Answers
Chapter 1: Measurement and
uncertainties
Exercise 1.1 – Measurement in physics
1 Fundamental unit: not made up of a combination of
any other units.
Derived unit: made up by a combination of
fundamental units.
2 a metres: distance or displacement
b kilograms: mass
c seconds: time
d amperes: electrical current
e kelvins: temperature
f moles: amount of a substance
3 a m s−2
b N (newtons)
c J (joules)
d kg m s−1
e C (coulombs)
kg m
2
4 P = F and F = ma, so 1 Pa = 1 2 ÷ m
s
A
= 1 kg m−1 s−2
5 a 1 N = 1 kg m s−2
b work done = F × s, so base units are:
kg m s−2 m = kg m2 s−2
c KE = 1 mv 2, so base units are:
2
kg (m s−1)2 = kg m2 s−2
6 Since the base units of energy are kg m2 s−2, the base
units of power (= E ÷ t) must be kg m2 s−3
7 a Q = I t so the base units of charge must be A s
b 1 J = 1 kg m2 s−2
c Since 1 V = 1 J ÷ C, the base units of a volt must
be: kg m2 s−2 ÷ A s ≡ kg m2 A−1 s−3
8 1 kW h = 1 × 103 J s−1 × 60 × 60 = 3.6 × 106 J
9 a milli (× 10−3)
b mega (× 106)
c micro (× 10−6)
d nano (× 10−9)
e giga (× 109)
10 a 5.7 × 10−2 m
b 1.3 × 10−5 m
11
12
13
14
15
16
17
c 7.4 × 105 m
d 5.3 × 10−7 m
e 5.4 × 10−2 kg
f 2.6 × 103 s
g 1.0 kg
h 4.6 × 10−8 s
a 106
b 1015
c 10−3
d 10−12
e 108
a 2.23 × 10−10 F
b 4.5 × 103 V
c 1.32 × 109 W
d 5.03 × 108 J
a 8460
b 8500
c 8000
a 1000 m
b 1359.5 m
c 1400 m
d 1359.46 m
One student may have a ruler that measures in
millimetres (mm), while the other student has a
ruler that measures in centimetres (cm). If they
both write their measurements in SI units, the first
student will give their answer to three decimal
places (d.p.) and the second student will give their
answer to two d.p.
a 4.12 × 102
b 3.200 000 000 00 × 1011
c 8.7 × 10−3
d 1 × 10−10
e 5 × 107
a 10−2 × 10−5 = 10(−2 + −5) = 10−7
b 106 ÷ 102 = 10(6 − 2) = 104
c 108 ÷ 10−3 = 10(8 − −3) = 1011
d 10−9 × 109 = 10(−9 + 9) = 100 = 1
e 106 × 103 × 10−5 = 10(6 + 3 + −5) = 104
f 106 × 103 ÷ 10−5 = 10(6 + 3 − −5) = 1014
g (10−5 ÷ 10−3) × 104 = 10(−5 − −3 +4) = 102
h 10−5 ÷ (10−3 × 104) = 10(−5 − −3 + 4) = 102
ANSWERS
1
18 a 4 × 104 × 2 × 10−2 = 8 × 102
b 6 × 10−5 × 3 × 10−3 = 18 × 10−8 = 1.8 × 10−7
c 5 × 104 × 4.1 × 104 = 20.5 × 108 = 2.1 × 109
(2 s.f.)
d 8 × 109 × 55 = 440 × 109 = 4 × 1011 (1 s.f.)
19 a 53 + 6.047 = 59 (2 s.f.)
b 2π × 0.65 = 4.1 (2 s.f.)
c 6.443 = 267 (3 s.f.)
d 360 ÷ 2π = 57.3 (3.s.f.)
e 32 × 2.1 = 20 (1 s.f.)
20 a (1.6 × 10−27)2 = 1.6 × 1.6 × 10−27 × 10−27
= 2.6 × 10−54 (2 s.f.)
b 1 ÷ (6 × 106)2 = 1 ÷ (6 × 6 × 106 × 106)
= 1 ÷ (36 × 1012) = 1 ÷ (3.6 × 1013)
= 0.28 × 10−13 = 3 × 10−14 (1 s.f.)
11
c (1.5 × 1011) ÷ 3 × 108 = 1 5 × 10 8 = 0.5 × 103
3
10
= 5 × 102
21 a 3.6 × 102 (2 s.f.)
b 4.400 × 10−3 (4 s.f.)
c 5.6 × 104 (2 s.f.)
d 5.9 × 10−6 (2 s.f.)
e 2.1 × 103 (2 s.f.)
22 a 3 × 108 × 3.5 × 107 = 10.5 × 1015
= 1 × 1016 (1 s.f.)
b 2 × 10−6 × 8 × 105 = 16 × 10−1
= 1.6 × 100 = 2 (1 s.f.)
c 4.0 × 1.7 × 10−27 = 6.8 × 10−27
d 5.0 × 106 × 1.6 × 10−19 = 8.0 × 10−13
e (8.0 × 1012) ÷ (4 × 104) = 2 × 108 (1 s.f.)
f (1.2 × 107) ÷ (3.0 × 106) = 0.4 × 101
= 4.0 (2 s.f and 1 d.p.)
23 a 1600 = 16 × 102 = 4 × 101 = 40
b (5 × 10−3)2 = 5 × 5 × 10−3 × 10−3 = 25 × 10−6
= 2.5 × 10−5 = 3 × 10−5 (1 s.f.)
c
(
)2 + (
(4
102
)2 =
) (3 × 10 )
2
2 2
=
(16 + 9)
104
= 5 × 102 = 500
d (3.0 × 10−15)3 = 3.03 × 10−45 = 27 × 10−45
= 2.7 × 10−44
e 3 8 10 −12 = 3 8 3 10 −12 = 2 × 10−4
24 a 106
b 104 (this is an important example: 4 × 103 is closer
to 1 × 104 than 1 × 103 on a logarithmic scale)
c 10−8
2
25 10−45 m3
26 a 104
b 1011
c 1024
d 1016
27 10−24 g or 10−27 kg
28 12 g of 126 C contains 6 × 1023 atoms, so one atom
12
g
has a mass of
6 1023
1 × 12
So, u must be
= 1.7 × 10−24 g or
12 6 1023
1.7 × 10−27 kg
29 a 1000 = 15.7
63.5
b 15.7 × 6 × 1023 = 9.42 × 1024
c 8900 × 9 42 1024 = 8.4 × 1028
30 3.0 × 108 × 365 × 24 × 60 × 60 = 9.5 × 1015 m
Exercise 1.2 – Uncertainties and errors
1 a ± 1 cm (because this is the least significant figure in
the measurement).
b 1
126
c 1 × 100% = 0.794%
126
2 a 0.1 A
01= 1
b
1 5 15
1 × 100%
c
= 6.7% (1 d.p.)
15
3 Perimeter = 2 × 12 + 2 × 5 = 34 cm
Δ perimeter = 2 × 1 + 2 × 1 = ± 4 cm
4 y = 61.5 − 30.2 = 31.3 °C
Δy = 0.2 + 0.4 = ± 0.6 °C
5 a Repeating measurements of the same thing.
b Zero error, parallax error.
6 a When a measurement is supposed to be zero
(because the independent variable has had no
effect yet on the dependent variable) but the
measurement shows a non-zero value.
This can often occur with a meter, such as an
ammeter.
b When a measurement is being taken by eye and
the eye is not level with what is being measured.
The angle between the eye and whatever is being
measured produces a discrepancy on the
measuring instrument, which is called the parallax
error.
7
8
9
a 5%
b 5% × 2 = 10%
c 5% × 3 = 15%
5% + 3% = 8%
a R = V = 4.0 = 16 Ω
I 0.25
b ΔR = 0.1 + 0.01 × 16 = 1.04
4.0 0.25
Therefore,
ΔR = ± 1.0 Ω (2 s.f.)
% uncertainty = 1 × 100% = 6%
16
6% + 3% = 9%
3% + 2 × 5% = 13%
1 × 10% = 5%
2
1 × 4% + 1 × 2% = 3%
2
2
a 0.1 × 100% = 2%
5.0
0.1 × 100%
b
= 2.5%
4.0
c 2% + 2.5% = 4.5%
d 1 × 5 × 4 = 10.0 cm2
2
e ΔA = 4.5% × 10 = 0.45 cm2 = 0.5 cm2
(1 d.p.)
a 1 × 100% = 0.2%
500
b 0.01 × 100% = 0.1%
9.81
c 0.2% + 0.1% = 0.3%
W = 500
= 51.0 kg (3 s.f.)
d m=
g 9.81
e Δm = 0.3% × 51 = 0.2 kg (1d.p.)
a Accurate: Close to, or equal to, what the value
actually is. This implies that any systematic error
will be very small.
b Precise: With a small value of random uncertainty;
i.e. when repeated, the same, or nearly the same,
value will be achieved.
a Example: A measurement is supposed to be 4.0
and the three repeated measurements made are:
4.0, 1.0 and 7.0. Their mean is 4.0, which is
accurate, but the random uncertainty in the three
measurements is large. So, the mean of
measurements is accurate, but the measurements
are not precise. 2.1 × 103
(
10
11
12
13
14
15
16
17
)
b Example: A measurement is supposed to be 4.0
and the three repeated measurements made are:
2.3, 2.2, 2.3. The mean is 2.3 (1 d.p.), which is not
accurate, but the random uncertainty in the three
measurements is very small. So, the measurements
are precise.
c Example: A measurement is supposed to be 4.00
and the three repeated measurements made are:
4.00, 3.95, 4.05. The mean is 4.00, which is
accurate, and the random uncertainty in the three
measurements is 0.05, which is very small. So, the
measurements are precise.
d Example: A measurement is supposed to be 4.0
and the three repeated measurements made are:
2.3, 1.7, 4.8. The mean is 2.9, which is not
accurate, and the random uncertainty in the three
measurements is 1.6, which is large. So, the
measurements are not precise.
18 Repeating measurements makes the calculated mean
more likely to be an accurate value and reduces the
effect of an anomalous measurement.
19 a
b
c
d
20 a
b
c
d
e
21 a
b
c
d
22 a
(12.2 + 12.5 + 11.8 + 12.0 + 11.9 + 12.2 + 12.0 + 12.3)
8
= 12.1 cm
12.5 − 11.8 = 0.7 cm
0.7
Δ mean =
= 0.4 cm (1 d.p.)
2
12.1 ± 0.4 cm
(3.24 + 3.22 + 3.16 + 3.26 + 3.23 + 3.28 + 3.22)
7
= 3.23 kg
3.28 − 3.16 = 0.12 kg
1 × 0.12 = 0.06 kg
2
Yes, 0.06 > 0.01
3.23 ± 0.06 kg
(4 1 + 4 2 + 4 1 + 4.1 + 4 2) = 4.1 s (1 d.p.)
mean =
5
Δ mean = 1 (4.2 − 4.1) = 0.1 s (1 d.p.)
2
The absolute uncertainty in the measurements is
larger than the random uncertainty in the repeats.
4.1 ± 0.2 s
mean height =
(1 72 1.84 + 1 66 + 1.88 + 1 70 1.75) = 1.76 m
6
(2 d.p.)
ANSWERS
3
b half the range of heights = 1 (1.88 − 1.66) =
2
0.11 m
c 1.76 ± 0.11 m (Note that the uncertainty in each
measurement is assumed to be ± 0.01, since this is
the least significant figure. Since the random
uncertainty is larger than the absolute uncertainty,
the random uncertainty is used to give the
uncertainty in the mean.)
d 0.11 × 100% = 6.25% (3 s.f.)
1.76
23 When drawing the maximum and the
minimum gradients, lines need to be drawn that
pass through ALL of the error bars. In some cases
(not this one) the size of the error bars might not be
the same; this will affect the possible maximum
and minimum gradients that can be drawn
(see Figure A1.1).
1.4
1.2
25 a
b
c
d
Graphs 2, 3, 4 and 5
Graphs 4 and 5
Graphs 1 and 6
Graph 6
Exercise 1.3 – Vectors and scalars
1 a Scalar: a quantity with magnitude only; i.e.
direction is not important. Examples: energy,
length, mass, time, speed.
b Vector: a quantity with both magnitude and
direction. Examples: force, momentum,
displacement, velocity.
2 Scalars: mass; speed; time; density; pressure;
temperature; kinetic energy; gravitational potential
energy; volume; area; length; power
Vectors: velocity; force; weight; acceleration;
momentum; displacement; current
3 See Figure A1.2.
a+b
b
max. gradient
a
1.0
Figure A1.2
4 a See Figure A1.3.
0.8
a
min. gradient
0.6
–b
a–b
0.4
0.2
Figure A1.3
b See Figure A1.4.
–a
0.0
Figure A1.1
24 a The student can do two things:
Method 1: process the length data by finding the
square root of l and then plotting a graph of T
against l .
Method 2: Process the time data by squaring T
and then plotting T 2 against l .
b For Method 1, % uncertainty in T is unchanged,
so 5%. The % uncertainty in l is 1 × 4%, so 2%.
2
For Method 2, % uncertainty in T is 2 × 5% =
10% and the % uncertainty in l is unchanged and
so is 4%.
b
b–a
Figure A1.4
5 a See Figure A1.5.
2a + 3b
3b
2a
Figure A1.5
4
b See Figure A1.6.
Exam-style questions
1
2
3
4
5
–3a
2b
2b – 3a
C
D
A
C
a, b, c and d See Figure A1.8. (Note: for part c this
represents speed.)
Distance against time
Figure A1.6
Δv = vfinal − vinitial (see figure A1.7).
Distance, x/m
6
vfinal
Δv
–vinitial
Figure A1.7
7
8
()
Δv = 5 m s−1 in a direction tan−1 4 = 53° from the
3
westwards direction.
a Using Pythagoras’ theorem: a b = 42
()
22 =
1
4.5 in a direction of tan−1
= 26.6° above the
2
horizontal.
2
2
b a b = 4 ( 2) = 4.5 in a direction 26.6°
below the horizontal.
c b a = 22 ( 4 )2 = 4.5 in a direction 153.4°
above the horizontal.
9 a phorizontal = 5 cos 30° = 4.3
b pvertical = 5 sin 30° = 2.5
10 a Magnitude is 62 32 = 6.7
3
= 26.6° above the
b Direction is tan−1
6
horizontal.
11 a Magnitude = 52 42 = 6.4 N
4
= 38.7°
b Angle = tan−1
5
12 a Horizontal component of P is 8 cos 40° = 6.1.
b Vertical component of P is 8 sin 40° = 5.1.
()
()
13 a Magnitude of the unspecified force = 62 32
= 6.7 N.
b Direction of the unspecified force = tan−1 6
3
= 63.4° below the right-to-left horizontal axis.
14 Weight, W = 2 × 60 cos 35° = 98 N (2 s.f.)
()
15 a v = 3002 + 402 = 303 m s−1
b Actual direction = tan−1 40 = 7.6° west of the
300
northwards.
c Distance travelled west = 40 × 60 × 60 = 144 km.
( )
35
30
25
20
15
10
5
0
–5
max. gradient = 2.24 m s–1
best-fit gradient = 2.0 m s–1
min. gradient = 1.81 m s–1
2
4
6
8
10
Time, t/s
12
14
16
Figure A1.8
e
f
g
h
Maximum gradient = 2.24 m s−1
Shown on graph
Minimum gradient = 1.81 m s−1
1 range of gradients = 1 (2.24 − 1.81)
2
2
= 0.22 m s−1
i x = (2.00 ± 0.22) t
6 a Plot s against t2 to produce a proportional graph.
Or, plot √s against t.
b For s against t2, % uncertainty in s will be
unchanged.
So, 3% and the % uncertainty in t2 will be
2 × 2% = 4%.
1
For √s against t, % uncertainty in √s is × 3%
2
= 1.5%, and the % uncertainty in t is unchanged,
so 4%.
7 a It is not a straight line that passes through the
origin.
b Method 1: Plot a new graph of t against √m;
Method 2: Plot a new graph of t2 against m.
c For Method 1, the uncertainties in t will
be unchanged, whilst the percentage
uncertainties in
1
√m will be the % uncertainties of m.
2
For Method 2, the % uncertainties in t2
will be twice the % uncertainties in t,
whilst the uncertainties in m will be
unchanged.
ANSWERS
5
f Δv3 = (3 × 0.09) × 1.13 = 0.36 m3 s−3
g Gradient of graph is A3 = 0.12.
8 a and c See Figure A1.9.
Force, F/N, ±2N
Force against extension
60
50
40
30
20
10
0
So, A = 3 0 12 = 0.5
max. gradient = 7.1 N mm–1
−1
best-fit gradient = 6.5 N mm–1
min. gradient = 5.8 N mm–1
0
1
2
3
4
5
6
Extension, x/mm
7
8
9
Figure A1.9
b Graph is a straight line passing through the origin,
suggesting a proportional relationship between
force and extension.
d Stiffness = 6.5 N mm−1
1
e Δ gradient = (7.1 − 5.8) = 0.7 (1 d.p.)
2
So, F = (6.5 ± 0.7)x
9 a i Independent variable: power supplied to the
hair drier.
ii Dependent variable: air speed.
iii Control variables: distance from the hair drier to
the anemometer – reducing this increases the air
speed measured; angle of the hair drier to the
anemometer – any change may affect air speed;
Keeping the same hair drier – any change would
not be a fair test; Keeping the same anemometer
– any change might measure a different air speed.
b Two error bars each ± 0.1 m s−1
c 10 W point: % uncertainty in v
= (0.1 ÷ 1.1) × 100% = 9%
80 W point: % uncertainty in v
= (0.1 ÷ 2.1) × 100% = 5%
1
d Since v = A P 3 then v3 = A3 P
So, a graph of v3 against P will produce a straight line
passing through the origin with a gradient of A3.
e See Figure A1.10.
v3/m3 s–3
v3 against P
14
12
10
8
6
4
2
0
–2
Figure A1.10
6
20
40
60
80
Power, P/W
100
120
h Units for A are m s−1 W 3
10 Part of the scientific method involves trying to find
an example that contradicts a given hypothesis. It
is also practically impossible to make all possible
observations of a certain phenomenon. Therefore
scientists have to make fairly random decisions about
when they have made enough observations. This
is affected by how complicated the observation is,
or by how many variables the observation requires
to be controlled, or by how many observations are
actually possible. There is no simple way of stating
how many observations will always be enough.
11 a 8.5% of 12 = 1.02. So, Δx = 1.02
b Yes, it is a sensible suggestion because if the
relationship is of this form, then a plot of y against
x2 will produce a straight line through the origin
since y ∝ x2
2
Δx x 2 = 2 × 1.02 × 122
c Δx2 =2 Δx ⇒ Δx
Δ 2 =2 Δx
x
x
12
x
= 24.48 (24.5 to 3 s.f.)
d Since the graph passes through the origin the
y
value of A must be 2
x
y
18
= 0.125
So, A = 2 =
144
x
835 + 840 + 847 + 841 + 837 + 844
12 a
6
= 841 mm (to 3 s.f.)
b Half the range of their measurements is
847 − 835 = 6 mm.
2
This is larger than the absolute uncertainty in
their measurements (± 1 mm), so they should
choose an uncertainty that is ± 6 mm.
c Sophie’s suggestion in a valid one. Both of Tim’s
measurements are 6 mm too large. This suggests
that he may have a zero error on his measuring
instrument – and this is a systematic uncertainty.
d Mean width =
595 + 593 + 600 + 594 + 604 + 597
= 597 mm
6
Their half range is: 604 −593 = 6 mm (to 1 s.f.),
2
and this is larger than ± 1 mm
So, the percentage uncertainty in their mean is
6 × 100% = 1%.
597
So, the students’ measurement of width is
597 ± 1%
s 12
= 6.0 s
13 a t = =
v
2
b i During this 6 s period the current drags the
swimmer a distance 6 × 3 = 18 m downstream.
c Direction = tan−1
d 500 m s−1 at 53° to the horizontal
7
a Acceleration
b Displacement
8
a Distance travelled = area under graph
1
= ×8×5
2
= 20 m
8
b a = gradient = = 1.6 m s−2
5
total distance travelled 20 + 40
c average speed =
=
time taken
10
= 6 m s−1
a s=ut
b See Figure A2.1.
ii So, actual distance travelled = 122 + 182
= 21.6 m (22 m to 2 s.f.)
21.6
3
iii v =
= 3.6 m s−1 at an angle of tan−1
6
2
= 56° to the straight-across direction.
()
Alternatively, v = 32 22 = 3.6 m s−1. Angle
calculated in the same way.
( 43 ) = 53° to the horizontal
9
Exercise 2.1 – Motion
1 Distance: scalar quantity, magnitude only.
Displacement: vector quantity, magnitude and
direction.
200 × 103
2 a v=s=
= 37 m s−1 (2 s.f.)
t
90 × 60
3
s 1 5 × 10
= 330 m s−1 (2 s.f.)
b v= =
t
45
6000 × 103
s
= 14 m s−1 (2 s.f.)
c v= =
t
5 × 24 × 60 × 60
117 × 103
= 20 minutes
3 t= s =
v
97
3 78 × 1016
4 a t= s =
= 1.26 × 108 s
8
v
3 × 10
⎛ 1.26 × 108
⎞
⎜⎝ = 3 15 × 107 = 4.01 years⎟⎠
b Proxima Centauri is 4.01 light years from the Earth.
distance travelled 12
5 a vaverage =
=
= 1.0 m s−1
time taken
12
b v = s = 12 = 1.5 m s−1
t
8
6 a Using Pythagoras’ theorem: displacement =
32
42 = 5
5 × 10 −2
s
b v= =
= 500 m s−1
t 100 × 10 −6
s = 12 at2
u
s = ut
0
0
Time/s
t
Figure A2.1
change in velocity
t
= v −u
time
t
d Area of triangular section = 1 (v u ) t = 1 at 2
2
2
e See Figure A2.1
1
f s = ut + at2
2
28.8 103
10 a 28.8 km h−1 =
= 8.0 m s−1
60 × 60
And v = u + at = 8 + 2 × 10 = 28 m s−1
1
1
b s = ut + at2 = 8 × 10 + × 2 × 102 = 180 m
2
2
11 a See Figure A2.2.
c a=
17.3
Velocity/m s–1
Chapter 2: Mechanics
Velocity/m s–1
v
s = 15 m
0
0
Time/s
1.73
Figure A2.2
ANSWERS
7
So, the percentage uncertainty in their mean is
6 × 100% = 1%.
597
So, the students’ measurement of width is
597 ± 1%
s 12
= 6.0 s
13 a t = =
v
2
b i During this 6 s period the current drags the
swimmer a distance 6 × 3 = 18 m downstream.
c Direction = tan−1
d 500 m s−1 at 53° to the horizontal
7
a Acceleration
b Displacement
8
a Distance travelled = area under graph
1
= ×8×5
2
= 20 m
8
b a = gradient = = 1.6 m s−2
5
total distance travelled 20 + 40
c average speed =
=
time taken
10
= 6 m s−1
a s=ut
b See Figure A2.1.
ii So, actual distance travelled = 122 + 182
= 21.6 m (22 m to 2 s.f.)
21.6
3
iii v =
= 3.6 m s−1 at an angle of tan−1
6
2
= 56° to the straight-across direction.
()
Alternatively, v = 32 22 = 3.6 m s−1. Angle
calculated in the same way.
( 43 ) = 53° to the horizontal
9
Exercise 2.1 – Motion
1 Distance: scalar quantity, magnitude only.
Displacement: vector quantity, magnitude and
direction.
200 × 103
2 a v=s=
= 37 m s−1 (2 s.f.)
t
90 × 60
3
s 1 5 × 10
= 330 m s−1 (2 s.f.)
b v= =
t
45
6000 × 103
s
= 14 m s−1 (2 s.f.)
c v= =
t
5 × 24 × 60 × 60
117 × 103
= 20 minutes
3 t= s =
v
97
3 78 × 1016
4 a t= s =
= 1.26 × 108 s
8
v
3 × 10
⎛ 1.26 × 108
⎞
⎜⎝ = 3 15 × 107 = 4.01 years⎟⎠
b Proxima Centauri is 4.01 light years from the Earth.
distance travelled 12
5 a vaverage =
=
= 1.0 m s−1
time taken
12
b v = s = 12 = 1.5 m s−1
t
8
6 a Using Pythagoras’ theorem: displacement =
32
42 = 5
5 × 10 −2
s
b v= =
= 500 m s−1
t 100 × 10 −6
s = 12 at2
u
s = ut
0
0
Time/s
t
Figure A2.1
change in velocity
t
= v −u
time
t
d Area of triangular section = 1 (v u ) t = 1 at 2
2
2
e See Figure A2.1
1
f s = ut + at2
2
28.8 103
10 a 28.8 km h−1 =
= 8.0 m s−1
60 × 60
And v = u + at = 8 + 2 × 10 = 28 m s−1
1
1
b s = ut + at2 = 8 × 10 + × 2 × 102 = 180 m
2
2
11 a See Figure A2.2.
c a=
17.3
Velocity/m s–1
Chapter 2: Mechanics
Velocity/m s–1
v
s = 15 m
0
0
Time/s
1.73
Figure A2.2
ANSWERS
7
1 2
g t ,t =
2
u = 0 − 30 =
3s
−g
−10
1
1
b s = ut − g t2 = 30 × 3 − × 10 × 32 = 45 m
2
2
13 a vv = v sin 40° = 10 × 0.64 = 6.4 m s−1
b v = u – gt
So, t = v u = 0 − 6.4 = 0.65 s
−g
−9.81
1 2
1
c s = ut − g t = 6.4 × 0.65 − × 9.81 × 0.652
2
2
= 2.1 m
d In the absence of air friction,
x = 2 × 0.65 × 10 cos 40° = 10 m (2 s.f.)
14 Equipment:
Object to drop: something small and dense.
Timing apparatus: stopwatch or data logger with
motion sensor.
Ruler.
Method:
Measure height of drop, s, with ruler.
Measure the time it takes to the fall to the ground.
Repeat several times, recording all measurements.
Change the height and repeat steps 1-4.
Repeat for about 7-10 different heights, ranging
from 1 m to as high as possible.
For each repeat, calculate t2.
Plot s against t2, to give a straight-line graph passing
through origin.
1
Determine gradient = g; multiply by two to find g.
2
v 2 u2
15 Using v2 = u2 + 2gs, s =
= 36 = 1.8 m
2g
2 × 9.81
12 a v = u − gt ⇒ t =
v
b For a body to be in equilibrium, it must not be
accelerating or rotating, so all the forces acting on a
body must add up to zero and all the turning
moments acting on a body must add up to zero.
F 20
3 Using Newton’s second law, a = =
= 4 m s−2
m 5
1500 − 0
4 F = ma = 0.05 ×
= 750 N
01
5 a Net force on paper cone = 0.12 − 0.08 = 0.04 N
So, a = F = 0.04 = 3.3 m s−2
m 0.012
b See Figure A2.3.
3.3
Accelaration/m s–2
2s = 2 × 15 = 1.73 s
g
10
−1
c v = gt = 10 × 1.73 = 17.3 m s
b Using s =
0
0
Time/s
Figure A2.3
c At the start, the forces acting on the cone are not
balanced (weight > air friction force), so the cone
accelerates. Increasing the speed of the cone
increases the air friction force acting upwards. The
resultant force on the cone decreases and so, using
Newton’s second law, the acceleration of the cone
decreases. When air friction force and weight are
balanced, the resultant force on the cone is zero
and the cone travels at a constant velocity.
6 a See Figure A2.4.
lift
air
friction
engine
thrust
Exercise 2.2 – Forces
1 Inertial mass: the property of a body to resist
an acceleration caused by an unbalanced force
acting on it.
Gravitational mass: the property of a body that defines
the effect of a gravitational field acting on it.
2 a Newton’s first law: A body will continue to move
with a constant velocity, or remain at rest unless it
is acted upon by an unbalanced force.
8
weight
Figure A2.4
b Horizontal and vertical forces must add to be zero.
So, lift + weight = 0
and air friction + engine thrust = 0.
7 a See Figure A2.5.
normal
reaction
force
weight
Figure A2.5
8
9
10
11
Note: In reality, forces are collinear, but have been
slightly displaced on diagram to make it clearer.
b i Newton’s first law: resultant force = 0 so box
remains stationary.
ii Newton’s second law: F = ma, resultant
force = 0 so acceleration = 0 and box remains
stationary.
iii Newton’s third law: the push of the box on
the table top is balanced by the push from the
table top on the box.
a Increases
b Decreases
c An upwards force equal to the weight of the box.
a Using Newton’s laws of motion, the sum of the
forces acting on the person must equal
the mass × acceleration of the person.
So, 60g + N = 0 ⇒ N = −60g = −60 × 9.81
= 590 N upwards = reading on weighing scales.
b Using Newton’s laws of motion:
60g + N = 60a ⇒ N = 60a − 60g
= 60 × 0.25 × 9.81 − 60 × 9.81 = −440 N
c Using Newton’s laws of motion:
60g + N = −60a ⇒ N = −60a − 60g
= − 60 × 0.2 × 9.81 − 60 × 9.81 = −710 N
a Along the slope: F = W sin θ
Perpendicular to the slope: N = W cos θ
F = W sin θ F = N tan θ
N W cos θ
b μs = tan θ when the book is not slipping along
the slope.
c As θ increases, W sin θ increases. When
W sin θ > μs N, the book will begin to slip.
For θ = 30° the component of weight down the
slope is W sin 30° = 0.5 W
The static frictional force is F = μs N = μs W
cos 30° = 0.45 × 0.866 × W = 0.39 W, less than
the component of weight down the slope, so the
container slides down the slope.
12 The box is in equilibrium (moving at a constant
velocity), the vertical and the horizontal forces must
balance.
Vertically, 150 sin 40° + N = mg so
N = mg − 150 sin 40° = 25 × 9.81 − 150 × 0.64
= 149 N
Horizontally, 150 cos 40°
= μd × N
150 cos 40° 0.77
⇒ μd =
=
149
Exercise 2.3 – Work, energy and power
1 2 1
mv = × 5 × 22 = 10 J, GPE = 0
2
2
So, Etotal = 10 J
b GPE = mgh = 4 × 9.81 × 2 = 78.5 J (80 J to 1 s.f.)
KE = 0
So, Etotal = 80 J
1
1
c KE = mv2 = × 3 × 42 = 24 J,
2
2
GPE = mgh = 3 × 9.81 × 5 = 147 J
So, Etotal = 24 + 147 = 171 J (200 J to 1 s.f.)
2 a i GPE = mgh = 0.45 × 9.81 × 1.5 = 6.6 J
ii when the cup of coffee falls, it transfers its GPE
into KE
1
So, 6.6 J = mv2
2
1 a KE =
Therefore, v =
2 × 6.6
= 5.4 m s−1
0 45
1
b No. mgh = mv2 so the mass of the object
2
falling
occurs on both sides of the equation and therefore
cancels out.
3 a A joule is defined as the amount of work done
when a force of 1 N moves through a distance
of 1 m.
b The principle of conservation of energy states that
energy can be transferred from one form into
another (or several), but it cannot be destroyed.
c i Yes. The book has gained GPE.
ii Yes, the pupil picking up the book has
applied a force upwards on the book and the
book has moved a distance upwards, so the
pupil has done work. The pupil must have ‘lost’
some energy.
ANSWERS
9
4
5
6
7
8
9
10
iii Some of the energy used by the pupil may
have transformed into other forms (such as
thermal energy), so the energy ‘lost’ by the
pupil will be more than the energy gained by
the book.
a work done = force × distance moved in direction
of force = 8.6 × 104 × 15 = 1.3 × 106 J
b GPE gained = m g Δh = 3 × 103 × 9.81 × 15 sin
20° = 1.5 × 105 J
c Some of the work done on the block has been
transformed into other forms, such as thermal
energy caused by the friction between the block
and the slope.
In stopping the car, the KE of the car is transformed
into thermal energy in the brakes. So we can use
the idea that work done (= energy transformed)
= average force × distance moved (in direction of
force).
1 mv 2 1 × 1400 × 202
1
2
=2
So, m v = Fs ⇒ s = 2
2
F
9 × 103
= 31 m
work done = area under graph
= 1 × 60 × 2 10 −3 + (60 × 3 × 10−3) = 0.24 J
2
a The spring constant of the spring (the force
required to stretch the spring by 1 cm).
b The elastic potential energy stored by the
stretched spring. (Alternatively, the work done in
stretching the spring.)
a Hooke’s law: the extension of a spring, x, is
proportional to the force, F, applied to stretch it,
providing that the elastic limit is not exceeded.
b i F = k x = 25 × 0.3 = 7.5 N
ii F = k x = 0.3 × 2.5 × 10−3 = 7.5 × 10−4 N
a In series, a force stretching the springs makes each
spring stretch by the same amount, so the overall
extension will be three times as much as for one
spring. So, the spring constant for one spring
must be 3 × 12 = 36 N m−1.
b In parallel, each spring stretches only half of what it
would on its own. So, the spring constant for one
1
spring is × 50 = 25 N m−1.
2
1
EPE = k x2 = mgh
2
(
b
(
)
2
power =
energy transformed 4.7 105
=
time taken
5
= 94 kW
0.55 E = 4.7 × 105
4.7 105
So, E =
= 8.5 × 105 J
0.55
15 a power =
c
energy transformed mg Δh 65 × 9.81 × 5
=
=
= 530 W
time taken
t
6
useful energy transformed 65 × 9.81 × 5
=
efficiency
0.2
= 16 kJ
b E=
)
5000 × 1 10 2
k x2
So, h =
= 1.0 m
=
2mg 2 25 10 3 9.81
10
11 a work done = GPE gained = mgΔh = 40 × 9.81 × 30
= 1.2 × 104 J
work done 1.2 104
b output power =
=
time taken
12
= 1.0 kW
power 15 × 103
= 750 N
12 power = F v ⇒ F =
=
v
20
13 Anton is the most powerful (3.61 W).
1
1
14 a ΔKE = mΔv2 = × 1.5 × 103 × (322 − 202)
2
2
= 4.7 × 105 J
Exercise 2.4 – Momentum and impulse
1
2
3
4
a Principle of conservation of linear momentum: in
any interaction involving no external forces, the
total momentum before the interaction is equal to
the total momentum after the interaction.
b As far as we know, the principle of conservation
of linear momentum is a universal law (in other
words, it applies to all interactions).
a p = mv = 50 × 6 = 300 kg m s−1 westwards
b p = mv = 9.1 × 10−31 × 2 × 107 = 1.82 × 10−23
= 1.8 × 10−23 kg m s−1 (2 s.f.)
c p = mv = 0.11 × 60 = 6.6 kg m s−1
pbefore = pafter
3×4
Therefore, 3 × 4 = (3 + 1) × v ⇒ v =
= 3 m s−1
3+1
a p = mv = 0.4 × 8 = 3.2 kg m s−1
b p = mv = 0.4 × -5 = −2.0 kg m s−1
c The ball has changed its momentum by
−2.0 −3.2 = −5.2 kg m s−1.
So, the Earth must have gained momentum of
5.2 kg m s−1.
d The bad guy will be pushed at a speed of 1.6 m s−1
in the direction of the bullet.
e It’s not usual to see a person in the movies being
shot and moving backwards with this kind of
speed. Frequently, people who are shot in movies
just collapse (or even fall forward). These are not
accurate portrayals.
Since the mass of the Earth is large (MEarth =
6 × 1024 kg), the speed at which the Earth moves
will be 5 2 24 = 8.7 × 10−25 m s−1.
6 10
This speed is too small for us to notice/measure ...
but that is what the Earth is doing!
4 × 1.6 × 107
5 a 4 × 1.6 × 107 = 237 v ⇒ v =
237
= 2.7 × 105 m s−1
(
)
2
1 ×4×
×
KEα
b
= 2
= 59.
2
KENp 1 × 237 ×
×
2
So, the alpha particle has 59 times more KE than
the Np nucleus. (It is 59 times heavier.)
1
6 a Impulse = area under graph = × 50 × 20
2
−1
= 500 kg m s
b See Figure A2.6.
)
Velocity/ m s–1
(
1
2
3
4
5
6
7
8
C
A
A
B
C
C
A
a See Figure A2.7.
5
0
20
Velocity/m s–1
0
Exam-style questions
0
highest point
reached is where
v=0
0.5
1.0
Time/s
Time/s
Figure A2.6
Since the force acting on the object increases, its
acceleration increases, so the gradient of the graph
of v against t increases.
c Δp = 500
Therefore, v = 500 = 200 m s−1 (1 s.f.)
3
7 Crumple zones extend the time during which a crash
occurs. This means that the impulse (the change in
momentum of the car and its passengers) occurs over
a longer time, making the force experienced by the
passengers smaller. This reduces the chance of injury.
m (v − u )
Δ
Δp
8 F=
= mv − mu =
= ma
t
t
t
9 a p = mv = 0.25 × 450 = 112.5 kg m s−1
(= 110 kg m s−1 to 1 s.f.)
b p = 112.5 = (70 + 0.25)v ⇒ v = 112.5 = 1.6 m s−1
70.25
Δ
Δp
70 × 1 6
=
c F=
= 1120 N (= 1100 N to 2 s.f.)
t
01
–5
Figure A2.7
b Area under the graph between t = 0 and t = 0.5
gives displacement.
1
= × 5 × 0.5 = 1.25 m
2
c total area = area first 0.5 s + area for second 0.5 s.
These two areas are equal in magnitude, but area
between 0.5 s and 1.0 s is negative. So, overall
area = 0. Therefore, displacement = 0.
2 50
9 a t = 2s =
= 3.2 s
g
9 81
b Horizontal distance = ut = 40 × 3.2 = 128 m
c Vertical component of v = gt = 9.81 × 3.2
= 31.4 m s−1
Horizontal component of v = 40 m s−1
Magnitude of v = 31.42 + 402 = 51 m s−1 (2 s.f.)
Direction of v = tan−1 31.4 = 38° from the
40
horizontal.
( )
ANSWERS
11
10 a Point A: The graph shows a constant acceleration
(g). The skydiver is not yet falling at a speed
fast enough for air friction to have a significant
effect.
Point B: There are two forces acting on the
skydiver: the skydiver’s weight (downwards) and
the force of air friction (upwards). The skydiver
continues to accelerate, but at a decreasing rate
due to the air friction force increasing with speed.
So, gradient decreases.
Point C: Forces acting on the skydiver are equal
and opposite so there is no resultant force.
According to Newton’s first law, the skydiver now
travels at a constant velocity.
b See Figure A2.8.
Velocity/m s–1
C
0
B
A
0
Time/s
Figure A2.8
c When the parachute opens, its larger area
produces a larger air friction force. This force
opposes and is larger than the weight. The
resultant force is now upwards. So, the skydiver
slows down.
As the skydiver slows, the air friction force
decreases and so the deceleration decreases, until
the two forces acting on the skydiver are balanced
and the skydiver reaches a new, smaller, terminal
velocity (the final horizontal section of the
graph).
11 a k = F = 3 = 30 N m−1
x 01
1
1
b E = F x = × 3 × 0.1 = 0.15 J
2
2
12 a Before the collision, p = mv = 6 × 6 = 36 kg m s−1
After the collision, p = (m1 + m2)v = (6 + 3)v
So, v = 36 = 4 m s−1
( + )
b F=
12
Δp 3 × 4
Δ
=
= 60 N
t
02
c F = 60 N in the opposite direction (Newton’s
third law).
1
d Before the collision, KE = × 6 × 62 = 108 J.
2
1
After the collision, KE = × 9 × 42 = 72 J.
2
There is less KE after the collision, so the collision
is inelastic.
13 a The two forces are the same in magnitude and
opposite in direction: Newton’s third law.
b Before the collision, p = 60 × 2 M − 60 × M
= 60 M.
After the collision, p = 3 Mv.
60 M
Therefore, v =
= 20 km h−1.
3M
c Before the collision:
1
1
KE = × 2 M × 602 + × M × 602
2
2
3
= × M × 3600 = 5400 M
2
1
d After the collision: KE = × 3 M × 202 = 600 M
2
This is less than the KE before the collision, so
the collision was inelastic.
e Some of the ‘lost’ KE will have been transformed
into energy to deform the two vehicles, thermal
energy and sound energy.
Δ p 15
=
= 0.75 N
14 a F =
t
20
2
p2
= 15 = 37.5 J (38 J to 2 s.f.)
b ΔKE =
2m 2 3
Chapter 3: Thermal concepts
Exercise 3.1 – Thermal concepts
1
2
3
4
5
Regularly arranged, close together.
Examples: children sitting in an old-fashioned
classroom in neat rows and columns, or people on
an aeroplane in rows, and so on.
Fairly close together, able to move about, but
clustered in groups with occasional atoms leaving
one group and moving to join another.
Example: people at a party, with small groups of
people having others join and leave.
Far apart, moving randomly with a range of speeds
and random directions. Occasionally colliding with
each other.
10 a Point A: The graph shows a constant acceleration
(g). The skydiver is not yet falling at a speed
fast enough for air friction to have a significant
effect.
Point B: There are two forces acting on the
skydiver: the skydiver’s weight (downwards) and
the force of air friction (upwards). The skydiver
continues to accelerate, but at a decreasing rate
due to the air friction force increasing with speed.
So, gradient decreases.
Point C: Forces acting on the skydiver are equal
and opposite so there is no resultant force.
According to Newton’s first law, the skydiver now
travels at a constant velocity.
b See Figure A2.8.
Velocity/m s–1
C
0
B
A
0
Time/s
Figure A2.8
c When the parachute opens, its larger area
produces a larger air friction force. This force
opposes and is larger than the weight. The
resultant force is now upwards. So, the skydiver
slows down.
As the skydiver slows, the air friction force
decreases and so the deceleration decreases, until
the two forces acting on the skydiver are balanced
and the skydiver reaches a new, smaller, terminal
velocity (the final horizontal section of the
graph).
11 a k = F = 3 = 30 N m−1
x 01
1
1
b E = F x = × 3 × 0.1 = 0.15 J
2
2
12 a Before the collision, p = mv = 6 × 6 = 36 kg m s−1
After the collision, p = (m1 + m2)v = (6 + 3)v
So, v = 36 = 4 m s−1
( + )
b F=
12
Δp 3 × 4
Δ
=
= 60 N
t
02
c F = 60 N in the opposite direction (Newton’s
third law).
1
d Before the collision, KE = × 6 × 62 = 108 J.
2
1
After the collision, KE = × 9 × 42 = 72 J.
2
There is less KE after the collision, so the collision
is inelastic.
13 a The two forces are the same in magnitude and
opposite in direction: Newton’s third law.
b Before the collision, p = 60 × 2 M − 60 × M
= 60 M.
After the collision, p = 3 Mv.
60 M
Therefore, v =
= 20 km h−1.
3M
c Before the collision:
1
1
KE = × 2 M × 602 + × M × 602
2
2
3
= × M × 3600 = 5400 M
2
1
d After the collision: KE = × 3 M × 202 = 600 M
2
This is less than the KE before the collision, so
the collision was inelastic.
e Some of the ‘lost’ KE will have been transformed
into energy to deform the two vehicles, thermal
energy and sound energy.
Δ p 15
=
= 0.75 N
14 a F =
t
20
2
p2
= 15 = 37.5 J (38 J to 2 s.f.)
b ΔKE =
2m 2 3
Chapter 3: Thermal concepts
Exercise 3.1 – Thermal concepts
1
2
3
4
5
Regularly arranged, close together.
Examples: children sitting in an old-fashioned
classroom in neat rows and columns, or people on
an aeroplane in rows, and so on.
Fairly close together, able to move about, but
clustered in groups with occasional atoms leaving
one group and moving to join another.
Example: people at a party, with small groups of
people having others join and leave.
Far apart, moving randomly with a range of speeds
and random directions. Occasionally colliding with
each other.
6
7
8
9
10
11
12
13
14
15
16
17
Example: small children in a playing area running
about randomly.
a −273 °C = −273 + 273 = 0 K
b −173 °C = −173 + 273 = 100 K
c 0 °C = 0 + 273 = 273 K
d 27 °C = 27 + 273 = 300 K
e 100 °C = 100 + 273 = 373 K
f 273°C = 273 + 273 = 546 K
a 0 K = 0 − 273 = −273 °C
b 50 K = 50 − 273 = −223 °C
c 173 K = 173 − 273 = −100 °C
d 273 K = 273 − 273 = 0 °C
e 323 K = 323 − 273 = 50 °C
f 600 K = 600 − 273 = 327 °C
Internal energy is the sum of the potential energies
and the kinetic energies of all the particles in the
sample of substance.
An ideal gas has no potential energy. In a real gas,
molecules are joined together by intermolecular
forces or bonds, so they must possess potential
energy. It cannot therefore be an ideal gas.
Pressure and volume remain constant and it does not
change state. If energy is added, the average KE of
the molecules will increase, producing an increase in
temperature.
By heating it up (the average KE of the particles
increases) and by doing work on it (the average KE
of the particles is increased).
a Steam contains latent heat of vaporisation as well
as the kinetic energy of the particles.
b A burn from steam can, therefore, transfer more
energy than a burn from the same mass of boiling
water.
Specific heat capacity: the energy required to heat
up 1 kg of a substance by 1 K. Heat capacity: the
energy required to heat up a sample/body by 1 K.
energy change
300
heat capacity =
=
change in temperature 0 5
= 600 J K−1
energy
= 3600 = 8 K
ΔT =
heat capacity 450
Object A, since heat capacity is the energy required
to warm up a body by 1 K, and the object with the
smallest heat capacity will be the one to heat up the
most.
20 × 103
18 a ΔT = E =
= 3.2 K
m c 1.5 × 4200
b E = m c ΔT = 0.3 × 4200 × (100 − 20) = 1 × 105 J
c E = m × c ΔT
t
t
m
3000
So, = E =
= 0.02 kg (or 20 g)
t t c ΔT 1 × 4200 × 35
E = −80 000 =
d ΔT = (Tfinal − Tinitial) =
−7.6
m c 2.5 × 4200
Therefore, Tfinal = 20 − 7.6 = 12.4 °C
e E = m c ΔT = 2.5 × 4200 × 60 = 6.3 × 105 J
19 a Energy lost by aluminium = m c ΔT
= m c (Tinitial − Tfinal) = 0.4 × 900 × (800 − Tfinal)
b Energy gained by water = m c ΔT
= m c (Tfinal − Tinitial) = 2.5 × 4200 × (Tfinal − 20)
c 0.4 × 900 × (800 − Tfinal)
= 2.5 × 4200 × (Tfinal − 20)
=(
((
×
×
×
Therefore,
(. ×
Tfinal =
(
)+( ×
)+(
×
. ×
)) × Tfinal
×
×
)
× )
)+( . ×
) + (0.4 × 900)
2 88 105 2.1 104
= 46°C (2 s.f.)
10 500 + 360
d No energy has been lost to the surroundings.
E = m c ΔT = 5 × 103 × 420 × (1540 − 15)
= 3.2 × 109 J
2 5 × 103
ΔT = E =
= 31.25
mc 50 × 10 −3 × 1600
Therefore, Tfinal = 31.25 + 15 = 46°C
E = m × L = 1.5 × 2.3 × 106 = 3.5 × 106 J (2 s.f.)
E = (mc ΔT) + (m L)
= (1.5 × 4200 × (100 − 15)) + (1.5 × 2.3 × 106)
= 5.355 × 105 + 3.45 × 106 = 3.99 × 106 J
3 99 × 106
= 2.7 × 103 s (= 45 minutes)
So, t = E =
3
P
1 5 × 10
Energy gained by ice
= m cice ΔT + m L + m c water ΔT
= m cice ΔT + m L + m c (Tfinal)
= (0.05 × 2100 × 18) + (0.05 × 3.3 × 105) +
(0.05 × 4200 × Tfinal)
Energy lost by water
= m c ΔT = 0.25 × 4200 × (Tfinal − 12)
=
20
21
22
23
24
ANSWERS
13
(0.05 × 2100 × 18) + (0.05 × 3.3 × 105) +
(0.05 × 4200 × Tfinal) = 0.25 × 4200 × (Tfinal−12)
((0.05 × 4200) − (0.25 × 4200)) Tfinal
= (0.25 × 4200 × 12) − ((0.05 × 2100 × 18) +
(0.05 × 3.3 × 105))
So, Tfinal =
(0 25 × 4200 × 12) − ((0 05 × 2100 × 18) + (0 05
((0 05 × 4200) − ( × ))
3.3 105
•
•
))
= 7 °C (1 s.f.)
25 a The human body’s working temperature is lower
than the temperature of its surroundings. As the
runner produces more energy, it cannot be
transferred away by the usual methods of
conduction, convection or radiation.
b Energy can only be lost by the evaporation of
water from the surface of the skin: sweating. This
evaporation causes cooling.
Exercise 3.2 – Modelling a gas
1 An ideal gas:
• obeys each of the three gas laws under all
conditions: Boyle’s law, Charles law and the
pressure law
• consists of particles whose volume is negligible
compared to the volume of space the gas occupies
• consists of particles that collide elastically
• has a density sufficiently low that the vast majority
of collisions are with the walls of the container, not
with each other
• has no potential energy – this means that that an
ideal gas cannot be composed of molecules.
2 For a real gas to approximate to an ideal gas, it should:
• have a low pressure/density
• not liquefy or solidify
• have a fairly high temperature.
This usually implies that:
• molecules are small compared with the volume
they occupy
• there are no forces between molecules
• molecules collide elastically.
3 Gases exert pressure because:
• large numbers of particles, moving quickly, collide
with the walls of their container
14
each collision exerts a force on the container wall
the sum of all the forces from all the collisions
divided by the area of the container walls gives the
pressure.
4 Boyle’s law: p ∝ 1 when the temperature is constant.
V
5 Example: a syringe of gas (see Figure A3.1).
pressure
gauage
plunger
syringe
scale for
volume of gas
Figure A3.1
Equipment: Syringe with a volume scale, pressure
gauge fitted to end of syringe.
Measurements: Volume of air in syringe, using scale
on side of syringe, pressure of air inside syringe, using
pressure gauge.
Method: Change the volume of the air inside the
syringe by moving the plunger inwards and outwards.
This is the independent variable. Measure the
resulting pressure. This is the dependent variable.
For each value of volume, record the volume and
pressure of the air. Then, plot a graph of pressure
against volume. This should show that p ∝ 1 .
V
6 p1 ×V
V1 p2 ×V2
pV
100 × 20
Therefore, p2 = 1 1 =
= 500 kPa
V2
4
7 Assuming that the temperature remains constant, as
the air bubble rises through the water, the pressure
around the bubble decreases. Boyle’s law states that
p1 ×V
V1 p2 ×V2 so, if the pressure reduces, the
volume of the air bubble must increase.
8 Pressure law: if the volume of a sample of ideal gas is
constant, then the pressure of the gas is proportional
to its absolute temperature: p ∝ T.
9 See Figure A3.2.
pressure
gauge
Therefore, P2 =
5
P1 T2 1 01 10 × ( 273 − 60 )
=
T1
(273 + 10)
= 8 × 104 Pa (1 s.f.)
12 Charles’s law: if the pressure of a sample of ideal
gas is constant then the volume of the gas is
proportional to its absolute temperature;V ∝ T .
13 See Figure A3.4
thermometer
air
water
thermometer
ruler
thin tube
thread of
sulfuric acid
heating
Figure A3.2
water
A fixed-volume container of gas is attached to a
pressure gauge. The temperature of the gas can be
varied by heating the beaker of water in which it
is immersed. The temperature of the water (and
hence the temperature of the gas) is measured with
a thermometer. The pressure of the gas is measured
with a pressure gauge.Values of temperature and
pressure are recorded in a table. A graph of pressure
(on the y-axis) against temperature (on the x-axis)
should produce a straight line. Extrapolating the
graph backwards to where the pressure is zero leads
to a value of absolute zero (see Figure A3.3).
p/atm 20
15
10
5
−300 −250 −200 −150 −100
−50
0
50
100
T/°C
Figure A3.3
Note: Immersing the container of gas in a pot of
liquid nitrogen provides a useful extra measurement
for the graph at a temperature of −196 °C. This extra
point ‘anchors’ the straight line graph and makes the
value extrapolated for absolute zero more accurate.
10 At 0 K, atoms do not have any kinetic energy; they
do not move around. This means they cannot exert
forces on the walls that give rise to pressure.
11
P1 P2
=
T1 T2
trapped
dry air
heating
Figure A3.4
A small amount of liquid in a thin, uniform capillary
tube traps a volume of air beneath it. If the top end of
the capillary tube is open, atmospheric pressure keeps
the pressure of the trapped volume of air constant.
The capillary tube is attached to a ruler, so that the
length of the trapped volume of air can be measured;
in a uniform tube this volume is proportional to
the length. The capillary tube and ruler are then
immersed in a beaker of water, which can be heated.
The temperature of the water (and hence the air in
the capillary tube) is measured with a thermometer.
