1 Quantities and units
Answers to Test yourself questions
Page 1 Test yourself on prior knowledge
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
= 1.3 m s-1
Average speed =
2
Time =
3
metres per second, per second; m s–2
4
Speed is a scalar quantity. It is the distance travelled (in any direction) divided by the time taken. Velocity is
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑠𝑝𝑒𝑒𝑑
=
=
1600 𝑚
1
20 𝑥 60 𝑠
200 𝑚
8.0 𝑚 𝑠 −1
= 25 s
a vector quantity. It is the distance travelled in a specific direction divided by the time taken.
Page 4 Test yourself
1
Force in newtons, N; work done in joules, J; power in watts, W
2
a)
N ≡ kg m s–2
b)
N s ≡ kg m s–2 × s ≡ kg m s–1
a)
240 mg ≡ 2.40 × 10−4 kg
b)
470 pF ≡ 0.470 µf
c)
11 GHz ≡ 1.1 × 1010 Hz
a)
T = 4l2f2µ
b)
Units: (m2) (s–1)2 (kg m–1) ≡ kg m s−2 ≡ N
3
4
Page 7 Test yourself
5
Vectors: displacement, velocity, acceleration and force; scalars: mass, distance, time and speed
6
a)
17 km
b)
13 km
c)
R = 8.0 N, θ = 62 °
a)
Horizontal component = 18 cos 25° = 16 m s−1
b)
Vertical component = 18 sin 25° =7.6 m s−1
7
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
1 Quantities and units
Answers to Exam practice questions
Pages 8–9 Exam practice questions
1
a)
Metre and second are base units and speed is a derived quantity; the only base quantity is
therefore length – the answer is A.
b)
Force is a derived quantity; metre and second are both base units; the newton, N, is the only
derived unit – the answer is C.
c)
[1]
Acceleration and force both have implied direction and so are vector quantities; metre is a unit;
speed has magnitude only, and so is a scalar quantity – the answer is D.
d)
[1]
[1]
Distance, mass and time have magnitude but not direction and are all scalar quantities; velocity
is the only vector quantity – the answer is D.
[1]
[Total 4 Marks]
2
F = PA = 10 × 106 Pa × 220 × 10−6 m2 [1] = 2.2 × 103 N = 2.2 kN [1]
3
Horizontal component = 120 cos 30 = 104 m s–1
–1
Vertical component = 120 sin 30 = 60 m s
[Total 2 Marks]
[1]
[1]
[Total 2 Marks]
4
From Table 1.2:
• The units of Q are: A s
[1]
• Those of V are: kg m2 A−1 s–3
[1]
Therefore the units of C =
As
= kg–1 m–2 A2s4
kg m2 A −1 s −3
[1]
[Total 3 Marks]
5
a)
11 km [1]; S 27°E [1]
b)
Average speed = 3.0 km h–1
[1]
Average velocity = 2.2 km h–1 in the direction S 27 °E
[1]
[Total 4 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
1 Quantities and units
Answers to Exam practice questions
6
[1]
v = 2.1 m s–1 [1] at 76 ° to the bank [1]
[Total 3 Marks]
=
7
 pr 4 t
8Vl
units =
Pa m4 s
= Pa s
m3 m
[1]
[1]
[Total 2 Marks]
8
F22 = (6.6 N)2 + (4.7 N)2
[1]
F2 = 8.1 N at 35 ° to the 6.6 N force
[1]
[Total 2 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
2 Practical skills
Answers to Test yourself questions
Page 24 Test yourself
1
a)
For example:
Average
l/mm
w/mm
t/mm
95
62
43
96
62
43
95(.5)
62
43
Volume, V = lwt
= 9.55 cm × 6.2 cm × 4.3 cm
= 255 cm3
Density =
=
𝑚
𝑣
250 𝑔
255𝑐𝑚3
= 0.98 g cm–3 (980 kg m–3)
b)
Percentage uncertainty in l =
1 𝑚𝑚
95.5 𝑚𝑚
Percentage uncertainty in w =
Percentage uncertainty in t =
1 𝑚𝑚
62 𝑚𝑚
1 𝑚𝑚
43 𝑚𝑚
× 100% = 1.0%
× 100% = 1.6%
× 100% = 2.3%
The overall percentage uncertainty is therefore in the order of 5% (probably more when also
taking into account any manufacturing tolerance and the mass of the wrapper). This means that
the density of the butter could lie between about 0.95 g cm–3 and 1.03 g cm–3.
Although the experimental value of 0.98 g cm–3 suggests that butter will float, the experiment
may not be sensitive enough to confirm this beyond all doubt. (You might like to check what
happens with a small piece of butter in a cup of water!)
2
a)
i) For example, mass of packet of paper, M = 2.52 kg
ii) Mass of single sheet, m =
b)
12520 𝑔
500
= 5.04 g
i) For example:
l/mm
297
297
Average: 297
w/mm
210
210
Average: 210
ii) Area A = 0.297 m × 0.210 m = 0.0624 m2
'gsm' =
5.04 𝑔
0.00624 𝑚2
= 80.8 g m–2
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
2 Practical skills
Answers to Test yourself questions
iii) Percentage difference
=
(80.8 −80)𝑔 𝑚−2
80 𝑔 𝑚 −2
This is acceptable experimental error, particularly when taking into account the mass of the
packet and the lack of sensitivity of the kitchen scales.
c)
i) For example, measured thickness of packet/mm
= 48, 47, 48, 49 = average of 48 mm
Thickness of single sheet, t =
Density of paper =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
48 𝑚𝑚
500
5.04 𝑔
29.7 𝑐𝑚
= 0.096 mm = 0.0096 cm
x 0.096 mm = 0.0096 cm
= 0.84 g cm–3 (840 kg m–3)
NOTE: The final answer can be quoted only to the number of significant figures of the least
precise of your measurements. In this case, the answer can only be stated to two significant
figures as t has only been measured to 2 s.f.
i) The thickness of a single sheet of paper could be checked as follows:
•
First check the micrometer screw gauge or digital calipers for zero error.
•
Fold the paper four times to get 16 thicknesses.
•
Compress to remove any air.
•
Measure 16t in four different places.
•
Take the average and hence find t.
NOTE: Notice the experimental techniques given here – multiple readings (16t) taken in different
places – and the use of bullet points. This is a good strategy as it helps you set out your answer in
a clear and logical manner. Examiners love bullet points.
3
a)
For example:
10d/mm
10d/mm
Mean d/mm297
203
203
20.3
10t/mm
10t/mm
Mean t/mm297
16.5
16.5
1.65
Note that d and t have been found by measuring the length of ten coins in a row and the height
of ten coins, respectively.
b)
V=
𝜋𝑑 2 𝑡
4
=
Density =
𝜋×(2.03 𝑐𝑚)2 × 0.165 𝑐𝑚
4
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
3.56 g
0.534 𝑐𝑚3
= 0.534 cm3
= 6.7 g cm–3
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
2 Practical skills
Answers to Test yourself questions
c)
i) Percentage difference =
(7.8 −6.7)g 𝑐𝑚3
7.8 g 𝑐𝑚3
ii) Percentage uncertainty in 10d =
Percentage uncertainty in 10t =
× 100% = 14%
1 mm
200 mm
0.5 mm
16.5 mm
× 100% = 0.5%
× 100% = 3%
The fact that the percentage difference between the experimental value for the density of
the coins and the given value for the density of mild steel differs by 14%, which is much more
than the experimental uncertainty, suggests that the coins are not made of mild steel.
However, the true value for the average thickness is considerably less than that measured.
When 10 coins are stacked on top of each other, the thickness measured is actually the
thickness of the rim of the coin and not its average thickness. A better average value could be
obtained by using a micrometer and measuring the thickness in several places.
d)
The percentage difference between the density of brass (8.5 g cm–3) and that of mild steel (7.8 g
cm–3) is:
(8.5 −7.8)g 𝑐𝑚 −3
8.15 𝑔 𝑐𝑚3−
× 100% = 9%
This is more than the estimated experimental uncertainties, so it should be possible to
distinguish between the two types of coin. In particular, the difference in 10 thicknesses would
be (16.5 – 15.2) mm = 1.3 mm, which can easily be detected with a rule. If only one coin were
available, the difference in thickness would be 0.13 mm, which could be detected easily with a
micrometer or digital callipers.
Note: the discussion and conclusions in parts (c) and (d) have been argued on the basis of
quantitative evidence. You must remember to do this wherever possible.
4
a)
If h = ½gt2, t2 =
2ℎ
𝑔
=
2ℎ
𝑔
xh
A graph of t2 against h therefore should be a straight line through the origin of
2
gradient equal to .
𝑔
b)
h/cm
40
60
80
100
120
t/s
0.30
0.38
0.42
0.47
0.51
t2/s2
0.090
0.144
0.176
0.221
0.260
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
2 Practical skills
Answers to Test yourself questions
Your graph should look like this:
c)
The graph is a straight line but does not pass through the origin. This suggests a small systematic
error. There has possibly been a systematic error in measuring the distance, h (perhaps caused
by measuring to the wrong place each time) or, more likely, there has been a systematic error in
t due to time delays in releasing the sphere or opening the trap door.
d)
Gradient = 0.250 s2 𝟐
𝒈
0.250 𝑠 2 −0.015 𝑠 2
1.14 m −0.00 𝑚
= 0.206 s2m–1
= 0.206
g=
2
0.206
= 9.7 m s–2
(9.8 −9.7)m 𝑠 −2
e)
Percentage difference =
f)
i) The values of h have been recorded only to the nearest centimetre when they would have,
9.8 m 𝑠 −2
× 100% = 1%
presumably, been set to a precision of 1 mm. The values should therefore have been
recorded as 40.0 cm, 60.0 cm, etc. to reflect this.
ii) Repeat timings should definitely have been recorded, with at least three values for each
height. More readings, with larger values of h if possible, would also improve the results.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Test yourself questions
Page 25 Test yourself on prior knowledge
1
A vector quantity has both magnitude and direction.
2
distance = average velocity × time = 4.5 m s–1 × 8.0 s = 36 m
3
36 km h–1 = 10 m s–1; 18 km h–1 = 5 m s–1
acceleration =
=
change in velocity
time interval
−5 m s−1
8s
= −0.63 m s −1
(note that the minus sign means that it is slowing down)
4
𝑎=
∆𝑣
∆𝑡
Δv = a × Δt = 9.8 m s–2 × 2.2 s = 22 m s–1
5
Straight line, drawn through the origin
Page 26 Activity 3.1
1
2
a)
δt1 = 0.62 s; δt2 = 0.16 s
b)
𝛥𝑡=4.92 −0.41=4.51 s (mid-point to mid-point)
a)
𝑢=
b)
𝑣ave =
=
5.0×10−2 m
0.62 s
= 0.081 m s −1 𝑣 =
5.0×10−2 m
0.16 s
= 0.313 m s −1
𝑢+𝑣
2
0.081 ms−1 +0.313 ms−1
2
= 0.197 m s −1
3
∆𝑥
𝑣ave =
=
∆𝑡
0.800 m
4.51 s
=0.177 m s−1
4
Reasons for different average velocities: the velocities are not instantaneous; the acceleration may not
have been uniform.
Page 29 Test yourself
1
Rectilinear means that the quantities are all along the same line, so they have a specified direction.
2
𝑎=
=
𝑣−𝑢
𝑡
5 m s−1 −2 m s−1
0.5 s
= 6 m s −2
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Test yourself questions
3
v=u+a
t =
=
v−u
a
6.0 m s−1 −4.0 m s−1
0.8 m s−2
= 2.5 s
4
𝑠 = 𝑢𝑡 +
1
2
𝑎𝑡 2
= 0 + ½ × 0.5 m s–2 × (20 s)2
= 100 m
5
v2 = u2 + 2as
= (5 m s–1)2 + 2 × 0.8 m s–2 × 12 s
= 44.2 m2 s–2
v = 6.6 m s–1
Page 30 Core Practical 1
1
Method 1; t2 values (s2) 0.084, 0.16. 0.24, 0.32, 0.42
Method 2; v2 values (m2 s–2) 4.00, 7.72, 11.1, 14.8, 20.7
2
The graph is a straight line through the origin; gradient in the range 0.20 to 0.21 m–1 s2
𝑔=
3
= 9.5 to 10 m s −2
The graph is a straight line through the origin; gradient in the range 19.2 to 19.6 m s–2
𝑔=
4
2
gradient
2
gradient
= 9.6 to 9.8 m s −2
The values of h are all too big by the same amount; i.e. there is a systematic error in the measurement of h.
This could be due to a zero error on the scale, or measurements taken to mid-point of the ball rather than
to the bottom.
5
The times measured in Method 2 are much smaller than those for Method 1. The precision of the timer is
the same so the percentage uncertainties will be larger for Method 2.
6
Light gates give a more precise start and stop reading. With the electromagnetic switch there may be a
delay between the current being switched off and the magnet losing its magnetism. It is also easier to
interface the light gates with a computer.
7)
a)
Five readings are taken so that an average value will improve the accuracy of the value and
eliminate any readings that are well outside the average.
b)
Average time = 0.64 s ± 0.06 s; % uncertainty = ±9%
c)
Advantage: the time is much longer and so the percentage uncertainty will be smaller;
Disadvantage: high speeds could lead to air resistance affecting the motion.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Test yourself questions
Page 32 Test yourself
6
a)
s = ut + ½at2u
= 0 m s–1a
= 9.8 m s–2
𝑡=√
2×30 m
9.8 m s−2
= 2.5 s
b)
v = u + at
= 0 m s–1 + 9.8 m s–2 × 2.5 s
= 24 m s–1
Or use v2 = u2 + 2as;
v = √(2 × 9.8 m s −2 × 30 m)
= 24 m s–1
7
c)
Air resistance has a negligible effect.
a)
v = u + at
= 8.0 m s–1 – 9.8 m s–2 × 0.5 s
= 3.1 m s–1
b)
v2 = u2 + 2as;
v = 0 m s–1;
a = −9.8 m s–2
0 m s–1 = (8.0 m s–1)2 – 2 × 9.8 m s–2 × s
s = 3.3 m
H = 1.8 m + 3.3 m
= 5.1 m
c)
s = ut + ½at2
−5.1 m = 0 + ½ × 9.8 m s–2 × t2
t = 1.0 s
Page 34 Activity 3.2
Although the bullet is moving horizontally, it is also falling vertically. As the can and the bullet are both
accelerating down with an acceleration of 9.8 m s–2 they will both have fallen the same distance in the same time.
Provided the bullet was correctly aimed at the can at the start, it will always hit the can.
Page 35 Test yourself
8
Horizontal and vertical components of the velocity are independent. In the vertical plane both coins will
accelerate downwards at 9.8 m s–2 over the same vertical displacement.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Test yourself questions
9
a)
In the vertical plane:
s = ut + ½at
60 m = 0 + ½ × 9.8 m s–2 × t2
t = 3.5 s
In the horizontal plane:
s = horizontal velocity × time
= 200 m s–1 × 3.5 s
= 700 m
10
b)
Air resistance is not taken into account.
a)
uh = 40 cos 40 ° m s–1 = 31 m s–1
b)
uv = 40 sin 40 ° m s–1 = 26 m s–1
c)
In the vertical plane, v = u + a t
0 = 26 m s–1 − 9.8 m s–2 × t
where t = time to reach maximum height = 2.7 s
Time in flight = 2t = 5.4 s
d)
Range = horizontal velocity × time = 31 m s–1 × 5.4 s = 167 m
Page 37 Test yourself
1
The graph for experiment 1 is a straight line through the origin; for experiment 2 the line curves upwards.
2
The gradient of the graphs represent the velocity of the trolleys. In experiment 1 the line has a constant
gradient and so the velocity is uniform; in experiment 2 the gradient is increasing so the velocity is
increasing – the trolley is accelerating.
3
The gradient of graph 1 is 0.68 m s–1. This is the same at all times. The velocity at 5.0 s for experiment 2 is
the gradient of the tangent drawn at 5.0 s. The value should be in the range of 1.0 to 1.2 m s–1.
Page 39 Test yourself
11
a)
Horizontal line parallel to the time axis
b)
Straight line passing through the origin
c)
Line through origin curving upward
12
The gradient represents the velocity
13
a)
Horizontal line parallel to the time axis
b)
Straight line passing through the origin
14
The gradient represents the acceleration.
15
The area underneath the graph represents the displacement.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Exam practice questions
Pages 42–43 Exam practice questions
1
2
a)
a=
v − u (40 − 0) m s −1
=
= 0.50 m s −2 – the answer is B
t
80 s
[Total 1 Mark]
b)
s=
u+v
(40 + 0) m s −1
t =
 80 s = 1600 m – the answer is B
2
2
[Total 1 Mark]
The gradient of a displacement–time graph represents
the velocity – the answer is D
3
[Total 1 Mark]
The area under a velocity–time graph represents
the displacement – the answer is B
4
Average velocity =
[Total 1 Mark]
total displacement
time
Instantaneous velocity =
[2]
x
at an instant
t
[2]
[Total 4 Marks]
5
s = ½(u + v) × t
[1]
v = u + at
[1]
s = u t + ½at2
[1]
2
2
v = u + 2as
[1]
[Total 4 Marks]
6
v
t
a)
Acceleration is the change of velocity in unit time: a =
b)
i) Average acceleration ≈ 3 m s–2
[1]
ii) Gear changes are likely to affect the acceleration.
[1]
[1]
[Total 3 Marks]
7
a)
v = 12 m s–1
[1]
b)
s = 160 m
[1]
[Total 2 Marks]
8
a)
s = ut + ½at2 = 0 + ½ 9.8 m s–2 × (2.2 s)2
= 24 m
b)
v = u + at = 0 + 9.8 m s–2 × 2.2 s
= 22 m s–1
[1]
[1]
[1]
[1]
[Total 4 Marks]
9
a)
7.3 m
[1]
b)
1.2 s
[1]
c)
7.6 m s–1 (downward)
[1]
[Total 3 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Exam practice questions
10
a)
Vertical component = 10 m s–1
Horizontal component = 17 m s–1
[1]
b)
t = 2.0 s
[1]
c)
x = 36 m
[1]
[Total 3 Marks]
11
a)
b)
A = stationary
[1]
B = uniform velocity
[1]
C = uniform velocity in the opposite direction to B
[1]
D = increasing velocity (acceleration)
[1]
A = constant velocity
[1]
B = uniform acceleration
[1]
C = uniform negative acceleration
[1]
D = increasing acceleration
[1]
[Total 8 Marks]
12
a)
−0.2 m s–2
[2]
b)
1800 m
[2]
c)
10 m s–1
[1]
[Total 5 Marks]
13
a)
Any six of the following:
labelled diagram
[1]
measure height with metre rule
[1]
measure time to fall
[1]
repeat several times at each height
[1]
details of timing method
[1]
range of heights
[1]
use of s = ut + ½at2
[1]
2
b)
plot graph of t against h
[1]
gradient = 2/g
[1]
For a height of about 2 m, the time is about 0.6 s
The % uncertainty in h is (0.001 m)/(2.000 m) × 100% = 0.05%
[1]
for t, the value is (0.01 s)/(0.60 s) × 100% = 1.7%
[1]
Therefore, time has the greater effect on the uncertainty in g.
[1]
[Total 9 Marks]
14
a)
i) Horizontal distance = 120 cos 36 ° = 97 m
ii) Horizontal velocity =
97 m
= 21.6 m s–1
4.5 s
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Exam practice questions
iii) Vertical displacement = 120 sin 36° m + 3 m = 74 m
iv) Use s = uv t + ½ a t
[1]
2
−74 m = uv × 4.5 s − ½ × 9.8 m s–2 × (4.5 s)2
uv = + 5.7 m s–1
[1]
The displacement and acceleration are downward and assigned negative values. The initial vertical
component of velocity is positive and so is upward.
v) θ = tan−1
b)
5.8
22
= 15 °
[1]
vi) u 2 = 21.62 + 5.82 ; u = 22.4 m s−1
[1]
The jumper is affected by air resistance
[1]
The position and shape of the body relative to the skis affects the trajectory of the jump.
[1]
[Total 8 Marks]
15
a)
i) The graph should be an approximate ‘S’ shape.
[3]
ii) The velocity is found by measuring the gradient at each time; e.g.
t/s
v/m s−1
0
0
0.10
0.95
0.20
2.63
0.30
4.30
0.40
1.25
0.50
0.33
0.60
0.00
[3]
iii) The graph should rise almost linearly up to 0.3 s and then fall in an exponential-like curve
[3]
iv) Over the first 0.3 s the graph is approximately linear, indicating a uniform acceleration. The
velocity decreases, rapidly at first and then more slowly until the car stops. The negative gradient
shows that the negative acceleration (deceleration) of the car is decreasing.
b)
[3]
The computer samples data at very small time intervals [1] whereas the gradient of a tangent to the
graph needs to be drawn to find the velocity
[1]
[Total 14 Marks]
16
a)
Horizontal component of velocity, uh = 12 cos 30 ° = 10.4 m s−1
Time to reach the plane of the net, t =
1.4 m
= 0.135 s
10.4 m s −1
Vertical component of velocity, uv = 12 sin 30 ° = 6.0 m s−1
Vertical displacement, s =ut + ½at
[1]
[1]
[1]
2
= 6.0 m s−1 × 0.135 s – ½ × 9.8 m s−2 × (0.135 s)2
[1]
= 0.72 m
[1]
Height of ball at the net = 0.20 m + 0.72 m
= 0.92 m
[1]
This is greater than the height of the net (0.90 m)
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
3 Rectilinear motion
Answers to Exam practice questions
b)
Apply the equation, s =ut + ½at2 for the time the ball leaves the racquet until it reaches the ground:
−0.20 m = 6.0 m s−1 × t – ½ × 9.8 m s−2 × t2
[1]
t = 1.26 s
[1]
Note: the solution of quadratic equations is not required in an AS or A-level examination. At home it is
easy to use a quadratic equation solver (e.g. on math.com) or you can do the calculation in two stages,
i.e. use v2 = u2 + 2as to find v and then use v = u + at to determine the time of flight.
Horizontal distance travelled = uh × t = 10.4 m s−1 × 1.26 s
= 13.1 m
[1]
As the horizontal distance from the striking point to the base line is 13.3 m
(11.9 m + 1.4 m), the ball will land in play.
c)
[1]
The effect of air resistance [1] will mean that the range of the ball will be reduced. A ball with
back-spin will also swerve upwards, increasing the height of the shot and further reducing its length.
[1]
[Total 12 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
4 Momentum
Answers to Test yourself questions
Page 45 Test yourself on prior knowledge
1
Distance, mass, speed and time are scalar quantities; acceleration, displacement and velocity are vector
quantities.
2
Mass is the amount of matter in a body. Weight is the gravitational force that acts upon a body.
3
19.6 N–12 N = 7.6 N
4
5.0 N
5
𝑎=
6
FR = 800 N – 200 N
14 m s−1
12 s
= 1.17 m s −2 F = 5.6 kg × 1.17 m s−2 = 6.5 N
= 600 N
= 1.2 × 103 kg × a
a = 0.50 m s−2
Page 46 Test yourself
1
Momentum = mass × velocity
2
Momentum is a vector quantity.
3
kg m s−1 or N s
4
0.640 kg × 25 m s−1 = 16 kg m s−1
5
Mass of rhino of the order of 1–3 tonne; speed up to 15 m s−1
Momentum ≈ 2000kg × 10 m s−1
= 2×104 kg m s−1
Page 46 Activity 4.1
1
t1/s
t2/s
u/m s−1
v/m s−1
Initial momentum/N s
Final momentum/N s
0.64
1.28
0.313
0.156
0.269
0.266
0.73
1.45
0.274
0.138
0.236
0.234
0.48
0.95
0.417
0.211
0.358
0.358
0.60
1.20
0.333
0.167
0.286
0.283
2
The values of the initial and final momentum are the same (±1%) in each case, indicating that the total
momentum has been conserved.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
4 Momentum
Answers to Test yourself questions
3
a)
There should be no resultant external forces acting on the system for the momentum to be
conserved. There will usually be resistive frictional forces acting on the trolley. These forces can
be compensated for by raising the end of the runway so that there is a component of the weight
of the trolley acting down the plane. When this force is equal to the frictional forces, the
resultant force on the trolley is zero. If the trolley is given a gentle push, it will continue to move
down the runway at constant speed, i.e. just as it would on a horizontal track with no friction.
b)
If the runway were horizontal, the frictional forces would lead to a change in momentum – the
total momentum after the collision would be less than that before the collision.
4
To reduce the danger of the trolleys falling off the bench, a means of stopping them safely at the bottom of
the runway is needed. An enclosed area or piece of sponge would suffice. The area around the equipment
should be kept free of bags, books etc.
Page 49 Test yourself
6
There are no external forces acting.
7
Initial momentum = 0.400 kg × 2.0 m s−1;
final momentum = 1.200 kg × v
v = 0.67 m s−1
8
Initial momentum = 0.400 kg × 2.0 m s−1;
final momentum = 0.400 kg × (−0.2 m s−1) + 0.800 kg × v
v = 1.1 m s−1
9
0.6 g × v = (119.4 g + 0.6 g) × 0.3 m s−1
v = 60 m s−1
Page 50 Test yourself
10
Unit of impulse = N s
11
Impulse = Δ(mv)
= 2.5 kg × 4.8 m s−1 – 2.5 kg × 3.2 m s−1
= 4.0 N s
12
𝐹=
=
∆𝑚𝑣
∆𝑡
600 N s −400 N s
20 s
= 10 N
13
Change in momentum = 0.400 kg × 6.0 m s−1 – (−0.400 kg × 4.0 m s−1) = 4.0 N s
14
Initial momentum = 0;
final momentum = −1200 kg × 0.25 m s−1 + 5 kg × v
v = 60 m s−1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
4 Momentum
Answers to Test yourself questions
Page 52 Test yourself
15
N s or kg m s−2
16
The rate of change in momentum of a body is directly proportional to the resultant force applied to it.
17
A closed system has no resultant external forces acting upon it.
18
𝐹=
=
∆𝑚𝑣
∆𝑡
4.8 N s −3.2 N s
0.50 s
= 3.2 N
19
Δ(mv) = 0.06 kg × 10 m s−1 –(–0.06 kg × 30 m s−1)
= 2.4 N s
𝐹=
2.4 N s
0.2 s
= 12 N
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
4 Momentum
Answers to Exam practice questions
Page 54-56 Exam practice questions
1
a)
Bullet’s momentum = 6.0 × 10−3 kg × 450 m s−1
= 2.7 kg m s−1
[1]
b)
Canal barge’s momentum = 12 000 kg × 1.5 m s−1 due East
[1]
c)
50 km h−1 = 50 000 m/3600 s = 13.9 m s−1, so the truck’s momentum is
8700 kg × 13.9 m s−1 = 1.2 × 105 kg m s−1
[1]
[Total 3 Marks]
2
B
[Total 1 Mark]
3
C
[Total 1 Mark]
4
B
[Total 1 Mark]
5
C
[Total 1 Mark]
6
A
[Total 1 Mark]
7
C
[Total 1 Mark]
8
Momentum is conserved. m × 390 km s−1 + 0 = (M + m) × 30 km s−1
360m = 30M i.e. M = 12m
So m is one-twelfth the mass of the carbon nucleus
[Total 1 Mark]
(perhaps it is a neutron)
9
D
[Total 1 Mark]
10
Initial momentum = 0.055 kg × 6 m s−1 down
= 0.33 kg m s−1 down
[1]
Final momentum = 0.055 kg × 4 m s−1 up
= 0.22 kg m s−1 up
[1]
 The change of momentum = 0.22 kg m s−1 up − (−0.33 kg m s−1 up)
= 0.55 kg m s−1 up
[1]
(A simpler solution would be to say that the change of velocity is 10 m s−1 up, etc.)
[Total 3 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
4 Momentum
Answers to Exam practice questions
11
a)
[2]
b)
R2 = (100 kg m s−1)2 + (50 kg m s−1)2
 R = 112 kg m s−1
[2]
[Total 4 Marks]
12 The boat will ‘recoil’ with a speed υ in such a way that the total momentum remains zero.
300 kg × υ = 50 kg × 1.2 m s−1
[1]
 υ = 0.20 m s
−1
[1]
[Total 2 Marks]
13 The area under the graph ≈ 10.5 squares [1], each of which represents 200 N × 0.002 s [1]
 The impulse exerted by the wall on the ball = 10.5 × 200 N × 0.002 s
= 4.2 N s or 4.2 kg m s−1 [1]
Using the impulse–momentum equation, FΔt = mΔυ, the change in velocity Δυ of the ball
Δυ = FΔt/m
= 4.2 kg m s−1/0.0575 kg
= 73 m s−1
[1]
[Total 4 Marks]
14 The rocket car ejects hot gases backwards. These gases are given backward momentum.
As momentum is conserved, the car gains forward momentum.
[1]
[1]
[Total 2 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
4 Momentum
Answers to Exam practice questions
15
[Total 3 Marks]
16
The impulse of the club on the ball is equal to the area under the graph.
[1]
Approximating the curved graph to a single triangle of height 2600 N and base 7.2 m s leads to an
approximate area of ½ × 2600 N × 0.0036 s = 4.68 N s
[1]
−1
The momentum of the 45 g golf ball as it leaves the club is therefore 9.36 kg m s , and so the initial speed
of the club = 4.68 kg m s−1/0.045 kg = 104 m s−1
[1]
[Total 3 Marks]
17
Yes, the Earth recoils because the principle of conservation of momentum always applies.
[1]
The total upward momentum of the Indian people in this experiment is
1 × 109 × 60 kg × 2 m s−1
10
[1]
−1
= 1.2 × 10 kg m s
[1]
so the Earth must recoil in the opposite direction with exactly this momentum.
If the Earth’s recoil speed is υ, then
6 × 1024 kg × υ = 1.2 × 1010 kg m s−1
[1]
giving υ = 2 × 10−15 m s−1 (so none of us on the other side of the Earth will feel the Earth move when one
billion Indians jump into the air)
[1]
[Total 5 Marks]
18
As the helicopter is stationary there is an upward force on it equal to its weight W – Newton’s first law. [1]
As the helicopter forces air down, the air pushes the helicopter up with an equal but opposite force F –
Newton’s third law.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
4 Momentum
Answers to Exam practice questions
The push of the helicopter on the air down, F = υΔm/Δt
 1200 kg × 9.8 N kg−1 = 20 m s−1 × Δm/Δt
[1]
 Δm/Δt = 590 kg s−1
[1]
[Total 4 Marks]
19
The volume of air that reaches the wall in 1 second = 120 m2 × 25 m = 3000 m3
3 −1
−3
[1]
−1
So the mass-rate of arrival of air Δm/Δt = 3000 m s × 1.3 kg m = 3900 kg s
[1]
 the force exerted by the wall on the air υΔm/Δt = 25 m s−1 × 3900 kg s−1
= 98 000 N or nearly 100 kN
[1]
By Newton’s third law the force exerted by the wind on the wall = 98 000 N
[1]
[Total 4 Marks]
20
a)
After the runner’s foot first touches the ground at 0.05 s, the horizontal force of the ground is
backwards, slowing the runner down. [1] This reaches a maximum at t ≈ 0.1 s. [1] At t = 0.175 s
the forward force of the ground on the jogger is zero [1] and after that there is a forward force
on him until his foot leaves the ground at t ≈ 0.34 s.
b)
[1]
The average backward force  120 N, [1] because a horizontal line drawn at this level splits the
area of the curve below the axis into two approximately equal areas.
c)
[1]
The area under the graph represents the jogger’s change of momentum. [1] The backward
change in momentum is the same as the forward change of momentum, so the runner continues
forward at the same velocity as a result of this stride.
d)
[1]
The area is nearly 2½ ‘big’ squares. [1] One big square is equivalent to a change of momentum of
200 N × 0.05 s = 10 N s = 10 kg m s−1. [1] So the area under the graph ≈ 25 kg m s−1.
[1]
[Total 11 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Test yourself questions
Page 57 Test yourself on prior knowledge
1
Horizontal: F = 50 cos 30 ° = 43 N; vertical: F = 50 sin 30 ° = 25 N
2
Resultant horizontal force = 480 N – 80 N
= 400 N
Total resultant force = √(4002 + 5002 )
= 640 N at an angle tan−1
3
500
400
(= 51 °) to the horizontal
Change in momentum = 0.12 kg × 8.0 m s−1 – (−0.12 kg × 6.0 m s−1)
a)
= 1.68 N s
𝐹=
b)
=
∆𝑚𝑣
∆𝑡
1.68 N s
0.04 s
= 42 N
4
𝑎=
=
𝐹
𝑚
100 N
12 kg
= 8.3 m s −2
Page 60 Activity 5.1
1
For example:
The resultant is 5.8 N at an angle of 10 ° to the vertical.
2
The resultant should be 6.0 N acting vertically in order to be in equilibrium with the 6.0 N downward
weight. The difference may be due to frictional forces acting on the pulley axles, errors in marking the
string positions or variations in the 100 g masses.
Page 62 Test yourself
1
Distant: electromagnetic, electrostatic and gravitation; Contact: elastic, friction, viscosity
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Test yourself questions
2
Newton’s first law of motion states that an object will remain in a state of rest or continue to move with a
constant velocity unless acted upon by a resultant external force.
3
When the ball is falling there are no horizontal forces acting upon it.
a)
Level. The initial horizontal velocity is zero so, by Newton’s first law, it will remain at zero during
the fall and hit the floor directly below.
b)
Level. The ball is initially moving at the speed of the train. It will continue to move at this speed
as there are no resultant horizontal forces acting on it. As the train continues to move at the
same speed it will hit the floor directly below the release point.
c)
Level – as b)
d)
Behind. The ball is moving at the same horizontal speed as the train when it is dropped. It will
continue to move at this speed while falling, but the train accelerates and so the floor will move
a greater horizontal distance in the time it takes the ball to fall.
4
The cycle experiences equal and opposite forces to the driving force. The resultant force is zero and so, by
Newton’s first law, the cycle will continue to move at constant speed.
5
A body is in equilibrium if, when acted on by a number of forces, the resultant force is zero.
