17
,
124,
22–5
specific
latent
heat
127
,
130–1
153,
proportionality
effect
93
specific
motion
140–5
40–1
sound
constants
154
54
specific
latent
heat of fusion
specific
latent
heat of vaporisation
127
thermometric
V
116
163
102
principle of
p
115,
134–5
phases
projectile
154–7
(pressure–volume)
diagrams
properties
110,
111
144–5
127
,
threshold of
hearing
threshold of
pain
100,
101
130–1
Q
100
spectra
timbre
quantities
2–7
electromagnetic
96
time
radiation
94
67
155–6
displacement–time
graphs
17
,
63
187
Index
force–time
graphs
measurement
velocity–time
torque
40,
28
V
W
10
graphs
vacuum flasks
158
vaporisation,
specific
latent
heat
127
,
internal
reflection
waves
76,
wavelengths
vector triangles
vehicles
manometers
16–19,
waves
waves
38
weight
48,
62–3
graphs
96,
callipers
72,
76–97
,
76,
90–1,
93
154
9,
29
wide
gap
diffraction
wind
energy
wind
instruments
83
43
18–19
8
work
42,
94
43
97
vertical
uncertainty
circular
motion
50
14–15
X
virtual
circular
motion
images
102
49
visible
electromagnetic
waves
96,
97
X-rays
analysis
96,
97
4–5
vitreous
units
72,
94
vernier
unit
159
159
165
velocity–time
uniform
6–7
52
velocity
ultraviolet
128,
78–81
U
ultrasound
heaters
73
vector quantities
transverse
U-tube
72,
water
41
130–1
total
water
18–19
humour
104
2–7
volts
unpolarised
waves
(V)
Y
10
80
volume
upper fixed
point
(steam
point)
111
Young
measurement
upthrust
modulus
Young’s
molar
p
V
heat
capacity
diagrams
166–9
9
36
double
slit
experiment
144
144–5
Z
thermodynamics
140
zero-order
188
diffracted
light
91
86,
90–1
Physics
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