Values of temperature of the trapped air (the
independent variable) and volume of the trapped air
(the dependent variable) are recorded in a table. A
graph of volume against temperature in kelvin should
produce a straight line passing through the origin,
showing that V ∝ T .
14 Extrapolate the graph backwards to produce a value
for absolute zero. Note: if liquid nitrogen is available,
measuring the volume of the trapped air at a
temperature of −196 °C will ‘anchor’ the graph and
produce a more accurate value for absolute zero.
15 a Since V ∝ T , the absolute temperature quadruples.
b Absolute temperature is a measure of the average
kinetic energy KE = 1 mv 2 of the molecules.
2
If the temperature is quadrupled then the average
speed of the molecules must have doubled.
(
)
ANSWERS
15
16 For constant pressure, V1 = V2
T1 T2
300 × ( 273 + 20)
V T
So, T2 = 2 1 =
= 586 K
V1
150
= (586 − 273) °C
= 313 °C (300 °C to 1 s.f.)
17 a One mole is the quantity of a substance that
contains the same number of particles as that of
12 g of 12C.
b Since 1 g of hydrogen atoms contains
6.023 × 1023 atoms, 2 g must contain
2 × 6.023 × 1023 atoms = 1.2046 × 1024 atoms.
(1.2 × 1024 to 2 s.f.)
18 a N = 20 × 6 02 × 1023 = 2.15 × 1023 atoms
56
20 × 6 02 × 1023
bN=
= 5.1 × 1022 atoms
235
c Number of molecules = 20 × 6 02 1023
18
23
= 6.7 × 10 .
So, number of atoms = 6.7 × 1023 × 3
= 2.0 × 1024 atoms (2 s.f.)
19 The molar mass of a substance is the mass in grams
of one mole of the substance.
1000
20 a number of moles =
= 34.5 moles
29
b number of molecules = 34.5 × 6.02 × 1023
= 2.1 × 1025 molecules
c number of molecules m−3 = 2.1 × 1025 × 1.3
= 2.7 × 1025 molecules m−3
12 × 10 −3
21 a One atom of 12C has a mass of
6 02 1023
−26
= 2 × 10 kg
2 × 10 −26
= 1.66 × 10−27 kg
12
22 a p = pressure, Pa
V = volume, m3
N = number of particles in the sample of gas
m = mass of particle of gas, kg
b u=
c 2 = the mean squared velocity of the gas particles,
m2 s−2.
b p V = 1 N m c 2 and N m = M. Since M = ρ,
V
3
we can write p = 1 ρ c 2
3
c In any sample of gas in a container, there are a
large number of particles moving around in
16
23
24
25
26
random directions and with a range of velocities.
The total velocity summed across all particles
must be zero, otherwise the gas sample would
move out of the container. So, to find an average
of the magnitudes of the velocities, we sum the
squares of the velocities and take the average of
this sum by dividing by the number of particles.
This is called the mean-squared-velocity, c2 .
p = 1 ρ c2
3
3× p
3 × 1.01 × 105
Therefore, c rms =
=
= 483 m s−1
ρ
1.3
≈ 500 m s−1
Since 1 m c 2 ∝ T , if the mass is 16 times larger, then
2
2
c will be 1 , so the average speed of the oxygen
16
1
atom will be × 1000 = 250 m s−1.
4
a pV = n RT
p V 100 × 103 × 5 × 10 −3
=
= 0.2 moles
So, n =
R T
8 31 × (
+ )
b N = n × A = 0.2 × 6.02 × 1023
= 1.2 × 1023 molecules
KEaverage = 3 kT
2
2 × KEaverage
2 × 5.0 × 10 −21
So, T =
=
3×k
3 × 1.38 × 10 −23
= 240 K (2 s.f.)
Exam-style questions
1
2
3
4
5
6
B
D
B
C
C
a heat capacity = specific heat capacity × mass
= 160 × 0.6 = 96 J K−1
b GPE lost = mg Δh = 0.6 × 10 × 0.8 = 4.8 J
c KE → internal energy
internal energy gained 50 × 4 8
d ΔT =
=
=25K
heat capacity
96
So, T = 20 + 2.5 = 22.5 °C
7 a The molecules of water have a range of values of
energy. Those molecules with the largest energies
are able to break free of the water surface and
become water vapour.
b The temperature of the water is a measure of the
average kinetic energy of the water molecules. The
molecules that evaporate are those with the largest
energies, so when these are ‘lost’, the average
kinetic energy decreases. The temperature decreases
and the water cools.
c power × time = mass × SHC × ΔT
m × S C × ΔT 0.163 × 4200 × 80
=
= 274 s
power
200
d power × time = SLHV × Δm
⇒ time =
⇒ Δm =
power × time 200 × 5 × 60
=
= 26.5 g
SLHV
2.26 × 106
So, the students’ reading will be
163 − 26.5 = 137 g (3 s.f.)
8 a Energy from the heater passes through the
aluminium by conduction (the passing of energy
from one atom to the next). This takes a long time
because there are so many atoms between the
heater and the thermometer. So, the thermometer
does not record an increase for a couple of minutes.
b In one minute, the energy supplied to the
aluminium is 60 × 50 = 3000 J.
During this time the temperature of the
aluminium rises by 3.3 °C.
So, using the energy equation:
SHC =
9
energy supplied
= 3000
mass × ΔT
1 × 3.3
= 909 ≈ 900 J kg−1 °C−1
c Once the temperature of the aluminium block is
greater than the ambient temperature, the
aluminium block will start to share its energy with
the surroundings. This causes the rate of
temperature increase to slow down. A constant rate
of 3.3 °C min−1 would produce a temperature of
103 °C, but the rate is not constant; it is decreasing.
a mass × SLHF = power × time
⇒t =
m × SLHF 0 5 × 3 3 × 105
=
power
100
= 1650 s (27.5 mins)
b mass × SHC × ΔT = power × t
⇒t =
m × S C × ΔT 0 5 × 4200 × 100
=
power
100
= 2100 s (35 mins)
c mass × SLHV = power × t
m × SLHV 0 5 × 2.26 × 106
=
power
100
= 11 300 s (3.1 hours)
⇒t =
10 a Using Boyle’s law:
p1V1
p2V2
p2 =
p1V1 1 0 × 105
=
1
V2
2
= 2.0 × 105 Pa
b Atoms of ideal gas move randomly and collide
with the syringe walls. Each collision exerts a
force on the syringe wall. The sum of all the
forces from all the collisions, divided by the area
of the syringe walls, gives the pressure.
c The gas is compressed slowly, so there is time
for the increase in internal energy of the gas –
because of the work done on the gas – to be
shared with the surroundings. So, the temperature
of the gas does not increase (a necessary condition
for Boyle’s law).
d The pressure would be greater. If the compression
occurs quickly, the work done on the gas will
increase the temperature of the gas. The atoms of
gas will move faster and collide with the syringe
walls more frequently and with greater force. This
makes the pressure inside the syringe greater than
that calculated in part a.
Chapter 4: Waves
Exercise 4.1 – Oscillations
1 a Time period, T, is the time taken to make one
complete oscillation.
b Frequency, f, is the number of complete oscillations
made in one second.
c Amplitude, A, is the maximum displacement from
the equilibrium position.
d Equilibrium position is the position of the
oscillator when there are no unbalanced forces
acting on it.
e Displacement, x, is the distance (and direction) of
the oscillator away from the equilibrium position.
2 10 −3
= 4 × 10−4 s
5
b Amplitude = maximum displacement = 3 cm
2 a Period =
ANSWERS
17
b The temperature of the water is a measure of the
average kinetic energy of the water molecules. The
molecules that evaporate are those with the largest
energies, so when these are ‘lost’, the average
kinetic energy decreases. The temperature decreases
and the water cools.
c power × time = mass × SHC × ΔT
m × S C × ΔT 0.163 × 4200 × 80
=
= 274 s
power
200
d power × time = SLHV × Δm
⇒ time =
⇒ Δm =
power × time 200 × 5 × 60
=
= 26.5 g
SLHV
2.26 × 106
So, the students’ reading will be
163 − 26.5 = 137 g (3 s.f.)
8 a Energy from the heater passes through the
aluminium by conduction (the passing of energy
from one atom to the next). This takes a long time
because there are so many atoms between the
heater and the thermometer. So, the thermometer
does not record an increase for a couple of minutes.
b In one minute, the energy supplied to the
aluminium is 60 × 50 = 3000 J.
During this time the temperature of the
aluminium rises by 3.3 °C.
So, using the energy equation:
SHC =
9
energy supplied
= 3000
mass × ΔT
1 × 3.3
= 909 ≈ 900 J kg−1 °C−1
c Once the temperature of the aluminium block is
greater than the ambient temperature, the
aluminium block will start to share its energy with
the surroundings. This causes the rate of
temperature increase to slow down. A constant rate
of 3.3 °C min−1 would produce a temperature of
103 °C, but the rate is not constant; it is decreasing.
a mass × SLHF = power × time
⇒t =
m × SLHF 0 5 × 3 3 × 105
=
power
100
= 1650 s (27.5 mins)
b mass × SHC × ΔT = power × t
⇒t =
m × S C × ΔT 0 5 × 4200 × 100
=
power
100
= 2100 s (35 mins)
c mass × SLHV = power × t
m × SLHV 0 5 × 2.26 × 106
=
power
100
= 11 300 s (3.1 hours)
⇒t =
10 a Using Boyle’s law:
p1V1
p2V2
p2 =
p1V1 1 0 × 105
=
1
V2
2
= 2.0 × 105 Pa
b Atoms of ideal gas move randomly and collide
with the syringe walls. Each collision exerts a
force on the syringe wall. The sum of all the
forces from all the collisions, divided by the area
of the syringe walls, gives the pressure.
c The gas is compressed slowly, so there is time
for the increase in internal energy of the gas –
because of the work done on the gas – to be
shared with the surroundings. So, the temperature
of the gas does not increase (a necessary condition
for Boyle’s law).
d The pressure would be greater. If the compression
occurs quickly, the work done on the gas will
increase the temperature of the gas. The atoms of
gas will move faster and collide with the syringe
walls more frequently and with greater force. This
makes the pressure inside the syringe greater than
that calculated in part a.
Chapter 4: Waves
Exercise 4.1 – Oscillations
1 a Time period, T, is the time taken to make one
complete oscillation.
b Frequency, f, is the number of complete oscillations
made in one second.
c Amplitude, A, is the maximum displacement from
the equilibrium position.
d Equilibrium position is the position of the
oscillator when there are no unbalanced forces
acting on it.
e Displacement, x, is the distance (and direction) of
the oscillator away from the equilibrium position.
2 10 −3
= 4 × 10−4 s
5
b Amplitude = maximum displacement = 3 cm
2 a Period =
ANSWERS
17
a i Where the gradient of the graph is a
maximum – this occurs when the displacement
is zero.
ii Where the gradient of the graph is zero – this
occurs when the displacement is a maximum
or minimum from the equilibrium position.
b The gradient of the graph of displacement against
time gives the velocity of the oscillating body.
10
4 a f = = 0.4 Hz
25
1
b f = 1 =
= 2 5 × 106 Hz (or 2.5 MHz)
T 4 × 10 −7
26
= 1.3 Hz
c f =
20
5 a T = 1 = 1 = 2 × 10 −2 s (or 0.02 s)
f 50
b T = 60 = 1.2 s
50
1
c T= 1 =
= 1 09 × 10 −10 s
f 9 19 × 109
6 Isochronous means taking the same time. So, an
oscillation that is isochronous takes the same time to
make an oscillation for every oscillation.
7 a θ = 2π radians
2π π
= radians
b θ=
2
4
2π = π
radians
c θ=
12 6
d θ = 1 radian. This shows how to convert between
degrees and radians.
a θ = 360°
b θ = π × 360 = 45°
4 2π
2π × 360 =
1° This is how to convert
c θ=
360 2π
between radians and degrees.
9 a When x is small, sin x ≈ x.
b At two decimal places, sin x ≈ x breaks down
when x > 0.31 radians (about 18°).
c The mathematical rule is: For x < 18°
(0.31 radians), sin x ≈ x.
10 Restoring force must be proportional to
displacement from the equilibrium position.
Restoring force must be directed towards the
equilibrium position.
11 GPE → KE → GPE, and so on.
8
18
12 a EPE → KE → EPE, and so on.
b The velocity of the mass will be a maximum
when the kinetic energy of the mass (KE) is a
maximum. This will occur when the elastic
potential energy (EPE) is a minimum. This occurs
when the mass is at the equilibrium position.
13 a and b See Figure A4.1
displacement/cm
velocity/cm s–1
acceleration/cm s–2
0
0.5
1
1.5
2
2.5
3
3.5
Time/s
Figure A4.1
c The graph for acceleration is the same shape as
the graph for displacement, but it is reflected in
the x-axis. The acceleration is proportional to the
displacement and the constant of proportionality
is negative. These agree with the definition of
simple harmonic motion.
14 a v = vmax sin ω t
b v = vmax cos ω t
15 a, b and c, see Figure A4.2.
Displacement/cm
3
D
0
D
P
O
O
P
Time/s
Figure A4.2
Any two points that are a whole number of waves
apart will be in phase.
Any two points that are an odd number of halfwaves apart will be out of phase.
Two points that are a quarter of a wave out of phase
will have a phase difference of π .
2
Exercise 4.2 – Travelling waves
1 a Wavelength, λ: the distance between two adjacent
points that are in phase.
b Frequency, f: the number of complete waves in one
second.
Displacement/cm
c Time period, T: the time it takes for one complete
wave to pass a certain point.
d Amplitude, A: the maximum displacement from
equilibrium.
2 A transverse wave has oscillations that occur at 90o
to the direction in which the wave is moving. A
longitudinal wave has oscillations that are in the same
(or opposite) direction as that of the wave’s motion.
3 a and b See Figure A4.3.
C
0
Time/s
T
Figure A4.3
c Phase difference is 1 of 2π = π radians.
2
4 a Echo
distance travelled
b speed of sound wave =
and
time taken
distance travelled = 2 × distance to wall.
So, distance to wall =
speed of wave × time 330 × 0 5
=
= 82.5 m
2
2
(83 m, 2 s.f.)
5 Equipment: an oscilloscope with a dual trace and two
microphones, as in Figure A4.4.
Speed of sound in air =
distance between two microphones
time interval between the two blipss on the oscilloscope
% uncertainty in speed = % uncertainty in distance
+ % uncertainty in time
3 × 108
6 a f = c =
= 1.5 × 106 Hz (1.5 MHz)
λ
200
3 × 108
b f = c =
= 6 × 1025 Hz
λ 5 × 10 −18
3 × 108
c f = c =
= 5.7 × 1014 Hz
λ 530 × 10 −9
7 Radio waves; microwaves; infra-red; visible light;
ultraviolet; X-rays; gamma rays
8 a Microwaves (or very short wavelength radio
waves).
8
c = 3 × 10
b fmin =
= 3 × 107 Hz (30 MHz)
λmax
10
c 11 GHz is a higher frequency than the minimum
frequency of 30 MHz, so will not be absorbed by
the ionosphere.
9 The wavelength of X-rays is of the same order of
magnitude as the size and spacing of atoms. This
allows X-rays to penetrate between atoms of solid
substances. Substances with greater densities have
atoms closer together and so absorb more X-rays,
making the image received less intense.
Exercise 4.3 – Wave characteristics
make sound
here
microphone
microphone
1m
Figure A4.4
Place the two microphones a set distance apart, say
1.0 m. Connect each microphone to the oscilloscope.
When each microphone receives a sound, the
oscilloscope trace will show a small blip.
Measurements: distance between the two
microphones, time between the two blips on the
oscilloscope screen.
1 Wavefront: line representing points of a wave that are
all of the same phase, perpendicular to the direction
in which the wave is travelling. Ray: line showing the
direction in which the wave is travelling.
2 Wavefronts are perpendicular to rays.
3 See Figure A4.5.
wavefront
Huygens secondary
wavelet source
With Huygens secondary wavelet sources closer
together, the combined wavefront becomes a
straight line.
Figure A4.5
ANSWERS
19
4 a Intensity: energy per second per unit area or power
per unit area.
1
b I ∝ 2 from a small emitter, so double distance
r
means ¼ intensity.
c I ∝ A2 so new intensity = 25 times initial intensity.
5 a Output power from source will be spread out over
a spherical surface of area 4πr2.
Intensity
i Intensity = P = 20 = 1.6 W m−2
4πr 2 4π12
P = 20
ii Intensity =
= 0.18 W m−2
4 π r 2 4 π 32
b i See Figure A4.6.
I~
1
(distance)2
Distance
Figure A4.6
Intensity
ii See Figure A4.7.
I ~ A2
Amplitude
Figure A4.7
P
6 a I=
4 πr 2
Therefore, P = 4π Ir
I 2 = 4 × π × 1370 × (1.5 × 1011 )2
= 3.9 × 1026 W
b At the surface of Mercury,
3 9 × 1026
3
−2
I= P2 =
2 = 9.6 × 10 W m
4 πr
4π ×
×
(
)
Constructive superposition (or constructive
interference).
8 a When the two waves overlap exactly there will be
no displacement – the two waves will cancel each
other out.
b Destructive superposition (or destructive
interference).
9 Answer depends on student’s selections.
10 Unpolarised means that the oscillations that form
the wave occur in all possible planes/directions.
11 a They will make the free electrons oscillate in the
vertical direction, up and down the metal rod.
b The free electrons require energy to oscillate.
They get this energy from the wave.
c They will try to make the free electrons oscillate
in the horizontal direction. Since the metal rod is
thin, the free electrons will not have sufficient
space in which to oscillate and will not absorb
energy from the wave.
d No
e The number of waves with their electric force
vector in the vertical direction will be reduced
after passing by the metal rod.
f No change
g Having passed the metal rod, it is likely that the
waves will have their electric force vectors in the
horizontal direction.
h Horizontal
12 a See Figure A4.9.
unpolarised
light
transmitted
electric
vector
long chain
polymer molecules
Figure A4.9
7 See Figure A4.8.
Figure A4.8
20
b Since half of the waves will have been absorbed,
the intensity after passing through the polarising
1
material will be .
2
13 a Malus’s law for a polariser states that the transmitted
intensity, I, is given by I I 0 cos2 θ, where θ is the
angle between the transmission axis of the polariser
and electric vector of the incident wave.
2 a See Figure A4.13.
b See Figure A4.10.
incident
ray
Intensity
Imax
reflected
ray
normal
i
air
0
water
0
90°
Angle, θ
r
180°
refracted
ray
Figure A4.10
Figure A4.13
14 See Figure A4.11.
Figure A4.11
Unpolarised light is polarised into one plane by the
first piece of polarising material. Light passing through
one piece of sticky tape has its plane of polarisation
rotated. Rotated polarised light is not transmitted very
well through the second piece of polarising material.
The eye sees a dark region where there is one piece
of sticky tape. Light passing thorough the overlapped
pieces of sticky tape has its plane of polarisation
rotated twice, effectively cancelling out the rotation of
the plane of polarisation.The eye sees a bright region
where the two pieces of sticky tape overlap.
b sin i = n , where n is the refractive index of the
sin r
water
c Snell’s law.
3
n is defined as the ratio of the speed of the waves
in air (or in a vacuum) to the speed of the waves in
v air
the substance: n =
.
v substance
4 a sin i = n
sin r
Therefore, r = sin −1
= 25° (2 s.f.)
3 108
b vglass = c =
= 2 × 108 m s−1
15
15
3 × 108
c
c λair = =
= 5 × 10−7 m
f 6 × 1014
d λglass =
Exercise 4.4 – Wave behaviour
1 a See Figure A4.12.
incident
ray
normal
i
reflected
ray
r
( )
i = siin −1 ⎛ sin 40o ⎞
⎜⎝ 1 5 ⎟⎠
n
v gglass 2 × 108
=
= 3.3 × 10−7 m
f
6 × 1014
e Green
f The same (green). This is an important idea: it is
the frequency of an electromagnetic wave that
dictates what colour we perceive, not the
wavelength.
c
ny
v
n
c
5 nx = c and ny =
So, x = x =
c
vy
vx
vy
nx
ny
plane mirror
Figure A4.12
b Angle of incidence equals the angle of reflection.
ANSWERS
21
6 a See Figure A4.14.
air
water
θc
incident
ray
normal
8
Repeat this several times to get a range of angles of
incidence.
Record values of i and r in a suitable table.
Process the values of i and r into two more columns,
sin i and sin r.
Plot a graph of sin i (on the y-axis) against sin r (on
the x-axis).
Obtain a best-fit line. It should be a straight line that
passes through the origin, with gradient = sin i
sin r
= refractive index of the glass block.
a See Figure A4.16.
Figure A4.14
sin i
sin r
i = 90° and r = θc the critical angle
So, n = 1
sin θ c
c For angles greater than θc, the ray undergoes total
internal reflection.
7 Equipment: Light box or laser to produce a thin beam
of light, rectangular glass block, protractor, pencil,
white paper.
Method: Place the glass block on white paper.
Set up the ray box or laser so that a thin beam of
light is incident on one side of the glass block, as in
Figure A4.15.
b Snell’s law gives: n =
normal
ray from ray
box/laser
i
glass
block
r
Figure A4.15
Draw around the glass block. Make marks where
the incident beam approaches and is incident on the
block, and where the beam leaves the opposite side of
the glass block.
Remove the glass block. Complete the paths of the
incident ray and the refracted ray through the glass
block. Draw in the normal where the ray is incident
on the glass block. Now measure, using the protractor,
the angle of incidence, r, and the angle of refraction, r.
22
outer core/cladding
inner core
outer core/cladding
Figure A4.16
An optical fibre uses the principle of total internal
reflection. An inner core of material with a high
refractive index is surrounded by an outer core of
a material with a smaller refractive index. Light
that is incident on the boundary between the
inner core and outer core (at an angle greater than
the critical angle between the two media) will
undergo total internal reflection. The light will
pass along the inner core without loss of energy.
b Information transfer, optical leads between
computers or hi-fi units, medical uses, endoscopy.
9 Diffraction: the spreading out of waves as they pass
by an object or through a gap.
10 Interference. The bright regions are occurring
because waves are arriving there in phase and adding
constructively. The dark regions in between the
bright regions are caused by waves arriving out of
phase and adding destructively.
11 a At B, waves from the two slits are arriving in
phase. They add together constructively and so
produce a wave with a large amplitude.
b At D, waves from the two slits are arriving out of
phase. They add together destructively and so
produce a wave with little or no amplitude.
c At C, you would expect to see a bright region.
This is because the waves from the two slits have
to travel exactly the same distance to get to C. If
these waves began in phase, they will still be in
phase when they get to C and will add together
constructively to produce a bright region.
12 s = λ D = 0.5 × 4 = 2 m
d
1
13 Path difference: the difference between how far
waves travel along one path and how far waves travel
along another path.
14 a Waves will interfere constructively, producing a
wave with a large amplitude.
b Waves will interfere destructively, producing little
or no amplitude.
15 There would be places within the oven where waves
would interfere destructively. This would mean that
no energy was arriving there, and so the food would
not be cooked at these points. In other places, waves
could arrive and interfere constructively. This would
mean that a large amount of energy was arriving
there, and the food would cook quickly (and
possibly burn). A rotating turntable ensures that food
is continually moving past any points of destructive
or constructive interference, allowing the food to
cook more evenly.
16 Equipment: a laser (or other coherent source of
light), a pair of Young slits, a screen and a ruler.
Allow the laser light to be incident on the Young
slits normally. Place the screen a few metres away
from the slits and observe the interference pattern
on the screen. Measure the separation of the bright
fringes, s, using the ruler and measure the distance
from the slits to the screen, D.
λ can then be found using the equation: λ = sd .
D
The uncertainty in the two distance measurements
will depend on the measuring device used, but
usually this will be about ± 1 mm. For s, the
percentage uncertainty might be about 5%, while
the percentage uncertainty for D will be very small;
less than 0.1%. The separation of the Young slits, d,
is usually stated by the manufacturer on a label
on the apparatus; it is likely to be stated to two
significant figures (for example, 1.0 × 10−4 m)
so the uncertainty in this will be about 10%.
The percentage uncertainty in λ is the sum of
the percentage uncertainties of s, d and D; this is
likely to be something like 5 + 10 + 0.01 = 15%.
(In practice this method produces very accurate
results.)
17 Young used a single slit. He ensured that the central
maximum of the single slit diffraction pattern from
it illuminated the pair of slits, thus making the light
incident on the double slits coherent.Young also
used a colour filter to produce monochromatic light.
λ D = 450 × 10 −9 × 8
18 a s =
= 2.4 cm.
d
0.15 × 10 −3
b Since s ∝ λ, if the new λ is 1.5 times larger, then
the separation of the maxima will be 1.5 times
larger.
So, snew = 1.5 × 2.4 = 3.6 cm.
Exercise 4.5 – Standing waves
1 In a progressive wave, energy is transferred from one
place to another. In a standing wave, no energy is
transferred.
2 A standing wave is formed by the superposition of
two waves. Where the superposition is constructive
an antinode is formed and where destructive
superposition occurs a node is formed.
3 a Since one end of the string is a node and the other
end of the string is an antinode, this must be ¼ of a
wave. Therefore, λ= 4 l.
b See Figure A4.17.
second harmonic
l
Figure A4.17
Now the string is ¾ of a wave. So, λ = 4 l.
3
c See Figure A4.18.
third harmonic
l
λ=
4
l
5
fourth harmonic
l
λ=
4
l
7
Figure A4.18
ANSWERS
23
d The set of possible wavelengths is given by
4
λ=
l
( − )
4 a Since there is an antinode at each end of the pipe,
the length of the pipe contains 1 a wave. So, for
2
the first harmonic, λ = 2l.
b See Figure A4.19.
4
5
6
7
8
D
A
C
C
a and c See Figure A4.20.
direction of sound wave
second harmonic
Pmin Pamb
Pmax
l
Figure A4.20
λ=l
third harmonic
b See Figure A4.21.
Air pressure
l
λ=
2
l
3
fourth harmonic
ambient
air pressure
2
4
Distance/m
l
λ=
l
2
Figure A4.19
c The set of possible wavelengths is given by: λ = 2 l
n
5 a No. The equation for the possible wavelengths
would be the same as that for a string that is held
at one end and oscillated at the other.
b No. The equation for the possible wavelengths
would be the same as that for the pipe that is open
at both ends. Since the separation of nodes is the
same as the separation of antinodes, the same
mathematical relationship will apply to both
situations.
c Yes, they are the same.
d Yes, they are the same.
Exam-style questions
1 B
2 D
3 C
24
Figure A4.21
d f = v = 330 = 165Hz
λ
2
9 a There are 1.25 wavelengths in the pipe.
So, 1.25 λ = 3 ⇒ λ = 3 = 2.4 m
1.25
v 330
b f = =
= 137.5 Hz (140 Hz to 2 s.f.)
λ 24
10 a A node is a place where there is no oscillation of
the medium. An antinode is where the oscillation
of the medium is a maximum.
b The wavelength of the waves is twice the node
separation, so λ = 2 × 8 cm = 16 cm.
c Since the tension in the string and the mass per
unit length of the string have not changed, the
speed of the waves along the string remains
constant. So, if the frequency of the standing wave
has increased, then the wavelength must have
decreased so that f × λ = constant.
11 a In water, the molecules are much closer together
than in air. The energy of an oscillation is
transmitted quickly from one molecule to the
next. Air molecules are much further apart, so an
oscillation is transmitted much less quickly.
v 1 5 × 103
b i λ= =
= 7.5 × 10−3 m
f 200 × 103
(or 7.5 mm)
v
330
= 1.65 × 10−3 m
ii λ = =
f 200 × 103
(or 1.65 mm)
c No. 200 kHz is above the range of human ears, so
cannot be heard.
12 a See Figure A4.22.
Intensity
Distance across screen
Figure A4.22
b The blue regions (as opposed the previous red
regions) will be closer together.
c With a smaller slit, the intensity of the pattern
would be reduced (because less light is able to
pass through), and the spacing of the bright
regions would increase (more diffraction).
Chapter 5: Electricity and magnetism
Exercise 5.1 – Electric fields
1
a Friction between the sweater and the ruler causes
some electrons from the atoms of the ruler to be
rubbed off . These electrons are collected by the
sweater, making the sweater negatively charged.
The loss of electrons from the ruler leaves it
positively charged.
b The woollen sweater will have become negatively
charged.
2 a Friction between your hair and the plastic comb
causes electrons from your hair to be rubbed off.
These electrons are collected by the plastic comb,
making the plastic comb negatively charged.
b The loss of electrons from your hair leaves your
hair positively charged.
3 a Negative charge.
b −1.6 × 10−19 C
1
c
= 6.25 × 1018 electrons
1 6 × 10 −19
4 The positively charged Perspex® rod attracts
electrons in the pieces of tissue paper. These
electrons move towards the top of the pieces of
paper, increasing the force of attraction between
them and the positively charged rod. The pieces
of paper move towards the rod when the force of
attraction is larger than the weight of the piece of
paper.
5 The toy doll is sitting on the top of a positively
charged metal dome. Electrons from the toy doll
have moved off the doll onto the dome. This leaves
the toy doll – and each strand of the toy doll’s hair –
positively charged. Since like charges repel each
other, every strand of the doll’s hair is repelling every
other strand. This makes the doll’s hair stick up and
out, as in Figure 5.2.
6 a An electrical conductor is a material that allows
electrons (or, in some cases, ions) to move freely
within it.
b An electrical insulator does not allow electrons
(or other charged particles) to move freely
within it. This is usually because the number of
free (delocalised) electrons per unit volume is
small compared to how many there are in a
conductor.
7 Metals have very large numbers of free electrons. For
example, the number of free electrons per m3 for
copper is of the order of 1029 m−3.
8 a A fundamental unit is one from which other units
are derived. Fundamental units have exact
definitions and can be experimentally verified.
b A coulomb is the amount of charge transferred
when a current of 1 A flows for one second.
c 1 C ≡ 1A s
ANSWERS
25
11 a In water, the molecules are much closer together
than in air. The energy of an oscillation is
transmitted quickly from one molecule to the
next. Air molecules are much further apart, so an
oscillation is transmitted much less quickly.
v 1 5 × 103
b i λ= =
= 7.5 × 10−3 m
f 200 × 103
(or 7.5 mm)
v
330
= 1.65 × 10−3 m
ii λ = =
f 200 × 103
(or 1.65 mm)
c No. 200 kHz is above the range of human ears, so
cannot be heard.
12 a See Figure A4.22.
Intensity
Distance across screen
Figure A4.22
b The blue regions (as opposed the previous red
regions) will be closer together.
c With a smaller slit, the intensity of the pattern
would be reduced (because less light is able to
pass through), and the spacing of the bright
regions would increase (more diffraction).
Chapter 5: Electricity and magnetism
Exercise 5.1 – Electric fields
1
a Friction between the sweater and the ruler causes
some electrons from the atoms of the ruler to be
rubbed off . These electrons are collected by the
sweater, making the sweater negatively charged.
The loss of electrons from the ruler leaves it
positively charged.
b The woollen sweater will have become negatively
charged.
2 a Friction between your hair and the plastic comb
causes electrons from your hair to be rubbed off.
These electrons are collected by the plastic comb,
making the plastic comb negatively charged.
b The loss of electrons from your hair leaves your
hair positively charged.
3 a Negative charge.
b −1.6 × 10−19 C
1
c
= 6.25 × 1018 electrons
1 6 × 10 −19
4 The positively charged Perspex® rod attracts
electrons in the pieces of tissue paper. These
electrons move towards the top of the pieces of
paper, increasing the force of attraction between
them and the positively charged rod. The pieces
of paper move towards the rod when the force of
attraction is larger than the weight of the piece of
paper.
5 The toy doll is sitting on the top of a positively
charged metal dome. Electrons from the toy doll
have moved off the doll onto the dome. This leaves
the toy doll – and each strand of the toy doll’s hair –
positively charged. Since like charges repel each
other, every strand of the doll’s hair is repelling every
other strand. This makes the doll’s hair stick up and
out, as in Figure 5.2.
6 a An electrical conductor is a material that allows
electrons (or, in some cases, ions) to move freely
within it.
b An electrical insulator does not allow electrons
(or other charged particles) to move freely
within it. This is usually because the number of
free (delocalised) electrons per unit volume is
small compared to how many there are in a
conductor.
7 Metals have very large numbers of free electrons. For
example, the number of free electrons per m3 for
copper is of the order of 1029 m−3.
8 a A fundamental unit is one from which other units
are derived. Fundamental units have exact
definitions and can be experimentally verified.
b A coulomb is the amount of charge transferred
when a current of 1 A flows for one second.
c 1 C ≡ 1A s
ANSWERS
25
9
a See Figure A5.1.
14 a F
k
Qq
50 × 10 −9 × −20 × 10 −9
9
=
9
×
1
0
×
2
r2
×
−4
Z
X
(
5
Figure A5.1
b F ∝ 12, where r is the distance between the
r
two charges.
10 a F is towards B (because the charges are opposite).
b Yes, sphere B experiences the same sized force,
but in the direction towards A.
c Newton’s third law.
d
Charge
on B
Distance
between
A and B
Force
experienced
by A
+50 mC
−50 mC
x
F
−F
+50 mC
+50 mC
x
−F
F
+50 mC
−200 mC
x
4F
−4F
+100 mC
−100 mC
x
4F
−4F
+50 mC
−50 mC
2x
1
F
4
1
− F
4
+100 mC
+100 mC
x
3
−36 F
36 F
11 a F ∝ q
b F∝Q
c F ∝ 12
r
12 Inverse square law means that something is
proportional to 12 , where r is the distance between
r
two particles/objects.
13 Coulomb’s law: the electrical force between two
charged particles, Q and q, separated by a distance,
r, is proportional the product of their charges and
inversely proportional to the square of their
Qq
separation, r. Algebraically, this is: F k 2 ,
r
where k is a constant.
)
= 2 × 10 N (1 s.f.)
(Note that the positive value shows this is a
repulsive force.)
Qq
−1 6 × 10 −19 × 1 6 × 10 −1
c F k 2 = 9 × 109 ×
2
r
×
(
)
= −2.3 × 10−8 N
dF
k
Qq
6 4 × 10 −19 × 1.26 × 10 −17
9
=
9
×
1
0
×
2
r2
×
(
)
3
= 8.1 × 10 N
1 Qqq =
1
15 F =
4πε r 2 4π × 7.8 × 10 −10
Force
experienced
by B
Table A5.1
26
)
= −1.0 × 10 N
(Note that the minus sign is showing this is an
attractive force.)
−6
Qq
10 −6
b F k 2 = 9 × 109 × 30 × 10 × 20 ×
2
r
×
Y
Charge
on A
(
× 1 6 × 10
= 6.5 × 10−13 N
16 a F
k
(
−19
× 1 6 × 10 −19
)
2
×
−3
Qq
10 −3
= 9 × 109 × 50 × 10 × 20 ×
2
2
r
×
12
(
)
= 2.3 × 10 N (2 s.f.)
The sign of the force is positive, so the force is
repulsive.
−3
−3
Qq
b F k 2 = 9 × 109 × −50 × 10 × 202× 10
r
×
(
12
c
17 a
b
c
)
= −2.3 × 10 N
Attractive.
A positive force is repulsive and a negative force is
attractive.
Positively. The sphere must be positively charged
to make the force repulsive on a unit positive test
charge.
i Horizontal from left to right, radially away from A.
ii Vertically downwards, radially away from B.
i Horizontally left to right, radially from C
towards the sphere.
ii Vertically downwards, radially from D towards
the sphere.
18 See Figure A5.2.
+
+
+
+
+
+
+
–
–
–
–
–
–
–
Figure A5.2
19 a Electric field strength: the force that acts on a unit
positive test charge in the field.
b E has units of N C−1
c E=F
q
20
Charge on object
creating the field / C
Distance from
charge / m
Field strength /
N C−1
40 × 10−6
3 × 10−2
4 × 108
1.6 × 10−19
1.1 × 10−10
1.2 × 1011
6.4 × 10−19
3 × 10−14
6.4 × 1018
−3 × 10−3
1.5 × 10−2
1.2 × 1011
c For each section, δQ1, the opposite section, δQ2,
would have an opposite contribution to the total
electric field strength at P. This suggests that the
total field strength at the centre of the sphere
would be zero.
d All opposing sections of charge on the surface of
the sphere would create a total field strength at P
of zero. Therefore, the electric field strength inside
a hollow sphere is zero in all places.
24 Using the same arguments for small charge volumes
within the inside of the sphere, we can conclude
that the electric field strength inside a solid charged
sphere is zero. (This assumes that the density of
charge is the same throughout the sphere.)
25 a See Figure A5.4.
R
X
Table A5.2
P
FY
Y
FX
Figure A5.4
−19
Q
6 1.6 10
= 9 × 109 ×
2
2
r
7 10 10
= 1.8 × 1010 N C−1
b F = E q = 1.8 × 1010 × −1.6 × 10−19
= −2.9 × 10−9 N
F 2 9 × 10 −9
= 3.2 × 1021 m s−2
c a= =
m 9 1 × 10 −31
22 The separation of electrons in a metal lattice is likely
to be of the order of 10−10 m, so each electron is
subject to a large repulsive force from the other
electrons. This results in the electrons being regularly
spaced within the metal lattice.
23 a and b See Figure A5.3.
21 a
=k
(
1
2
b Resultant force on 1 C particle is zero.
c See Figure A5.5.
FY
60°
FX
R
X
Y
P
Figure A5.5
The resultant force on the 1 C particle at R is
given by:
Qq
F = 2 × k 2 cos 30°
r
50 × 10 −6 × 1 × 0.866
= 2 × 9 × 109 ×
2
2 10 2
= 1.9 × 109 N vertically upwards.
(
P
δQ2
)
δQ1
)
Figure A5.3
ANSWERS
27
=
(3
9 x 109
10
)
2 2
30 a See Figure A5.7.
(40 + 20) × 10 −66
Therefore, E = 6 × 108 N C−1 horizontally, right
to left.
b Total field strength
Q
Q
= k 12 + k 22 = 9 109 ×
(r1 )
(r2 ) ⎛
⎞
5
20
⎜
⎟
−
2
2
⎜⎝ 3 10 2
6 10 2 ⎟⎠
(
) (
)
=0
Therefore, E = 0.
Q
Q
c Total field strength = k 2 left to right + k 2 top
r
r
to bottom.
Using Pythagoras’ theorem:
4
= 9 × 109 × 2 ×
3 10 2
at −45° to the horizontal.
27 See Figure A5.6.
(
)
2
A
Distance
Figure A5.7
The work done is the area under the graph
between A and B.
B
b work done = ∫ F dx
A
31 a work done = q × ΔV = 1.6 × 10−19 × 1
= 1.6 × 10−19 J
6 4 × 10 −19
= 4 eV
b i 6.4 × 10−19 J =
1 6 × 10 −19
+
+
32 a
b
c
15 – x
33 a
Figure A5.6
5 −
12
⇒ 5(
2
x
(15 − x )2
−
)2 = 12x 2
Therefore, 5 (152 − 30x + x2) = 12x2.
2
2
Therefore, x = 150 ± 150 + 4 × 7 × 5 × 15
−2 × 7
The only viable solution to this is: x = 5.9 cm from X.
28 a F = E q
b Because F is a constant, energy used
= force × distance = E q x
c Work done
= E q x = 2 × 10−7 × 3 × 10−3 × 5 × 10−2
= 3 × 10−11 J
29 a F = E q = 4 × 108 × 3.2 × 10−19
= 1.3 × 10−10 N
F
1 3 × 10 −10
= 2.0 × 1016 m s−2
b a= =
m 4 × 1.66 × 10 −27
28
B
= 5.7 × 1013 N C−1
+12μC
E=0=k
area = work done
ii 3.2 × 10−13 J =
+5μC
x
Force
26 a Total field strength
Q1
Q2
= k2 (Q1 Q2 )
= k 2 +k
2
(r1 )
(r2 ) r
b
c
34 a
b
c
35
3 2 × 10 −13
= 2 × 106 eV
1 6 × 10 −19
= 2 MeV
2 10 −15
iii 2 × 10−15 J =
1 6 × 10 −19
= 1.25 × 104 eV (10 keV 1 s.f.)
3 eV = 3 × 1.6 × 10−19 = 4.8 × 10−19 J
200 keV = 200 × 103 × 1.6 × 10−19 J
= 3.2 × 10−14 J
7.4 MeV = 7.4 × 106 × 1.6 × 10−19
= 11.8 × 10−13 = 1.2 × 10−12 J
E = q × ΔV = 4.8 × 10−19 × 200 = 9.6 × 10−17 J
(OR: E = 3 × 200 = 600 eV)
E = V I t = 4 × 2.5 × 60 = 600 J
E = V I t = 240 × 40 × 10−3 × 60 × 60
= 3.5 × 104 J
V =A v
Q (in one second) = n A v e
I = nA v e
Electron
number
density
Cross
sectional
area
1.6 × 1029 m−3
29
−3
Drift speed
Charge
on charge
carrier
Current
1.0 × 10−6 m2 1.5 × 10−5 m s−1 1.6 × 10−19 C 0.38 A
1.6 × 10 m
4.0 × 10−6 m2 4.9 × 10−6 m s−1 1.6 × 10−19 C 0.5 A
1.6 × 1029 m−3
1.3 × 10−8 m2
2.5 × 10−4 m s−1 1.6 × 10−19 C 80 mA
Table A5.3
36 a Number of moles per kg =
= 15.7 moles kg−1
1000
63.5
b Number of atoms per kg = 15.7 × 6.02 × 1023
= 9.5 × 1024 atoms kg−1
c Number of atoms per m3 = 9.5 × 1024 × 8900
= 8.4 × 1028 atoms m−3
d n = 2 × 8.4 × 1028
= 1.6 × 1029 free electrons m−3
1000 × 6 02 1023 × 2700 × 1
37 a n =
27
= 6.0 × 1028 free electrons m−3
1000 × 6 02 1023 × 19300 × 1
b n=
197
= 5.9 × 1028 free electrons m−3
1000 ×6.02
6 02
0 10023 ×7850 ×1
c n=
56
= 8.4 × 1028 free electrons m−3
I =
100 × 10 −3
38 a v =
nAe 1 6 × 1029 × 1 × 10 −6 × 1.6 × 10 −19
= 3.9 × 10−6 m s−1
08
bt= l =
= 2.0 × 105 s (about 57 hours).
v 3 9 × 10 −6
c The wires are full of free electrons. As soon as the
circuit is switched on, all of the electrons in the
wire (and in the light bulb filament) start to move.
This allows energy to be transferred immediately
in the light bulb filament.
dQ 4 × 10 −3
39 a I =
=
= 8 × 10−5 A
dt
50
b Q = It = 30 × 10−3 × 60 = 1.8 C
−3
Q
= 1 × 10 −6 = 40 s
I
25 × 10
Q = I t = 250 × 10−3 × 60 = 240 C
Electrical potential energy → kinetic energy of
electrons → thermal energy in resistor
E = V Q = 6 × 240 = 1440 J (1.4 kJ 2 s.f.)
P = E = 1440 = 24 W
t
60
c t=
40 a
b
c
d
Exercise 5.2 – Heating effect of electric currents
1 a Current. (Note that it is usual to plot the
independent variable on the x-axis.) Changing the
variable resistor varies the current, and the student
then measures the voltage across the resistor as a
dependent variable.
b Voltage across the component. The dependent
variable would then be the current flowing
through the component.
c See Figure A5.8.
6V
V
A
component
here
Figure A5.8
d In this circuit, the potential divider allows a voltage
to be selected that can be 0 V (this occurs when
the slider on the potential divider is at the far
left-hand side of the resistor) or 6 V (this occurs
when the slider of the potential divider is at the far
right-hand side of the resistor).
2 a Resistance: the ratio of the voltage across a
component to the current flowing through it,
R = V , measured in ohms, Ω.
I
b Volts are J/C, so base units for this are:
kg m2 s−2 / A s = kg m2 s−3 A−1
Base unit for current is A.
Therefore, base unit for resistance = kg m2 s−3 A−2
V
05
3 a R= =
=2Ω
I 250 × 10 −3
V
5
= 1 × 105 (100 kΩ)
b R= =
I 50 × 10 −6
V
120
= 4 × 103 (4 kΩ)
c R= =
I 30 × 10 −3
4 Suppose a current, I flows into the cube. Using
Kirchhoff ’s first law:
I flows through one of the three resistors at the first
3
junction. V1 = IR
3
At the next junction, current flowing through one of
IR
the resistors is I , so V2 =
6
6
ANSWERS
29
5
6
I flows through the last resistor before the opposite
3
IR
corner, so V3 =
3
Using Kirchhoff ’s second law: Vtotal = V1 + V2 + V3
IR IR IR 5IR
+
+
=
=
6
3
6
3
Vtotal 5IR 5
=
= R
Rtotal =
I
6I
6
a Resistivity: the resistance of a sample of material
that has a cross-sectional area of 1 m2 and a
length of 1 m.
1 10 −2
l
b R = ρ = 4 × 10−8 ×
= 2 × 10−4 Ω
A
2 10 −6
a Length is 100 × shorter, cross sectional area is
100 × 100 times smaller, so resistance will be
100 times larger.
R = 100 × 1.7 × 10−8 = 1.7 × 10−6 Ω
b i 10 km
ii Rod is 104 times longer and has crosssectional area 104 times smaller.
So, new resistance is:
R = 104 × 104 × 1.7 × 10−8 = 1.7 Ω
(
π×
×
7 ρ = R A = 2.2 × 103 ×
l
2 × 10 −2
8
9
A
)
2
= 0.5 Ω m
−2
r l = 4 × 10 −4 × 3 × 10 = 1.2 × 10−7 m2
R
100
a B is ½ as long and ¼ cross-sectional area, so its
resistance will be:
½ × 4 × 100 = 2 × 100 = 200 Ω.
b ½ the radius means ¼ the cross-sectional area.
But this also means length is four times longer.
So, R = 4 × 4 × 100 = 1.6 kΩ.
10 a Ohm’s law: voltage is proportional to current for
a component at a constant temperature.
b The light bulb will not stay at a constant
temperature if the voltage across the bulb is
varied. As the voltage across a light bulb is
increased, the temperature of the light bulb will
increase (as more current flows, more energy is
transferred to thermal energy), meaning Ohm’s
law cannot be obeyed.
30
c See Figure A5.9.
V
I
Figure A5.9
d i
As current increases, the resistance of the light
bulb increases.
ii As the current increases, electrons collide
more frequently (and harder) with atoms of
the filament. Each collision transfers energy in
the form of thermal energy to the atoms,
making them vibrate more violently. This
makes it more difficult for the electrons to
pass along the filament of the light bulb. We
measure this difficulty as the increase in
resistance of the light bulb.
Since R is defined as V , an increase in V
I
does not produce a corresponding similar
increase in I , so the ratio of V to I increases,
thus increasing the resistance.
11 First bullet point: a negative thermal coefficient
thermistor (a thermally sensitive resistor whose
resistance decreases when its temperature increases).
Second bullet point: a light dependent resistor, LDR.
V
3
12 a R = =
= 7500 Ω (7.5 kΩ)
I 0 4 × 10 −3
b A = 0.2 mA; B = 0.2 mA; C = 0.6 mA;
D = 0.6 mA
3
= 5000 Ω (5 kΩ)
c Either: R = V =
I 0.6 × 10 −3
or:
1 = 1 + 1 = 1 +
1
Rtotal R1 R2 7500 7500 + 7500
3 = 1 ∴R = 5000
0 Ω
ttotal
t l
15000 5000
13 Rtotal + R1 + R2 + R3 = 3.0 + 5.2 + 0.3 = 8.5 Ω
=
14
1 = 1 + 1 = 1 + 1 = 3 = 1 ∴R = 2 Ω
total
Rtotal R1 R2 3 6 6 2
15 These are the possible combinations:
10 Ω one resistor on its own OR two parallel sets
of two resistors in series
20 Ω two of the resistors in series
30 Ω three of the resistors in series
40 Ω all four resistors in series
15 Ω one resistor in series with two resistors in
parallel
13.3 Ω one resistor in series with three resistors in
parallel
25 Ω two resistors in series with two resistors in
parallel
5Ω
two resistors in parallel
3.3 Ω three resistors in parallel
2.5 Ω all four resistors in parallel
16 a The total resistance in the circuit is:
3 6
R=1+
=3Ω
3 6
So, the current flowing through the 1 Ω resistor is
I =V = 6 = 2A
R 3
b Either: The voltage across the 6 Ω resistor is:
V 4
67 A
V = 6 − (1 × 2) = 4 V, so I = = = 0.67
R 6
or: 2 A splits in the ratio 1:2 for the resistors 6:3,
1
so I = × 2 = 0.67 A for the 6 Ω resistor.
3
c Either: The voltage across the 3 Ω resistor is:
V 4
V = 6 − (1 × 2) = 4 V, so I = = = 1 33 A
R 3
or: 2 A splits in the ratio 1:2 for the resistors 6:3,
so I = 2 × 2 = 1.33 A for the 3 Ω resistor.
3
17 See Figure A5.10.
i2
1Ω
3Ω
i1
A
6V
5Ω
3V
i3
Figure A5.10
For the left-hand loop: 6 = i2 + 5i3 (i) (Kirchhoff ’s
second law)
For the right-hand loop: 3 = 3i1 + 5i3
Now, i1 = i3 − i2 (Kirchhoff ’s first law)
So, 3 = 3i3 − 3i2 + 5i3 = 8i3 − 3i2 (ii)
3(i) + (ii) : 18 + 3 = 3i2 + 15i3 + 8i3 − 3i2
21
A
= 23i3 ⇒ i3 =
23
And, from (i), i2 = 6 − 5i3
= 6 − 5 × 21 = 138 − 105 = 33 A (= 1.43 A)
23
23
23
18 a This is a Wheatstone bridge circuit – a kind of
bridge in which all the components are resistors.
R2
b Voltage across R2 = ε
R1 R2
R4
R3 R4
d If the ammeter reads zero, there is no voltage
across the ammeter. So,
R
R
R4
R2
E
=ε
⇒ 1 = 3
R1 R2
R3 R4
R 2 R4
c Voltage across R4 = ε
19 a The potential at X is 2 × 6 2 V.
2 4
The potential at Y is 5 × 6 5V.
1 5
So, the voltage across the 3 Ω resistor is
5 − 2 = 3 V.
V 3
Therefore, the ammeter will read I = = = 1 A.
R 3
b The current flows from high potential to lower
potential, so it flows from Y to X.
20 a Kirchhoff ’s first law: the sum of currents flowing
into a junction must equal the sum of currents
flowing out of the junction.
Kirchhoff ’s second law: the sum of voltages in a
closed loop of a circuit must equal the emf
supplied to that loop.
b i For the right-hand loop: Let the current
flowing through the 3 Ω resistor be i1, the
current flowing through the 1 Ω resistor be i2
and the current flowing through the 2 Ω
resistor be i3.
Then: 2 = 3 i1 + 2 i3
For the left-hand loop: 3 = i2 + 2 i3
And i2 = i3 − i1
So, 3 = i3 − i1 + 2i3 and 2 = 2i3 + 3i1
9 = 9i3 − 3i1 and 2 = 2i3 + 3i1
11 = 11i3 ⇒ i3 = 1A
ii 2 = 3i1 + 2 ⇒ i1 = 0
ANSWERS
31
The voltmeter is in parallel with the 1 kΩ
resistor, making the total resistance of this
combination of the two resistors 500 Ω.
This 500 Ω is in series with the other 1 kΩ
resistor.
So, the voltage across the combination is
1 × 6 = 2 V.
3
ii Total resistance of the combination is:
1 × 100
= 0.99 kΩ.
Rtotal =
1 + 100
So, the voltage across the combination is:
0 99 × 6
= 2.98 V.
V=
0 99 1
b An ideal voltmeter has such a high resistance that
when it is placed in parallel with another resistor,
the total resistance of the combination is no
different from the resistance of the resistor. The
voltmeter does not change the circuit in which it
has been put.
c In the case above, a ‘perfect’ voltmeter would read
3.00 V. The difference the voltmeter in part ai has
made is only about 1%. So, as long as a voltmeter
has a resistance that is at least 100 times larger
than the resistor across which it is placed, the
value of the voltage it reads will be within 1% of
what it ought to be. A larger ratio of resistances
will reduce the difference further.
0.8
22 a E = VI t ⇒ I = E =
= 0.35 A
Vt 230 × 10 × 10 −3
(2 s.f.)
b As the filament of the lamp heats up, its resistance
increases. This will make the current flowing
through it decrease.
E
08
= 80 W
c P= =
t 10 × 10 −3
23 a P = E = 1 0 −9 = 6.7 × 107 W (67 MW)
t 15 × 10
b This is a very high level of energy use per second.