6
a)
b)
In the vertical plane: 𝑇 cos 30° = 𝑚𝑔 → 𝑇
=
24 kg × 9.8 m s−2
cos 30°
= 272 N
In the horizontal plane: F = T sin 30 ° = 272 sin 30 ° N = 136 N
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Test yourself questions
Page 63 Activity 5.2
1
F/N
t1 /s
t2/s
u/m s–1
v/m s–1
a/m s–2
0.10
0.48
0.29
0.42
0.69
0.38
0.20
0.32
0.20
0.62
1.00
0.76
0.30
0.25
0.16
0.80
1.25
1.15
0.40
0.16
0.12
1.25
1.67
1.53
0.50
0.15
0.11
1.33
1.81
1.92
2
The gradient of the graph should be in the range 3.6–4.0 kg−1. Using 𝑎 =
of
1
𝑚
𝐹
𝑚
the graph will have a gradient
so the mass of the trolley plus the 5 10 g weights is between 0.24 and 0.28 kg. The mass of the trolley
will therefore be about 0.21 kg.
3
The track must be level.
Page 65 Test yourself
7
Newton’s second law of motion states that the rate of change of momentum of a body is proportional to
the resultant force applied to the body (or ΣF = ma if m is fixed).
m = kg; a = m s−2 → N ≡ kg m s−2
8
ΣF = ma
9
Mass is a measure of the quantity of matter of a body while weight is the gravitational force acting upon
the body. The mass of a body remains the same regardless of its position, whereas the weight will change in
different regions of gravitational field strength.
10
v2 = u2 + 2as 0 = (20 m s−1)2 − 2 × a × 40 m → a = 5.0 m s−2
F = ma = 12 × 103 kg × 5.0 m s−2 = 6.0 × 104 N
11
a)
Use Pythagoras: F = 50 N at an angle of 37 ° (tan−1 0.75) to the 40 N force
b)
𝑎=
𝐹
𝑚
=
50 N
50 kg
= 1.0 m s −2
Page 69 Activity 5.4
The position of the centre of gravity of the card can be tested by balancing the card on a pencil point at that point.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Test yourself questions
Page 70 Test yourself
12
Newton’s third law states that if body A exerts a force on body B, then body B will exert an equal and
opposite force on body A.
13
Newton’s third law pairs must always act on two separate bodies; be of the same type; act along the same
line; be equal in magnitude; act in opposite directions.
14
The Earth exerts a downward gravitational force on the object; the object exerts an equal, upward
gravitational force on the Earth. The object exerts a downward elastic (stretching) force on the spring; the
spring exerts an equal, upward elastic force on the object.
15
When walking, the foot exerts a backward contact (frictional) force on the ground. By Newton’s third law,
the ground will exert an equal and opposite frictional force on the foot, pushing it forward. If the ground is
icy the frictional force between the ground and the foot is greatly reduced so the foot slips backwards.
16
The force of the tap on the water, 𝐹 = 𝑣
∆𝑚
∆𝑡
= 0.60 m s −1 ×
12
60
kg s −1
= 1.2 N
17
Clockwise moment = 10 N × 1.2 m = 12 N m; anticlockwise moment = 15 N × 0.9 m = 13.5 N m
Resultant moment = 1.5 N m anticlockwise
18
Balance the soup ladle on your finger. The centre of gravity will be directly above the point of contact.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Exam practice questions
Pages 72–75 Exam practice questions
1
a)
The weight is a downward gravitational force of the Earth on the man. By Newton’s third law
there must be an equal upward gravitational force of the man on the Earth – the answer is C.
[Total 1 Mark]
b)
Only two forces act on the man: his weight pulling him down and the upward push of the table
on his feet. For the man to be in equilibrium these must be equal and opposite – the answer is B.
[Total 1 Mark]
2
Examples of distant forces are: gravitational, electrostatic and magnetic
(electromagnetic) and nuclear.
[2]
Examples of contact forces are: friction, air resistance and normal reaction forces.
[2]
[Total 4 Marks]
3
Examples of gravitational forces are: Sun and planets, satellites, weight of objects on Earth
[1]
Examples of electromagnetic forces are: motors, electrons orbiting the nucleus, forces on static electrical
charges
[1]
Examples of nuclear forces are: strong forces binding protons and neutrons, weak forces involved in beta
decay
[1]
[Total 3 Marks]
4
a)
The resultant force acting on the system = 7000 N − (1000 + 1000) N
= 5000 N.
Using Newton’s second law: a =
b)
F 5000 N
=
= 1.25 m s −2 – the answer is A
m 4000 kg
Resultant force on the trailer = (T − 1000) N = 2500 kg × 1.00 m s−2
T = 2500 N + 1000 N = 3500 N – the answer is B
5
[Total 1 Mark]
a)
A body is in equilibrium if the vector sum of all the forces acting on it is zero.
b)
The student needs:
[Total 1 Mark]
[1]
• A diagram showing weights and hangers and pulley wheels (see figure 5.5)
[1]
• Pencil lines drawn to show the direction of the two outer forces
[1]
• To choose a scale and mark off the appropriate lengths of the force lines
[1]
• To construct a parallelogram
[1]
• To measure the length of the diagonal to determine the magnitude
of the resultant of the two forces
[1]
• The resultant should be equal and opposite to the third weight for the system to be in
equilibrium
[1]
[Total 7 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Exam practice questions
6
a)
Free body force diagram should include all of the 4 forces stated.
[4]
b)
Weight
[1]
c)
i) F = (4.0 kN) cos 15
[1]
= 3.86 kN ≈ 3.9 kN
ii) R + (4.0 kN) sin 15 = 800 kg × 9.8 m s−2
[1]
[1]
⇒ R = 6.81 kN ≈ 7 kN
[Total 8 Marks]
7
a)
[2]
b)
Diagram showing T and W acting at a point
[1]
c)
Any of the above diagrams [1]; R = mg tan 30 ° [1] = 0.28 N [1]; or by scale drawing (± 0.02 N)
[Total 6 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Exam practice questions
8
a)
Newton’s third law states that if body A exerts a force on body B, then body B will exert an equal
and opposite force on body A.
b)
[1]
i) Only one force on each body i.e a force from the centre of the Moon towards the centre of
the Earth and a force from the centre of the Earth towards the centre of the Moon.
[2]
ii) The forces are both gravitational, of the same magnitude, co-linear and act in opposite
directions. (any three)
[3]
[Total 6 Marks]
9
a)
The nozzle exerts a force on the water to shoot it out of the rear of the jet-ski [1], so the water
will exert an equal and opposite force on the jet-ski [1], pushing it forwards.
b)
F= v
F= v
c)
m
t
m
t
= 5.0 m s−1 × 150 kg s−1 = 750 N
[1]
= 6.0 m s−1 × 160 kg s−1 = 960 N
[1]
The resistive forces in part (b) must equal the driving force for the velocity to be constant.
Assuming that these remain the same during the acceleration [1], the resultant force on the jetski will be:
FR = 960 N – 750 N = 210 N
Hence acceleration, a =
F
[1]
=
m
210 N
380 kg
= 0.56 m s−2
[1]
Using v = u + at, final velocity = 15 m s−1 + 0.56 m s−2 × 5.0 s = 18 m s−1
[1]
[Total 8 Marks]
10
a)
N + F – W = 0; or N + F = W
[1]
b)
8.0 kg × 9.8 m s−2 × 0.10 m = F × 0.50 m [1]
c)
Moving the centre of gravity to a lower position would reduce the perpendicular distance of the
F = 16 N
[1]
line of action of the weight from the pivot [1] and so reduce the clockwise moment. The value of
the anticlockwise moment will also reduce [1] and so the size of force F needs to be less to
maintain equilibrium.
[1]
[Total 6 Marks]
11
a)
Resultant force = 4000 N – (2250 + 500) N = 1250 N [1] = 3750 kg × a
a=
b)
c)
1250 N
3750 kg
[1] = 0.33 m s−2
[1]
For the plough alone: T – 2250 N = 1250 kg × 0.33 m s−2
[1]
T = 2670 N
[1]
Coupling force = 500 N [1]; assuming that the resistive force has remained constant
[1]
[Total 7 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Exam practice questions
12
a)
The force on truck B is given by the rate of change of momentum of truck B.
FB =
( mv) B
t
=
45000 kg m s
−1
− 15000 kg m s
−1
[2]
(1.0 s − 0.64 s)
 FB = 83 000 kg m s−2 or 83 000 N
(b)
[1]
The force on truck A is given by the rate of change of momentum of truck A.
FA =
 ( mv ) A
t
=
30000 kg m s
−1
− 60000 kg m s
−1
[2]
1.0 s − 0.64 s
 FA = −83 000 kg m s−2 or −83 000 N
(c)
[1]
The two forces are equal in size but opposite in direction, as we would expect from Newton’s
third law.
13
[2]
The argument is not valid because Newton’s third law cannot be applied to a single body.
[1]
There are two pairs of forces involving the book:
1
the downward gravitational pull of the Earth on the book and the upward gravitational pull of
the book on the Earth
2
[1]
the elastic contact force of the book pushing down on the table and the elastic contact force of
the table pushing upwards on the book
[1]
The book is in equilibrium because the forces acting on the book alone (the downward
gravitational force of the Earth and the upward elastic force of the table) are equal and opposite.
[1]
The resultant force acting on the book is therefore zero and so it is in equilibrium.
[1]
[Total 5 Marks]
14
a)
The Earth pulls the teacher plus suitcase down with a gravitational force of 800 N [1]; the
teacher/case pull the Earth upward with a gravitational force of 800 N. [1]
The suitcase pulls down on the spring with a force of 98 N [1]; the spring pulls the suitcase up
with an elastic force of 98 N [1]
b)
To accelerate upwards the teacher and case need a resultant upward force FR = upward push of
scales – weight of teacher plus case = ma
Scale reading is equal to upward push = ma + mg =m (a + g) = 82 kg × (9.8 + 2.0) m s−2 = 970 N [1]
Spring balance reading = 10 kg × (9.8 + 2.0) m s−2 = 118 N
[1]
At constant velocity the resultant force is zero
Scale reading = 800 N [1]
Spring balance reading = 98 N [1]
Using v2 = u2 + 2as for the initial stage
v2= 0 + 2 × 2 m s−2 × 4 m → v = 4.0 m s−1
[1]
Using v2 = u2 + 2as for the final stage
0 = (4.0 m s−1)2 – 2 × a × 8 m → a = −1 m s−2
[1]
When slowing down there is a resultant downward force = weight – scale reading
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
5 Forces
Answers to Exam practice questions
Scale reading = mg –ma =82 kg × (9.8 − 1.0) m s−2 =720 N
[1]
−2
Spring balance reading = 10 kg × (9.8 − 1.0) m s = 88 N
[1]
[Total 12 Marks]
15
a)
Using v2 = u2 + 2as
b)
v2 = 0 + 2 × 9.8 m s−2 × 0.20 m
c)
→ v = 2.0 m s−1
d)
F=
e)
When the sphere enters the water it experiences an upward force from the water [1]. By
mv
t
=
[1]
0.020 kg  2.0 m s
0.50 s
−1
-0
[1] = 0.080 N [1]
Newton’s third law the sphere will exert an equal downward push on the water [1] that will
result in an increase in the reading on the balance. The average force of the water on the sphere
is 0.080 N, which would show an increase of about 80 g [1] on the balance, but, because this is
an average value, the balance reading will momentarily show a higher value during the first 0.50
s [1].
At the instant that the sphere comes to rest in the water it will be experiencing an ‘upthrust’
from the water [1]. This will produce an equal downward push on the water [1]. The upthrust
must be greater than the weight of the sphere because it causes it to move up towards the
surface. The balance reading at this point will be greater than 820 g [1].
As the sphere moves upwards there will be downward resistive forces [1] that will reduce the
resultant upward force by the water on the ball and so the balance reading will fall [1].
When the sphere is floating, the upthrust equals the weight of the sphere [1]. The sphere
therefore pushes down on the water with a force equal to its weight and so the balance will give
a steady reading of 820 g [1].
[Any 6 marks]
The beaker could be placed on a pressure/force sensor interfaced with a computer. Suitable
software could be used to take readings of pressure or force at very short time intervals and a
display of force against time produced on the screen. [1]
[Total 10 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
6 Work, energy and power
Answers to Test yourself questions
Page 76 Test yourself on prior knowledge
1
Work is done when the point of application of a force is moved through a distance.
2
Work done = 10 N × 25 m = 250 J
3
Work done = 15 kg × 9.8 m s−2 × 3.0 m = 440 J
4
a)
Acceleration =
change in velocity
time
=
4.0 ms−1
8.0 s
= 0.5 m s −2
b)
F = ma
= 5.0 kg × 0.5 m s−2
= 2.5 N
c)
Distance = average velocity × time
= 2.0 m s−1 × 8.0 s
= 16 m
d)
Work = 2.5 N × 16 m = 40 J
5
Electrical energy → thermal energy + radiant energy (light)
5
power =
=
work done
time
120 N × 2.5m
0.5s
= 600 W
Page 81 Test yourself
1
a)
Work = F cos θ × Δx
= 14 N × cos 30 × 40 m
= 480 J
b)
Pushing the pram will create a downward push onto the ground, and so the frictional force
opposing the motion will increase.
2
GPE = mgh
=2.0 × 103 kg × 9.8 m s−2 × 0.80 m
= 16 kJ
3
EPE = ½FΔx
= ½ × 3.0 N × 0.16 m
= 0.24 J
4
KE = ½mv2
= ½ × 7500 kg × (40 m s−1)2
= 6.0 MJ
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
6 Work, energy and power
Answers to Test yourself questions
Page 82 Activity 6.1
1
Δh/m
ΔGPE/J
t/s
v/m s−1
ΔKE/J
0.05
0.20
0.22
0.90
0.16
0.10
0.39
0.16
1.25
0.31
0.15
0.59
0.13
1.54
0.47
0.20
0.78
0.11
1.81
0.66
0.25
0.98
0.10
2.00
0.80
a)
The graph is a straight line through the origin.
b)
ΔKE is proportional to ΔGPE.
c)
The gradient equals 0.80 so 80% of the GPE is converted to KE. This can never be greater than
100% by the law of conservation of energy.
2
a)
t = 0.22 s ± 0.02 s;
percentage uncertainty = ±
0.02 s
0.22 s
× 100%
= ±9%
b)
Δh = 0.250 m ± 0.005 m;
percentage uncertainty = ±
0.005 m
0.250 m
× 100%
= ±2%
Page 84 Test yourself
5
a)
Initial KE of puck = ½ × 0.17 kg × (80 m s−1)2 = 540 J
b)
Work done = 2.0 N × 20 m = 40 J;
residual KE of puck = 540 J – 40 J = 500 J
6
c)
½ × 0.17 kg × v2 = 500 J; v = 77 m s−1
a)
GPE at the highest point is converted to KE as the bob swings towards the mid-point, where the
KE is a maximum. As the bob rises, KE is converted back to GPE.
b)
mgΔh = ½mv2 → v = √(2gΔh) → v = √(2 × 9.8 m s−2 × 0.10 m) = 1.4 m s−1
c)
Both KE and GPE are dependent on the mass.
d)
Spheres of different masses could be of different sizes. The loss of energy from the pendulum to
the surrounding air would be greater for larger bobs.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
6 Work, energy and power
Answers to Test yourself questions
Page 87 Activity 6.2
1
Power output =
0.100 kg × 9.8 m s−2 × 1.20 m
0.8 s
= 1.47 W
Power input = 3.0 W
Efficiency =
2
1.47 W
3.0 W
× 100% = 49%
The low efficiency is due to some electrical energy in the motor being transferred as internal energy in the
coils and some of the output is used as work against resistive forces.
3
The times are quite short and reaction speeds are significant. By repeating several times a lower
uncertainty can be attained by averaging the times. Also, any readings that are clearly different from the
majority can be ignored.
Page 88 Test yourself
7
Power is represented by the width of the bar.
Input width = 16 mm; Output width = 5 mm Efficiency
= 5/16 × 100% = 31%
8
Efficiency =
a)
=
useful work output
work input
100kg × 9.8ms−1 × 6.0 × 10−2 m
40 N × 2.0m
× 100%
= 74%
b)
The system is not 100% efficient because some work is done against frictional forces and in
raising the weight of the pulley blocks.
9
Power = F × v
= 15 N × 12 m s−1
= 180 W
10
P=
=
F × d 1200 N × 186m
t
5 × 60s
1200 N × 186m
5 × 60s
= 744 W
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
6 Work, energy and power
Answers to Exam practice questions
Pages 90–93 Exam practice questions
1
Work done by each force = Fs cos θ
= 100 N × 100 m × cos 60 = 5000 J
Total work done by both forces = 10 000 J – the answer is C
2
[1]
At the lowest point the jumper is stationary and the kinetic energy must be zero. The elastic cord will have
elastic strain energy stored within it, but some of the initial gravitational potential energy will have been
converted to internal energy as the molecules are displaced within the cord – the answer is B.
3
Power =
work done mgh 1000 kg  9.8 m s −2  2.4 m
=
=
= 2940 W ≈ 3000 W – the
time
t
8.0 s
answer is C
4
[1]
Useful power output = 75% of 120 W = 90 W
P=F×v F =
5
[1]
P
90 W
=
= 45 N – the answer is A
v 2.0 m s −1
[1]
Work is the product of the force and the distance moved in the direction of the applied force [1]. Work is
measured in joules (J).
[1]
[Total 2 Marks]
6
a)
Potential energy is the ability of a body to do work by virtue of its position or state.
b)
The gravitational potential energy (GPE) of a body is the ability of the body to do work by
virtue of its position in a gravitational field.
[1]
[1]
An object of mass m raised through a height Δh will gain GPE of mgΔh. However, the elastic
potential energy (EPE) of an object is the energy stored in the object when it is stretched or
compressed. For a cord extended Δx by an average force Fave, the gain in EPE will be FaveΔx. [1]
[Total 3 Marks]
7
a)
b)
Kinetic energy (KE) is the energy of a body by virtue of its motion. The KE of a body of mass m
moving with a velocity v is ½mv2.
[1]
KE of the golf ball = ½ (0.05 kg)(20 m s−1)2 = 10 J
[2]
[Total 3 Marks]
8
Efficiency =
useful work output
× 100%
work input
[Total 1 Marks]
9
a)
W = FΔx= 60 N × 100 m = 6000 J
[2]
b)
W = FΔxcos θ = 80 N × 50 cos 30 m [1] = 3500 J (to two significant figures) [1]
[Total 4 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
6 Work, energy and power
Answers to Exam practice questions
10
Draw the diagram. [3]
After drawing the diagram, measure the vertical height, h, of the trolley [1] above the base of the ramp.
Record the time, t, for the interrupter card to cut through the light beam [1] and find the velocity:
v=
length of card
[1]
t
Repeat for a range of values of h. [1]
ΔGPE = mgh and ΔKE = ½mv2 [1]
Plot a graph of v2 against h. The percentage of GPE converted to KE is found
2g
by
× 100%. [1]
gradient
[Total 9 Marks]
11
a)
Power is the rate of doing work [1]. It is measured in watts (W). [1]
b)
Work done = force × distance = (70 × 9.8) N × (20 × 0.25) m = 3430 J
Power =
c)
[1]
work done 3430 J
=
= 690 W
time
5.0 s
[1]
The student will do more work moving limbs, contracting muscles, pumping blood, etc.
[1]
[Total 5 Marks]
12
a)
Using v2 = u2 + 2as where v = 0; u = 5.0 m s−1; s = 20 m
−1 2
b)
−2
[1]
−2
0 = (5.0 m s ) + 2a(20 m) ⇒ a = −0.625 m s ≈ −0.6 m s
[1]
i) F = ma = (100 kg) × (0.6 m s−2) = 60 N
[1]
−1
ii) P = Fv= (60 N) × (5.0 m s ) = 300 W
[1]
[Total 4 Marks]
13
a)
ΔGPE = ΔKE
mgΔh = ½mv2 → v = √(2gΔh) → v = √(2 × 9.8 m s−2 × 18 m) [1] = 19 m s−1 [1]
b)
Work done on car = gain in KE [1]
200 × 103 N × 2.0 m = ½ × 1760 kg × v2 [1] → v = 21 m s−1 [1]
c)
ΔKE = ΔGPE
½m (u2 – v2) = mgΔh → 212 – v2 = 2 × 9.8 m s−2 × 18 m [1] → v = 10 m s−1 [1]
d)
Initial KE = work done against resistive forces + ΔGPE
400 × 103 J = 300 N × 35 m + m × 9.8 m s−2 × 18 m [1]
Total mass, m = 2200 kg [1]
Maximum mass of passengers = 1400 kg [1]
e)
The equation only applies to rectilinear motion [1]. The quantities u, v, a and s are vectors. The
rollercoaster car is continuously changing direction as it climbs the slope, but the acceleration, g,
is always acting downwards [1]. KE and GPE are scalar quantities. [1]
The application of the equation gives the correct magnitude of the initial speed but is an error of
physics and would not gain credit in an examination.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
6 Work, energy and power
Answers to Exam practice questions
[Total 13 Marks]
14
a)
Nitrogen molecules have a mass of 28 u. The unified mass unit, u, is 1.66 × 10−27 kg and will be
defined later in the course.
b)
KE = ½ × 28 × 1.66 × 10−27 kg × (500 m s−1)2 (1) = 5.8 × 10−21 J
[1]
ΔGPE = 28 × 1.66 × 10−27 kg × 9.8 m s−2 × 2.5 m (1) = 1.1 × 10−24 J
[1]
There is no change in g over the height of the room and the variations of gravitational potential
energy are negligible compared with the kinetic energy of particles at this temperature.
[2]
[Total 4 Marks]
15
a)
ΔGPE = ΔKE
m1g Δh – m2g Δh =½ (m1 + m2) v2
[1]
m1 = mass of top car + passengers + mass of water (mw)
m2 = mass of bottom car + passengers
Δh = change in vertical height = 5.0 sin θ = 2.8 m
{(3500 kg + 400 kg + mw) – (3500 kg + 700 kg)} × 9.8 m s−2 × 2.8 m
b)
= ½ (7000 kg + 1100 kg + mw) × (0.70 m s−1)2
[1]
mw = 380 kg
[1]
The volume of added water will be 380 litres.
[1]
In practice some work will be done against resistive forces and the mass of the cable (0.50
tonnes!) has been ignored.
c)
d)
[1]
At constant speed the braking forces must equal the component of the resultant weight along
the track.
[1]
F = (m1 – m2) g sin θ = 200 kg × 9.8 m s−2 × 30/53 (1) = 1100 N
[1]
W = F × x = 1100 N × 43 m = 48 kJ
[1]
Stage 2 loss of GPE of cars = gain in KE (plus some work done against resistive forces that leads to
an increase in internal energy in the moving parts)
[1]
Stage 3, change in GPE of the cars (= work done against the braking forces) [1] = gain in elastic PE
in fluid and compressed gas + gain in internal energy (thermal energy) in the brakes
[1]
Stage 4 loss of GPE + loss of KE [1] = gain in internal energy in the brakes + EPE in the
fluid/gas
[1]
After the journey, some of the EPE of the gas is used to pump water to the lake, so increasing its
GPE.
[1]
(Student only required a maximum of 4 of these)
[Total 12 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
7 Charge and current
Answers to Test yourself questions
Page 94 Test yourself on prior knowledge
1
Number of electrons =
=
total charge
charge on an electron
10 × 10−6 C
1.6 × 10−19 C
= 6.25 × 1013 electrons
2
ΔQ = lΔt
= 8.0 A × (3.0 × 60) s
= 1440 C
3
ΔQ = IΔt  Δt
=
Δt =
𝛥𝑄
𝐼
1.8 × 105 A s
500 × 10−3 A
= 360 000 s
= 100 h
Page 96 Test yourself
1
a)
ΔQ = IΔt
= 500 × 10−3 A × (24 × 60 × 60) s
= 43 200 C
b)
ΔQ = IΔt  Δt
=
=
𝛥𝑄
𝐼
43200 A s
3.0 A
= 14 400 s
= 4.0 h
2
a)
total charge = number of electrons × electron charge
Q = 1 × 1020 × 1.6 × 10−19 C
= 16 C
b)
ΔQ = IΔt  Δt
=
=
𝛥𝑄
𝐼
16 A s
10 × 103 A
= 1.6 ms
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
7 Charge and current
Answers to Test yourself questions
Page 96 Activity 7.1
1
By the conservation of charge, the rate of flow of charge (i.e. the current) must be the same throughout a
series circuit because there is nowhere else for the charge to go. Therefore the current must be the same in
each resistor.
2
If the resistors have a tolerance of 5%, the 100 Ω resistor would have a resistance of 100 ± 5 Ω and the 47 Ω
resistor would have a resistance of 47 ± 2 Ω. The combined resistance should therefore be written as 147 ±
7 Ω.
Page 97 Activity 7.2
1
IZ/mA
IX/mA
IY/mA
IX + IY/mA
5.0
1.6
3.4
5.0
15.0
4.9
10.3
15.2
25.0
8.1
16.8
24.9
35.0
11.2
23.7
34.9
45.0
14.5
30.6
45.1
As can be seen, IX + IY = IZ within experimental error, showing that current is conserved at a junction.
2
Your graph should be a straight line through the origin, like Figure 7.5 in your textbook.
3
You should find that the gradient is 2.11 ± 0.02.
The ratio of RX to RY = 100 : 47 = 2.13
As the gradient of the graph is the ratio of IY to IX, the experiment shows that the current splits in a ratio
that is equal to the inverse ratio of the resistors – there is more current, proportionally, in the smaller
resistance.
Page 98 Test yourself
3
a)
As the 5 Ω resistor is in series with ammeter A1, the current in it is 180 mA.
b)
The 10 Ω resistor is in series with ammeter A2 and so the current in it is 60 mA.
c)
When the 180 mA current splits, with 60 mA in the 10 Ω resistor, the remaining 120 mA must be
in the 2 Ω resistor.
d)
As the 3 Ω resistor is in series with the 2 Ω resistor, the current in the 3 Ω resistor must also be
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
7 Charge and current
Answers to Test yourself questions
120mA.
4
a)
I1 = I 2 + I3
b)
I2 = I 1 + I3  I1 = I 2 − I3
c)
I3 = I 2 + I1  I1 = I 3 – I2
Page 101 Activity 7.3
1
As unlike charges attract, the copper ions must be positive, as they are attracted to the negative terminal.
The permanganate ions must be negative as they are attracted towards the positive terminal.
2
Start a stopclock at the same time as the power supply is switched on. When the stains in the soaked paper
are long enough to measure, switch off the power supply and stop the clock. Record the time, t, and
𝑙
measure the length, l, of each stain. The drift speed, v, can then be calculated from v = .
𝑡
3
It is advisable to wear plastic gloves and goggles when handling chemicals such as these. Care should also
be taken not to ‘short’ the power supply, for example by letting the crocodile clips or pins touch each other
when the power is on.
Page 102 Test yourself
𝐼
5
I = nAvq  v =
6
As the wires are in series, I will be the same in each.
𝑛𝐴𝑞
 v ∝ I (Answer B)
As n is a property of the material and both wires are copper, n must be the same
for each.
As q is the electron charge, it will be the same for both.
This means that v is NOT the same – it will be greater in the thinner wire.
(Answer D)
7
a)
I is in A, n is in m−3, A is in m2, q is in C
b)
n for copper is very much greater than n for silicon and so more charge will flow per second in
the copper (assuming the same order of magnitude for v). This means that copper is a better
conductor.
c)
n for an insulator, such as porcelain, is theoretically zero, but in practice may be very small.
d)
Plastic and rubber are good insulators and are used to insulate wires. Also switch components,
other than the electrical contacts, are often made of plastic.
8
a)
𝐼 = 𝑛𝐴𝑣𝑞 ⟹ 𝑣 =
𝐴=
=
1
𝑛𝐴𝑞
𝜋𝑑 2
4
π×(0.50×10−3 m)2
4
= 1.96 × 10−7 m2
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
7 Charge and current
Answers to Test yourself questions
𝑉=
220×10−3 C s−1
8.5×1028 m−3 ×1.96×10−7 m2 ×1.6×10−19 c
= 8.2 × 10−5 m s −1
b)
𝐼 = 𝑛𝐴𝑣𝑞 ⟹ 𝑣 =
1
𝑛𝐴𝑞
As the pencil lead and connecting wires are in series, I must be the same in each. The electron
charge is the same for both and the areas are of the same order of magnitude. As n is much less
in the pencil lead, v in the pencil lead must be much greater so that the rate of flow of charge, I,
is the same through the pencil lead and connecting wires.
c)
An answer like this can often be expressed in bullet points (this can often make it easier for an
examiner to spot the relevant data):
•
Use either a digital vernier or a micrometer.
•
Check for zero error before taking measurements.
•
Measure the diameter in four different places.
•
Rotate the instrument to measure the diameter at different angles.
•
Find the average of the measurements.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
7 Charge and current
Answers to Exam practice questions
Pages 104–106 Exam practice questions
1
ΔQ = IΔt = 32 × 10−3 C s−1 × 3000 s
= 115.2 C
Number of electrons =
115.2 C
total charge
charge on an electron
= 7.2 × 1020 electrons (The answer is B)
[Total 1 Mark]
2
n for copper is much greater than n for silicon. (The answer is A)
[Total 1 Mark]
3
a)
=
1.6 × 10−19 C
I=
Δ𝑄
Δ𝑡
=
8C
[1]
0.5 × 10-3 s
= 16 000 A (= 16 kA)
b)
[1]
Number of electrons =
total charge
charge on an electron
=
8C
[1]
1.6 × 10-19 C
= 5.0 × 1019 electrons
[1]
[Total 4 Marks]
4
a)
I = current
n = number of charge carriers per unit volume
A = area of cross-section normal to current
v = drift velocity of charge carriers
q = charge on each charge carrier
b)
I = nAvq  v =
v=
[3]
𝐼
𝑛𝐴𝑞
10 × 10-3 C s-1
[2]
7.0 × 1022 m-3 × (6 × 10-3 × 0.5 ×10-3 ) m2 ×1.6 × 10-19 C
= 0.298 m s−1 ≈ 0.3 m s−1
c)
v=
[1]
𝐼
𝑛𝐴𝑞
If I, A and q are the same, then v ∝
1
[1]
𝑛
As n for copper is in the order of 106 times greater than n for a semiconductor, v for copper will
be in the order of 106 less (‘very much greater’ and ‘very much less’ would be sufficient in an
exam answer).
[1]
[Total 8 Marks]
5
Time =
distance
speed
=
21 × 10-2 m
4.2 × 107 m s-1
Number of electrons =
=
6
I = nAvq  v =
= 5.0 × 10−9 s
charge
charge on an electron
=
𝐼𝛥𝑡
𝑒
32 × 10-3 C s-1 × 5.0 × 10-9 s
1.6 × 10-19 C
= 1.0 × 109 (The answer is A)
[Total 1 Mark]
𝐼
𝑛𝐴𝑞
DX = ½DY  AX = ¼ AY
vX = 4vY  vX : vY = 4 : 1 (The answer is D)
[Total 1 Mark]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
7 Charge and current
Answers to Exam practice questions
7
a)
A=
𝜋𝐷2
4
D=√
4𝐴
𝜋
4 × 0.20 ×10-6 m2
=√
π
= 5.05 × 10−4 m ≈ 0.5 mm
[2]
i) A micrometer or digital vernier having a precision of 0.01 mm would be suitable for
measuring a diameter in the order of 0.5 mm.
ii) Proceed as follows:
• Check for zero error before taking measurements.
• Measure the diameter in four different places.
• Rotate the instrument to measure the diameter at different angles.
• Find the average of the measurements.
[3]
iii) Assuming the diameter is measured to the precision of the instrument, 0.01 mm:
% uncertainty =
0.01 mm
0.50 mm
× 100% = 2%
[1]
[1]
iv) As the area is given by πD2/4 [i.e. (π × D × D)/4] and the % uncertainty in the diameter is 2%,
the percentage uncertainty in the area of cross-section will be 2% + 2% = 4%.
[1]
[Total 8 Marks]
8
a)
I = 25 mA = 25 × 10−3 A
i) ΔQ = IΔt = 25 × 10−3 C s−1 × (8 × 60) s = 12 C
ii) Number of electrons =
total charge
charge on an electron
=
[2]
12 C
1.6 × 10-19 C
= 7.5 × 1019 electrons
b)
[2]
The copper leads are likely to be much thicker than the tungsten filament, so they will have a
much larger area of cross-section. From I = nAvq:
• as the leads and filament are in series, I will be the same for both
• q, the electron charge, is the same for copper and tungsten
• if A is larger for the leads, v will be less in the leads for the same current
• n is larger for copper (it is a better conductor than tungsten)
• which will also make v less in the leads.
[4]
[Total 8 Marks]
9
a)
I = nAvq  v =
v=
𝐼
𝑛𝐴𝑞
0.25 C s-1
8.5 × 1028 m-3 × 0.50 × 10-6 m2 ×1.6 × 10-19 C
= 3.7 × 10−5 m s−1
t=
distance
speed
=
[2]
[1]
3.0 m
[1]
3.7 × 10-5 m s-1
81 600 s
= 81 600 s= (60
× 60)s h-1
= 23 hours
b)
[1]
29
As there are in the order of 10 conducting electrons per cubic metre, a very large number will
pass through the filament in a very short time, so the filament will appear to light almost
immediately a potential difference is applied by switching on, even though each electron is only
drifting slowly.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
7 Charge and current
Answers to Exam practice questions
[Total 7 Marks]
10
a)
density =
mass
volume
 volume =
mass
density
=
63.5 × 10-3 kg
8.9 × 103 kg m-3
= 7.13 × 10−6 m3
[2]
If 7.13 × 10−6 m3 contains 6.0 × 1023 atoms
1.0 m3contains
6.0 × 1023 atoms
7.13 × 10-6 m3
= 8.4 × 1028 atoms m−3
[1]
Assuming that there is one conduction electron per atom of copper
n = 8.4 × 1028 m−3
b)
I = nAvq  v =
v=
𝐼
𝑛𝐴𝑞
[1]
where A =
𝜋𝑑 2
4
v=
4𝐼
𝜋𝑑 2 𝑛𝑞
4 × 5.0 C s-1
[2]
2
π × (0.2 ×10-3 ) m2 × 8.4 × 1028 m-3 ×1.6 × 10-19 C
= 0.012 m s−1 = 12 mm s−1
c)
In the equation v =
[1]
𝐼
𝑛𝐴𝑞
• the fuse is in series with the connecting leads, so I is the same for each
• both are made of copper, so n is the same for each
• q is the electron charge, which is the same for each.
v ∝
1
[2]
𝐴
The diameter of the connecting wire (1.0 mm) is 5 times that of the fuse wire (0.2 mm), so the
area of cross-section of the connecting wire is 25 times that of the fuse wire. The drift velocity in
the connecting wire is therefore 25 times less than in the fuse wire.
d)
28
[2]
3
From part (a) we found that there were 8.4 × 10 atoms in 1.0 m of copper. If we assume that
copper has a simple cubic structure, with x atoms along each 1.0 m side of a 1.0 m3 cube, then
there will be x3 atoms in the cube.