If this were continuous, the laser would overheat
and stop working/possibly cause a fire.
Exercise 5.3 – Electric cells
1 a Chemical energy is transformed into electrical
energy.
2
3
4
5
6
b No. The laws of thermodynamics state that, in any
energy transformation, some energy is wasted as
thermal energy. So, as the cell transforms chemical
energy into electrical energy, some is transformed
into thermal energy that heats up the cell.
a In a resistor, electrical energy is transformed into
thermal energy.
b Since some energy is transformed into thermal
energy when a cell is used in a circuit – much like
the effect of a resistor – it is sensible to consider
that the cell itself has a resistance.
Once the stored energy in a primary cell has been
used up, the cell is of no use. It is not possible to
‘re-energise’ the primary cell. In a secondary cell,
once the energy has been used up, it is possible to
re-energise the cell by rearranging the distribution of
charges within the cell – this is what happens when
a rechargeable cell is attached to a charger. So, the
secondary cell can be used many times.
1200 mA-hours is really a measure of how much
charge can flow through the cell before it has used up
all of its energy. So, 1200 mA-hours
= 1200 × 10−3 × 60 × 60 = 4.32 × 103 C
With a constant current of 1.6 mA, the cell will last
3
for 4 32 10−3 = 2.7 × 106 s (750 hours).
1 6 × 10
a emf means how many joules of energy are
transformed by the cell/battery for every coulomb
of charge that flows through the cell/battery.
b The total resistance is now:
Rtotal = 0.4 + 5.6 = 6.0 Ω.
So, the current is: I = ε = 1 5 = 0.25 A.
R
6
c The voltage across the internal resistance will be
V = I r = 0.25 × 0.4 = 0.1 V.
So, the terminal voltage will be 1.5 − 0.1 = 1.4 V.
d Rtotal is now 0.6 + 0.4 = 1 Ω. Terminal voltage will
be 0.6 × 1.5 V = 0.9 V.
a See Figure A5.11.
Terminal voltage
21 a i
Current
Figure A5.11
32
b The equation for this graph is:V = ε − ir , where V
is the terminal voltage, ε is the emf of the cell, I is
the current in the circuit and r is the internal
resistance of the cell. The internal resistance of the
cell, r, is the gradient of the graph.
c The value of V when i = 0 is ε. So, the terminal
voltage when no current flows is the emf of the
cell.
7 See Figure A5.12.
Exercise 5.4 – Magnetic effect of electric currents
1 See Figure A5.13.
N
S
Figure A5.13
V
2 a See Figure A5.14.
A
Figure A5.12
Equipment required: cell, leads, variable resistor,
ammeter, voltmeter.
Measure the terminal voltage using the voltmeter.
Measure the current flowing in the circuit using the
ammeter.Vary the current in the circuit by changing
the resistance of the variable resistor. Make a table of
values of current and voltage.
Plot a graph of terminal voltage (y-axis) against current
(x-axis.) The internal resistance, r, can be found from
the gradient of the graph. The emf of the cell, ε, is the
y-axis intercept.
ε = 6
8 a I=
= 1A
Rtotal 1 + 5
b V = I R = 1 × 5 = 5V
c P = I 2 R = 12 × 5 = 5 W
9 a
Resistance of
variable resistor / Ω
Current in
circuit / A
Power dissipated in
variable resistor / W
1.0
1.25
1.56
2.0
1.00
2.00
3.0
0.83
2.08
4.0
0.71
2.04
5.0
0.63
1.95
Figure A5.14
b Use the right-hand grip rule: point the thumb in
the direction of the current and allow the fingers
to naturally curl. The direction in which the fingers
curl shows the direction of the magnetic field lines.
c More current means that the magnetic field around
the wire will be stronger. To show this, the diagram
should have more field lines – and these field lines
will be closer together.
d B ∝I
e B ∝ 1 where r is the radial distance from the wire.
r
3 a See Figure A5.15.
S
N
S
N
Table A5.4
b The power dissipated in the resistor is a maximum
when the resistance is the same value as the
internal resistance of the cell.
Figure A5.15
ANSWERS
33
b Because the magnetic field lines are parallel, the
magnetic field pattern is usually described as a
uniform field.
c tesla (T)
4 a See Figure A5.16.
N
S
Figure A5.16
b A catapult field.
c The effect of this catapult field is to exert a force
on the conductor (in this case in the upwards
direction.)
d Fleming’s left-hand rule: point your first finger
ahead, then point your thumb and your second
finger perpendicular to the first finger. The first
finger represents the field, the second finger
represents the current and the thumb represents
the force on the conductor.
e The force can be made larger by using stronger
magnets, a larger current in the conductor, or a
longer length of the conductor in the catapult
field. (Also: making sure that the angle between the
conductor carrying the current and the magnetic
field of the two magnets is a right angle.)
5 a F BI l sin θ = 4 × 10−5 × 250 × 10−3 ×
5 × 10−2 × sin 90° = 5 × 10−7 N
b F BIl sin θ = 4 × 10−5 × 250 × 10−3 ×
5 × 10−2 × sin 30° = 2.5 × 10−7 N
6
power
supply
+
–
variable
resistor
retort stand
N
S
0:00
retort stand
Figure A5.17
34
electronic
balance
Figure A5.17 shows the experimental set up. Make
sure that the horizontal wire carrying the current is
above the top of the electronic balance, between the
two magnets and cannot move. Then, the current
(the independent variable) can be varied using
the variable resistor and the force (the dependent
variable) caused by the catapult field can be measured
using the electronic balance. The force from the
catapult field acts on the wire and on the two
magnets. Since the wire cannot move, the force acting
on the magnets will be registered by an increase in
the value measured by the electronic balance.
Control variables: the length of the wire that is in
between the two magnets; the angle between the wire
and the magnetic field; the magnetic field strength.
7 a F∝B
b F∝I
c F∝l
d F ∝ sin θ
8 B is defined using the equation:
F = BIl i θ⇒ B = F for θ = 90°, the force exerted
Il
on a 1 m length of conductor carrying a 1 A current
perpendicular to the magnetic field.
9 If the catapult force is sufficient to balance out the
weight of the wire then the wire will be suspended
in mid-air.
So, BIl = mg, where m = π 2l ρ
(
)
2
3
× 8900 × 9 8
π 2l ρ g π × 1 0 × 100
=
So, I =
Bl
02
= 1.4 A (2 s.f.)
10 a The two wires would move towards each other.
Each wire produces a magnetic field that
interacts with the current flowing in the other
wire to produce a catapult force. The direction
of these two catapult forces is such that each
wire experiences a force acting towards the
other wire.
b With opposite currents, the two catapult forces
exert a force on each wire away from the other
wire. The two wires move away from each
other.
c The SI unit of current, the ampere, is defined by
the force experienced by each wire due to the
currents flowing in the two wires:
•
An ampere is that constant current, which
flowing in two infinitely long, straight, parallel
conductors, of negligible cross section, 1 m apart
in a vacuum, such that the force experienced by
each wire is 2 × 10-7 N per metre length of the
wire.
11 a See Figure A5.18
the proton’s path will have a much larger radius
(by a factor of about 1800, because the mass of
the proton is this much larger, meaning that its
acceleration – and hence how much it curves –
is much less).
Exam-style questions
I
I
1
2
3
4
5
6
C
B
A
A
B
a and b See Figure A5.19.
electron
6V
Figure A5.18
B would be larger: B ∝ N
l
ii B would be larger: B ∝ I
iii No change to B.
b i
−3
12 a B = μ 0 NI = 4π × 10 −7 × 150 × 30 ×−10
l
20 × 10 2
−5
= 2.8 × 10 T
b At the edge of the solenoid, the field is half the
strength of the field in the centre.
So, B = 1.4 × 10−5 T at the edge.
l
13 a t =
v
q
bI =
t
q
q
c F BIl = B l = B
=Bqv
t
l
v
14 a F = B q v = 0.5 × 1.6 × 10−19 × 2 × 106
= 1.6 × 10−13 N
b The direction of the force is perpendicular to the
velocity, v, and perpendicular to the magnetic
field, B.
c The electron will follow a circular path.
15 a Both paths will be circular.
b The paths will differ because:
• the proton’s path will be in the opposite
direction to that of the electron (because their
charges are opposite, so the force acting on
them will be opposite), and
0V
e
I
Figure A5.19
c E = V = 6 = 30 V m−1
d 02
−19
d a = F = Ee = 30 × 1 6 × 10
= 5.3 × 1012 m s−2
−31
m m
9 1 × 10
1 at 2 we can deduce that
2
2×01
t = 2s =
= 1.9 × 10−7 s
a
5 3 × 1012
f The electron will collide with a vibrating atom
within a very short distance of its travel. This will
slow down the electron. This process will happen
many, many times as the electron moves through
the conductor, making its actual journey time
very long.
7 a See Figure A5.20.
e Using s
6V
A
V
Figure A5.20
ANSWERS
35
b The graph of V against I is a straight line that
passes through the origin, so the resistor obeys
Ohm’s law and is ohmic.
c Using the point on the graph at I = 50 mA and
5
V = 5 V, R =
= 100 Ω.
50 × 10 −3
d For there to be a 0 V voltage across the resistor,
there would have to be zero current flowing. This
would require the variable resistor to have infinite
resistance. It cannot have this, so some current will
flow. Therefore, there will be some voltage across
the resistor.
For there to be a 6 V voltage across the resistor, the
variable resistor would require a resistance of 0 Ω.
Since this is not possible, the voltage across the
resistor cannot be 6 V.
8 a Ammeter: connected ion series;Voltmeter:
connected in parallel.
b 6 V. The resistance of the voltmeter is high
compared to the resistance of the resistor. Since the
voltmeter and resistor are in series, the ratio of the
voltages across them will be the same as the ratio
of their resistances. This is so high that the voltage
across the voltmeter will be very nearly 6 V.
c The ammeter has been connected in parallel. So,
the voltage across the resistor and the voltage across
the ammeter will be the same (6 V). The resistance
of the ammeter is very nearly zero, so the current
flowing through it will be very large I = V . The
R
resistance of the resistor is larger than that of the
ammeter, so the current flowing through the
resistor will be almost zero.
9 a i See Figure A5.21.
( )
6V
Figure A5.21
36
ii See Figure A5.22.
6V
Figure A5.22
iii See Figure A5.23.
6V
Figure A5.23
b The circuit in part ai has the highest total
resistance, so the smallest current will flow. The
cell will be able to provide this current for the
longest time of the three circuits.
10 a Rtotal = 30 + 60 × 30 = 50 Ω
60 + 30
V
6
bI= =
= 0.12 A
R 50
c Power dissipated in X = I 2R = 0.122 × 30
= 0.43 W
Power dissipated in Y = I 2R = 0.082 × 30
= 0.19 W
PX 0 43
So, P = 0 19 = 2.3
Y
d E = VQ = 6 × 0.12 × 60 × 60 = 2.6 kJ
−3
11 a I = P = 6 × 10 = 1.2 mA
V
5
b Q = It = 1.2 × 10−3 × 30 × 60 = 2.16 kC
c E = V Q = 5 × 2.16 × 103 = 1.1 × 104 J
(11 kJ 2 s.f.)
d While the cell is transforming chemical energy
into electrical energy, some of the chemical
energy transformed is wasted as thermal energy in
the cell itself.
8.3 × 14
12 a B = μ 0 NI ⇒ N = Bl =
l
μ 0 I 4π × 10 −77 × 11850
= 7800 ≈ 8000 turns.
14.3 =
1.8 mm.
8000
c The conducting wire is made from a specially
designed superconductor. This superconductor is
kept at a temperature of just a few kelvin to
maintain its near-zero resistance.
13 Because the ammeter reads 0 A, there cannot be
a voltage across the 4 Ω resistor. So, the mid-way
point between the 3 Ω and the 4 Ω resistors must
have a potential of ε. (Alternatively, the voltage
across the 1 Ω resistor must be ε.) So, in the lefthand loop of the circuit, the emf supplied (6 V) is
shared in the ratio 3:1 by the 3 Ω resistor and the
1 Ω resistor. The voltage across the 1 Ω resistor is
1.5 V. Therefore, ε = 1.5 V
14 See Figure A5.24.
b Diameter of conducting wire =
i2
2Ω
2Ω
i3
A
3V
2Ω
4V
Chapter 6: Circular motion and
gravitation
Exercise 6.1 – Circular motion
1 Time period, T: the time it takes for the object to
make one complete revolution of the circle.
2 a 60 s
b 12 hours (= 4.32 × 104 s)
c One year (= 3.15 × 107 s)
3 T = 3 × 60 = 0.45 s
400
4 15 minutes is ¼ of a circle, so angle
= ¼ × 2π = π radians.
2
5 a 20 minutes is 20 revolutions of the second hand.
So, angular displacement is:
20 × 2π = 40π radians = 126 radians (3 s.f.)
1
b 20 minutes is of a complete revolution of the
3
minute hand. So, angular displacement is:
1 × 2π = 2π radians = 2.1 radians (2 s.f.)
3
3
c 20 minutes is 1 × 1 of a complete revolution. So,
3 12
angular displacement is:
6 a
i3
b
Figure A5.24
For the left-hand loop: 3 = 2i2 + 2i3 (i) (Kirchhoff ’s
second law)
For the right-hand loop: 4 = 2i1 + 2i3
Now, i1 = i3 − i2 (Kirchhoff ’s first law)
So, 4 = 2i3 −2i2 + 2i3 = 4i3 − 2i2 (ii)
7
(i) + (ii) : 7 = 2i2 + 2i3 + 4i3 − 2i2 = 6i3 i3 = A
6
3 2i3
= 1 5 − i3 = 1 5 − 7 = 1 A
And, from (i), i2 =
2
6 3
7 a
b
8 a
b
c
1 × 1 × 2π = π radians = 0.17 radians (2 s.f.)
3 12
18
60
T=
= 5.5 × 10−4 s
110 000
110 000
× 2π
angular displacement =
60
4
= 1.2 × 10 radians.
ω = 2π
T
Since f = 1 , ω = 2πff
Τ
2π
ω = 2π =
= 1.99 × 10−7 radians s−1
T
3 15 × 107
ω = 2π = 0.10 radians s−1
60
ω = 13 × 2π 7 = 2.6 × 10−6 radians s−1
3 15 × 10
ANSWERS
37
d While the cell is transforming chemical energy
into electrical energy, some of the chemical
energy transformed is wasted as thermal energy in
the cell itself.
8.3 × 14
12 a B = μ 0 NI ⇒ N = Bl =
l
μ 0 I 4π × 10 −77 × 11850
= 7800 ≈ 8000 turns.
14.3 =
1.8 mm.
8000
c The conducting wire is made from a specially
designed superconductor. This superconductor is
kept at a temperature of just a few kelvin to
maintain its near-zero resistance.
13 Because the ammeter reads 0 A, there cannot be
a voltage across the 4 Ω resistor. So, the mid-way
point between the 3 Ω and the 4 Ω resistors must
have a potential of ε. (Alternatively, the voltage
across the 1 Ω resistor must be ε.) So, in the lefthand loop of the circuit, the emf supplied (6 V) is
shared in the ratio 3:1 by the 3 Ω resistor and the
1 Ω resistor. The voltage across the 1 Ω resistor is
1.5 V. Therefore, ε = 1.5 V
14 See Figure A5.24.
b Diameter of conducting wire =
i2
2Ω
2Ω
i3
A
3V
2Ω
4V
Chapter 6: Circular motion and
gravitation
Exercise 6.1 – Circular motion
1 Time period, T: the time it takes for the object to
make one complete revolution of the circle.
2 a 60 s
b 12 hours (= 4.32 × 104 s)
c One year (= 3.15 × 107 s)
3 T = 3 × 60 = 0.45 s
400
4 15 minutes is ¼ of a circle, so angle
= ¼ × 2π = π radians.
2
5 a 20 minutes is 20 revolutions of the second hand.
So, angular displacement is:
20 × 2π = 40π radians = 126 radians (3 s.f.)
1
b 20 minutes is of a complete revolution of the
3
minute hand. So, angular displacement is:
1 × 2π = 2π radians = 2.1 radians (2 s.f.)
3
3
c 20 minutes is 1 × 1 of a complete revolution. So,
3 12
angular displacement is:
6 a
i3
b
Figure A5.24
For the left-hand loop: 3 = 2i2 + 2i3 (i) (Kirchhoff ’s
second law)
For the right-hand loop: 4 = 2i1 + 2i3
Now, i1 = i3 − i2 (Kirchhoff ’s first law)
So, 4 = 2i3 −2i2 + 2i3 = 4i3 − 2i2 (ii)
7
(i) + (ii) : 7 = 2i2 + 2i3 + 4i3 − 2i2 = 6i3 i3 = A
6
3 2i3
= 1 5 − i3 = 1 5 − 7 = 1 A
And, from (i), i2 =
2
6 3
7 a
b
8 a
b
c
1 × 1 × 2π = π radians = 0.17 radians (2 s.f.)
3 12
18
60
T=
= 5.5 × 10−4 s
110 000
110 000
× 2π
angular displacement =
60
4
= 1.2 × 10 radians.
ω = 2π
T
Since f = 1 , ω = 2πff
Τ
2π
ω = 2π =
= 1.99 × 10−7 radians s−1
T
3 15 × 107
ω = 2π = 0.10 radians s−1
60
ω = 13 × 2π 7 = 2.6 × 10−6 radians s−1
3 15 × 10
ANSWERS
37
9
1
= 5.6 × 10−4 Hz
30 × 60
b ω = 2πf = 2 × π × 5.6 × 10−4
= 3.5 × 10−3 radians s−1
c v = r ω = 75 × 3.5 × 10−3 = 0.26 m s−1
a f =
2π R = 2π × 1 5 × 1011
10 v =
= 3 × 104 m s−1
T
3 15 × 107
11 ω = 2π × 250 = 26.2 radians s−1
60
So, r = v = 330 = 12.6 m.
ω 26.2
2π =
2π
= 7.3 × 10−5 radians s−1
12 a ω =
T
24 × 60 × 60
r = 42 × 106 × 7.3 × 10−5 = 3.1 km s−1
b v = rω
13 A body moving in a circular path is changing its
direction of motion. This means that its velocity is
changing, so it must be accelerating.
14 a and b See Figure A6.1.
c See Figure A6.4.
a
r
Figure A6.4
17 a See Figure A6.5.
F
m
Figure A6.5
b See Figure A6.6.
F
a
v
v
Figure A6.1
Figure A6.6
15 a Towards the centre of the circle.
b The centripetal force.
16 a See Figure A6.2.
c See Figure A6.7
F
a
r
m
Figure A6.2
b See Figure A6.3.
a
v
Figure A6.3
38
Figure A6.7
mv 2
18 F =
, where m is the mass of the object moving
r
in the circular path, v is the linear speed at which
the mass is moving, and r is the radius of the circular
path.
19 a Gravitational force on the Moon.
b Electrical attractive force on electron.
c Magnetic force due to the proton moving in a
magnetic field.
d Friction between the car’s tyres and the road.
e Tension in the string.
M SM E
R2
2
M v
ii F = E
R
b Equating the two equations (ignoring the minus
sign):
20 a i F
G
(
2 3
4π 2 1 5 1011
M M
M v2
G S 2 E = E ⇒ M S = 4π R
=
2
R
R
T2
3 15 107
(
= 2.0 × 1030 kg
21 a F
k
Qqq
= 9 × 109 ×
2
r
(
(
×
×
)
)
)
)
T
3
2
2
ii See Figure A6.8.
= 8 23 × 1100 −77 N
mv 2 ⇒ v = 8 23 × 10 −7 × 5 29 × 10 −11
bF=
r
9 1 × 10 −31
= 6.9 × 106 m s−1
22 a F μN , where μ is the coefficient of static
friction (as long as the car is not skidding).
2
b mv = μN μm
mg ⇒ v = μrrg which is
r
independent of mass, m
c When the road is wet or icy, the value of μ will be
smaller – the friction is less – and so the value for v
must be smaller. If the car exceeds this speed, there
will not be sufficient friction force and the car will
not be able to move in a circular path: it will skid.
23 a The maximum speed on the circular part of the
road depends on the square root of the coefficient
of friction between the tyres and the road, the
radius of the circular arc and the value of g. This
value will be smaller than the motorway speed
limit because the radius of the curve is a relatively
small value.
b v = μrrg = 0 75 × 80 × 9.81 = 24.3 m s−1
24 If m is halved, then v will change by a factor
⎛
1 ⎞
1
of
= 0.707 ⎜ or
⎟.
2
⎝
2⎠
25 a i v = r ω 2π rrff = 2π × 0.6 × 2.5 = 9.4 m s−1
ii T = mr ω 2 mr π 2 f 2 = 4π 2 × 0.1 × 0.6 × 2.52
= 14.8 N
b i The ball on the end of the string has mass. The
tension in the string must have a component
that balances the weight of the mass. This
component must be in the vertical plane, so
the string cannot be horizontal.
w = mg
Figure A6.8
c Larger
d The horizontal component.
e The vertical component is balancing the weight
of the mass.
26 a See Figure A6.9.
N
θ
w = mg
θ
Figure A6.9
mg and
b N cos θ= W m
2
sin θ mv 2
N sin q mv ⇒ mgg
=
r
cos θ
r
Therefore, θ = tan −1 ⎛⎜ v ⎞⎟
⎝ gr ⎠
2
c We would expect an Olympic cyclist to cycle faster
than an amateur. So, if v is larger, the equation for θ
shows that θ is larger: the banking is steeper.
27 a See Figure A6.10.
B
centre
of circle
T
A
w = mg
Figure A6.10
ANSWERS
39
b The tension must balance the weight, so the
centripetal force must be the sum of these two
vectors:
2
mv 2 = T m
mg T = mv + mg
r
r
c Now, the centripetal force must equal the sum of
the tension and the weight forces:
d
e
f
g
h
2
mv 2 = T m
mg T = mv − mg
r
r
The tension in the string is maximum at the
bottom of the vertical circle and becomes less as
the mass moves towards the top of the circle.
Work done by centripetal force = 0
Since work done = force × distance moved in direction
of force, and the force is always perpendicular to
the distance moved, the work done must be zero
(no movement occurs in the same direction as the
force).
GPE gained = 2mgr
The total energy of the mass is given by:
Etotal = KE + GPE.
At the top of the circle, the mass has gained GPE,
so to keep Etotal a constant, KE must be reduced.
Therefore, the speed of the mass must be less at
the top of the circle.
1 A gravitational field is a region in space in which a
mass experiences a gravitational force.
2 a G is the universal gravitational constant;
G = 6.67 × 10−11 N m2 kg−2
M and m are the masses of the two bodies,
measured in kg.
R is the separation of the two bodies,
measured in m.
b The minus sign is there to show that the force
between the two masses is attractive.
G
30
24
M Sun M Earth
= −6 67 × 10 −11 × 2 × 10 × 6 × 10
2
2
R
×
= −3.6 × 1022 N
b F
2
G m 2 = −6 67 × 10 −11 ×
R
−35
= −4.8 × 10
40
5
G
N
(
(
)
×
(
×
)
2
)
2
(
−1
)
= 9.8 N kg
b The Earth is not a perfect sphere; it is a sphere
that is ‘squashed’ from top-to-bottom. So, its
radius is not the same everywhere. We usually
quote g as a globally averaged value to eliminate
the difficulty caused by this changing radius.
c g is largest at the poles, because the radius of the
Earth is smallest there – and since the
gravitational field strength is an inverse square law,
a smaller radius gives a bigger field strength.
6
7
2
N = J = kg m = m
kg m kg m kg s2 s2
Physicists use the word weight to mean the
gravitational force acting on a mass.
M
−G 2Earth
2
g Earth
REarth
M
RM
oon
=
= Earth
M Moon M Moon × RE2arth
g Moon
−G 2
RMoon
=
6 × 1024 ×
(
7 3 × 1022 ×
= 5.1
×
(
×
)
2
)
2
9 G M Mars = 0 38G M Earth
(RMars )2
(REarth )2
⇒ M Mars = 0 38
= 6.5 × 1023 kg.
3 a
F
4
8
Exercise 6.2 – Newton’s law of gravitation
m1m2
= −6 67 × 10 −11 × 60 ×2 70
2
R
1
= −2.8 × 10−7 N
a Gravitational field strength, g: the amount of
gravitational force experienced by a unit mass in
the field.
F
b g=
m
M
6 × 1024
a g = −G 2 = −6 67 × 10 −11 ×
2
R
×
c F
(
9 81 × 3.4 × 106
6 67 × 10 −11
)
2
30
10 a gSun = G M2 = 6 67 10 −11 × 2 10
R
1 5 × 1011
= 5.93 × 10−3 N kg−1
(
)
2
22
b gMoon = G M2 = 6 67 10 −11 × 7 3 × 10 2
R
3 8 × 108
−5
= 3.37 × 10
−1
N kg
(
)
c The gravitational field strength of the Sun where
the Earth is in its orbit is greater than the
gravitational field strength of the Moon where
the Earth is. However, if one considers the
relatively small distance that the Earth is away
from the Moon, the size of the force acting on
the face of the Earth closest to the Moon is
significantly greater than the size of the force
acting on the face of the Earth farthest from the
Moon. It is this difference in force from the
gravitational field of the Moon that causes the
tides. The extra distance that the opposite face of
the Earth adds to the distance between the Earth
and the Sun means that there is little difference in
the gravitational force from the Sun on the
opposite faces of the Earth.
4π R 3 ρ
M
11 a i g = −G 2 and M =
so
3
R
4πR ρ
4πρR
= −G
2
3
3R
ii See Figure A6.11.
g = −G
3
c v=
6.67 × 10 −11 × 2 × 1030
1.5 × 1011
G M Sun
=
R
= 3 × 104 m s−1 (1 s.f.)
13 a When g = 0, the two gravitational field strengths
from the two bodies have to be equal:
M
M Moon
G Earth
=G
, where x is the distance
2
x
(R x )2
from the Earth to point X and R is the separation of
the Earth and the Moon.
So,
M Earth
M Moon
2
=
⇒ M Earth ( R − x )
2
2
x
(R x )
M Moon x 2
Therefore,
( M Earth
M Moon ) x 2
2M EEarthth R
Rx + M EEarthth R 2 = 0
Solving:
x=
Radial distance from centre of Earth
0
2R
3R
R
2M
M Earth R ±
(
R ) − 4 (M
M
) M EarthR 2
(
×
×
2
M
2 ( M Earth
M Moonn )
× ×
(
)
2
)
−
2 × 6 × 1024 × 3 8 × 108 ± 4 6 × 1024 − 7 3 × 1022 ×
=
2
6 × 1024 ×
(
×
−
(
×
)
)
2
×
= 3.5 × 10 m (~92% of the distance between the
Earth and the Moon).
b See Figure A6.12.
8
R
Earth
Moon
Figure A6.11
iii g = G
4πρR
3g
⇒G =
3
4πR ρ
3 × 9.81
4π × 6.4 × 106 × 5.5 × 103
= 6.65 × 10−11 N m2 kg−2
So, percentage difference is: 6 67 6.65 × 100%
6 67
= 0.3%
M Earth v 2
M M
12 a
= G Sun 2 Earth
R
R
bG=
b Rearranging the above equation:
M Earth v 2
M M
G M Sun
= G Sun 2 Earth ⇒ v =
R
R
R
g
g
Figure A6.12
( ) = 4π M
M Earth 2 R
M Earth v
T
14 a
=
R
R
2
2
2
Earth
2
R2
RT
4π M Earth R
M S n M Earth
=
= G Su
2
T
R2
2
Therefore, R 3 =
GM Sun 2
T
4π 2
ANSWERS
41
b One year = 60 × 60 × 24 × 365 = 3.15 × 107 s
Then,
M Sun
2
3
4π 2
= 4 × R2 =
×
G T
6 67 × 10 −11
(
(
3
b The gradient of this graph is R 2 = GM2
T
4π
= 3.221 × 1015 m3 s−2
)
)
3
×
4π 2
× 3.221 × 1015
G
4π 2
=
× 3.221 × 1015 = 1.91 × 1027 kg
−11
6 67 10
8 a Since the vertical component of the tension must
balance the weight of the skater
T sin 30° = 60g ⇒ T = 60 × 9 81 = 1177 N
sin 30
(1200 N to 2 s.f.)
2
×
So, M =
= 2.0 × 1030 kg
c For Jupiter orbiting the same Sun,
R Jupiter 3 =
GM Sun
T Jupiter 2
4π 2
So,
R Jupiter =
=
3
GM Sun
T Jupiter 2
4π 2
−11
3
6 67 × 10 × 2 × 10
4π 2
30
(
× 11.86
86 × 33.15
1 × 10
= 7.78 × 1011 m
R
15 M Earth = 4π × Moon2
G
TMoon
(
)
×
4π 2
=
×
−11
6 67 × 10
27.3 × 8.64 ×
(
3
)
2
= 5.84 × 10 kg
24
D
A
C
B
C
B
a See Figure A6.13.
R3 (× 1025)/m3
R3 against T 2
800
700
600
500
400
300
200
100
0
y = 3.221x
0
Figure A6.13
42
50
⇒ ω = T cos 30° = 1177 × 0.866
mr
60 × 2.2
Chapter 7: Atomic, nuclear and
particle physics
Exercise 7.1 – Discrete energy and radioactivity
Exam-style questions
1
2
3
4
5
6
7
b T cos30° = mr ω 2
= 2.8 radians s−1
3
2
)
7 2
100 150 200
T 2/(× 1010) s2
250
1 a Excited gas at a low pressure emits electromagnetic
radiation as electrons of the atoms of gas fall from
higher energy levels to lower energy levels, each
time emitting a photon. The term spectrum is used
to describe the different wavelengths present in
these emissions. Each wavelength corresponds to a
different change in energy levels.
b Place a sample of gas in a glass tube and close the
tube. Excite the gas by placing a strong electric
field across it. The gas will glow. A student may
then observe the emission spectrum by looking at
the gas through a diffraction grating.
2 Emission spectra consist of a set of discrete lines. Each
line in the emission spectrum represents a wavelength.
Each different wavelength is associated with a
different amount of energy. These different amounts
of energy are due to different energy level transitions
of electrons in atoms. Because the emission lines are
discrete, the energy level transitions of the electrons
must correspond to discrete amounts. This can only
be true if the energy levels themselves exist in discrete
energy values.
b One year = 60 × 60 × 24 × 365 = 3.15 × 107 s
Then,
M Sun
2
3
4π 2
= 4 × R2 =
×
G T
6 67 × 10 −11
(
(
3
b The gradient of this graph is R 2 = GM2
T
4π
= 3.221 × 1015 m3 s−2
)
)
3
×
4π 2
× 3.221 × 1015
G
4π 2
=
× 3.221 × 1015 = 1.91 × 1027 kg
−11
6 67 10
8 a Since the vertical component of the tension must
balance the weight of the skater
T sin 30° = 60g ⇒ T = 60 × 9 81 = 1177 N
sin 30
(1200 N to 2 s.f.)
2
×
So, M =
= 2.0 × 1030 kg
c For Jupiter orbiting the same Sun,
R Jupiter 3 =
GM Sun
T Jupiter 2
4π 2
So,
R Jupiter =
=
3
GM Sun
T Jupiter 2
4π 2
−11
3
6 67 × 10 × 2 × 10
4π 2
30
(
× 11.86
86 × 33.15
1 × 10
= 7.78 × 1011 m
R
15 M Earth = 4π × Moon2
G
TMoon
(
)
×
4π 2
=
×
−11
6 67 × 10
27.3 × 8.64 ×
(
3
)
2
= 5.84 × 10 kg
24
D
A
C
B
C
B
a See Figure A6.13.
R3 (× 1025)/m3
R3 against T 2
800
700
600
500
400
300
200
100
0
y = 3.221x
0
Figure A6.13
42
50
⇒ ω = T cos 30° = 1177 × 0.866
mr
60 × 2.2
Chapter 7: Atomic, nuclear and
particle physics
Exercise 7.1 – Discrete energy and radioactivity
Exam-style questions
1
2
3
4
5
6
7
b T cos30° = mr ω 2
= 2.8 radians s−1
3
2
)
7 2
100 150 200
T 2/(× 1010) s2
250
1 a Excited gas at a low pressure emits electromagnetic
radiation as electrons of the atoms of gas fall from
higher energy levels to lower energy levels, each
time emitting a photon. The term spectrum is used
to describe the different wavelengths present in
these emissions. Each wavelength corresponds to a
different change in energy levels.
b Place a sample of gas in a glass tube and close the
tube. Excite the gas by placing a strong electric
field across it. The gas will glow. A student may
then observe the emission spectrum by looking at
the gas through a diffraction grating.
2 Emission spectra consist of a set of discrete lines. Each
line in the emission spectrum represents a wavelength.
Each different wavelength is associated with a
different amount of energy. These different amounts
of energy are due to different energy level transitions
of electrons in atoms. Because the emission lines are
discrete, the energy level transitions of the electrons
must correspond to discrete amounts. This can only
be true if the energy levels themselves exist in discrete
energy values.
3 a The planetary model of the atom has a nucleus at
the centre of the atom and electrons that orbit
around the nucleus, like planets orbit around a star.
The different radii of the orbits of the electrons help
us to visualise the different electron energy levels.
b i The electromagnetic force keeps the electrons
in orbit around the nucleus.
ii Towards the nucleus.
iii Yes. The electron in its orbit is changing
direction all the time, so it must be accelerating.
c A constantly accelerating electron should be
constantly emitting radiation, which would cause its
energy to decrease and it would spiral in towards the
nucleus. Experiments show that this does not occur,
so the planetary model must be incorrect.
4 a Since the electron is attracted to the positive
charge in the nucleus, it takes energy to pull the
electron away. By convention, this is given a
negative value.
b The level labelled n = 1 is the ground state. This is
the lowest energy state of the atom.
c In level n = 1
d The electron would jump up to the next energy
level, n = 2.
e This process is called excitation.
5 a In the levels n = 2 or higher.
b Excited: the electrons are in an energy level higher
than they would normally be.
c An excited electron is most likely to fall down to a
lower energy level, emitting the difference in
energy between the two levels as a photon.
d The process of falling to a lower energy level is
called spontaneous emission.
e The energy of the atom has decreased. It has
become more negative.
−13.6 eV
6 a En =
then for n = 2,
n2
−13.6 eV
E2 =
= −3.4 eV
22
−13.6 eV
then for n = 3,
b En =
n2
−13.6 eV
E3 =
= −1.5 eV
32
−13.6 eV
then for n = 4,
c En =
n2
−13.6 eV
E4
= −0.85 eV
42
7 a and b See Figure A7.1.
n=5
n=4
–0.5 eV
–0.9 eV
n=3
–1.5 eV
n=2
–3.4 eV
n=1
–13.6 eV
Figure A7.1
c i The highest frequency photon is produced by
the transition from n = 4 to n = 1.
ii The lowest frequency photon is produced by
the transition from n = 4 to n = 3.
8 a The four visible lines are due to transitions from
n = 6, 5, 4 and 3 to n = 2.
hc 6 64 × 10 −34 × 3 × 108 = 3.04 × 1100 −19 J
b Ε= =
λ
656.3 × 10 −9
−19
3 01 10 = 1.88 eV ≈ 1.9 eV
=
1 6 × 10 −19
hc 6 64 × 10 −34 × 3 × 108 =
435 nm.
c γ = =
E
2 55 × 1.6 × 10 −19
The turquoise coloured line.
−34
8
9 a E = hc = 6 64 × 10 × −39 × 10
λ
630 × 10
−19
= 3 16 × 10 −19 J = 3 16 × 10−19
1 6 × 10
= 1.98 eV
−34
8
b Ε = hc = 6 64 × 10 × −39 × 10 = 3.07
0 × 10
10 −19 J
λ
532 × 10
−19
3
74
10
=
= 2.34 eV
1 6 × 10 −19
−34
8
c E = hc = 6 64 × 10 × −39 × 10
λ
430 × 10
−19
= 4 63 × 10 −19 J = 4 63 × 10−19
1 6 × 10
= 2.89 eV
10 A photon of wavelength 500 nm has energy,
−34
8
E = hc = 6 64 × 10 × −39 × 10 = 3 98 × 10 −19 J
λ
500 × 10
So, the number of photons s−1 =
= 1.5 × 1020 s−1
11 n = 2
12 Ultraviolet.
60
3 98 10 −19
ANSWERS
43
13 The emission spectrum from a filament light bulb is
continuous: it contains a wide range of wavelengths
from the deep red colours to the violet end of the
spectrum. The emission spectrum from an excited
gas consists of a limited number of discrete lines at
particular wavelengths.
14 a Taking an average wavelength of 589.3 nm,
−34
8
E = hc = 6 63 × 10 × 3−9× 10 = 3 38 × 10 −19 J
λ
589.3 × 10
=
3 38 10 −19
= 2.11 eV
1 6 × 10 −19
b See Figure A7.3.
long
wavelengths
short
wavelengths
Figure A7.3
c You would be able to see all the emission lines
that have energies between 1.8 eV and 3.1 eV
(Figure A7.4).
So, the difference in energy between these two levels
must be 2.11 eV.
b The energy level n = 2 is actually split into two
energy levels, which are about 3.4 × 10−22 J apart
(or about 2 meV apart).
15 a Red
−34
b i E = hc = 6 63 × 10 × 3−9× 10 = 3 15 × 10 −19 J
λ
632.8 × 10
8
−19
ii E = 3 15 × 10−19 = 1.97 eV
1 6 × 10
16 A photon of wavelength 500 nm has energy,
−34
8
E = hc = 6 63 × 10 × −39 × 10 = 4 43 × 10 −19 J
λ
450 × 10
−3
So, the number of photons s−1 = 1 0 × 10 −19
4 43 10
15 −1
= 2.3 × 10 s
8
17 a λ = c = 3 × 10 17 = 9.4 × 10−10 m
f 3 2 × 10
b E = h f = 6.63 × 10−34 × 3.2 × 1017
= 2.1 × 10−18 J
c Since this energy must have been KE of the
electron,
−18
2E = 2 × 2.1 × 10
m
9 1 × 10 −31
18 a See Figure A7.2.
v=
= 2.1 × 106 m s−1
These are UV
Figure A7.4
19 a A nucleon is any of the two kinds of particle in
the nucleus (a neutron or a proton).
b An isotope is a form of an atom that has the same
number of protons in the nucleus but a different
number of neutrons.
c A nuclide is a description of a particular kind of
nucleus: it expresses the nucleon number and the
proton number (from which it is possible to
calculate the neutron number).
20
Isotope
Number of protons
in the nucleus
Number of neutrons
in the nucleus
3
2
2
1
He
12
6
C
6
6
13
6
C
6
7
Fe
26
30
Na
11
10
56
26
21
11
–0.1 eV
–0.3 eV
–1.4 eV
n=2
–3.1 eV
–5.6 eV
21 a Beta-minus decay occurs because there are too
many neutrons in the nucleus.
b Beta-plus decay occurs because there are not
enough neutrons in the nucleus.
13
12
−
22 a 6 C
7 N + β + νe
Th →
b
234
90
c
21
11
Na →
d
11
6
C
Figure A7.2
44
These are
visible
Table A7.1
n=5
n=4
n=3
n=1
These are IR
234
91
21
10
11
5
P + β − + νe
Pa
N + β+ + ν e
Ne
B + 01β + + ν e
23 a
A
Z
X→
b ZA X
c
A
Z
X→
A 4
Z −2
Y + 42 α
Y + −01β − + ν e
A
Z +1
Y + 01β + + ν e
A
Z −1
d ZA X → ZA X + γ
24 a An alpha particle is a helium nucleus that has
come from an unstable nucleus.
b A beta-plus particle is a fast moving anti-electron
(or positron) that has come from an unstable
nucleus.
25 a The both have the same mass.
b They have opposite charges (and they have
opposite electron lepton numbers).
26 a Ionise means that they can remove electrons
from an atom, leaving the atom as a positively
charged ion.
b They have a double positive charge (that will
attract electrons); They are quite large (by subatomic standards); They move relatively slowly –
and so spend a relatively large amount of time in
the proximity of atoms, which enables them to
pull electrons from atoms.
c i β− particles have a negative charge, so do not
attract electrons.
ii They move very quickly, so do not spend
much time near to electrons in atoms.
iii They are small (compared with alpha
particles) and so the likelihood of them
interacting directly with electrons is low.
27 Number of ion pairs produced = 5 0 × 10 = 170 000
30
28 They are approximately inversely proportional: If the
ionising ability is large, they will lose their energy
quickly and so not pass very far through a material.
29
6
Emission type
What stops the emission?
α particle
A piece of paper – or 2–3 cm of air
−
β particle
A few mm of aluminium plate
γ ray
Several cm of lead plate (not really
stopped completely by anything)
Table A7.2
−5
30 a i 8 10 = 8 × 105 atoms thick.
1 10 −10
6
ii 5 10 = 1.67 × 105 ionising events.
30
b To pass through the paper, the alpha particle
would have to make at least 8 × 105 ionising
events. With an initial kinetic energy of only
5.0 MeV, the alpha particle can make only
1.67 × 105 ionising events. So, the alpha particle
will have run out of energy before it passes
through the piece of paper.
31 a It will attract two electrons from nearby and
become a helium nucleus.
b Rutherford and Royds were able to trap the
emissions from an alpha emitter in a glass jar. After
leaving the jar for some time, for the alpha
particles to attract electrons from the nearby air
and become helium atoms, they were able to test
the gas in the jar. They found the gas to be
helium and so were able to show that alpha
particles were helium nuclei.
32 Gamma ray emissions occur at discrete energies.
This suggests that the nucleus has energy levels of
discrete values and, each time the nucleus shifts from
a higher energy level to a lower one, the emitted
gamma ray emitted carries energy equal to the
difference between the two energy levels. Though
the concept is the same as that for electrons in
atoms, the energies involved are considerably higher.
33 a Using the constant ratio rule for every 5 mm:
400 = 1.87;
87; 214 = 1.86; 115 = 1.90;
90; 61 = 1 85
214
115
61
33
Since these values are all approximately the same,
the graph is exponential.
b i Reading from the graph, half the intensity
occurs for a thickness of 5.5 ± 0 2 mm.
ii 1/e = 0.37. 0.37 × 400 = 148. The thickness
required for this is about 8.0 ± 0 2 mm.
34 a Half-life, t1/2 : the average time it takes for half of
the nuclei present in a sample to decay. Or, the
time it takes for the activity of a sample of
radioactivity to halve.
b The activity of a sample is the number of decay
events occurring in one second.
ANSWERS
45
Plot the graph of average count for 10 seconds
against time for about half an hour.
Use the graph to find the half-life in the usual way.
40 a The time for which the radioactive isotope is
active is long enough for the medical personnel
to complete the procedure, but it is not so long
that the radioactive isotope remains in the
patient’s body continuing to decay.
No. of undecayed nuclei
35 a 1
6
1 1 1
b 1+ + =
6 6 6 2
c N sixes.
6
36 a N λ
b and c See Figure A7.5
b Since A
N
N
2
N
4
0
0
t 1/2
2t 1/2
Time
Figure A7.5
d The gradient of the graph is
37 a
b
c
38 a
dN
= activity, A
dt
80 g
40 g
10 g
Background radiation: emissions from radioactive
atoms that occur naturally.
b The Sun, cosmic rays, rocks (especially granite).
c Corrected count means that the count due to
background radiation has been subtracted from
any count made on a sample, so the corrected
count is due to the sample only.
39 Equipment required: GM tube and counter,
radioactive sample, two stopwatches, tongs to hold
the sample.
Method: 1: Measure the background radiation for
a period of 10 seconds several times and find the
average count due to background radiation over a 10
second period. Measure the time with the stopwatch.
2: Place the sample against the GM tube. 3: Start one
of the stopwatches and use the other stopwatch and
GM tube and counter to measure the count over
a 10 second period three times. 4: Find the average
count over the 10 second period. 5: Subtract the
background count to get a corrected count. 6:
Repeat step 2 every minute and record all results
in a table.
46
λN, N = A =
λ
(
5 × 1015
×
×
)
= 3.4 × 1020 atoms
41 a The radioactive isotope 146 C decays by β−
emission. In living tissue, the proportion of 146 C to
12
6 C remains constant but, in dead tissue, the
proportion decreases as the radioactive isotope
decays. So, if a known quantity (say, 5 g) of
fossilised organic material is measured for its
count rate and then compared to the corrected
count rate for a sample of 5 g of the same kind of
living organic material, an approximate age of the
fossilised material can be found, using half-life.
144 = 12
144
ii 18 = 3 so the material will be three
2
half-lives old = 3 × 5700 years = 17 100 years.
Gravitational; Weak nuclear; Electromagnetic;
Strong nuclear
Gravity and the electromagnetic force have an
infinite range.
The force with the shortest range is the weak
nuclear force (about 10−18 m).
Weak nuclear
Gravitational
Electromagnetic
Strong nuclear
Three of the four fundamental forces act on the
nucleons in the nucleus. Of these, the gravitational
force is too weak to have any appreciable effect on
holding the nucleons together.The electromagnetic
force causes the positively charged protons to be
repelled from each other – suggesting that they
should fly apart. However, the strong nuclear force
is about 100 times stronger than the
electromagnetic force and is attractive between all
nucleons. This overcomes the electromagnetic force
and holds the nucleons together.
b i
42 a
b
c
43 a
b
c
d
44 a
b The strong nuclear force acts most strongly on
nucleons that are closest together. So, in a nucleus
with a large number of nucleons, some of the
nucleons do not ‘feel’ the strong force from many
of the other nucleons; they only ‘feel’ the strong
force from the nucleons immediately next to
them. This allows a tightly bound alpha particle to
break free of the nucleus if the nucleus is heavy
enough.
45 a Alpha particles approaching the gold nuclei were
repelled by the protons in the nucleus. Since the
alpha particles did not have enough energy to get
close enough for the strong nuclear force to
overcome the electromagnetic force of repulsion,
the alpha particles were deflected.
b If the alpha particles had had more energy, so that
they could approach the nucleus to within about
3 fm, the strong nuclear force would have been
able to overcome the electromagnetic force and
cause the alpha particle to be absorbed by the
gold nucleus. Had this been the case, Rutherford
would not have seen the very large-angle
deflections that occasionally occurred.
Exercise 7.2 – Nuclear reactions
1 1 of the mass of a 126 C atom.
12
2
Name of particle
Mass / u
Mass / kg
proton
1.007 276
1.672621 × 10−27
neutron
1.008 665
1.674928 × 10−27
electron
0.00055
9.11 × 10−31
Table A7.3
3 a Mass of 1 u =
1 × 12.0 10 −3
12 6.023 × 1023
= 1.660 539 × 10−27 kg
E = m c2 = 1.66 × 10−27 × 9 × 1016 = 1.49 × 10−10 J
−10
= 1 49 10−19 = 931 MeV
1 6 × 10
b The number of significant figures quoted in most
data books for quantities such as Avogadro’s
number, the masses of protons and neutrons and
the speed of light are not sufficient when
calculating the value of u in MeV accurately. If
4
5
6
each of the values are quoted to ten or more
significant figures, then the value for the energy
equivalence of u is correctly given as 931.5 MeV.
a 4.002 604 u = 4.002 604 × 1.66 × 10−27
= 6.64 × 10−27 kg
b Mass of particles = 2 (1.007276 + 1.008665)
= 4.031 882 u
c The mass of the particles is greater than the mass
of the alpha particle
d Mass defect = 4.031 882 − 4.002 604
= 0.029 278 u
e Energy equivalence = 0.029 278 × 931.5 MeV
= 27.3 MeV
Mass defect
= (6 × 1.007 276 + 9 × 1.008 665 − 15.010 60) u
= 0.111 041 u
= 0.111 041 × 931.5
= 103.435 MeV c−2
4200 J =
4200 = 2.6 × 1016 MeV
1 6 × 10 −13
16
= 2 6 × 10 = 2.79 × 1013 u
931.5
= 2.79 × 1013 × 1.66 × 10−27 = 4.63 × 10−14 kg
So, percentage increase in mass = 20 × 4.63 × 10−12
= 9.3 × 10−11 %
7 Binding energy: the amount of energy required to
break apart all of the nucleons in a nucleus.
13
8 6 C has a greater binding energy than 126 C because it
has more nucleons to hold together. So, it requires
more energy to break apart.
9 a E = 9.11 × 10−31 × 9 × 1016 = 8.2 × 10−14 J
8 2 × 10 −14
= 0.51 MeV
1 6 × 10 −13
2
b KE = 1 m v 2 = 1 × 9 11 10 31 5 × 106
2
2
= 1.14 × 10−17J
c The energy associated with the mass of the
electron is much greater than the KE of the
electron.
10 Mass deficit
= (11 × 1.007276 + 13 × 1.008665 − 23.99096)
u = 0.2017 u
So, binding energy
= 0.2017 × 931.5 MeV =
187.9 MeV
Therefore the binding energy per nucleon = 187.9
24
= 7.8 MeV nucleon−1
=
(
)
ANSWERS
47
11 Mass deficit
= (26 × 1.007 276 + 30 × 1.008 665 − 55.934 939) u
= 0.514187 u
So, binding energy = 0.514 187 × 931.5 MeV
= 478.965 MeV
Therefore, the binding energy per nucleon
478.965
=
= 8.55 MeV nucleon−1
56
Exercise 7.3 – The structure of matter
1 a A positively charged mass of atom in which
negatively charged electrons were embedded.
b Geiger and Marsden fired alpha particles at gold
nuclei and observed where they were scattered. As
a result of their observations, they were able to
construct a new model for the atom.
2
12 a, c and d See Figure A7.6.
Experimental observation
Conclusion
The vast majority of alpha
Most of the space taken up
particles passed through the by the atom is empty – i.e it
gold foil undeflected.
does not contain anything.
9
B.E. per
nucleon
Some alpha particles were
deflected through such
large angles that they
bounced backwards.
fission
region
fusion
region
There was a small, positively
charged and dense nucleus
at the centre of the atom.
Table A7.4
0
0
56
238
Nucleon number
MeV / nucleon
Figure A7.6
b The most stable nuclei are those that are held
together with the most binding energy per
nucleon. So those nuclei at the peak of the curve
are the most stable.
13 a The joining together of two smaller mass particles
to form a larger mass particle.
b The breaking apart of a larger mass particle to
produce two smaller mass particles.
14 a Number of moles of 235
in 1 kg = 1000 = 4.26
92 U
235
So, number of atoms = 4.26 × 6.023 × 1023
= 2.57 × 1024 atoms
b Specific energy of 235
92 U
= 2.57 × 1024 × 173 × 1.6 × 10−13
= 7.11 × 1013 J
15 a The moderator does two things: It slows down
the fast neutrons, so that they have a better chance
of colliding with uranium nuclei; It heats up,
providing the thermal energy that the heat
exchanger can pass on.
b The control rods are used to absorb excess
neutrons, so that a sustainable chain reaction is
produced.
3 a A fundamental particle is one that has no internal
structure; i.e. it is not composed of other particles.
b Quarks, leptons and bosons.
4 The standard model has not been able to reconcile
quantum physics with gravity.
5
Generation
Name of
quark
First
up
u
+2/3
1/3
down
d
−1/3
1/3
Second
Third
Charge
Baryon
number
charm
c
+2/3
1/3
strange
s
−1/3
1/3
top
t
+2/3
1/3
bottom
b
−1/3
1/3
Table A7.5
6
Generation
Name of
lepton
Symbol
First
electron
e
−1
1
ve
0
1
Second
electron
neutrino
muon
μ
−1
1
vμ
0
1
Third
muon
neutrino
tau
τ
−1
1
tau
neutrino
vτ
0
1
Table A7.6
48
Symbol
Charge
Lepton
number
7
a A hadron is a particle that is made of quarks.
b A baryon is a hadron that is made of three quarks.
A meson is a hadron that is made of one quark
and one anti-quark.
c i uud
ii udd
iii uud
iv ud
d Quarks are confined to existing in hadrons,
combined with other quarks/anti-quarks, so that
it is not possible to observe a quark on its own.
8 a Charge: 1 + −1 → 0 √
Baryon number: 1 + 0 → 1 √
Electron lepton number: 0 + 1 → 0 ✗
So, this interaction cannot occur.
b Charge: 1 + 1 → 1 + 1 + 0 √
Baryon number: 1 + 1 → 1 + 1 + 1 ✗
Electron lepton number: 0 + 0 → 0 + 0 + 0 √
So, this interaction cannot occur.
c Charge: 1 + 1 → 1 + 1 + −1 + 0 ✗
Baryon number: 1 + 1 → 1 + 1 + 0 + 0 √
Electron lepton number: 0 + 0 → 0 + 0 + 1 + −1 √
So, this interaction cannot occur.