[1]
So, x3 = 8.4 × 1028 m−3  x = √8.4 × 1028 = 4.4 × 109 m−2
[1]
1
1
𝑥
4.4 × 109 m-1
As x = number of atoms in a 1.0 m length, atomic spacing d = =
[1]
d = 2.3 × 10−10 m = 0.23 nm
[1]
The structure of copper is actually more complex than the simple cubic structure that we
assumed and therefore this is only a very approximate value.
[Total 15 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Test yourself questions
Page 107 Test yourself on prior knowledge
1
ΔE = PΔt
= 60 J s−1 × (60 × 60) s
= 216 kJ
2
𝑃 = 𝐼𝑉
⟹𝐼=
=
𝑃
𝑉
60 W
240 W
= 0.25 A
3
a)
ΔGPE = mgΔh
= 840 kg × 9.8 J kg−1 × 15 m
= 1.23 × 105 J
b)
𝑃=
=
∆𝐸
∆𝑡
1.23×105 J
3.1 s
= 39.8 kW ≈ 40kW
c)
In practice, the car would have to generate more power than this in order to do work against air
resistance and rolling resistance. The latter is the energy transferred when a tyre is rolling due to
the constant deformation of the tyre – see hysteresis on page 214.
(NOTE: You should note that an answer such as ‘due to friction’ would not get any marks.)
Page 110 Test yourself
1
a)
ΔE = PΔt
= 3.0 J s−1 × (60 × 60) s
= 10.8 kJ
b)
ΔE = PΔt
= 0.13 J s−1 × (23 × 60 × 60) s
= 10.8 kJ
If, after charging the phone, the charger is left on standby for the rest of the day it will use as
much energy as charging the phone. This is wasteful.
2
a)
𝑃 = 𝐼𝑉
⟹𝐼=
=
𝑃
𝑉
1.0 W
4.5 V
= 0.22 A
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Test yourself questions
b)
ΔQ = IΔt
= 0.22 C s−1 × (15 × 60) s
= 200 C
c)
ΔE = VΔQ
= 4.5 V × 200 C
= 900 J
Page 111 Activity 8.1
1
2
a)
We can assume p.d. across lamp ≈ 6 V if lamp is shining brightly.
b)
So p.d. across resistor ≈ (9 – 6) V = 3 V
a)
𝑃 = 𝐼𝑉
⟹𝐼=
=
𝑃
𝑉
300×10−𝑠 𝑊
6𝑉
= 0.05 A = 50 mA
b)
The resistor is in series with the lamp. By conservation of charge, therefore, the rate of flow of
charge (current) in the resistor must be the same as that in the lamp.
3
For the variable resistor:
P = IV
= 50 mA × 3.0 V
= 150 mW
This 150 mW of power dissipated in the resistor is likely to make it slightly warm.
4
It is essential to start with the variable resistor at its maximum value so that the p.d. across (or the current
in) the lamp is a minimum. The resistance of the variable resistance can then be gradually reduced until the
current in the lamp is sufficient to make it glow brightly. If it is not done this way round, a p.d. of more than
6 V could be applied across the lamp, which would cause it to burn out.
Page 112 Test yourself
3
a)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Test yourself questions
b)
An e.m.f of 12 V means that the battery converts 12 J of chemical energy into electrical energy
for every 1 C of charge that passes through the battery.
A p.d. of 9 V means that the resistor is converting 9 J of the electrical energy into thermal energy
for every 1 C of charge passing through the resistor.
c)
The voltmeter should be connected as shown in the diagram in part (a).
d)
If the p.d. across R1 is 9 V, the p.d. across R2 will be (12 – 9) V = 3 V
Page 113 Test yourself
4
a)
i) charge ii) s iii) V iv) C v) power vi) W vii) A viii) W
b)
volt-ampere  V × A
 J C−1 × C s−1
 J s−1
W
Page 113 Activity 8.2
1
V/V
I/A
Power in, IV/W
Force, mg/N
Height, h/m
Time, t/s
Power out,
Efficiency,
𝒎𝒈𝒉
/𝐖
𝒕
𝑷𝒐
× 𝟏𝟎𝟎%
𝑷𝒊
6.0
0.24
1.44
0.98
0.95
4.1
0.227
15.8
5.9
0.28
1.65
1.96
0.95
5.8
0.321
19.5
5.8
0.31
1.80
2.94
0.95
7.5
0.372
20.7
5.6
0.36
2.02
3.92
0.95
9.3
0.400
19.8
5.4
0.40
2.16
4.90
0.95
11.6V
0.401
18.6
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Test yourself questions
2
(Note that the y-axis does not begin at zero. If it did, the plots would be very compressed.)
3
The graph shows that the efficiency is dependent on the load and that the motor has maximum efficiency
for a load of about 3.0 N (i.e. when it is lifting a mass of about 300 g).
Page 116 Test yourself
5
a)
P = IV
= 5.1 V × 2.1 A
= 11 W
b)
ΔE = PΔt
= 11 J s−1 × (4 × 60 × 60) s
= 150 kJ
c)
𝑃 = 𝐼𝑉
⇒𝐼=
=
𝑃
𝑉
5W
5V
=1A
d)
If the current delivered by the iPhone charger is only half that from the iPad charger, it would
take twice as long to charge the iPad.
6
a)
ΔQ = IΔt
= 2.0 C s−1 × (4.5 × 60 × 60) s
= 32 400 C
b)
ΔE = VΔQ
= 3.6 V × 32 400 C
= 1.2 MJ
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Test yourself questions
c)
ΔE = PΔt
= 324 J s−1 × (1 × 60 × 60) s
= 1.2 MJ
This suggests that the manufacturer’s claim is correct.
7
a)
i) 𝑃 = 𝐼𝑉
⇒𝐼=
𝑃
𝑉
=
40 W
240 V
= 0.17 A = 170 mA
ii) P = IV
= 240 V × 150 × 10-3 A
= 36 W
iii) 𝑃 = 𝐼𝑉
⇒𝑉=
=
𝑃
𝐼
4.5 W
500×10−3 A
= 9V
b)
i) ΔE = PΔt
= 40 J s−1 × (5 × 60 × 60) s
= 720 kJ
ii) ΔE = PΔt
= 4.5 J s−1 × (5 × 60 × 60) s
= 81 kJ
c)
Nearly all the LED power is converted into visible light but only about 10% of the filament lamp’s
power is converted into energy in the visible spectrum – the rest is mainly in the infra-red region.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Exam practice questions
Pages 118–121 Exam practice questions
1
ΔQ = IΔt  C = A s (The answer is B)
2
a)
Rearranging P = VI
𝐼 =
b)
[Total 1 Mark]
𝑃
𝑉
=
1500 W
230 V
= 6.5 A
[2]
Substituting into Q = It, and remembering to put the time in seconds:
𝑄 = 6.5 A × 20 × 60 s = 7.8 × 103 C (to 2 Significant Figures)
c)
[2]
Rearranging the definition that power = work done/ time taken:
Thermal energy = power × time of operation
= 1500 J s −1 × 20 × 60 s = 1.8 MJ
[2]
[Total 6 Marks]
3
a)
ΔE = PΔt = 50 J s-1 × (12 × 60 × 60) s = 2.16 MJ
[2]
b)
1 kW h  1000 J s−1 × (1.0 × 60 × 60) s  3.6 MJ
[2]
c)
Number of units of electricity used in 1 year =
(365 × 2.16) MJ
3.6 MJ
Cost = 219 × 15 p
= £32.85
≈ £33
4
a)
= 219 units
[2]
[1]
[Total 7 Marks]
Energy saved over 8000 hours = (60 − 18)W × 8000 h
= 42 × 8 kWh = 336 kWh
[2]
Cost of energy saved will be = 336 × 15 p
= £50.40
[1]
Allowing for £2 cost of lamp, total cost saved will be £48.40, which comfortably exceeds the
manufacturer’s claim.
b)
5
[2]
Apart from monetary cost, the saving in energy is important in conserving the world’s energy
resources as well as reducing the carbon footprint (less electrical energy consumed means less
carbon dioxide is created in its production). Against this, there may be a greater carbon cost in
manufacturing a low-energy lamp compared with a filament lamp, although it will almost
certainly last longer. From a safety aspect, the low-energy lamp will probably not get as hot. [2]
[Total 7 Marks]
W = J s −1 (power = rate of doing work)
J = N m (work = force × distance)
N = kg m s −2 (from F = ma)
W = kg m s −2 × m × s −1 = kg m2 s −3
The answer is D.
[Total 1 Mark]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Exam practice questions
6
Quantity or quantities
Which quantity is the product of two other
quantities?
Which quantity is one of the quantities divided
by one of the other quantities?
(3 possible answers)
Which quantity, when divided by time, gives
another quantity in the table?
(2 possible answers)
Power = p.d. × current (from P = VI)
[1]
Current = power ÷ p.d. (from P = VI)
P.D. = power ÷ current (from P = VI)
P.D. = energy ÷ charge (from V = W/Q)
Current = charge ÷ time (from I = Q/t)
Power = energy ÷ time (from P = W/t)
[3]
[2]
[Total 6 Marks]
7
a)
ΔE = PΔt = VIΔt = 800 × 103 V × 500 × 10−3 A × 1.0 s = 400 kJ
𝛥𝐸
400 × 103 J
1000 J s-1
[2]
b)
ΔE = PΔt  Δt =
c)
It has been assumed that all the energy radiated by the fire is received by the person.
𝑃
=
= 400 s ≈ 7 minutes
[2]
[1]
[Total 5 Marks]
8
a)
Work done = force × distance moved
= 𝑚𝑔 × Δℎ = 0.600 kg × 9.8 N kg −1 × 0.94 m = 5.5 J
b)
Power =
=
[2]
work done
time taken
5.5 (3)J
7.7 s
= 0.72 W
9
[2]
c)
𝑃 = 𝑉𝐼 = 3.2 V × 0.75 A = 2.4 W
d)
Efficiency =
e)
The rest of the energy will be converted into thermal energy by the work done by the motor
against friction as it rotates and by the masses against air resistance. Also thermal energy will be
generated in the coil of the motor and in the connecting wires due to the current in them having
to overcome their electrical resistance.
[2]
[Total 10 Marks]
a)
The spread of the measurements for t is from 7.5 s to 7.9 s = 0.4 s.
useful power taken out
total power put in
=
[2]
0.718 W
× 100% ≈ 30%
2.40 W
[2]
The average value is therefore 7.7 ± 0.2 s.
[1]
This gives the percentage uncertainty in t as
b)
0.2 s
7.7 s
The percentage uncertainty in the height Δh is
× 100% = 3%
0.01 m
0.94 m
[1]
× 100% = 1%
[1]
The useful output power of the motor is given by mgΔh/t and so the percentage uncertainty in
its value is 3% (for t) + 1% (for Δh) = 4%.
The uncertainty in the power is therefore
[1]
4
100
× 0.72 W = 0.03 W.
The value for the power should therefore be written as 0.72 ± 0.03 W
[1]
[Total 5 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Exam practice questions
10
a)
Power delivered by alternator = 𝑉𝐼 = 14V × 70 A = 980 W
b)
Current taken by starter motor is given by rearranging 𝑃 = 𝑉𝐼:
𝐼 =
c)
𝑃
𝑉
=
1500 W
12 V
= 125 A
For each headlight:
𝐼 =
𝑃
𝑉
=
60 W
12 V
= 5A
so the next fuse up, i.e. 8 A would be suitable.
d)
Energy = 𝑉𝐼𝑡 = 12 V × 1 A × (62 × 60 × 60)s ≈ 2.7 M J
e)
If the four sidelights are left on for 12 hours:
energy used = 4 × 5 W × (12 × 60 × 60) s = 864 000 J
[Total 12 Marks]
11
Using 𝑃 = 𝐹𝑣 power to drive at 90 km per hour up a gradient of 10% is given by:
P = 0.1 × 𝑚𝑔 × 𝑣 = 0.1 × 1740 kg × 9.8 N kg −1 ×
90 × 103 m
60 ×60 s
= 43 kW (to 2 Significant Figures)
Maximum power is stated as 180 kW, so fraction of maximum power
=
43 kW
1
= 0.24 ≈
180 kW
4
[Total 5 Marks]
12
a)
b)
c)
[3]
P = VI = 0.53 V × 1.8 A = 0.954 J s−1
I = nAvq  v =
𝐼
𝑛𝐴𝑞
=
[2]
1.8 C s-1
8.0 × 1028 m-3 × 0.059 × 10-6 m2 × 1.6 × 10-19 C
= 2.38 × 10−3 m s−1
d)
From P = Fv  F =
[1]
𝑃
𝑣
Total force on all the electrons in the wire =
= 400 N
[1]
0.954 N m s-1
2.38 × 10-3 m s-1
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Exam practice questions
Volume of wire = length × area of cross-section = 1.0 m × 0.059 × 10−6 m2
e)
[1]
Number of electrons in wire = n × volume
= 8.0 × 1028 m−3 × 1.0 m × 0.059 × 10−6 m2
= 4.72 × 1021 electrons
Force on each electron =
total force
number of electrons
=
[1]
400 N
[1]
4.72 × 1021 electrons
= 8.5 × 10−20 N per electron
[1]
[Total 13 Marks]
13
J
Nm
C
As
V= =
=
kg m s−2 × m
As
= kg m2 s −3 A−1
The answer is B.
[Total 1 Mark]
14
a)
ΔQ = IΔt = 4.5 C s−1 × (4.0 × 60 × 60) s = 64 800 C
[2]
b)
Assuming p.d. is constant at 12 V
[1]
ΔE = VΔQ = 12 V × 64 800 C = 780 kJ
[2]
Charge stored = area under graph ≈ 12¼ squares
[1]
c)
Each square  1.0 A × 2.0 h  2.0 A × (60 × 60) s  7200 C
4
[1]
5
Charge after 10 hours = 12.25 × 7200 C = 8.8 × 10 C ≈ 1 × 10 C
d)
As 3600 C  1 A h  8.8 × 104 C 
8.8 × 104
3600
[1]
A h = 24 A h
[2]
Allowing for the difficulty in getting an accurate value for the area (any value between 12 and
12½ squares is reasonable), this is a good approximation to the manufacturer’s stated value of 25
A h.
[2]
[Total 12 Marks]
15
a)
Difference in energy between 50-inch plasma and 50-inch LED
= (300 – 71) J s−1 × (365 × 12 × 60 × 60) s = 3.61 × 109 J per year
=
3.61 × 109 J per year
3.6 × 106 J unit-1
= 1000 units per year
Cost difference per year = 1000 × 15 p = £150
[2]
[1]
[1]
Difference in purchase price of televisions £950 − £650 = £300
The payback time would therefore be 2 years.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Exam practice questions
b)
(50-inch)2
(42-inch)2
= 1.42
[2]
For plasma set:
For LED set:
300 W
210 W
71 W
50 W
= 1.43
= 1.42
[2]
This strongly suggests that the power taken is proportional to the screen size squared. As the
area of the screen is proportional to the square of the screen size, the data suggest that the
power taken is proportional to the area of the screen. As most of the power will be used
illuminating the screen, this is not unexpected.
[2]
[Total 11 Marks]
16
a)
Ek = ½mv2
As density =
mass
volume
 mass = density × volume
The air travels a distance v in 1 second, so volume of air stopped in 1 s = Av
Mass of air stopped in 1 s =
P=
c)
𝑑𝑡
= density × volume stopped per second = ρ × Av
𝑑(½𝑚𝑣 2 )
𝑑𝑚
𝑑𝑡
[1]
[1]
𝑑𝑡
= ½v2 ×
b)
𝑑𝑚
[1]
= ½v2 × ρAv = ½ρAv3
[2]
Energy will also be expended doing work against rotor blade friction and air resistance and in the
gearbox, generator and converter.
[2]
P = 0.3 × ½ρAv3
[1]
= 0.3 × 0.5 × 1.3 kg m−3 × π × (
1.82
) m2 × (12.53) m3 s−3 = 970 W
[2]
This is therefore fairly close to the manufacturer’s claim of 1 kW.
[1]
4
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
8 Potential difference, electromotive force and
power
Answers to Exam practice questions
d)
If the average UK wind speed is 6 m s−1, the average power will be
Pav = 970 W × (
6 m s-1
12.5 m s-1
3
) = 107 W
[2]
As the turbine does not operate for speeds below 4 m s−1 or above 13 m s−1, the actual output
would be less than this, probably somewhat less than 100 W. The turbine was shown, therefore, not to be a
viable proposition and so was withdrawn from sale.
[2]
[Total 15 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Test yourself questions
Page 122 Test yourself on prior knowledge
1
2
a)
V = IR = 20 × 10−6 A × 220 × 103 Ω = 0.044 V = 44 mV
b)
𝑉 = 𝐼𝑅 ⟹ 𝐼 =
c)
R =
a)
P = 𝑉𝐼 ⟹ 𝐼 =
b)
R=
𝑉
𝐼
𝑉
𝐼
=
=
𝑉
𝑅
=
3.0 V
0.15×10−6 A
12 V
3.0 A
𝑃
𝑉
=
240 V
10×106 Ω
= 2.4 × 10−5 A = 24 μA
= 20 MΩ
36 W
12 V
= 3.0 A
= 4.0 Ω
Page 123 Activity 9.1
1
The second circuit – the potential divider – is the more useful circuit as it can provide a variable p.d. from
0.0 V to 1.5 V and not just between 0.50 V and 1.50 V as in the first circuit.
2
Setting the rheostat to its maximum resistance will ensure the current is at its minimum value and so
prevent the possibility of damage to the lamp or ammeter.
3
A switch would make the circuit safer, as would a resistor in series with the lamp to provide a current
limiting resistance.
Page 124 Activity 9.2
1
Your graph should look like Figure 9.3 in the textbook.
2
As the graph is a straight line through the origin we can deduce that the current is proportional to the p.d.
The wire therefore obeys Ohm’s law, irrespective of the direction of the current.
Page 126 Test yourself
1
a)
Current in X = 500 μA
p.d. across X = 3.35 V
R =
b)
𝑉
𝐼
=
2.35 V
500×10−6 A
= 4.7 kΩ
V10k = IR = 500 × 10−6 A × 10 × 103 Ω = 5.0 V
V3.3k = IR = 500 × 10−6 A × 3.3 × 103 Ω = 1.65 V
c)
Sum of potential differences = 2.35 V + 5.00 V + 1.65 V = 9.0 V
The p.d. across a resistor is the electrical energy (in J) per unit charge (in C) passing through the
resistor. Conservation of charge tells us that the rate of flow of charge must be the same through
each resistor. The conservation of energy tells us that the electrical energy per coulomb
transferred into thermal energy in the three resistors (9.0 J C−1) must be equal to the chemical
energy per coulomb transferred into electrical energy by the cell (9.0 J C−1). This is what we mean
by the e.m.f. of the cell.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Test yourself questions
2
The statement that ‘the resistance is equal to voltage divided by current’ is the correct definition of
resistance. However, it is not Ohm’s law. This states that the current is proportional to the potential
difference (for which we can, perhaps, accept ‘voltage’) or, in other words, voltage divided by current is
constant for a metallic conductor and provided that physical conditions, such as temperature, do not
change.
3
a)
R =
𝑉
𝐼
=
1.5 V
31×10−8 A
= 48.4 Ω
As 5% of 47 Ω is 2.35 Ω and the difference is only (48.4 – 47) Ω = 1.84 Ω, the resistor can be said
to be 47 Ω within ±5%.
b)
Maximum total resistance =
𝑉𝑚𝑖𝑛
𝐼𝑚𝑖𝑛
=
0.5 𝑉
2.5 × 10−3 𝐴
= 200𝛺
Maximum rheostat resistance = (200 – 48) Ω = 152 Ω
c)
d)
With the potential divider circuit, a full range of p.d. from 0.00 to 1.50 can be achieved. This is
not possible with the first circuit, where no values of p.d. below 0.50 V are possible – the range is
only ⅔ of that of the potential divider circuit.
Page 127 Activity 9.3
1
a)
Start with the power supply set to its minimum value so that the lamp starts with the lowest
current and is then gradually heated up as the current is increased. It also prevents the possibility
of the voltage exceeding 12 V and blowing the filament.
b)
Leave at least 5 minutes before repeating the experiment with the connections reversed to give
the filament time to cool down.
2
As the current increases, the filament heats up and its resistance increases. This can be deduced from the
graph by observing that the rate at which the current increases with voltage (the gradient,
d𝑙
d𝑉
) gets less as
the voltage increases – this means that the resistance must be getting greater.
Page 129 Test yourself
4
a)
𝑃 = 𝑉𝐼 ⟹ 𝐼 =
𝑃
𝑉
=
24 W
12 V
= 2.0 A
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Test yourself questions
b)
p.d. across resistor = (16 – 12) V = 4.0 V
Therefore the 2.2 Ω resistor should be used as it is the nearest value and the 0.2 Ω difference is
within the 10% tolerance.
d)
Figure 9.12, when p.d. = 6.0 V the current can be read off as 1.4 A.
𝑉
𝐼
=
12 V
1.4 A
𝐼
=
12 V
For the lamp: 𝑅 =
𝑅=
5
𝑉
c)
2.0 A
= 6.0 𝛺
= 4.3 Ω
Ohm’s law does apply to metals, but only if the temperature remains constant. The temperature of a
tungsten filament will typically vary from 20 °C when it is off to perhaps 2000 °C when it is on. So, clearly,
the bulb will not obey Ohm’s law.
Page 129 Activity 9.4
1
Your graph should look like Figure 9.12 in your textbook.
2
a)
The point at which the diode starts to conduct is never clear cut, so a value of between 0.40 V
and 0.50 V is acceptable.
b)
From the graph, when the current is 6.5 mA, the p.d. is 0.65 V.
R =
c)
𝑉
𝐼
=
0.65 V
6.5×10−3 A
= 100 Ω
Draw a straight line from the origin passing through the point (0.50 V, 10 mA) and extend it to
cross the diode curve. All points on this line correspond to a resistance of:
R =
𝑉
𝐼
=
0.50 V
10×10−3 A
= 50 Ω
Therefore, where this line intersects the diode curve will be the point at which The diode has a
resistance of 50 Ω. This occurs at a current of 13.7 mA.
Check: When the current is 13.7 mA, the p.d. is 0.69 V
R =
𝑉
𝐼
=
0.69 V
13.7×10−3 A
= 50 Ω
Page 130 Test yourself
6
a)
b)
The diode starts to conduct at a p.d. of about 0.05 V.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Test yourself questions
c)
From the graph, when V = 0.20 V, I = 300 μA
𝑅=
7
𝑉
𝐼
=
0.20 V
300×10−6 A
= 670 Ω
a)
Forward bias is when the diode is connected with its positive terminal to the positive terminal of
the cell so that the diode conducts. In reverse bias, the diode is connected with its negative
terminal (cathode) to the positive terminal of the cell so that it does not conduct.
b)
If the diode is forward biased, the lamp will light. If the diode is reverse biased, the lamp will not
light.
c)
Page 131 Activity 9.5
1
Your graph should look like Figure 9.15 in your textbook.
2
From the graph:
a)
When I = 3.00 mA, V = 1.50 ± 0.05 V
R =
b)
𝑉
𝐼
=
1.50 V
3.00×10−3 A
= 500 Ω
When I = 15.00 mA, V = 6.57 ± 0.05 V
R =
𝑉
𝐼
=
6.57 V
15.00×10−3 A
= 438 Ω
A reasonable estimate of the uncertainty of reading off the p.d. values from the graph would be
±0.05 V as indicated. This represents a percentage uncertainty of:
a)
b)
0.05 V
1.50 V
0.05 V
6.57 V
× 100% = 3.3% uncertainty in R = 3.3% × 500 Ω = 16 Ω  R = 500 ± 16 Ω
× 100% = 0.77% uncertainty in R = 0.76% × 438 Ω = 3 Ω  R = 438 ± 3 Ω
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Test yourself questions
3
These resistance values indicate that the resistance of the thermistor gets less as the current increases. This
is because the current causes the thermistor to warm up and the resistance of a thermistor decreases with
temperature (see Activity 10.4 on page 148).
This is illustrated by the shape of the graph: as the current increases, the line curves upwards. This means
that the rate of increase in current is increasing, i.e. the resistance is decreasing.
4
It is essential to wait at least 5 minutes before taking measurements with the current reversed to allow the
thermistor to cool down to room temperature again.
5
It is not advisable to use currents significantly larger than the 20 mA suggested because there is a danger
that the thermistor may over-heat and get damaged. When the current increases, the resistance gets less
and so the current will get even bigger. The temperature will therefore increase, causing a further decrease
in the resistance, and so an even greater increase in the current … and so on. This can produce what is
called ‘thermal runaway’.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Exam practice questions
Page 133–135 Exam practice questions
1
A typical semiconductor diode will not start to conduct until a potential difference of about 0.5 V is applied
– the answer is C.
2
[Total 1 Mark]
The filament will have resistance even when there is no potential difference applied – its resistance is a
property of the wire from which it is constructed. This resistance will increase when a potential difference is
applied as the current heats the filament – the answer is D.
[Total 1 Mark]
3
From P =
𝑉2
𝑅
 P α V2 as R is constant, since the resistor obeys Ohm’s law - the answer is B
[Total 1 Mark]
4
If the resistor obeys Ohm’s law, its resistance will remain constant irrespective of the current - the answer
is B
[Total 1 Mark]
𝑉
1
𝐼
𝐼
5
R =  R α if V is constant – the answer is D
6
a)
[Total 1 Mark]
Graph A, for the carbon resistor, is a straight line through the origin. This indicates that the
current is proportional to the potential difference so that the resistor obeys Ohm’s law and has a
constant resistance. As graph B, for the filament lamp, is a curve, it shows that the lamp does not
obey Ohm’s law. As the curve gets less steep, it indicates that the resistance of the lamp
increases with increased potential difference due to the heating effect of the current in the
tungsten filament wire.
b)
(4)
i) As the resistor is ohmic, 𝑅 =
𝑉
𝐼
= constant = inverse gradient.
Using a large triangle:
𝑅 =
(8.0 − 0.0) 𝑉
(0.25 −0.0)𝐴
= 32 Ω
ii) Percentage difference =
=
[2]
difference in values
stated value
(33 −32)Ω
33 Ω
× 100%
× 100% = 3%
which is within the stated 5% tolerance.
or
The nominal value is 33 Ω ± 5% = (33 ± 1.65)Ω, so the measured value of 32 Ω is within
the stated tolerance.
c)
[2]
i) Reading off from graph B, when the potential difference across the lamp is 12 V, the current
in it is 0.375 A, so:
𝑃 = 𝑉𝐼 = 12 V × 0.375 A = 4.5 W
[2]
ii) Percentage difference between this value and the stated value of 5 W
=
(5.0 −4.5)𝑊
5.0 𝑊
× 100% = 10%
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Exam practice questions
d)
i) From graph B, when the potential difference across the lamp is 12.0 V, the current in it is
0.375 A, so:
𝑅 =
𝑉
𝐼
=
12.0 V
0.375 A
= 32 Ω
[2]
ii) From graph B when the potential difference across the lamp is 1.0 V, the current in it is 0.10
A, so:
𝑅 =
𝑉
𝐼
=
1.0 V
0.10 A
= 10 Ω
[1]
[Total 14 Marks]
7
a)
From the graph, the current in the lamp at 12 V = 0.37 A
𝑉
12 V
𝐼
0.37 A
Resistance of lamp at 12 V: R = =
[1]
= 32 Ω
[1]
The student therefore mistakenly thinks that the circuit resistance will be
(32 + 33) Ω = 65 Ω.
[1]
𝑉
12 V
𝑅
65 Ω
This would give a current of I = =
b)
= 0.18 A
[1]
What the student has forgotten is that there will no longer be 12 V across the lamp and so the
lamp will be cooler and have less resistance.
[2]
If the current is actually 0.24 A, the p.d. across the resistor will be:
V = IR = 0.24 A × 33 Ω = 7.9 V
[1]
The p.d. across the lamp will therefore be (12.0 – 7.9) V = 4.1 V
[1]
Check: From the graph, the current in the lamp at a p.d. of 4.1 V is 0.24 A, which is the stated
circuit current.
(Remember: It is always a good idea, time permitting, to check answers. If this can be done by
working out the answer by a different method this is even better.)
c)
‘Quantitatively’ means we must justify our answer with some numbers, so:
At 12 V: Power developed in lamp P = VI = 12 V × 0.37 A = 4.4 W
[2]
At 4.1 V: Power developed in lamp P = VI = 4.1 V × 0.24 A = 0.98 W
[1]
As the power is less than ¼ of the normal working power, the lamp will only glow dimly.
[1]
(Remember: If possible, any argument or discussion should be backed-up with numerical
evidence.)
d)
We have just found that the lamp consumes 0.98 W.
The power consumed by the resistor is P = I2R = (0.24 A)2 × 33 Ω = 1.9 W
[2]
The resistor therefore consumes about twice as much power as the lamp.
[1]
[Total 15 Marks]
8
a)
i) Ohm’s law states that for a metallic conductor at a constant temperature the current in the
conductor is proportional to the potential difference across the conductor.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Exam practice questions
ii) The resistance of any conductor is defined as the potential difference across the conductor
divided by the current in the conductor.
b)
𝐼 =
𝑉
𝑅
=
2.0 V
11.3 Ω
= 0.177 A = 177 mA
[1]
[2]
c)
[3]
d)
[3]
The current sensor and voltage sensor are connected to the data logger (analogue to digital
converter), the output of which is fed into a computer. The potential divider is used to vary the
potential difference from zero to 2.0 V (note that a series variable resistor cannot do this – see
page 117) and the current and potential difference are recorded at set intervals. This data is
stored in the data logger and can then be used to plot a graph via the computer.
The main advantage of using a data logger is that a large amount of data can be collected and
processed in a relatively short time, thus giving better average values.
[1]
The only real disadvantage is the complexity of the set-up compared with just using digital
meters.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Exam practice questions
e)
The resistance of the whole ribbon when the toaster is operating at 240 V, 1000 W is given by
𝑃 =
𝑉2
𝑅
⟹𝑅 =
𝑉2
𝑃
=
(240 V)2
1000 W
= 57.6 Ω
[2]
If a length of 1.00 m has a resistance of 11.3 Ω, then the total length of the ribbon will be:
length of ribbon =
57.6 Ω
11.3 Ω
× 1.00 m = 5.10 m
[1]
This assumes that the resistivity of the nichrome is the same when it is at the operating
temperature of the toaster as it is at 2.0 V. In practice, the resistivity will increase with
temperature and so the length of ribbon will be less than the calculated value. An estimate of
‘about 5 m’ would not be unreasonable.
[2]
[Total 18 Marks]
9)
Reading from the graph (figure 9.20 on page 127), when the current in the diode is 25 mA, its resistance is
30 Ω.
[1]
The potential difference across the diode is then given by V = IR:
𝑉 = 25 × 10−3 A × 30 Ω = 0.75 V
[1]
The potential difference across the resistor will therefore be
(1.58 − 0.75)V = 0.83 V.
[1]
𝑉
The required value of resistance is given by 𝑅 = :
𝐼
𝑅 =
0.83 V
25 ×10−3 A
= 33.2 Ω
[1]
The most suitable value would therefore be the 33 Ω resistor.
[1]
[Total 5 Marks]
10)
a)
[3]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Exam practice questions
b)
[3]
[3]
c)
i) See part (i) of figure above. Make sure you label the p.d. axis so that the peak p.d. is 4.0 V
and that you label the time axis to show that one wave takes
1
50 s-1
= 0.02 s = 20 ms
[2]
ii) See part (ii) of figure above. Your graph should have three distinct features as discussed in
part (iii).
[2]
iii) You should explain the following:
•
The diode will not start to conduct until the p.d. across rises to 0.7 V. This means that
there will be a short time delay before it starts to conduct and that it will stop
conducting 0.7 V before the p.d. falls to zero. This means the positive half-cycles are not
quite a full half-cycle.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
9 Current-potential difference relationships
Answers to Exam practice questions
• As 0.7 V is dropped across the diode, the output p.d. will only be:
Vout = (5.0 – 0.7) V = 3.3 V
Your peak should therefore be drawn at 3.3 V instead of 4.0 V
• The diode will not conduct at all during the negative half-cycle.
[2]
[1]
[Total 15 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
Page 136 Test yourself on prior knowledge
1
Copper has a much greater number of charge carriers per cubic metre (n) than carbon, so from the formula
I = nAvq it will be a much better conductor.
2
Rmin = 0.95 × 33 Ω = 31 Ω
a)
P = I2R  I2 =
𝑃
𝑅
𝑃
𝐼=√
𝑅
1.0W
=√
33Ω
= 0.17 A (170 mA)
b)
If the current exceeds this value, the resistor will heat up and could burn out.
c)
𝐼=
𝑉
𝑅
6.0 V
=
33Ω
= 0.18 A (180 mA)
d)
Although this is just above the maximum current previously calculated and the resistor will get
warm, it would probably not get dangerously hot or burn out. It should therefore be safe to use.
However, it would be advisable to try to find a resistor having a higher power rating.
Page 139 Activity 10.1
1
𝑉
The resistance of the filament is given by 𝑅 = .
𝐼
a)
When it is operating normally:
V = 2.40 V (from the graph; 2.5 V nominally)
I = 0.30 A (from the graph; 0.3 A nominally)
𝑅=
b)
2.4 V
0.30 A
= 8.0 𝛺
When the current in it is at its peak:
V = 1.65 V (from the graph)
I = 0.96 A (from the graph)
𝑅=
2
1.65 V
0.96 A
= 1.7 𝛺
The power developed in the filament is given by P = VI.
a)
When it is operating normally:
P = 2.4 V × 0.30 A = 0.72 W
b)
When the current in it is at its peak:
P = 1.65 V × 0.96 A = 1.58 W
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
Page 139 Test yourself
1
a)
‘Ohmic’ means that the resistor obeys Ohm’s law, i.e. the current in it is proportional to the p.d.
across it. This means that its resistance is constant. Above the stated power rating of 250 mW,
the resistor will gradually heat up and so its resistance will change – it will therefore no longer be
‘ohmic’.
b)
Rmin = 0.95 × 150 Ω = 143 Ω;
Rmax = 1.05 × 150 Ω = 158 Ω
c)
P = I2R
𝑃
 I2 =
𝑅
𝑃
250 × 10−3 W
𝑅
150 𝛺
𝐼= √ =√
d)
𝐼=
=
= 0.041 A (41 mA)
𝑉
𝑅
6.0 V
150 𝛺
= 0.040 A (40 mA).
This is right on the limit of 41 mA, but should be safe.
e)
If a p.d. of 12 V were to be applied, the current would be about twice the safe limit and the
power about four times the 250 mW rating (these values are only approximate as the resistor
will heat up and its resistance will increase). The large increase in current and power would make
the resistor very hot and probably burn it out. This would be potentially dangerous.
2
a)
If the motor uses 100 W, the element takes (1200 – 100) W = 1100 W
𝑃 = 𝐼𝑉
⟹𝐼=
=
𝑃
𝑉
1100 W
220 V
= 5.0 A
b)
𝑅=
=
𝑉
𝐼
220 V
5.0 A
= 44 𝛺
c)
If connected to a 110 V supply, the element will still function but the power dissipated in it will
be much less and so it will not heat up anywhere near as much. As the resistance of the element
will change slightly with temperature, it is not possible to determine the exact reduction in
power, but as 𝑃 =
𝑉2
𝑅
, the power will only be about ¼ of the 220 V power.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
d)
However, connecting a 110 V rated hair dryer to a 220 V mains would be potentially dangerous.
The current in the element would be about twice as much as it should be and the power about
four times the specified value. If the dryer is correctly fused, the extra current should blow the
fuse and render the dryer safe. If not, the element would overheat and burn out.
Page 140 Activity 10.2
1
Before taking any readings, the moveable crocodile clip should be touched firmly on the crocodile clip at
the zero end of the rule to check for any contact resistance or any zero error in the meter.
2
The crocodile clip should be pressed firmly on the wire to ensure a good electrical contact, but not too
firmly or else the wire could get damaged and its resistance changed as a consequence.
3
Taking readings at, say, 10 cm intervals would give 10 values, which would be adequate for plotting a graph.
Page 142 Core practical 2
1
A micrometer or digital calipers would be suitable as they can both measure a diameter of about 2 mm to a
precision of 0.01 mm.
2
The micrometer or calipers should be checked for any zero error before measurements are taken.
3
Readings should be taken at different places to average out any non-uniformity along the length of the
pencil lead and at different angles to average out any non-uniformity in the circular cross-section of the
lead.
4
You should have found values for the resistance as shown in the table below:
l/mm
V/V
I/A
R/Ω
20
1.48
0.925
1.60
40
1.51
0.786
1.92
60
1.53
0.648
2.36
80
1.55
0.587
2.64
100
1.58
0.527
3.00
120
1.60
0.465
3.44
Your graph should look like Figure 10.7 in your textbook, with a gradient of somewhere between 18.0 to
18.5 Ω m−1.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
To determine a value for the resistivity ρ you need to use:
𝑅=
⟹𝜌
𝑅𝐴
𝑙
𝜌𝑙
𝐴
where
𝑅
𝑙
= gradient of your graph of R against l and A =
𝜋d2
4
The average diameter is:
𝑑 = 2.24 mm
= 2.24 × 10−3 m
π × (2.24 × 10−3 m)2
𝜌 = 18.0 𝛺 m−1 ×
= 7.1 × 10
−5
4
𝛺m
Page 143 Test yourself
3
𝑅=
a)
⟹
𝑅
𝑙
=
=
𝜌𝑙
𝐴
𝜌
𝐴
1.72 × 10−8 𝛺 m
0,1642 × 10−6 m2
= 0.105 𝛺 m−1
b)
𝐷2
𝐴=π
2
⟹𝐷 =
4
4𝐴
𝜋
4𝐴
𝐷=√
𝜋
4 ×(0.0779 mm)2
=√
π
= 0.315 mm
c)
𝑅=
⟹𝜌
𝜌𝑙
𝐴
𝑅𝐴
𝑙
= 6.29 𝛺 m−1 × 0.0779 × 10−6 m2
= 4.90 × 10−7 𝛺 m
d)
𝐴=π
=
𝐷2
4
𝝅 × (𝟎.𝟐𝟑𝟑𝟕 𝐦𝐦)𝟐
𝟒
= 0.0429 mm2
e)
𝑅=
⟹𝜌
𝜌𝑙
𝐴
𝑅𝐴
𝑙
= 25.2 𝛺 m−1 × 0.0429 × 10−6 m2
= 1.08 × 10−6 𝛺 m
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
4
𝑅=
a)
⟹𝜌
𝜌𝑙
𝐴
𝑅𝐴
𝑙
=
𝑅 × 𝜋 × 𝐷2
𝑙 ×4
2
=
15.8 𝛺 × 𝜋 ×(0.69 × 10−3 m)
17.5 × 10−2 m ×4
= 3.38 × 10−5 𝛺 m
𝑅=
⟹𝜌
𝜌𝑙
𝐴
𝑅𝐴
𝑙
=
𝑅 × 𝜋 × 𝐷2
𝑙 ×4
2
=
33.8 𝛺 × 𝜋 ×(2.61 × 10−3 m)
17.0 × 10−2 m ×4
= 1.06 × 10−3 𝛺 m
b)
This would suggest that clay is a much poorer conductor than carbon (graphite).
Page 144 Activity 10.3
1
Heat the water gently. Just before taking each reading, turn the flame down, stir well and record the
temperature and meter readings.
2
You should have found values for the resistance as shown in the table below:
θ/°C
20
40
60
80
100
V/mV
114
122
130
137
144
I/mA
143
142
141
140
139
R/Ω
0.797
0.859
0.922
0.979
1.036
Your graph should then look like Figure 10.11 in your textbook.
3
The gradient should be about 3.0 × 10−3 Ω K−1 and the intercept about 0.74 Ω.
4
R0α = gradient = 3.0 × 10−3 Ω K−1
R0 = intercept = 0.74 Ω
Dividing:
α =
3.0 × 10-3 Ω K-1
0.74 Ω
= 4.1 × 10−3 K−1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
Page 148 Activity 10.4
1
Your graph of R against θ should look like Figure 10.15 in your textbook.
2
It can be deduced from the graph that the resistance initially falls very rapidly as the temperature increases.
The rate of decrease of resistance (i.e. the gradient –dR/dθ) then gradually gets less and less.
3
As the resistance gets less as the temperature increases (i.e. the gradient is negative) the thermistor must
have a negative temperature coefficient of resistivity.
4
θ/°C
20
30
40
50
60
70
80
90
100
R/Ω
706
491
350
249
179
135
105
87
74
ln(R/Ω)
6.56
6.20
5.86
5.52
5.19
4.91
4.65
4.47
4.30
From the graph we can see that:
5.20 -7.20
Gradient = (60
- 0) °C
= − 0.033 K−1 = −k
 k = 0.033 K−1
Intercept (when θ = 0°C) = 7.20 = ln(R0/Ω)
 R0 = 1.3 kΩ
Note:
The y-axis does not start at zero. If it did, the scale would be very compressed and a mark would be lost.
The intercept is ln(R0/Ω) and so R0 is given by the ‘inverse ln’ of 7.20.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
Page 149 Test yourself
5
a)
A positive temperature coefficient means that the resistivity of copper increases as the
temperature increases, whereas a negative temperature coefficient means that the resistivity of
silicon decreases as the temperature increases.
b)
PTC: any other metal (e.g. aluminium, iron, silver or gold); NTC: any other semiconductor (e.g.
carbon, graphite or germanium)
c)
Referring to the equation I = nAvq, in a metal n does not change with temperature but v gets
slightly less with a rise in temperature. This is because the lattice ions vibrate with more energy
and offer greater resistance to the ‘drift’ of the conduction electrons. From the equation, if v gets
less, the current I will decrease, which means the resistance will increase.
In a semiconductor, n increases exponentially with temperature. Although there is a small
decrease in v, the very much greater increase in n means that the current increases with
temperature, i.e. the resistance gets less.
6
a)
If the temperature coefficient of resistivity of constantan is ‘not excessive’ it means that its
resistivity (and hence the resistance of a resistor made from constantan) will not change very
much with temperature.
b)
𝑅=
⟹𝑙=
=
𝜌𝑙
𝐴
𝑅𝐴
𝜌
1.0 𝛺 ×1.0 × 10−3 m × 0.0125 × 10−3 m
49.4 × 10−8 𝛺 m
= 0.0253 m ≈ 25 mm
c)
Pure copper has a much lower resistivity ( 1.7 × 10−8 Ω m) than constantan, so a much longer
length would be needed for a given value resistor. Copper also has an appreciable positive
temperature coefficient of resistivity, which means that the resistance of a resistor made of
copper would vary with temperature.
7
a)
If the resistance decreased exponentially with temperature, a log graph such as that shown
would be a straight line of negative gradient. The graph is fairly linear for lower temperatures but
then has a more pronounced curve as the temperature increases, suggesting that the decrease in
resistance is not quite a simple exponential function of temperature.
b)
θ/°C
−20
−15
−9
−5
10
30
55
R/kΩ
100
70
50
40
20
8
2
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
c)
Your graph should look like Figure 10.15 on page 148.
d)
i) 90 kΩ
ii) 32 kΩ
iii) 6 kΩ
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
Page 151–153 Exam practice questions
1
The number of charge carriers per unit volume will remain virtually constant with temperature (it would
actually reduce very slightly as the copper expanded, but this would be almost negligible over the
temperature range of 0°C−100°C) – the answer is B.
2
The resistance would increase linearly with temperature from a finite value at 0°C (not zero) – the answer is
C.
3
[Total 1 Mark]
The resistance of a thermistor decreases with temperature due to the significant increase in the number of
charge carriers – the answer is D.
4
[Total 1 Mark]
a)
[Total 1 Mark]
Between the longer edges of the chip:
𝑙 = 5 mm = 5 × 10−3 m
[1]
𝐴 = 10 mm × 1 mm = 10 × 10−3 m × 1 × 10−3 m = 10 × 10−6 m2
[1]
Resistance is:
𝑅 =
b)
ρ𝑙
A
=
3.0 × 10−5 Ω m ×5 ×10−3 m
= 15 × 10−3 Ω
10 × 10−6 m2
[2]
Between the faces of the chip:
𝑙 = 1mm = 1 × 10−3 m
𝐴 = 10 mm × 5 mm = 10 × 10−3 m × 5 × 10−3 m = 50 × 10−6 m2
[1]
Resistance is:
𝑅 =
𝜌𝑙
𝐴
=
3.0 × 10−5 Ω m ×1 × 10−3 m
50 × 10−6 m2
= 6 × 10−4 Ω
[1]
[Total 6 Marks]
5
a)
From Figure 9.1b on page 129, the potential difference is about 0.7 V for a current of 40 mA.
Using R = V/I gives:
𝑅 =
b)
0.7 V
40 × 10−3 A
= 17.5 Ω ≈ 18 Ω
[3]
Draw a straight line through the origin and the point (0.8 V, 16 mA). This represents a resistance
of 50 Ω. This cuts the curve at about 0.6 V, so the component will have a resistance of 50 Ω when
the potential difference across it is about 0.6 V.
As a check, the current at 0.6 V is about 12 mA, giving:
𝑅 =
𝑉
𝐼
=
0.6 𝑉
12 ×10−3 𝐴
= 50 Ω
[3]
[Total 6 Marks]
6
a)
i) Rearranging P = VI,
𝐼 =
P
V
=
3000 W
240 V
= 12.5 A
[2]
So the kettle will operate safely with a 13 A fuse.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
ii) From 𝑅 =
240 V
𝑅 =
b)
i)
𝑉
12.5 A
𝐼
= 19.2 Ω ≈ 19 Ω
At 110 V, from 𝐼 =
110 V
𝐼 =
19.2 Ω
[2]
𝑉
𝑅
= 5.7 A
[2]
ii) The power dissipated is given by 𝑃 =
(110 V)2
𝑃 =
c)
19.2 Ω
𝑉2
𝑅
:
= 630 W
[2]
The assumption is reasonable. As a kettle is used to boil water, the element will heat up the
water until both reach a temperature of 100°C whether the voltage is 240 V or 110 V – it will just
take longer at 110 V. As the temperature of the element is the same in both cases, its resistance
will be the same.
[2]
[Total 11 Marks]
7
a)
The resistivity ρ of a material is defined by the equation
ρ=
𝑅𝐴
[1]
𝑙
where R is the resistance of a length l of the material, having an area of cross-section A
[1]
perpendicular to the length.
b)
𝜌𝑙
i) R =
π𝑑 2
ii) A =
iii) ρ =
4
𝑅𝐴
𝑙
v) ρ =
𝐴
4𝐴
π
=√
= 0.0701 Ω m−1
4 × 0.1110 mm2
π
= 0.3759 mm
𝑅
= × A = 4.41 Ω m−1 × 0.1110 × 10−6 m2 = 4.90 × 10−7 Ω m
𝑙
=
4
𝑅𝐴
𝑙
𝜌
𝑙
d=√
π𝑑 2
iv) A =
𝑅
 = =
𝐴
1.72 × 10-8 Ω m
0.2453 × 10-6 m2
π × (0.2743 mm)2
4
= 0.0591 mm2
𝑅
= × A = 18.3 Ω m−1 × 0.0591 × 10−6 m2 = 1.08 × 10−6 Ω m
𝑙
[2]
[2]
[2]
[2]
[2]
[Total 12 Marks]
8
a)
As the graph is a straight line through the origin, it shows that the current is proportional to the
potential difference, which is Ohm’s law.
b)
[2]
Resistivity is given by ρ = RA/l. We are given 𝑙 (= 2.00 m), A can be found from the diameter (D =
0.25 mm) and R can be found from the inverse gradient of the graph:
𝐴 =
𝑅 =
𝜌 =
πD2
4
=
π × (0.25 ×10−3 m)2
4.0 𝑉
90 × 10−3 𝐴
𝑅𝐴
𝑙
=
4
= 4.91 × 10−8 m2
= 44.4 Ω
44.4 Ω ×4.91 × 10−8 m2
2.00 m
[2]
[2]
= 1.1 × 10−6 Ω m
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
c)
A negligible temperature coefficient of resistivity means that the resistivity, and therefore the
resistance, changes very little with temperature (over the range of temperatures being
considered).
[2]
[Total 10 Marks]
9
a)
(Remember: you should put the relevant data on the graph. The current will be 0.42 A when the
p.d. is 12 V if the lamp is rated at 12 V, 5 W.)
b)
Rearranging 𝑃 = 𝑉 2 / 𝑅:
𝑅=
c)
[3]
𝑉2
𝑃
=
(12 V)2
5W
= 29 𝛺
[2]
When the lamp is ‘off’ it is at a much lower temperature than when it is ‘on’ and glowing white
hot. At a lower temperature, the lattice ions vibrate much less and therefore impede the flow of
electron charge carriers much less. The drift velocity of the electrons is therefore much greater.
[3]
This means that in the equation I = nAvq, v is much greater whilst n, A and q remain the same.
The current is therefore much greater, which means the resistance is much less.
[2]
[Total 10 Marks]
10
a)
𝜌𝑙
R=
𝐴
A=
b)
A=
𝜋𝑑 2
4
𝜌𝑙
𝑅
=
49 × 10-8 Ω m × 1.0 m
d=√
6.29 Ω
4𝐴
𝜋
= 7.79 × 10−8 m2
4 × 7.79 ×10−8 m2
π
=√
= 3.15 × 10−4 m = 0.315 mm
[2]
[2]
i) A systematic error would be caused if the vernier had a zero error. Before taking any
measurements this should be checked.
Random errors could be the result of the wire not being uniform along its length, so
measurements for the diameter should be taken at different points along the length of the
wire. Also, the wire may not be uniformly circular, so two measurements should be taken at
right angles to each other at each point along the wire.
[4]
(Note that the question says ‘Explain’, so you need to explain why the measurements should
be taken in this way, not just how they would be taken.)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
ii) % uncertainty in diameter =
c)
d)
𝑅
𝑙
= 6.29 Ω m−1  l =
R
6.29 Ωm-1
=
0.01 mm
0.315 mm
2.2 Ω
6.29 Ω m-1
× 100% = 3%
[1]
= 0.350 m = 35 cm
Tolerance in cutting wire = 2% of 35 cm =
2
100
[3]
× 35 cm = 0.7 cm = 7 mm
[2]
[Total 14 Marks]
11
a)
i) From the graph, when 𝑉 = 1.0 V, 𝐼 = 20 mA:
𝑅 =
𝑉
𝐼
=
1.0 V
20 × 10−3 A
= 50 Ω
[3]
ii) From the graph, when 𝑉 = 3.0 V, I = 100 mA:
𝑅 =
b)
𝑉
𝐼
=
3.0 V
100 × 10−3 A
= 30 Ω
𝑃 = 𝑉𝐼 = 1.0 V × 20 × 10−3 A = 20 mW
𝑃 = 𝑉𝐼 = 3.0 V × 100 × 10
c)
[2]
−3
[2]
A = 300 mW
[1]
i) The above answers suggest that as energy is being converted at a much greater rate at 3.0 V,
then the temperature of the thermistor will be greater at 3.0 V than at 1.0 V.
[2]
ii) A thermistor is a semiconductor. In semiconductors the number of charge carriers increases
exponentially with temperature. This means that in the equation I = nAvq, n increases
significantly with temperature whilst the quantities A and q remain the same.
Although v is slightly reduced due to increased lattice vibrations, this is nowhere near as
significant as the increase in n. The overall effect is that the current increases, and therefore
the resistance gets less, as the temperature rises.
[5]
[Total 15 Marks]
12
a)
‘Nano’ means 10−9, which suggests that the magnitude of structures in nanotechnology is in the
order of 10−9 m.
b)
[2]
Metals are, mostly, good electrical conductors and much better conductors than semiconductors. In general, the resistivity of metals increases with temperature (positive
temperature coefficient) whilst the resistivity of semi-conductors decreases with temperature
(negative temperature coefficient). There are exceptions to this depending on the composition of
some metal alloys and the doping of semi-conductors.
c)
[4]
The tensile strength of a material is the maximum stress (i.e. force per unit area of cross-section)
that a material can withstand when stretched along its length.
[1]
The elastic modulus (Young modulus) for a material is defined as the ratio of tensile stress/tensile
strain, where tensile strain is the ratio of extension/original length.
d)
𝐹
6422 kg × 9.8 N kg-1
𝐴
1 × 10-6 m2
Tensile strength = =
= 6.26 × 1010 N m−2 = 63 GPa
[1]
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
10 Resistance and resistivity
Answers to Test yourself questions
e)
Tensile strain is the ratio of extension/original length (or reduction in length/original length if the
sample is under compression).
[1]
Plastic deformation means that the sample will not return to its original shape or size when the
applied forces are removed. The sample will be permanently stretched.
[1]
As CNTs have a long, hollow construction they are liable to buckle under compression and so
they are not as strong under compression as they are under tension – think of how easy it is to
buckle a drinking straw by compressing it compared with how difficult is it to stretch it.
f)
[1]
In graphite the atoms of carbon have a hexagonal structure. The sheet in Figure 14.20 in your
textbook is 15 hexagons wide. Each graphite hexagon is about 0.2 nm across, so the width w of
the sheet is about 15 × 0.2 nm = 3 nm.
[2]
When the sheet is rolled into a cylinder, w = πd and so d = w/π = 3 nm/π = 0.95 nm.
We can therefore say that the diameter is about 1 nm.
[2]
(If you had assumed anything between 0.1 nm and 1 nm for the dimension of the hexagons, this
would be a reasonable estimate and get full credit.)
g)
It is not possible to use the formula I = nAvq because the nanoscale cross-section means that the
transport of electrons involves quantum effects.
h)
As units of h = J s and units of e = C  units for
i)
Rmin =
ℎ
4𝑒 2
=
6.63 × 10-34 J s
4 × (1.6 × 10-19 C)
2
= 6.5 kΩ
[2]
ℎ
4𝑒 2
=
Js
C2
=
J C-1
Cs-1
V
= =Ω
A
[3]
[2]
[Total 24 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
Page 154 Test yourself on prior knowledge
1
a)
In 2.2 Ω resistor: 𝐼 =
=
𝑉
𝑅
6.6 V
2.2 𝛺
= 3.0 A
b)
In 3.3 Ω resistor: 𝐼 =
=
𝑉
𝑅
6.6 V
3.3 𝛺
= 2.0 A
2
c)
In ammeter: I = 3.0 A + 2.0 A = 5.0 A
a)
P = I2R
= (3.0 A)2 × 2.2 Ω
= 19.8 W
b)
P = I2R
= (2.0 A)2 × 3.3 Ω
= 13.2 W
c)
Rate at which cell is converting energy = (19.8 + 13.2) W = 33 J s−1
d)
In 5 minutes: ΔE = PΔt
= 33 J s−1 × (5.0 × 60) s
= 9.9 kJ
Page 158 Core practical 3
1
Your graph should look like Figure 11.7 from the textbook.
2
The e.m.f. is the intercept on the V-axis (assuming that you have started the I-axis at zero). You should find
this to be about 1.56 V.
Note that the gradient of the graph is negative (gradient = −r). The internal
resistance therefore is given by the negative of the gradient. You should find this
to be about 0.60 Ω.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
Page 159 Test yourself
1
a)
b)
Total circuit resistance = 2.2 Ω + 0.3 Ω = 2.5 Ω
c)
Circuit current =
=
1.50 V
2.5 𝛺
e.m.f.
total circuit resistance
= 0.60 A
Voltmeter registers p.d.: V = ε – Ir
= 1.50 V – (0.60 A × 0.3 Ω)
= (1.50 – 0.18) V = 1.32 V
d)
In resistor: P = I2R
= (0.60 A)2 × 2.2 Ω
= 0.79 W
e)
2
In cell: P = I R
= (0.60 A)2 × 0.3 Ω
= 0.11 W
2
a)
She probably expected the current to be given by:
𝐼=
b)
𝑉
𝑅
=
9.0 V
4.7 𝛺
= 1.9 A
Total circuit resistance: (𝑅 + 𝑟) =
=
𝑉
𝐼
9.0 V
1.50 A
= 6.0 𝛺
Internal resistance: r = (6.0 – 4.7) Ω = 1.3 Ω
3
a)
𝐼=
𝜀
𝑅+𝑟
= (0.04
12 V
+ 0.01) 𝛺
= 240 A
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
P = I2(R + r)
b)
= (240 A)2 × (0.04 + 0.01) Ω
= 2880 W
c)
V = ε – Ir
= 12 V – (240 A × 0.01 Ω)
= (12 – 2.4) V = 9.6 V
d)
If the internal resistance increased to r = 0.1 Ω:
Starting current =
𝜀
𝑅+𝑟
= (0.04
12 V
+ 0.01) 𝛺
= 86 A.
This would be nowhere near large enough to start the motor.
Page 166 Activity 11.1
1
There are seven possible values of resistance (in kΩ) :3.3, 5.0, 6.7, 10.0, 15.0, 20.0 and 30 Ω.
If you couldn’t get all seven values, look back at the example on page 164 of your textbook.
2
For values below 20 kΩ, changing scales from 200 kΩ to 20 kΩ would give an extra decimal place of
precision. For example, if you had a resistance of 6.73 kΩ it would register as just 6.7 kΩ on the 200 kΩ
scale (maximum resistance 199.9 kΩ). On the 20 kΩ scale (maximum resistance 19.99 kΩ) it would register
as 6.73 kΩ.
Page 166 Activity 11.2
1
Your graph should be a curve of negative gradient like Figure 11.23 on page 166 in your textbook.
2
You can get an experimental value for the resistance by reading off the value of resistance for a current of
0.51 mA from your graph. You should find this to be 3.05 ± 0.10 kΩ.
3
For the parallel combination:
1
𝑅
=
1
1.0 k𝛺
+
1
4.7 k𝛺
= 1.00 kΩ−1 + 0.213 kΩ−1
= 1.213 kΩ−1
𝑅=
1
1.213 k𝛺 −1
= 0.82 kΩ
Adding the 2.2 kΩ resistance in series, we get
Circuit resistance = 2.2 kΩ + 0.82 kΩ = 3.02 kΩ
As the nominal values of the resistors are only to two significant figures, the resistance should be stated as
3.0 Ω.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
4
If we take the experimental value from the graph as 3.1 kΩ, to two significant figures:
% difference =
(3.1 − 3.0) k𝛺
3.0 k𝛺
× 100% = 3%
Note: the theoretical value should be taken as the denominator although, in this particular case, the values
are so close together that it makes little difference to the answer.
5
The difference is likely to be due to the difficulty drawing a curve of best fit and other sources of
experimental error, such as
•
resistor values being only nominal (there is commonly a manufacturer’s tolerance of perhaps 2%,
sometimes as much as 5%)
•
meter errors
•
contact resistance
•
the cell running down (especially if the current taken is more than a few milliamps).
Page 169 Test yourself
4
Figure 11.19 in the textbook shows the seven possible combinations.
a)
Obviously simply 15 Ω!
b)
Two in series: 15 Ω + 15 Ω = 30 Ω
c)
Three in series: 15 Ω + 15 Ω + 15 Ω = 45 Ω
d)
1
𝑅
=
𝑅=
e)
1
𝑅
=
𝑅=
f)
1
15 𝛺
+
15 𝛺
1
=
2
15 𝛺
= 7.5 𝛺
2
15 𝛺
1
15 𝛺
+
15 𝛺
1
15 𝛺
+
1
15 𝛺
=
3
15 𝛺
=5𝛺
3
The two parallel resistors have a combined resistance of 7.5 Ω (see (d)).
This is now added to the resistor in series with them:
R = 7.5 Ω + 15 Ω = 22.5 Ω
g)
The two series resistors have a combined resistance of 30 Ω (see (b)).
They are in parallel with the third resistor, so:
1
𝑅
=
1
30 𝛺
+
1
15 𝛺
= 0.033 Ω-1 + 0.067 Ω-1
= 0.10 Ω-1
𝑅=
1
0.10 𝛺 −1
= 10𝛺
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
5
a)
i) 𝑃 = 𝐼𝑉
⟹𝐼=
=
𝑃
𝑉
150 mW
3V
= 50 mA
ii) 𝑅 =
=
𝑉
𝐼
3V
50×10−3 A
= 60 Ω
b)
Circuit current = sum of currents in the two lamps
= 2 × 50 mA
= 100 mA
c)
Across resistor: p.d. V = (4.5 – 3.0) V = 1.5 V
d)
If one lamp were to blow, at that instant the circuit resistance would be that of the other bulb
plus the 15 Ω resistor as they are in series:
R = (15 + 60) Ω = 75 Ω
At this point, as we now have a simple series circuit:
𝐼=
𝑉
𝑅
=
4.5 V
75 𝛺
= 60 mA
This exceeds the 50 mA rating of the bulb, so the filament will glow very brightly.
At the same time, its resistance will increase slightly as it will be hotter. The circuit resistance will
therefore be slightly increased, reducing the current.
However, it is possible that this will not be enough to stop the filament blowing, even though
bulbs are manufactured to withstand a current slightly larger than the rated value.
6
a)
𝑃=
⟹𝑅=
=
𝑉2
𝑅
𝑉2
𝑃
(14 V)2
60 W
= 3.27 Ω
This differs from 3.3 Ω by only (3.3 – 3.27) Ω = 0.03 Ω
As a 2% tolerance on 3.3 Ω is 0.02 × 3.3 Ω = 0.066 Ω, this resistor would be suitable.
b)
If the cell is only 15% efficient, the total power incident on it must be
𝑃=
100%
15%
× 60 W = 400 W
Area of solar panel needed =
400 W
500 W m−2
= 0.80 m2
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
Page 174 Core practical 12
1
Your graph should look like Figure 11.48 b) on page 180 of your textbook.
2
At 37 °C: output voltage = 2.28 ± 0.05 V
3
The water should be heated gently. Just before taking each reading you should be turn down the flame, stir
well and then record the temperature and meter readings.
4
To get temperatures below room temperature small pieces of ice should be added and the water should be
stirred well until all the ice has melted. The temperature and resistance should then be recorded. This
should be repeated until the temperature approaches 0 °C.
5
For safety, make sure:
•
the beaker is secure so that it is not knocked over
•
bare wires do not touch anywhere and cause a short circuit
•
the burner is placed on a suitable mat to protect the bench.
Page 176 Test yourself
7
a)
For parallel combination:
1
𝑅
=
=
𝑅=
1
2.2 k𝛺
+
1
2.2 k𝛺
2
2.2 k𝛺
2.2 k𝛺
2
= 1.1 k𝛺
1.1 k𝛺
𝑉out = 𝑉in × (1.1
= 9.0 V ×
+ 3.3) k𝛺
1.1
4.4
= 2.25 V
b)
P in each 2.2 kΩ resistor =
=
𝑉2
𝑅
(2.25V)2
2.2 k𝛺
= 2.3 mW
P in 3.3 kΩ resistor =
=
=
𝑉2
𝑅
(9.0 V − 2.25 V)2
3.3 k𝛺
(6.75 V)2
3.3 k𝛺
= 13.8 mW
Therefore resistors rated at 0.1 W (100 mW) would be entirely suitable.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
c)
As the circuit is effectively a 3.3 kΩ resistor in series with a 1.1 kΩ resistor (the resistance of the
parallel combination found in part (a)), the total circuit resistance is 4.4 kΩ. Thus the effect of the
battery having an internal resistance of 0.5 Ω would be completely negligible (0.5 Ω is only one
hundredth of 1% of 4.4 kΩ!).
d)
With a 2% tolerance:
the 3.3 kΩ resistor could have a maximum value of
1.02 × 3.3 kΩ = 3.366 kΩ
the 2.2 kΩ resistors could each have a minimum value of
0.98 × 2.2 kΩ = 2.156 kΩ
so the parallel combination could have a minimum value of
½ × 2.156 kΩ = 1.078 kΩ
1.078 k𝛺
Minimum 𝑉out = 𝑉in × (1.078+3.366)
= 9.0 V ×
k𝛺
1.078
4.444
= 2.18 V
The output voltage should therefore be stated as 2.25 ± 0.07 V
8
a)
We need to drop 6.0 V across the thermistor so:
𝑉out = 9.0 V ×
450 Ω
Total circuit resistance
= 6.0 V
Rearranging:
6 × total resistance = 9 × 450 Ω
Total resistance =
9 × 450 Ω
6
= 675 𝛺
Extra resistance required = (675 – 450) Ω = 225 Ω
This would be about mid-range for a 500 Ω potentiometer, so a 500 Ω potentiometer would be a
suitable choice.
Clearly, a 250 V rating would be more than enough when there will be only about 3 V dropped
across it.
The power developed in the potentiometer is given by
𝑃=
=
𝑉2
𝑅
(3 V)2
225 𝛺
= 0.04 W ≪ 0.5 W
The potentiometer chosen would therefore be entirely suitable.
b)
The student would need to immerse the thermistor in a water bath and adjust the temperature
to 20 °C, which should be checked with a thermometer. The potentiometer would then be
adjusted until the switch just comes on and activates the heater.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Test yourself questions
5
a)
i) At twilight, illumination = 1.0 lux
From the graph, the resistance of the LDR = 100 kΩ
𝑉out = 𝑉in ×
𝑅2
𝑅1 +𝑅2
100 k𝛺
= 9.0 V × (10
= 9.0 V ×
+100) k𝛺
100
110
= 8.2 V
ii) In the laboratory, illumination = 500 lux.
Interpolating from the graph (which is not easy for a log/log graph),
LDR resistance ≈ 1 kΩ
𝑉out = 𝑉in ×
𝑅2
𝑅1 +𝑅2
1 k𝛺
= 9.0 V × (10
= 9.0 V ×
+1) k𝛺
1
11
= 0.82 V
b)
As above, at twilight R2 = 100 kΩ
We require Vout to be 6.0 V
Substituting in 𝑉out = 𝑉in ×
6.0 V = 9.0 V × (𝑅
𝑅2
𝑅1 +𝑅2
100 k𝛺
1 +100) k𝛺
6.0 (R1 + 100) = 9.0 × 100
6R1 + 600 = 900
6R1 = 300
R1 = 50 kΩ
If you are good at working out ratios, you would probably spot that the 9.0 V input needs to be
divided in the ratio 3.0 V : 6.0 V across R1 and the LDR, respectively. This means that R1 must be
half the LDR resistance of 100 kΩ, i.e. 50 kΩ.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
Pages 178–182 Exam practice questions
Start by working out the resistance of each of the combinations:
W: In series 𝑅 = 𝑅1 + 𝑅2 + 𝑅3 = 15 Ω + 15 Ω + 15 Ω = 45 Ω
X: Start by adding the two series resistors: 15 Ω + 15 Ω = 30 Ω
Then combine this with the parallel resistor:
1
𝑅
=
1
𝑅1
+
1
𝑅2
1
⟹
1
=
𝑅
30 Ω
1
+
1 +2
=
15 Ω
30 Ω
3
=
⟹ 𝑅 = 10 Ω
30 Ω
Y: Combining the two parallel resistors:
1
𝑅
=
1
𝑅1
+
1
𝑅2
1
⟹
1
=
𝑅
15 Ω
1
+
2
=
15 Ω
15 Ω
⟹ 𝑅 = 7.5 Ω
Then adding the 15 Ω in series gives 15 Ω + 7.5 Ω = 22.5 Ω
Z: Combining the three parallel resistors:
1
𝑅
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
1
⟹
𝑅
=
1
15 Ω
+
1
15 Ω
+
1
15 Ω
3
=
15 Ω
⟹𝑅 = 5Ω
1
We can now see that Z = W/9 − the answer is A.