9 a Charge: 1 + 1 → 1 + 1 + 0 √
Baryon number: 1 + 1 → 1 + 1 + 0 √
Electron lepton number: 0 + 0 → 0 + 0 + 0 √
So, this interaction can occur.
b Charge: −1 + 1 → 0 + 0 √
Baryon number: 0 + 1 → 1 + 0 √
Electron lepton number: 0 + 0 → 0 + 0 √
So, this interaction can occur.
c Charge: 1 + 0 → 1 + 0 √
Baryon number: 1 + 1 → 0 + 0 ✗
Electron lepton number: 0 + 0 → 0 + 0 √
So, this interaction cannot occur
10 a A strange quark has −1 strangeness.
b No. The conservation of strangeness does not
apply to processes involving the weak interaction.
c 1
11 Charge: 0 → 1 + −1 √
Baryon number: 0 → 1 + −1 √
Electron lepton number: 0 → 0 + 0 √
So, this interaction is viable.
12
Force
Range
Relative
strength
Acts on
Exchange
particle
Strong nuclear
3 × 10−15 m
100
quarks and
baryons
gluon and
meson
Electromagnetic
infinite
1
charged
particles
photon
Weak nuclear
10−18 m
10−11
quarks and
leptons
W, Z
bosons
Gravitational
infinite
10−36
particles
with mass
graviton?
Table A7.7
13 a The W and the Z bosons
b The Higgs boson does not mediate a force; it
mediates the concept of mass.
14 a This shows an electron being repelled by another
electron by the exchange of a photon (the boson
that mediates the electromagnetic force).
b The electromagnetic force.
15 See Figure A7.7.
νe–
p
W
–
e–
n
Figure A7.7
16 a β+ decay.
b W+ boson.
17 a There are two possible answers, shown in Figures
A7.8 and A7.9.
n
νe–
W+
p
e
Figure A7.8
n
W
p
–
νe–
e
Figure A7.9
ANSWERS
49
b W boson. In Figure A7.8 it is the W+ boson and
in Figure A7.9 it is the W− boson.
18 a There is no exchange of charge; i.e. the boson
responsible has no charge.
b See Figure A7.10.
e–
W+
νe–
p
n
Figure A7.10
c There is a transfer of charge between the electron
anti-neutrino and the neutron, via the W+ boson;
i.e. in this case the W+ boson is charged. (It is also
worth pointing out that there is a transfer of mass,
too.)
Exam-style questions
1
2
3
4
C
C
D
a B
b A
5 a The dark lines are caused by absorption of those
wavelengths that correspond to the energy level
transitions of the electrons in the atoms of the
hydrogen gas. The excited electrons then re-emit
the photons but in all directions. Since the
intensity of the light in the direction being viewed
has been reduced, the observer sees dark lines.
hc 6 64 × 10 −34 × 3 × 108
b E= =
λ
434 × 10 −9
4 59 × 10 −19
= 4 59 × 10 −19 J =
1 6 × 10 −19
= 2.87 eV
c The transition causing this must have an energy
level difference of 2.87 eV. So, this must be
from n = 2 to n = 5
d The absorption spectra from stars show discrete
energy level transitions within the atoms in the star.
Scientists have catalogued the spectra of elements
and compounds in laboratory experiments. The
star spectra can be matched up to the catalogued
spectra, giving us information about which
elements and compounds are present in the star.
50
6 a Two protons; one neutron
b Mass deficit
= (2 × 1.007276 + 1.008665 − 3.01603) u
= 0.0072 u
So, binding energy = 0.0072 × 931.5 MeV
= 6.72 MeV
Therefore the binding energy per nucleon
6 72
=
= 2.24 MeV nucleon−1
3
c 21 D + 21 D → 32 He + 01 n
d The binding energy of the He-3 is more than the
sum of the binding energies of the two D nuclei.
The difference in these is the energy available from
the interaction.
233
229
4
7 a 92 U
90Th + 2 He
b KE available = (4.00787 − 4.00260) = 0.00527 u
= 0.00527 × 931.5 MeV = 4.91 MeV
c The conservation of momentum states that the
total momentum before the interaction must equal
the total momentum after the interaction. If the
alpha particle moves off in one direction then the
daughter nucleus left behind must move off in the
opposite direction. This will require some of the
energy available from the alpha decay process. What
remains is the actual KE of the alpha particle.
8 a Mass available
= (220.01140 − 216.00192 − 4.002604) u
= 0.00688 u
So, KE available is 0.00688 × 931.5 MeV =
6.41 MeV
b The Po nucleus is 54 times more massive than the
α-particle, so its velocity will be
1
1
of that of the
54
α-particle. So, its KE will be
of that of the
54
α-particle.
c KEα = 54 × 6 41 MeV = 6.29 MeV
55
14
0 −
9 a 146 C
7 N + −1 β + νe
b Mass available for conversion into energy
= 14.003241 − 13.999231 − 0.00055 = 0.00346 u
This energy is 0.00346 × 931.5 MeV = 3.22 MeV.
(Note that in this calculation, the mass of the
electron anti-neutrino has been ignored. Though it
is in fact non-zero, the mass of the anti-neutrino is
likely to be too small to have an effect here.)
So, the KE of the electron must be less than 3.22
MeV.
c Some of this KE is required for the mass of the
electron anti-neutrino. Some of this energy is also
required for the KE of the electron anti-neutrino.
So the maximum KE of the electron is never
observed to be as high as 3.22 MeV.
10 a Nuclear fission
b Mass defect for energy production
= ((236.0526 − (143.92292 + 88.91781 +
(3 × 1.008665))) u
= 0.185875 u
So, energy available = 0.185875 × 931.5 MeV
= 173.14 MeV
c Energy is released as KE of the fission fragments.
11 a Nuclear fusion.
b Δm = ((2.014102 + 3.01605) − (4.002604 +
1.008665)) u = 0.01888 u
So, E = 0.01888 × 931.5 = 17.59 MeV
(= 17.59 × 1.6 × 10−13 = 2.81 × 10−12 J)
c Energy is released in the form of KE of the alpha
particle and the neutron.
b Non-renewable sources of energy are finite
sources, which are being depleted much faster than
they can be produced and so will run out. They
include fossil fuels (e.g. oil, natural gas and coal)
and nuclear fuels (e.g. uranium).
6
Energy
source
Advantages
Disadvantages
Fossil Fuels
Relatively high
energy density;
Easy to source;
Relatively cheap.
Non-renewable; Cause
excessive pollution
create greenhouse
gases.
Nuclear fuels
High energy
density; Do
not produce
greenhouse
gases; Relatively
plentiful.
Radioactive waste;
Difficult to mine;
Consequences of
malfunction of powers
stations are severe;
Ethical issues involving
misuse of by-products.
Solar energy
Free; Renewable;
Does not cause
excessive
pollution/create
greenhouse
gases.
Variable supply;
Energy produced is
in relatively small
amounts – so requires
large areas; Expensive
to set up.
Wind energy
Free; Renewable;
Does not
produce
greenhouse
gases.
Highly dependent
on meteorological
conditions; Causes
environmental issues
/ noise; Geographical
conditions make
transportation of
energy a problem; Can
create problems with
maintenance.
Hydroelectric
energy
Free; Renewable;
Does not
produce
greenhouse
gases.
Highly dependent on
geographical location;
Causes environmental/
ethical issues;
Expensive to set up.
Chapter 8: Energy production
Exercise 8.1 – Energy sources
1 a A primary energy source is one that has not
undergone any kind of processing to transform the
energy from one form into another; a fossil fuel
such as coal, for example.
b A secondary energy source is one that has
undergone some kind of processing to change the
energy from one form into another; electrical
energy, for example.
2 a The energy available from 1 kg of a source,
measured in J kg−1.
b The energy available from 1 m3 of a source,
measured in J m−3.
3 1: uranium-235; 2: natural gas; 3: petrol; 4: coal
4 a density = mass , so energy density
volume
= specific energy × density.
b Energy density of U-235
= 7.0 × 1013 × 1.9 × 104 = 1.3 × 1018 J m−3.
5 a Renewable sources include solar energy (and the
other forms indirectly dependent on solar energy,
such as wind energy and wave energy) and tidal
energy. In principle, they will be available as long as
the Sun shines and that means billions of years.
Table A8.1
7 Energy in a form that cannot be transformed easily
into a useful kind of energy.
useful energy transferred
35
8 Efficiency =
=
= 35%
100
total energy
g used
ANSWERS
51
c Some of this KE is required for the mass of the
electron anti-neutrino. Some of this energy is also
required for the KE of the electron anti-neutrino.
So the maximum KE of the electron is never
observed to be as high as 3.22 MeV.
10 a Nuclear fission
b Mass defect for energy production
= ((236.0526 − (143.92292 + 88.91781 +
(3 × 1.008665))) u
= 0.185875 u
So, energy available = 0.185875 × 931.5 MeV
= 173.14 MeV
c Energy is released as KE of the fission fragments.
11 a Nuclear fusion.
b Δm = ((2.014102 + 3.01605) − (4.002604 +
1.008665)) u = 0.01888 u
So, E = 0.01888 × 931.5 = 17.59 MeV
(= 17.59 × 1.6 × 10−13 = 2.81 × 10−12 J)
c Energy is released in the form of KE of the alpha
particle and the neutron.
b Non-renewable sources of energy are finite
sources, which are being depleted much faster than
they can be produced and so will run out. They
include fossil fuels (e.g. oil, natural gas and coal)
and nuclear fuels (e.g. uranium).
6
Energy
source
Advantages
Disadvantages
Fossil Fuels
Relatively high
energy density;
Easy to source;
Relatively cheap.
Non-renewable; Cause
excessive pollution
create greenhouse
gases.
Nuclear fuels
High energy
density; Do
not produce
greenhouse
gases; Relatively
plentiful.
Radioactive waste;
Difficult to mine;
Consequences of
malfunction of powers
stations are severe;
Ethical issues involving
misuse of by-products.
Solar energy
Free; Renewable;
Does not cause
excessive
pollution/create
greenhouse
gases.
Variable supply;
Energy produced is
in relatively small
amounts – so requires
large areas; Expensive
to set up.
Wind energy
Free; Renewable;
Does not
produce
greenhouse
gases.
Highly dependent
on meteorological
conditions; Causes
environmental issues
/ noise; Geographical
conditions make
transportation of
energy a problem; Can
create problems with
maintenance.
Hydroelectric
energy
Free; Renewable;
Does not
produce
greenhouse
gases.
Highly dependent on
geographical location;
Causes environmental/
ethical issues;
Expensive to set up.
Chapter 8: Energy production
Exercise 8.1 – Energy sources
1 a A primary energy source is one that has not
undergone any kind of processing to transform the
energy from one form into another; a fossil fuel
such as coal, for example.
b A secondary energy source is one that has
undergone some kind of processing to change the
energy from one form into another; electrical
energy, for example.
2 a The energy available from 1 kg of a source,
measured in J kg−1.
b The energy available from 1 m3 of a source,
measured in J m−3.
3 1: uranium-235; 2: natural gas; 3: petrol; 4: coal
4 a density = mass , so energy density
volume
= specific energy × density.
b Energy density of U-235
= 7.0 × 1013 × 1.9 × 104 = 1.3 × 1018 J m−3.
5 a Renewable sources include solar energy (and the
other forms indirectly dependent on solar energy,
such as wind energy and wave energy) and tidal
energy. In principle, they will be available as long as
the Sun shines and that means billions of years.
Table A8.1
7 Energy in a form that cannot be transformed easily
into a useful kind of energy.
useful energy transferred
35
8 Efficiency =
=
= 35%
100
total energy
g used
ANSWERS
51
9
a See Figure A8.1.
90 000 J
120 000 J
10 000 J
20 000 J
Figure A8.1
b efficiency =
90000
= 75%
120000
90 000
= 0.27 kg
c m= E =
cT 4 200 × (
− )
Efficiency = 100 – (30 + 25 + 12) = 33%
Efficiency = 100 – (25 + 17 + 4) = 54 %
chemical energy in fuel → thermal energy when
fuel is burnt → thermal/kinetic energy of steam →
rotational kinetic energy of turbines → electrical
energy produced by generators.
a chemical energy → thermal energy
b thermal/kinetic energy of steam → rotational
kinetic energy of turbine
c rotational kinetic energy → electrical energy
a Nuclear fission
b Neutrons
c Kinetic energy of the fission fragments.
a 235
92 U is a nucleus; it absorbs a slow neutron and
undergoes fission quite easily. The nucleus 238
92 U
does not readily undergo fission; it absorbs
neutrons without further nuclear processes.
b Specific energy of natural uranium
= 0.6% × 8.0 × 1013 = 4.8 × 1011 J kg−1
c This value is about 104 greater than the specific
energy for fossil fuels.
d Enriched nuclear fuel is uranium that has had its
235
content increased. A greater percentage of
92 U
the uranium fuel rod can undergo fission,
increasing the specific energy of the fuel. This
improves the efficiency of the nuclear power
station.
=
10
11
12
13
14
15
52
useful energy transferred
total energy
g used
16 200 MeV = 200 × 1.6 × 10−13 = 3.2 × 10−11 J from
the fission of one nucleus.
In 1 kg there are: 1000 × 6.023 × 1023
235
= 2.6 × 1024 nuclei.
So the specific energy = 3.2 × 10−11 × 2.6 × 1024
= 8.3 × 1013 J kg−1, which is about 8 × 1013 J kg−1
17 Nuclear binding energy in the nucleus of uranium
→ kinetic energy of fission fragments → thermal
energy of moderator → thermal energy in heat
exchanger → thermal energy of steam → rotational
energy of turbines → electrical energy produced by
generator.
18 a Fast-moving neutrons (from the fission process)
collide with atoms/molecules of the moderator,
slowing down the neutrons so they are more likely
to be absorbed by uranium-235 nuclei. This
transfer of energy heats the moderator. This
thermal energy is transferred via a heat exchanger
to a more conventional system that produces steam.
Typically, a moderator can be water or graphite.
b Control rods absorb neutrons. This reduces the
number of neutrons that are able to collide with,
and be absorbed by, uranium-235 nuclei. This
allows a controlled chain reaction to occur; i.e. a
sequence of fission processes that do not increase
in number, but keep a steady output of energy.
Inserting (or withdrawing) the control rods
reduces (or increases) the number of fission
reactions occurring, thus controlling the amount
of energy produced. Control rods are usually
made from boron.
c The heat exchanger takes the thermal energy
from the moderator and uses it to produce steam
for the turbines. The heat exchanger is a closed
system, so that if it is contaminated in any way by
radioactive material, it will not affect the
surrounding power station. Common heat
exchanger materials include pressurised water and
carbon dioxide gas.
19 a Removing the moderator would reduce or stop
the output of the nuclear power station:
1 It would not slow down the fast neutrons –
so fewer fission reactions occur.
2 There would be no facility to transfer the
energy from the kinetic energy of the fast
neutrons to the heat exchanger in order to
produce steam for the turbines.
20
21
22
23
24
25
b Without the control rods, it would not be possible
to sustain a chain reaction. Too many neutrons
will be available to produce further fission
reactions – which go on to produce even more
neutrons. The nuclear power station would
probably overheat, meltdown and explode.
The nuclear waste from a fission reactor is highly
radioactive with long half-lives, and chemically
reactive. It must be disposed of somewhere where
it will not be a risk to living things for thousands of
years. It is an expensive technological challenge.
kinetic energy in the wind → rotational kinetic
energy in the rotor blades → electrical energy
produced by the generator.
In one second, volume of air that passes through the
wind turbine = π r2 v
So, mass of air passing through = π r2 v ρ
Kinetic energy available from this volume of air
1
= m v2 = 1 × π r2 v ρ × v2 = 1 πρr2v3
2
2
2
If all this energy is transformed, maximum power
from the wind turbine = 1 πρr2v3
2
1
1
a Powermax = 0.3 × πρr2v3= π × 1 3 × 152 × 153
2
2
= 4.7 × 105 W
b Since 15 m s−1 = 3 × 5 m s−1 maximum power
= 33 times min power = 27 times as much energy
per second.
c Energy from a wind-powered generator depends
on the speed of the wind, which changes with
location, time and weather.
gravitational potential energy (GPE) of the water in
the higher reservoir → kinetic energy as it falls →
rotational kinetic energy of turbines → electrical
energy produced by generators.
a GPE = mgh and m = Vρ
So, GPE = Vρgh
b For 1 kg flowing through the generators each
second, power available = gh
So, for Q m3 flowing per second, power
available = Qρgh
So, power output = power available × efficiency
= εQρgh
c Power = εQρgh
power
22 × 109
So, Q =
=
ερ ggh 0.52 × 1 × 103 × 9.81 × 110
= 3.9 × 104 m3 s−1
26 At times of low energy consumption, when the cost
of energy is low, water can be pumped back up from
the lower reservoir to the higher reservoir. This can
then provide energy when the demand (and hence
the price) is high, making it highly cost-effective.
27 a electromagnetic wave energy (in the form of IR)
→ thermal energy of the solar panel → thermal
energy of water in the panel’s pipes.
b energy of photons of light (and some UV) →
increased potential energy of electrons in the solar
cell (which allows them to migrate from the
n-type semiconductor to the p-type
semiconductor) → electrical energy of the
moving electrons.
28 Energy available from sunlight per second
= 30 × 650 = 1.95 × 104 W
Useful energy produced by the solar panel
= 0.3 × 1.95 × 104 = 5.85 kW
In one second mass of water heated from 25 °C to
35 °C = 5850 = 0.14 kg
10 × 4200
This is about eight litres per minute – just about
enough for a shower.
29 a In series, the emfs add up, so that a larger emf is
available.
b Putting the solar cells in series also makes their
internal resistances add up, creating a larger internal
resistance and a smaller terminal voltage when a
current is being drawn. Putting several of these
series of cells in parallel reduces the overall internal
resistance of the combination, so that less energy is
lost as thermal energy in the combined system.
Exercise 8.2 – Thermal energy transfer
1 Heat is the movement of thermal energy caused by a
difference in temperature.
2 a Conduction is the transfer of heat (internal energy)
by collisions of particles and movement of
electrons within a body. Since particles in solids
and liquids are relatively close together, particles
collide more frequently. Particles in gas are much
further apart, and collisions much less frequent.
b Energy is passed on from one particle to the next
in the process of conduction. Since there are many
particles along the path of the energy transfer,
energy must be passed many times from particle to
particle, which takes time.
ANSWERS
53
3
4
5
6
54
c Conduction is enhanced by the motion of free
electrons. Metals have a high density of free
electrons, which can move freely along a conductor
carrying energy with them, and which can transfer
that energy in collisions with particles. This
increases the rate of conduction.
a In electrical conduction, energy is transferred
because of a difference in electrical potential. In
thermal conduction, energy is transferred because
of a difference in temperature. For electrical energy
being transferred by an electrical current, resistance
is a measure of how difficult it is for the current to
flow. In a similar way, the path along which
thermal energy flows has a thermal resistance.
b Thermal conduction of energy: R = c l ,
A
where c is thermal resistivity. In fact, materials are
usually given a thermal conductivity value, k,
which is how well thermal energy can move
through a material. This makes the equation for
thermal conductivity R = 1 × l .
k A
Convection is the main way in which thermal energy
is transferred in liquids and gases. It is driven by
the density difference between where a substance
is hot and where it is cold. Hotter substances are
generally less dense and so rise into regions where the
substance is cooler and denser. This sets up convection
currents that allow thermal energy to be transferred
quickly around the substance.
Forced convection is a process where atoms are made
to move away from a hot surface at a greater rate than
they would otherwise do because of the influence of
an external force. This increased rate of energy loss
means that the surface cools faster.
When a liquid or gas is heated from below it
becomes less dense, and rises. This allows cooler parts
of the liquid or gas to move in and take its place. The
hot liquid or gas then interacts with the cooler liquid
or gas above it, shares its energy, spreads out and cools
and a convection current is formed.
Land mass during a sunny day absorbs solar energy
better than the sea, and its specific heat capacity is
lower than that of water. This makes the land heat
faster than the sea. So, warmed air above the land
rises (because it is less dense) and cooler air from the
sea moves in to take its place. A convection current
is produced. The motion of the cooler air from the
sea onto the beach is what is known as an onshore
breeze.
7 a Conduction and convection both require a
medium to act as the transfer mechanism for the
thermal energy to move. Radiation is energy in
the form of electromagnetic waves, which do not
require a medium to travel through.
b The Sun emits its energy in the form of
electromagnetic waves, which can travel without
a medium (i.e. they can travel through empty
space) to the Earth.
8 A body that absorbs all of the wavelengths of the
electromagnetic radiation incident on it.
9 a The surface area, A, from which the radiation can
emanate and the temperature, T, of its surface.
b Stefan–Boltzmann law: power radiated = σAT 4,
where σ is the Stefan–Boltzmann constant, A is
the surface area, and T is the absolute temperature
of the radiator surface.
c For a body that is not perfectly ‘black’, the law is
modified by introducing a factor called the
emissivity, e, a unit-less factor that describes how
‘black’ an object’s surface is.
e is defined as:
power emitted by body
power emitted by a black body of the
sam
a e physical size and temperature
So, the equation becomes:
power radiated = e σAT 4
10 P = σAT 4 = 5.7 × 10−8 × 4.0 × 10−2 × (400 + 273)4
= 470 W
11 P = σAT 4 = σ4πr2T 4
P
3 8 × 1026
=
4
4πσT
4π × 5.70 −8 × 57784
= 6.9 × 108 m
12 P = eσAT 4 = 0.85 × 5.7 × 10−8 × 2.0 ×
(273 + 32)4 = 841.5 W (840 W to 2 s.f.)
13 a The wavelength that produces the highest
intensity, λmax, is related to the temperature of the
radiator by Wien’s displacement law:
λmax T = 2.9 × 10−3 m K
⇒r =
b See Figure A8.2.
Intensity
higher temperature
18 a
lower
temperature
λ
Figure A8.2
c
c The area under the curve (the total amount of
energy radiated) is smaller for the lower
temperature radiator.
14 a Wien’s displacement law:
λmax =
(
b
19 a
)
×
T
where T is the absolute temperature of a black
body and λmax is the wavelength at which the
largest intensity is being radiated.
−3
b i λmax = 2 9 × 10 = 4.8 × 10−7 m
6000
(480 nm – a greenish/turquoise colour of light).
−3
ii λmax = 2 9 × 10 = 9.7 × 10−6 m (infrared).
300
c A human is ~300 K. The surface of the Sun is
~6000 K. The Sun’s peak emission is in the visible
region of the electromagnetic spectrum.
A human’s peak emission is in the infrared region,
which we cannot see.
−3
15 a T = 2 9 × 10 7 = 5800 K
5 0 × 10
−3
b T = 2 9 × 10 −3 = 2.7 K (the average temperature
1.063 × 10
of deep space).
3.8 × 1026
= 1300 W m−2
16 I = L 2 =
2
4π d
4π ×
×
(
)
17 a The solar constant: the intensity of solar radiation
(at all wavelengths) at a distance equal to 1 AU
from the Sun. Or, the solar constant is the amount
of solar energy incident at the top of the Earth’s
atmosphere per second per square metre of area.
b The luminosity of the Sun varies. There are
long-term variations (over periods of thousands
or millions of years), medium-term variations
(over the solar cycle of 11 years) and short-term
variations (due to solar flares, prominences,
and so on).
b
20 a
The actual distance between the Sun and the
Earth changes over the period of a year.
(The Earth’s orbit is elliptical, not circular.)
Albedo: the amount of radiation scattered from
the surface of a body compared to the amount of
radiation incident on it.
Zero: a black body absorbs all of the radiation
incident on it.
Different places on the Earth’s surface are
different colours. The darker the colour, the
smaller the albedo.
Equatorial regions are dominated by
darker colour vegetation or ocean surfaces. Both
of these have low albedos and are able to absorb
solar radiation more effectively to warm the
surface.
The solar radiation is more concentrated at the
equator, so the intensity is greater.
Convection
Power available = 0.7 × 1370 × πr2
= 0.7 × 1370 × π × (6.4 × 106)2 = 1.23 × 1017 W
This is spread out over the Earth’s surface, so
power absorbed per m2
17
= 1.23 10
4π r 2
=
1.23 1017
4π
(6.4
−2
10
)
6 2
= 239 W m−2,
which is about 250 W m
P =4
250
= 257 K
σ
5 7 × 10 −8
Some of the radiation from the Earth’s surface is
absorbed by gases in the atmosphere. These gases
then radiate some of this energy back towards the
Earth’s surface, which increases the surface
temperature.
The gases primarily responsible for this are carbon
dioxide (CO2), water vapour (H2O), methane
(CH4) and nitrous oxide (N2O).
Greenhouse gases.
The greenhouse effect is the absorption of
radiation emitted from the Earth’s surface by
certain constituents in the Earth’s atmosphere,
which then re-radiate some of this energy back
towards the Earth’s surface.
The enhanced greenhouse effect is an increase in
the amount of greenhouse gases in the
atmosphere, which causes more heating of the
Earth’s surface.
b P = σT 4 ⇒ T =
c
d
e
21 a
b
4
ANSWERS
55
22 a Different molecules have different modes of
vibration. For example, a molecule of water may
vibrate by allowing its hydrogen atoms to move
further from and closer to the oxygen atom, or it
may allow the hydrogen atoms to vibrate at right
angles to their bonds. Since different molecules
are composed of different atoms with different
bond strengths and different modes of vibration,
different amounts of energy will be required to
allow these vibrations to occur. This leads to
different values of photon energies that can be
absorbed, so that each molecule will have its own
set of photon energies it can absorb.
b The peak of the radiated waves from the Earth’s
−3
surface will be: λmax = 2 9 × 10 = 1.0 × 10−5 m.
288
The wavelengths shown are just either side of this
value and so will be prominent in the emission
spectrum of the Earth. Once absorbed, this
radiation can then be re-radiated, some of which
will return to the Earth’s surface.
23 a Global warming refers to the increase in the
surface temperature of the Earth over the last
250 years. This has been attributed, in large part,
to the measured increase in the amount of carbon
dioxide in the atmosphere – the enhanced
greenhouse effect. Although significant amounts
of carbon dioxide are produced naturally, there
seems little doubt that since the industrial
revolution of the late 18th century, increased
industrialisation and deforestation have
contributed to greater quantities of carbon
dioxide in the atmosphere.
b i The burning of fossil fuels has been a major
contributor to the increase in carbon dioxide
in the atmosphere. It is also thought that the
deforestation of large parts of the world (for
example in South America) has contributed a
significant amount to the increase in carbon
dioxide in the atmosphere.
ii The photosynthesis of light to produce energy
in plants and trees removes carbon dioxide
from the atmosphere. Since large areas of the
world’s surface are covered with vegetation,
this is an effective way of removing carbon
dioxide from the air.
56
iii Governments are informed by scientists and
regularly meet to discuss ways in which their
countries can help to prevent further increases
of greenhouse gases. Among their
recommendations are the reduction in burning
of fossil fuels (and the associated need to
provide energy by alternative methods) and a
more sustainable programme of supplying fuel
for energy (by planting faster-growing trees,
for example). International agreements have
been made to limit the production and use of
greenhouse gases (such as CFCs and HCFCs).
Exam-style questions
1
2
3
4
5
6
7
D
D
C
A
D
B
a Rate at which fossil fuel power station uses fuel
1 5 × 109
=
= 150 kg s−1
0 25 4.0 107
b Rate at which nuclear power station uses fuel
9
= 7.5 × 10−5 kg s−1
= 1 5 × 10
0 25 8 × 13
8 a The power station transforms thermal energy
into mechanical and then electrical energy; it is a
form of heat engine. The efficiency is given by the
term:
T
ε = 1 − cold , where Tcold is the temperature of the
Thot
steam after it has been used to transfer its energy
into useful forms and Thot is the temperature of the
steam before it has transferred its energy. So, the
biggest influence on the efficiency of the power
station is the ratio of the two temperatures at
which the power station operates: Tcold and Thot.
b Fossil fuels are relatively easily mined and have a
high specific energy. It is technologically simple to
extract their energy.
c A gas-fuelled power station uses fuel with a higher
specific energy, it has a higher efficiency (almost
twice that of a coal fired power station) and it
produces smaller volumes of greenhouse gases.
9
a See Figure A8.3.
useful electrical
energy 30%
100%
5%
25%
which means the surroundings are more
aesthetically pleasing. But, this must be kept in a
secure location within the power station to avoid
any environmental contamination.
c
3 × 108
12 a λ = =
= 8.6 × 10−6 m (or 8.6 μm)
f 3 5 × 1013
b The peak wavelength from the Sun will be:
electrical
losses
friction
40%
hydrodynamical
losses
Figure A8.3
c
b efficiency = 100 – (40 + 25 + 5) = 30 %
10 a 200 MeV = 200 × 1.6 × 10−13 = 3.2 × 10−11 J
Then, using the equation KE = 3 kT,
2
−11
2
×
KE
2
×
3
.
2
×
1
0
T=
=
= 1.5 × 1012 K
3×k
3 × 1.38 × 10 −23
b 3 kT = 3 × 1 38 10 23 ( 20 + 273)
2
2
6 1 × 10 −21
= 3.8 × 10−2 eV
= 6.1 × 10−21 J =
1 6 × 10 −19
13 a
b
c 0.7n × 3.2 × 10−11 = 6.1 × 10−21
−21
⇒ 0 7n = 6 1 × 10 −11
3 2 × 10
−10
= 1.9 × 10
So, n =
(
log 1.9 10
10
) = 63.
log (0.7)
So, the neutron will need to make about 60
collisions for its energy to be reduced to that of a
thermal neutron.
1 109 3.15 × 107
11 a Volume of coal required =
0 3 × 7 1010
6
3
= 1.5 × 10 m
b Volume of uranium-235 required
1 109 3.15 × 107
= 7.5 × 10−2 m3
=
0 3 × 1 4 × 1018
c The volume of coal required for the power
station needs to be stored, which makes the
surroundings unsightly and dirty. The small
volume of uranium required can be stored easily,
2 9 × 10 −3
= 5.1 × 10−7 m
5700
The wavelength of the radiation absorbed by the
N2O molecule is over ten times longer than this;
there is not a significant amount of radiation at
this wavelength for the N2O molecule to absorb.
−3
For the Earth, λmax = 2 9 × 10 = 1.0 × 10−5 m.
288
This is very close to the wavelength at which
N2O molecules absorb. So, there will be
significant absorption by N2O of the radiation
emitted by the Earth’s surface. This contributes to
the greenhouse effect.
Your brain tells you that the wooden desk top is
quite warm (it does not feel cold for more than a
couple of seconds) and the metal legs are colder
(they feel cold for longer than the wooden
desk top).
If both parts of the desk have been in the
same environment for a long time they
will be at the same temperature. Your brain’s
interpretation of the sensation of touch
gives you misleading information about the
temperature (or relative temperature) of the
two materials. The sensation of feeling does
not always provide accurate or reliable
knowledge of the temperature of things.
The metal desk legs feel colder because they are
better conductors of thermal energy. The
sensation of feeling seems to be an indicator of
how much thermal energy is being transferred,
rather than an indicator of temperature. The
wooden desk top is at the same temperature but,
because it is a poorer conductor, the rate of
transfer of thermal energy is less – and this is the
sensation we feel that our brain translates as being
warmer.
λmax =
c
ANSWERS
57
Chapter 9: Wave phenomena
Exercise 9.1 – Simple harmonic motion
1 The restoring force is proportional to the
displacement from the equilibrium position. The
restoring force is directed towards the equilibrium
position.
2 ω = 2πf
Units: radians per second.
3 a ω = 2π f = 2 × π × 25 = 160 radians s−1 (2 s.f.)
b ω = 2π f = 2 × π × 400 = 42 radians s−1 (2 s.f.)
60
c ω = 2π f = 2 × π × 6.0 × 1014
= 3.8 × 1015 radians s−1
4 a is the acceleration (m s−2), ω is the angular
frequency (radians s−1) and x is the displacement
from the equilibrium position (m).
5 a x0 = 0.4 m (the maximum value that x can have
is 0.4 m).
b ω = 20 radians s−1
c f = ω = 20 = 3.2 Hz (2 s.f.)
2π 2π
d T = 1 = 1 = 0 31
31s (2 s.f.)
f 32
e i x = 0.4 cos (20 t) = 0.4 cos (20 × 2) = − 0.27
m from the equilibrium position.
ii v = 20 × 0.4 × sin (20 × 2) = 6.0 m s−1 (2 s.f.)
iii a = −202 × 0.4 × cos (20 × 2)
= 110 m s−2 (2 s.f.)
6 T = 2π l = 2 π 22.45 = 9.51 s (3 s.f.)
g
9 81
2π = 2π
= 3.1 radians s−1 (2 s.f.)
7 a ω=
T
2
b x = 0.15 cos 3.1 t
c i x = 0.15 cos (3.1 × 0.25) = 0.11 m.
ii x = 0.15 cos (3.1 × 0.5) = 0.00 m
(i.e. at the equilibrium point).
iii x = 0.15 cos (3.1 × 1.0) = − 0.15 m from the
equilibrium position.
8 a No change: T does not depend on mass.
b Since T ∝ l , if the length doubles, the period
increases by a factor of 2 .
c No change: T does not depend on the amplitude
of the oscillations.
58
9
1
d ∝ . On the Moon, the value of g is about 1 of
6
g
that on the Earth, so the period will increase by a
1
factor of 1 = 2.5.
6
a f ∝ 1 so, if mass is doubled, f will change by a
m
factor of 1 or 0.71.
2
b f ∝ k so, if k is doubled, f will change by a
c
10 a
b
c
d
e
f
g
h
i
factor of 2 or 1.41.
No change: f does not depend on A.
F = k x, where k is the spring constant of
the spring.
The spring constant, k, is the amount of force is
required to stretch the spring by a unit amount.
EPE = 1 kx 2
2
k = m ω2
So, area = EPE = 1 mω 2 x 2
2
The maximum EPE will occur when x is a
1
2 2
maximum, EPEmax = mω A .
2
At maximum displacement from the equilibrium
position, A.
At the equilibrium position.
1
When EPE = 0, KEmax = mω 2 A2
2
1 mv 2 = 1 mω 2 A2
2 max 2
Therefore, v max A ω
11 a ω = 2π = 2π = 12.6 radians s−1
T
05
b vmax = ωA = 12.6 × 0.4 = 5.04 m s−1
c The equation of motion for this simple harmonic
motion is: x = 0.4 sin (12.6 t), so the equation
for the velocity is: v = 12.6 × 0.4 cos (12.6 t).
When t = 0.2 s, v = 12.6 × 0.4 cos (12.6 × 0.2)
= − 4.1 m s−1
d amax = ω 2 A = (12.6)2 × 0.4 = 63.5 m s−2
12 Equipment required: light string (~2 m),
mass (~100 g), stop clock, ruler.
Method: Tie one end of the string to the mass and
the other end to something else so that the mass is
suspended above the floor. Measure the length of the
string using the ruler. (The expected uncertainty in
the length of the string is going to be about 1 cm or
so.) Allow the mass to oscillate by pulling the string
to one side a short distance (20 cm or so) and then
letting go. Use the stop clock to time ten swings.
The expected uncertainty is going to be human
uncertainty, ± 0.2 s. Divide the value of time for ten
swings by ten to get the period, T.
2
Rearrange T = 2π l to give g = 4π2 l . Substitute
g
T
in values for l and T to find g.
Alternative method: Repeat the first method, but
for a range of values of l , and plot a graph of T 2
against l . This graph should be a straight line passing
4π 2
through the origin with a gradient given by
.
g
Find the gradient from the graph and from this the
value of g.
This method makes it is relatively easy to plot error
bars from T 2 and l , which will allow maximum and
minimum gradients to be plotted. This will give the
uncertainty in your value for g.
Exercise 9.2 – Single-slit diffraction
1 a Path difference =
b
c
2 a
b
3 a
b sinθ
2
1
λ
2
b sinθ λ ⇒ b i θ = λ. Showing the angle at which
2
2
the first minimum on the single-slit diffraction
pattern occurs.
Since the first minimum of the diffraction pattern
occurs when sin θ = λ , there would be very little
b
diffraction observed.
There would be a maximum amount of diffraction
– the wavefronts would be semi-circular once they
have passed through the slit.
See Figure A9.2.
Intensity
−3
13 a a = F = kx = 400 × 15 × 10 = 60 m s−2
m m
01
(
b ω 2 = 2π
2
) = mk ⇒ f =
k
4π 2
=
400
π 2 0.1
= 10 Hz (2 s.f.)
1
c Total energy = mω 2 A2
2
1 × 0 1 × 400 ×
=
(15 × 10−3)2 = 4.5 × 10−2 J
2
01
14 a and b See Figure A9.1.
1 mω2A2
2
A
0
A
Displacement from equilibrium position
EPE
KE
Figure A9.1
c The total energy equals the sum of EPE and KE;
this is a constant amount.
Distance across screen
Figure A9.2
b i Smaller separation of the maxima.
ii Larger separation of the maxima.
−9
⎛
⎞
4 b sin θ = λ ⇒ θ = sin−1 λ = sin−1 600 × 10−4
⎝
b
1 5 × 10 ⎠
−3
= 4 × 10 radians.
So, the angular width of the central maximum is
twice this: 8 × 10−3 radians (0.46 degrees).
5 Sound waves of frequency 300 Hz have a wavelength
in air of: λ = v = 330 = 0.89m
f 370
So, the angle at which the first minimum of the
single-slit diffraction pattern would occur is:
sin−1 λ = sin−1 0 89 = 90°
09
b
This means that the central maximum of the single-slit
diffraction pattern covers all angles possible between
−90o and +90°. All pupils within the classroom will be
within this range and can hear the sound from the shoes.
()
()
( )
ANSWERS
59
−2
6 a i θ = sin−1 λ = sin−1 ⎛ 2 10−2 ⎞ = 23.6°
b
⎝ 5 10 ⎠
(0.412 radians)
ii Second-order minimum occurs at
−2
⎛
⎞
θ = sin−1 2λ = sin−1 ⎜ 2 2 × 10
⎟⎠ = 53.1°
−2
b
⎝ 5 10
(0.927 radians)
So, at a distance of 1.5 m away, these minima will be
separated by:
(0.927 − 0.412) × 1.5 = 0.77 m (77 cm).
b The third-order minimum should occur at an
angle
6 a See Figure A9.3.
Intensity
( )
( )
5 × 10–3
Angle (radians)
5 × 10–3
Figure A9.3
−2
⎞ = sin−1 (1.2)
θ = sin−1 3λ = sin−1 ⎛⎜ 3 2 × 10
b
⎝ 5 10 −2 ⎟⎠
Since this is not possible, there will be four minima
altogether: two either side of the central maximum.
7 The width of the central maximum would be
infinitely large.
b See Figure A9.4.
Intensity
Exercise 9.3 – Interference
1 The Young double-slit experiment demonstrates
interference, which is proof of the wave nature of
light.
2 The infinitesimal width of each of the two slits means
that their separate diffraction patterns would have a
central maximum that is infinitely wide. Since the
double-slit interference pattern occurs within this
central maximum ‘envelope’, each maximum would
have the same intensity.
3 s = λD
d
s is the spacing of the interference maxima on the
screen, λ is the wavelength of the waves being used to
create the interference pattern, D is the distance from
the two slits to the screen and d is the separation of
the two slits.
4 s = λ D = 0 8 × 4 = 2.7 m
d
1.2
−9
3 0 = 12.6 × 10−3 m
5 s = λ D = 630 × 10 ×
−4
d
1 5 × 10
(13 mm to 2 s.f.)
60
5 × 10–3
Angle (radians)
5 × 10–3
Figure A9.4
c The single-slit diffraction pattern forms an
‘envelope’ inside which the interference pattern
must occur. This is why, if each of the two slits is
considered to be infinitesimally thin, the maxima
of the interference pattern will be equally bright
– they are all inside the central maximum of the
diffraction pattern.
7 a In the single-slit diffraction pattern, n λ = b sin θ n
tells us where the minima of the diffraction pattern
occur.
b In the two-slit interference pattern, m λ d sin θ m
tells us where the maxima of the interference
pattern occur.
nλ
c sin θn =
= sin θm = mλ
b
d
n
m
Therefore, =
b d
d
m
So, =
b n
8
12 s sin θ = 2λ ⇒
−3
⎛
⎞
λ = s × sin θ = 10
× sin tan −1
⎝
⎠
2
2 × 600
= 640 nm (2 s.f.)
13 a The first maximum of an interference pattern
would occur at an angle given by:
d The missing maximum from the interference
pattern is given when n = 1, so it is the mth
maximum that is missing: this is the value of d .
b
When n = 2, it will be the (m × 2)th maximum
that is missing: 2d .
b
e The number of maxima of the interference
pattern inside the central maximum of the
diffraction pattern is given by 2d −1.
b
a See Figure A9.5.
( )
θ = sin−1
many slits
pattern
2 slit
pattern
Figure A9.5
b The intensity of the maxima has increased. There
are more slits, so more energy is able to pass
through onto the screen.
c The spacing of the maxima has remained the
same, because the spacing of the slits is the same.
d The width of the maxima has decreased. With
more slits, it is less likely that waves can arrive in
phase over a small range of distances on the
screen. Waves that do arrive in phase – and
therefore produce a maximum – only do so for a
limited region on the screen, leading to thinner
maxima.
9 A diffraction grating is a set of many, very narrow,
slits. Each slit is separated from the one next to it
by a very small distance. Light passing through a
diffraction grating will then produce an interference
pattern similar to that shown in Figure A9.5.
10 The light would be a different colour: blue rather
than orange. The spacing of the maxima would be
450
smaller by a factor of
.
590
10 −3
= 1.7 × 10−6 m.
11 s =
600
The angle between the central maximum and the
first-order maximum is given by:
−9
sin−1 λ = sin−1 630 × 10−6 = 21.8° (0.38 radians).
s
1 7 × 10
Therefore, the separation on the screen
= 0.38 × 5 = 1.9 m.
(
)
( λs )
= sin−1
1 0 × 10 −10 = 1 0 × 10 −10
1 7 × 10 −6
⎛ 10 −3 ⎞
⎝ 600 ⎠
= 5.9 × 10−5 radians (≈ 3.4 × 10−3 degrees).
This angle is too small to be resolved by an X-ray
detector.
b The spacing of the atoms is now much smaller,
so now the angle for the first maximum
of an interference pattern would occur at:
1 0 × 10 −10
λ
θ = sin−1
= sin−1
= 19.5°
s
0 3 × 10 −9
This is easily resolved.
c Using X−rays with crystals produces interference
patterns that can be easily observed. This allows
scientists to make accurate measurements of the
spacing and orientation of atoms within a crystal.
14 a The reflected ray, A, will undergo a phase change
of π radians.
b Its phase is not changed: the far side of the
material has a refractive index that is smaller than
the material it is already in.
c 2t.
d Within the material, the speed of the waves has
decreased by a factor 1 . Since the frequency
n
remains constant, the wavelength of the waves has
also decreased by a factor of 1 .
n
2
t
2
tn
=
e m=
λ
n
f No change in phase.
g m = 1 or a whole number of λ.
h 2tn = mλ for destructive interference.
15 For constructive interference, 2tn = m + 1 λ,
2
where m is an integer.
16 For no reflected light, we need destructive
interference. So, 2tn = λ for the minimum thickness.
−9
Therefore, t = λ = 500 × 10 210 nm.
2n
2 × 1.2
()
()
( )
ANSWERS
61
17 a The coloured patterns are formed when
constructive interference occurs. Rays of light
that are reflected from the front surface of the
thin film superpose with rays of light that have
been transmitted, reflected from the rear surface
and then transmitted back into the air. When this
superposition is constructive, a large amplitude of
light waves is produced, producing a bright
coloured region.
b In between the coloured patterns there will be
regions where there is destructive interference
occurring. This will make the coloured patterns
have a spacing. This spacing will change when the
thickness of the thin liquid film changes.
18 a Since the refractive index of the film is greater
than that of air, there will be a phase change of π
radians at this reflection.
b π radians.
c 2tn
d Destructive interference will occur. There will be
little or no light reflected from the surface of the
film.
e This addition of a thin film is called ‘lens
blooming’. The bloom on a lens can usually be
recognised by its purple/red colour. It reduces the
amount of light that is reflected from the front
surface of a lens – and so increases the amount of
light that is transmitted. This is useful in any
optical device that uses a lens, for example a
camera, binoculars, microscope or a telescope.
19 The thickness of the lens bloom is set to maximise
the destructive interference caused by reflections
from the front and back surfaces of the bloom. Since
there is a range of wavelengths of light (from red at
the long wavelength part of the spectrum, to purple
at the short end of the spectrum) this thickness is
set to the average wavelength of light, which is a
green colour at a wavelength of about 500 nm. So,
those wavelengths either side of this value will not
suffer destructive interference as much as the green
light. This will make the lens bloom appear to be a
mixture of purple and red colours.
20 2tn = λ
9
So, t = λ = 520 × 10 = 186 nm (about 190 nm to
2n
2 × 1.4
2 s.f.)
62
Exercise 9.4 – Resolution
1 a A series of concentric circles.
b The first minimum will be given by:
1.22 × λ = sin θ
−9
⎛
⎞
Therefore, θ = sin −1 ⎜ 1.22 × 500 −×3 10 ⎟
⎝
⎠
4 × 10
= 1.5 × 10−4 radians.
x
c Angle subtended = (in radians).
D
d The angle subtended by the two sources must be
greater or equal to 1.5 × 10−4 radians for the two
sources to be resolved.
Angle subtended by the two sources is
−2
2
3
4
5
6
θ = 2 × 10 = 4 × 10−3 radians.
5
Since this angle is greater than 1.5 × 10−4, the two
sources are well resolved.
Rayleigh’s criterion for resolution: Two sources will
be just resolved when the angle they subtend with
an aperture is equal to the angle at which the first
minimum of the diffraction pattern will occur. This
means that the first maximum of the diffraction pattern
from one of the sources coincides exactly with the first
minimum of the diffraction pattern of the other source.
−9
a θmin = sin−1 1.22 × λ = sin−1 ⎛ 1.22 × 500 × 10 ⎞
−
2
⎝
⎠
b
1 6 × 10
= 3.8 × 10−5 radians
b x = 50 × 3.8 × 10−5 = 1.9 mm
−9
1.22 × λ
= 2 5 × 10 −7 ⇒ b = 1.22 × 550 ×−710 = 2.7 m
b
2 5 × 10
22λ = 1.22 × 500 × 10 −9
a θmin = 1.22
b
24
−7
= 2.5 × 10 radians
b separation = 4 × 9.46 × 1015 × 2.5 × 10−7
= 9.5 × 108 m
1.22 λ 1.22 × 500 × 10 −9
=
= 2.0 × 10−4 radians
a θmin =
b
3 10 −3
x
1
= 5 km
b D= =
θ 2 × 10 −4
c i In blue light, the wavelength is smaller so the
minimum angle for resolution is smaller. This
makes it easier to resolve.
ii At night, there is less ambient light. This makes
our pupils enlarge. A larger diameter pupil will
make the minimum angle for resolution smaller.
This makes it easier to resolve at night (or in
low light levels). (However, the lower levels of
(
)
light will mean that less energy is being
received by the eyes. This may counteract the
eye’s ability to see objects clearly.)
−3
7 a s = 1 × 10 = 3.3 × 10−6 m
300
b N = 300 × 15 = 4500 slits
c R = mN = 2 × 4500 = 9000 = 9 × 103
d λ = N ⇒ Δ λ ≥ λ for resolution.
Δλ
mN
−9
λ = 500.1 × 10 = 5.6 × 10−11 m = 0.056 nm.
mN
9000
Since Δλ = 0.2 nm > 0.056 nm, these two
wavelengths can be resolved.
8 Number of slits required, N = λ
= 589.3 = 491
m × Δ λ 2 0.6
Therefore, number of slits mm−1 required = 491 = 123
4
−3
Therefore, spacing of slits = 1 10 = 8 × 10−6 m
123
5 a As the train approaches the bridge, the trainspotter will hear the whistle from the train as a
higher frequency sound than normal. When the
train has passed under the bridge and recedes from
the train-spotter, the train-spotter will hear the
whistle from the train at a lower frequency than
normal.
′
330
b i f = f
= 800
= 978 Hz
+
330 − 60
(980 Hz to 2 s.f.)
′
330
= 800
= 677 Hz
ii f = f
+
330 − 60
(680 Hz to 2 s.f.)
6 a Since the measured wavelength is larger than the
actual wavelength, the distant star must be moving
away from the Earth.
3 × 108 × (580.9 − 527.0 )
b Δλ = v ⇒ v = c × Δλ =
c
λ
527.0
λ
7
−1
= 3.1 × 10 m s away from the Earth.
Exercise 9.5 – Doppler effect
Exam-style questions
1 a The frequency observed will decrease, so the
observer will hear a lower pitch sound.
330
b f = f
= 1.5 × 103
+
330 + 15
= 143 kHz (140 kHz to 2 s.f.)
1 9 × 103
2 a % of speed of light =
= 6.3 × 10−4 %
3 108
b Δ λ = 6 3 × 10−6 ⇒
λ
νλ = 6.3 × 10−6 × 656.28 × 10−9
= 4.1 × 10−12 m
c −4.1 × 10−12 m
3 A Doppler radar gun measures small changes in
frequency of radio waves. Radio waves are transmitted
at a known frequency. These waves reflect from a
moving car to the receiver in the Doppler gun.
The received waves show a change in frequency if
there is a relative motion between the observer and
the car. Since the speed of electromagnetic waves is
Δff
known, the Doppler gun uses the expression v =
c
f
to calculate an accurate value for v, the speed of the
moving car.
3
⎛
⎞
330 + 40 × 10
⎝
⎠
60
×
60
′
4 f = f
= 450 ×
330
+
1 B
2 B
3 D
4 D
5 A
6 a The graph shows that α = x because it is a straight
line that passes through the origin, with a negative
gradient.
b The gradient of the graph is ω2, so ω = 64 =
16
2 radians s−1 . The gradient of the graph is ω2 and
this is seen to be 4. So ω = 2 radians s−1.
c f = ω = 2 = 0.32 Hz and T = 3.1 s (2 s.f.)
2π 2π
1 g = 1 9 81
= 0.6 Hz (1 s.f.)
7 a f =
2π l
2π 0 8
b Distance travelled in one hour
= 0.6 × 60 × 60 × 1.7 = 4 km (1 s.f.).
c Most published suggestions are around
5 km h−1, so this is quite a good approximation.
8 a ω2 = k
m
k = 25
Therefore, ω =
= 7.9 radians s−1
m
04
b vmax = ω A = 7.9 × 0.1 = 0.79 m s−1
c When EPE and KE are equal, the EPE will be:
1 k x2 = 1 1
1 1 k 2
m A
m 2 A2 =
2
2 2
2 2 m
So, x 2 = 1 A2
2
Therefore, x 2 = 1 A = 0.71 × 0.1 = 0.07 m
2
(7 cm) from the equilibrium position.
( )
( )
= 465 Hz (470 Hz to 2 s.f.)
( )
( )
(
(
)
)
( )
(
) (
)
ANSWERS
63
9
a Angular width of the central maximum
−9
⎛
⎞
−11
= 2 × sin 1 530 × 10
= 2 × sin
⎝ 1 10 −4 ⎠
= 1.06 × 10−2 radians
So, width of the central maximum on the
screen = 1.06 × 10−2 × 8 = 8.5 cm.
b 650 = 1.23.
530
So, the width of the new central maximum will
be 1.23 × 8.5 cm = 10.5 cm.
c With white light:
The central maximum will be white.
The higher order maxima will each be a
spectrum of colours, with violet and blue closest
to the central order maximum and red furthest
way from the central maximum.
However, beyond about the second order, it is
likely that these maxima will overlap and so other
colours will be seen where this occurs.
10 a 300 lines mm−1 means that the spacing of the
1
⎞ = 3.3 × 10−6 m.
slits is: ⎛
−1
⎝ 300 mm ⎠
()
−9
b θ = sin −1 λ = siin −1 590 × 10−6 = 10.3° (0.18 radians)
s
3.3 × 10
c Maxima would also be produced as long as n λ ≤ 1.0
s
So, there would be a second-order maximum at:
λ
sin−1 2 = 21°
s
λ
A third-order maximum at: sin−1 3 = 32.4°
s
−1 λ
A fourth-order maximum at: sin 4 = 45.7°
s
λ
A fifth-order maximum at: sin−1 5 = 63.4°
s
λ
But, there is no sixth-order maximum because 6 > 1.
s
4
5
6
7
8
b Gravitational field strength: the gravitational force
acting on a unit mass.
F = Eq = 4 × 107 × 1.6 × 10−19 = 6.4 × 10−12 N
F = mg = 3.5 × 104 × 25 = 8.8 × 105 N
V
300
a E= =
= 1.3 × 104 V m−1
d 2 4 × 10 −2
b F = Eq = 1.3 × 104 × (2 × 1.6 × 10−19)
= 4.2 × 10−15 N
W = mg = 80 × 9.81 = 780 N (2 s.f.)
5 97x1024
a ρ=M =
= 5.49 × 103 kg m−3
3
4π×
V
×
3
G 4 πρE RE3 4G πρ R
GM
3
E E
=
bg= 2 =
2
3
RE
RE
(
)
3g
4πρE RE
3g
3 × 9.81
ρE =
=
4πGRE 4π × 6.67
67 × 10 −11 × 6 38 × 106
3
−3
= 5.50 × 10 kg m
Percentage difference = 1 × 100% = 0.18%
549
E = Vq = 50 × 1 = 50 J
E = Vq = 10 × 1 = 10 J
Work done = qΔV = 1 × (50 − 10) = 40 J
No. The path does not matter, only the start and
end points matter.
See Figure A10.1.
⇒G=
c
9
d
a
b
c
d
10 a
Chapter 10: Fields
Exercise 10.1 – Describing fields
1
a A region in space in which a charged particle
experiences an electrical force.
b A region in space in which a mass experiences a
gravitational force.
c A region in space in which a magnet, or something
that can be magnetised, experiences a magnetic force.
2 Uniform field: where the field strength is the same
magnitude and in the same direction everywhere.
3 a Electric field strength: the electrical force acting
on a unit positive charge.
64
Figure A10.1
b The field lines are getting further apart.
c At an infinite distance from the sphere.
d At the surface of the sphere, the 1 C charge will
have electrical potential energy of:
E = qV = 1 × 400 = 400 J
At infinity, the 1 C charge will have electrical potential
energy of zero. So, the work done will be 400 J.
e Electric potential at a point in a field: the work
done in moving a unit positive test charge from
infinity to that point.
9
a Angular width of the central maximum
−9
⎛
⎞
−11
= 2 × sin 1 530 × 10
= 2 × sin
⎝ 1 10 −4 ⎠
= 1.06 × 10−2 radians
So, width of the central maximum on the
screen = 1.06 × 10−2 × 8 = 8.5 cm.
b 650 = 1.23.
530
So, the width of the new central maximum will
be 1.23 × 8.5 cm = 10.5 cm.
c With white light:
The central maximum will be white.
The higher order maxima will each be a
spectrum of colours, with violet and blue closest
to the central order maximum and red furthest
way from the central maximum.
However, beyond about the second order, it is
likely that these maxima will overlap and so other
colours will be seen where this occurs.
10 a 300 lines mm−1 means that the spacing of the
1
⎞ = 3.3 × 10−6 m.
slits is: ⎛
−1
⎝ 300 mm ⎠
()
−9
b θ = sin −1 λ = siin −1 590 × 10−6 = 10.3° (0.18 radians)
s
3.3 × 10
c Maxima would also be produced as long as n λ ≤ 1.0
s
So, there would be a second-order maximum at:
λ
sin−1 2 = 21°
s
λ
A third-order maximum at: sin−1 3 = 32.4°
s
−1 λ
A fourth-order maximum at: sin 4 = 45.7°
s
λ
A fifth-order maximum at: sin−1 5 = 63.4°
s
λ
But, there is no sixth-order maximum because 6 > 1.
s
4
5
6
7
8
b Gravitational field strength: the gravitational force
acting on a unit mass.
F = Eq = 4 × 107 × 1.6 × 10−19 = 6.4 × 10−12 N
F = mg = 3.5 × 104 × 25 = 8.8 × 105 N
V
300
a E= =
= 1.3 × 104 V m−1
d 2 4 × 10 −2
b F = Eq = 1.3 × 104 × (2 × 1.6 × 10−19)
= 4.2 × 10−15 N
W = mg = 80 × 9.81 = 780 N (2 s.f.)
5 97x1024
a ρ=M =
= 5.49 × 103 kg m−3
3
4π×
V
×
3
G 4 πρE RE3 4G πρ R
GM
3
E E
=
bg= 2 =
2
3
RE
RE
(
)
3g
4πρE RE
3g
3 × 9.81
ρE =
=
4πGRE 4π × 6.67
67 × 10 −11 × 6 38 × 106
3
−3
= 5.50 × 10 kg m
Percentage difference = 1 × 100% = 0.18%
549
E = Vq = 50 × 1 = 50 J
E = Vq = 10 × 1 = 10 J
Work done = qΔV = 1 × (50 − 10) = 40 J
No. The path does not matter, only the start and
end points matter.
See Figure A10.1.
⇒G=
c
9
d
a
b
c
d
10 a
Chapter 10: Fields
Exercise 10.1 – Describing fields
1
a A region in space in which a charged particle
experiences an electrical force.
b A region in space in which a mass experiences a
gravitational force.
c A region in space in which a magnet, or something
that can be magnetised, experiences a magnetic force.
2 Uniform field: where the field strength is the same
magnitude and in the same direction everywhere.
3 a Electric field strength: the electrical force acting
on a unit positive charge.
64
Figure A10.1
b The field lines are getting further apart.
c At an infinite distance from the sphere.
d At the surface of the sphere, the 1 C charge will
have electrical potential energy of:
E = qV = 1 × 400 = 400 J
At infinity, the 1 C charge will have electrical potential
energy of zero. So, the work done will be 400 J.
e Electric potential at a point in a field: the work
done in moving a unit positive test charge from
infinity to that point.
9
−9
11 a V = kQ = 9 × 10 × 3 −×2 10 = 90 V (= 90 J C−1)
r
30 × 10
−11
30
b V = − GM = − 6 67 × 10 × 211.0 × 10
r
1 5 × 10
= − 8.9 × 108 J kg−1
12 a See Figure A10.2.
0
16 a See Figure A10.3.
Distance from the Earths surface
3RE
2RE
RE
g/N kg–1
9.81
Figure A10.2
Figure A10.3
b −200 V
c The potential varies inversely with increasing
distance from the surface of the sphere.
d Figure A10.4.
b The area under the graph from infinity to the
surface of the Earth.
c The area under the graph from the surface of the
Earth to the where the distance is twice the
radius of the Earth.
−11
24
13 a Vg = − GM = − 6 67 × 10 × 6.0 × 10
6
R
6 4 × 10
= − 6.3 × 107 J kg−1
b At twice the distance away, the gravitational
potential will be halved.
So, Vg = −3.15 × 107 J kg−1
(= −3.2 × 107 J kg−1 to 2 s.f.)
c E = m ΔVg = 2500 × (−3.15 − −6.3) × 107
= 7.9 × 1010 J
14 a A line or surface on which the value of the
electrical potential is the same everywhere.
b A line or surface on which the value of the
gravitational potential is the same everywhere.
c Equipotentials and field lines are always
perpendicular.Yes, this is always true for electric
and gravitational fields.
15 a Zero. X and Y are on an equipotential, so no
work is required to move along from X to Y.
b E = q ΔV = 4 × 10−6 × (12 − 3) = 3.6 × 10−5 J
c 3.6 × 10−5 J
Figure A10.4
e When the electric field strength is large, the
equipotentials are close together. When the
electric field strength is small, the equipotentials
are far apart.
17 a and b See Figure A10.5.
+
+
Figure A10.5
c The electric field strength is smallest where the
equipotentials are furthest apart.
ANSWERS
65
18 a and b See Figure A10.6.
2 a and b See Figure A10.8.
+
Earths surface
Figure A10.8
negatively charged plate
Gravitational potential/×101 J kg–1
c The equipotentials are equally spaced.
3 ΔV = g Δh = 9.81 × 8.8 × 103 = 8.6 × 104 J
4 a i g is given by the gradient of the graph
(see Figure A10.9).
Figure A10.6
c The equipotentials and the field lines are
perpendicular to each other.
19 a and b See Figure A10.7.
Distance from the Earth’s surface/Mm
0
–1
–2
–3
–4
–5
–6
–7
5
10
15
20
25
30
Figure A10.9
+
–
Figure A10.7
Field lines have arrows to show the direction of
the field. Equipotentials have no direction, since
potential is a scalar quantity.
c The electric field strength is greatest where the
equipotentials are closest together.
7
At a distance of 10 Mm, g = − 6 10 6
15 × 10
= −4.0 N kg−1
−7
ii At a distance of 20 Mm, g = − 2 10 −6
20 × 10
= −1.0 N kg−1
b Yes. The value of g at 20 Mm is ¼ of the value of g
at 10 Mm.
5 a ΔV =
b
c
Exercise 10.2 – Fields at work
1 a E = mg Δh = 1000 × 9.81 × (4 × 4) = 1.6 × 105 J
E 1000 × 9 81 × ( ×
b ΔV = =
m
1000
)
= 39 J kg−1
c Given that, over a height difference of 4 m, the
ΔV
gravitational field is approximately uniform, g =
Δh
66
6 a
b
⎛
⎞
1
1
⎞
kQ ⎜ 1 − 1 ⎟ = 9 109 6 × 10 3 ⎛
−
⎝ 5 10 −22 10 × 10 2 ⎠
⎝ r1 r2 ⎠
= 5.4 × 108 V
E = q ΔV = 2.0 × 10−6 × 5.4 × 108 = 1.1 kJ
Since the force acting on it will be to repel it from
the 6 mC charge, the 2.0 μC charge will be
repelled and it will accelerate radially away.
F=Eq
EPE = q VE
c E=−
d F=−
7
dVE
dr
2
d(
⎛ GM E ⎞
GM Em
12 a KE = 1 mv 2 = 1 m ⎜
=1
⎟
2
2 ⎝
r ⎠
2 r
)
a F=mg
b GPE = m Vg
dV
Vg
c g=−
dr
d F =−
d(
c
)
d
dr
e
8
a The 1 kg mass would have to go from a
GM
to zero. So,
gravitational potential energy of −
R
the energy it would require is GM = 6.3 × 107 J
R
2 6.3 107
b 1 mv 2 = 6 3 × 107 ⇒ v =
2
1
= 1.1 × 104 m s−1
−11
22
2GM = 2 × 6.67 × 10 × 7 35 × 10
9 v=
6
R
1 74 × 10
= 2.4 × 103 m s−1 (2.4 km s−1)
2GM ⇒ R = 2GM
R
c2
−11
30
= 2 × 6.67 × 10 × 2 42 × 2 0 × 10
×
10 a c =
(
)
= 7.1 × 10 m (7.1 km)
b The actual size of the black hole will be smaller
than this. The value of the gravitational field
strength at the surface of the black hole will be
too great to allow radiation to leave, but the field
strength reduces with distance, so there will be a
distance away where the gravitational field
strength is just strong enough to prevent radiation
from escaping. Any distance greater than this and
radiation can escape.
c The Schwarzschild radius.
3
−11
24
GM E
= 6 67 × 10 × 66 × 10
r
6 8 × 10
= 7.7 km s−1
6
b T = 2π = 2π × 6 8 × 310 = 5.5 × 103 s
v
7 7 × 10
(about 90 minutes).
24 × 60 × 60
So, number of orbits in one day =
5 5 × 103
= 16 orbits.
11 a v =
GM Em
r
Etotal = KE + GPE
GM Em GM Em
GM Em
=1
,
−
=−1
2 r
r
2 r
which is less than zero.
KE = 1 GPE
2
Etotal = − KE
GM E
6 67 × 10 −11 × 6 × 1024
v=
=
= 1.0 km s−1
r
3 8 × 108
b GPE = −
dr
13 a
8
b T = 2π = 2π ×3 8 × 10
= 2.4 × 106 s.
3
v
1 0 × 10
7
So, number of orbits in one year = 3 15 106
2 4 × 10
= 13 orbits.
−11
30
GM Sun
= 6 67 × 10 × 211.0 × 10
14 a v =
r
1 5 × 10
4
−1
= 3.0 × 10 m s
2π = 2π ×1 5 × 1011
bT=
= 3.1 × 107 s = one year.
v
3 0 × 104
kQe
15 a F = 2
r
2
Q
kQe
b mv = kQe
⇒ 1 mv 2 =
r
2
2r
r2
kQe
c KE of electron =
and EPE of the electron
2r
kQe
because the electron’s charge, e, is negative.
=−
r
kQe kQe
Q
kQe
−
=−
, which is less than
So, Etotal =
2r
r
2r
zero.
(
)
2
9 × 109 ×
×
kQe
dr=−
=
2 × (−
eV )
2 × 13 6 × 1 6 × 10 −19
= 5.3 × 10−11 m
16 a Decreased
b The total energy of the electron is now more
negative than it had been. This means that it must
be in an orbit that is closer to the nucleus. So, its
kinetic energy has increased because it is inversely
proportional to r.
ANSWERS
67
c The EPE of the electron has become more
negative.
d Closer
17 2440 × 0.52 = 610; 612 × 12 = 612;
270 × 1.52 = 607.5; 153 × 22 = 612;
98 × 2.52 = 612.5; 68 × 3.02 = 612. Since all these
values are about 612, the light intensity and the
distance are related by an inverse-square law.
Exam-style questions
1
2
3
4
5
6
B
B
A
C
B
D
7
a E = V = 300 −3 = 1.0 × 104 V m−1
d 30 × 10
b See Figure A10.10.
11 a An inverse-square law is one where Y = k2 ,
where k is a constant (i.e. if X doubles inXvalue,
then Y becomes ¼ of the value).
30 mm
Figure A10.10
c F = Eq = 1.0 × 104 × 1.6 × 10−19 = 1.6 × 10−15 N
d Towards the top plate.
−15
e a = F = 1 6 × 10 −31 = 1.8 × 1015 m s−2
m 9 1 × 10
a The gravitational force between the satellite and
the Earth.
2
GM Em
GM E
b mv =
⇒v=
2
r
r
r
2 3
GM ET 2
2
r
4
π
r
3
⇒
= GM E ⇒ r =
c v=
T
T2
4π 2
=
3
6.67 × 10 −11 × 6 × 1024 × ( 24 × 60 × 60 )2
4π 2
= 4.2 × 107 m
(This is about six Earth radii from the Earth’s
surface.)
68
a E = V = 100 −2 = 500 V m−1
d 20 × 10
b i EPE = Vq = 100 × 6 × 10−6 = 6 × 10−4 J
ii EPE = Vq = 50 × 6 × 10−6 = 3 × 10−4 J
iii EPE = Vq = 0 × 6 × 10−6 = 0 J
c i F = Eq = 500 × 6 × 10−6 = 3 × 10−3 N
ii Since the electric field is the same everywhere
between the two parallel plates, F = 3 × 10−3 N
iii F = 3 × 10−3 N
d Work done = q ΔV = 6 × 10−6 × (100 − 0)
= 6 × 10−4 J
e Because the force is a constant, we can write:
ΔEPE = F × d
12
10 a E = V =
= 20 V m−1
d 60 × 10 −2
b Because the field is uniform, E = F × d = Eq × d
= 20 × 1.6 × 10−19 × 0.6 = 1.9 × 10−18 J
c EPE → KE of electrons → thermal energy of
atoms
300 V
8
9
b Since Y = k2 , then YX2 = k. So, if pairs of values
X
2
Y and X are multiplied together, they should all
come to the same constant, k.
c 302 × 22 = 1208, 72 × 42 = 1216, 34 × 62 = 1224,
19 × 82 = 1216, 12 × 102 = 1200, 8 × 122 = 1152.
So, to 2 s.f., all of these give k = 1200. This
confirms that X and Y are related by an inversesquare law.
Chapter 11: Electromagnetic
induction
Exercise 11.1 – Electromagnetic induction
1 a The needle on the galvanometer will kick to one
side of the scale and then return to zero.
b The needle on the galvanometer will kick to the
other side of the scale (same amount as before)
and then return to zero.
c The needle will not move.
d The needle will kick further.
e The needle will not move.
c The EPE of the electron has become more
negative.
d Closer
17 2440 × 0.52 = 610; 612 × 12 = 612;
270 × 1.52 = 607.5; 153 × 22 = 612;
98 × 2.52 = 612.5; 68 × 3.02 = 612. Since all these
values are about 612, the light intensity and the
distance are related by an inverse-square law.
Exam-style questions
1
2
3
4
5
6
B
B
A
C
B
D
7
a E = V = 300 −3 = 1.0 × 104 V m−1
d 30 × 10
b See Figure A10.10.
11 a An inverse-square law is one where Y = k2 ,
where k is a constant (i.e. if X doubles inXvalue,
then Y becomes ¼ of the value).
30 mm
Figure A10.10
c F = Eq = 1.0 × 104 × 1.6 × 10−19 = 1.6 × 10−15 N
d Towards the top plate.
−15
e a = F = 1 6 × 10 −31 = 1.8 × 1015 m s−2
m 9 1 × 10
a The gravitational force between the satellite and
the Earth.
2
GM Em
GM E
b mv =
⇒v=
2
r
r
r
2 3
GM ET 2
2
r
4
π
r
3
⇒
= GM E ⇒ r =
c v=
T
T2
4π 2
=
3
6.67 × 10 −11 × 6 × 1024 × ( 24 × 60 × 60 )2
4π 2
= 4.2 × 107 m
(This is about six Earth radii from the Earth’s
surface.)
68
a E = V = 100 −2 = 500 V m−1
d 20 × 10
b i EPE = Vq = 100 × 6 × 10−6 = 6 × 10−4 J
ii EPE = Vq = 50 × 6 × 10−6 = 3 × 10−4 J
iii EPE = Vq = 0 × 6 × 10−6 = 0 J
c i F = Eq = 500 × 6 × 10−6 = 3 × 10−3 N
ii Since the electric field is the same everywhere
between the two parallel plates, F = 3 × 10−3 N
iii F = 3 × 10−3 N
d Work done = q ΔV = 6 × 10−6 × (100 − 0)
= 6 × 10−4 J
e Because the force is a constant, we can write:
ΔEPE = F × d
12
10 a E = V =
= 20 V m−1
d 60 × 10 −2
b Because the field is uniform, E = F × d = Eq × d
= 20 × 1.6 × 10−19 × 0.6 = 1.9 × 10−18 J
c EPE → KE of electrons → thermal energy of
atoms
300 V
8
9
b Since Y = k2 , then YX2 = k. So, if pairs of values
X
2
Y and X are multiplied together, they should all
come to the same constant, k.
c 302 × 22 = 1208, 72 × 42 = 1216, 34 × 62 = 1224,
19 × 82 = 1216, 12 × 102 = 1200, 8 × 122 = 1152.
So, to 2 s.f., all of these give k = 1200. This
confirms that X and Y are related by an inversesquare law.
Chapter 11: Electromagnetic
induction
Exercise 11.1 – Electromagnetic induction
1 a The needle on the galvanometer will kick to one
side of the scale and then return to zero.
b The needle on the galvanometer will kick to the
other side of the scale (same amount as before)
and then return to zero.
c The needle will not move.
d The needle will kick further.
e The needle will not move.
2
3
4
5
6
7
a The needle will show an increasing current
flowing in the coil.
b The galvanometer reading would be greater.
c The needle will show a negative current that
decreases.
a Towards the left (Fleming’s left-hand rule).
b The left-hand side of the conductor will have a
build-up of electrons, making it negatively
charged. The right-hand side of the conductor
will have lost some of its electrons, making it
positive charged.
c Yes, there will be an electric field across the conductor
because one end of the conductor is negatively
charged and the other end is positively charged.
d In equilibrium, the electrical force acting on each
electron is the same magnitude as the magnetic force
acting on each electron but in the opposite direction.
e FE + Fm = 0 ⇒ ΔVe = Bev
B v ⇒ ΔV = Blv
l
ε = B l v = 4.0 × 10−2 × 12 × 10−2 × 2.0 = 9.6 × 10−3
= 96 mV
ε = B l v = 0.14 × 5 × 10−2 × 60 × 10−2 = 4.2 × 10−3
= 4.2 mV
ε = Bl i θ ⇒ θ = sin −1 ε
b ε = ∝N
ε=
perpendicular to the area and the field lines.
Magnetic flux linkage is the product of the
magnetic flux and the number of turns of
conductor: N Φ
8 a N Φ = BAcos θ (N = 1)
b N Φ is a maximum when θ = 0 (i.e. the plane of
the coil is perpendicular to the field lines).
c When θ = 90°, N Φ = 0
9 N Φ = BAcos θ
= 4.87 × 10−5 × 91.4 × 55.0 × cos 24o = 0.22 Wb
10 a ε is the induced emf (measured in volts), d is
dt
the operator that signifies the rate of change, N is
the number of turns of the conductor and Φ is
the magnetic flux associated with the conductor
(measured in webers).
(
+
)
c An emf can be induced by changing A; this usually
means moving something so that an area A is
swept out in one second. Or, an emf can be
induced by changing B; this usually means
producing a magnetic field using an electromagnet
and changing the strength of the electromagnet by
changing the current flowing in it.
11 a The moving wire sweeps out an area, A, per
second. This induces an emf in the wire, according
to Faraday’s laws. Since the wire is connected to a
galvanometer, there is a complete electrical circuit
and so the induced emf causes a current to flow.
b The kinetic energy of the moving wire.
c Since the wire has lost kinetic energy, the speed
of the wire must be reduced.
2 s = 2 × 2 0 = 0.64 s
g
9 81
As the magnet falls, an emf is induced in the
copper tube. A current flows around the tube.
This current requires energy, which is provided by
some of the kinetic energy from the moving
magnet. So, the magnet moves slower and takes
longer to reach the bottom of the tube.
In this case, no current can flow in the copper tube
because there is not a complete electrical pathway.
Since no current flows, no kinetic energy is lost by
the magnet and so the magnet falls with the same
speed as it had with no tube present.
Lenz’s law: the direction of the induced emf is
such that its effect will be to oppose the flux
change that caused it.
Conservation of energy.
There is an induced emf in the coil, which causes
a current to flow.
Lenz’s law states that the direction of the induced
emf (and hence the direction of the current) is
such that its effect will be to oppose the original
flux change. This means that the left-hand side of
the coil will have to be a north pole, so that it
repels the north pole of the approaching magnet.
When the switch is pressed, the current in the
coil tries to change from zero to its proper value
in a very short time. The changing current during
12 a t =
b
( )
⎛
⎞
97 × 10 −3
= sin −1 ⎜
⎝ 0 3 × 25 × 10 −2 × 1 5 ⎟⎠
= 60° (2 s.f.)
Magnetic flux, Φ, is the product of the magnetic
flux density, B, and the perpendicular area through
which it passes, A:
Φ = BAcos θ , where θ is the angle between the
N
d( )
and Φ = BA, so
dt
c
13 a
b
14 a
b
15 a
ANSWERS
69
b
16 a
b
c
d
e
f
70
this time causes a changing magnetic field in the
solenoid. Faraday’s laws means that an induced
emf occurs in the solenoid itself. But Lenz’s law
tells us that the effect of this induced emf must be
to oppose its cause – i.e. to oppose the change
from 0 to 10 V. So, the actual voltage across the
solenoid will be less than 10 V. However, as soon
as there is a potential difference across the
solenoid, the rate of change of flux becomes less
and so the induced emf becomes less, the voltage
across the solenoid increases and so the current
flowing through the solenoid increases. In this
case, after about 100 ms, the voltage across the
solenoid will be 10 V and the current flowing
through the solenoid will be 1.0 A.
Twice the number of turns means twice the
induced voltage. So, the trace would take twice as
much time to rise to 1.0 A.
The iron rod will become magnetised, but the
direction of the magnetic field in the iron rod
will be changing periodically with the frequency
of the supply.
The changing magnetic field around the
aluminium ring will induce an emf in the ring,
which will make a current flow in the ring.
However, since the rate of change of the magnetic
field is changing (because the field itself is
changing sinusoidally) the induced emf – and
hence the current – is also changing.
The current in the aluminium ring creates a
magnetic field that opposes the magnetic field
that is causing it. So, two opposing magnetic fields
occur.
The opposing magnetic fields exert a repulsive
force on the ring, which pushes the ring upwards.
The change of magnetic flux experienced by the
ring is not as great as it had been when the switch
was first pressed. So, the opposing magnetic fields
do not exert as large a force on the ring. In fact,
the ring then floats because the upwards force
from the opposing magnetic fields is balancing
the weight of the ring.
For the ring with a cut through it, no induced
current can flow around the ring and so no
magnetic field is produced by the ring. This
means no repulsive forces will occur and the ring
will stay where it is.
Exercise 11.2 – Power generation and
transmission
1 a N Φ = NBA
B cos θ
b θ ωt
c N Φ = NBA
B
ωt
d(NBA
N
cos ωt )
d ε=−
= ωNBA
B
(ω )
dt
e ε 0 = ωNBA
B
π
f 90° or radians
2
2 a ε ∝ ω, so if ω is doubled then ε will double;
so, ε = 120 V
b ∝ N , so if N is reduced to 150 from 200 (i.e. it is
now ¾) then ε will now be ¾; so, ε = 45 V
3 a Zero
b Zero
c P = IV = ( I 0 i t ) × ( E 0 i ω t ) = I 0 E 0 sin 2 ω t
1
d Since the average value of sin2 ωt is , the mean
2
1
power = I 0 E 0
2
4 a The RMS value of an alternating current is the
value of a direct current that would have the same
power dissipation.
b I RMS =
I0
2
c V0 = VRMS × 2 = 240 × 1.414 = 339 V
5 Paverage = 0V0 = 600 × 10−3 × 5.0 = 3 W
6 a VRMS =
V0
= 18 = 13 V (2 s.f.)
2
2
1 (VRMS ) = 1 132
b P=
= 7.0 W
2 R
2 12
2
7 a See Figure A11.1.
Figure A11.1
b Alternating current
c Iron has a high value of relative permeability
(about 1000) and so is very receptive to magnetic
fields. Iron is a magnetically soft material, so it
will magnetise and de-magnetise quickly and
easily.
8
9
d To prevent eddy currents from flowing in the
core. Faraday’s laws state that an induced emf will
occur in any conductor present where there is a
changing magnetic flux. So, there will be an
induced emf in the iron core. Without laminating
(which increases its resistance) this induced emf
would make a current flow, which wastes energy
heating the core.
e The alternating current in the primary coil sets
up an alternating magnetic field in the iron core.
The iron core links the secondary coil with the
primary coil via the changing magnetic flux.
Faraday’s law states that the changing flux induces
an emf in the secondary coil. The value of the
induced emf depends on the number of turns on
the coils. If the number of turns on the two coils
are different, then the voltage across them will be
different.
a A step-up transformer has more turns on the
secondary coil than on the primary coil, so the
voltage across the secondary coil is higher than
the voltage across the primary coil.
b A step-down transformer has more turns on the
primary coil than on the secondary coil, so the
voltage across the secondary coil is lower than the
voltage across the primary coil.
c An ideal transformer does not waste any energy;
the power supplied to the primary coil is the same
as the power delivered by the secondary coil.
d This kind of transformer is called a parity
transformer. It is used to isolate the actual power
supply from the device that is going to be used. It
is a safety feature.
a Step-down transformer.
b
Vs N s
=
⇒ Vs
Vp N p
Vp ×
Ns
= 110 × 45 = 14 V
Np
360
c This may be used in several household appliances
where a working voltage of 14 V is required, such
as a power supply for a computer or a children’s
train set.
10 a
Vs N s
=
⇒ Ns
Vp N p
= 5 × 104 turns
Np ×
b i P = IV = 150 × 2 × 103 = 3 × 105 W
11
12
13
14
P = 3 × 105
ii I s =
= 0.75 A
Vs 400 × 103
P
20
= 83 mA
a i I= =
V 240
ii I s = 20 I p = 20 × 83 mA = 1.7 A
b Q = It = 1.7 × 10 × 60 = 1.0 kC
The power loss in transmission lines depends on
I 2. Reducing the current reduces energy losses. The
step-up transformer changes the output voltage of
the power station from about one thousand volts up
to about 400 kV. In turn, this makes the current in
the transmission wires 103 times smaller – and so the
power losses will be 106 times smaller than without
the transformer.
a A half-wave rectifier allows only half of one
complete cycle of an alternating current to flow.
The other half of the cycle (the part that is
reverse biased) produces zero current.
b A full-wave rectifier makes the negative parts of
an alternating current into positive currents as
well as allowing the positive parts of the current
to flow.
a A diode is a semiconductor device that only
allows current to flow in one direction.
b A diode is forward biased when it is placed in a
circuit in an orientation that allows current to
flow.
c See Figure A11.2.
Figure A11.2
15 a See Figure A11.3.
V
Time
Vs
= 250 × 400 kV
Vp
2 kV
Figure A11.3
ANSWERS
71
b See Figure A11.4.
V
Time
Figure A11.4
c Half of the wave is transmitted; the other half of
the wave produces a zero output.
16 a See Figure A11.5.
flow. The diodes between B and C and between
D and A are in reverse bias and so do not allow
current to flow. When the current is negative,
diodes between B and C and between D and A
are in forward bias and so allow the current to
flow; the current then flows from C to B, through
the load, and from D to A. Thus, the current
always flows through the load in the same
direction in both the negative and the positive
parts of the cycle.
b See Figure A11.7.
I
output current
Time
load
input current
Figure A11.7
Figure A11.5
c i See Figure A11.8.
b See Figure A11.6.
V
A
smoothed output
B
D
Time
non-smoothed output
C
Figure A11.6
c Discharging the capacitor must take longer than
the alternating current takes to flip from positive
to negative. Once the capacitor is fully charged, it
will begin to discharge. As it discharges, the
voltage across it slowly reduces. When the output
from the rectifier is positive again, the capacitor
charges again – but the voltage across the
capacitor starts above zero. So, the regular
charging and discharging of the capacitor occurs
in times shorter than it takes for the capacitor to
fully discharge. This keeps the voltage across the
capacitor – and hence the voltage of the output
of the rectifier circuit – more of a constant value.
17 a When the current is positive, it will flow from A
to B, through the load to D and from D to C. The
diodes between A and B and between D and C
are in forward bias and so allow the current to
72
load
Figure A11.8
ii See Figure A11.9.
V
smoothed output
non-smoothed output
Time
Figure A11.9
Exercise 11.3 – Capacitance
1 a Electrons will flow off plate X and towards the
positive electrode of the emf supply. This will leave
plate X positively charged.
b Electrons will flow on to plate Y, which will make
plate Y negatively charged.
c Because the current in a series circuit is the same
throughout, the positive charge on plate X will be
the same as the negative charge on plate Y. So, the
capacitor hasn’t gained any charge at all. Rather,
the charge has been separated.
d Energy.
2 a The material between the capacitor plates must be
an insulator. If it were a conductor then the charge
would be able to flow through the material and
prevent the plates from gaining charge.
b A large permittivity produces a large electric field
in the dielectric, which will help the build-up of
charge on the plates.
c The area of overlap of the plates, A; the distance
apart of the plates, d; the voltage across the two
plates, V
VA
dQ=ε
d
Q
e C= =εA
V
d
−2
−2
3 C = ε 0 A = 8 85 × 10 −12 × 5 × 10 × 1−3× 10 =
d
1 × 10
4.4 × 10−12 F (4.4 pF)
4 Commercially produced capacitors exploit all three
of the terms in this equation: C = ε A
d
They use a dielectric that has a high value of
permittivity (usually a kind of specially treated paper)
The thickness of the dielectric – which is the distance
apart of the two plates – is very thin; only the
thickness of a thin piece of paper.
The area of overlap of the plates is maximised by
having the plates rolled up in a cylinder. In this way a
large area can be contained in a small volume.
Cd = 0 5 × 1 × 10 −2
5 A=
= 3.3 × 10−7 m2
ε
1 5 × 104
6 a C ∝1
d
bC A
7 a i Electrons from plate A move towards the
positive side of the battery, leaving plate A
positively charged.
ii Electrons move to plate B and make it
negatively charged.
iii Electrons from plate D move to plate B.
iv D becomes positively charged.
v Electrons move on to plate E to make it
negatively charged.
bC=
Q
V
c V
2
d C = Q = 2Q
V
V
2
e The capacitance of two identical capacitors in
series is half that of one of the capacitors. This
suggest that capacitors in series have a combined
capacitance given by the equation:
1 = 1 + 1
C total C1 C 2
8
1 = 1 + 1 =1+1=3 2= 5
C total C1 C 2 4 6
12
12
Therefore, C total = 2 4 μF
9
a
1
C total
= 1 + 1 = 1+1= 2 1= 3
C1 C 2 4 8
8
8
Therefore, C total = 2 7 μF (2 s.f.)
b Q = C V = 2.7 × 10−6 × 6 = 1.6 × 10−5 C
c Across the 4 μF capacitor V =
Q 1 6 × 10 −5
=
= 4V
C
4 × 10 −6
−5
Across the 8 μF capacitor V = Q = 1 6 × 10−6 = 2 V
C
8 × 10
10 Since both capacitors will have the same amount
of charge on their plates, the voltage across each
capacitor will be dictated by the capacitance. The
bigger the capacitance, the smaller the voltage
required to separate charge. So, the 60 μF capacitor
will require half the voltage of the 30 μF capacitor.
The two voltages must add to be 12 V. Therefore the
voltage across the 60 μF capacitor must be 4 V (and
the voltage across the 30 μF capacitor must be 8 V).
11 a 6 V
b Q = C V = 5 × 10−6 × 6 = 3.0 × 10−5 C
c Total charge separated = 2 × 3.0 × 10−5
= 6.0 × 10−5 C.
−5
Q
So, Ctotal = = 6 × 10 = 10 μF
V
6
d Ctotal = C1 + C2
12 a For the two capacitors in parallel, their total
capacitance = 3 + 5 = 8 μF.
So, the voltage across these must be 4 V. The
voltage across the 3 μF capacitor is, therefore, 4 V.
b Q = C V = 3 × 10−6 × 4 = 1.2 × 10−5 C
1
1
c E = Q V = × 1.2 × 10−5 × 4 = 2.4 × 10−5 J
2
2
ANSWERS
73
13 a i Q = C V = 220 × 10−6 × 6 = 1.3 × 10−3 C
ii E = Q V = 1.3 × 10−3 × 6 = 7.8 mJ
1
iii Energy stored by capacitor = Q V
2
1
= × 1.3 × 10−3 × 6 = 3.9 mJ
2
b The capacitor stores only half of the energy used
by the battery. The other half of the energy is
transformed into thermal energy in the rest of the
circuit.
14 a See Figure A11.10. After one time constant, the
current will have become 1/e = 0.37 of its initial
value. From the graph this takes 20 s. So, τ = 20 s.
Current/μA
120
100
80
60
40
37 20
0
0
20
40
60
80
Time/s
100
120
d See Figure A11.12.
Q0
Q
0
− τ ⎞
⎛
e Q = Q0 1 − e RC ⎟
⎝
⎠
16 a Q = C V = 470 × 10−6 × 12 = 5.6 × 10−3 C
(5.6 mC)
b i τ = RC = 22 × 103 × 470 × 10−6 = 10.34 s.
So, after 10.34 s the charge on the plates
will be:
Q = 0.37 Q0 = 0.37 × 5.6 × 10−3 = 2.1 mC
ii At t = 2τ, Q = 0.372 × 5.6 × 10−3
= 7.7 × 10−4 C
iii Q = Q0 e
Figure A11.10
b The initial current is 100 mA, so the resistance in
the circuit must be
6
R =V =
= 60 kΩ
I 100 × 10 −6
c
15 a
b
c
τ
20
If the time constant is 20 s, then C = =
R 6 × 104
= 330 μF (2 s.f.)
The charge is given by the area under the graph
of current against time.
When t = RC, I = 0.37 I 0
Time required for I to fall to less than 1% of I 0
= 5RC.
See Figure A11.11.
I0
0
0
Figure A11.11
74
t
t
0
Figure A11.12
−t
τ
5 6 × 10 −3 × e
− 15.0
10.34
= 1.3 × 10−3 C
Exam-style questions
1
2
3
4
5
D
D
C
B
C
6 ε = − N d Φ so ε = −200 × gradient of graph
dt
09
= −30 V
= −200 ×
6
7 a The rate of magnetic flux linkage change is
increasing. This is due to the speed of the magnet
increasing, and the magnetic flux linked to the coil
increasing because the magnet is getting closer.
b The magnet is leaving the bottom of the coil and so
the change in magnetic flux is opposite to what it
had been when the magnet was approaching the coil.
c The speed at which the magnet is falling is
increasing all the time, so the rate of change of
magnetic flux is increasing and the induced emf is
increasing.
12 a See Figure A11.14.
Charge/μC
8
d Twice the number of turns would mean twice the
magnetic flux linkage, and so twice the induced
emf. So, the trace on the oscilloscope would show
twice the voltage.
a N Φ = 80 Wb
b and c See Figure A11.13.
Magnetic flux/×10–2 Wb
c
50
O
40
30
20
10
0
–10
–20
–30
–40
–50
O
M
M
M
50
100
150
200
Time/ms
O
−2
10 a
b
c
11 a
b
c
20
40
60
80
Time/s
100
120
−6
Current = gradient of graph = 60 × 10 = 5 μA
12
b V = I R = 5 × 10−6 × 100 × 103 = 0.5 V
Q 60 × 10 −6
=
= 1.2 × 10−4 F
V
05
d τ = RC = 100 × 103 × 1.2 × 10−4 = 12 s
An interesting corollary from this graph is:
The equation for Q against t is:
c C=
d Using Figure A11.13, ε 0 = −200 × 100 × 10−3
53 × 10
= 3.8 kV
a
b
c
d
0
Figure A11.14
Figure A11.13
9
70
60
50
40
30
20
10
0
Larger B means a larger ε.
Larger N means larger ε.
Smaller A means smaller ε.
Larger ω means larger ε (it also means that the
frequency of E increases.)
The two capacitors in parallel have a combined
capacitance given by:
Ctotal = C1 + C2 = 1 + 3 = 4 μF
So, the voltage across the 2 μF capacitor is 4 V.
Q = C V = 2 × 10−6 × 4 = 8 μC
1
1
E = Q V = × 8 × 10−6 × 4 = 1.6 × 10−5 J
2
2
The graph is a straight line that shows that Q ∝ V.
If the constant of proportionality is C, then
Q = C V.
28 × 10 −6
C = gradient =
= 3.5 μF
8
Energy stored by the capacitor is the area under
the graph.
1
For V = 4 V, the area = × 14 × 10−6 × 4
2
= 2.8 × 10−5 J
Q = Q0 e
− t
RC
= CV
Ve
− t
RC
Differentiating this to give the current:
I=
− t
dQ
= − CV e RC
dt
RC
Q
When t = 0, I = I 0 = − CV = − 0
RC
RC
From the diagram, ΔQ = Q0 giving Δt = RC.
So, the time constant of the circuit is where
the tangent to the curve at t = 0 meets the
time axis.
13 a Using the constant ratio rule and reading voltage
values from the graph with times of 0, 10 s, 20 s,
30 s, 40 s, 50 s:
5 = 1.67; 3 0 = 1.67; 1 8 = 1.64
64;; 1.1 = 1.57
57;; 0 7 = 1 75
30
18
1.1
07
04
So, there is evidence that this is an exponential
decay.
ANSWERS
75
Voltage/V
b See Figure A11.15.
0.37 × 5 = 1.85, so τ = 20 s.
1.85
6
5
4
3
2
1
0
c
0
20
40
60
80
Time/s
100
120
140
4 a
Figure A11.15
Chapter 12: Quantum and nuclear
physics
Exercise 12.1 – The interaction of matter with
radiation
−34
8
1 a E = hf = hc = 6 63 × 10 × 3 × 10 = 8 0 × 10 −26 J
λ
25
−26
8
0
×
10
=
= 5 0 × 10 −7 eV
1 6 × 10 −19
b
hc 6 63 × 10 −34 × 3 × 108 = 3 3 × 10 −20 J
b E = hf = =
λ
6 0 × 10 −6
−20
= 3 3 × 10−19 = 0 21eV
1 6 × 10
1
hc 6 63 × 10 −34 × 3 × 108
c E = hf = =
λ
623 × 10 −9
−17
= 3 2 × 10 −19 J = 3 2 × 10−19 = 2 0eV
1 6 × 10
1
−34
3 × 108
d E = hf = hc = 6 63 × 10 ×−10
λ
1 5 × 10
5 a
b
−15
= 1 3 × 10 −15 J = 1 3 × 10 −19 = 8.1 keV
1 6 × 10
−34
8
2 a E = hf = hc = 6 63 × 10 × −39 × 10 = 3 2 × 10 −19 J
λ
630 × 10
5 0 × 10 −3
16
b Number of photons s−1 =
−19 = 1.6 × 10
3
2
×
10
photons s−1.
3 a Einstein was able to show that light behaved like
particles.
b Until Einstein’s work with the photoelectric effect,
it was an accepted fact that radiation – and light in
particular – was made up of waves. The behaviour
76
c
of waves had been well understood for more than
100 years and so it was a very big step to consider
that light could show particle-like properties, and
that the accepted wave properties of light could
not explain the observations of the photoelectric
effect.
The photoelectric effect is the beginning of
quantum physics, which introduces the idea that
radiation can behave like particles.
i The photons of light do not have enough
energy for them to eject an electron free of the
metal surface. The metal cannot lose any of its
negative charge.
ii Over time, the electrons in the metal would
absorb energy from waves. Eventually, they
would have enough energy to break free of the
metal surface and the negatively charged plate
would discharge. Since this does not happen,
light cannot be behaving like a wave.
i A photon of light with this shorter wavelength
has enough energy to give to an electron so
that it can break free of the metal surface,
without having to wait for energy to build up.
ii This suggests that the energy needed to break
an electron free of the surface must be
contained in a small space over a small time – a
discrete packet of energy that we call a photon.
A rapid, complete discharge of the metal plate
suggests that there are numerous electrons
released, in turn suggesting a stream of photons.
The metal plate will lose electrons at an increased
rate, so the coulombmeter will show a faster
decrease in charge.
Higher intensity means that there are more
photons of light per second. This allows more
electrons per second to break free of the metal
surface, so the charge decreases at a faster rate.
It increases the number of photons per second. It
does not change the energy that each individual
photon has. So, the maximum kinetic energy of the
photoelectrons would not change.
−19
6 a f 0 = 3 2 × 1 6 × 10
= 7.7 × 1014 Hz
6 63 × 10 −34
c = 3 × 108
= 3.9 × 10−7 m (390 nm)
b λ=
f 0 7 7 × 1014
Voltage/V
b See Figure A11.15.
0.37 × 5 = 1.85, so τ = 20 s.
1.85
6
5
4
3
2
1
0
c
0
20
40
60
80
Time/s
100
120
140
4 a
Figure A11.15
Chapter 12: Quantum and nuclear
physics
Exercise 12.1 – The interaction of matter with
radiation
−34
8
1 a E = hf = hc = 6 63 × 10 × 3 × 10 = 8 0 × 10 −26 J
λ
25
−26
8
0
×
10
=
= 5 0 × 10 −7 eV
1 6 × 10 −19
b
hc 6 63 × 10 −34 × 3 × 108 = 3 3 × 10 −20 J
b E = hf = =
λ
6 0 × 10 −6
−20
= 3 3 × 10−19 = 0 21eV
1 6 × 10
1
hc 6 63 × 10 −34 × 3 × 108
c E = hf = =
λ
623 × 10 −9
−17
= 3 2 × 10 −19 J = 3 2 × 10−19 = 2 0eV
1 6 × 10
1
−34
3 × 108
d E = hf = hc = 6 63 × 10 ×−10
λ
1 5 × 10
5 a
b
−15
= 1 3 × 10 −15 J = 1 3 × 10 −19 = 8.1 keV
1 6 × 10
−34
8
2 a E = hf = hc = 6 63 × 10 × −39 × 10 = 3 2 × 10 −19 J
λ
630 × 10
5 0 × 10 −3
16
b Number of photons s−1 =
−19 = 1.6 × 10
3
2
×
10
photons s−1.
3 a Einstein was able to show that light behaved like
particles.
b Until Einstein’s work with the photoelectric effect,
it was an accepted fact that radiation – and light in
particular – was made up of waves. The behaviour
76
c
of waves had been well understood for more than
100 years and so it was a very big step to consider
that light could show particle-like properties, and
that the accepted wave properties of light could
not explain the observations of the photoelectric
effect.
The photoelectric effect is the beginning of
quantum physics, which introduces the idea that
radiation can behave like particles.
i The photons of light do not have enough
energy for them to eject an electron free of the
metal surface. The metal cannot lose any of its
negative charge.
ii Over time, the electrons in the metal would
absorb energy from waves. Eventually, they
would have enough energy to break free of the
metal surface and the negatively charged plate
would discharge. Since this does not happen,
light cannot be behaving like a wave.
i A photon of light with this shorter wavelength
has enough energy to give to an electron so
that it can break free of the metal surface,
without having to wait for energy to build up.
ii This suggests that the energy needed to break
an electron free of the surface must be
contained in a small space over a small time – a
discrete packet of energy that we call a photon.
A rapid, complete discharge of the metal plate
suggests that there are numerous electrons
released, in turn suggesting a stream of photons.
The metal plate will lose electrons at an increased
rate, so the coulombmeter will show a faster
decrease in charge.
Higher intensity means that there are more
photons of light per second. This allows more
electrons per second to break free of the metal
surface, so the charge decreases at a faster rate.
It increases the number of photons per second. It
does not change the energy that each individual
photon has. So, the maximum kinetic energy of the
photoelectrons would not change.
−19
6 a f 0 = 3 2 × 1 6 × 10
= 7.7 × 1014 Hz
6 63 × 10 −34
c = 3 × 108
= 3.9 × 10−7 m (390 nm)
b λ=
f 0 7 7 × 1014
−34
=
hc = 6 63 × 10 × 3 × 10
λ
6 5 × 10 −8
8
−18
= 3 06 × 10 −18 J = 3 06 × 10−19
= 19.1 eV
1 6 × 10 1
So, maximum KE of photoelectrons
= 19.1 − 3.2 = 15.9 eV.
7 KE of photoelectron = ½ m v2
= ½ × 9.1 × 10−31 × (4.5 × 105)2
= 9.2 × 10−20 J
So, photon energy
= 9.2 × 10−20 + (1.5 × 1.6 × 10−19)
= 3.32 × 10−19 J.
hc = 6 63 10 −34 3 × 108
= 6.0 × 10−7 m
E
3 32 10 −19
(600 nm).
So, λ =
E 4 2 × 1 6 × 10 −19
8 a f0 = =
= 1.0 × 1015 Hz
h
6 63 × 10 −34
b Ultraviolet
c Photon energy = hf = 6.63 × 10−34 × 2.2 × 1015
1 47 10 −18
9.19 eV
1 6 × 10 −19
So, KEmax = 9.19 − 4.2 = 4.99 eV
Therefore,
= 1.47 × 10−18 J =
2E = 2 × 4 99 × 1.6 × 10 −19
= 1.3 × 106 m s−1.
m
9 1 × 10 −31
9 a KEmax is the maximum kinetic energy of the
photoelectron emitted; h is Planck’s constant; f is
the frequency of the incident radiation; φ is the
work function of the metal surface, from which the
photoelectrons are emitted.
b See Figure A12.1.
v=
KEmax
0
f0
f
Figure A12.1
c Planck’s constant, h
d The threshold frequency, fo (see Figure A12.1), the
minimum frequency of radiation that will cause
the emission of photoelectrons from the metal
surface.
10 a i See Figure A12.2. The intercept on the x-axis
is at 7.5 × 1014 Hz.
So, the work function must be
φ = hf0 = 6.63 × 10−34 × 7.5 × 1014
= 4.98 × 10−19 J
4 98 =
.1 eV)
(=
16
ii See Figure A12.2. h is the gradient of the graph.
30 × 10 −20
= 6.7 × 10−34 J s
4 5 × 1014
(The accepted value for h is 6.63 × 10−34 J s).
b See Figure A12.2.
So, h =
KEmax/× 10–20 J
c Photon energy
45
40
35
30
25
20
15
10
5
0
30 × 10–20 J
0
f0 = 7.5 × 1014 Hz
2
4
6
8
10 12
4.5 × 1014 Hz
Frequency/× 1014 Hz
14
Figure A12.2
11 a Energy gained = e V
b p = mv, so p2 = m2v2 and KE
1 2 2 2
p2
= 2 mv m v =
= KE.
2m
2m
2meV
If KE = eV, then p = 2m
12 De Broglie proposed that, if light could exhibit
particle properties, then particles should be able to
exhibit wave properties. By considering mass and
energy, de Broglie was able to show that momentum
could be linked with a wavelength.
13 Davisson and Germer illuminated a zinc crystal surface
with electrons and measured the intensity of electrons
that were reflected at various angles.They observed a
pattern that showed the same characteristics as that of
a diffraction pattern, suggesting that the electrons were
exhibiting wave-like properties.
h
6 63 × 10 −34
14 a i λ = =
= 3.6 × 10−11 m
p 9 1 × 10 −31 × 2 × 107
140 × 103 = 38.9ms −1
60 × 60
h
6
63
×
1
0 −34 = 1.1 × 10−34 m
λ= =
p 0 16 × 38 9
ii 140 km hr−1 =
−34
iii λ = h = 6 63 × 10 = 9.5 × 10−64 m
p
70 × 1 0
ANSWERS
77
b i The electron of wavelength 3.6 × 10−11 m is
going to behave like a wave – because it will
have the opportunity to show diffraction effects.
The wavelength is too short for the cricket
ball to show diffraction effects within its
surroundings.
The human cannot show diffraction effects
within its surroundings, and so cannot show
wave behaviour.
ii For a human to show diffraction effects, their
momentum would have to be ~10−35 kg m s−1.
Humans do not move this slowly so they do
not show wave properties; they behave like
particles.
15 a i
λ =h =
p
h
=
2meV
6.63 × 10 −34
2 × 9.1 × 10 −31 × 1.6 × 10 −19 × 400
= 6.1 × 10−11 m
ii 100 is ¼ of 400 and λ ∝ 1
V
−10
So, new λ = 1.2 × 10 m
λ
10 −10
b sinθ = =
= 0.33
d 3 × 10 −10
So, yes, there should be diffraction effects for
either of these electrons.
16 a Pair production
b Q: 0 = 1 + −1 √
B: 0 = 0 + 0 √
L: 0 = −1 + 1 √
S: 0 = 0 + 0 √
c i Emin = Eelectron + Epositron
= 0.511 MeV + 0.511 MeV = 1.022 MeV
= 1.022 × 1.6 × 10−13 = 1.64 × 10−13 J
ii Using E = 3 kT ,
2
−13
T = 2E = 2 × 1.64 × 10−23 = 9 × 109 K
3k 3 × 1.38 × 10
17 a E = 2 × 938 MeV = 2 × 938 × 1.6 × 10−13
= 3.0 × 10−10 J
b i Pair production requires the intervention of a heavy
nucleus. During the production process, some of
the energy of the photon is given to the nucleus.
Excess energy is transformed into KE of the two
particles produced. So, the actual energy required
by the photon is greater than this calculated value.
ii Most will be transformed into KE of the proton
and anti-proton, a small amount will be
transformed into KE of the nearby heavy nucleus.
18 a Annihilation
b i E = 2 × 938 MeV = 2 × 938 × 1.6 × 10−13
= 3.0 × 10−10 J
−34
3 × 108
ii E = hc ⇒ λ = hc = 6 63 × 10 ×−10
λ
E
3 0 × 10
−16
= 6.6 × 10 m
c This photon will be in the gamma ray section of
the electromagnetic spectrum.
19 a Standing/stationary wave
b λ = 2π r
2π rn
c λn =
n
d λ = h ⇒ p = h = hn
p
λ
πrn
p = mv So, the angular momentum, mvrn = h
2π
Therefore, angular momentum is quantised in
units of h
2π
2
2
2
mv
k
ke
20 a
= 2 ⇒ 1 mv 2 = ke
r
2
2r
r
2 2
2
b mvr = nh ⇒ m 2v 2r 2 = n h2 and 1 mv 2 = ke
2π
2
2r
4π
2
So, m 2v 2 = ke m
r
2
2 2
2 2
Therefore, ke m = n h2 2 ⇒ r = n2 h 2
r
4π r
4π kke m
c i
KE + PE =
ke 2 − kke 2 = − ke 2
2r
r
2r
2
n 2h 2
ii Etotal = − ke and r = 2 2
2r
4π kke m
2 4
2
So, Etotal = − 2π e2 mk
h n2
All the terms in this equation are constants
except n, which is an integer; the total energy
is quantised.
iii for n = 1,
2 4
2
E total = 2π e 2mk
h
=
2π ×
(
×
= 2.17 × 10−18 J
78
)
(
4
(
× 9.109 × 10 −31 × 8.988 × 109
×
)
2
)
2
2.17 × 10 −18 =
3.6 eV
1.602 × 10 −19
21 a Ψ 2 gives the probability density function / the
probability for finding the electron in a given
volume, ΔV.
b i At the value of r for which the value of Ψ 2 is
largest.
ii The value of Ψ 2 is not zero at values of r
beyond the edge of the atom. This means that
the probability is greater than zero of finding
the electron there. Since the value of Ψ 2 is
quite small, the probability of finding the
electron outside the atom is also small.