[Total 1 Mark]
2
W = 2Y − the answer is D.
[Total 1 Mark]
3
X = 2 × Z − the answer is D.
[Total 1 Mark]
4
a)
Each ‘arm’ has two 4.7 Ω resistors in series, giving 9.4 Ω in total for each ‘arm’. These
‘arms’ are in parallel.
[1]
So:
1
𝑅
b)
=
1
𝑅1
+
1
𝑅2
⟹
1
𝑅
=
1
9.4 Ω
+
1
9.4 Ω
=
2
9.4 Ω
⟹ 𝑅 = 4.7 Ω
[2]
This network of four resistors has a resistance equal to that of each of the single
resistors. It might be preferable to use this network rather than a single resistor as the current
will split, with half going through each ‘arm’. This means that the current in each resistor is only
half what it would have been for a single resistor. As the power depends on the square of the
current (P = I2 R), the power in each resistor will be a quarter of what it would be in a single
resistor. This means that the resistors will not heat as much.
[2]
Note that the total power developed in the network is the same as for a single resistor –
it is shared equally by the four resistors.
5
a)
[Total 5 Marks]
For the bulb:
𝑉
6V
𝐼
0.5 A
i) R = =
= 12 Ω
ii) P = IV = 0.5 A × 6 V = 3 W
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
b)
As the battery consists of four cells in series:
Total internal resistance r = 4 × 0.15 Ω = 0.60 Ω
I=
e.m.f.
Total circuit resistance
= (12
6.0 V
+ 0.60) Ω
[1]
6.0 V
= (12.60) = 0.476 A
[2]
Ω
V = ε – Ir = 6.0 V – (0.476 A × 0.60 Ω) = (6.0 – 0.29) V = 5.71 V ≈ 5.7 V
c)
[2]
Power wasted = power developed in the internal resistances
P = I2r = (0.476 A)2 × 0.60 Ω = 0.14 W (140 mW)
d)
[2]
The original 3 W of power dissipated in the filament is only reduced by 0.14 W as a result of the
increase in internal resistance. This will not significantly reduce the temperature of the filament
so, although the resistance of the filament will decrease slightly, the assumption that it remains
constant is not unreasonable.
[2]
[Total 11 Marks]
6
If the battery is short-circuited, the only resistance will be the internal resistance of the battery.
𝑉
Using 𝐼 = :
𝑅
𝐼 =
9V
0.50 Ω
= 18 A
[2]
This is a large current and will generate a power of 𝑃 = 𝐼 2 𝑅 = (18A)2 × 0.50 Ω = 162 W inside the
battery, which will make the battery hot.
[2]
[Total 4 Marks]
7
a)
A 50 MΩ resistor in series with the output will keep the current very small. As the severity of an
electric shock depends on the current passing through your body to earth, the resistor acts as a
safety device.
b)
From 𝐼 =
𝐼 =
c)
𝑉
𝑅
[2]
the maximum current will be:
5 × 103 V
50 × 106 Ω
= 1.0 × 10−4 A = 0.10 mA
[2]
The total resistance between the terminal of the supply and the ground will be 50 MΩ + 10 kΩ.
As 10 kΩ is only 0.01 MΩ, the girl’s resistance has very little effect and so the current in the girl
would be virtually the same as in part b).
d)
[2]
2
Using 𝑃 = 𝐼 𝑅, the power dissipated in the girl would be:
𝑃 = (1.0 × 10−4 A)2 × 10 × 103 Ω = 1.0 × 10−4 W = 0.10 mW
This is considerably less than the 14 mW from the car battery, and so shows the effectiveness of
the 50 MΩ resistor as a safety device.
[3]
Warning: high voltage supplies, even when safety protected, are still very dangerous and must
be treated with the utmost caution.
[Total 9 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
8
The table is completed by adding values for the resistance and power, for example V = 1.44 V and I = 0.20 A
gives:
𝑅=
𝑉
1.44 V
=
= 7.2 Ω
𝐼 0.20 mA
and
P = VI = 1.44V x 0.20A = 0.29W
I/A I/A
V/V
R/Ω
P/W
0.20
1.44
7.20
0.29
0.40
1.32
3.30
0.53
0.60
1.20
2.00
0.72
0.80
1.09
1.36
0.87
1.00
0.95
0.95
0.95
1.20
0.84
0.70
1.01
1.40
0.73
0.52
1.02
1.60
0.59
0.37
0.94
Note the maximum power generated in the load resistor is when its resistance is equal to the internal
resistance of the cell (0.60 ohm)
[Total 7 Marks]
9
a)
I/mA
V/V
R/kΩ
P/mW
0.10
6.07
60.7
0.61
0.15
5.90
39.3
0.88
0.20
5.72
28.6
1.14
0.30
5.35
17.8
1.60
0.40
4.97
12.4
1.99
0.50
4.20
8.4
2.10
0.60
3.00
5.0
1.80
0.65
2.01
3.1
1.31
[3]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
b)
i)
For small currents the graph, as shown above, is linear, indicating a constant internal
resistance. For larger currents the graph clearly curves downwards, showing that the
internal resistance increases as the current gets larger.
ii)
The e.m.f. of the cell will be the voltage when the current is zero, i.e. the intercept on the
voltage axis. From the graph this is 6.5 V. The internal resistance for low current values
is given by the numerical value of the gradient of the linear part of the graph.
Extending the linear part to give a large triangle:
𝑟=
c)
(6.50 –4.25) V
(0.60 – 0.00) mA
= 3.7 kΩ
[9]
i)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
ii)
From the graph, the maximum power is approximately 2.1 mW.
[4]
[Total 16 Marks]
10
a)
i) Combining the 22 Ω and 33 Ω parallel resistors:
1
𝑅
1
=
𝑅1
+
1
𝑅2
⟹
1
𝑅
=
1
22 Ω
+
1
33 Ω
=
3+2
66 Ω
=
5
66 Ω
⟹𝑅 =
66 Ω
5
= 13.2 Ω
The total circuit resistance is therefore 47 Ω + 13.2 Ω = 60.2 Ω.
The circuit current will be given by:
𝐼=
𝑉
𝑅
=
6.02 V
60.2 Ω
= 0.10A
The current through the 47 Ω resistor is therefore 100 mA. When this current comes to the
parallel network, it will split in the inverse ratio of the resistances, i.e. 22/55 (= 40 mA)
through the 33 Ω resistor and 33/55 (= 60 mA) through the 22 Ω resistor.
Alternatively, the potential difference across the parallel network is:
V = IR = 0.10 A × 13.2 Ω = 1.32 V
giving the current in the 22 Ω resistor as:
𝐼=
𝑉
𝑅
=
1.32 V
22 Ω
= 0.060 A = 60 mA
and the current in the 33Ω resistor as
𝐼=
b)
𝑉
𝑅
=
1.32 V
33 Ω
= 0.040 A = 40 mA
The power generated is given by P = I2 R in each case:
47 Ω: (0.10 A)2 × 47 Ω = 0.47 W
22 Ω: (0.06 A)2 × 22 Ω = 0.079 W
33 Ω: (0.04 A)2 × 33 Ω = 0.053 W
[8]
[Total 10 Marks]
11
a)
Before the voltmeter is connected, the total circuit resistance is
R = 22 kΩ + 33 kΩ = 55 kΩ
[1]
This gives a circuit current of:
𝐼=
𝑉
𝑅
=
7.5 V
55𝑘 Ω
= 0.136 mA
[1]
The potential difference across the 33 kΩ resistor is therefore:
V = IR = 0.136mA × 33kΩ = 4.5 V
[1]
Alternatively, if you are good at ratios you might spot that the 7.5 V will split up in the ratio of
the two resistors:
V across 33 kΩ = 33/55 × 7.5 V = 4.5 V
b)
i) When the switch is closed, the voltmeter is connected in parallel with the 33 kΩ resistor. The
resistance of this parallel combination will be less than 33 kΩ, so the voltage dropped across
the combination will be less than the previous 4.5 V.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
ii) If the voltmeter reads 4.0 V, the potential difference across the 22 kΩ resistor must be (7.5 –
4.0) V = 3.5 V.
[1]
The current in the 22 kΩ resistor, which is also the circuit current, will be:
𝐼=
𝑉
=
𝑅
3.5V
22𝑘Ω
= 0.159mA
[1]
The combined resistance of the parallel arrangement of the voltmeter and the 33 kΩ resistor
will therefore be:
𝑅=
𝑉
𝐼
=
4.0 V
0.159 mA
= 25.1 kΩ
[1]
For this parallel arrangement:
1
𝑅
1
𝑅𝑉
=
1
𝑅33
+
1
𝑅𝑉
⟹
1
𝑅𝑉
=
1
𝑅
1
−
𝑅33
=
1
25.1 kΩ
−
1
33 kΩ
= 0.0398𝑘𝛺 −1 − 0.0303𝑘𝛺 −1 = 0.0095𝑘𝛺−1
RV = 106 kΩ
[2]
If the voltmeter is rated as 10 V/100 μA, its resistance should be:
𝑅=
𝑉
𝐼
=
10V
100×10−6 A
= 1.0 × 105 Ω = 100 kΩ
[2]
This differs by 6% from the experimental value. As each resistor has a tolerance of 5%, the
experimental value is within the overall tolerance and so is compatible with the stated rating
of the voltmeter.
[2]
[Total 14 Marks]
12
a)
As the temperature of the thermistor falls, its resistance increases as there will be less charge
carriers per unit volume. If the resistance of the thermistor increases, the proportion of the
supply voltage dropped across it will also increase and so the proportion of the supply voltage
b)
across the resistor R will decrease and Vout will fall.
[3]
i) Reading from the graph, at 0 °C the thermistor has a resistance of 18.0kΩ.
[1]
ii) When Vout is 5.0 V, the voltage across the thermistor and the potentiometer will be
(9.0 − 5.0) V = 4.0 V. Assuming that the potentiometer is set at zero, the voltage across the
thermistor will be 4.0 V.
[1]
The current in the thermistor (and therefore the circuit current as it is a series circuit) will be
given by:
𝐼=
4.0V
18.0kΩ
= 0.22mA
[1]
The value of R is then given by:
𝑅=
𝑉𝑜𝑢𝑡
𝐼
=
5.0 V
0.22 mA
= 22.5 kΩ
The best value to use would therefore be the 22 kΩ resistor.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
iii) The purpose of the potentiometer is to fine tune the circuit. It is adjusted so that the lamp is
switched on at exactly 0C.
[2]
[Total 10 Marks]
13
a)
The electromotive force of an electrical source is defined as the energy per unit charge converted
into electrical energy by the source, in this case the cell, whilst the potential difference between
two points in a circuit is the electrical energy per unit charge converted into other forms of
energy, in this case in the resistor.
In the equation V = ε – Ir:
• ε is the chemical energy per coulomb converted into electrical energy by the cell
• Ir is the energy per coulomb used up by the charge doing work against the internal resistance
of the cell
• V (= ε – Ir) is the energy per coulomb that will be dissipated in the resistor R.
b)
i) I =
e.m.f.
Total circuit resistance
ii) V = ε – Ir = ε –
c)
From I =
𝜀
𝑅+𝑟
𝜀𝑟
=
𝑅+𝑟
𝜀(𝑅 + 𝑟) − 𝜀𝑟
𝑅+𝑟
[4]
𝜀
=
=
𝑅+𝑟
𝜀𝑅
[3]
𝑅+𝑟
 ε = I(R + r) so:
ε = 616 mA × (2.2 Ω + r)
ε = 308 mA × (4.7 Ω + r)
Dividing we get:
1=2×
2.2 Ω + 𝑟
4.7 Ω + 𝑟
 1 × (4.7 Ω + r) = 2 × (2.2 Ω + r)
 4.7 Ω + r = 4.4 Ω + 2r
 r = 0.3 Ω
Substituting into ε = I (R + r):
ε = 616 mA × (2.2 + 0.3) Ω = 1.54 V
d)
From (b) we have:
I=
𝜀
𝑅+𝑟
P = IV =
e)
[4]
and V =
𝜀
𝑅+𝑟
×
𝜀𝑅
𝑅+𝑟
𝜀𝑅
𝑅+𝑟
𝜀2𝑅
= (𝑅
[2]
+ 𝑟)2
To find a maximum we need to differentiate with respect to R:
d𝑃
d𝑅
=
d
d𝑅
𝜀2𝑅
((𝑅
+ 𝑟)2
) = (𝑅
𝜀2
+ 𝑟)2
2𝜀 2 𝑅
– (𝑅
+ 𝑟)3
𝜀 2 (𝑅 + 𝑟) −2𝜀 2 𝑅
=
(𝑅 + 𝑟)3
=
𝜀 2 (𝑟 − 𝑅)
(𝑅 + 𝑟)3
This will be zero (i.e. the power will be a maximum) when R = r
[3]
[Total 16 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
15
a)
Conservation of charge means that we cannot gain or lose charge. Consider the 3-way junction
shown in the diagram.
As current is the rate of flow of charge, by conservation of charge I2 + I3 = I1. If this were not so,
charge would be gained or lost at the junction which, by the conservation of charge, is not
possible.
b)
[3]
This question is really about symmetry, something we come across a lot in physics. The first thing
to do is draw a diagram showing the current in each resistor.
If a current I enters the network at point X, by symmetry it will split equally in each of the three
resistors connected to point X. We can now put ⅓I in each of these resistors.
[1]
By the same argument, the current in each of the three resistors connected to point Y will also be
⅓I so that a current I leaves at point Y.
[1]
But what about the remaining six resistors? Again, by symmetry, the ⅓I splits equally so that the
current in the remaining six resistors is ⅙I.
[1]
Your distribution of current should now look like the figure below.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
What we now have is three resistors in parallel at X (combined resistance R/3) connected in series with the
middle six resistors in parallel (combined resistance R/6), which in turn are connected in series the three
parallel resistors at Y (combined resistance R/3). This is shown in the diagram below.
𝑅
𝑅
𝑅
5𝑅
3
6
3
6
The network resistance is therefore + + =
16
a)
[3]
[1]
From the graph of R against L we can take two convenient values of R and L and substitute into
the equation
logR = nlogL + loga
The obvious points to take are: L = 1.0, R = 100 and L = 10 000, R = 0.1.
We then have:
log(100) = n log(1.0) + log a  2 = 0 + log a  log a = 2
log(0.1) = n log(10 000) + log a  −1 = 4n + log a
Substituting that ln a = 2 into the second equation gives us
−1 = 4n + 2  4n = −3  n = −¾
If log a = 2  a = 100
3⁄
4
From R = aLn  R = 100 𝐿−
b)
100
R=4
√𝐿3
[4]
When L = 50 lux:
100
R=4
√𝐿3
=4
100
√503
= 5.3 kΩ
[2]
(You should check from the graph that this looks about right.)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
11 Internal resistance, series and parallel
circuits and the potential divider
Answers to Exam practice questions
c)
If the LDR is 0.5 m from the lamp (i.e. its distance from the lamp is halved), then four times as
much light will fall on the LDR, giving L = 200 lux.
100
R=4
√𝐿3
d)
=
100
4
√2003
= 1.9 kΩ
[2]
We have to assume that the lamp acts as a point source and that the illumination is inversely
proportional to the square of the distance from the lamp. We also have to assume that the only
light falling on the LDR is the light from the lamp, i.e. there is no background illumination.
[2]
[Total 10 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
12 Fluids
Answers to Test yourself questions
Page 183 Test yourself on prior knowledge
1
density =
mass
volume
0.250 kg
=
2.3 × 10−4 m3
= 1100 kg m−3
2
mass = volume × density
= (5.2 m × 4.0 m × 2.3 m) × 1.3 kg m−3
= 62 kg
3
force
pressure =
area
=
kg m s−2
m2
= kg m−1 s −2
4
force = pressure × area
= 1.0 × 105 Pa × 11.0 m2
= 1.1 × 106 N
5
Liquid molecules have forces (bonds) with neighbouring molecules but are able to move relative to each
other. Gas molecules experience no forces from other molecules except during collisions and so move
freely.
Page 184 Activity 12.1
1
Mass of air removed = 421.38 g – 420.80 g = 0.58 g
Volume of air removed = 450 ml
mass
Density =
volume
=
5.8 × 10−4 kg
4.50 × 10−4 m3
= 1.29 kg m−3
Page 185 Test yourself
1
Length = 50 m, width = 25 m, depth between 2 m and 3 m
Mass = volume × density = (50 m × 25 m × 2.0 m) × 1000 kg m−3 to
(50 m × 25 m × 3.0 m) × 1000 kg m−3
= 2.5 × 106 kg to 3.8 × 106 kg
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
12 Fluids
Answers to Test yourself questions
2
Mass of carbon dioxide removed = 180.39 g – 179.60 g
= 0.79 g
Volume of carbon dioxide removed = 0.80 × 500 ml
= 400 ml
density =
7.9 × 10−4 𝑘𝑔
400 × 10−6 𝑚3
= 2.0 kg m−3
3
ΔP = ρgΔh
= 13 600 kg m−3 × 9.81 m s−2 × 0.772 m
= 1.03 × 105 Pa
4
ΔP = ρgΔh; 1.03 × 105 Pa = 1000 kg m−3 × 9.81 m s−2 × Δh → Δh = 10.5 m
5
Hold a piece of paper in any plane in still air. The paper does not move, so it is in equilibrium in all of the
planes. The force due to the air pressure must be the same on each side of the paper and the same for all
orientations.
Page 186 Activity 12.2
Estimations of volume for adult man: head = sphere of diameter 20 cm;
volume ≈ 4 × 10−3 m3
Torso = cylinder of diameter 24 cm and length 60 cm; volume ≈ 3 × 10−2 m3
Arms = 2 × cylinders of diameter 6 cm and length 70 cm; volume ≈ 2 × 10−2 m3
Legs = 2 × cylinders of diameter 10 cm and length 90 cm; volume ≈ 6 × 10−2 m3
Total volume ≈ 0.1 m3
Mass of displaced water = volume × density ≈ 0.1 m3 × 1000 kg m−3 ≈ 100 kg
Mass of displaced air = volume × density ≈ 0.1 m3 × 1.3 kg m−3 ≈ 0.1 kg
Page 187 Test yourself
6
Archimedes’ principle states that if a body is wholly or partially immersed in a fluid, it will experience an
upthrust equal to the weight of the fluid displaced.
7
Weight of stone = 0.125 kg × 9.81 m s−2 = 1.23 N
a)
F=W–U
→U=W–F
= 1.23 N – 1.00 N
= 0.23 N
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
12 Fluids
Answers to Test yourself questions
b)
Mass of displaced water = 25 g
Volume of displaced water =
=
mass
density
2.5 × 10−2 kg
1000 kg m−3
= 2.5 × 10−5 m3
= volume of stone
c)
8
Density of stone =
0.125 kg
2.5 × 10−5 m3
= 5000 kg m−3
When floating: weight of body = upthrust = weight of displaced fluid = volume × density of fluid
a)
The density of ethanol (800 kg m−3) is less than water, so a greater volume needs to be displaced
for flotation. The tube will float lower in the ethanol.
b)
Water at 4.0 °C is denser than water at 20 °C, so less cold water needs to be displaced for
flotation. The tube will float higher in the cold water.
9
Weight of the displaced air is greater than the combined weight of the hot air and the fabric of the balloon.
The upthrust will be greater than the total weight giving a resultant upward force that will make the
balloon accelerate upwards.
Page 189 Activity 12.3
1
Graph shows that the time increases (possibly) exponentially as the concentration increases.
2
The increase in time for a fixed volume to flow indicates that the rate of flow is decreasing with increased
concentration, and so the viscosity of the fluid must also be increasing.
3
For the fluid to pass through in 2.1 s, the concentration should be 27–28%.
4
The volume of fluid and the temperature must be the same for every reading.
Page 190 Test yourself
10
Laminar flow occurs when there are no abrupt changes in the velocity of the fluid, and layers of the fluid
flow over each other without crossing.
11
Turbulence occurs when layers of moving fluid cross and eddy currents are formed.
12
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13
At a critical speed (depending on the density and viscosity of the fluid) the flow becomes turbulent. The
resistance to the flow increases so the rate of flow is reduced.
Page 192 Core practical 4
1
•
Apparatus: tall measuring cylinder, oil, ball bearings (diameter 1–5 mm), plastic tweezers, 30 cm rule and
stopwatch.
•
Diagram: see Figure 12.10.
•
Measure the diameter of the ball bearings using a micrometer (or, if not available, lay 20 or more balls in
a paper groove and find the length of the row with the 30 cm rule).
•
Make two marks on the cylinder, one near the top and the other near the bottom. Fill the cylinder with
the oil so that the surface is a few cm above the upper mark (so that the ball bearings will reach terminal
velocity above the mark). Measure the distance between the marks using the rule.
•
Use the tweezers to hold a ball just above the surface and then release it into the liquid.
•
Use the stopwatch to measure the time for the ball to fall between the two marks. Repeat this twice
more for the same ball bearing and find the average time interval.
•
Repeat the experiment using balls of different diameters.
•
The independent variable is the diameter of the balls and the dependent variable is the time.
•
The temperature of the oil must be kept constant throughout. The cylinder should be securely placed
away from other items and safety goggles worn throughout the experiment.
•
The terminal velocity is calculated by dividing the distance between the marks by the average time for
each size of bearing to fall between them.
•
2
The value of the viscosity of the liquid is found using the equation 𝜂 =
2(𝜌𝑠 −𝜌𝑓 )𝑔𝑟 2
9𝑣
.
There are no units given for the terminal velocity in the headings. This is significant as they have been
expressed in cm s−1. The number of significant figures is inconsistent with the precision of the readings;
average values of d should be 4.00 and 6.10 mm. It would be helpful to include values of the radius.
2(𝜌𝑠 −𝜌𝑓 )𝑔𝑟 2
3
𝑣=
4
Plot a graph of v against r2. This gives a straight line through the origin showing that v is directly
9𝜂
proportional to r2.
5
The value of v at r2 = 4.00 mm2 is below the line of best fit of the other plots.
The experiment using 2 mm balls should be repeated.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
12 Fluids
Answers to Test yourself questions
6
The gradient of the graph =
2(𝜌𝑠 −𝜌𝑓 ) 𝑔
9𝜂
= 2.4 × 104 m−2 s −1
Using the values of density given 𝜂 =
2(7800 kg m−3 − 820 kg m−3 ) ×9.8 m s−2
9 ×2.4 × 104 m−2 s−1
= 0.63 Pa s
7
Smallest diameter d = 1.98 ± 0.01 mm;
percentage uncertainty = ±
0.01 mm
1.98 mm
× 100% = ±0.5%
Smallest time = 0.9 ± 0.1 s; percentage uncertainty = ±
8
0.1 s
0.9 s
× 100% = ±11%
The greatest uncertainty is in the measurement of the time (particularly for the larger balls). To increase
the time interval, and so reduce the percentage uncertainty, the distance between the marks could be
made bigger (or smaller ball bearings used).
Page 193 Test yourself
14
See Figure 12.9 from textbook.
15
When the terminal velocity is reached the resultant force on the body is zero.
Weight = upthrust + viscous drag
16
For Stokes’ law to apply, the object must be spherical and the flow must be laminar.
17
Rearranging Stokes’ law, 𝜂 =
18
Any three from: molecular structure of the oils is different so the viscosities differ; the higher temperature
𝐹
6𝜋𝑟𝑣
=
N
m × m s−1
= N s m−2
of the oil in the Middle East will reduce its viscosity; the pressure across the pipes may differ; the diameter
and/or lengths of pipe may be different.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
12 Fluids
Answers to Exam practice questions
Pages 195–198 Exam practice questions
1
a)
At the instant of release, the viscous force is zero; the ball will begin to fall at the rate of the
acceleration due to gravity, 9.8 m s−2 – the answer is D.
b)
At point 3 the sphere is travelling at constant velocity; the acceleration is
zero – the answer is A.
c)
[Total 1 Mark]
[Total 1 Mark]
Between points 1 and 2 the sphere is still accelerating downwards; there must be a resultant
downward force, so the weight is greater than the sum of the two upward forces, U and F – the
answer is D.
d)
[Total 1 Mark]
At point 3 the sphere is moving at the terminal velocity; the resultant force is zero, so the weight
must equal the sum of the two upward forces – the answer is C.
[Total 1 Mark]
2
Stokes’ law force does not depend on the mass – the answer is B.
[Total 1 Mark]
3
h × 13 600 kg m−3 × 9.8 m s−2 = 1.01 × 105 Pa
[1]
→ h = 0.76 m of mercury
[1]
[Total 2 Marks]
4
mass
= 1.5 kg m−3
𝜌=
b)
The air is compressed so its density inside the balloon is greater than that at normal atmospheric
volume
=
50 × 10−3 kg
a)
4
𝜋(0.20 m)3
3
[2]
pressure.
[1]
[Total 3 Marks]
5
a)
𝑉 × 𝜌 × 𝑔 = 600 N
𝑉=
b)
600 N
1200 kg m−3 × 9.8 m s −2
= 5.1 × 10−2 m3
[2]
The above volume is less than the volume displaced by the less dense water in the pool.
[1]
[Total 3 Marks]
6
a)
The temperature [1] and the place of origin [1] will both affect the viscosity of the oil. The rate of
flow is inversely proportional to the viscosity – stickier fluids move more slowly through the
pipeline, so the rate of flow is greater at higher temperatures. [1]
Increasing the diameter of the pipe will greatly increase the rate at which the oil flows through it
(at the same pressure). [1]
b)
If the flow rate is too fast, turbulence occurs [1] and much more energy is needed to transport
the oil.
[Total 5 Marks]
7
a)
Laminar at A and B [1]; turbulent at C [1]
b)
Uplift, F = ΔP × A = 4.0 × 103 Pa × 120 m2 [1] = 4.8 × 105 N [1]
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12 Fluids
Answers to Exam practice questions
c)
For level flight: uplift = weight
4.8 × 105 N = m × 9.8 m s−2 → m = 49 000 kg (49 tonnes)
[1]
[Total 5 Marks]
8
a)
mg = 6πηrv
𝑣=
b)
[1]
4 × 10−12 kg × 9.8 m s −2
6 𝜋 (2 × 10−5 N s m−2 )(0.1 × 10−3 𝑚)
≈ 1 mm s−1
[2]
The larger raindrops have a much bigger terminal velocity.
[1]
[Total 4 Marks]
9
a)
U + F upward; W or mg down
[2]
b)
W = 34  r 3 v = 34  (1.5  10−3m)3 (7800 kg m−3 )(9.8 m s −2 ) = 1.08 × 10−3 N
[2]
c)
U+F=W
[1]
d)
0.20 × 10−3 N + 6πηrv = 1.1 × 10−3 N
𝑣=
e)
0.9×10−3 𝑁
6𝜋×0.35 Pas×1.5×10−3 m
[1]
= 9.1 × 10−2 m s −1
[2]
i) 0.16 m s−1
[1]
ii) A rise in the terminal velocity indicates a reduction in the viscosity of the liquid. [1] As the
temperature rises, the viscosity decreases [1] until, after a temperature of about 70 °C, the
viscosity stays the same. [1]
iii) Viscostatic means that the viscosity does not vary with temperature.
[1]
[Total 13 Marks]
10
a)
Larger particles have a higher terminal velocity than the smaller ones [1]. The smaller particles
reach their terminal velocity much more quickly [1] than the larger ones; in such a short distance
the pebbles will not reach their terminal velocity [1] and will fall quickly to the bottom.
The reason for this is that the viscous drag is proportional to the radius whereas the weight is
proportional to the radius cubed [1]. This leads to the expression:
𝑣=
2(𝜌𝑠 −𝜌𝑓 )𝑔 𝑟 2
9𝜂
[1]
Showing that terminal velocity is proportional to the radius squared. The layers will therefore
have the largest particles at the bottom with those of decreasing size settling in order on top. [1]
b)
For the fine sand particles, 𝑣 =
=
2(𝜌𝑠 −𝜌𝑓 )𝑔 𝑟 2
9𝜂
2 × (1200 kg m−3 - 1000 kg m−3 ) × 9.8 m s −2 × (1.0 × 10−4 m)2
9 × 0.8 N s m−2
= 5.4 × 10−6 m s−1
[1]
[1]
Assuming terminal velocity is reached instantaneously,
time to settle =
distance
terminal velocity
=
0.22 m
5.44 × 10−6 m s −1
= 40 000 s = 11 hours
[2]
[Total 10 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
12 Fluids
Answers to Exam practice questions
11
a)
The non-streamlined shape of the obstruction means that there will be abrupt changes in the
velocity of the fluid as it moves past the object [1]. This causes turbulence [1] and the formation
of vortices (eddies) in the fluid beyond the obstruction. The vortices move along at the speed of
the fluid [1], and, if the diameter of the tube is known, the rate of flow can be calculated.
b)
Rate of flow =
volume
Speed of flow =
c)
time
=
𝑡
= 𝜋 𝑟2 𝑣 =
2.0 × 10−4 m3 s −1
𝜋(1.1 × 10−2 m)2
12 × 10−3 m3
60 s
[1] = 2.0 × 10−4 m3 s−1
[1]
[1] = 0.53 m s−1
[1]
Area of pipe = π × (0.30 m)2 = 0.283 m2
Av =
39
60
m3 s−1 = 0.65 m3 s−1 [1] → v =
Frequency of pulses =
d)
𝜋 𝑟2 𝑙
2.3 m s −1
0.53 m s −1
0.65 m3 s −1
0.283 m2
= 2.3 m s−1
[1]
× 20 Hz [1] = 87 Hz
[1]
Assume that the flow is laminar [1] and the velocity of the flow is constant across the crosssection of the pipe. [1]
e)
If the diameter is halved, the pressure across the ends will need to be eight times bigger for the
same rate of flow. [1]
The area of cross-section is four times less than that of the 60 cm pipe. This means that the
speed of flow will be four times as big as that of the wider pipe [1] (9.2 m s−1).
Although the higher pressure could cause problems, it is likely that the flow will become
turbulent at such high speeds of flow. [1]
[Total 16 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Test yourself questions
Page 199 Test yourself on prior knowledge
1
12.5 N
a)
∆I =
b)
Each spring extends by 4.0 cm. Total extension = 8.0 cm.
c)
The tension in each spring = half of the weight = 2.5 N. So Δl = 1.0 cm.
5.0 N
× 2.0 cm = 5.0 cm
2
For equilibrium; tension = weight = 0.120 kg × 9.8 m s−2 = 1.2 N
3
Work = force × distance = 4.0 N × 2.5 m = 10 J
4
EPE = average force × extension = ½ (5.0 kg × 9.8 m s−2) × 2.0 × 10−3 m = 4.9 × 10−2 J
Page 202 Test yourself
1
Tough; brittle
2
Malleable
3
Stiff; brittle
4
Ductile
5
Stronger
Page 204 Activity 13.3
1
The extension of the wire depends on its length in addition to the load. A short wire will give very small
extensions even for large loads. These will be difficult to measure and lead to large percentage
uncertainties in the values. Typically, a length of two to three metres of copper wire is needed to give
extensions of several cm for the maximum applied load.
2
If there is a gap between the marker and the scale there is a possibility of parallax error. If the marker is
viewed from different angles, the scale reading will vary. If the marker touches the scale there will be no
parallax error.
3
a)
The wire still has some elasticity and contracts when the load is removed. However there has
been some plastic flow resulting in a permanent deformation of the wire.
b)
When the load is removed the wire stays the same length. The wire is plastic and shows no
elastic properties.
Page 205 Test yourself
6
Hooke’s law states that, up to a certain limit, the extension is directly proportional to the load.
7
a)
k=
b)
i) k =
𝐹
∆𝑙
=
ii) k =
10 kg × 9.8 m s−1
5.0 × 10−2 m
10 kg × 9.8 m s−1
10 × 10−2 m
𝐹
∆𝑙
=
= 1.96 × 103 N m−1
=
𝐹
∆𝑙
10 kg × 9.8 m s−1
2.5 × 10−2 m
= 0.98 × 103 N m−1
= 3.92 × 103 N m−1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Test yourself questions
8
a)
Limit of proportionality is the point where Hooke’s law ceases to be obeyed.
b)
Elastic limit is the point beyond which the wire will not regain its original shape when the
deforming force is removed.
c)
Yield point is at the onset of plastic flow.
d)
Plastic flow is the region where the wire deforms plastically. Layers of atoms slide over each
other and when the load is removed there will be no change in the length of the wire.
Page 210 Test yourself
9
a)
i) Stiffer material will have a steeper gradient.
ii) Tougher material has a greater area under the curve.
10
b)
The brittle line will show little or no plastic deformation when it breaks.
a)
k=
=
𝐹
∆𝑙
12 kg × 9.8 m s−1
2.5 × 10−3 m
= 4.7 × 104 N m−1
b)
EPE = average force × extension
= ½ (12 kg × 9.8 m s−2) × 2.5 × 10−3 m
= 0.15 J
11
A polymer is made up of molecules consisting of many atoms (mainly carbon and hydrogen) in long chains.
12
Work done = area under the graph. This can be estimated by calculating the energy represented by each
square and counting the squares, or by drawing triangles, rectangles or trapeziums that give an
approximate fit of the curve and finding their areas.
Page 211 Core practical 5
1
b)
Force/N
Extension/mm
Stress/Pa
Strain/%
0
0
0
0
5.0
0.5
3.0 × 106
0.017
10.0
1.0
6.0 × 106
0.033
15.0
1.5
9.0 × 106
0.050
20.0
2.0
12 × 106
0.067
25.0
2.5
15 × 106
0.083
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Test yourself questions
30.0
3.0
18 × 106
0.100
35.0
4.0
21 × 106
0.133
40.0
6.0
24 × 106
0.200
35.0
300
21 × 106
10.0
30.0
750
18 × 106
25.0
2
The independent variable is the load and the dependent variable is the extension.