22 a Δp Δx ≥ h
4π
=−
b i
−11
λ = 25 × 10 6.3 × 10−11 m
4
−34
p = h = 6 63 × 10−11 1.1 × 10−23 m
λ 6 3 × 10
iii That the uncertainty in the position of the
electron is infinite.
iv This suggests that the free electron occurs in
all positions in space.
c ΔE Δt ≥ x ≥ h
4π
d i The probability density function, Ψ 2, for the
electron in the hydrogen atom shows a
non-zero value at distances greater than the
edge of the atom.
ii The total energy of the electron in the
ground state is −13.6 eV. So, for it to exist
outside the atom, its uncertainty in energy
must be greater than or equal to 13.6 eV:
ii
h =
6 63 × 10 −34
4π ΔE 4π × 13 6 × 1 6 × 10 −19
= 2.4 × 10−17 s
iii To observe something, we need some kind of
interaction to occur and for that interaction to
occur in a time long enough for us to observe
it. The time of 2.4 × 10−17 s is too small for
us to be able to observe anything.
b Δt =
h =
6 63 10 −34
= 3.2 × 10−17 s
4πΔE 4π 10 2 × 1 6 × 10 −19
24 a KE = 3 kT = 3 × 1 38 × 10 −23 × 1.0 × 107
2
2
= 2.1 × 10 −16
(
)
2
×
e2
9
b E p = k = 9 × 10 ×
= 7.7 × 10−14 J
r
3 0 × 10 −15
−16
c The factor is: 2.1 × 10 −14 ≈ 1000 ≈ 360 times
2 73
7 7 × 10
smaller
d No. The kinetic energy is too small to overcome
the electrical potential energy.
e The probability density function, Ψ 2, for the
proton is non-zero at a distance smaller than 3 fm
from the other proton. This means that, although
the probability of existing there is very small, it is
not zero. Thus, the proton can approach to within
3 fm of the other proton and fuse.
f Quantum (mechanical) tunnelling.
g Uses of this process include: scanning tunnelling
microscopes, quantum tunnelling composites for
the screens of smart phones and tablets and in
monitors and cameras.
25 a i Ψ2 becomes smaller as ΔE increases Ψ 2 ∝ e − ν E
ii Ψ2 becomes smaller as d increases Ψ 2 ∝ e − d
iii Ψ2 becomes smaller as m increases Ψ 2 ∝ e −
m
b See Figure A12.3.
Ψ
2
Δt =
23 a For n = 1, E = −13.6 eV.
13.6
For n = 2, E = − 2 = −3.4 eV
2
So, ΔE = 10.2 eV
0
d
Distance through energy barrier
Figure A12.3
Exercise 12.2 – Nuclear physics
1
a i V = x3
m
ii ρ = 3
x
b i V = 8x 3
8 x = 2x
8
m
m
iii ρ = 3 = 3
8x
x
ii l
3
ANSWERS
79
c i V = 27x 3
5
7 x = 3x
27m = m
iii ρ =
27x 3 x 3
d i V = Ax 3
ii l
ii l
3
b r = r 0 3 197 = 1.2 × 10−15 × 5.82 = 7.0 × 10−15 m
c The first minimum will occur when
3.1 × 10 −15 ⎞
λ
−1 ⎛
sin θ = 1.22 ⇒ θ = sin ⎜ 1.22
⎟ = 33°
⎝
d
7.0 × 10 −15 ⎠
(If the correction of 1.22 for a circular aperture is
−15
⎞
−1 ⎛ 3.1 × 10
= 26°)
not applied, then θ = sin ⎜
⎝ 7.0 × 10 −15 ⎟⎠
3
Ax
Am = m
iii ρ =
Ax 3 x 3
e i l 3A
ii ρ is a constant; it does not vary with A.
2
a i
ii
r = r 0 3 12 = 1.2 × 10−15 × 2.29
= 2.7 × 10−15 m
−27
ρ = m = 12 × 1 7 × 10 3 = 2.5 × 1017 kg m−3
V 4π
×
3
(
)
iii r = r 0 3 235 = 1.2 × 10−15 × 6.17
= 7.4 × 10−15 m
3
r = 9 × 10 ×
9
4
80
6
−27
iv ρ = m = 235 × 1 7 × 10 3 = 2.4 × 1017 kg m−3
V 4π
×
3
b Their densities are approximately the same.
a i Most of the space taken up by an atom is
empty.
ii There is a small, very dense charged region
within the atom.
iii The charged region in the atom is positively
charged.
2e × 79e ⇒
b i KE = k
r
(
)
2 × 79 ×
(
×
)
2
7 7 × 1 6 × 10 −13
= 2.95 × 10−14 m
ii r = r 0 3 197 1.2 × 10−15 × 5.82
= 7.0 × 10−15 m
iii The closest approach distance is about four
times the radius of the nucleus. So, because it
is within a factor of ten, it is a reasonable
estimation for the radius of the nucleus of
gold.
a The strong nuclear force.
b With higher energy, alpha particles were able to
approach the nuclei to within about 3 fm. At this
distance, the strong nuclear force overcomes any
electrical repulsion from the nucleus and the
alpha particles are absorbed by the nuclei.
a Electrons are members of the lepton family of
particles. Leptons do not feel the strong force, so
the strong force cannot be responsible for
modifying the electrical forces on the electrons.
7
d At much smaller electron energies, the de Broglie
wavelength will be larger than the diameter of the
nucleus. It is not possible to calculate sin−1 of a
number greater than one. The electron
wavelength is too large to give observable
diffraction effects from a gold nucleus.
a Gamma radiation emitted by energetic nuclei
exhibits a spectrum of discrete lines in a similar
way to the emission spectrum of light from
excited atoms. This suggests that the nucleus must
exist in discrete energy levels. When the nucleus
moves from one of these energy levels to a lower
energy level, it emits a gamma ray of energy equal
to the difference between the two energy levels.
b KEmax must be the difference in mass-energy
between the parent nucleus and the daughter +
α-particle.
So: (211.946 − (207.937 + 4.0015)) u
= 0.0075 × 931.5 MeV = 6.99 MeV
c Because of the conservation of momentum, the
208
81Tl nucleus must recoil with some velocity.
This means it must take a small amount of the
6.99 MeV, leaving the α-particle with slightly less.
−34
8
a i λ = hc = 6 63 × 10 × 3 −×1610 = 0.1 × 10−11 m
E
59 × 1 6 × 10
−34
8
ii λ = hc = 6 63 × 10 × 3 ×−1610 = 0.2 × 10−11 m
E
102 × 1 6 × 10
−34
8
iii λ = hc = 6 63 × 10 × 3 −×1610 = 2.9 × 10−11 m
E
43 × 1 6 × 10
b
241
95
Am →
237
93
Np + 42 α
c When the 241
95 Am nucleus decays, it will change
237
from a single energy state to become a 93 Np
nucleus in one of the three possible energy states
shown in Figure 12.7. The energy carried away by
the α-particle must take on one of three values.
90
0 −
a 90
38 Sr → 39Y + −1 β + ve
b The total energy available is equal to the massenergy difference between the parent nucleus and
the three product particles. This is a constant. This
energy is shared between the α-particle and the
ν e , so the electron has a spectrum of energies
between zero and the total energy available.
c The conservation of electron lepton number
suggests that there must be an anti-lepton
produced as well as the α-particle, since the
parent nucleus had zero electron lepton number
count. Since the conservation of charge shows
that this anti-lepton must have zero charge, it
must be an electron anti-neutrino.
64
0
64
0
9 a Cu + −1 e → 28 Ni + 0 v e
b The conservation of electron lepton number
requires the right-hand side of the equation to
have an electron lepton number of one. The
conservation of charge requires the right-hand
side of the equation to have a charge of 28. So,
the particle emitted with the Ni nucleus cannot
have any charge; it must be an electron neutrino.
10 a The mass of the neutron is larger than the
combined masses of the particles into which it
decays. This allows the neutron to decay whether
it is isolated from a nucleus or not.
b The mass of a proton is less than the mass of a
neutron. This means that it cannot decay into a
neutron, a positron and an electron neutrino
without intervention from a nucleus. When the
proton is inside a nucleus, it can ‘borrow’ some
energy from the nucleus, which it can then use to
produce the decay products.
11 a i Half-life: the average time it takes for half of
the nuclei present to decay. Or, the time it
takes for the activity of a given sample of
radioactive material to halve.
ii Decay constant, λ, is the probability that a
given nucleus will decay in a given period of
time – usually one second.
iii Activity is the number of decay events
occurring per second.
ln ( ) 0.693
b λ=
=
t1
t1
8
2
2
c i
λ=
ln (
t1
)=
0.693
1.25 × 109 × 3.15 × 107
2
= 1.8 × 10−17 s−1
ii
λ=
ln (
t1
)=
0.693 = 1.2 × 10−3 s−1
9.96 × 60
2
iii λ =
ln (
t1
)=
0.693
5.27 × 3.15 × 107
2
= 4.2 × 10−9 s−1
ln ( 2)
12 a t 1 =
= 0.693 −5 = 5.4 × 104 s
λ
1.28 10
2
5 4 × 104
= 15.0 hours.
60 × 60
i After 15 hours, the activity of a sample of
24
11 Na will be ½.
ii After 30 hours, the activity of a sample of
24
11 Na will be ¼.
b 5.4 × 104 s =
13 λ =
ln (
t1
)=
0.693
= 7.9 × 10−10 s−1
7
28 × 3.15 × 10
2
N=
3 10 −3 × 6 02 1023
= 1.3 × 1019 nuclei.
137
Therefore, A = λ N = 7.9 × 10−10 × 1.3 × 1019
= 1.0 × 1010 Bq
14 a The count rate that the GM tube has measured
has the value of the background radiation count
rate subtracted from it, so that the corrected
count rate (CCR) is due only to the sample.
b CCR =
CCR 0
− λt
( )
⎛
⎞
ln ⎜ CCR ⎟ − ln
⎝ CCR 0 ⎠
⇒ −λ =
=
t
20 × 60
= 1.8 × 10−4 s−1
0.693
c t1 =
= 3.85 × 103 s = 1.07 hours.
1 8 × 10 −4
2
d This method has two major problems with a
half-life as long as this:
• The activity of a sample is likely to be very small.
• The activity may not be significantly different
from (or smaller than) the background activity.
So, trying to make a corrected count rate would
be meaningless.
ANSWERS
81
There will not be any appreciable change in the
activity of the sample over a period of time in
which the measurements might be made. So, a
calculation of this kind would produce a
ln (1) = 0, meaning that t 1 cannot be calculated.
2
15 Measure the mass of the sample accurately. Use the
mass of the sample and the relative atomic mass
of the nuclide to calculate the number of nuclei
present. Measure the corrected count rate with
a GM tube placed 1 cm away from the sample.
Assuming that the sample emits its decay products in
all directions, find the fraction of the area of the GM
tube window to the area of a sphere of radius 1 cm.
Divide the corrected count rate by this fraction to
get the activity of the sample.
Find the decay constant using λ = A
N
7
8
Determine the half-life by using t 1 = 0.693
λ
2
16 A = Ao e − λt
⎛ ⎞
⎛ ⎞
− ln ⎜ A ⎟
− ln ⎜ A ⎟
⎝ A0 ⎠
⎝ A0 ⎠
⇒t =
= t1 ×
−λ
−
0
.693
2
− ln (0..92)
−0.693
= 690 years.
= 5730 ×
B
A
B
B
D
a Work function: the minimum amount of energy
that an electron requires to break free from the
surface.
b Threshold frequency: the minimum frequency
of radiation that will cause the emission of
photoelectrons.
c Photons have energy,
−34
8
E = hc = 6 63 × 10 × −39 × 10
λ
450 × 10
−19
= 4 4 × 10 −19 J = 4 4 × 10−19
= 2 75 eV
1 6 × 10 1
82
(
)
( )
Exam-style questions
1
2
3
4
5
6
So, the maximum KE is 2.75 − 1.5 = 1.25 eV
(1.3 eV to 2 s.f.)
−3
d Number of incident photons s−1 = 3 10 −19
4 4 × 10
15 −1
= 6.8 × 10 s
1
Only of these eject photoelectrons,
8
so current = 1 × 6 8 × 1015 × 1 6 × 10 −19 = 0.14 mA
8
a KEmax
hc − φ = 6 64 × 10 −34 × 3 × 108 −
×
×
=
λ
260 × 10 −9
= 8.6 × 10−20 J
b The right-hand terminal will have to be the
negative terminal in order to inhibit the flow of
electrons to it.
−20
c Minimum terminal voltage = 8 6 × 10−19 = 0.54 V.
1 6 × 10
a To remove atoms that might obstruct the
movement of electrons from the graphite crystal
to the fluorescent screen.
b The pattern observed on the fluorescent screen is
a diffraction pattern. Displaying interference in
this way is a property that we associate with wave
behaviour.
p2
⇒ p = 2meV
c i KE = eV =
2m
h
ii λ = h =
p
2meV
d λ = s sin θ and θ = tan−1 3 5 = 9.9°
20.0
h
So, s = λ =
sinθ sinθ × 2meV
6.63 × 10 −34
=
9
sin 9.9° × 2 × 9.1 × 10 −31 × 1.6 × 10 −19 × 1000
= 2.3 × 10−10 m
a ΔE = mgΔh = 70 × 9.81 × 3 = 2.1 kJ
h = 6 63 × 10 −34
b Δt =
= 2.5 × 10−38 s
4π ΔE 4π × 2.1 × 103
c This time is insufficient for the athlete to be able
to move over the distance of 3 m.
10 a λ =
ln (
t1
2
)=
0.693
= 5.1 × 10−11 s−1
433 × 3.15 × 107
−6
10
A = λ N ⇒ N = A = 5 × 10 × 3.7−11× 10
λ
5 1 × 10
= 3.6 × 1015 nuclei.
ii M = 3.6 × 1015 × 241 × 1.67 × 10−27
= 1.4 × 10−9 kg (= 1.4 μg)
c In one year, the activity of the sample will not
drop appreciably because its half-life is so long;
the advertised activity will be very close to the
actual activity.
b i
11 a λ =
ln (
t1
)=
0.693
= 5.1 × 10−11 s−1
433 × 3.15 × 107
2
−6
b N = 11 × 10 × 6 02 × 1023 = 6.7 × 1016 nuclei
99
c A = λ N = 5.1 × 10−11 × 6.7 × 1016
= 3.4 × 106 Bq
Option A: Relativity
Exercise A.1 – The beginnings of relativity
1 a Newton’s first law of motion: A body at rest, or in
uniform motion, will remain at rest, or in uniform
motion, unless it is acted upon by a resultant force.
b A frame of reference is a co-ordinate system and a
means of measuring time that can provide a value
for the position and time for a particle, anywhere
and at any time.
c An inertial frame of reference is a frame of
reference in which Newton’s first law of motion is
obeyed.
2 a 6 m s−1
b No. In the frame of reference of the moving bus,
Ellie (and the two boys) are stationary. This would
be the case whatever the speed of the bus is.
c Oscar sees the bus travelling forwards at 15 m s−1
and the chocolate bar travelling forwards at
15 + 6 = 21 m s−1.
d Yes. The bus is moving relative to Oscar in his
frame of reference. So, the speed of the bus will
affect how fast Oscar sees the chocolate bar
moving.
e 3 m.
f t = s = 3 = 0.5 s
v 6
g s = vbus t + vchoc t = (15 + 6) × 0.5
= 21 × 0.5 = 10.5 m.
h Yes. Because Oscar and Ellie measure the same
time for the event to occur, their laws of physics
explain their observations in the same way.
i Newton said that whatever the frame of reference
is, an observer must see the same event occurring
in the universe as any other observer.
j The laws of physics are the same for all inertial
frames of reference.
3 a i v = vcar − vtruck = 18 − 12 = 6 m s−1 in the same
direction as the velocity of the truck.
ii v = vtruck − vcar = 12 − 18 = −6 m s−1 in the
direction opposite to that of the car.
b i v = vnitrogen − voxygen = 500 − (−438)
= 938 m s−1 in a direction upwards.
ii v = voxygen − vnitrogen = −438 − 500
= −938 m s−1 in a direction downwards.
4 a i The observer would see an electrical force,
2
F k e 2 , where r is the separation of the two
r
electrons and e is their charge.
ii The force on e1 is vertically upwards, away
from e2
b i Yes
ii Yes
c The two electrons and the observer are in the same
frame of reference in a and b. In both cases there is
no relative motion between the electrons and the
observer or between one electron and the other
electron.
d i Inside the wire there are the same number of
electrons as there are positive ions, so the
overall charge on the wire is zero.
ii Although the electron is negatively charged,
the wire is electrically neutral, so there is no
force between them.
iii A magnetic field.
iv A charged particle would experience a
magnetic force (F B q v sin θ) due to the
magnetic field around the current carrying
wire only if the charged particle has a velocity
component that is perpendicular to the current.
Since the electron, e1, is not moving, there
cannot be a magnetic force on e1.
ANSWERS
83
−6
10
A = λ N ⇒ N = A = 5 × 10 × 3.7−11× 10
λ
5 1 × 10
= 3.6 × 1015 nuclei.
ii M = 3.6 × 1015 × 241 × 1.67 × 10−27
= 1.4 × 10−9 kg (= 1.4 μg)
c In one year, the activity of the sample will not
drop appreciably because its half-life is so long;
the advertised activity will be very close to the
actual activity.
b i
11 a λ =
ln (
t1
)=
0.693
= 5.1 × 10−11 s−1
433 × 3.15 × 107
2
−6
b N = 11 × 10 × 6 02 × 1023 = 6.7 × 1016 nuclei
99
c A = λ N = 5.1 × 10−11 × 6.7 × 1016
= 3.4 × 106 Bq
Option A: Relativity
Exercise A.1 – The beginnings of relativity
1 a Newton’s first law of motion: A body at rest, or in
uniform motion, will remain at rest, or in uniform
motion, unless it is acted upon by a resultant force.
b A frame of reference is a co-ordinate system and a
means of measuring time that can provide a value
for the position and time for a particle, anywhere
and at any time.
c An inertial frame of reference is a frame of
reference in which Newton’s first law of motion is
obeyed.
2 a 6 m s−1
b No. In the frame of reference of the moving bus,
Ellie (and the two boys) are stationary. This would
be the case whatever the speed of the bus is.
c Oscar sees the bus travelling forwards at 15 m s−1
and the chocolate bar travelling forwards at
15 + 6 = 21 m s−1.
d Yes. The bus is moving relative to Oscar in his
frame of reference. So, the speed of the bus will
affect how fast Oscar sees the chocolate bar
moving.
e 3 m.
f t = s = 3 = 0.5 s
v 6
g s = vbus t + vchoc t = (15 + 6) × 0.5
= 21 × 0.5 = 10.5 m.
h Yes. Because Oscar and Ellie measure the same
time for the event to occur, their laws of physics
explain their observations in the same way.
i Newton said that whatever the frame of reference
is, an observer must see the same event occurring
in the universe as any other observer.
j The laws of physics are the same for all inertial
frames of reference.
3 a i v = vcar − vtruck = 18 − 12 = 6 m s−1 in the same
direction as the velocity of the truck.
ii v = vtruck − vcar = 12 − 18 = −6 m s−1 in the
direction opposite to that of the car.
b i v = vnitrogen − voxygen = 500 − (−438)
= 938 m s−1 in a direction upwards.
ii v = voxygen − vnitrogen = −438 − 500
= −938 m s−1 in a direction downwards.
4 a i The observer would see an electrical force,
2
F k e 2 , where r is the separation of the two
r
electrons and e is their charge.
ii The force on e1 is vertically upwards, away
from e2
b i Yes
ii Yes
c The two electrons and the observer are in the same
frame of reference in a and b. In both cases there is
no relative motion between the electrons and the
observer or between one electron and the other
electron.
d i Inside the wire there are the same number of
electrons as there are positive ions, so the
overall charge on the wire is zero.
ii Although the electron is negatively charged,
the wire is electrically neutral, so there is no
force between them.
iii A magnetic field.
iv A charged particle would experience a
magnetic force (F B q v sin θ) due to the
magnetic field around the current carrying
wire only if the charged particle has a velocity
component that is perpendicular to the current.
Since the electron, e1, is not moving, there
cannot be a magnetic force on e1.
ANSWERS
83
e i A magnetic force.
ii Around the current-carrying wire there is a
magnetic field. In the region where e1 is, this
magnetic field is directed into the page (using
the right-hand grip rule).
Since e1 is moving perpendicularly to this
magnetic field, there will be a magnetic force
on the electron given by
F = μ 0 I × e × v = μ 0 Iev
2π r
2π r
iii Using Fleming’s left-hand rule, this magnetic
force will be directed vertically downwards
towards the wire.
f i Although there is a magnetic field around the
wire, e1 has no velocity component
perpendicular to the magnetic field; e1 is
stationary with respect to the moving electrons
in the wire and so there is no magnetic force
on e1.
ii If the force on e1 is to be directed towards the
wire, then the wire must have gained a positive
charge so that the positively charged wire can
attract the negatively charged electron
towards it.
iii The electrons in the wire, the observer and e1
are all stationary with respect to each other.
The positive ions in the wire are not; they are
moving in the opposite direction to the
electrons. If the motion of the positive ions
causes them to become closer together, then
there will be more positive charge per unit
length of the wire than negative charge caused
by the electrons in the wire. This will make
the wire positively charged and explain why
the observer sees the force between the wire
and e1.
5 a i An electrical force.
2
ii F k e 2 , where e is the charge on a proton, r is
d
the separation of the two protons, and k is the
Coulomb constant.
iii Away from the other proton.
b i Yes. The two protons are positively charged and
so must experience a Coulomb force between
them.
84
2
k e 2 , where e is the charge on a proton,
d
r is the separation of the two protons, and k is
the Coulomb constant.
iii Away from the other proton.
iv Yes. A moving proton is a flow of charge. So,
the moving protons will form a current in the
same direction.
v The magnetic force on one proton will be
directed towards the other proton – in the
opposite direction to the electrical force.
The observer in S sees only an electrical force
repelling the protons. The observer in S’ sees the
same electrical force but also a magnetic force in
the opposite direction, which partly counteracts
the electrical force. The observers seem to be
seeing the event differently.
If both observers are to see the event similarly, they
must both see the same overall force acting on the
two protons. This suggests that the electrical force
observed in S’ must be larger than the electrical
force observed in S, so that the sum of the
electrical force and the magnetic force equals the
electrical force observed in S.
The battery pulls electrons from the right-hand
plate, leaving it positively charged. These electrons
are fed to the left-hand plate, which makes it
negatively charged.
It will travel an extra distance of vt.
ΔA xvt
The charge flowing onto the extra area in time t is
ΔQ IIt .
ΔQ
= It = I
So, the charge density is σ=
ΔA xvt xv
E=σ = I
ε 0 ε 0 xv
ii F
c
d
6 a
b
c
d
e
f x=
g Δ
ΔA
I
, so B = μ0 I = μ0
x
ε 0vE
dvt
h ΔΦ = B νdt = B νd
Δt
t
i E = V = Bvd = Bv
d
d
I =μ 0ε0Ev
I
ε 0 vE
j E = Bv and B = μ0ε0Ev
So, v 2 = 1 ⇒ v = 1
μ0 ε 0
μ0ε 0
k v=
1 =
μ0 ε 0
7
a i x′ = γ (x − vt)
ii x = γ (x′ + vt)
⎛ vx ⎞
iii t ′ = γ ⎝ t − 2 ⎠
c
1
4π ×10 × 8.85 × 10 −12
−7
= 3 × 10 m s
a All of the terms referring to the parallel plates
have cancelled out. This suggests that
electromagnetic waves do not require atoms for
their propagation; hence they can propagate
through a vacuum.
b Yes. The equation for the speed of the waves
depends on the permeability and the permittivity
of the medium through which the waves
propagate. A medium with larger values of either
of these will cause the speed of light to be slower.
c No. There is no term in the equation for the
speed of motion of the parallel plates.
d No. There is no term in the equation for the
speed of motion of the medium.
e The Galilean transformation ideas had implied
that the observed speed of light would be
dependent on the speed of an observer. Maxwell
showed that the speed of light was independent of
the speed of the observer.
f Lorentz suggested that, if time and distance were
not absolute (i.e. that they themselves depended
on the motion of an observer), then the
established Newtonian mechanics could become
compatible with Maxwell’s speed of light. This
would require a new set of transformations that
altered the observed values of time and distance
according to the speed at which the observer was
moving: the Lorentz transformations.
8
2
−1
3
iv t = g ⎛ t ′ + vx2 ′ ⎞
⎝
c ⎠
b When the clocks in both frames of reference
show zero, (i.e. t = t′ = 0) the origins of the two
frames of reference coincide (i.e. x = x′ = 0).
a The time for light to travel from the teacher’s
waving hand to each of the students is so small
that it will not significantly alter perception of
when t = 0.
b Δt − 100 − 10
= 3 × 10−7 s
3 × 108
(Since a human’s reaction time is about 0.2 s, this
delay is insignificant.)
4
c Each student should set their stopwatch to a value
of t = x and start it when they see the teacher
c
wave her hand.
a x1’ = γ ( x1 vt1 ) and x2’ = γ ( x 2 vt 2 )
b Δx’ = x2’ − x1’ = γ ( x 2
= γ (Δ
5
v
(
c2
a See Figure AA.1.
a To detect the effect of the aether on the speed of
light.
b The results from the experiment were null. This
suggested that there was:
i no aether
ii no difference measured for the speed of light
in one direction compared to the speed of
light in another direction.
c The second postulate of relativity: that the speed
of light is the same for all inertial observers.
γ
Exercise A.2 – Lorentz transformations
1
γ ( Δx
Δx
vt1 )
v Δt )
v ⎞
⎛
v ⎞
⎛
a t1’ = γ ⎝ t1 − 2 x1 ⎠ and t2’ = γ ⎝ t 2 − 2 x 2 ⎠
c
c
v ⎞
⎛
v ⎞
⎛
b Δt ’ = t2’ − t1’ = γ ⎝ t 2 − 2 x 2 ⎠ − γ ⎝ t1 − 2 x1 ⎠
c
c
⎛
= γ ⎝ (t 2
6
))
(
vt 2 ) − γ ( x1
60
50
40
30
20
10
0
t1 )
0
0.2
2
0.4
1
)⎞⎠
0.6
v/c
γ ⎛ νt − v2 νx ⎞
⎝
⎠
c
0.8
1
1.2
Figure AA.1
b At about v = 0.4c
c i 1.0001
ii 1.01
ANSWERS
85
iii
iv
d i
ii
iii
b i
1.33
5.26
0.58c
0.71c
0.895c
ii
a u = v + u ′ = 0.7c + 0 3c = c = 0.83c
1 + vu2 ′ 1 + 0.7c ×2 0 3c 1.21
c
c
v + u ′ = 0.7c + c = 1.7c = c
b u=
which is what
1 + vu2 ′ 1 + 0.7c2 × c 1.7
c
c
the second postulate says: the speed of light is the
same for all observers in all inertial frames of
reference.
8 u = v + u ′ = 0.7c + 0 6c = 1.3c = 0.92c
1 + vu2 ′ 1 + 0.7c ×2 0 6c 1.42
c
c
v
+
u
0
.
5
c
+
0 5c = c = 0.8c
′
=
9 u=
vu
0
.
5
c
×
0 5c 1.25
′
1+ 2
1+
2
c
c
10 a Invariant: a quantity is the same for all observers
in all inertial frames of reference.
b Δs2 = (cΔt)2 − Δx2
7
⎛ ⎛
vx ′ ⎞
vx ′ ⎞ ⎞
⎛
c cΔt = c (t2 − t1) = c ⎜ γ t 2′ + 22 − γ t1′ + 21 ⎟ ⎟
⎝
⎝ ⎝
c ⎠
c ⎠⎠
= γ c Δt ′ + v Δx ′
c
(
So, (cΔt)2 = γ
2
(
)
c Δt ′ + v Δx ′
c
)
2
And, Δx = (x2 − x1) = γ ( ′ + ′ ) γ ( ′ + ′ )
= γ(Δ ′ Δ ′)
So, Δx2 = Δ2 ( Δx ′
2
Therefore, Δs2 = (cΔt)2 − Δx2
2
v Δx ′
2
= γ c Δt ′ + c
− γ 2 (Δx ′
= γ2 ×
v2Δ x 2
t 2 22cc Δt ′ v Δx ′
Δx ′ 2
c
c2
(
⎛ 2
⎜⎝ c
v Δt ′ )
)
⎛
= γ 2 ⎜ c2
⎝
(
v Δt ′)2
⎞
2Δx vtt − v 2 Δt ′ 2 ⎟
⎠
2
⎞
⎛
⎞
v 2 Δt ′ 2 − 1 − v2 ⎟ Δ x ′ 2 ⎟
⎝
⎠
c ⎠
)
= ( c Δt ′ ) Δ x ′
2
= Δ s′
So, the spacetime interval is invariant.
11 a Spacetime interval, rest mass, proper time and
proper length.
2
86
2
iii
12 a i
ii
b i
ii
Rest mass: the mass of an object in the frame
of reference in which it is stationary.
Proper length: the length of an object
measured by an observer in a frame of
reference in which the object is stationary
with respect to the observer.
Proper time: the time interval between two
events that occur in a reference frame in
which both events occur at the same
position.
t=H
c
Yes. The container and the both observers are
in the same frame of reference.
Yes. The observer inside the container is in
the same frame of reference as the light
source, so this observer still measures the time
for the light beam to reach the top of the
container as t = H
c
(distance travelled)2 = (ct)2 + (vt’)2
iii ( ct ′ ) = ( ctt ) + (vtt ′ ) ⇒
2
2
2
⇒ ′2 =
(
t ′2 c − v
c
2
) =t
2
t2
2
1 − v2
c
So, t ′= γ t
c It is called time dilation because the time t ′ is
longer than the time t, since γ > 1
13 a L = x2 − x1 = γ ( x 2′ + vtt
= γ ( x 2′ + vtt
)
γ ( x1′ + vtt
)
x1′ − vtt ′ ) = γ (x 2′ x1′ )
b (x 2′ x1′ ) = L ′ so L ′ = L
γ
c This is called length contraction because L′ is
shorter than L, since γ > 1.
14 a The student measures the two events, the kettle at
the start and the kettle when it has boiled, at the
same point in space in the student’s inertial frame
of reference. So, the time measured is a proper
time interval.
bγ =
1
=
1
= 1.67
2
v
1
−
(
)
1− 2
c
c γ × proper time = 1.67 × 2 minutes = 3.3 minutes
2
15 γ =
1
2
=
1
=19
1 − 0.7225
1 − v2
c
So, the observer in the fast car measures the length
of the building to be 100 = 53 m (2 s.f.).
19
16 a γ =
1
2
=
1
= 1 51
1 − 0.5625
1 − v2
c
So, Δt ′ = γ t = 1 51 × 5.00 × 10 −3 = 7.56 × 10−3 s
b Light from the signal light ahead of the car will
reach the car before light from the signal light
behind the car reaches the car. This is because the
speed of light is independent of the motion of the
observer, but in this case the observer is moving
towards the signal light ahead of it, making the
distance that the light has to travel shorter –
hence it takes less time.
17 a Distance to Proxima Centauri is
4.0 × 3 × 108 × 3.15 × 107 = 3.78 × 1016 m
16
Therefore, t = s = 3 78 × 10 8 = 1.4 × 108 s
v 0 9 × 3 × 10
(= 4.44 years)
1
1
=
= 2.29
bγ =
2
1 − 0.81
v
1− 2
c
So, distance to Proxima Centauri as measured by
2 a Gradient of the worldline for s is 1/c, so s is
travelling at speed c.
b The gradient of the worldline for r suggests that r
would be travelling faster than c. This is not
possible.
3 a x = 2.5 m
b t = 6 = 6 8 = 2 × 10−8 s
c 3 × 10
4 a The speed of the photon is c. So, x = ct.
Gradient = ct = 1.
ct
b The speed of B is v. So, x = vt
Therefore, θ = tan −1
()
= tan −1
= tan −1
()
c Since tan θ = v , this allows us to find v in units of c.
c
d θmax = 45° because this is the angle that gives a
value of v as c. Any larger value of v (i.e. any value
of tan θ greater than one) is not possible because it
would mean that v > c.
5 a-d See Figure AA.2.
θ – = tan–1(0.8) = 38.7° θ + = tan–1(0.3) = 16.7°
P–
ct
Q–
θ
–
θ
Q+
+
16
Anand = s = 3 78 × 10 = 1.65 × 1016 m
γ
2.29
Therefore, t = s ′ = 1 65 × 10 8 = 6.1 × 107 s
v 0 9 × 3 × 10
(= 1.94 years)
c Anand’s measurement is a proper time because he
is measuring both leaving the Earth and arriving
at Proxima Centauri at the same place: his rocket
ship.
()
P+
x
16
Figure AA.2
()
v
c
b i See Figure AA.3.
6 a θ = tan−1
ct
ct9
θ
Exercise A.3 – Spacetime diagrams
1 a p is stationary.
b q is travelling at a constant speed.
c w is accelerating.
M
x9
x
Figure AA.3
ANSWERS
87
ii See Figure AA.4.
ct
ct9
θ
M
x9
9
b The length of the car in S′ is shorter than
it is in S.
c Because of relativistic length contraction, the scale
of the axes for S and S′ have to be different.
Relative to S, the scale for S′ is different by a
factor of γ.
a See Figure AA.6.
ct
x
ct9
Figure AA.4
x′ = γ x
ct′ = γ ct
A and B.
A and B cannot occur simultaneously in S′
because they do not lie on a line that is parallel
to the x′-axis
i B and E.
ii B and E cannot occur in the same place in S′
because they do not lie on a line that is parallel
to the ct′-axis.
i B and C.
ii B and C cannot occur simultaneously in S
because they do not lie on a line that is parallel
to the x-axis.
i E and D.
ii E and D cannot occur in the same place in S
because they do not lie on a line that is parallel
to the ct-axis.
i Between E and B in S, and between E and D in
S′, it is possible to measure a proper time. This is
because E and B occur in the same place in S
and E and D occur in the same place in S′.
ii Between A and B in S and between B and C in
S′, it is possible to measure a proper length. This
is because A and B occur at the same time in S
and B and C occur at the same time in S′.
See Figure AA.5.
c i
ii
7 a i
ii
b
c
d
e
8 a
ct9
ct
worldline for
stationary car in S
length of
car in S9
x9
10 m
E2
time
measured
in S
time measured
in S9
E1
5m
88
10 m
x
Figure AA.6
b The time in S ′ is longer than it is in S.
10 a and b See Figure AA.7.
ct
worldline
for photon
emitted
backwards
X
worldline for
photon emitted
forwards
x
Figure AA.7
c The region in between the two worldlines for the
photons shows where photons emitted at X can
be observed. They form a kind of cone.
d Any event within the cone can be considered
to be affected by – or caused by – what occurred
at X.
11 a See Figure AA.8.
ct
photon worldlines
from event Y
Z
Y
5m
Figure AA.5
x9
x
x
Figure AA.8
b Looking at Figure AA.8, event Z is outside the
cone of influence from event Y. So, it is not
possible that event Z is caused by event Y.
12 a Two identical twins age by different amounts
because they move at a relative speed to each
other. One twin travels to a distant place at a
relativistic speed, whilst the other twin stays at
home on Earth. Because each twin sees their
sibling as moving relative to themselves, both
twins should exhibit the same time dilation and
so each twin should consider their sibling to be
younger than they are. That is why this is called a
paradox.
b The twin that stays at home on the Earth.
c The twin that stays at home on Earth has
remained in the same frame of reference for the
whole of the other twin’s journey. For the twin
that has made the journey, on arrival at the distant
place, the twin has changed their frame of
reference because they have changed their
velocity (which requires an acceleration and
hence unbalanced force). This change of reference
frame breaks the symmetry of the observations of
the two twins and so allows both twins to agree
that it is the twin who stays at home that ages the
most.
10 ly
13 a t = s =
= 11.1 years
v
0 9c
1
1
=
= 2.29
bγ =
2
1
−
0.81
v
1− 2
c
So, Som has aged 11.1 = 4.85 years.
2.29
c Som’s clock shows 4.85 years, but she sees Minky
to have aged 4 85 = 4.85 = 2.12 years.
γ
2.29
d When Som turns around at Fazer, she changes her
frame of reference because she has changed her
relative motion to Minky.
e See Figure AA.9.
t
Minky measures
that she has aged
22.2 years
When Som turns
around she sees the
Earth clocks
measuring 20.1
years
Som returns to
Minky. Som measures
that she has aged
by 9.7 years
t0
Som arrives at Fazer
t9 = 4.85 years
Minky measures
time of Som’s
arrival at Fazer
t = 11.1 years
Som thinks
Minky has
aged 2.12
years
Figure AA.9
t9
x0
x9
x
f Som sees that the Earth clocks now show a time
of 2.12 + (22.2 − 2 × 2.12) = 20.1 years.
g Minky is older by 22.2 − 9.7 = 12.5 years
Exercise A.4 – Relativistic mechanics
1 a Rest energy: the energy required to produce a
particle that is not moving.
b E = KE + E0
c Using E0 = m0 c2, E0 = 9.11 × 10−31 × 9 × 1016
= 8.199 × 10−14 J
−14
This is 8.199 × 10−13 = 0.51 MeV
1 6 × 10
d E = γ E0
1
1
γ=
=
= 2.29
2
1 − 0.81
v
1− 2
c
Therefore, E = 2.29 × 0.51 = 1.168 MeV46
2 a E = 3 E0 ∴γ = 3
2
⎛
⎞
∴ ⎜ 1 − v2 ⎟ = 1 ⇒ v 2 = 8 c 2
⎝
9
c ⎠ 9
∴v =
8 c = 0.943 c
9
ANSWERS
89
b E = 5 E0 γ = 5
6 a i
⎛
⎞
∴⎜ 1 − v2 ⎟ = 1 ⇒ v 2 = 24 c 2
⎝
25
c ⎠ 25
2
24 c = 0.9798 c
25
c E = 10 E0 ∴γ = 10
2
⎛
⎞
∴ ⎜ 1 − v2 ⎟ = 1 ⇒ v 2 = 99 c 2
⎝
100
c ⎠ 100
99 c = 0.995 c
100
E = 7 × 106
3 E = γ E0 ∴γ =
= 7462.7
E0
938
∴v
2
⎛
⎞
1
∴ ⎜ 1 − v2 ⎟ =
= 1 7 × 10 −8
⎝
c ⎠ 7462.72
⇒ v 2 = 0 999 999 983 c 2
999 999 983 c 0 999 999 991 c
∴ v = 0.999
4 a KE = (γ − 1) E0
So, KE = (2.5 − 1) × 0.511 = 0.767 MeV
b KE = (γ − 1) E0
So, KE = (3.0 − 1) × 938 = 1876 MeV
c KE = (γ − 1) E0 = (5 − 1) × 1.5 × 10−3 × 9 × 1016
= 5.4 × 1014 J
5 4 × 1014
MeV= 3.375 × 1027 MeV
=
1 6 × 10 −13
5 a At a speed of 0.99c,
1
1
γ =
=
= 7 09
2
1 − 0.9801
v
1− 2
c
So, KE = (γ − 1) E0 = (7.09 − 1) × 0.511
= 3.11 MeV
b The Newtonian KE
2
= 1 mv 2 = 1 × 9 11 100 31 0 99 × 3 108
2
2
= 4.018 × 10−14 J
(
−14
This is 4.018 × 10−13 = 0.251 MeV
1 6 × 10
KErel
= 3.11 = 12.4
∴
KENewt 0.251
90
1
ii γ =
)
2
=
1
= 3 20
1 − 0.9025
1 − v2
c
So, KE = (γ − 1) E0
So, KE = (3.20 − 1) × 0.511 = 1.126 MeV.
1
1
=
= 7 09
iii γ =
2
1 − 0.9801
v
1− 2
c
So, KE = (γ − 1) E0
So, KE = (7.09 − 1) × 0.511 = 3.11 MeV.
b See Figure AA.10.
Energy required
∴v =
Energy required is the KE gained by the
electron.
1
1
γ =
=
= 2.29
2
1
−
0.81
v
1− 2
c
So, KE = (γ − 1) E0
So, KE = (2.29 − 1) × 0.511 = 0.659 MeV
So, energy required is 0.659 MeV.
8
7
6
5
4
3
2
1
0
0
0.2
0.4
0.6
v/c
0.8
1
1.2
Figure AA.10
c As the speed of the electron increases, the amount
of energy required increases at an increasing rate,
such that it quickly becomes a very large value
when the speed approaches that of the speed of
light. The graph shows that the energy required
becomes asymptotic to the v/c = 1 line, meaning
that an infinite amount of energy would be
required to travel at the speed of light, which is
not possible.
7 a Newtonian momentum p = m0v
m0 v
Relativistic momentum p =
= γm0v
2
v
1− 2
c
b E = γ E0
2 2 4
2
c E2 − p2c2 = γ m0 c − γ
d E2 − p2c2 =
(
γ m c c
2
2 2
0
2
v
2
)=
(
γ 2m02 c 2 c 2 − v 2
2 2 2
0
(
m02 c 2 c 2
v2
2
⎛
⎞
1 − v2
⎝
c ⎠
)
2
⎛
⎞
m02 c 4 1 − v2
⎝
c ⎠
=
= m02 c 4
2
v
1− 2
c
e E2 = p2c2 +
8 a γ =
1
2
2 4
0
c
=
1
= 1.898
1 − 0.7225
1 − v2
c
So, E = γ E0 = 1.898 × 0.511 = 0.970 MeV
b KE = E − E0 = 0.970 − 0.511 = 0.459 MeV
c Using p2c2 = E2 − E 02, p =
0.9702 − 0.5112
c2
= 0.8244 MeV c−1
Alternatively, p = γ m0v = 1.898 × 0.511 × 0.85c
= 0.8244 MeV c−1
9 a 400 keV
b Using KE = 1 2 ⇒ v = 2 × KE
2
m
2 × 400 × 1 6 × 10 −16
9 11 × 10 −31
= 3.75 × 108 m s−1
This is unreasonable because it is faster than the
speed of light.
c E = E0 + KE = 0.511 + 0.411 = 0.911 MeV
=
2
d Using p2c2 = E2 − E 0 ⇒ p =
0.9112 − 0.5112
c
= 0.754 MeV c−1
E = 0.911
e E = γ E0 ⇒γ =
= 1.783
E 0 0.511
p = γm0v ⇒ v =
p
0.754
=
= 0.828c
γ m0 1.783 × 0.511
= 2.48 × 108 m s−1
10 a E = E0 + KE ⇒ g E0 − E0 = eV
∴ γ − 1 = eV ⇒ γ = 1 + eV
E0
E0
)
b i
γ = 1 + eV = 1 + 400 = 1.426
E0
938
So, E = γ E0 = 1.426 × 938 = 1338 MeV
13382 − 9382
ii Using p2c2 = E2 − E 02 ⇒ p =
c
−1
= 954 MeV c
p
954
=
iii p = γm0v ⇒ v =
= 0.713c
γ m0 1.426 × 938
= 2.14 × 108 m s−1
1
1
=
= 2.29
11 a At v = 0.9c, γ =
2
1 − 0.81
v
1− 2
c
γE0 = E0 + eV ⇒ V =
E0 ( −
e
) = 938 (
−
)
e
= 1.21 GV
E 0 ( − ) 0.511(
− )
=
bV =
= 659 kV
e
e
The two voltages are in the ratio of their rest
masses: 1.21 GV = 938 × 659 kV.
0.511
Exercise A.5 – General relativity
1 The equivalence principle: the effects of a
gravitational field are the same as the effects caused
by a constant acceleration.
2 a i The cup of tea would stay floating in space in
front of the observer.
ii Since there is no gravitational field, the cup of
tea and the observer are weightless, and will not
be accelerated in any direction.
b i The observer would still expect the cup of tea
to remain where it is in front of him.
ii No.
iii Again, in the absence of a gravitational field,
there is nothing to accelerate the cup of tea.
c i He would expect the cup of tea to fall towards
the floor of the container.
ii No.
iii The observer could say: the cup of tea falls
because of a gravitational pull on it or the cup
of tea falls because I am in a container that is
accelerating.
ANSWERS
91
d Either of the observer’s two explanations for the
behaviour of the cup of tea would be acceptable
physics. This confirms that the two situations are
equivalent.
3 a i The opposite wall exactly horizontally from
where the laser is.
ii A horizontal straight line.
b i The opposite wall exactly horizontally from
where the laser is.
ii No.
c i The opposite wall below the point that is
directly horizontal from where the laser is.
ii Curving downwards.
iii In the time it takes for the light to reach the
opposite wall, the container’s speed has
increased. This means that the container has
travelled further upwards than the light beam
has, causing the light beam to hit the opposite
wall below where it would if the container
were moving at a constant speed.
iv No.
v When a beam of light passes near to a large
mass, the gravitational field of the large mass
acts on the beam of light to bend it towards
the mass.
4 a The measurements of the front and back of the
space rocket are made at the same time in the same
frame of reference.
l
b Δt =
c
c Δv a Δt
d Δv a Δt
Δt = al
c
e From the Doppler shift equation,
Δ f Δv
= Δv ⇒ Δ f
f al2
f
c
c
f No.
g When a photon is emitted upwards from the
Earth’s surface, it undergoes a gravitational red shift
when observed by an observer above the Earth’s
surface.
8
5 a f = c = 3 × 10 −9 = 6.0 × 1014 Hz
λ 500 × 10
f al2 = 6 0 × 1014 × 9 81 × 555
= 36.3 Hz
So, Δ f
c
9 × 1016
92
b This is a gravitational red shift. This implies that
the wavelength of the photon has become longer
and so its energy must have decreased.
c Using the same relationship for frequency,
ΔT T al2 = 8 64 × 104 × 9 81 × 555
= 5.23 × 10−9 s.
c
9 × 1016
d Time will pass more slowly in the presence of a
gravitational field.
e Satellites must account for two aspects of relativity:
First, there is a time dilation effect caused by the
relative motion of the satellite and the Earth’s
surface (in fact this causes the time to be shorter by
about 7 μs every day). Second, the gravitational
time dilation causes the time on the satellite to be
ahead of the time on the ground by about 45 μs
every day. A difference of 38 μs would mean a
difference in distance of 38 × 10−6 × 3 × 108
= 11.4 km. This is well outside the operational
tolerance of the GPS systems.
6 Gamma rays of known frequency (and energy) were
fired from a source to a detector. In the first part of
the experiment, the gamma rays were fired upwards
from the source and detected by the detector, as
in Figure AA.11a. The frequency shift observed
confirmed the red shift of the gamma rays in the
Earth’s gravitational field. Then, the experiment was
repeated by firing the gamma rays downwards, as
in Figure AA.11b. Once again, the frequency shift
observed confirmed the blue shift of the gamma rays
in the Earth’s gravitational field. The values of both
the red- and blue-shifts caused by the gravitational
field of the Earth agreed with what Einstein had
predicted in his general theory of relativity.
detector
a)
γ-source
γ-source
b)
detector
Earth’s surface
Figure AA.11
7
a
b
8
a
b
c
d
9
a
b
c
Δf
g Δh
= 2 = 9 81 × 2162 6 = 2.46 × 10−15
f
c
9 × 10
To detect the effect of the Earth’s gravitational
field on the red shift of a photon, it is necessary to
detect fractional changes of this order of
magnitude. For a gamma ray of frequency, say,
1026 m, this would mean being able to detect a
difference of 1011 Hz.
Yes. In two dimensions a straight line is the
shortest distance between two points.
Curved.
Einstein suggested that the very fabric through
which the path of light travelled, which he called
spacetime, could be curved and that large masses
caused spacetime to curve. The shortest path that
light followed would curve along with spacetime.
A gravitational field causes spacetime to become
curved and the path of a ray of light becomes
bent. This is analogous to how the path of a ray of
light bends when it passes through a lens, even
though the light is not being refracted.
Eddington wanted to observe the distant star
when it was just behind the Sun, hence
confirming that the path of light from the star
would have to be bent by the gravitational field
of the Sun. This would provide empirical
evidence in support of Einstein’s warping of
spacetime.
Light from the Sun needed to be eliminated from
the observational region. By observing when the
light from the Sun was blocked completely by the
Moon, it would be possible to see the distant (and
much fainter) star.
See Figure AA.12.
apparent
position
Earth
Moon
Figure AA.12
Sun
distant
star
d Einstein said that the spacetime through which
the light from the distant star passed was curved
by the presence of the gravitational field of the
Sun. This curvature of spacetime caused the light
to bend around the Sun, enabling the star to be
seen even though it was, in fact, behind the Sun.
10 a A black hole is a massive star that has collapsed
under its own gravitational force into a singularity.
The large mass of the star, in such a small volume,
has warped the spacetime around it into an
infinitely curved gradient.
b The event horizon of a black hole is the surface
of a sphere around the black hole, outside which
it is possible for light (and other matter) to move
outwards along the curved spacetime. Inside the
event horizon, light and matter will move inwards
towards the singularity.
c The Schwarzschild radius is the distance from the
event horizon to the point of the singularity.
2 × 6 67 × 10 −11 × 6 × 1024
d i Rs = 2GM
=
c2
9 × 1016
= 8.89 mm
2GM 2 × 6 67 × 10 −11 × 2 × 1030
ii Rs = 2 =
c
9 × 1016
= 2.96 km
2GM = 2 × 6 67 10 11 3 × 2 1030
c2
9 1016
= 8.89 km
b Since the warping of spacetime suggests that
matter and light will be forced to travel towards
the black hole, this will have to increase the mass
of the black hole. Since RS M, the
Schwarzschild radius will increase in time.
11 a Rs min =
2 × 6 67 × 10 −11 × 30 × 2 × 1030
12 a Rs = 2GM
=
c2
9 × 1016
= 88.9 km
t0
1
bt=
=
= 1.005 s
Rs
1
1−
1−
100
d
c t=
t0
R
1− s
d
=
1
1− 1
10
= 1.054 s
ANSWERS
93
d The time between the two signals observed at the
space station will become longer and longer.
Eventually, when the space ship reaches the
Schwarzschild radius the time interval will
become infinitely long.
e The black hole is not rotating.
13 a Einstein’s original solution to one of his general
relativity equations for the universe suggested that
the size of the universe would change in time.
Einstein introduced the cosmological constant to
provide a mathematical way of showing that his
universe could be static, whilst conforming to his
solution.
b Hubble showed that the universe was expanding.
This contradicted Einstein’s view that the
universe was static. Interestingly, once Hubble’s
findings had been experimentally verified by a
number of other means, Einstein’s cosmological
constant took on a greater significance – one that
eventually led to the concepts of dark energy and
dark matter.
c Solutions to Einstein’s equation showed that the
density of the universe was crucial in determining
its fate. If the actual density of the universe were
greater than the critical density, then the universe
would stop expanding and begin to collapse in on
itself. On the other hand, if the actual density of
the universe were less than the critical density
then the universe would expand forever.
d Measurements of the mass and energy in the
universe have not agreed with what astrophysicists
expect them to be. This has led to the idea that
the cosmological constant of Einstein’s equations
might be related to the presence of dark matter
and dark energy. The effect of this dark matter
and dark energy is to increase the rate at which
the universe is expanding. Recent measurements
of the expansion of the universe have confirmed
that this is happening – and so have supported the
idea of the existence of dark matter and dark
energy.
Exam-style questions
1
a See Figure AA.13.
moveable mirror
half-silvered
mirror
light
source
fixed
mirror
observer
Figure AA.13
b If the aether existed, then a rotation of the
apparatus would introduce a change in the speed
of light from one direction to another. This
change in speed would produce a shift in the
interference pattern observed.
c The moveable mirror would introduce a change
in the path length of one of the rays of light. This
would result in a shift in the interference pattern.
d There was no shift in the interference pattern.
e Because there was no shift in the interference
pattern, Michelson and Morley concluded that
there was no aether (or that the aether had no
effect on the speed of light). This was in
agreement with what Maxwell and Einstein had
predicted: that the speed of light was independent
of any motion of the source of light or of the
observer.
2
a γ =
1
2
=
1
= 1.51
1 − 0.752
1 − v2
c
b i x′ = γ (x − vt)
= 1.51 (500 − 0.75 × 3 × 108 × 3.0)
= −1.0 × 109 m
⎛ vx ⎞
ii t ′ = γ ⎝ t − 2 ⎠
c
8
⎛
⎞
51 ⎜ 3 0 − 0 75 3 × 1016 × 500 ⎟ = 4.5 s
= 1.51
⎝
⎠
9 10
94
3
a Taking S to be the frame of reference of the
observer on the ground and S′ to be the frame of
reference of the observer on the train,
Δt ′ = 400 8 = 1.3 × 10−6 s
3 × 10
1
1
=
= 1.25
bγ =
2
1 − 0.36
v
1− 2
c
c Δx = γ (Δx′ + v Δt′)
= 1.25 (400 + 0.6 × 3 × 108 × 1.3 × 10−6)
= 790 m (2 s.f.)
v
d Δt = γ Δt′ + 2 Δx ′ = 1.25
c
⎛
0 6 × 3 × 108 × 400 ⎞
6
⎜ 1.3 10 +
⎟ = 2.6 × 10−6 s
2
×
⎜⎝
⎟⎠
(
905
giving v = s =
= 0.97 c
t 1 3 1 × 10 −6
2
)
Alternatively, having found the length of the train
in S, and knowing the second postulate, the time
taken for the light to reach the front of the train
Δx = 790
must be Δt =
= 2.6 × 10−6 s
c
3 × 108
a No
b John sees the flag at the front of the carriage fall
before the flag at the rear of the carriage falls.
Although both John and Mary agree on the speed
of light, the light from the front of the carriage
takes less time to travel to John than the light
from the rear of the carriage.
c i John is making the measurement at the same
time in a frame of reference in which both
events occur, with no motion relative to John.
6
2
b Distance travelled = vt
= 0.97 × 3 × 108 × 1.28 × 10−5 = 3.72 km
ct9
ct
B
x9
A
C
x
Figure AA.14
1
5
1
=
= 1 67
ii γ =
2
1 − 0.64
v
1− 2
c
So, L ′ = L = 25 = 15 m
γ 1 67
1
1
=
= 4 11
a γ =
2
1
−
0.94
v
1− 2
c
So, t ′1 = γ t = 4.11 × 3.1 × 10−6 = 1.28 × 10−5 s
d Number of half-lives = 15 = 4.0
3 72
e After 4.0 half-lives, the number of muons
4
reaching the Earth’s surface will be 1 = 1 of
2
16
the number produced. Since the number
produced is very high, there will be significant
numbers of muons reaching the surface.
(In fact there will be fewer than 1 of the
16
number produced, because some of the muons
will interact with atoms along their path through
the atmosphere and never get as far as the Earth’s
surface.)
See Figure AA.14.
()
)
(
4
c In the muon’s frame of reference, the distance
they travel is less but the time is also less.
t 1 = 3.1 × 10−6 s and s = s ′ = 3720 = 905 m
γ 4 11
2
7
a i A
ii B
b i C
ii A
a Prior to the pair production, the photon had
momentum. After the production of the electronpositron pair, momentum must be conserved. If
the electron and the positron travelled in opposite
directions, their total momentum would be zero.
This would not satisfy the conservation of
momentum.
ANSWERS
95
bγ =
1
2
=
b i ω f = ω i + αt = 40 + 5 × 6 = 70 radians s−1
ii θ = ωi t + ½ α t2 = 40 × 6 + ½ × 5 × 62
= 240 + 90 = 330 radians.
330
= 52.5.
So, number of rotations =
2π
1
= 2.29
1 − 0.81
1 − v2
c
So, KE = (γ − 1) E0 = (2.29 − 1) × 0.511
= 0.659 MeV
c p = γ m0 v = 2.29 × 0.511 × 0.9c = 1.053 MeV c−1
4
⎛
⎞
2π
ωS ⎜⎝ 24.47 × 86 400 ⎟⎠
a
=
= 1 = 4.08 × 10−2
ωE
24.47
⎛ 2π ⎞
⎜⎝ 86 400 ⎟⎠
5
vS rSω S 6.96 × 105
=
=
× 4.08 × 10 −2 = 4.46
v E rEω E 6.37 × 103
15 − 20
= −1.25 radians s−2
a α=
4
b t = Δω = 600 = 40 s
α
15
Option B: Engineering physics
Exercise B.1 – Rigid bodies and rotational
dynamics
1 a
b
2 a
b
3 a
i ω = θ = 2π = 0.10 radians s−1 (2 d.p.)
t
60
π = 1.7 × 10−3 radians s−1
ii ω = θ =
t
60 × 60
(2 s.f.)
π
= 1.5 × 10−4 radians s−1
iii ω = θ =
t 60 × 60 × 12
(2 s.f.)
i
= r ω = 2.5 × 10−2 × 0.1 = 2.5 mm s−1
ii v = r ω = 2.0 × 10−2 × 1.7 × 10−3
= 3.4 × 10−2 mm s−1
iii v = r ω = 1.5 × 10−2 × 1.5 × 10−4 = 2.3 μm s−1
Angular acceleration is the rate at which the
angular speed is changing: α = dω
dt
i ω i = 300 × 2π = 31.4 radians s−1
60
120 × 2π
ii ω f =
= 12.6 radians s−1
60
31 4 = −3.76 radians s−2
iii α = 12.66 − 31
5
See Figure AB.1.
b
ω s2 402 − 152
=
= 344 radians (3 s.f.)
2α
2×2
a ω i = v = 10.0 = 25.0 radians s−1
r 0.400
distance travelled
b number of rotations =
circumference of wheel
e
50
.
0
=
0 40 × 2π
= 19.9.
c θ=
6
c α=
d
7
a
ω/rads s–1
b
c
70
8
40
0
0
Figure AB.1
96
6
t/seconds
a
ω f2
ω f2
ω i2
=
0 − 25 02 = 2.50 radians s−2
π × 19.9
2θ
ω − ω i 0 − 25.0
t= f
=
= 10.0 s
α
2.50
Torque is the product of the perpendicular force
and the distance from the axis of rotation where it
is applied: Γ = Fd sin θ, where F is the force
applied, d is the distance from the axis of rotation
and θ is the angle between the force and the line
joining where the force is applied to the axis of
rotation.
N m (not joules).
i G = 250 × 0.6 = 150 N m
ii G = 400 × 3 × sin 30° = 600 N m
A couple is a pair of forces of the same magnitude
acting in opposite directions; they are not
collinear, so they cause a rotation of the object
they are applied to.
bγ =
1
2
=
b i ω f = ω i + αt = 40 + 5 × 6 = 70 radians s−1
ii θ = ωi t + ½ α t2 = 40 × 6 + ½ × 5 × 62
= 240 + 90 = 330 radians.
330
= 52.5.
So, number of rotations =
2π
1
= 2.29
1 − 0.81
1 − v2
c
So, KE = (γ − 1) E0 = (2.29 − 1) × 0.511
= 0.659 MeV
c p = γ m0 v = 2.29 × 0.511 × 0.9c = 1.053 MeV c−1
4
⎛
⎞
2π
ωS ⎜⎝ 24.47 × 86 400 ⎟⎠
a
=
= 1 = 4.08 × 10−2
ωE
24.47
⎛ 2π ⎞
⎜⎝ 86 400 ⎟⎠
5
vS rSω S 6.96 × 105
=
=
× 4.08 × 10 −2 = 4.46
v E rEω E 6.37 × 103
15 − 20
= −1.25 radians s−2
a α=
4
b t = Δω = 600 = 40 s
α
15
Option B: Engineering physics
Exercise B.1 – Rigid bodies and rotational
dynamics
1 a
b
2 a
b
3 a
i ω = θ = 2π = 0.10 radians s−1 (2 d.p.)
t
60
π = 1.7 × 10−3 radians s−1
ii ω = θ =
t
60 × 60
(2 s.f.)
π
= 1.5 × 10−4 radians s−1
iii ω = θ =
t 60 × 60 × 12
(2 s.f.)
i
= r ω = 2.5 × 10−2 × 0.1 = 2.5 mm s−1
ii v = r ω = 2.0 × 10−2 × 1.7 × 10−3
= 3.4 × 10−2 mm s−1
iii v = r ω = 1.5 × 10−2 × 1.5 × 10−4 = 2.3 μm s−1
Angular acceleration is the rate at which the
angular speed is changing: α = dω
dt
i ω i = 300 × 2π = 31.4 radians s−1
60
120 × 2π
ii ω f =
= 12.6 radians s−1
60
31 4 = −3.76 radians s−2
iii α = 12.66 − 31
5
See Figure AB.1.
b
ω s2 402 − 152
=
= 344 radians (3 s.f.)
2α
2×2
a ω i = v = 10.0 = 25.0 radians s−1
r 0.400
distance travelled
b number of rotations =
circumference of wheel
e
50
.
0
=
0 40 × 2π
= 19.9.
c θ=
6
c α=
d
7
a
ω/rads s–1
b
c
70
8
40
0
0
Figure AB.1
96
6
t/seconds
a
ω f2
ω f2
ω i2
=
0 − 25 02 = 2.50 radians s−2
π × 19.9
2θ
ω − ω i 0 − 25.0
t= f
=
= 10.0 s
α
2.50
Torque is the product of the perpendicular force
and the distance from the axis of rotation where it
is applied: Γ = Fd sin θ, where F is the force
applied, d is the distance from the axis of rotation
and θ is the angle between the force and the line
joining where the force is applied to the axis of
rotation.
N m (not joules).
i G = 250 × 0.6 = 150 N m
ii G = 400 × 3 × sin 30° = 600 N m
A couple is a pair of forces of the same magnitude
acting in opposite directions; they are not
collinear, so they cause a rotation of the object
they are applied to.
9
b The two forces act in opposite directions; they
combine to give zero resultant translational force.
The two forces contribute an equal torque, so the
resultant torque is not zero and is not in rotational
equilibrium.
c A constant couple will produce a constant angular
acceleration.
a The lid will require a couple that is equal to (or
greater than) 15 N m. So, each force must be at
Γ =
15
least F =
= 214 N (210 N 2 s.f.)
2r 2 × 3.5 × 10 −2
b The handles of the gadget increase the distance
from the axis of rotation, so the force required
now is F = Γ =
2r 2 × (
15
+
) × 10 −2
= 40.5 N
(41 N 2 s.f.)
c Someone may not be able to apply a force of
214 N to open the jar with their hands. The
gadget reduces the required force to 40.5 N.
10 a I ∑mi ri 2 , where m is the mass of a small part of
i
b
c
11 a
b
c
d
12 a
b
c
an object and r is its distance from the axis of
rotation. When all contributions from all parts of
an object are summed, this gives the moment of
inertia.
Mass is the property of a body that resists being
accelerated. The moment of inertia, , of a body is
the property of a body that resists a body’s angular
acceleration.
Γ = Iα
I = m r2 = 0.20 × 0.402 = 3.2 × 10−2 kg m2
I = 2 × m r2 = 2 × 0.20 × 0.402 = 6.4 × 10−2 kg m2
I = 2 × m r2 = 2 × 0.20 × 0.802 = 25.6 × 10−2
= 0.256 kg m2
i I ∝m
ii I ∝ r2, so doubling r makes I four times larger.
i I = 5 × 22 + 8 × 22 = 20 + 32 = 52 kg m2
ii I = 5 × 0.52 + 8 × 4.52 = 1.25 + 162
= 163.5 kg m2
I = 8 × 4.52 = 162 kg m2
The contribution to the moment of inertia of the
mass close to the axis of rotation is very small; the
answer to aii is less than 1% different to the
answer to b. So, it is mass some distance from the
axis of rotation that dominates the moment of
inertia.
1 MR
M 2 = 1 × 0.300 × 0 22 = 6.0 × 10−3 kg m2
2
2
So, α = Γ = 20 × 0 2−3 = 667 radians s−2
I 6 0 × 10
b Now the moment of inertia is
I = 6.0 × 10−3 + 0.12 × 0.122
= 7.7 × 10−3 kg m2
So, the new angular acceleration is
13 a I
α = Γ = 20 × 0 2−3 = 519 radians s−2
I 7 7 × 10
14 a Γ = F r = 20 × 0.25 = 4.0Nm
40
1
2
b α = Γ = × 5 0.25 = 25.6 radians s−2
2
I
c After five seconds,
θ = 1 αt 2 = 1 × 25.6 × 52 = 320 radians.
2
2
320
∴ number of rotations =
= 51
2π
αt Γ t = FRt = 2Ft = 2 × 4 0 × 8 0
15 ω =α
1 MR 2 MR
I
6 × 0.3
2
= 35.6 radians s−1
16 a Γ = mgR
b No
c total 1 MR
MR 2 mR 2
2
mgR
mg
Γ =
=
dα=
1 MR 2 + mR
2
I total
R
R
+
2
(
e a
Rα
Rmg
m
mg
=
M +m
R M +m
(
) (
)
)
v = Rω
vtotal = Rω + v
vtotal = v − Rω
Since the coin is not slipping, vtotal = 0
vtotal at Q = v + Rω and Rω = v, so vtotal = 2v
18 a Number of rotations s−1 = 20.0 = 2.12.
2π 1.5
b ω = 2.12 × 2π = 13.3 radians s−1
c The fastest part of the cylinder will be the
part on the edge directly above the cylinder’s
centre of mass:
v = 20.0 cm s−1 + ωR = 20.0 + 13.3 × 1.5
= 40.0 cm s−1
17 a
b
c
d
e
ANSWERS
97
19 a i KE = 1 Mv 2
2
ii KErot = 1 Iω
2
d The larger mass cylinder has a larger moment of
inertia. When it rolls down the slope, more of its
energy is transferred into rotational kinetic energy,
allowing its translational kinetic energy to be the
same as the smaller cylinder’s.
2
b KEtotal = 1 ( Mv 2 + I
2
⎛
= 1 ⎜ Mv 2 + 2 MR 2
2⎝
5
2
()
2
)
⎞ 1
2
⎟⎠ = 2 Mv
( )
+
22
= 7 Mv 2
10
= 7 × 163 × 10 −3 × 4 02
10
= 1.83 J
2st 2s = 2 × 1.5 = 0 5
20 a v = at = 2 =
m s−1
t
60
t
v
05
bω= =
= 14.3 radians s−1
R 3 5 × 10 −2
c i Mgh
(
)
()
∴ gh = 7 v 2
10
2
2
ii h = 7 v = 7 0 5 = 1.8 cm.
10 g 10 9 81
(
2
)
1⎛M
Mvv 2
2⎝
(
()
2
1 MR 2 v ⎞
2
R ⎟⎠
= 1 M v2 + 1 v2
2
2
)
Rotational motion
F=ma
Γ= Iα
work done = force × perpendicular
distance
work done =Γ Δ ν
power = force × velocity
power = Γ ω
Table AB.1
1 M
Mvv I
2
2
⎛
⎞
= 1 Mv 2 + 2 MR 2 v ⎟ = 7 Mv 2
2⎝
5
R ⎠ 10
21 a Mgh = 1 Mv 2 + I
2
Linear motion
3 Mv 2
4
Mgh
4 gh
So, v = 4
=
3 M
3
Since the mass, M, and the radius, R, do not
appear in this expression, the final speed of the
cylinder is independent of the mass and radius of
the cylinder; so both cylinders will arrive at the
bottom of the sloping surface with the same
speed.
4 gh
= 4 × 9.81 × 0 20 = 1.6 m s−1
3
3
c This is reminiscent of the story of Galileo
dropping objects of different masses from the
Leaning Tower of Pisa and showing that they hit
the ground at the same time – implying that they
were travelling at the same speed.
2π × 300
23 a i ω =
= 31.4 radians s−1
60
ii angular momentum,
L I ω = 2 × 20.0 × 0.402 × 2π × 300
5
60
= 40.2 kg m2 s−1
b The principle of conservation of angular
momentum states that, when the overall torque
acting on a body is zero, its angular momentum
remains constant.
c The new moment of inertia is
2 MR 2 + mR 2 = 2 × 20.0 0.402 + 4 0 × 0 402
5
5
= 1.92 kg m2
40.2
So, the new angular velocity =
1 92
= 20.9 radians s−1
So, number of rotations per minute
= 20.9 × 60 = 196.
2π
Exercise B.2 – Thermodynamics
1
bv=
98
2
a Internal energy is the sum of all the potential and
the kinetic energies of all the particles in the gas.
b U 3 nRT = 3 × 3 0 × 8 31 × 300 = 11.2 kJ
2
2
c Since the gas is ideal, there is no potential energy.
So, in this case U ∝ T. If T is halved then U is
halved.
a An open system is one that allows mass to enter
or leave.
b A closed system is a system that prevents mass
from entering or leaving.
c An isolated system is a system that prevents any
form of energy from entering or leaving.
3
Thermodynamic
process
Details of relevant quantity
Isothermal
Temperature remains constant
Isovolumetric
Volume remains constant
Isobaric
Pressure remains constant
Adiabatic
No heat is exchanged with the
surroundings
Table AB.2
4 a In an isothermal change of state, p1V1 = p2V2
pV
25 × 2 5
So,V2 = 1 1 =
= 62.5 cm3
p2
10
b i
5
5
In an adiabatic process, p1V1 3
p2V2 3
5
So,V2 =
3
5
p1V1 3
=
p2
3
5
5
2 5 × 253 = 43.3 cm3
10
2
2
3
ii In an adiabatic process, T
TV
1V1
T2V2 3
2
( )
2
3
⎛V ⎞ 3
So, T2 = T1 ⎜ 1 ⎟ = 300 × 25 = 163 K
62 5
⎝ V2 ⎠
NkT , but N = nA (where A is Avogadro’s
5 a U 3N
2
constant) and kA = R.
N = 3 nAkT = 3 nRT 3 pV
So, U 3 NkT
2
2
2
2
3
If p is constant, ΔU
p ΔV
2
3 pΔ
ΔV
V = 3 × 2 0 × 106 × (
−
)
b ΔU
2
2
= − 3.0 × 105 J
c Work done = p ΔV = 2.0 × 106 × ( .
− .
)
= − 2.0 × 105 J
d Q = ΔU + W = (−3.0 − 2.0) × 105 = −5.0 × 105 J
(the gas loses heat to the surroundings).
6 a For an adiabatic process there must be no heat
exchanged. To replicate this process, we have to do
it very quickly so there is insufficient time for heat
to be exchanged with the surroundings.
b If we perform a process on some gas very slowly,
there will be sufficient time for heat to be
exchanged between the gas and its surroundings to
keep the gas at a constant temperature.
c When gas is sprayed out of an aerosol can, it
expands very quickly (because the atmospheric
pressure is much less than the pressure inside the
can). So, the gas does work on the surroundings
(W > 0) but there is insufficient time for heat to be
exchanged (Q = 0) so ΔU must be < 0 and the gas
cools to a lower temperature.
7 a Because the process is isothermal, ΔU = 0.
b Q = 0 + W, so Q = W. Therefore, the gas gains heat
of 2.5 kJ from its surroundings.
8 a Q = ΔU + W. For an isothermal process, ΔU = 0.
So, Q = W = − 4.5 kJ (the gas loses heat to its
surroundings).
b For an adiabatic process, Q = 0. So, 0 = ΔU + W.
Since work is being done on the gas, the value of
W is negative and so the value of ΔU must be
positive. This means that the gas will be gaining
internal energy; its temperature must increase.
9 a i A is an isobaric process: the pressure is
remaining constant.
ii B is an isovolumetric process: the volume is
remaining constant.
b i Work done = area under graph =
2 × 105 × 15 × 10−3 = 3.0 kJ
ii At Z, the gas must be at a lower temperature.
Points X and Z must lie on different isotherms
and, because Z will be on an isotherm that is
lower than X, it must be at a lower
temperature.
iii ΔU must be negative because its temperature
has fallen. But, work has been done on the gas,
so W is also negative. Therefore, Q = ΔU + W
must be negative. This means that heat has
been lost to the surroundings from the gas.
c i No. Work done = p ΔV and, here, ΔV = 0. So,
process B does not involve any work being
done on or by the gas.
ii At point Z, the gas had lost internal energy
compared to point X. At Y, it has the same
internal energy as at point X. So, during
process B, the gas must have gained heat from
the surroundings.
ANSWERS
99
Work done = area under graph
= 5 × 105 × 15 × 10−3 = 7.5 kJ
ii Z is at a higher temperature than Y.
iii ΔU > 0 and work done is on the
surroundings by the gas, therefore W > 0, so
Q > 0. This means that heat must be absorbed
by the gas from the surroundings.
i No. There is no change in volume, so there is
no work done on or by the gas.
ii Since X must be at a lower temperature than
Z and no work is done, ΔU < 0 so Q < 0.
This means that heat is being lost by the gas
to its surroundings.
Work done = area under graph from W to X
= 5 × 105 × 15 × 10−3 = 7.5 kJ
No.
Work done = area under the graph from Y to Z
= 2 × 105 × −15 × 10−3 = − 3.0 kJ
Net work done = 7.5 − 3.0 = 4.5 kJ
The area of the rectangle enclosed by the four
thermodynamic processes.
There must be an input of energy from an
external source. Such a source might be, for
example, the fuel in a car engine.
Otto cycle.
The area enclosed by the four processes.
i D
ii B
D
See Figure AB.2.
10 a i
b
11 a
b
c
d
e
f
12 a
b
c
d
e
heat source
T = TH
QH
useful work
QC
heat sink
T = TC
Figure AB.2
f Some of the heat taken from the heat source is
given to the heat sink. So, not all of the heat from
the heat source is available to produce useful work.
100
13 a Carnot cycle.
b A and C are isothermic processes. B and D are
adiabatic processes.
c A
d i A
ii C
e A
f See Figure AB.2.
Useful work done = QH − QC
Total energy used = QH
Q − QC
Q
= 1− C
So efficiency = η = H
QH
QH
14 a η = 1 −
TC
= 1 − 273 + 27 = 0.5 or 50%
TH
273 + 327
b In a real power station, there are other sources of
energy loss, such as lost heat in the processes of
conduction, convection and radiation or
mechanical energy losses due to friction.
c i TH could be increased or TC could be
decreased.
ii The value of TH is limited by engineering and
thermodynamic issues. TC is usually
determined by the temperature of the ambient
conditions (this is why it is usual to build
nuclear power stations near large masses of
cold water, such as the sea).
15 a Heat, QC, is extracted from the heat sink by the
mechanical work done. Heat, QH, is then
deposited into the heat source. In this case,
conservation of energy gives QH = QC + W.
b This is what happens in a household refrigerator.
16 a i This is the Kelvin interpretation of the
second law.
ii As the engine works to transform the thermal
energy produced in the combustion of the fuel
into useful mechanical work (kinetic energy of
the car), the engine becomes hot. This shows
that some of the energy is exchanged with the
engine and the surroundings, meaning that not
all of the energy from the combusted fuel is
available for transformation into useful
mechanical work.
b i The Clausius interpretation of the second law
states that it is not possible to exchange heat
from a colder body to a hotter one without the
use of mechanical work input.
ii A refrigerator requires mechanical work (the
changes of state of a volatile substance) to
extract heat from its colder inside to the
warmer surroundings of the kitchen.
17 a ΔS = Q , where ΔS is the entropy change, Q is the
T
heat exchanged and T is the (constant)
temperature at which the heat is exchanged.
(Note how this equation makes the assumption
that, in extracting heat from a body, the body’s
temperature does not change.)
b i The ice cube melts quickly because the metal
plate is a good conductor of thermal energy,
allowing heat to be exchanged with the metal
plate at a high rate.
ii The loss of energy from the surroundings
causes a negative entropy change. But, the
gain of energy by the ice cube causes a larger
positive entropy change (because the value of
Q is the same for both processes, but T is
lower for the ice cube). So, the net entropy
change is greater than zero.
−3
3
Q
iii ΔS = = 10 × 10 × 334 × 10 = 12.2 J K−1
T
273
18 a i
The cooling tea produces a negative entropy
change. The warming surroundings produced
a positive entropy change. The temperature of
the tea is warmer than that of the
surroundings, so the magnitude of the entropy
change for the tea is smaller than that of the
surroundings. So, the net entropy change is
positive.
ii The atoms of sodium and chlorine are losing
energy (atomic bonds are being created) so
the entropy change for the crystal is negative.
The energy lost by these atoms is gained by
the surroundings, making the entropy change
of the surroundings positive. This entropy
change of the surroundings is greater than
that of the crystal, making the net entropy
change positive.
iii During sweating, the most energetic atoms of
sweat break free of the atoms around them
(evaporating). The average energy of the
atoms left behind is lower – hence the
cooling. This will create a negative entropy
change. However, the addition of those high
energy atoms into the surroundings creates a
greater positive entropy change. This allows
the net entropy change to be positive,
following the second law of thermodynamics.
b Calculate the net entropy change when 2 kJ
of heat are exchanged from a heat source at a
temperature of 700 K to a heat sink at a
temperature of 300 K.
Exercise B.3 – Fluids and fluid dynamics
1 a weight = mg = Ahρg
Ahρ g
=h g
b p liquid = F =
A
A
c p = p0 + pliquid = 1.0 × 105 + (0.3 × 1000 × 9.81) =
1.03 × 105 Pa
2 a pliquid = p0 = hρg ⇒ h =
p0
1 0 × 105
=
ρ g 1 36 × 104 × 9.81
= 0.75 m
5
p0
= 1 0 × 10 = 10.2 m
ρ g 1000 × 9 81
3 a Pascal’s principle states that, for an incompressible
liquid, a pressure applied to any part of the liquid
will be transmitted through all parts of the liquid
and to its container.
b To lift the mass will require a force of
1.5 × 103 × g = 1.47 × 104 N
b pliquid = po = hρg ⇒ h =
This force is given by F = 1 47 × 10
4
= 3.69 × 103 N
c The work done on the mass will be
1.47 × 104 × 1.0 = 1.47 × 104 J
So, the work done by the force, F, must be equal
to this.
4
4
∴ F × d = 1 47 × 104 ⇒ d = 1 47 × 10 3 = 4.0 m
3 69 × 10
4 a For an object that is partly or fully submerged in a
fluid, the buoyancy force acting on the object is
equal to the weight of the fluid displaced by the
object.
ANSWERS
101
b i p = p0 + ρgh
ii Upwards force on the underside of the
cylinder is F = p A = A(p0 + ρgh).
iii Downwards force from atmospheric
pressure = p0A.
iv Net force upwards = buoyancy force
= A(p0 + ρgh) − p0A = Aρgh.
v Weight of liquid displaced by the cylinder
= Ahρ × g = Aρgh. This verifies Archimedes’
principle.
vi Its density must be the same as that of the
liquid around it.
5 a Volume of the block = mass = 100 3
density 1.1 × 10
−2
3
= 9.1 × 10 m
Weight of the water it displaces is
W = 9.1 × 10−2 × 1000 × 9.81 = 893 N
Weight of block itself = 100 × 9.81 = 981 N
So, net force on the block = 981 − 893 = 88 N
downwards.
b Sink
3
6 a i Volume of the block = mass = 68000 × 10
density
7 8 × 103
3
3
= 8.72 × 10 m
Weight of the water it displaces is
W = 8.72 × 103 × 1000 × 9.81 = 8.55 × 107 N
By Archimedes’ principle, this is the buoyancy
force.
ii The weight of the block itself
= 68 000 × 103 × 9.81 = 6.67 × 108 N
Net force on the block
= 6.67 × 108 − 8.55 × 107 N
= 5.82 × 108 N downwards.
The block will sink.
b i Buoyancy force = weight of water displaced
= 68 × 380 × 25 × 1000 × 9.81 = 6.34 × 109 N
ii For the ship to float, the weight of the ship and
its cargo force must not exceed the buoyancy
force.
So, weight of cargo = buoyancy force − weight
of empty ship = 6.34 × 109 − 6.67 × 108
= 5.67 × 109 N
(This is about 8.5 times the weight of the
empty ship.)
7 a i The fluid undergoes laminar flow; i.e. at any
place within the fluid, the speed at which the
fluid is moving is constant.
102
ii The fluid maintains a constant density; i.e. it is
incompressible.
iii No frictional forces act on the fluid during its
flow; i.e. it is non-viscous.
b The volume flow rate is the volume of fluid that
flows past a point in 1 second.
Volume flow rate = A v, where A is the crosssectional area of the fluid flowing and v is the
speed at which the fluid is flowing.
c Volume flow rate = 20.0 × 10−4 × 1.5
= 3.0 × 10−3 m3 s−1
d i The volume flow rate must remain constant
(because the ideal fluid is incompressible), so it
is still 3.0 × 10−3 m3 s−1.
ii A1 v1 = A2 v2 . Since the radius is halved, the
cross-sectional area is quartered, so the speed of
the fluid = 4 × 1.5 = 6.0 m s−1.
8 a i F = p1 A1
ii It moves, Δx1 = v1 Δt
iii W1 = F Δx1 = p1 A1 v1 Δt
b i F = p2 A2
ii Δx2 = v2 Δt
iii W2 = F Δx2 = p2 A2 v2 Δt
c Wnet = W1 − W2 = p1 A1 v1 Δt − p2 A2 v2 Δt
= V (p1 − p2)
d i Fluid moves through a height Δz = z2 − z1
ii Work done = mgΔz = VΔg (z2 − z1)
e ΔKE = 1 V ρ
−
2
f i Wtotal = ΔKE + ΔGPE
(
)
⇒ V (p1 − p2) = VΔg (z2 − z1)
(
+ 1 V ρ v 2 2 − v12
2
)
(
)
1ρ v 2 v2
2
1
2
1
2
⇒ p1 + 1 ρ v12 + ρ gz1 = p2 + ρ v 2 + ρ gz2
2
2
9 Applying Bernoulli’s equation:
p1 + 1 ρ v12 + ρ gz1 = p2 + 1 ρ v 2 2 + ρ gz2
2
2
⇒ p1 = p2 + ρ gz2
∴ p2 = p1 − ρgz
= 2.5 × 105 − 1.0 × 103 × 9.81 × 5.0 = 2.0 × 105 Pa
10 a 80 × 4.0 = 20 v
So, vx = 16.0 m s−1
ii p1 − p2 = ρg (z2 − z1) +
b Applying Bernoulli’s equation:
p1 + 1 ρv12 + ρ gz1 = p2 + 1 ρv 2 2 + ρ gz2
2
2
5
∴ px = 2.0 × 10 +
(
)
1 × 1000 2 − 2 + 1000 × 9 81 5.0
1
2
2
= 2.0 × 105 − 1.2 × 105 + 4.9 × 104
= 1.3 × 105 Pa.
11 a p = p0 + hρg = 1.0 × 105 + 3.0 × 1000 × 9.81
= 1.29 × 105 Pa (1.3 × 105 Pa 2 s.f.)
1 2
1
2
b p1 + ρv1 + ρ gz1 = p2 + ρv 2 + ρ gz2
2
2
p1 and p2 are equal (to the atmospheric pressure)
because the bung has been removed, and v1 = 0
(because the surface of the water is stationary), so:
1
2
1000 × 9.81 × 3.0 = × 1000 × v + 0
2
⇒ v = 2 × 9.81 × 3 = 7.7 m s−1
12 Air flows faster past the shiny side of the cricket
ball; so, the streamlines are closer together and the
pressure of the air is lower than on the rough side
of the ball. A force acts on the ball that pushes it
towards the shiny side. This is what cricketers call
‘reverse’ swing.
13 a An aeroplane’s wing is more curved on the top
than on the bottom. This makes air travel faster
along the top of the wing than it does along the
underside. Bernoulli’s equation shows that, where
speed is higher, pressure is lower. So, the difference
in pressure between the top and the bottom of
the wing creates an upwards force (lift) on the
wing. This force keeps the aeroplane in the air.
b See Figure AB.3.
air flow
Δh
Figure AB.3
The Pitot tube is a thin tube with an opening at
the front, facing the motion of the air it measures.
When the air hits the opening, it is brought to a
stop. Along the length of the tube there are lots of
small holes. One part of a manometer is
connected to the opening of the tube and the
other to the tube itself, some way along. The
difference in pressure measured by the
manometer (as a difference in the height, Δh, of
the liquid in the manometer) can be used to give
the speed of the air at the opening to the Pitot
tube.
14 a Example a is non-viscous laminar flow;
example b is turbulent flow; example c is viscous
laminar flow.
b i Stokes’ law gives: viscous drag force = 6πηrv
ii mg = 6π ηrv + 4 π r 3 ρf g and m = 4 π 3 ρs
3
3
4 3
So, 6πηrv = 4 π r 3 ρs g − π r ρf g
3
3
4 3
= π g ( ρs − ρf )
3
4 πr 3 g −
(
⇒v = 3
6πηr
c i
2r 2 g ( −
v=
9η
)
=
2r 2 g ( −
9η
)
)
2 × 0.012 × 9 81 × ( − ) × 103
9 × 8.9 × 10 −4
= 36 m s−1
ii The Reynolds number would be
v ρr 36 × 1000 × 5 × 10 −2
R=
=
= 2.0 × 106.
η
8 9 × 10 −4
So, yes, this would create turbulent flow since
this value is significantly greater than 1000.
iii If the water was warmer, its viscosity would
significantly decrease. This would increase its
terminal velocity.
v ρr 2 0 × 750 × 5 0 × 10 −2
=
15 a R =
= 2143
η
3 5 × 10 −2
So, the flow would be turbulent.
ηRmin 3 5 × 10 −2 × 1000
=
b vmax =
= 0.93 m s−1
ρr
750 × 5 0 × 10 −2
=
ANSWERS
103
Exercise B.4 – Forced vibrations and resonance
1 a Free oscillation: when a system is displaced from its
equilibrium point and left to oscillate without the
effect of any additional forces.
b Forced oscillation: when a system is acted upon by
a periodic external force.
c Resonance: when the frequency of an external
periodic force applied to a system is the same as
the natural (or resonant) frequency of the system.
This will result in large amplitudes of oscillation.
2 a Damping means that energy is removed from the
oscillator, leading to a reduction in the oscillator’s
amplitude.
b A system that is overdamped will move towards its
equilibrium position over a long period of time,
but it will not oscillate. An underdamped system
will continue to oscillate but will move towards its
equilibrium position over a long period of time.
A critically damped system will move towards its
equilibrium position before the oscillator has been
able to make another oscillation.
c See Figure AB.4.
Amplitude
O
C
Time
Figure AB.4
3 a
Q = 2π ×
b The maximum amplitude will occur at the natural
frequency of the mass–spring system:
f 0 = 1 k = 1 10.0 = 1.13 Hz
2π m 2π 0.200
c i One possibility: add a piece of card to where
the spring is attached to the mass so that when
the mass oscillates the card has to push air out
of its way.
Another possibility: make the mass oscillate in
a container filled with water or oil. Friction
and viscous forces in the water/oil will
dissipate energy.
ii See Figure AB.6.
Amplitude
with
heavy
damping
Forcing frequency
Figure AB.6
iii The maximum amplitude now occurs at a
slightly smaller frequency.
d i The mass–spring system is in phase with the
forcing frequency.
ii The mass–spring system lags the forcing
π
frequency by .
2
iii The mass–spring system lags the forcing
frequency by π.
Exam-style questions
2
E
= 2π × A 2 = 2π × 1
E lost per cycle
0 64
( A)
= 9.8
b 10
4 a See Figure AB.5.
Amplitude
1 a ω = 2π × 7200 = 754 radians s−1
60
b v = r ω = 3.00 × 10−2 × 754 = 22.6 m s−1
c Reading rate = 22.6−6 = 5.65 Mb s−1
4 10
2 a Work done = Γ θ =
θ = 150 × 0 4 × 2π = 377 J
b Using ω2 = 2αω
ω
Forcing frequency
Figure AB.5
104
2α θ = 2 × Γ × θ = 2 × FR × θ
1 MR 2
I
2
= 2 × 2F × θ
MR
= 4×
150 × 2π = 21 7 radians s −1
20 × 0 4
work done 377
=
= 151 W
time taken 2 5
3 a A is an isobaric process: the pressure remains
constant; B is an isovolumetric process: the volume
remains constant; C is an isothermal process: the
temperature remains constant – this is shown by
p1V1 = p2V2; D is an adiabatic process: it is steeper
than the isothermal process and is shown by
c average power =
5
p1V1 3
c Work done = area under graph = (2 × 105 ×
15 × 10−3) + (½ × 3 × 105 × 15 × 10−3)
= 3000 + 2250 = 5250 J
d Since the end state is at the same temperature,
ΔU = 0. So, Q = W = 5250 J.
e i See Figure AB.7
P/Pa × 105
6
5
4
3
2
1
0
5
p2V2 3 .
b Work done = area under graph = pΔV =
4 × 105 × (40 − 10) × 10−6 = 12 J
pV 4 × 105 × 10 × 10 −6
=
= 241 K
nR
2 0 × 10 −3 × 8 31
pV 1 × 105 × 40 × 10 −6
=
= 241 K
ii TX =
nR
2 0 × 10 −3 × 8 31
d Since process C is isothermal, the work done by the
gas expanding must be equal to the amount of heat
gained from the surroundings. A rough estimation of
the area under the graph for process C is:
WD = (30 × 10−6 × 1 × 105)
+ (½ × 30 × 10−6 × 3 × 105) = 3.0 + 4.5 = 7.5 J
4 a Q is the amount of heat exchanged by a system
with its surroundings. A positive Q means that heat
is supplied to the gas from the surroundings.
ΔU is the change in internal energy of the gas. ΔU
will be positive if the internal energy of the gas
increases – i.e. its temperature increases.
W is the work done by the gas on its surroundings.
W is positive if the gas expands, doing work on the
surroundings.
b Since the surroundings are hotter, heat will be
absorbed by the system, making Q positive. The
system then does work on its surroundings, so W is
also positive.
∴ 50 = ΔU + 30 ⇒ ΔU = 50 − 30 = 20 J
c i In an isothermal process, ΔU = 0, so Q = 0 + W
Q = W. So, the system must receive 25 J of heat
from its surroundings.
ii Heat is transferred from the surroundings to
the system.
c i
TX =
−3
pV 5 × 10 × 10 × 10
=
= 2.0 moles
RT
8 31 × 300
b pV = nRT, so, for a fixed amount of gas, if
pV = constant, then T is constant. AtY, pV = 2 × 25 = 50
and at X, pV = 5 × 10 = 50. So, the temperature at
Y must be the same as the temperature at X.
5 a n=
5
X
Y
0
10
20
30
V/× 10–3m3
40
Figure AB.7
6 a
b
7 a
b
ii It will be less. The work done during the
expansion is now less – the area under the curve
is smaller. Since ΔU = 0 Q = W and so smaller
W means smaller Q.
Simple harmonic motion occurs without any loss
of energy from the system – leading to a constant
amplitude without the addition of extra energy to
the system. In practice, this is not possible because
all oscillating systems will suffer some kind of
energy loss, usually because of the presence of
frictional forces.
i In this context the term, exponential refers to
the fact that the amplitude of the oscillations
decreases by the same factor, or fraction, on
each oscillation.
A
A
A
ii Use the constant ratio rule: i.e. 1 = 2 = 3
A2 A3 A4
… and so on.
iii 0.5 = 0.9n
log 0.5
log 0.5 = n log 0.9 ⇒ n =
= 6.6.
log 0.9
So, it would take seven oscillations for the
amplitude to have fallen below a half of its
initial value.
Energy ∝ A2
0.001 = 0.9n
log 0.001
log 0.001 = n log 0.9 ⇒ n =
= 66
log 0.9
So, 66 oscillations would be required for the
energy to be 0.1% of its initial value.
ANSWERS
105
c Q = 2π ×
E
= 2π × E
E lost per cycle
0 1E
= 2π × 10 = 63
d The answers to b and c are about the same.
So, Q is the number of oscillations required for the
energy of the system to be reduced to about 0.l %
of its original value. In practice, engineers will
assume that, when the energy in an oscillating
system has fallen to such a low amount compared
to its initial value, then the oscillations have ceased.
e One
c No
1
= 1 = −0.17 m
power of lens −6
ii The minus sign means that the focal point is on
the same side of the lens as the incident rays.
4 See Figure AC.2.
f =
d i
focal
point
Option C: Imaging
Exercise C.1 – Introduction to imaging
1 a
b
c
d
Convex/converging
Principal axis of the lens.
Focal point/principal focus of the lens.
Focal length of the lens.
1
2 a f =
power of lens
power = 1 = 1 = 1 dioptre
f 10
ii power = 1 = 1 = 1.67 dioptres
f 06
iii power = 1 = 1 = 4.0 dioptres
f 0 25
Figure AC.2
5 See Figure AC.3.
lens
R1
P
R3
b i
c i
ii
1
= 1 = 0.10 m
power of lens 10
1
f =
= 1 = 5.9 cm
power of lens 17
f =
R1
Figure AC.3
6 a and b See Figure AC.4.
lens
u = 2f
1
= 1 = 0.25 m
power of lens 4
3 a Concave/diverging.
b i and ii See Figure AC.1.
iii
f =
f
2f
2f
f
v
Figure AC.5
focal point
v = 2f
Yes, the rays would be incident on a screen to
produce an image.
ii Real
iii Same size as the object
iv Inverted
b i
c i
f
Figure AC.1
106
c Q = 2π ×
E
= 2π × E
E lost per cycle
0 1E
= 2π × 10 = 63
d The answers to b and c are about the same.
So, Q is the number of oscillations required for the
energy of the system to be reduced to about 0.l %
of its original value. In practice, engineers will
assume that, when the energy in an oscillating
system has fallen to such a low amount compared
to its initial value, then the oscillations have ceased.
e One
c No
1
= 1 = −0.17 m
power of lens −6
ii The minus sign means that the focal point is on
the same side of the lens as the incident rays.
4 See Figure AC.2.
f =
d i
focal
point
Option C: Imaging
Exercise C.1 – Introduction to imaging
1 a
b
c
d
Convex/converging
Principal axis of the lens.
Focal point/principal focus of the lens.
Focal length of the lens.
1
2 a f =
power of lens
power = 1 = 1 = 1 dioptre
f 10
ii power = 1 = 1 = 1.67 dioptres
f 06
iii power = 1 = 1 = 4.0 dioptres
f 0 25
Figure AC.2
5 See Figure AC.3.
lens
R1
P
R3
b i
c i
ii
1
= 1 = 0.10 m
power of lens 10
1
f =
= 1 = 5.9 cm
power of lens 17
f =
R1
Figure AC.3
6 a and b See Figure AC.4.
lens
u = 2f
1
= 1 = 0.25 m
power of lens 4
3 a Concave/diverging.
b i and ii See Figure AC.1.
iii
f =
f
2f
2f
f
v
Figure AC.5
focal point
v = 2f
Yes, the rays would be incident on a screen to
produce an image.
ii Real
iii Same size as the object
iv Inverted
b i
c i
f
Figure AC.1
106
10 a See Figure AC.7.
7 a See Figure AC.5.
lens
lens
f
2f
f
2f
f
Figure AC.5
Figure AC.7
The image is between f and 2f from the lens. It is
inverted, real and diminished.
b See Figure AC.6.
lens
f
2f
f
2f
2f
f
Figure AC.6
The image is further away than 2f, inverted, real
and magnified.
8
Position
of object
Position Real or
of image virtual
image?
Upright or
inverted
image?
Size of
image of
object
u > 2f
f < v < 2f
real
inverted
diminished
u = 2f
2f
real
inverted
same size
f < u < 2f
v > 2f
real
inverted
magnified
Table AC.1
9 a For a lens of power +10 D, the focal length is 10
cm. Here, the object is placed at twice the focal
length from the lens. This means that the image
will be formed at twice the focal length from the
lens. So, v = 20 cm.
b Real
c Same size as the object.
d Inverted
The image would be located at infinity – the rays
that have passed through the lens are parallel and so
will only come to a focus at an infinite distance away.
b i Inverted
ii Magnified
iii Real
11 a i 1 = 1 + 1
f v u
ii f is the focal length of the lens, u is the distance
of the object from the lens and v is the
distance of the image from the lens.
b i 1 = 1 + 1 ⇒ 1 = 1 − 1 = 1 − 1 = 20 − 15 = 1
f v u v
f u 15 20 20 × 15 60
So, v = 60 cm from the lens.
ii m = − v = − 60 = −3.0
u
20
So, the image is 3.0 times taller than the object,
inverted and real.
12 a 1 = 1 + 1 ⇒ 1 = 1 − 1 = 15 − 20 = − 1
f v u v 20 15 20 × 15
60
So, v = −60 cm (60 cm from the lens on the same
side of the lens as the object).
b i Upright
ii Virtual
− 60
iii m = − v = −
= 4 (the image is four times
u
15
larger than the object).
1
1
= + 1 ⇒ 1 = 1 − 1 = 1 − 1 = 14 − 10 = 1
13
f v u
v
f u 10 14 10 × 14 35
So, v = 35 cm from the lens.
m = − v = − 35 = −3
u
10
So, the image is 3.5 cm tall and inverted.
ANSWERS
107
14 a 1 = 1 + 1 ⇒ 1 = 1 − 1 = 1 − 1
f v u
v
f u 15 10
= 10 − 15 = −1
10 × 15 30
So, v = −30 cm (30 cm from the lens on the same
side of the lens as the object).
b i Virtual
ii Magnified
iii Upright
v
−
−30
c m=− =−
= 3 (the image is three times
u
10
bigger than the object).
15 See Figure AC.8.
18 a See Figure AC.10.
r
f=
r
2
Figure AC.10
b f =r
2
19 a See Figure AC.11.
concave
mirror
Figure AC.8
16 See Figure AC.9.
r
f
image
Figure AC.11
b i
Real. If a screen were used, the rays of light
would hit the screen and form an image.
ii Inverted
iii Magnified
20 a See Figure AC.12.
concave
mirror
Figure AC.9
17 a At the principal focus/focal point.
b The diverging wavefronts come to a focus further
from the lens than the principal focus.
c i When you look at something close by, the lens
has to change its power. The cilial muscles
‘squash’ the lens so it becomes thicker (has
greater power). This gives more curvature to
the wavefronts so they come to a focus at the
correct distance from the lens – on the retina.
ii You can feel your muscles doing the work to
make the lens change shape because, just for a
moment, your eyes hurt a little.
108
r
Figure AC.12
b Inverted
c Real
f
1
1 + 1 ⇒ 1 = 1 − 1
21 a f = d
dobject
dimage 0.125
125 0 08
image
0 08 0.125
= −4 5
0 08 × 0.125
So, the image is 1 = −0.22 m (0.22 m behind
−44 5
the mirror).
dimage
= − −0.22 = 2.8
b m=−
dobject
0.08
So, the image is 2.8 times larger than the object.
22 a See Figure AC.13.
=
convex
mirror
r
26 a
b
27 a
f
b
Figure AC.13
b f =r
2
23 a 1 = 1 + 1 ⇒ 1 = 1 − 1
f dimage dobject
dimage
f dobject
1 − 1 = −12 − 7.55 = −0 22
−7 5 12
7 5 × 12
So, the image is 1 = −4.5 m (4.5 m ahead of
− 0 22
the mirror/car).
dimage
= − −4.5 = 0.38
b m=−
dobject
12
=
So size of image = 0.38 × 1.5 = 0.57 m.
1
1 = f1 f 2
24 Dtotal = D1 + D2 = +
f1 f 2
f1 f 2
1
But Dtotal =
f total
∴ f total =
25 a f total =
b f total =
f1 f 2
f1 + f 2
2
f1 f 2
= 25 = 12.5 cm
f1 f 2
50
f1 f 2
= 10 × 15 = 6 0 cm
f 1 f 2 10 + 15
f1 f 2
12 × −15
=
= −60.0 cm
f1 f 2
12 − 15
If rays are incident on a spherical mirror, more
than a very small distance from the principal axis,
they will reflect and come to a focus at different
places. This means that there is no longer a single
focal point and the image formed will be blurred.
The best, and probably the most practical, way of
avoiding spherical aberration is to use a parabolic
mirror, rather than a spherical mirror. Such
mirrors will produce a single focal point for all
rays parallel to the principal axis no matter how
far from the principal axis they are.
Different frequencies of light (i.e. different colours)
refract through lenses by slightly different amounts –
the refractive index of a material is slightly frequency
dependent. If several different frequencies (colours of
light) are present, they will come to a focus at
different places.This makes the image slightly
blurred with coloured edges. It also changes the
magnification, causing a distortion of the image.
It is possible to reduce the chromatic aberration
by using a concave lens next to the convex lens.
This will remove the aberration from one
wavelength. If there is no range of colours present
from the object, then no aberration will occur.
The angle of reflection from a mirror does not change
depending on the frequency of the incident light.
Near point: the closest distance from the eye that
a healthy eye can focus on without straining.
25.0 cm
i Virtual
ii Upright
Angular magnification: the ratio of the angle
subtended by the image to the angle subtended
by the object to the eye at the near point. In this
case, M = D
f
i 1 =1+1
f u v
Here, v = −D
f D
So, 1 = 1 + 1 =
u f D
fD
fD
∴u =
f +D
ii In this case, M = 1 + D
f
c f total =
c
28 a
b
29 a
b
c
ANSWERS
109
30 a u =
fD
= 6 0 × 25.0 = 4.8 cm from the lens.
f D 6 0 + 25.0
b M = 1 + D = 1 + 25.0 = 5.17
f
60
So, size of image is 5.17 × 1.5 mm = 7.8 mm.
31 a See Figure AC.14.
f
2 a 1 = 1 + 1⇒ 1 = 1 − 1 = 1 − 1
f u v
v
f u 0 75 0 80
= 1 ⇒ v = 12.0 cm
12
v
12
b mo = − = −
= −15
u
08
c The image from the objective must become the
object of the eyepiece and its image must be
25 cm away.
So,
f
1 = 1 − 1 = 1 − −1 = 26 ⇒ u
= 3.85 cm.
u f v 4 25 100
Figure AC.14
b The gentleman wants to see a magnified, upright
image of the letter P. What he actually sees is a
diminished, inverted letter P.
c The gentleman needs to put the newspaper closer
to the lens than its focal length. Then an upright,
magnified (and virtual) image will be produced.
Exercise C.2 – Imaging instrumentation
( )
⎛
⎞
mo 1 + D ⎟ = −15 × +
= −109
fe ⎠
⎝
3 a The image will be formed at infinity.
b The rays will be parallel.
c The larger the objective lens, the more light it can
receive from the object. This makes the image
brighter. It also implies that the telescope will allow
the observer to see objects as far away as possible.
d fo + fe
e Virtual
f Inverted
4 See Figure AC.16.
d M
1 a i
The objective lens is on the left-hand side and
the eyepiece lens is on the right-hand side.
ii The objective lens.
b See Figure AC.15.
L
fo
fe
virtual
image
Figure AC.16
Figure AC.15
c M
110
M e × M o = 25 × L = 25L
fe
fo
fe fo
5 a X is the region in which the observer looks
through the telescope. This is called the eye ring.
b In this region, the image of the object is brightest
and this region provides the biggest field of view
for the observer, allowing all of the light entering
the objective lens to be seen.
6 a See Figure AC.17.
8
Figure AC.17
b Possible advantages: Larger apertures for the
objective lens are possible (this increases the
amount of light collected by the telescope and so
improves the range of observations possible);
Reflecting telescopes do not suffer from chromatic
aberration; Reflecting telescopes are easier to
build – only one optical surface needs to be
ground, rather than two (and making large lenses is
difficult because of the problems with the quality
of glass and its homogeneity); Reflecting telescopes
do not always have to be as long as refractors
(Cassegrain telescopes reduce the overall length of
the telescope without sacrificing the need for a
large objective focal length).
7 a See Figure AC.18.
fo
fe
which will improve the magnification of the
telescope.
a i The resolution of the telescope is its ability to
distinguish clearly between two objects that
subtend a small angle with the telescope’s
objective. When the angular resolution of the
telescope is smaller than the angle subtended
by two distant objects, then the two objects are
easily resolved and seen clearly as two separate
objects.
ii Rayleigh’s criterion states that, when the first
minimum of the diffraction pattern from one
object coincides with the centre of the central
maximum of the diffraction pattern from the
other object, the two objects will be just
resolved.
b i θ = sin −1
9
a
b
Figure AC.18
b i Parabolic
ii A parabolic mirror reflects all parallel rays to the
same focal point, eliminating problems of
spherical aberration.
c i Hyperbolic
ii The secondary mirror reflects light to the
small gap in the primary mirror, from which
an image is produced using a small eyepiece
lens. The hyperbolic mirror effectively increases
the focal length of the objective mirror,
10 a
b
( λ)
−9
⎛
⎞
= sin
i −1 ⎜ 1.22 × 500 × 10 ⎟
⎝
⎠
6.0
= 1.02 × 10−7 radians.
ii l = 1.02 × 10−7 × 3.8 × 108 = 39 m.
i Radio waves have a much larger wavelength
than light waves.
ii The dish is prone to distortion because of its
own weight. Also, a very large dish is difficult
to steer. The mounting for a dish of 100 m
diameter needs to be big and strong and is,
therefore, difficult to move.
For the optical telescope:
1.22 λ 1.22 × 500 × 10 −9
θ≈
=
d
10
-7
= 6.1 × 10 radians.
For the radio telescope:
1.22 λ 1.22 × 10 × 10 −2
θ≈
=
d
50.0
−3
= 2.4 × 10 radians.
The optical telescope is much better at resolving
images because its Rayleigh criterion angle is
much smaller than the radio telescope’s.
An interferometer is an array of telescopes with
signals that can be added together, with any
appropriate time delays, to produce an
amalgamated signal from a distant source. In this
case, the total length taken up by the
interferometer acts as the effective diameter of the
telescope.