3
A metre rule is suitable for measuring the length of the wire. The uncertainty of the reading is
± 0.005 m
3.000 m
× 100% = ± 0.2%
A micrometer is needed to measure the diameter of the wire. It needs to measure a thickness of about 0.5
mm to a precision of ±0.01 mm. The uncertainty is
± 0.01 mm
0.46 mm
× 100% = ± 2%
The extension needs to be measured over a range of 750 mm to a precision of ±0.5 mm, so a metre rule is
used. The uncertainty in the measurements up to 6 mm is
4
± 0.5 mm
6.0 mm
× 100% = ±8%.
When measuring the diameter of the wire, check the zero error of the micrometer and take pairs of
readings at right angles at several positions along the wire, to ensure that the wire is uniform.
When measuring the extension, ensure that the edge of the marker tape is in contact with the scale so that
parallax errors are avoided.
5
Safety goggles must be worn at all times in case the wire snaps and damages the eyes; a padded box should
be placed under the load so that if it falls, it will not cause injury.
6
The graph has a linear region for loads up to 30 N and then levels and slopes downward (as Figure 8.4). The
gradient of the graph represents the Young Modulus of copper. E = 1.8 × 108 Pa
7
To improve the accuracy of the experiment, the uncertainty in the extension reading could be reduced by
using a longer wire or by using a Vernier scale to measure the initial extensions.
Page 215 Test yourself
13
i) stress =
a)
ii) strain =
b)
force
cross−sectional area
extension
original length
i) Stress – unit Pa
ii) Strain is a ratio of two lengths and so has no unit
14
𝐸=
15
a)
𝐹𝑙
𝐴 ∆𝑙
=
5.0 kg × 9.8 m s−2 × 2.20 m
𝜋 (0.28 × 10−3 m)2 × 2.2 × 10−3 m
= 2.0 × 1011 Pa
When a rubber band is loaded and then unloaded, the unloading curve ‘lags behind’ the loading
curve forming a loop. This is known as hysteresis.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Test yourself questions
b)
The area between the loading and unloading curves represents the energy transferred to the
band (increasing its internal energy) during each cycle.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Exam practice questions
Pages 218–221 Exam practice questions
1
a)
The toughest material has the largest area beneath the curve – the answer is C.
[Total 1 Mark]
b)
The strongest material has the greatest breaking stress – the answer is B.
[Total 1 Mark]
c)
A polymer stretches easily as the chain molecules untangle, and then stiffens (the gradient rises)
when they are aligned – the answer is D.
2
[Total 1 Mark]
d)
A brittle material has little or no plastic extension before it breaks – the answer is B. [Total 1 Mark]
a)
Hooke’s law states that, up to a certain limit, the extension is directly proportional to the load. [1]
b)
𝑘=
𝐹
𝑥
=
8.0 N
0.125 m
= 64 N m−1
[2]
[Total 3 Marks]
3
4
Ductile materials can be drawn into wires
[1]
Malleable metals can be readily hammered into thin sheets
[1]
Copper can be drawn into wires for electrical work and gold can be hammered into very thin leaves for
decoration.
[Total 2 Marks]
If the plastic kettle is dropped, or receives a sharp blow, it is more likely to crack or shatter than the steel
one. This is because the plastic is a brittle material. The steel is much tougher, and is capable of absorbing
much more energy, and may only suffer from minor indentations due to plastic deformation.
The surface of the steel is harder than the plastic and so is less likely to be scratched.
The plastic kettle will retain its heat better than the steel one. It consists of long-chain molecules called
polymers, which give it a low thermal conductivity. The plastic is also less dense than steel and so the kettle
is lighter.
(Six correct [3] 4/5 correct [2], 2/3 correct [1])
[Total 3 Marks]
5
a)
i) C
[1]
ii) O–A
[1]
iii) C–D
[1]
b)
Calculate the area under the line.
[1]
c)
i) The wire returns to its original length.
[1]
ii) The wire will be permanently extended.
[1]
d)
The wire regains its original stiffness [1] – it will follow an identical loading curve. [2]
[Total 9 Marks]
6
a)
i) Stress (σ) =
F
A
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Exam practice questions
l
ii) Strain (ε) =
[1]
l
iii) Young modulus E =
b)
σ=
= 1.8  10 Pa
8
−3
 (0.3  10 m)
−3
[1]

50 N
2.5  10
ε=

m
2
= 1.25  10
−3
[1]
2.00 m
1.8  10 Pa
8
E=
1.25  10
= 1.4  10 Pa
11
−3
[1]
[Total 5 Marks]
7
a)
Hysteresis occurs when the unloading curve ‘lags’ behind the loading curve.
[1]
b)
The blue (upper) line is the loading curve.
[1]
c)
The area enclosed by the loop represents the energy per unit volume (energy density) transferred
to internal energy of the rubber during each cycle.
d)
[1]
The band is initially slightly stiff until the weak cross-links between the tangled chains are broken
[2]. The chains are then uncoiled, giving a large increase in strain for a little extra stress [2] until
the molecules become aligned. The band now becomes stiff [1] as the strong covalent bonds
between the atoms are stretched. On releasing the stress, the chains recoil until the initial
amorphous state is regained. [1]
[Total 9 Marks]
8
a)
Tension 𝑇 =
𝑇Cu
𝑇St
⇒ 𝑇Cu =
=
𝐸Cu 𝐴Cu
𝐸St 𝐴St
𝐸Cu 𝐴Cu
𝐸St 𝐴St
𝐸 𝐴 ∆𝑙
𝑙
as Δ𝑙 and 𝑙 are the same for both wires
× 𝑇St =
1.2 × 1011 Pa × 𝜋 ×(0.28 × 10−3 m)2
2.0 × 1011 Pa × 𝜋 ×(0.23 × 10−3 m)2
× 𝑇St = 0.89 𝑇St
4.00 kg × 9.8 N kg−1 = TCu + TSt = 0.89 TSt + TSt = 1.89 TSt
TSt =
b)
4.00 kg × 9.8 N kg −1
[2]
[2]
= 20.8 N
[1]
TCu = 39.2 N – 20.8 N = 18.4 N
[1]
∆𝑙 =
1.89
𝐹𝑙
𝐸𝐴
=
18.4 N ×2.20 m
1.2 × 1011 Pa × π (0.28 × 10−3 m)2
= 1.4 × 10−3
[2]
[Total 8 Marks]
9)
a)
E = gradient of linear section =
= 14 GPa
b)
Max. stress = 150 MPa
80 GPa
0.56 × 10−2
[1]
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Exam practice questions
c)
v2= u2 + 2as = 0 + 2 × 9.8 m s−2 × 2.00 m → v = 6.3 m s−1
𝐹=
Δ 𝑚𝑣
Δ𝑡
=
[1]
70 kg × 6.3 m s −1
[1]
0.30 s
= 1500 N
d)
[1]
Compressive stress =
1460 N
[1]
2 × 𝜋(12.5 × 10−3 m)2
= 1.49 × 106 Pa
[1]
(Assuming that the impact is shared equally by both tibia.) This is much less than the maximum
compressive stress that the bone can withstand (150 MPa).
[1]
e)
[2]
[Total 11 Marks]
10
a)
Consider one half of the band: length = 0.75 m, extension = 0.40 m,
cross-sectional area = 0.15 m × 0.4 × 10−3 m = 6.0 × 10−5 m2,
tension in each half of the band = 100 N
Average E =
𝐹𝑙
𝐴 Δ𝑙
=
100 N × 0.75 m
[1]
6.0 × 10−5 m2 × 0.40 m
= 3.1 × 106 Pa ≈ 3 MPa
b)
[2]
Initially the band extends easily because the long chain molecules are untangling [1], but, as the
force increases the chains become aligned [1] and require greater forces to stretch the
intermolecular bonds. The band gets stiffer as the force increases, so its Young modulus gets
bigger [1].
c)
If Hooke’s law were obeyed, F = kx would apply over the whole range of extension [1]. Two
bands in parallel would have twice the stiffness of a single band [1], and would therefore extend
by half [1] if the same force were applied. So we would expect an extension of 20 cm in this
situation.
With two bands, the tension in each section will be half that in the single band, but, because of the
untangling of the molecules, the extension will be more than half of that of the single band with
twice the tension. [1]
d)
Work done in stretching = average force × extension [1]. Here, the average force is the same but
the extension is smaller. So less work is done. [1]
e)
Each time the band is stretched and released, hysteresis occurs. More work is done on the band on
stretching than is done by the band when it contracts [1]. The difference in work done is
transferred as internal energy in the band [1]. This increase in internal energy over many cycles
results in a rise in temperature in the band. [1]
[Total 15 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
13 Solid materials
Answers to Exam practice questions
11
a)
i) Change in length = length after adding extra force – length before
C4=B4–B3
(1.80 × 10−2 m = 1.50 × 10−1 m − 1.32 × 10−1 m)
[1]
ii) Work done = force × change in length
D6=A6×C6
[1]
(1.76 J = 16 N × 0.11 m)
iii) Total work done = work done before load added + work done after load added
E8=E7+D8
[1]
(2.72 J = 2.53 J + 0.192 J)
b)
The error is in column D. The work done should equal the average force times the extension [1],
but here the maximum force is used [1]. The model would have been better if an average was
found by adding the initial force to the final force and dividing by two.
e.g. D 6 = (A 5 + A 6)/2 × C 6 [1]
Note: this assumes that the force varies in a linear fashion for this extension, which is not always
true.
c)
Estimate the area under the curve by counting the number of squares [1]. Calculate the value of
work done for each square e.g. if 1 division on the force axis ≡ 2.0 N and 1 division on the
extension axis ≡ 1.0 × 10−2 m, one square ≡ 2.0 × 10−2 J [1]. A graph using these scale values will
cover an area of about 150 squares [1], representing 3.0 J of work done.[1]
d)
The area under the curve is the sum of the products F δx, where δx represents a tiny
increase in length for a given constant force (A = ∫ F dx). Drawing a smooth curve
gives a better representation of the work done than using the model [1]. The value will
be smaller than that in cell E 12 because the values calculated in column D will always give
a value that is too high. [1]
[Total 12 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
14 Nature of waves
Answers to Test yourself questions
Page 222 Test yourself on prior knowledge
1
2
a)
Amplitude is the maximum displacement from the mean position.
b)
Frequency is the number of cycles per second.
c)
Period is the time taken for one complete cycle.
a)
Sound waves; compression waves
b)
Water waves; waves in strings
c)
Radio waves; microwaves; infrared radiation; visible light; ultraviolet radiation; X-rays;
gamma rays
3
𝑓=
=
𝑐
𝜆
20 m s−1
0.80 m
= 25 Hz
4
Similarity: both travel through a vacuum at 3 × 108 m s−1; difference: microwaves have a longer wavelength
(lower frequency) than X-rays
Page 223 Activity 14.1
Two good sites for wave simulations are: www.animations.physics.unsw.edu.au and
http://phet.colorado.edu/en/simulations/category/physics.
Page 225 Test yourself
1
The particles in a longitudinal wave oscillate along the line of propagation of the waves, whereas those in a
transverse wave oscillate at right angles to the direction of propagation.
2
a)
i) A = 0.20 m
ii) λ = 2.0 m
b)
i) 𝑐 =
0.50 m
50×10−3 s
= 10 m s −1
𝑐
ii) 𝑓 =
=
𝜆
10 m s−1
2.0 m
= 5.0 Hz
iii) 𝑇 =
=
1
𝑓
1
5.0 Hz
= 0.20 s
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
14 Nature of waves
Answers to Test yourself questions
Page 227 Test yourself
3
4
a)
180 ° = π rad
b)
270° =
c)
57.3° = 1.0 rad
d)
120° =
a)
b)
c)
d)
5
a)
𝜋
6
𝜋
2
2
2𝜋
3
rad
rad
rad = 30°
rad = 90°
5𝜋
3
rad = 300°
3.5 rad = 201°
𝜋
2
rad
b)
2π rad
c)
π rad
d)
3𝜋
3𝜋
2
rad
Page 230 Test yourself
6
7
8
a)
X-rays
b)
visible light
c)
Microwaves
d)
radio waves
a)
Red light has a longer wavelength than blue light.
b)
Green light has a higher frequency than yellow light.
c = fλ;
909 × 103 Hz × 330 m = 3.0 × 108 m s−1;
693 × 103 Hz × 433 m = 3.0 × 108 m s−1
9
a)
𝑓=
=
𝑐
𝜆
3 × 108 m s−1
3 × 10−2 m
= 10 GHz
b)
𝑓=
=
𝑐
𝜆
3 × 108 m s−1
583 × 10−9 m
= 5.1 × 1014 Hz
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
14 Nature of waves
Answers to Test yourself questions
c)
𝑓=
=
𝑐
𝜆
3 × 108 m s−1
4.2 × 10−12 m
= 7.1 × 1019 Hz
10
a)
𝜆=
=
𝑐
𝑓
1540 m s−1
1.4 × 106 Hz
= 1.1 mm
b)
𝜆=
=
𝑐
𝑓
4080 m s−1
1.4 × 106 Hz
= 2.9 mm
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
14 Nature of waves
Answers to Exam practice questions
Page 231–233 Exam practice questions
1
Waves are in antiphase when they are half a cycle out of phase. One cycle represents a phase difference of
2π radians; the phases of oscillations in antiphase therefore differ by π radians – the answer is B.
[Total 1 Mark]
2
Sound waves are longitudinal waves that transfer energy from vibrating sources.
The speed of sound in air is about 340 m s−1 at 20 °C, but it varies with air temperature – the answer is A.
[Total 1 Mark]
3
X-rays are produced by electron bombardment, travel at the speed of light and can produce images on a
photographic plate. X-ray photons are much more energetic than those of visible light, and as E =
their wavelength is lower than that of light – the answer is C.
4
Using c = fλ; f =
c

3  10 m s
8
=

,
[Total 1 Mark]
−1
0.12 m
= 2.5 × 109 Hz = 2.5 GHz – the answer is C.
5
hc
[Total 1 Mark]
Amplitude is the maximum displacement from the mean position. Its SI unit is metre.
[1]
Frequency is the number of complete oscillations per second. Its SI unit is hertz.
[1]
Period is the time taken for one oscillation. Its SI unit is second.
[1]
[Total 3 Marks]
6
The particles in longitudinal waves oscillate along the line of the direction of propagation of the wave [1],
e.g. sound waves [1]. The particles in a transverse wave oscillate at right angles to the direction of
propagation of the wave [1], e.g. waves on a stretched wire. [1]
[Total 4 Marks]
7
Mechanical waves require particles to oscillate and so must have a medium for transmission [1].
Electromagnetic waves are associated with variations of electric and magnetic fields and they travel
through vacuum with a speed of 3 × 108 m s−1. [1]
[Total 2 Marks]
8
a)
T = 40 ms; f = 25 Hz
[1]
b)
Amplitude of the upper trace is 4.0 cm [1]; the lower trace 2.0 cm [1]
c)
The oscillations are of a cycle out of phase; a phase difference of
1
4

radians.
[1]
2
[Total 4 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
14 Nature of waves
Answers to Exam practice questions
9
Wavelength is the distance between two adjacent points that are in phase.
[1]
v = distance/time; for one oscillation distance = 1 wavelength, time = 1 period
[1]
v=

1
𝑇
but f = ; so v = fλ
T
[1]
[Total 3 Marks]
10
a)
b)
i) 0.75 m s−1
[1]
ii) 6.0 × 10−8 m
[1]
iii) 11 GHz
[1]
The waves in ii) are in the ultraviolet (UV) region.
[1]
[Total 4 Marks]
11
a)
b)
λ=
c 340 m s −1
=
= 1.80 m
f
189 Hz
[1]
λ=
c 340 m s −1
=
= 1.20 m
f
283 Hz
[1]
i) The waves will be in phase when both have completed a whole number of cycles
N1 × λ1 = N2 × λ2 i.e. first occurs when N1 = 2 and N2 = 3
[1]
2 × 1.80 m = 3 × 1.20 m = 3.60 m
[1]
ii) The waves will be in antiphase when one has completed one extra half cycle than the other.
[1]
N1 × λ1= (N2 + ½) × λ2 i.e. first occurs when N1 = 1 and N2 = 1
c)
1 × 1.80 m = 1½ × 1.20 m = 1.80 m
[1]
i) 4.5 m = 3.75 × 1.20 m = 2.50 × 1.80 m
[1]
So the shorter wave has completed 1.25 cycles more than the longer one. It will therefore be
5𝜋
2
𝜋
radians ahead ( radians out of phase)
[1]
2
ii) 9.9 m = 8.25 × 1.20 m = 5.50 × 1.80 m
[1]
So the shorter wave has completed 2.75 cycles more than the longer one and will be
radians ahead (
3𝜋
2
11 
2
radians out of phase)
[1]
[Total 10 Marks]
12
a)
P waves are longitudinal waves [1]; S waves are transverse [1]. The particles in the Earth’s crust
oscillate along the direction of propagation of the P waves whereas the particles oscillate
perpendicular to the direction of propagation of the S waves. [1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
14 Nature of waves
Answers to Exam practice questions
b)
Time =
distance
speed
P wave: t =
S wave: t =
Δt = x (
;
𝑥
[1]
7800 m s −1
𝑥
[1]
4200 m s −1
1
4200 m s −1
−
1
7800 m s −1
) = 88 s
[1]
[1]
x = 800 800 m = 801 km
c)
The readings from one station give the distance but not the direction of the epicentre [1]. If
the distances from two other stations are known and circles are drawn onto a map around each
station, each having a radius equal to the distance from the station, the point of intersection of
all three circles will give the location of the epicentre. (This is known as ‘triangulation’.) [1]
d)
Frequency of waves 𝑓 =
1
𝑇
=
1
20 × 60 s
= 8.33 × 10−4 Hz
[1]
Deep water:
c = √(g d) = √(9.8 m s−1 × 2000 m) [1] = 140 m s−1 [1]
𝜆=
𝑐
𝑓
=
140 m s −1
8.33 × 10−4 Hz
= 168 km [1]
Shallow water:
c = √(g d) = √(9.8 m s−1 × 20 m) = 14 m s−1 [1]
𝜆=
e)
𝑐
𝑓
=
14 m s −1
8.33 × 10−4 Hz
= 17 km [1]
Energy of the particles within a single cycle of the wave was initially spread across a distance of
140 km but as the wave approaches land, this energy is concentrated along 14 km of the water
surface [1]. The energy of a wave depends on the amplitude of the oscillations so the shortened
wave will increase in amplitude [1]. This means that the wave that had the small amplitude of
about a metre in the deep water can have an amplitude of 10–20 metres as it approaches land
[1].
[Total 18 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Test yourself questions
Page 234 Test yourself on prior knowledge
1
The particles in a longitudinal wave oscillate along the direction of propagation of the wave; the particles in
a transverse wave oscillate perpendicularly to the direction of propagation.
2
Sound waves are transmitted by the transfer of energy from one particle to the next. There are no particles
in a vacuum so no sound can be transmitted.
3
Light travels at 3.0 × 108 m s−1 and so is seen instantly, but sound travels much more slowly in air (340 m s−1)
and can take several seconds to travel from the source of the lightning to the listener.
4
Ultrasound is sound that has a frequency greater than the threshold for human hearing (20 kHz).
5
𝜆=
=
𝑐
𝑓
340 m s−1
256 s−1
= 1.33 m
6
𝑓=
=
𝑐
𝜆
3.0 × 108 m s−1
360 × 109 s−1
= 8.3 × 1014 Hz
7
As the frequency remains constant, and c = fλ, a reduction in c will lead to a reduction in λ.
Page 235 Activity 15.1
1
Two springs in parallel give a spring constant that is twice that for a single spring.
2
a)
Speed of pulse = 1.00 m s−1 for 1 trolley and 1 spring
= 1.41 m s−1 for 1 trolley and 2 springs
= 0.69 m s−1 for 2 trolleys and 1 spring
b)
The spring constant is doubled and the speed increases from 1.00 m s−1 to 1.41 m s−1, i.e. by a
factor of √2 . This suggests that the speed is proportional to the square root of the spring
constant.
c)
The uncertainties are: 1.2 ± 0.1 s = ±8%; 0.85 ± 0.05 s = ±6%; 1.73 ± 0.07 s = ±4%. The first set of
times has the greatest uncertainty.
Page 236 Core practical 6
1
Time period =
1
800 s−1
= 1.25 ms;
10 divisions = 4 × 1.25 ms;
1 division = 0.5 ms = 500 μs
2
3
1
𝑓
/ms 2.50
2.00
1.67
1.40 1.25
The graph is a straight line through the origin.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Test yourself questions
4
Gradient = speed of sound in air = 340 m s−1
5
At higher frequencies the wavelength is shorter. Two or more wavelengths will mean that a longer distance
will be measured, and the percentage uncertainty will be less. The time base may have to be adjusted so
that several waves can be seen, and the voltage sensitivity will need to be increased as the intensity of the
sound will decrease as the microphone moves away from the speaker.
Page 237 Activity 15.2
Time taken = 2.6 × 0.10 ms = 0.26 ms
Speed =
=
21
𝑡
2×0.78 m
2.6 × 10−4 s
= 6000 m s −1
A graph of c2 against d will give a straight line of gradient g
1
2
d/cm
2.0
4.0
6.0
8.0
10.0
t/s
3.7
2.6
2.1
1.8
1.7
c/m s-1
0.44
0.63
0.78
0.91
0.96
c2/m2 s-2
0.20
0.40
0.61
0.83
0.93
The graph is a straight line through the origin of gradient 10.1 m s−2.
Page 240 Test yourself
1
𝑇
𝑐=√
𝜇
=√
kg m s−2
kg m−1
= √m2 s 2
= m s −1
𝑐 = √𝑔 𝑑
= √(m s −2 )(m)
= m s −1
2
The speed of a wave through a solid is proportional to the square root of modulus of elasticity (or the
‘spring constant’ of the bonds between the atoms) and inversely proportional to the square root of the
mass of the particles. Steel has a higher elastic force between the atoms and the mass of each atom is
smaller than lead. Either of these indicate that the speed of sound in steel will be greater than that in lead.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Test yourself questions
3
Page 244 Activity 15.4
sin 48°
1
𝑛𝑔 =
2
For example:
sin 29°
1.53 =
= 1.53
sin 31°
sin 𝑟°
𝑛1 = 𝑛𝑔 sin 61° = 1.34
⇒ 𝑟 = 20° ⇒ 𝐶 = 70°
𝑛1 = 1.53 sin 61° = 1.44
Page 244 Test yourself
4
Snell’s law states that the refractive index for a wave travelling from one medium to another is given by the
expression: 1𝑛2 =
5
a)
𝑛=
iii) 𝑐 =
𝑛2
=
=
𝑣1
𝑣2
speed of light in a vacuum
ii) 𝑐 =
𝑛1
sin 𝜃2
speed of light in the medium
i) 𝑐 =
b)
sin 𝜃1
3.0×108 m s−1
1.37
3.0×108 m s−1
1.46
3.0×108 m s−1
2.39
= 2.2 × 108 m s −1
= 2.1 × 108 m s −1
= 1.3 × 108 m s −1
sin 𝜃1
sin 𝜃2
i) sin 𝜃2 =
=
ii) sin 𝜃2 =
=
𝑛1
𝑛2
sin 𝜃1
1.00
1.46
𝑛1
𝑛2
sin 30° → 𝜃2 = 20°
sin 𝜃1
1.37
1.46
sin 30° → 𝜃2 = 28°
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Test yourself questions
𝑛1
iii) sin 𝜃2 =
𝑛2
sin 𝜃1
2.39
=
1.37
sin 30° → 𝜃2 = 61°
6
The critical angle is the angle of incidence above which total internal reflection occurs.
7
a)
sin𝐶 =
=
b)
sin𝐶 =
=
c)
sin𝐶 =
=
𝜇2
𝜇1
1.00
2.39
→ 𝐶 = 25°
𝜇2
𝜇1
1.37
1.46
→ 𝐶 = 70°
𝜇2
𝜇1
1.37
2.39
→ 𝐶 = 35°
Page 248 Activity 15.5
1
The graph is a straight line of negative gradient.
2
The intercepts = =
1
1
1
𝑢
𝑣
𝑓
.
From the graph, the intercepts are at 5.0 m−1 so f = 0.20 m (20 cm).
3
Focus the image by moving the screen away from the lens and then by bringing it towards the lens. Use the
average value of the two distances for v.
Page 249 Test yourself
8
Focus light from a distant object onto a sheet of paper. As the rays from the object will be approximately
parallel, the image will be at the focal point. Use the rule to measure the distance between the lens and the
paper.
9
A real image is formed by the actual intersection of rays of light. Virtual images are where light entering our
eyes appears to come from. Real images can be projected onto a screen; virtual images are on the same
side of the lens as the object and so cannot be formed on a screen.
10
a)
i) 𝑃 =
=
1
𝑓
1
+0.20 m
= +5.0 dioptres
ii) 𝑃 =
=
1
𝑓
1
−0.50 m
= −2.0 dioptres
b)
i) Power of combination = + 5.0 dioptres – 2.0 dioptres = + 3.0 dioptres
ii) The combination is a converging lens of focal length 0.33 m (33 cm).
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Test yourself questions
11
12
a)
b)
1
𝑣
1
𝑣
=
=
1
𝑓
1
𝑓
=
=
1
𝑢
1
𝑢
=
=
1
15 cm
=
1
15 cm
=
1
10 cm
1
20 cm
=
=
2−3
30 cm
4−3
60 cm
=−
1
30 cm
=−
1
60 cm
⇒ 𝑣 = −30 cm ; negative, so virtual
⇒ 𝑣 = +60 cm ; positive, so real
Page 251 Activity 15.6
The graph is a straight line through the origin. The concentration for a rotation of 33 ° is 0.2 g ml−1.
Page 252 Test yourself
13
Plane polarised waves are transverse waves in which the oscillations occur in a single plane. For light the
variations of the electric (or magnetic) field are all in the same plane if it is polarised.
14
Sound waves cannot be polarised because they are longitudinal and so the particles have no transverse
oscillations.
15
An optically active medium rotates the plane of polarisation when polarised light passes through it.
16
Examples of uses of polarisation include: reduction of glare, LCDs, stress analysis and measuring the
concentration of sugar solutions.
Page 255 Test yourself
17
Pulse echo methods are used by bats and dolphins, for ultrasound imaging, radar and sonar.
18
Distance from cliff = ½ speed × time = ½ × 340 m s−1 × 1.2 s = 204 m
19
Ultrasound is sound that has a frequency greater than the threshold of human hearing (about 20 kHz).
20
Advantage of ultrasound: unlike X-rays it is not an ionising radiation and does not damage living cells.
Disadvantage: the images are low resolution and so less detail is seen compared to X-ray images.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Exam practice questions
Page 257–260 Exam practice questions
1
Blue light has a shorter wavelength and higher frequency than red light. Blue light is slowed down more
than red light (it has a higher refractive index) and so it is deviated more towards the normal when it enters
the glass – the answer is D.
2
[Total 1 Mark]
Passing light through a narrow slit may produce diffraction but the amplitude of oscillations within photons
is so small that vibrations in all planes will be transmitted. Reflection, scattering and Polaroid films can all
create polarised light – the answer is A.
3
[Total 1 Mark]
Ultrasound is non-ionising radiation; it is much less likely to kill or mutate cells than X-rays – the answer is
C.
4
sin θ =
[Total 1 Mark]
1.52
[1] ⇒ θ = 72 ° [1]
1.60
[Total 2 Marks]
5
Advantage: better resolution [1]; disadvantage: shorter range [1]
[Total 2 Marks]
6
The image needs to be real for the CCD to detect it; it must be smaller to fit inside the camera. The ray
diagram for an object greater than 2f from a converging lens always gives an inverted image – the answer is
B.
7
[1]
−1
Sound travels more slowly than light [1]. It takes the sound 0.3 s to travel 100 m so the speed is 300 m s
[1] (estimate to one significant figure).
[Total 2 Marks]
8
a)
T = 4 × 200 ns = 800 ns; f = 1/T = 1250 Hz
[1]
Y1: Vp = 1.0 V; Y2: Vp = 50 mV
[1]
b)
Sketch with the peaks of trace 2 directly below the troughs of trace 1.
[2]
c)
Wavelength = 49.0 cm – 22.0 cm = 27.0 cm
[1]
Speed of sound, v = fλ = 1250 Hz × 0.270 m = 338 m s−1
[1]
d)
The intensity of the sound received by the microphone will be reduced when it is moved away
from the speaker, so the Y2 voltage sensitivity should be adjusted to give a larger trace [1]. The Y
position controls should be adjusted so that the trough of the upper trace just touches the peak
of the lower trace [1].
[Total 8 Marks]
9
a)
A wavefront is a line, or surface, along which the oscillating particles of the wave are in phase [1].
Wavelength is the distance between adjacent wavefronts. [1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Exam practice questions
b)
Light travels faster in water than it travels in glass, so in your diagram you need to show that the
wavelength will increase [1]. The direction of the wave moves away from the normal. [1]
[Total 4 Marks]
10
a)
b)
11
1.55 =
sin 40
sin r
[1]
⇒ r = 24.5 °
[1]
1.55 sin 24.5 ° = n sin 28 °
[1]
⇒ n = 1.37
[1]
Depth of the shoal = ½ speed × time = ½ × 1500 m s−1 × 1.6 s [1] = 1200 m [1]
[Total 2 Marks]
12
a)
b)
c)
1
1
1 1 1
−
 v = +100 cm
= − =
+
20
cm
25
cm
v f u
[1]
Linear magnification = 5.0; the image is real
[1]
1
1
1 1 1
−
 v = −67 cm
= − =
+40 cm 25 cm
v f u
[1]
Linear magnification = 2.7; the image is virtual
[1]
1
1
1 1 1
−
 v = −17 cm
= − =
−50 cm 25 cm
v f u
[1]
Linear magnification = 0.7; the image is virtual
[1]
[Total 6 Marks]
13
a)
Power at the far point P =
1 1 1 1
1
= + = +
= +40 dioptres
f u v  0.025 m
Power at the near point P =
b)
1 1 1
1
1
= + =
+
= +44 dioptres
f u v 0.25 m 0.025 m
[1]
[1]
Maximum power of the lens for a near point of 1.25 m
P=
1 1 1
1
1
= + =
+
= +40.8 dioptres
f u v 1.25 m 0.025 m
[1]
To focus on an object at 25 cm requires a total power of 44 dioptres.
Total power = power of eye lens + power of spectacle lens
[1]
44 dioptres = 40.8 dioptres + power of spectacle lens  power of lens = +3.2 dioptres
[1]
(This is a converging lens of focal length 31 cm.)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Exam practice questions
c)
To focus on a distant object the focal length of the combined lens must be 27 mm. The minimum
power must therefore be
1
= +37 dioptres
0.027 m
[1]
The eye lens has a minimum power of +40 dioptres so the required spectacle lens must have a
power of –3 dioptres. (This is a diverging lens of focal length 33 cm). [1]
[Total 7 Marks]
14
a)
Diagram showing illuminated object (e.g. cross-wire), lens in lens-holder, screen with scale (e.g.
graph paper) and metre rule (diagram [1] all labels correct [1])
Measurements: size of object (e.g. with clear rule), size of image (e.g. graph paper or translucent
scale) [1] and image distance (metre rule) [1]
Plot graph of m (=
size of image
) against v
size of object
Gradient of graph =
b)
1
[1]
f
[1]
[max. 5 marks]
i) Light is reversible. So if the object and screen were changed around, the lenses would occupy
the same two positions.
[1]
ii) In position 1, v > u so the image is magnified; in position 2, v < u so the image is diminished.
[1]
c)
The magnification method uses the gradient of a graph to find f, which is more reliable than the
use of an intercept.
[1]
The two-position method does not require the exact position of the centre of the lens to be
known, as long as l is measured between the same positions (on the lens-holder), therefore the
method can be used for a lens fitted into a tube.
[1]
[Total 9 Marks]
15
a)
When the plane of polarisation of the wave is at an angle θ to the analyser, the component of
the amplitude transmitted will be A0 cos θ [1], where A0 is the amplitude of the incident light.
The intensity of the wave received by the light meter is proportional to the square of the
amplitude, so:
I
A2 = (A0 cos θ)2 = A02 cos2 θ → I = I0 cos2 θ
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Exam practice questions
b)
In a darkened room, set up a polarimeter (see Figure 15.25) or use two polarising filters and a
360 ° protractor so that the intensity of light passing from the source through both filters can be
measured using a light meter [1]. Arrange the filters so that their planes are aligned; that is when
the intensity value is a maximum. [1]
Rotate the analyser by 10 ° and record the intensity of the transmitted light [1]. Repeat the
readings at 10 ° intervals until the planes of the filters are at right angles.[1]
Plot a graph of I against cos2 θ. [1] This should give a straight line through the origin if the
relationship is correct. [1]
c)
[max. 5 marks]
I0 can be found using the gradient of the graph, or the value of I when θ = 0 ° (cos θ = 1). [1]
When the planes of the filters are aligned, not all of the incident light is allowed through, so the
value of the intensity of the polarised light will be less than when the analyser is removed. [1]
[Total 9 Marks]
16
a)
Polarisation occurs when the planes of vibration (or field variation) of a transverse wave all lie in
one plane [1]. Diagram e.g. Figure 15.21 or 15.23
b)
[1]
Light reflected from the surface of the water is (partially) polarised in the parallel to the plane of
the surface [1]. The lenses of the spectacles have the plane of transmission that is perpendicular
to the polarised light [1] so very little/none of the reflected light passes through them. [1]
c)
i) Refractive index of water =
sin i sin 53o
=
= 1.33  r = 37 °
sin r
sin r
[2]
ii)
The reflected ray is perpendicular to the refracted ray shown on diagram [1]. When the light is
refracted, no energy can be transmitted at right angles to the direction of propagation [1] so
none of the light with planes in the direction of the reflected light will be transmitted. [1] The
reflected light is therefore polarised in the plane parallel to the surface [1]. (All marks can be
gained from labelled diagram.)