The effective diameter of an interferometer is
very large. A large diameter makes the Rayleigh
ANSWERS
111
criterion angle small and so improves the
resolving power.
1.22 λ = 1.22 × 12 × 10 −2
c θ≈
= 1.1 × 10−8 radians
d
2 × 6.4 × 106
11 a Twinkling mainly occurs because of the
inhomogeneity of the Earth’s atmosphere,
through which light has to pass from the distant
stars to our eyes. Any motion or changes in
density of the atmosphere introduces small
changes in scattering and transmission of light –
this is why stars appear to twinkle. The Hubble
telescope is outside our atmosphere, in orbit
around the Earth, and so its images do not suffer
from the effects of the atmosphere.
b The atmosphere around the Earth absorbs and
scatters different wavelengths of the
electromagnetic spectrum by very different
amounts. This means that some wavelengths are
not able to penetrate through the atmosphere to
be received by telescopes located on the Earth’s
surface. Outside the Earth’s atmosphere, this is not
a problem and so the whole range of the
electromagnetic spectrum is available.
c Satellite-borne telescopes are very expensive to
build and put into orbit.
Exercise C.3 – Fibre optics
1
8
a v = c = 3 × 10 = 2 × 108 m s−1
n
1 50
ncladding
= sin −1 1.40 = 69°
b θ = sin −1
ncore
1.50
c i See Figure AC.19.
cladding
n = 1.45
70°
core
cladding
Figure AC.19
ii Total internal reflection
112
n = 1.50
n = 1.45
2 a sin x =
n1
sin θ A
b x = 90 − θc ⇒
c cos θ c
2
So, sin (90
d sin θ c =
(
x=
−
c
) = cos θc
θc
)
1−
c
2
θc
n2
n1
So, sinθ A
n22
n12
θ A = n12
n22
e θ A represents the largest angle that the ray can
enter the optical fibre for it to be totally internally
reflected inside the fibre.
3 The maximum angle for total internal reflection is:
θ = sin−1 n12
n2 2 = sin
1
1 532
1.472 = 25.1°
All rays incident at angles less than this will propagate
through the optical fibre.
So, yes, this ray will propagate.
8
4 a Speed of light in the core = 3 10
1 52
= 1.97 × 108 m s−1
25 × 103
So, time taken to travel along the fibre =
1 97 108
−4
= 1.27 × 10 s
b The distance travelled by ray B is 25km = 26.6 km
sin 70°
26.6 103
The time taken for ray B =
1 97 108
−4
= 1.35 × 10 s.
So, the time delay is (1.35 – 1.27) × 10-4
= 8.0 × 10−6 s (8.0 μs) with ray B arriving after ray A.
5 a i Light of different wavelengths travels along the
optical fibre with slightly different speeds, so
that they do not all arrive at the other end of
the optical fibre at the same time.
ii If light rays take slightly different paths through
the optical fibre they take slightly different times
to reach the other end of the fibre, because the
number of times each ray has to reflect off the
core-cladding boundary will be different.
b Because the output signal pulses become widened
(in time) when one pulse overlaps with the next
pulse, it is not possible to distinguish between the
two. This limits the frequency at which pulses can
be sent, so that no two pulses can overlap after the
effect of dispersion.
c i A smaller area means less energy, which shows
that some energy has been lost during the
transmission of the pulse along the optical fibre:
the pulse has been attenuated.
ii The wider output pulse shows the effect of
material dispersion; it takes some of the
wavelengths longer to travel along the fibre than
others, hence the spreading of the pulse on the
time axis. Some spreading of the pulse may also
be caused because the paths of light rays are
different. This is called waveguide dispersion.
d One way is to use a monomodal optical fibre.This
consists of a very thin core, which prevents the range
of possible paths that light can take along the fibre,
thus reducing the effects of waveguide dispersion.
Another way is to use a graded refractive index
core. The core of the optical fibre is made
from a material with a refractive index that
is not constant throughout the core; it
increases closer to the central axis. This
allows light to travel faster in the part of the
core closest to the core–cladding boundary.
6 a Twisted pairs reduce the problems associated with
electromagnetic induction. If two wires are parallel,
a varying signal in one wire induces a signal in the
other wire (and hence introduce noise to the
signal). Twisting the wires can help to reduce this.
b The core of the optical fibre is an insulator, so it
cannot produce any induced currents from varying
emfs. This makes the optical fibre immune to
external electromagnetic interference.
7 a See Figure AC.20.
in an optical fibre, meaning more signals at once;
They can be very thin in diameter, so that more of
them can be used in the space available than
electrical wires; There is very little chance of
detecting the signal being transmitted along the fibre
optic cable from outside, so they are more secure
than electrical cables.
9 a Decibels, dB
⎛P ⎞
b Number of decibels = 10 log ⎜ o ⎟
⎝ Pi ⎠
c i
⎛P ⎞
Attenuation = 10 log ⎜ o ⎟ = 10 log
⎝ Pi ⎠
= −20 dB
⎛P ⎞
ii Attenuation = 10 log ⎜ o ⎟ = 10 log
⎝ Pi ⎠
= −10 dB
⎛P ⎞
iii Attenuation = 10 log ⎜ o ⎟ = 10 log
⎝ Pi ⎠
insulator
Po
= 0.13
Pi
10 a Attenuation
∴
⎛P ⎞
= 10 log ⎜ o ⎟ ⇒ Po
⎝ Pi ⎠
Pi × 10
( attenuation
)
10
b Coaxial cables are bulky and expensive to produce.
Optical fibres are less bulky (so more can be used
in the same space to carry more signals), and are
cheaper to produce.
8 They are immune to electromagnetic interference;
They suffer less from attenuation than metal wires;
More frequencies can be transmitted at the same time
Pi × 10
( −0 105×8 0 )
( attenuation
)
10
= 50 mW × 10
conducting care
Figure AC.20
( )
P
⎛P ⎞
So, −8.75 = 10 log ⎜ o ⎟ ⇒ 10 −0 875 = o
Pi
⎝ Pi ⎠
∴ Po = 12 mW
b Attenuation
⎛P ⎞
= 10 log ⎜ o ⎟ ⇒ Po
⎝ Pi ⎠
metallic shield
( )
= −3.0 dB
d Specific attenuation: the amount of attenuation
per unit length of the optical fibre.
e Total attenuation = 0.35 × 25 = 8.75 dB
= 30 mW × 10
plastic cover
( )
( −0.410×2.5)
∴ Po = 40 mW
c Attenuation
⎛P ⎞
= 10 log ⎜ o ⎟ ⇒ Po
⎝ Pi ⎠
Pi × 10
( attenuation
)
10
= 20 mW × 10
( −1 510×12.0 )
∴Po = 0.32 mW
ANSWERS
113
⎛P ⎞
11 a Gain = 10 log ⎜ o ⎟ , used when Po > Pi
⎝ Pi ⎠
P
⎛P ⎞
b 5 = 10 log ⎜ o ⎟ ⇒ o = 100 5 = 3 16
P
Pi
⎝ i⎠
c Attenuation along fibre = 8 × −0.3 = −2.4.
Gain from repeater = 2.4.
So, net gain = 0
⎛P ⎞
0 = 10 log ⎜ o ⎟ ⇒ Po Pi
⎝ Pi ⎠
So, there is no net loss of signal.
Exercise C.4 – Medical imaging
1 a Electrons are accelerated thorough high potential
differences and then directed at a heavy metal
target, such as tungsten. The rapidly decelerated
electrons provide the required energy for X-rays.
b Most of the energy from the fast-moving electrons
is transferred to internal energy of the target. By
rotating the target, the energy spreads around the
target ensuring that a particular section does not
become too hot and melt or distort.
c i Soft X-rays have a lower frequency (higher
wavelength) and lower energy. They cannot
penetrate materials as well as hard X-rays.
ii Hard X-rays have a higher frequency (lower
wavelength) and higher energy. They can
penetrate further into a material than soft X-rays.
2 a From any source of X-rays, diffraction causes the
beam to diverge. This reduces the intensity of the
beam because it is more spread out.
b A high-energy X-ray can be absorbed by an
electron in one of the close orbits to the nucleus. If
this absorbed energy is sufficient, the electron will
be ejected from the atom. This removes X-rays
from the beam – and is the most significant
contributor to attenuation.
c Soft X-rays may interact with atoms of a material.
Such interactions can alter the direction in which
the X-ray photons travel, removing some of the
X-rays from the beam.
d In this process, X-rays can interact with electrons in
the outer shells of atoms, losing only part of their
energy to enable the electron to be ejected from
the atom. Conservation of momentum dictates that
the direction of the X-ray photon is altered,
removing some of the X-rays from the beam.
114
e If an X-ray passes near a nucleus of one the
material’s atoms, the X-ray photon energy may be
used to create a particle and its anti-particle pair.
This will also remove some of the X-rays from the
beam.
3 a i μ is the linear absorption coefficient – a
measure of how much attenuation will occur
for each metre length of material that the
X-rays pass through.
ii Units for μ are m−1, cm−1 or mm−1
b μ = ρ μm, where μm is the mass absorption
coefficient and ρ is the density of the material.
4 a I = Io
b μm =
− μx
⇒ I = e −0 04 ×10 = 67%
Io
μ
= 65 = 1.4 × 10−2 m2 kg−1
ρ 4500
c i I = Io
− μx
⇒ x 1 = ln 2 = 0.693 = 23.1 cm
μ
0 03
2
ii 2 × 23.1 = 46.2 cm
5 a μ = ln 2 = 0.693 = 0.17 mm-1
x1
4.0
2
b I = e − μ x = e −0 17 × 5.0 = 43%
I0
6 a To see the radiogram clearly, X-rays must be
absorbed by the bones (because they are denser
and so have a larger linear absorption coefficient)
but not absorbed by the tissue around the bones.
This will give good contrast.
b The photoelectric effect.
c i The various different tissues in the stomach
have linear absorption coefficients that are
almost the same value. This means that X-rays
will be absorbed about the same amount by
each different tissue in the stomach. This will
not provide any contrast on the radiogram and
so no detail will be seen by the radiographer.
ii Better contrast is achieved if the patient has a
barium meal (usually a barium liquid that they
drink). The barium absorbs the X-rays well (it
has a large linear absorption coefficient)
compared to the tissue around where the
barium is able to go. This will provide good
contrast on the radiogram for the radiographer
to see any problems.
7
a Monochromatic means that the X-rays all have
the same energy.
b From the graph, the half thickness is x 1 = 3.5 mm.
2
c μ = ln 2 = 0.693 = 0.20 mm−1
x1
3.5
2
8
a An intensifying screen consists of one or two
layers of fluorescent material placed next to the
image-capturing film. When hit by X-rays, the
layers emit light. Since the image-capturing film
is more sensitive to light than to X-rays, a
brighter image is produced.
b X-rays scattered by the tissue in a human body
will blur the image on the radiogram. To avoid
this, a two-dimensional collimating grid of lead
strips is placed between the patient and the
image-capturing film. Any X-rays not travelling
perpendicularly to the lead strips are absorbed,
improving the quality of the radiogram. To avoid
the radiogram showing the collimating grid, the
grid is oscillated in the horizontal direction.
9
In a CAT scan, an X-ray source provides images
which are captured by a set of photodetectors
linked to a computer. The X-ray source and the
detector can be rotated about the axis of the
patient, so that a series of images can be produced
at different angles. Then, by moving the whole
system along the axis of the patient, this can be
repeated. These images are then combined by the
computer imaging system to provide a threedimensional picture of the patient.
10 a Ultrasound is any sound waves with a frequency higher
than the human audible range (higher than ~20 kHz).
b In medical diagnostics, ultrasound is usually in the
range of 1-10 MHz.
c Advantages: ultrasound does not dump energy
that can ionise atoms of vital organs, as X-rays do
(this makes ultrasound safer). It is non-invasive,
quick and inexpensive.
Disadvantages: the higher wavelengths of
ultrasound makes them more susceptible to
diffraction and so the images are likely to be less
detailed; It cannot image bone material; It cannot
image the lungs or digestive organs because they
contain gases that have very different acoustic
impedances compared to the organs around them,
causing large reflective loss.
11 a Z = ρ c, where ρ is the density of the material and
c is the speed of the ultrasound waves through the
material.
b Z has units of kg m−3 × m s−1 ≡ kg m−2 s−1
c At a boundary between two materials, some of
the ultrasound energy will be reflected. To
minimise how much energy is reflected, the
acoustic impedance of the two materials is made
to be as similar as possible – thus making it appear
as if there is no boundary there.
d The acoustic impedances of air and human skin
are different. So, by using a gel between the
ultrasound emitter/receiver and the skin, the
ultrasound waves do not have to travel through
air. They travel through the gel, which has an
acoustic impedance very similar to that of skin.
This minimises the loss of energy by reflection.
12 An A scan provides a one-dimensional picture. It is
produced as a graph of signal strength against time. This
shows the relative position of features within the body
because of the relative reflections of the ultrasound.
A B scan is a two-dimensional picture of the body.
The ultrasound emitter/receiver is moved across the
body back and forth, so that the ultrasound waves
are reflected off features within the body at different
angles. By combining these reflections, the computer
can build up a detailed two-dimensional image.
13 a Protons
b Any parts of the body where water molecules are
present.
c Spin
d The application of the radio frequency signal
causes the protons to flip from their up-spin state
(the lowest energy state) to their down-spin state
(a higher energy state).
e When the radio frequency signal is switched off ,
the protons return to their up-spin state and emit
a photon of energy, which is the difference
between the two states. These photons have a
frequency equal to the frequency of the signal
and can be detected.
f The Larmor frequency is the forcing frequency of
the radio frequency signal at which the flip of the
protons from one spin state to the other occurs. It
is a resonant frequency for the spin states of the
protons.
ANSWERS
115
g The Larmor frequency is linearly dependent on
the strength of the magnetic field. So, the Larmor
frequency is largest where the magnetic field
strength is largest. When protons emit photons
during their return flips to the low energy spin
states, different magnetic field strengths cause
different Larmor frequencies. Different frequency
photons are emitted by the protons. Detecting
these different frequencies gives information
about where the photons were emitted.
h By detecting the different frequency photons, it is
possible to gain information about where the
photons were emitted in the body. The number of
photons emitted is an indication of the amount of
material in that place. Such information can be
combined to give a detailed two-dimensional
picture of a slice through a body.
14 See Table AC.2.
Imaging
technique
Resolution Advantages
Disadvantages
X-ray
0.5 mm
Cheap to run
Radiation risk;
Cannot image
all organs
CT scan
0.5 mm
Distinguishes
between
different
tissue types
Radiation risk
Ultrasound 2 mm
scan
No radiation
risk; Portable
Cannot image
bones
MRI scan
No radiation
risk;
Distinguishes
between
different
tissue types;
Very high
quality images
Expensive;
Cannot be used
with all patients;
Requires long
exposures
Table AC.2
116
1 mm
Exam-style questions
1 a See Figure AC.21.
concave
mirror
virtual
image
r
f
Figure AC.21
b i
Virtual. Rays do not actually come from
where the image occurs, so it is not possible to
place a screen there to form an image.
ii Magnified
iii Upright
c This kind of mirror is used to produce an enlarged
image of someone’s face – perhaps for shaving or
putting on make-up.
2 a See Figure AC.22.
convex
mirror
r
f
virtual
image
Figure AC.22
b i
Virtual. The rays do not actually come from the
image, so it is impossible to place a screen there
and form an image.
ii Upright
iii Diminished
b i
3 a See Figure AC.23.
Spherical aberration occurs because rays of
light passing through all of a spherical lens do
not come to a focus in the same place. This
produces a blurred image and can also cause
distortion of the image.
See Figure AC.24.
B
A
ii
O
fe
fe
OR
Figure AC.23
Figure AC.24
b 1 = 1 + 1⇒ 1 = 1 − 1 = 1 − 1
f u v
v
f u 1 80 2 00
= 1 ⇒ v = 18.0 cm
18
c The image from the objective must become the object
of the eyepiece and its image must be 25 cm away.
c Chromatic aberration. This is caused by different
colours of light having slightly different refractive
indices, making a blurred and coloured image.
6 a The objective lens.
b 84.0 + 12.0 = 96.0 cm
f
c M = o = 84.0 = 7.0
f e 12.0
d Angle subtended by image of Moon
= M × angle subtended by Moon = 7 × 0.4 = 2.8°
7 a Absorption and scattering (mostly Rayleigh
scattering, because the size of the atoms of the glass
fibre are smaller than the wavelength of the light
being transmitted).
b See Figure AC.25.
Attenuation/dB km–1
The distance of the intermediate image from the
objective lens to the eyepiece needs to be less than
the focal length of the eyepiece.
b Normal adjustment means the lenses are arranged
so that the image is formed at the near point from
the eyepiece – that is, the image will be formed 25
cm from the eyepiece.
c Virtual. Rays do not actually come from the image,
so it would not be possible to put a screen where the
image appears to be and see the image on the screen.
4 a The shorter of the two focal lengths should be the
objective lens, so use the lens with the focal length
of 1.80 cm as the objective.
1 = 1 − 1 = 1 − −1 = 6 ⇒ u
= 4.2 cm.
u f v 5 25 25
So, the two lenses must be 18.0 + 4.2 = 22.5 cm
apart.
( )
⎛
⎞
mo 1 + 25 ⎟ = − 18 +
= −54
fe ⎠
2
⎝
5 a The size of the objective lens is the most
significant factor in limiting the ability of
the telescope to see the most distant objects.
This is because as the objective is made larger and
larger, it suffers from distortion due to its own
weight. It is also more difficult to mount
accurately.
d M
1.5
1.0
Rayleigh absorption
Scattering
0.5
0
1.0
1.2
1.4
1.6
λ/μm
1.8
Figure AC.25
c i
Small inhomogeneity in the structure of the
glass fibre that forms the core of the optical
fibre will cause Rayleigh scattering.
ii The amount of scattering is proportional
to λ−4.
iii The glass fibre making up the core has to be
as pure as possible to reduce the effects of
scattering – as well as absorption.
ANSWERS
117
d The repeater must:
• re-shape the pulse to eliminate the effects of
dispersion and restore the step function in
power, and
• increase the amplitude of the signal to restore
the amount of energy being transmitted to its
original value.
8
9
3
a λ = v = 1 55 × 160 = 0.52 mm
f
3 × 10
b To resolve a feature clearly requires the size of the
feature to be larger than twice the wavelength of
the ultrasound waves. So, a feature that is about
1 mm in size will be the smallest feature that can
be clearly resolved.
c Using the equation f = 200 v ,
d
3
d = 200 × v = 200 ×11 55 6 10 = 10 cm
f
3 10
a i Zair = ρ c = 1.3 × 330 = 429 kg m−2 s−1
ii Zflesh = ρ c = 1000 × 1550
= 1.55 × 106 kg m−2 s−1
b Fraction of intensity transmitted
=
Z flesh )
2
(
)
c The acoustic impedance of the gel is the same as
that of the flesh. This makes the fraction of
4Z 2 = 1.
(Z Z )2
So, all of the ultrasound is transmitted.
transmitted intensity:
Option D: Astrophysics
Exercise D.1 – Stellar quantities
1 a Planet: celestial body that is in orbit around a star,
has sufficient hydrostatic equilibrium to render it a
spherical (or nearly spherical) shape and has cleared
its orbit of any material that might obstruct its
motion around the star.
b Asteroid: small rocky body orbiting the Sun in
between the orbits of Mars and Jupiter.
c Comet: celestial body orbiting the Sun in a highly
elliptical orbit, composed of frozen ice, debris and
rock and exhibiting a tail of material forced off the
118
Sun
1 parsec
distant star
1 AU
6
4Z air Z flesh
(Z air
= 4 × 429 × 1 55 102 = 0.11%
429 + 1 55 106
comet by the Solar wind when the comet
approaches the Sun.
d Moon: natural satellite of a planet; a body orbiting
around the centre of mass of the planet–moon
system.
e Galaxy: collection of a large number of stars
together with some gas and dust, held together by
their mutual gravitational attraction.
f Cluster: collection of galaxies held in relatively close
proximity by their mutual gravitational attraction.
g Constellation: recognisable pattern of stars in the
sky that is often identified with a mythological
figure.
h Nebula: large cloud of interstellar gas and dust.
2 A light year is the distance that light can travel in one
year through space.
So, 1 ly = 3 × 108 m s−1 × (365 × 24 × 60 × 60)
= 9.46 × 1015 m
3 A parsec is the distance away of a star that subtends a
parallax angle of one second of arc. See Figure AD.1.
Earth
1 second
of arc
Figure AD.1
4 From simple geometry, 1 AU = 1 parsec × 1 second
of arc (expressed in radians)
So, 1 pc
11
1 AU
=
= 1 5 × 10 m
1 × 1 × 2π
1 second of arc ( in radians )
60 60 360
16
= 3.09 × 1016 m = 3 09 1015 = 3.26 ly
9 46 10
5 800 pc = 800 × 3.09 × 1016 = 2.47 × 1019 m
6 Since the speed at which light travels through space is
3 × 108 m s−1, and using the equation, v = s , it will
t
11
take light a time, t = s = 1 5 × 108 = 500 s = 500
v
60
3 × 10
light minutes = 8.33 light minutes
7 1 pc is 3.09 × 1016 m, and 1 AU = 1.5 × 1011 m, so
the difference between the distance from a star to
the Sun and the distance from a star to the Earth is
11
maximum of 1 5 × 10 16 = 5 × 10−4 %. This is too
3 09 10
small a difference to be of concern.
d The repeater must:
• re-shape the pulse to eliminate the effects of
dispersion and restore the step function in
power, and
• increase the amplitude of the signal to restore
the amount of energy being transmitted to its
original value.
8
9
3
a λ = v = 1 55 × 160 = 0.52 mm
f
3 × 10
b To resolve a feature clearly requires the size of the
feature to be larger than twice the wavelength of
the ultrasound waves. So, a feature that is about
1 mm in size will be the smallest feature that can
be clearly resolved.
c Using the equation f = 200 v ,
d
3
d = 200 × v = 200 ×11 55 6 10 = 10 cm
f
3 10
a i Zair = ρ c = 1.3 × 330 = 429 kg m−2 s−1
ii Zflesh = ρ c = 1000 × 1550
= 1.55 × 106 kg m−2 s−1
b Fraction of intensity transmitted
=
Z flesh )
2
(
)
c The acoustic impedance of the gel is the same as
that of the flesh. This makes the fraction of
4Z 2 = 1.
(Z Z )2
So, all of the ultrasound is transmitted.
transmitted intensity:
Option D: Astrophysics
Exercise D.1 – Stellar quantities
1 a Planet: celestial body that is in orbit around a star,
has sufficient hydrostatic equilibrium to render it a
spherical (or nearly spherical) shape and has cleared
its orbit of any material that might obstruct its
motion around the star.
b Asteroid: small rocky body orbiting the Sun in
between the orbits of Mars and Jupiter.
c Comet: celestial body orbiting the Sun in a highly
elliptical orbit, composed of frozen ice, debris and
rock and exhibiting a tail of material forced off the
118
Sun
1 parsec
distant star
1 AU
6
4Z air Z flesh
(Z air
= 4 × 429 × 1 55 102 = 0.11%
429 + 1 55 106
comet by the Solar wind when the comet
approaches the Sun.
d Moon: natural satellite of a planet; a body orbiting
around the centre of mass of the planet–moon
system.
e Galaxy: collection of a large number of stars
together with some gas and dust, held together by
their mutual gravitational attraction.
f Cluster: collection of galaxies held in relatively close
proximity by their mutual gravitational attraction.
g Constellation: recognisable pattern of stars in the
sky that is often identified with a mythological
figure.
h Nebula: large cloud of interstellar gas and dust.
2 A light year is the distance that light can travel in one
year through space.
So, 1 ly = 3 × 108 m s−1 × (365 × 24 × 60 × 60)
= 9.46 × 1015 m
3 A parsec is the distance away of a star that subtends a
parallax angle of one second of arc. See Figure AD.1.
Earth
1 second
of arc
Figure AD.1
4 From simple geometry, 1 AU = 1 parsec × 1 second
of arc (expressed in radians)
So, 1 pc
11
1 AU
=
= 1 5 × 10 m
1 × 1 × 2π
1 second of arc ( in radians )
60 60 360
16
= 3.09 × 1016 m = 3 09 1015 = 3.26 ly
9 46 10
5 800 pc = 800 × 3.09 × 1016 = 2.47 × 1019 m
6 Since the speed at which light travels through space is
3 × 108 m s−1, and using the equation, v = s , it will
t
11
take light a time, t = s = 1 5 × 108 = 500 s = 500
v
60
3 × 10
light minutes = 8.33 light minutes
7 1 pc is 3.09 × 1016 m, and 1 AU = 1.5 × 1011 m, so
the difference between the distance from a star to
the Sun and the distance from a star to the Earth is
11
maximum of 1 5 × 10 16 = 5 × 10−4 %. This is too
3 09 10
small a difference to be of concern.
9
10
11
12
13
14
Since a star with a parallax of one second of arc is
1 pc away, a star with a parallax five times bigger than
this must be 1 the distance away. So, this star is
5
0.2 pc away.
A star with a parallax angle of 1 of a second of arc
100
will be 100 pc away from us. This forms the basis for
the maximum distance a star can be for us to use the
parallax method for finding its distance.
a Luminosity: the total amount of energy being
radiated by a star per second – or the total
radiated power.
b Apparent brightness: the amount of energy
received at the Earth per second per unit area –
or the received power per unit area.
Its surface area, A, where A = 4 π r2, and the fourth
power of its temperature, T4.
So, we may write, L = σ A T4
a L = σ A T4
= 5.67 × 10−8 × 4 π × (7 × 108)2 × 57004
= 3.7 × 1026 W
b L = σ A T4
= 5.67 × 10−8 × 4 π × (8.2 × 1011)2 × 35004
= 7.2 × 1031 W
c L = σ A T4
= 5.67 × 10−8 × 4 π × (4.9 × 1010)2 × 11 2004
= 2.7 × 1031 W
L Procyon σ AProcyonTP4rocyon 4 × 65304
=
=
=69
LSun
σ ASunTS4un
57004
LSirius A
σ ASi us ATSi4rius A
= 25.4 =
LSun
σ ASunTS4un
=
So,
15
σ 4 π (rSiriius
4
)2 TSSirius
ius A
2
4
σ 4 π (rSSun ) TSSun
4
rSirius A
TSu4 n
= 25.4 × 4Sun
= 25.4 × 57004 = 1.7
rSun
TSirius A
9940
LX σ 4 π r T
T
=
= 500 and X = 20
LY σ 4 π r T
TY
2
X
2
Y
So,
rX
=
rY
4
X
4
Y
500 = 0 06
204
(
)
)
2
4
σ 4 π rBetelgeuse TBete
L
lgeuse
16 Betelgeuse =
2
4
L Rigel
σ 4 π rRRigel TRRigel
(
=
(
)2 ( × SSun )4
( × SSun )2 (2 × TSSun )4
×
SSun
2
4
= 11002 × 04 6
70 × 2
=20
26
17 b = L 2 = 3.8 × 10
4π d
4π
×
(
)
2
= 1.3 × 103 W m−2
5.0 × 1028
18 b = L 2 =
4πd
4π ×
×
(
Lα
4 π (dα )
b
19 α =
LSun
bSun
2
4 π (d
= 152 ×
(
)2
)
2
= 2.8 × 10 −6 Wm−2
Lα (d
2
)
=
2
LSun (dα )
)
2
×
×
×
= 2.1 × 10 −9
Exercise D.2 – Stellar characteristics and stellar
evolution
1 a See Figure AD.2.
Spica
Betelgeuse
Intensity
8
Wavelength
Figure AD.2
b A hotter star will have the peak in its emission
spectrum at a lower wavelength.
2 The dark lines are caused by the absorption of light
by elements in the outer layers of the star. Absorption
and re-radiation by such elements will result in a
lower intensity – hence the dark line – because the
re-radiation occurs in all directions, removing energy
from the observer’s line of sight.
ANSWERS
119
3 a These wavelengths correspond to the spectrum of
wavelengths of hydrogen.
b This series of lines is called the Balmer series for
hydrogen.
c Absorption occurs because electrons in the n = 2
energy level of the hydrogen atom absorb energy
and jump up to n = 3 (for the wavelength 656 nm),
to n = 4 (for the wavelength 486 nm), to
n = 5 (for the wavelength 434 nm) and to n = 6
(for the wavelength 410 nm).
d The hotter surface temperature of Vega means that
more electrons in hydrogen atoms are likely to
exist in the n = 2 energy level. In the case of the
Sun, with a lower surface temperature, there will
be some electrons that have enough energy to exist
in the n = 2 level, but not as many as in Vega. So,
the absorption lines in Vega will be more
pronounced.
e See Figure AD.3. Since the peak of the emission
spectrum occurs at a smaller wavelength for Vega
than for the Sun, Wien’s law shows that it must be
at a higher temperature than the Sun.
Vega
Sun
So, a star having the peak wavelength in its
emission spectrum at 650 nm will have a surface
temperature
Intensity
−3
given by: T = 2 9 × 10 −9 = 4500 K
650 × 10
Wavelength
Figure AD.3
4 a The star that is hotter than Vega may be hot
enough for the electrons in its hydrogen atoms to
exist in energy levels above the n = 2 level. Any
absorption that would then occur would be at
wavelengths that are too long to be observed (i.e.
in the infrared part of the electromagnetic
spectrum).
b Stars were once classified by the absorption lines
from hydrogen that occurred in their spectra.
Those stars with the strongest absorption lines
were classified as Type A stars and stars with
120
progressively weaker absorption lines were
classified as Type B, Type C, and so on. Because we
understand that the strength of the absorption lines
is a function of the stars’ temperatures, stars are now
classified according to their surface temperatures.
This is why we now have the progression of stars
classes as O, B, A, F, G, K, M where Type O stars are
the hottest.
5 a The surface temperature of the Sun is cooler than
that of Vega. This means that electrons in hydrogen
atoms are less likely to exist in excited energy
levels, such as n = 2 or above. Fewer electrons in
energy level n = 2 means less absorption in the
visible wavelength range.
b Since a large number of electrons in hydrogen
atoms in the outer layers of the Sun will exist in
the ground state (n = 1) then absorption will
mostly occur for wavelengths corresponding to the
energy differences of levels n = 1 and n = 2,
n = 3, n = 4, and so on. These energy differences
correspond to ultraviolet wavelengths.
6 Wien’s displacement law states that;
λ peakT = 2 9 × 10 −3 m K
−3
7 a λ peak = 2 9 × 10 = 6 7 × 10 −7 m
4300
b With the peak in the spectrum at this
wavelength (almost right at the far red end of
the visible spectrum), there will be a large amount
of energy at wavelengths that are too long to be
visible. These infrared wavelengths will add to
the visible part of the emission to make the
luminosity (the total emitted power) larger than
one would expect by considering the visible
wavelengths only.
8 a A Hertzsprung–Russell diagram is a graph showing
the relationship between the luminosity of stars
and their surface temperatures.
b, c and d See Figure AD.4.
supergiants
Luminosity relative to Sun
106
Betelgeuse
104
red
giants
102
Vega
main
sequence
1
Sun
10–2
Sirius B
white
dwarfs
10–4
0
40 000 20 000 10 000 5 000 2 500
Temperature/K
Figure AD.4
9 a Main sequence stars have a range of temperatures
and luminosities that, on the HR diagram, form a
band from top left to bottom right.
b Red giant stars have quite small surface
temperatures and large luminosities, putting them
in a clump above and to the right of the main
sequence.
c Supergiant stars have extremely large luminosities
and a mid-range of temperatures, putting them in
a clump above the red giants on the HR diagram.
d White dwarf stars have high surface temperatures
and small luminosities, putting them in a region
below and to the left of the main sequence on the
HR diagram.
10 A main sequence star’s luminosity is proportional to
its mass to the power 3.5 ( ∝ M3 5)
11 (10M Sun )
35
103 5
M Sun 3 5 = 3162M Sun 3 5, which is
about 3200 LSun .
12 53.5 ≈ 280. So, the luminosity of the star will be
280 × 3.8 × 1026 = 1.1 × 1029 W.
13 Gravitational potential energy is transformed into
thermal energy as the cloud of gas contracts and
heats up.
14 a Thermonuclear fusion means that small nuclei
fuse to form heavier particles.
b The end products are helium nuclei and energy.
15 When the core of the star has converted 12% of the
star’s hydrogen into helium.
16 After moving off the main sequence, the star will
become a red giant star. After that, it will become
a planetary nebula and leave behind a small core. If
the mass of this core is less than 1.4 solar masses, it
will become a white dwarf. Since the white dwarf
cannot continue to produce energy, it will gradually
cool and fade until it becomes a brown dwarf and
eventually it will become too cold to radiate in the
visible part of the electromagnetic spectrum; it will
be a black dwarf.
17 In a white dwarf star, the gravitational forces are
balanced by the electron degeneracy pressure. This is
a consequence of the Pauli exclusion principle.
18 C 109 kg m−3
19 It will become a supergiant star once it leaves the
main sequence. After a period of continued fusion
of heavier elements, it undergoes a supernova
explosion, leaving behind a dense core. If the mass
of this core is less than 1.4 solar masses (this is called
the Chandrasekhar limit) then the core will become
a white dwarf. If the mass of the core is 1.4-3.0 solar
masses (called the Oppenheimer-Volkoff limit), it
will become a neutron star. If the mass of the core
is greater than three solar masses, it will become a
black hole.
20 The Chandrasekhar limit is the maximum mass
that a white dwarf star can have. For a mass below
this limit, electron degeneracy pressure can oppose
gravitational forces and allow the star to be a stable
white dwarf. For a mass larger than this, electron
degeneracy pressure is no longer sufficient to oppose
the gravitational forces and the star will contract
further into a neutron star.
21 Neutron degeneracy pressure.
22 The Oppenheimer–Volkoff limit states that for
masses greater than three solar masses, the core of
the star will not be able to provide sufficient
neutron degeneracy pressure to keep the star in
hydrostatic equilibrium. The core will collapse
into a black hole.
ANSWERS
121
23 a See Figure AD.5.
Luminosity relative to Sun
106
supergiants
104
red
giants
102
1
Sun
10–2
white
10–4 dwarfs
0
40 000 20 000 10 000 5 000 2 500
Temperature/K
Figure AD.5
b As the star moves off the main sequence it
expands. This causes its luminosity to increase and
its surface temperature to decrease. After a period
of time in the red giant region of the HR
diagram, the star undergoes a planetary nebula
event and the core left behind will move from the
red giant region into the white dwarf region
because it is now smaller and hotter. The effect of
the reduced surface area is greater than the effect
of the increased temperature and so the
luminosity of the star decreases.
24 At some distance away from the black hole, the
escape velocity becomes smaller than the speed
of light, allowing radiation to escape the huge
gravitational field. This is often called the event
horizon. Matter (from, for example, a binary
companion star) that has been attracted by the huge
gravitational field of the black hole emits X-rays
as it speeds up on its path towards the black hole.
Astronomers can observe this X-ray emission from
the region outside the event horizon and infer the
existence of a black hole.
Exercise D.3 – Cosmology
1 a Newton considered that:
• the universe was infinite
• the universe was static
• stars were uniformly distributed.
122
b Olbers’ paradox asks the question: ‘Why is the sky
dark at night?’ If Newton’s model of the universe
was correct then there would be, in every direction
one observed, a star emitting radiation towards us.
This would not allow the sky to be dark at night.
So, the fact that the sky is dark at night suggests
that Newton’s model is wrong.
2 The spectral lines observed from a distant star occur
at wavelengths that are longer than they would be
if observed in a laboratory. The lengthening of the
wavelength makes the colours of the spectral lines
move towards the red end of the visible spectrum.
It occurs because of the Doppler effect. If the star is
moving away from us, then the radiation it emits will
be redshifted.
3 v = z c ⇒ v = 0.04 × 3 × 108 = 1.2 × 107 m s−1
4 Hubble’s law states that the recessional speed of a
galaxy is proportional to its distance from the Earth.
v = H d, where v is the recessional speed of the galaxy,
H a constant of proportionality (called Hubble’s
constant) and d is the distance from the galaxy to the
Earth.
This leads to the Big Bang model of the universe
because almost all galaxies observed from the Earth
display redshift, so are seen to be moving away from
us. Extrapolating backwards suggests that the universe
began as something very small and expanded.
( − ) × 10 −7 = 0.075
5 a z = Δλ =
λ
5 3 × 10 −7
b z = v ⇒ v z c = 0.075 × 3 × 108 = 2.3 × 107 m s−1
c
23 000
c v = H d ∴d = v =
= 320 Mpc
H
72
v = zc = Δλ × c
6 d=
H H
λ
H
−9
( − ) × 10 × 3 × 105 = 570Mpc
=
72
660 × 10 −9
7 a v = H d, so if we consider v to be c and d to be ct,
we have: c = H ct or = 1 , where t is the time for
H
which the universe has been expanding; i.e. its age.
b H = 72 km s−1 Mpc-1
4
1
= 7 2 × 10 ms
= 2 33 10 −18 s.
22
3 09 10 m
1 =
1
= 4 29 1017 s
H 2 33 10 −18
17
= 4 29 10 7 = 1 36 1010 years
3 16 10
(a little less than 14 billion years).
8 H = v . So, any uncertainty in H is caused by
d
uncertainty in v and d. The uncertainty in v is caused
by the difficulty in measuring the redshift from very
distant galaxies, since their observed spectra are so
dim. The uncertainty in d is caused by the difficulty
in locating a Cepheid variable star within the galaxy
to be used as a standard candle, and hence being able
to measure accurately the star’s luminosity in order to
find d for its galaxy.
9 The cosmic microwave background (CMB) radiation.
This radiation is observed in all directions and
is considered to be the remnants of much more
energetic radiation that has cooled. This supports
the idea that the early universe was dominated by
very hot and energetic radiation and that, during the
expansion of the universe, this radiation has cooled
and moved to longer wavelengths.
The ratio of hydrogen to helium in the universe is
about 3:1 in terms of mass. This is consistent with
calculations that show that the early universe would
have had about this ratio of hydrogen to helium if the
universe had been created by a Big Bang event.
10 a The spectrum of the CMB is that of a black
body with a temperature, given by its peak
wavelength, of about 2.7 K. This is a cooler
temperature than the universe would have had a
long time ago.
b This spectrum is observed to be more or less the
same in every direction of observation. This
supports the idea that the universe looks the same
in all directions of observation: i.e. that it is
essentially isotropic.
c Fine detail observed in the CMB distribution
shows that some regions of the universe show
greater intensity of CMB and some regions
of the universe show smaller intensities of
CMB than the mean value. This supports
the idea that, in some regions of the universe, the
density of material is different to the mean
density.
So,
11 Wien’s displacement law gives:
−3
−3
T = 2 9 × 10 = 2 9 × 10−3 = 2 6 K
λpeak
1.1 × 10
12 Distant galaxies exhibit redshift. This redshift is
explained by the space between the galaxy and the
Earth expanding (and, by inference, that any relative
motion of the galaxy in that space is negligible
compared to the expansion of the space itself), rather
than the fact that the galaxy is moving away from
us. It is the rate at which the space is expanding
that is causing the z factor in the cosmological
redshift, whereas the z factor in the Doppler effect
explanation for redshift is the relative motion of the
observed object away from us.
13 The cosmic scale factor, R, is the factor by which
space in the universe is being stretched as the
universe expands. For example, if radiation of
wavelength λ1 had been emitted at time t1 when the
scale factor was R1 , and is observed at a time t2 to
have a wavelength λ2 , then the scale factor will have
changed to a value R2 such that:
λ 2 − λ1
R
or z = Δλ = 2 − 1
λ1
λ
R1
If z = 3.5, then v = 3.5 c = 3.5 × 3 × 108
= 1.1 × 109 m s-1
Rnow
− 1 = 3.5 ∴ Rnow 4.6 Rthen. So, the universe
Rthen
is 4.6 times bigger now than it was when the
photons were emitted.
Type 1a supernovae are used as standard candles
because their luminosity is well known. When
observations of distant type 1a supernovae showed
that they appeared less bright than expected, it
was concluded that they were further away than
expected. This suggested that the universe had
expanded more than had been expected – leading
to the idea that the universe was expanding at an
increasing rate.
Because scientists have linked the force required
to accelerate the universe with some kind of
energy source, it has been proposed that dark
energy must exist, which is invisible, and which
must be present in the universe to oppose the
force of gravity caused by the universe’s mass.
R2
R1
R1
14 a
b
15 a
b
=
ANSWERS
123
Exercise D.4 – Stellar processes
1 The cloud’s energy consists of: negative gravitational
potential energy (GPE) and positive kinetic energy
(KE) of all the atoms in the cloud. For the cloud to
collapse into a star that will be able to produce energy
by thermonuclear fusion, the sum of the GPE and
the KE must be less than zero. This is called the Jeans
criterion for a star.
A cold cloud of gas will have low KE, so does not
require much mass for its GPE to overcome the KE of
the particles and make the total energy less than zero.
A hot cloud of gas will likely have a low density and
a large amount of KE, so requires a large mass for its
GPE to overcome the KE and make the total energy
less than zero.
1
1
2
0 + 0
2 a 1 p + 1 p → 1 H + 1β + 0 ν
1
b 1p
2
1
H → 32 He +γ
c 32 He + 32 H
He → 42 He + 2 11 p
3 Two possible answers:
The mass of the 42 He nucleus is less than the mass of
the four protons that make it. The difference in mass
has been converted into energy, as given by E = m c2.
4
2 He has more binding energy (a negative quantity)
per nucleon than protons do. This negative value is
due to a release of positive kinetic energy, the massenergy required to produce the other particles in the
p-p chain and some electromagnetic energy in the
gamma rays emitted.
4 a 11 p 126 C → 137 N + γ
b 137 N
1
c 1p
C + 01β + + 00 ν
13
14
6C → 7N + γ
d 11 p
14
7
e
15
8
1
1
O
13
6
N → 158 O + γ
15
7
N + 01β + + 00 ν
f p 157 N → 126 C 42 He
5 In the carbon–nitrogen–oxygen (CNO) cycle, the
electrical potential energy between protons and
nuclei of, for example, carbon and nitrogen, is much
higher than it is between two protons. So, protons
fusing with nuclei have more kinetic energy to
transfer into electrical potential energy. More kinetic
energy means the temperature of the core must be
around ten times higher.
6 a 42 He + 42 He
H → 48 Be
Two helium nuclei fuse together to make unstable
beryllium.
124
b 42 He + 48 Be
B → 126 C
If a helium nucleus can fuse quickly enough with
an unstable beryllium nucleus, carbon can be
formed.
c Oxygen can then be formed by another helium
nucleus fusing with a carbon nucleus:
4
12
16
2 He + 6 C
8O
7 Iron-56 and nickel-62 have the highest binding
energy per nucleon of all nuclei. Energy must be
put into the system to produce a nucleus that is
heavier than either of these; they cannot fuse simply
into heavier nuclei than iron or nickel and produce
energy.
8 The s–process requires a small flux of neutrons
(these are by products of other processes involving
carbon, oxygen and silicon), which allows nuclei to
absorb a neutron, which then decays by β− decay.
Because the process happens slowly, there is enough
time for the neutron to decay. This produces nuclei
of higher atomic number.
9 In the r–process, a large neutron flux means that
the capture of neutrons by nuclei happens relatively
easily in very short times. Neutrons do not have
time to decay (by the usual β− decay) so the nucleus
formed will be a heavier isotope of the same
element.
10 A type 1a supernova occurs from a binary star
system, one star of which is a white dwarf with
a large gravitational field. Material from the
companion star is attracted to the white dwarf,
which increases the mass of the white dwarf. When
the mass of the white dwarf becomes larger than the
Chandrasekhar limit (1.4 solar masses), the white
dwarf will collapse (because the electron degeneracy
pressure is insufficient to balance the gravitational
force). Further fusion of carbon and oxygen
produces such a large amount of radiation pressure
that the star explodes into a supernova.
Since the supernova occurs because the mass of
the white dwarf has become 1.4 solar masses, the
resulting luminosity of the star will be a constant
value – hence the idea of it being a standard candle.
This constant value can then be used to find the
distance of the supernova from the Earth.
11 A type 2 supernova occurs after several stages of
fusion of successively heavier nuclei in stars heavier
than the Sun. Nuclei fuse in the core of a star until
the supply of nuclei runs out. This makes the star
collapse (because the hydrostatic equilibrium has
been lost and gravitational force is dominant). The
result of the collapse is a rapid heating of the core,
which enables heavier nuclei to fuse together until
the supply of these nuclei runs out. The process
repeats itself, producing shells of successively heavier
nuclei around the core until the innermost shell
is made from iron. Once iron is reached, no more
fusion can occur and hydrostatic equilibrium is
lost for the last time. The star collapses under its
gravitational force, which is too strong for electron
degeneracy pressure, and neutrons are produced. The
resulting neutron degeneracy pressure is extremely
large and the star explodes, producing elements
heavier than iron.
Type 1a supernovae have a luminosity that is higher
than type 2 supernovae (about 10 times higher) and
this luminosity decreases at a decreasing rate over
the next year or so.
Type 2 supernovae have a luminosity that decreases
sharply for a few days and then levels out for a
month or so before decreasing sharply again. After
about three months, the luminosity decreases
gradually for about a year.
Exercise D.5 – Further cosmology
1 a Einstein suggested that the universe was
homogeneous (its composition was essentially the
same everywhere) and isotropic (the universe looks
the same when observed in all directions). These
two conditions form the cosmological principle.
b Observations of the CMB have shown that, on a
large scale, the density of radiation is more or less
the same everywhere, from all directions. This
supports the idea of the universe being
homogeneous and isotropic.
On a smaller scale, however, structures are observed
in the CMB. These support the idea that smallscale fluctuations in the density of material in the
universe are necessary for the formation of stars
and galaxies.
2 If we consider the universe to be a homogeneous
sphere of radius, r, and density, ρ, and that there is a
galaxy of mass, m, on the edge of the universe, then
we have:
GPE of the galaxy
3
2
M = − G π r ρm = − 4Gπρr m
= − GMm
r
3r
3
KE of the galaxy = 1 mv 2 = 1 mH 0 2r 2
2
2
(because v = H 0 r ).
When these two energies are equal, the total energy
of the galaxy is zero.
2
2
So, 4Gπρr m = 1 m H 2r 2 ⇒ ρ = 3H 0
0
c
3
2
8Gπ
2
⎛ 72 × 103 ⎞
⎝ 3 09 × 1022 ⎠
3H 0 2
3 ρc =
=
= 9 7 × 10 −27 kg m
8Gπ 8 × 6.67 × 10 −11 × π
If we take the mass of a nucleon to be about u, then
this critical density corresponds to:
3×
3
9 7 × 10 −27 ≈ 6 nucleons m −3
1 67 10 −27
4 Density, ρ = m
V
It is difficult to obtain reliable measurements of:
• the mass of extremely distant galaxies
• extreme distances (and hence volumes) because the
objects observed are so dim.
ρ
5 The current density parameter, Ω 0 = , where ρ is
ρc
the actual density of the universe and ρc is the critical
density. In the absence of dark matter (Ω0 = 0) there
are three possible fates for the universe:
• Ω0 >1: Expansion of the universe will slow down,
stop and then contract into a Big Crunch; this is
the ‘closed universe’ model.
• Ω0 = 1: The universe will continue to expand, but
at a decreasing rate; this is the ‘flat universe’ model.
• Ω0 < 1: The universe will continue to expand at
a constant rate forever; this is the ‘open universe’
model.
6 See Figure AD.6.
3
R
Ω0 < 1
2
Ω0 = 1
1
0
Ω0 > 1
Now
Time
Figure AD.6
ANSWERS
125
200
150
with dark matter
100
50
no dark matter
0
0
10
20
30
40
50
Radial distance/kpc
Figure AD.7
c The rotation curve for the galaxy shows a flat
region outwards from the central core of the galaxy
instead of a curve downwards showing a decreasing
rotational speed. This flat region suggests the
presence of dark matter in a halo around the galaxy.
8 a Massive compact halo objects (MACHOs) and
weakly interacting massive particles (WIMPs).
b MACHOs are likely to be black holes, neutron
stars and brown dwarfs.
c WIMPs are likely to be sub-atomic, non-baryonic
particles or possibly neutrinos.
Exam-style questions
1 a See Figure AD.8. [2]
Force 1 = Gravitational force
Force 2 = Thermal or radiation
Figure AD.8
b Hydrostatic equilibrium means that these two
forces are balanced, so that there is no overall force
acting on the star. [2]
c If the gravitational force is bigger than the thermal
force then the star will contract. If the thermal
force is bigger than the gravitational force then the
star will expand. [2]
2 a Supergiant region. [1]
b More massive. [1]
c 1.4 solar masses. [1]
126
d Eventually the material ejected from the star will
form a new nebula and allow the formation of
new stars. [1]
3 a A Cepheid variable star is a star with luminosity
that varies in time in a regular pattern over a
period of several days.
b See Figure AD.9.
Intensity
Rotational velocity (km s–1)
7 a and b See Figure AD.7.
Time/days
Figure AD.9
c The luminosity varies in this way because the star
is not in hydrostatic equilibrium. Its surface area
varies because it expands and contracts. Since
luminosity is proportional to the star’s surface area,
a changing surface area will produce a changing
luminosity.
d For Cepheid variable stars, there is a relationship
between the peak luminosity of the star and the
period of its changing luminosity. (This
relationship is linear on a graph of log (L) against
log (time).
e Method:
Use the period of the changing brightness to find
the peak luminosity of the star from the graph of
log(L) against log(t).
Use the equation linking brightness and
luminosity
to find the distance: b =
L .
4πd 2
Measurements:
The brightness of the star – measured with
a telescope and a CCD device.
The period of its changing luminosity – by
measuring its brightness over a period of several
days.
4 a The strong nuclear force.
q2
b EPE = 1
= 9 × 109 ×
4πε o r
= 7 7 × 10 −14 J
(
×
3 × 10 −15
)
2
c Each proton would need about 3.85 × 10−14 J of
kinetic energy.
Using the equation 1 mv 2 = 3 kT , the temperature,
2
2
T is given by:
2
T = 2 × 3 85 × 10 −14 =
× 3 85 × 10 −14
−23
3k
3 × 1.38 × 10
9
= 1 86 × 1100 K
d This temperature is not high enough to account
for the protons being able to approach each other
within 3 fm; there must be another process that
allows them to get close enough for the strong
force to make them fuse.
e Two possible answers:
Consider the distribution of kinetic energies
associated with a temperature of 10 MK. There will
be a small number of protons with energy
significantly higher than that of the modal energy;
high enough to satisfy the requirement to
overcome the EPE of the protons.
Some of the protons will ‘quantum tunnel’
through/past the potential barrier posed by the
EPE. Because of the Heisenberg uncertainty
principle, a proton can increase its energy by the
amount required to pass the potential barrier, as
long as this increase in energy occurs for a small
enough time to satisfy the equation: ΔE Δt ≈ h
4π
5 a This equation suggests that the universe is
composed of 32% matter and 68% energy; but of
this matter, about 27% is dark matter, which has
not been observed and identified. The 68% dark
energy has also not been observed directly. The
equation also suggests that the overall density of
the energy-plus-matter in the universe is equal to
the critical density, and so the universe will follow
a flat universe model and expand forever at a
decreasing rate.
b if 84% of Ωm is dark matter then only 16% of 32%
of the matter in the universe has been observed so
far. This is about 5% of the universe. Therefore, so
far, we are able to observe only about 5% of the
universe.
6 a i A main sequence star is one that is in
hydrostatic equilibrium and converting protons
into helium by the process of nuclear fusion.
ii Luminosity means the total amount of energy
that is radiated by the star.
⎛
⎞
b L =⎜ M ⎟
LSun ⎝ M Sun ⎠
35
1
⎛
⎞35
⇒ M = ⎜ L × M Sun 3 5 ⎟
⎝ LS
⎠
1
⎛ ⎞35
= ⎜ L ⎟ × M Sun
⎝ LS ⎠
1
= 3000 3 5 × M Sun = 9.85 MSun ≈ 10 M Sun
c The star will expand to become a red supergiant.
After that it will undergo a supernova event
leaving behind a neutron star or a black hole.
2 9 × 10 −3 = 2 9 × 10 −3 = 2 6 K
7 a T=
λ peak
1.1 × 10 −3
b The Big Bang model states that the universe began
in a very hot state and, as it expanded, its
temperature decreased. The high energy photons in
the early universe have been ‘stretched’, changing
them into longer wavelength photons associated
with lower temperatures.
c The Doppler shift of spectral lines shows a
lengthening of their observed wavelengths. This is
consistent with the source of the photons receding
from us – as one would expect from a universe that
is expanding.
ANSWERS
127