[Total 11 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Exam practice questions
17
a)
A = r + r = 2r
[1]
Total deviation = deviation at first surface + deviation at second surface
D = (I – r) + (I – r) = 2I − 2r = 2I – A
Hence i =
A+D
2
sin i
=
=
sin r
b)
r=
[1]
A
2
[1]
A+D
2
A
sin
2
sin
[1]
D = (n − 1)A = (1.518 – 1) × 7.0 ° [1] = 3.6 °
[1]
c)
Diagram showing red and blue rays. [1]
Higher frequency light travels more slowly through glass than lower frequency light [1]. Hence
the refractive index for violet light is greater than that for red light and so it is deviated more by
the prism. [1]
d)
Dispersion = deviation of violet light – deviation of red light = (1.523 – 1) × 7.0 ° − (1.513 – 1) ×
7.0 °
[1]
= 0.14 °
[1]
e)
Diagram showing rays dispersing [1] and recombining [1].
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
15 Transmission and reflection of waves
Answers to Exam practice questions
f)
For an achromatic doublet, the dispersion of the second prism must be equal and opposite to
that of the first:
(1.667 – 1) × A – (1.639 – 1) × A = 0.14 ° [1] → A = 5.0 ° [1]
g)
Total deviation for yellow light = deviation of first prism − deviation of second prism
= (1.518 – 1) × 7.0 ° − (1.654 – 1) × 5.0° [1] = 0.36 ° [1]
[Total 17 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Test yourself questions
Page 261 Test yourself on prior knowledge
1
a)
i) 2π rad = 360°
5𝜋
ii)
2
rad = 450°
iii) 1.2π rad = 216°
b)
i) 90° =
𝜋
2
𝑟𝑎𝑑
ii) 57.3° = 1.0 𝑟𝑎𝑑
iii) 150° =
2
3
5𝜋
6
rad
a)
A = (+0.20 mm) + (+3.00 mm) = +5.0 mm
b)
A = (+2.0 mm) + (−3.0 mm) = −1.0 mm
a)
A = 2E
b)
A = E cos θ + E = E(1 + cos θ)
4
A difference of one wavelength if equivalent to a phase difference of 2π radians
a)
b)
c)
𝜆
2
= π rad
3𝜆
4
=
3𝜋
2
rad
2λ ≡ 4π rad
Page 262 Activity 16.1
1
The path difference = 10.23 m – 9.10 m = 1.13 m. At the first position of constructive interference the path
difference must equal one wavelength. Therefore the wavelength must be 1.13 m.
2
v = fλ = 300 s−1 × 1.13 m = 339 m s−1
Page 267 Test yourself
1
Superposition occurs when two similar waves meet at a point. The resultant amplitude equals the vector
sum of the amplitudes of the waves. See diagram below.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Test yourself questions
2
Coherent sources are of the same type, have the same frequency and have a constant phase relationship.
3
For constructive interference to take place the path difference from the two coherent sources must be zero
or an integral number of wavelengths.
i.e. p.d. = nλ, where n = 0 or an integer.
For destructive interference p.d. = (n + ½)λ
4
a)
Path difference = 6.20 m – 4.20 m = 2.00 m
For the third maximum the path difference = 3λ; λ = 0.67 m
b)
c = fλ
= 510 Hz × 0.67 m
= 340 m s−1
𝜆=
5
⇒𝑆=
=
𝑆𝑥
𝐷
𝜆𝐷
𝑥
640 × 10−9 m × 5.40 m
2.9×10−3 m
= 1.2 mm
Page 268 Activity 16.2
1
a)
λ = 3.2 cm
b)
𝑓= =
𝑐
3.0×108 m s−1
𝜆
0.032 m
= 9.4 GHz
2
f/Hz
400
600
800
1000
1200
2λ/m
1.70
1.14
0.86
0.68
0.56
λ/m
0.85
0.57
0.43
0.34
0.27
λ−1/m−1
1.18
1.75
2.33
2.94
3.70
The graph of f/Hz against λ−1/m−1 is a straight line through the origin of gradient 330 m s−1.
Page 271 Core practical 7
1
Measure the length of each piece of wire using a metre rule and its mass on a digital balance. Calculate the
mass per unit length of the wires. Clamp one end of one wire to the bench and feed the other over a pulley
wheel and hold taut with a mass of 5.0 kg fixed to the end. Place two bridges under the wire with the poles
of a magnet on each side of its mid-point. Connect the ends of the wire to a signal generator. Adjust the
frequency of the signal generator until the wire vibrates with a large amplitude. Measure l and f. Repeat for
the other wires keeping the length and tension the same. A graph of frequency squared against mass per
unit length should give a straight line through the origin, showing that f is proportional to 1/õ.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Test yourself questions
2
a)
There is a systematic error either in the measurement of the length or on the scale of the signal
generator.
b)
1
𝑇
1
5.0 kg × 9.8 m s−2
Gradient = √
2 𝜇
= √
2
𝜇
−1
= 60 m s ;
μ = 3.40 × 10−3 kg m−1 ≈ 3.5 g m−1
Page 274 Test yourself
6
Standing waves store energy, whereas travelling waves transfer energy from one point to another.
The amplitude of standing waves varies from zero at the nodes to a maximum at the antinodes, but the
amplitude of all the oscillations along a progressive wave is constant.
The oscillations are all in phase between nodes, but the phase varies continuously along a travelling wave.
7
a)
The distance between successive nodes is half a wavelength, so five gaps represent 2.5λ.
2.5λ = 80 mm → λ = 32 mm
b)
𝑓=
=
𝑐
𝜆
3.0 × 108 m s−1
32 × 10−3 m
= 9.4 GHz
𝑓=
8
9
10
1
2𝑙
𝑇
√𝜇
𝑓2
𝑇
1
𝑇1
2
= √ 2 ⇒ 𝑓2 = √ × 480 Hz = 340 Hz
a)
When l and µ are constant,
b)
When T and µ are constant,
c)
When µ is constant,
a)
For the fundamental frequency, 𝐼 =
b)
Frequency of the third harmonic = 3 × f0; and so λ =
a)
Length of air column at the first position of resonance =
𝜆
4
𝑓2
𝑓1
=
𝑙1
𝑙2
𝑓1
𝑓2
=
𝑓1
𝑇
𝑙1
𝑙2
2
⇒ 𝑓2 = × 480 Hz = 960 Hz
1
1
2
2
√𝑇 = 2 √1 × 480 Hz = 340 Hz
1
𝜆
2
so λ = 2l = 1.50 m
𝜆0
3
= 0.50 m
(antinode at top, node at water surface).
λ = 4 × 8.2 cm = 32.8 cm
b)
Next resonant length =
3
4
3
λ (ANAN) → × 32.8 cm = 24.6 cm
4
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Test yourself questions
Page 278 Core practical 8
Experiment 1
1
When the slit is narrowed the central maximum becomes wider and less intense. The other maxima shift
away from the centre. The reverse happens when the slit widens until it becomes so wide that no
diffraction is observed.
2
sin 𝜃 ≈ tan 𝜃 =
0.022 m
3.40 m
= 0.0065 ; 𝑎 =
𝜆
sin 𝜃
=
633 × 10−9 m
0.0065
= 9.8 × 10−5 m ≈ 0.10 mm
Experiment 2
1
The central maximum is white (all wavelengths). Continuous spectra occur on each side of the central
maximum, with violet end nearest to the centre.
2
tan 𝜗
0.255 m
2.000 m
⇒ 𝜃 = 7.3° ⇒ sin 𝜃 = 0.1265
𝑛𝜆 = 𝑑 sin 𝜃 =
1
0.1265
sin 𝜃 ; 𝑓𝑜𝑟 𝑛 = 1; 𝑛 =
= 2.0 × 105 m−1
𝑁
633 × 10−9 m
= 2000 lines per cm
3
The measurement of larger angles means that the percentage uncertainty is smaller leading to a more
accurate result. The measured angle can be made even bigger by measuring 2θ, i.e. the angle between the
spectra on both sides of the central maximum.
Page 280 Activity 16.4
The resolving power of the eye varies from person to person and in different light conditions as the size of the
pupil varies. It is generally accepted that a person with normal (20/20) vision can resolve to an angle of about 1
minute of arc, i.e. about 3 × 104 radians. This would enable a viewer to distinguish the lines 1 mm apart at a
distance of about 3.5 metres.
Page 281 Test yourself
11
Diffraction is the spreading of a wavefront when it is obstructed by an object or an aperture.
12
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Test yourself questions
13
The microwaves of the phone network have a much shorter wavelength (of the order of a few cm)
compared with that of the radio signal (of the order of 300 m for AM transmission) and so undergo much
less diffraction. The longer radio waves can spread over the edges of the valley.
14
The resolution depends on the angular separation of the central maximum and first minimum of the
diffraction pattern. For a circular aperture this is proportional to the wavelength and inversely proportional
to the diameter of the aperture. The smaller this angle is, the better the resolution. Hence
a)
For light, the telescope with the larger diameter will give the better resolution.
b) For radio telescopes, although the diameter is 100 times bigger than the optical telescope, the
wavelength of radio waves is of the order of 10 7 times longer than light waves. (Greater resolution can
be obtained using an array of radio telescopes spread over thousands of kilometres.)
15
In order to obtain diffraction patterns, the wavelength of the radiation must be of the same order of
magnitude as that of the spacings in the crystal lattice. The wavelength of X-rays (10−12–10−11 m) is similar to
most atomic separations.
16
16𝑛λ = 𝑑 sin 𝜃 3 × 599 × 10−9 m =
1
2.5×105 m
sin 𝜃
θ = 27 °
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Exam practice questions
Page 283–288 Exam practice questions
1
Although a constant phase relationship is needed for coherence, it is not essential that the sources are
always in phase; it is essential that they have the same frequency – the answer is B.
2
To form a minimum on an interference pattern the waves must be in antiphase at that point. The path
difference must therefore be half a wavelength – the answer is B.
3
[Total 1 Mark ]
[Total 1 Mark]
A guitar string vibrating in the fundamental mode has a node at each end with a single antinode in
between. The length is equal to half a wavelength, so wavelength = 1.24 m – the answer is D. [Total 1Mark]
4
The second harmonic has a frequency twice that of the fundamental – the answer is D.
5
A stationary transverse wave is set up in the string. This causes the air molecules to vibrate generating a
progressive, longitudinal sound wave – the answer is D.
6
[Total 1 Mark]

. The only situation that will cause it to widen will be an
a
The angular width of the central maximum =
increase in wavelength – the answer is D.
8
[Total 1 Mark]
At the central point the path difference from the speakers is zero, so if there is destructive interference the
speakers must be in antiphase – the answer is D.
7
[Total 1 Mark]
[Total 1 Mark]
The engine noise is picked up by a microphone, electronically processed [1] and delivered to the pilot’s
headphones exactly out of phase with the noise [1] so that destructive superposition [1] occurs. Other
signals to the headphones are not processed.
9
a)
[Total 3 Marks]
The reflected ray travels back and forth between the surfaces. The path difference between this
ray and the one passing straight through is therefore 2 × 450 nm = 900 nm. [1] The path difference is equal
to three complete wavelengths of the blue light in the solution, [1] so the two waves emerge in phase and
constructive superposition occurs for blue light. [1]
b)
Gravitational force pulls the liquid down making the film thicker at the bottom.
[1]
c)
When the path difference is a whole number of wavelengths, constructive interference occurs;
when the path lengths differ by an odd number of half wavelengths, destructive interference
takes place [1]. So, as the film gradually gets thicker alternate lines of constructive and
destructive interference appear as a series of horizontal stripes [1].
The minimum thickness for destructive interference to occur is when the path difference is half a
wavelength [1]. The path difference is twice the thickness of the film, so the minimum thickness
will be a quarter of a wavelength = 125 nm. [1]
10
a)
i)
f =
ii)  =
b)
i)
=
c

=
3.0  108 m s −1
= 4.69  1014 Hz
640  10−9 m
[Total 3 Marks]
[1]
cv
c
3.0  108 m s −1
 cm = v =
= 1.97  108 m s −1
cm

1.52
[2]
c
1.97  108 m s −1
=
= 420 nm
f
4.69  1014 Hz
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Exam practice questions
ii) The path difference between the wave reflected from the top edge of the bump and that
from the bottom edge is: 2 × 105 nm = 210 nm [1] =
𝜆
2
The two reflections are out of phase and so destructive interference takes place. [1]
[Total 6 Marks]
11
a)
b)
i) See Figure 16.19
[2]
ii) Fundamental: λ = 2l
[1]
First overtone: λ = l
[1]
1 −1
/m 2.06; 2.30; 2.60; 2.90; 3.08; 4.08; 4.65 [1]
l
i)
Graph: axes correctly labelled [1], appropriate scales [1], line of best fit through
points [1]
𝜆
ii) For an open-ended pipe l = in the fundamental mode; hence λ = 2l.
[1]
2
f=
𝑣
𝜆
=
𝑣
[1]
2𝑙
iii) Gradient = 170 m s−1 =
𝑣
[1]
2
v = 340 m s−1
[1]
The speed of sound in air increases as the temperature rises [1]. As the wavelength is fixed
the frequency is proportional to the speed, and will therefore rise and fall with the
temperature [1].
[Total14 Marks]
12
a)
Hot spots occur when the path difference between the direct and reflected waves equals a
whole number of wavelengths [1]. There will be a more intense ‘antinode’ at these points and
the food will be cooked more quickly than at the ‘nodes’ [1] where the direct and reflected
beams arrive out of phase and create a cold spot.
b)
The distance between adjacent nodes or antinodes in a standing wave is equal to half a
wavelength. As the hot spots are 6 cm apart, the wavelength of the microwaves must be about
12 cm.
𝑓=
c)
𝑐
𝜆
[1]
=
3 × 108 m s −1
0.12 m
= 2.5 × 109 Hz
[1]
The length of the reflected wave = 450 mm; the length of the direct wave = 270 mm
Path difference = 450 mm (use Pythagoras) – 270 mm = 180 mm. [1]
This is 1.5 wavelengths, but there is a phase change equivalent to half a wavelength at the
reflecting surface. The effective path difference becomes one wavelength [1] so the beams are in
phase at X so producing a ‘hot spot’. [1]
[Total 7 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Exam practice questions
13
a)
The aim of the experiments is to check the validity of the equation:𝑓 =
1
2𝑙
𝑇
√𝜇 .
The same wire is to be used so µ is constant. We need to show that when the tension is constant,
f is inversely proportional to l, and when the length is constant, f is proportional to√𝑇.
i) The equipment needed is a sonometer (or a length of wire, two ‘bridges’, a clamp and pulley),
weights plus hanger, metre rule and a microphone connected to an oscilloscope (or a digital
waveform device with sensor and computer).
[1]
ii) Figure 16.14 drawn.
[1]
iii) With a fixed mass on the hanger adjust the bridges so that they are 20 cm apart. The length
can be measured using a metre rule with a precision of 1 mm [1]. The rule should be placed
on the knife edges of the bridges and read from directly above to reduce parallax errors [1].
Pluck the wire at the central point and observe the waveform produced on the CRO or the
computer. Measure the period of the wave using the time base/scale [1] and calculate the
frequency using 𝑓 =
1
𝑡
[1]. Repeat this three times and find the average value of f [1]. Repeat
the experiment for a further five values of l up to the maximum length of wire available. [1]
iv) To ensure that the experiment is safely performed, safety goggles must be worn at all times,
the bench around the equipment uncluttered and a tray or mat placed beneath the load. [1]
For the second experiment, the length is fixed and the load is varied by adding weights to the
hanger to give a suitable range of tensions and the corresponding frequencies are found.
1
v) To verify the equation a graph of f against is plotted for the first experiment, and one of f2
𝑙
against T (or f against √𝑇 ) for the second experiment [1]. Both graphs should be straight lines
through the origin. [1]
b)
[max. 8 marks)
To determine the value of the mass per unit length of the wire, the gradient of either graph can
be found.
[1]
1
𝑇
For experiment 1 the equation can be written 𝑓 = √ ×
2 𝜇
so the gradient =
1
2
1
𝑙
𝑇
[1]
√𝜇
OR
For experiment 2 the equation can be written 𝑓 2 =
so the gradient =
1
4 𝑙2 𝜇
× 𝑇
1
[1]
4 𝑙2 𝜇
[Total 10 Marks]
14
a)
i) Narrower slit gives more diffraction and so the central maximum should be wider on your
sketch. [1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Exam practice questions
ii) and iii) Red light has a longer wavelength than blue [1] so the central maximum will be
wider for the red light sketch. [1]
b)
The peak of the central maximum of the blue pattern falls within the first minimum [1] and so
the two images can be resolved [1]. The central maximum of the red light lies outside the first
minimum and so only one image is detected [1].
[Total 6 Marks]
15
a)
The central maximum is white [1]. A continuous spectrum is seen on each side of the central
maximum [1] with the red end closest to the centre [1].
b)
The distances between the grating end of the screen [1] and between the central maximum and
the extreme positions of the spectrum (or, more accurately, between the ends of both first order
spectra) using a metre rule [1]. The angles are calculated using
tan𝜃 =
c)
distance from central max to first max
distance from the grating to the screen
nλ = d sin θ =
1
𝑁
[1]
sin θ where n = number of lines per metre
i) For first order red light, n = 1, λ = 700 nm, sin θ = 500 × 103 m−1 × 700 ×10−9 m, θ = 20 °
[1]
ii) For second order red light, sin θ = 2 × 500 × 103 m−1 × 700 × 10−9 m, θ = 44 °
[1]
iii) For third order violet light, sin θ = 3 × 500 × 10 m × 400 × 10 m, θ = 37 °
[1]
−1
3
d)
[1]
For first order violet light, sin θ = 500 × 10 m × 400 × 10 m, θ = 12 °
3
−9
−1
−9
The second and third order spectra have larger angles, so the measurements would have a
smaller percentage uncertainty [1]. In this case the second and third order spectra will overlap
(the angle of second order red maximum, 44 °, is bigger than that for the third order violet, 37 °,
so the second and third order are not complete.
[1]
[Total 12 Marks]
16
a)
Diffraction occurs at both the single slit and the double slit. The light from the single slit spreads
out so that light from the same wavefront [1] is incident on the double slit. This provides a pair of
coherent sources which are diffracted from each of the double slits [1]. When the wavefronts
from the double slits overlap, interference patterns are produced. [1]
𝑆𝑥
= 5.9 × 10−7 m
𝜆=
c)
The speed of light in glass is less than that in air, so the wavelength will be shorter [1]. This
means that there will be more waves in the glass strip than in an equal thickness of air [1]. This
effectively reduces the path difference needed to produce constructive or destructive
interference [1] and the pattern will shift to the side covered by the glass. (Another way of
explaining this is to consider the position of the ‘central’ maximum. Without the glass the
distance from each slit to the maximum is the same and therefore each path has the same
number of wavelengths. When the glass is introduced it effectively shortens the path length for
this number of waves and so the position where both are in phase will be moved.)
See Figure 16.25.
[2]
d)
𝐷
=
0.80 × 10−3 m × 3.7 × 10−4 m
b)
0.50 m
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
16 Superposition of waves
Answers to Exam practice questions
e)
The angular separation of the interference fringes from the central maximum,𝜃 =
n𝜆
𝑆
the angular separation of the first minimum of the diffraction pattern is given by 𝜃 =
[1], and
𝜆
𝑎
[1].
When the fourth fringe of the interference pattern coincides with the first minimum of the
diffraction pattern it will disappear. From the above relationships we can see that this occurs
when 𝑎 =
f)
𝑆
𝑛
=
0.80 mm
4
= 0.20 mm [1]
The central maximum will appear white because the path difference is zero for all wavelengths.
[1] The wavelength of blue light is less than that of red light, so the fringe separation for blue is
less than that for red.
The fringes close to the central maximum will appear as spectra with the violet end inside and
the red end at the outside [1]. After several sets of spectra, fringes of different orders will
overlap and the pattern will become indistinct. [1]
[Total 16 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Test yourself questions
Page 289 Test yourself on prior knowledge
1
𝑐
3.00 × 108 m s−1
𝑓
909 × 103 s−1
From c = fλ  λ = =
= 330 m
This is a radio wave (actually BBC Radio 5 on AM).
2
𝑐
3.00 × 108 m s−1
𝜆
12.25 × 10−2 m
From c = fλ  f = =
= 2.45 × 109 Hz (2.45 GHz)
This is in the microwave region (actually the frequency of domestic microwave ovens).
3
From c = fλ
λ=
𝑐
𝑓
=
3.00 × 108 m s−1
7.5 × 1018 s−1
= 4.0 × 10−11 m
This of the same order of magnitude as the lattice spacing in a graphite crystal (actually about one tenth)
and so the X-rays will exhibit diffraction effects.
4
For Electrical work done on electron = kinetic energy gained by electron
eV = ½mv2
Hence:
2𝑒𝑉
𝑣=√
𝑚
2 × 1.6 × 10−19 C × 1000 V
=√
9.1 × 10−31 kg
= 1.9 × 107 m s −1
Page 291 Test yourself
1
a)
By counting across ten fringes:
• an average value can be obtained
• to a greater precision.
Distance across 10 fringes = 49 mm
Fringe width x = (49 ÷ 10) mm = 4.9 mm
𝑥𝑠
𝜆=
c)
This is in the visible red part of the electromagnetic spectrum, which is in accordance with the
𝐷
=
4.9 × 10−3 m × 0.25 × 10−3 m
b)
1.95 m
= 6.28 × 10-7 m ≈ 630 nm
fringes in the photograph being red.
Page 294 Test yourself
2
a)
An efficiency of 5% means that only 5% of the 60 W power is converted into visible light – the
rest is converted into thermal energy.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Test yourself questions
b)
Intensity, or radiation flux density, is the power of radiation received per unit area of a surface.
c)
Light power = 5% × 60 W = 3.0 W
Intensity =
Power
4π𝑟 2
=
3.0 W
= 0.041 W m−2
4π × (2.4 m)2
We have had to assume that the lamp:
• is a point source
• radiates equally in all directions.
3
a)
Energy of each photon = ℎ𝑓 =
ℎ𝑐
𝜆
=
6.63 × 10−34 J s ×3.00 × 108 m s−1
600 × 10−9 m
−19
= 3.315 × 10
J
Laptop emits 5.0 W = 5.0 J s−1
Number of photons emitted per second =
b)
5.0 J s−1
3.315 × 10−19 J
= 1.5 × 1019 s −1
Total number of pixels = 1024 × 768 = 7.86 × 105 pixels
Average number of photons emitted per second per pixel
=
1.5×1019 s−1
7.86 × 10−5 pixels
= 1.9 × 1013 per second per pixel.
This is an average value because some pixels will be brighter (i.e. emit more energy) than others,
depending on the picture on the screen.
Page 297 Test yourself
4
𝑐
3.00 × 108 m s−1
𝜆
633 × 10−9 m
a)
From c = fλ  f = =
b)
E = hf = 6.33 × 10−34 J s × 4.74 × 1014 s −1 = 3.14 × 10−19 J
Energy in eV =
𝑐
3.14 × 10−19 J
1.6 × 10−19 J eV−1
= 4.74 × 1014 Hz
= 1.96 eV
3.00 × 108 m s−1
c)
c = fλ  λ =
d)
E = hf = 6.63 × 10−34 J s × 6.55 × 1014 s −1 = 4.34 × 10−19 J
𝑓
Energy in eV =
e)
=
6.55 × 1014 s−1
4.34 × 10−19 J
1.6 × 10−19 J eV−1
= 2.71 eV
hf = 2.34 eV = 2.34 eV × 1.6 × 10−19 J eV−1 = 3.74(4) × 10−19 J
hf = 3.74(4) J  f =
f)
= 4.58 × 10-7 m = 458 nm
3.744 × 10−19 J
6.63 × 10−34 J s
𝑐
3.00 × 108 m s−1
𝑓
5.65 × 1014 s−1
c = fλ  λ = =
= 5.65 × 1014 Hz
= 5.31 × 10−7 m = 531 nm
Colours:
633 nm = red
458 nm = blue
531 nm = green
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Test yourself questions
5
a)
From Einstein’s equation:
= ϕ + ½mvmax2
if hf > ϕ, photoelectrons will be emitted.
b)
Maximum kinetic energy is given by:
½mvmax2 = hf − ϕ = (2.34 – 2.28) eV = 0.06 eV
= 0.06 eV × 1.6 × 10−19 J eV−1
= 9.6 × 10−21 J
The minimum energy a photon must have for emission is when ½mvmax2 = 0,
c)
i.e. hf = ϕ.
The minimum frequency for emission is therefore fmin = ϕ/h and is called the threshold
frequency.
For the sodium surface:
𝑓min =
d)
𝜙
ℎ
=
2.28 eV × 1.6 × 10−19 J eV−1
6.63 × 10−34 J s
= 5.50 × 1014 Hz
For Laser 2: hf = 2.71 eV (see answer (d) in Question 4)
If ϕ = 2.28 eV for the sodium surface, the reverse voltage would have to be:
(2.71 – 2.28) V = 0.43 V
e)
For Laser 1: hf = 1.96 eV (see answer (b) in Question 4)
As 1.96 eV < 2.28 eV (the work function), each photon will have insufficient energy to release an
electron from the sodium surface.
Page 298 Activity 17.1
1
A 50 MΩ resistance is connected in series with high-voltage supplies to limit the current. This is a safety
feature to prevent the user getting a dangerous electric shock in the event of accidently touching a live
terminal. The intensity of an electric shock is determined by the current flowing to Earth through the
person’s body. If this current is small, the effect of the shock is reduced. Nevertheless, even the shock from
a protected high-voltage power supply is still dangerous and such devices should always be handled with
the utmost care.
2
The central maximum is formed by all the spectral lines overlapping and therefore has the same colour as
the hydrogen discharge tube when viewed without the grating.
3
Referring to the grating formula nλ = dsin θ:
a)
Blue light has a shorter wavelength than red light. As n and d are constant for a given order, sin
θ  λ, which means that the angle of diffraction will be less for the shorter wavelength blue light.
The blue lines are therefore nearer the centre for each order.
4
The grating is labelled as 600 lines per millimetre. The grating spacing is therefore:
d=
1
600 mm−1
= 1.67 × 10−3 mm = 1.67 × 10−6 m ≈ 1.7 × 10−6 m
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Test yourself questions
5
In the grating formula nλ = dsin θ, n = 1 and d = 1.67 × 10−6 m
 λ = 1.67 × 10−6 m × sin θ
 sin θ =
𝜆
1.67 × 10−6 m
 θb = sin−1 (
θr = sin−1 (
400 × 10−9 m
1.67 × 10−6 m
660 × 10−9 m
1.67 × 10−6 m
) = 13.9 ° (don’t forget to put your calculator in ‘degree’ mode!)
) = 23.3 °
θr – θb = 23.3 ° − 13.9 ° = 9.4 °
Page 303 Test yourself
6
a)
Ionisation energy is the energy that has to be given to an electron in the ground state so that it
can just escape from the atom.
For a mercury atom this is ΔE = [0 – (−10.43)] eV = 10.43 eV
= 10.43 eV × 1.6 × 10−19 J eV−1 = 1.67 × 10−18 J
b)
If an electron is excited to a higher energy (level 1) and then returns to a lower energy (level 2)
the difference in energy, E1 – E2, is emitted as a quantum of energy hf. As the energy levels for a
given element have discrete (quantised) values, as shown in Figure 16.15b) in your textbook, the
frequencies emitted will have fixed values and form a (line) emission spectrum. For example, one
such line (yellow) at a wavelength of about 580 nm can be seen in Figure 16.15a) in your
textbook.
c)
For transition Y:
hf = E1 – E2 = [−2.69 – (−4.97)] eV
= 2.28 eV = 2.28 eV × 1.6 × 10−19 J eV−1 = 3.64(8) × 10−19 J
f=
∆𝐸
ℎ
=
3.64(8) × 10−19 J
6.33 × 10−34 J 𝑠
= 5.50 × 1014 Hz
𝑐
3.00 × 108 m s−1
𝑓
5.50 × 1014 s−1
From c = fλ  λ = =
= 5.45 × 10−7 m = 545 nm
This corresponds to the green line in the spectrum in Figure 17.14a in your textbook.
d)
Transition W has the greatest energy difference of the four transitions shown and so must
correspond to the highest frequency line. This will be the violet line (≈ 410 nm).
Transition X is the next largest energy difference and so must correspond to the blue line (≈ 440
nm).
Transition Z has a smaller energy difference than Y and so must emit a line of lower frequency
(i.e. greater wavelength) than Y. This will be the yellow line (≈ 580 nm).
e)
The visible spectrum ends at about 400 nm, so a line of wavelength 254 nm would be well into
the ultra-violet region. It will not, therefore, be visible.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Test yourself questions
f)
ΔE = hf =
ΔE =
ℎ𝑐
𝜆
=
6.63 × 10−34 J s × 3.00 × 108 m s−1
254 × 10−9 m
7.83 × 10−19 J
= 7.83 × 10−19 J
= 4.89 eV
1.6 × 10−19 J eV−1
By inspection, the electron must return to its ground state (any other transition would be
< 4.89 eV), so
From ΔE = E1 – E2  E1 = ΔE + E2 = [4.89 + (−10.43)] eV = −5.54 eV
Therefore the transition must be from the −5.54 eV level to the ground state.
g)
When a photo of energy 2.14 eV interacts with a mercury atom it will excite an electron from the
−3.73 eV level to the −11.59 eV level.
When light from a filament lamp passes through mercury vapour, the wavelength (yellow)
corresponding to this change in energy will be selectively absorbed to excite the electrons in the
mercury vapour. When the electrons drop down to lower energy levels, they re-emit the quanta
of radiation randomly in all directions, so the corresponding wavelength is not observed. A dark
line will therefore appear in the white light spectrum of the filament lamp at about 580 nm,
where the yellow line has been absorbed.
Page 305 Test yourself
7
a)
V = J C−1 and the electron charge = C
Therefore eV = C × J C−1 = J = unit of energy
b)
30 keV = 30 × 103 V × 1.6 × 10−19 J eV−1 = 4.8 × 10−15 J
c)
½mv2 = 4.8 × 10−15 J
𝑣=√
2 × 4.8 × 10−15 J
= 1.03 × 108 m s-1 ≈ 1 × 108 m s-1
9.11 × 10−31 kg
d)
Momentum = mv = 9.11 × 10−31 kg × 1.03 × 108 m s−1 = 9.35 × 10−23 kg m s−1
e)
From de Broglie: λ =
f)
Because of this very short wavelength, electron microscopes:
ℎ
𝑚𝑣
=
6.63 × 10−34 J s
9.35 × 10−23 kg m s−1
= 7.10 × 10-12 m ≈ 0.007 nm
• can produce much larger magnification
• have a much larger resolving power (i.e. the ability to see detail) than optical microscopes.
The wavelength of 0.007 nm means that they can observe structures of less than a
nanometre without diffraction effects occurring.
8
𝑐
3.00 × 108 m s−1
𝑓
5.09 × 1014 s−1
a)
c = fλ  λ = =
b)
λ=
ℎ
𝑚𝑣
 𝑚𝑣 =
ℎ
𝜆
=
= 5.89 × 10−7 m = 589 nm
6.63 × 10−34 J s
5.89 × 10−7 m
= 1.12 × 10−27 kg m s−1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Test yourself questions
c)
If the atom is initially at rest, the momentum of the system will be zero and, by the conservation
of momentum, must remain so. Therefore the momentum of the recoil atom (Mv) will be equal
and opposite to that of the photon emitted.
Mv = −1.12 × 10−27 kg m s−1
𝑣=
−1.12 × 10−27
23 ×1.66 × 10−27
= -0.029 m s-1 = -29 mm s-1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
Page 307–311 Exam practice questions
1
First of all, 5% of 60 W = 3 W. We then need to use the relationship
that intensity I at a distance r is given by:
𝐼=
power
3W
=
= 0.11Wm−2
2
4𝜋𝑟
4𝜋 × (1.5m)2
The answer is A.
2
[Total 1 Mark]
Rearranging 𝜑 = ℎ𝑓0 and remembering to convert eV into J:
𝑓0 =
𝜑 2.28eV×1.6×10−19 JeV −1
=
= 5.5 × 1014 Hz
ℎ
6.63 × 10−34 Js
The answer is C.
3
[Total 1 Mark]
We will need to use 𝐸 = ℎ𝑓, so we must find the frequency corresponding to a wavelength of 447 nm.
Rearranging 𝑐 = 𝑓𝜆:
𝑓=
𝑐 3.00 × 108 ms −1
=
= 6.71 × 1014 s −1
𝜆
447 × 10−9 m
Then:
𝐸 = 4.45 × 10−19 Js × 6.71x1014 𝑠 −1 = 4.45 × 10−19 J
Converting to electron-volts using 1eV = 1.6 × 10−19 J:
𝐸=
4.45 × 10−19 J
= 2.78eV
1.6 × 10−19 JeV −1
The answer is C.
4
[Total 1 Mark]
In question 3 we found that the energy of a photon of blue light is 2.78eV. As the work function is 2.28eV,
the reverse potential difference that would have to be applied to just prevent photoemission would be:
𝑉 = (2.78 − 2.28)V = 0.5V
The answer is A.
5
a)
[Total 1 Mark]
A photon is a small, discrete amount, or quantum, of energy associated with electromagnetic
radiation.
b)
The energy to move an electron through a potential difference of 0.48V is given by:
𝐸 = 𝑄𝑉 = 1.6 × 10−19 C × 0.48V=7.68×10−20 J≈8×10−20 J
c)
The efficiency of energy conversion is therefore:
efficiency=
d)
7.68×10−20 J
4×10−19 J
× 100% = 19%
Rearranging E = hf:
𝑓=
𝐸
ℎ
=
4.0×10−19 J
6.63×10−34 Js
= 6.03 × 1014 s −1
Rearranging c = fλ:
𝑐
3.00×108 ms −1
𝑓
6.03×1014 s −1
λ= =
= 4.97 × 10−7 m=500nm (to 2 significant figures)
[Total 9 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
6)
a)
The work function of a surface is the amount of energy that is needed to just remove an electron
from the surface.
b)
We will need to use E = hf, so we must find the frequency corresponding to a wavelength of 532
nm. Rearranging c = fλ:
𝑐
3.00×108 ms −1
λ
532×10−9 m
𝑓= =
= 5.64 × 1014 s −1
Then:
E = hf = 6.63 × 10−34Js × 5.64 × 1014s−1 = 3.74 × 10−19J
Converting to electron-volts using 1eV = 1.6 × 10−19J:
𝐸=
c)
3.74×10−19 J
1.6×10−19 JeV −1
= 2.34eV ≈ 2.3eV
1
2
Rearranging ℎ𝑓 = 𝜑 + 𝑚𝑣𝑚𝑎𝑥
gives:
2
1
2
2
𝑚𝑣𝑚𝑎𝑥
= ℎ𝑓 − 𝜑 = (2.34 − 1.90)𝑒𝑉 = 0.44𝑒𝑉
= 0.44eV × 1.6 × 10−19 JeV − 1 = 7.0 × 10−20 J
d)
The frequency of the red light will be less than that of the green light, so a quantum of the red
light may not have sufficient energy (hf) to overcome the work function and release
photoelectrons. A quick way of calculating the wavelength corresponding to the work function is
to say that 2.34eV corresponds to 532 nm (from part b), therefore 1.90eV (the work function)
will correspond to:
𝜆=(
2.34
1.90
) × 532nm = 655nm
This is in the red region of the spectrum and so if the laser has a wavelength greater than this, it
will not cause any photoelectrons to be emitted.
[Total 10 Marks]
7
a)
Green light has the shortest wavelength and therefore the highest frequency. The green LED will
therefore emit photons of the highest energy as the photon energy is given by E = hf.
b)
i) We will need to use E = hf, so we must find the frequency corresponding to a wavelength of
630 nm. Rearranging c = fλ:
𝑐
3.00×108 ms −1
𝜆
630×10−9 m
𝑓= =
= 4.76 × 1014 s −1
Then:
𝐸 = ℎ𝑓 = 6.63 × 10−34 Js × 4.76 × 1014 s−1
= 3.16 × 10−19 J ≈ 3 × 10−19 J
ii) As 1eV=1.6×10−19 J:
𝐸=
3.16×10−19 J
1.6×10−19 JeV −1
= 2.0eV
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
c)
i) We then need use the relationship that intensity I at a distance r is given by:
𝐼=
power
4𝜋𝑟 2
=
18mW
4𝜋×(0.30m)2
= 15.9mWm−2 ≈ 16mWm−2
ii) The area of the pupil is given by:
𝐴=
𝜋𝐷2
4
2
=
𝜋(5×10−3 m)
4
= 1.96 × 10−5 m2
The energy of the photons entering the eye per second is given by:
energy per second = intensity × area
= 15.9mWm−2 × 1.96 × 10−5 m2
= 3.12 × 10−7 Js −1
From part b) i), the energy of each photon is 3.16 × 10−19J, so:
number of photons per second =
3.12×10−7 Js −1
3.16×10−19 J
= 1.0 × 1012 s −1
iii) As you move further away, the number of photons entering your eye per second will get less
and so the intensity will get less.
d)
Efficiency =
=
e)
18mW
120mW
visible light emitted
power consumption
× 100%
× 100% = 15%
Even at 15% efficiency, LEDs are far more efficient than filament lamps, which typically have an
efficiency of less than 5%. In addition, the coloured glass necessary for filament lamps absorbs a
certain amount of the light, and LEDs tend to last much longer than filaments. It is therefore
much cheaper and far more energy-efficient to use LEDs for traffic lights.
8
a)
Diffraction and interference suggest that light can behave as a wave.
b)
Monochromatic literally means ‘one colour’, in other words light all having the same frequency
or wavelength.
[2]
c)
d)
[1]
[3]
The term eVs is equal to the maximum kinetic energy ½mv2 max of the photoelectrons. It is a
measure of the work that has to be done to just stop the most energetic photoelectrons.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
e)
i) The threshold frequency f0 is the frequency when the stopping potential Vs = 0. From the
graph, f0 = 4.3 × 1014Hz.
ii) The work function φ is given by:
φ = hf0 = 6.63 × 10−34 J s × 4.3 × 1014 s−1 = 2.9 × 10−19 J
As 1 eV = 1.6 × 10−19 J:
φ=
f)
2.9 × 10−19 J
1.6 × 10−19 J eV–1
= 1.8 eV [4]
The energy to release a photoelectron comes from one photon. Below the threshold frequency f0
no electrons are emitted because the energy of each photon, hf0, is not sufficient to provide an
electron near the surface with enough energy to escape.
9)
a)
[2]
Using hf = E2 − E1 and remembering to convert the energy from eV to J:
f=
(69.6 – 1.8) × 103 eV × 1.6 × 10−19 J eV–1
6.63 × 10–34 J s
= 1.64 × 1019 s−1
Rearranging c = fλ:
λ=
10
3.00 × 108ms−1
1.64 × 1019 s−1
= 1.83 × 10−11m ≈ 0.02nm
[4]
b)
This is in the X-ray region of the electromagnetic spectrum.
[1]
a)
i) Excited means that the electrons have been given energy to raise them to higher energy
levels than their normal lowest energy level stable state.
ii) The electrons in the excited mercury atoms are in an unstable state. In order to achieve
stability, they emit energy in the form of quanta of electromagnetic radiation, thus returning
to lower, more stable, energy levels.
iii) According to quantum theory, the electrons can only exist in certain allowed discrete energy
levels. The frequency of the emitted radiation corresponds exactly to the energy released
when an electron drops from one of these allowed energy levels to a lower energy level,
given by hf = E2 – E1. Therefore only certain wavelengths of radiation are emitted.
b)
[6]
i) The energy levels for the phosphor electrons are different from those of mercury and so the
energy transitions will give rise to different wavelengths from the mercury.
ii) Using Q = It and remembering to convert 1½ hours to seconds:
Q = 200 × 10−3 A × 1.5 × 60 × 60s = 1080C
iii) Fluorescent tubes are more efficient, insomuch as they give out more light than a tungsten
filament light for the same amount of electrical energy supplied. They therefore cost less to
run and are more environmentally friendly. They are, however, more costly to manufacture,
both in terms of money and carbon footprint.
[8]
[Total 14 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
11
a)
Consider a simple model of an atom consisting of a nucleus surrounded by electrons in some
form of 'orbits' . An electron in a particular 'orbit' will have a certain amount of energy associated
with it, made up of its kinetic energy of rotation and its potential energy due to the electric field
of the nucleus. This is called the 'energy level' of the electron.
b)
[2]
A photon is a small, discrete amount, or quantum, of energy associated with electromagnetic
radiation. Although γ-rays could have the same wavelength, γ-rays come from the nucleus and
not from transitions of electron energy levels as we have here. An answer of γ-radiation would
therefore be wrong.
c)
[2]
The energy of the photon in the absorption diagram is:
hf = E2 – E1
d)
[1]
When a photon is absorbed by an electron, the increase in the energy level of the electron is
exactly equal to the energy of the photon. When the electron returns to its lower energy level
the photon emitted has an amount of energy exactly equal to the difference between the energy
levels. Therefore the laser light emitted by the stimulated emission process must have the same
wavelength as the photon in the spontaneous emission diagram.
e)
[2]
In general, 'coherent' means that there is a constant phase relationship between two waves.
In the case of a laser, it means that the emitted photons are all of the same frequency and in
phase.
[2]
[Total 9 Marks]
12
a)
The distance 2x between the two first order maxima either side of the central maximum is
2x = 18.0 ÷ 0.5 mm  x = 9.0 ± 0.25 mm
[2]
(Remember: It is good practice to make measurements as large as possible to reduce the
percentage uncertainty. You should therefore measure the distance 2x and halve it to find a
value for x.)
𝑥
9.0 mm
𝐷
330 mm
sin θ ≈ tan θ = =
= 0.02727 (as θ is small)
Applying nλ = dsin θ, where n = 1 and d =
1
50 mm-1
[2]
= 0.02 mm = 2.0 × 10−5 m
λ = 2.0 × 10−5 m × 0.02727 = 5.45 × 10−7 m = 545 nm
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
b)
% uncertainty in D =
% uncertainty in x =
1 mm
330 mm
0.25 mm
9.0 mm
× 100% = 0.3%
× 100% = 2.7% (or
0.5 mm
18 mm
× 100% = 2.7%)
% uncertainty in sin θ = 0.3% + 2.7% = 3.0%
Assuming that the uncertainty in the grating spacing is negligible, this is also the % uncertainty in
the value for λ.
[4]
Uncertainty in λ = ±3% of 545 nm = ±16 nm
Therefore λ = 545 ± 16 nm
c)
[2]
The experimental value of 545 ± 16 nm for λ agrees with the stated value of 543.4 nm within the
estimated experimental uncertainty.
d)
[2]
The precision of the value for the wavelength could be improved by moving the screen further
away from the grating, which would make both D and, more importantly, x larger. The %
uncertainty in these measurements would therefore be reduced. For example, if the distance D
were to be doubled, then the % uncertainty in both D and x would be halved. The % uncertainty
in λ would thus be reduced by a factor of 4 so that it was less than 1%. In addition, the distances
between the second and third order maxima could also be measured and two further values for λ
could be found. An average of the three values for λ could then be obtained.
[4]
[Total 18 Marks]
13
a)
Combining EK = ½mv2 and p = mv
2mEK = 2m × ½mv2 = m2v2 = p2
p = √2𝑚𝐸K
b)
[3]
We now need to use the equation λ =
ℎ
𝑝
−31
where m = electron mass = 9.11 × 10
kg
Ek = (15 × 103) eV × 1.60 × 10−19 J eV−1 = 2.40 × 10−15 J
ℎ
6.63 × 10−34 J s
𝑝
√2 × 9.11 × 10−31 kg × 2.40 × 10−15 J
λ= =
c)
= 1.003 × 10−11 m ≈ 0.01 nm
[3]
The layers of atoms in a crystal are arranged in a regular pattern, like the lines of a diffraction
grating, with a lattice spacing typically in the order of 10−10 m (or 0.1 nm). As this is about 10
times greater than the electron wavelength, a crystal could be used to produce electron
diffraction.
[3]
[Total 9 Marks]
14
a)
14 Area of panel: A = 120 cm × 60 cm = 1.2 m × 0.6 m = 0.72 m2
Energy per second incident on panel = 600 W m−2 × 0.72 m2 = 432 W
b)
[2]
Electrical energy produced per second = 20% × 432 W = 86 W
The article states that, when under load, the panel produces 75 to 100 W, so the calculated value
of 86 W falls comfortably within this range.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
c)
This suggests that the solar cell has an appreciable internal resistance r. When under load, i.e. a
current I is being taken, the e.m.f. of 17 V falls to 12–14 V as a result of between 3–5 V being
dropped across the internal resistance: V = ε – Ir.
d)
[3]
Initially let us consider the entire solar panel.
Let us assume a power is 85 W under load (i.e. approximately half way between the
75 W–100 W stated) and a mid-range value of 13 V for the p.d.
[3]
Then the current would be:
𝑃
85 W
𝑉
13 V
I= =
=5A
From V = ε – Ir we have: Ir = ε – V = (17 – 13) V = 4 V
𝑉
4V
𝐼
5A
r= =
= 0.8 Ω
[3]
The e.m.f. of the panel will be the 17 V peak output as the e.m.f. is the value of the voltage when
the panel is not taking any current.
[1]
As the panel is made up of 36 cells in series, the e.m.f. of each cell will be:
17 V ÷ 36 = 0.47 V and the internal resistance will be:
0.8 Ω ÷ 36 = 0.02 Ω
e)
[2]
i) From part (a) the energy incident on panel per second = 432 W = 432 J s−1
Each photon has energy:
hf =
ℎ𝑐
𝜆
=
6.63 ×10−34 J s × 3.00 × 108 m s-1
550 × 10−9 m
= 3.62 × 10−19 J
Number of photons incident on panel per second =
=
432 J s-1
3.62 × 10-19 J
[2]
Energy incident on panel per second
Energy of each photon
= 1.19 × 1021 s−1
[2]
ii) A wavelength of 550 nm is in the yellow region of the electromagnetic spectrum.
[1]
iii) This wavelength was chosen because at the Earth’s surface, the Sun appears yellow, which
suggests that this is the predominant wavelength. The answer is only very approximate as the
wavelength is an educated guess and doesn’t make any allowance for the non-uniform
distribution of energy across the Sun’s spectrum.
[3]
[Total 24 Marks]
14
a)
From:
density =
mass
volume
 volume =
mass
density
=
Number of atoms per cubic metre =
=
6.02 × 1026 atoms
5.29 × 10-3 m3
29
12 kg
= 1.14 × 10 atoms m
‘Volume’ occupied by one atom =
= 5.29 × 10−3 m3
2.27 × 103 kg m-3
number of atoms in 12 kg of carbon
volume of 12 kg of carbon
−3
1
1.14 × 1029 m-3
= 8.79 × 10−30 m3
[
3]
Assuming a simple cubic structure with the atoms a distance d apart:
d3 = 8.79 × 10−30 m3
3
d = √8.79 × 10−30 m = 2.06 × 10−10 m ≈ 0.2 nm
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
17 Particle nature of light
Answers to Exam practice questions
b)
Combining EK = ½mv2 and p = mv
2mEK = 2m × ½mv2 = m2v2 = p2
p = √2𝑚𝐸K
Energy of 4 kV electron = 4000 J C−1 × 1.6 × 10−19 C = 6.4 × 10−16 J
p = √2 × 9.11 × 10-31 kg × 6.4 ×10−16 J = 3.14 × 10−23 kg m s−1
ℎ
6.63 × 10-34 J s
𝑝
3.14 × 10-23 kg ms−1
λ= =
c)
= 1.94 × 10−11 m ≈ 0.02 nm
[5]
As the wavelength of the 4 kV electrons (0.02 nm) is about 10 times smaller than the spacing of
the carbon atoms (0.20 nm) it should be possible to diffract the electrons by means of a carbon
crystal.
d)
i) D =
[2]
𝑎
𝑑√𝑉
d=
𝑎
𝐷√𝑉
Measuring the diameters of the two rings by taking two readings at right angles and
averaging gives:
D1 = 25 mm and D2 = 42 mm
d1 =
d2 =
𝑎
𝐷1 √𝑉
𝑎
𝐷2 √𝑉
ii) Ratio =
=
=
3.32 × 10−10 m2 V½
25 × 10−3 m × √4000V½
3.32 × 10−10 m2 V½
42 × 10−3 m × √4000V½
0.210
0.125
[4]
= 2.10 × 10−10 m = 0.21 nm
= 1.25 × 10−10 m = 0.12 nm
[3]
= 1.68
[1]
Percentage difference from √3 = (
√3 −1.68
√3
) × 100% = 3%
(Note that we use the given value of √3 as the denominator.)
As there is some uncertainty is measuring the diameter of the rings, probably ±1 mm at best,
then 3% is within the limitation of the measurements, confirming that the ratio of the two
lattice spacings is probably √3.
[2]
iii) The atomic spacing in part (b) was found to be about 0.2 nm, which is of the same order of
magnitude as the spacings found in (i) above (0.21 nm and 0.12 nm). This suggests that the
simplifying assumption that the carbon atoms are arranged in a cubic structure was
reasonable as it enabled us to get a fairly good estimate of the atomic spacing.
[2]
[Total 24 Marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Test yourself questions
Page 312 Test yourself on prior knowledge
1
Momentum p = mʋ contains a single vector ʋ, but KE contains ʋ2, which is a scalar.
2
a)
velocity and acceleration
b)
mass and GPE. (You may have thought of others!)
3
4
1J1N×m
 1 kg m s−2 × m
 1 kg m2 s−2
5
Momentum p = mʋ,
 p2 = m2ʋ2
= 2m × ½ mʋ2
= 2m × KE, as KE
= ½mʋ2
6
Newton’s second law states that ‘The resultant force acting on a body is equal to the rate of change of
momentum of the body’.
7
The downward gravitational force on the book is
mg = 3.2 kg × 9.8 N kg−1 = 31 N.
The book is in equilibrium
 the upward push of the table on the book = 31 N
8
ΔGPE = mgΔh is equal here to a negative quantity as Δh is negative.
ΔGPE = − (1600 kg × 9.8 N kg−1 × 2.8 m)
 the loss of GPE is 44 000 J or 44 kJ
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Test yourself questions
Page 318 Core practical 9
1
2
F/N
Δt/s
FΔt/N s
t1/s
t2/s
v1/m s−1
v2/m s−1
Δv/m s−1
0.10
0.90
0.090
0.51
0.36
0.39
0.56
0.17
0.20
0.64
0.128
0.38
0.26
0.53
0.77
0.24
0.29
0.48
0.139
0.29
0.21
0.69
0.95
0.26
0.39
0.42
0.164
0.25
0.18
0.80
1.11
0.31
0.49
0.37
0.181
0.22
0.16
0.91
1.25
0.34
0.59
0.34
0.200
0.19
0.14
1.05
1.43
0.38
Looking at the values of F in the table, for F = 0.10 N, the mass required to produce this force must have
been 10 g (0.01 kg × 9.8 N kg−1 = 0.098 N = 0.10 N).
There are therefore 6 × 10 g masses altogether (in Figure 1.8 there are 2 masses on the trolley and 4
masses making up the ‘load’).
The total mass is therefore:
m = mass of trolley + 6 × (10 g)
= 450 g + 60 g = 510 g or 0.510 kg
3
Your graph of FΔt on the y-axis against Δv on the x-axis should be a straight line through the origin, as
shown the figure below.
4
Gradient = 0.21 N s/0.40 m s-1 = 0.53 kg
5
Percentage difference from total mass = 0.53 -0.51 kg, 0.51 kg × 100 % = 4 %
As the data is only recorded to 2 SF, the two values can be considered equal within the uncertainties of the
experiment. This shows that FΔt = mΔv or F =mΔv/Δt = rate of change of momentum
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Test yourself questions
Tip: In a question like this, you need to justify your answer with some data.
As you are comparing an experimental value (the gradient) with a stated value (m), you should take the
stated value as the denominator.
Page 318 Test yourself
1
a)
Impulse = force × time
= (200 × 103 ) N × (50 × 10−3) s
= 1.0 × 104 N s or 10 000 N s
b)
Force = impulse/time
= 250 N s / (100 × 10−3 s)
= 2 500 N or 2.5 kN
c)
Momentum before impact = mv = 1250 kg × 24 m s−1
= 30 000 kg m s−1
= 3.0 × 104 N s
Momentum after impact = 0
Change in momentum = 3.0 × 104 N s
From impulse = force × time  time = impulse / force
= 3.0 × 104 N s / 300 × 103 N
= 0.10 s
2
a)
Both passengers will have the same velocity, but as passenger A has a slightly larger mass,
her/his impulse on collision will be a bit more than that of passenger B. However, her/his time of
impact (0.09 s) is less than half that of passenger B (0.22 s).
From impulse = FΔt  F = impulse / Δt, so the maximum force experienced by passenger A will
be much greater than that on passenger B.
b)
The evidence suggests that passenger B was in the front seat as her/his time of impact was much
longer. This was probably because the airbag inflated and, hopefully, a seat belt was being worn.
Passenger A was likely to have been in the back, where there was no airbag and, possibly, no seat
belt was being worn. This passenger could possibly have been stopped by hitting the seat in
front, giving a much shorter impact time.
c)
The area under each graph represents the impulse on the passengers. As both passengers will
have the same velocity on impact (that of the car), passenger B, having slightly less mass, will
have less momentum than passenger A. The change of momentum (= impulse) of passenger B on
impact will therefore be slightly less than that for passenger A and so the area under the graph
will be less for passenger B.
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18 Momentum and energy
Answers to Test yourself questions
d)
To a good approximation we can take the graph for passenger B to be a triangle of height 16 kN
and base (0.32 – 0.11) s = 0.21 s. This gives an area of:
Area = ½ × 16 kN × 0.21 s = 1680 N s = impulse
Impulse = Δp = Δ(mv) = mΔv = 1680 N s = 1680 kg m s−1
This gives Δv = 1680 N s / mass of B
= 1680 kg m s−1/ 94 kg
= 18 m s−1
As the passenger’s velocity after impact = 0, the passenger’s velocity, and therefore the speed of
the car before collision must have been 18 m s−1.
(The NCAP testing speed is 40 mph, which you might like to show is about 18 m s −1)
Page 322 Test yourself
3
KE = ½mʋ2 = ½ × 85 kg × (10 m s−1)2 = 4250 J
4
Gravitational potential energy, chemical energy, radiant energy
5
5½mʋ2 = ½kΔx2: the half cancels, then dividing each side by k and taking the square root of each side gives
Δx = √(mʋ2/k).
6
5.0 GeV = 5.0 × 109 eV = (5.0 × 109 eV) × (1.6 × 10−19 J eV−1)
= 8.0 × 10−10 J or 0.80 nJ
7
N s  kg m s−2 × s  kg m s−1
Page 325 Test yourself
8
p = mʋ = 440 kg m s−1 and m = 220 kg.
 ʋ = p/m = 440 kg m s−1 ÷ 220 kg = 2.00 m s−1.
KE = ½mʋ2 = ½ × 210 kg × (2.00 m s−1)2
= 420 kg m2 s−2
= 420 N m
= 420 J
9
10
The event must be an explosion releasing kinetic energy as the original nucleus was at rest and the emitted
α-particle will have kinetic energy. Therefore this was an inelastic collision.
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18 Momentum and energy
Answers to Test yourself questions
11
Glider P has a mass of 0.20 kg. Before the collision it is moving to the right at 0.14 m s−1 and after the
collision it is moving at − 0.26 m s−1 (i.e. to the left as positive is to the right).
The change of momentum:
Δp = 0.20 kg × [− 0.26 − (+ 0.14)] m s−1
= − 0.080 kg m s−1, which is correct.
The change of KE is:
ΔEk = ½ × 0.20 kg × (0.26 m s−1)2 − ½ × 0.20 kg × (0.14 m s−1)2
= 0.0048 J, which is is also correct.
Page 327 Core practical 10
1
a)
The error in the flash rate is  1 s−1/30 s−1 × 100% ≈ 3%
b)
The error in the size of the photo is about  1/8× 100% ≈ more than  10%
So the percentage errors in the momentums will also be these figures, i.e. the possible error in the size of
the photo overwhelms the possible error in the flash rate.
2
In this table suffix l refers to the large ball and suffix s refers to the small ball, while suffix b refers to before
the collision and suffix a refers to after the collision:
3
ml = 0.100 kg
ms = 0.043 kg
ml = 0.100 kg
ms = 0.043 kg
ʋb = 3.6 m s−1
ʋb = 5.1 m s−1
ʋa = 4.0 m s−1
ʋa = 3.5 m s−1
mlʋb = 0.36 kg m s−1
msʋb = 0.22 kg m s−1
mlʋa = 0.40 kg m s−1
msʋa = 0.15 kg m s−1
The key here is that momentum is a vector and so we must take into consideration its direction. The figure
to the left shows that (vector mlʋb) + (vector msʋb) is equal to (vector mlʋa + (vector msʋa) though the
numerical values of the momentums given in the table do not give the same value when simply added.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Test yourself questions
4
KE = ½mʋ2 so KEbefore = ½(0.100 kg)(3.6 m s−1)2 + ½(0.043 kg)(5.1 m s−1)2
= (0.648 + 0.559) kg m2 s−2 = 1.2 J
KEafter = ½(0.100 kg)(4.0 m s−1)2 + ½(0.043 kg)(3.5 m s−1)2
= (0.800 + 0.263) kg m2 s−2 = 1.1 J
So, to 2 SF (the precision of the velocities) a little (about 10%) KE appears to be lost in the collision,
suggesting it is an inelastic collision.
Page 328 Test yourself
12
When two vectors are added, they can be represented as two adjacent sides of a parallelogram. The
resultant vector is then represented by the diagonal of the parallelogram.
13
a)
21 kg
b)
0
c)
8.0 s
14
Page 330 Test yourself
15
The unit of ʋ2Aρ is (m s−1)2 × m2 × (kg m−3)  kg m s−2 = N
16
F = ʋ2Aρ = (10 m s−1)2 × 200 m2 × 1.2 kg m−3 = 24 000 N or 24 kN, a BIG force!
17
A helicopter can ‘hover’ in the air by pushing the air downwards with a force F equal to its weight W. The
helicopter blades do this by giving air a total downward force equal to W and, by Newton’s third law, this
air will give the helicopter an upward force W.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Exam practice questions
Pages 332–334 Exam practice questions
1
Answer D
[Total 1 mark]
2
No kinetic energy is lost in elastic collisions.
[1]
In inelastic collisions some kinetic energy is lost, usually to internal energy.
[1]
In both types of collision total energy is conserved.
[1]
[Total 3 marks]
3
a)
As linear momentum is conserved:
(21 000 kg) × u = (28 000 kg) × 3.5 m s−1
[1]
 u = 4.66 m s−1 = 4.7 m s−1 to 2 SF
b)
[1]
ΔKE = ½(21 000 kg)(4.7 m s ) − ½(28 000 kg)(3.5 m s )
−1 2
−1 2
= (232 – 172) kJ = 60 kJ
 Percentage loss of KE =
60 kJ
232 kJ
[1]
[1]
× 100 %
= 26 %
[1]
[1]
[Total 6 marks]
4
Answer A
[Total 1 mark]
5
When the trolleys are released (in order to conserve momentum) the lighter trolley will move away at 4 ×
the speed of the heavier trolley.
[1]
Both trolleys gain kinetic energy, all of which comes from the elastic potential energy
[1]
stored in the compressed spring.
[1]
[Total 3 marks]
6
Suppose the woman’s velocity after separation is ʋ. Applying the principle of conservation of momentum:
(65 + 75) kg × 5.8 m s−1 = 75 kg × 3.8 m s−1 + 65 kg × ʋ
[2]
 ʋ = 8.1 m s ≈ 8 m s
[1]
Initial KE = ½(65 + 75) kg × (5.8 m s−1)2 = 2355 J
[1]
Final KE = ½ × 75 kg × (3.8 m s−1)2 + ½ × 65 kg × (8.1 m s−1)2 = 2674 J
[1]
 There is a gain in KE of 320 J
[1]
This comes from the work done by the skaters in pushing each other apart.
[1]
−1
−1
[Total marks 7]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Exam practice questions
7
Total KE before collision = ½ × (6.65  10−27 kg)  (1.50  107 m s−1)2
[1]
= 7.48 × 10−13 J
[1]
Total KE after collision
= ½ × (6.65  10−27 kg)  (1.23  107 m s−1)2 + ½ × (6.65  10−27 kg)  (0.86  107 m s−1)2
= 5.03 × 10−13 J + 2.46 × 10−13 J = 7.49 × 10−13 J
[2]
Therefore kinetic energy is conserved.
[Total 4 marks]
(The difference of 0.01 × 10−13 J comes from the fact that the speed of the helium nucleus after the collision
was only given to 2 SF. As the masses were both the same, showing that uα2 = ʋα2 + ʋHe2 would be enough –
and easier!)
8
Rockets ‘work’ by ejecting material (hot gases) backwards.
[1]
Suppose the force needed to do this is F.
By Newton’s third law the ejected material will exert a force F on the rocket car in the
forward direction.
[1]
The force F is equal to the rate of change of momentum of the ejected material, Newton’s
second law.
[1]
[Total 3 marks]
(Notice here that a rocket car is not driven forward by the push of the road on the car.)
9
a)
A speed of 180 kph = 180 000 m / 3 600 s = 50 m s−1
[1]
Therefore the ball leaves the racket with a momentum of
mv = 57 × 10−3 kg × 50 m s−1 = 2.85 kg m s−1
[1]
So the change of momentum Δp =2.85 N s = impulse
[1]
From impulse = FΔt  F = impulse/Δt
From Figure 18.24 we can estimate the contact time to be 4 ms giving:
F = 2.85 N s / 4.0 × 10−3 s = 700 N
b)
[1]
[2]
The force on the ball will not be constant – it will gradually increase as the ball deforms, reach a
maximum and then decrease again as the ball recovers its shape, rather like Figure 18.1b of the
Student Book The maximum force will therefore be considerably more than the calculated force
of 700 N – probably about twice as much.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Exam practice questions
c)
‘Elastic’ for a material means that the material will return to its original dimensions when the
deforming force is removed. Clearly, we want both the ball and strings to do this after each
shot and so they should be made of elastic materials.
[1]
We also want the collision to be as ‘elastic’ as possible – we want as much kinetic energy to
be conserved as possible. The elastic materials used for the ball and strings help to do this
too.
[1]
[Total 10 marks]
10
KE before the explosion = ½ × 2m × ʋ2 = mʋ2
[1]
2
KE after the explosion = 2 × mʋ which can be written as ½m(2ʋ)
2
[1]
If one part had a velocity of 2ʋ after the explosion and the other part had a velocity 0, then the
principle of conservation of energy would be satisfied.
[1]
But these numbers also have to meet the need to satisfy the principle of conservation of linear
momentum.
[1]
As the original momentum was 2m × ʋ, and the final momentum is m × 2ʋ, linear momentum is
also conserved.
[1]
[Total 5 marks]
11
The rotation of the helicopter’s blades, e.g. clockwise as seen from below, will tend to make the body of the
helicopter rotate in an anticlockwise direction.
[1]
(In physics this is described in a law called the conservation of angular momentum, which is not
discussed in this book as it is not an A Level requirement.)
To counter or balance this rotation, the helicopter has a set of small blades rotating in a vertical plane.
[1]
These blades exert a horizontal force on the air.
[1]
By Newton’s third law, the air exerts an equal and opposite horizontal force on the small blade
that prevents the helicopter body rotating.
[1]
[Total 4 marks]
(Some helicopters now have two supporting blades that operate in different directions, so doing Away with
the need to exert a horizontal force using a small rear rotor.)
12
a)
pshell = 12 kg × 320 m s−1 = 3840 kg m s−1 horizontally
pA = 2.0 kg × 450 m s−1 = 900 kg m s−1 at 45 above the horizontal
pB = 6.0 kg × 400 m s−1 = 2400 kg m s−1 horizontally
[3]
Note that, as momentum is a vector quantity, its direction must be stated in each case
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Exam practice questions
b)
pC has a mass m = 12 kg − 8.0 kg = 4.0 kg and a velocity ʋ at an angle θ
From this figure:
• PR measures 21 mm and thus represents 1050 kg m s−1
• θ = 38 
 vC =
1050 kg m s−1
4.0 kg
= 260 m s−1 at an angle of 38 ° below the horizontal
[5]
[Total 8 marks]
13
a)
Conservation of linear momentum gives:
4m × 5ʋ = 4m × Nʋ + 16m × 2ʋ
[1]
20mv = 4Nmv + 32mv
b)
which gives N = −3
[1]
 The velocity of the helium nucleus after the collision is 3ʋ backwards.
[1]
Calculating the KE before and after the collision gives:
KE before collision = ½ × 4m × (5ʋ)2 = 50mʋ2
[1]
KE after collision = ½ × 4m × (−3ʋ)2 + ½ × 16m × (2ʋ)2
[1]
= 18mʋ2 + 32mʋ2 = 50mʋ2
[1]
The collision is thus an elastic collision, as kinetic energy is conserved.
[1]
[Total 7 marks]
14
a)
Referring to Figure 1.26 in the question: suppose the molecule’s velocity after absorbing the
photon is ʋ2 to the right so that
Δu = ʋ1 − ʋ2
[1]
Applying the principle of conservation of momentum to this interaction gives
ℎ
𝜆
mʋ1 + ( ) = mʋ2
 mΔu =
b)
Here Δu =
ℎ
𝜆
[1]
or Δu =
ℎ
𝑚𝜆
=
ℎ
[1]
𝑚𝜆
6.6 × 10−34 J s
3.8 × 10−26 kg × 0.025 m
= 6.9 × 10−7 m s−1
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
18 Momentum and energy
Answers to Exam practice questions
c)
From the equation above, the units of Δu are
J×s
kg × m
.
We can write J  N m  kg m s−2 × m
This gives the unit of Δu to be
kg m s−2 × m × s
kg × m
[1]
= m s−1
[1]
[Total 7 marks]
15
Assume the two particles after the explosion have velocities ʋ1 and ʋ2.
Conservation of linear momentum gives:
2mʋ = mʋ1 + mʋ2
 2ʋ = ʋ1 + ʋ2 …………………………………...(1)
[1]
As KE is doubled, this gives:
2 × ½2mʋ2 = ½mʋ12 + ½mʋ22
[1]
 4ʋ2 = ʋ12 + ʋ22 …………………………………(2)
[1]
Squaring equation (1) gives:
4ʋ2 = ʋ12 + 2ʋ1ʋ2 + ʋ22 ………………………...(3)
[1]
Equations (2) and (3) yield that 2ʋ1ʋ2 = 0, so either ʋ1 or ʋ2 must be 0.
[1]
Equation (1) means that the other particle must be moving at 2ʋ.
[1]
[Total 6 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
19 Motion in a circle
Answers to Test yourself questions
Page 335 Test yourself on prior knowledge
1
π/2 rad
2
180
3
0.288 rad
4
150
5
A vector quantity because it always has a direction associated with it.
6
Fres =
7
600 N
8
12.0 kg
9
No, they are vector quantities because they act in a particular direction.
𝑆𝑝
𝑆𝑡
(The resultant force acting on a body is equal to the rate of change of momentum of the body.)
Page 338 Test yourself
1
The radius of the circle in which a place at latitude 60 moves in a year,
is r = (6.4 × 106 m)cos 60 = 3.2 × 106 m,
The time for one revolution, T = 1 year
= (365 days) × (24 h day−1) × (3600 s h−1)
= 3.15 × 107 s
 the rotational speed of such a point
ʋ=
2𝜋𝑟
𝑇
= (2π × 3.2 × 106 m)  (3.15 × 107 s)
= 0.64 m s−1
2
The angle subtended by the arc about 26 mm above the Pole Star is approximately 77. As a fraction of a
day this represents (77/360) × 24 hours = 5.1 hours, so that the exposure was about five hours.
3
50 r.p.m. means 50 revolutions per minute.
 50 r.p.m.  (2π × 50 rad)/60 s = 5.2 rad s−1
4
27.3 days  (27.3 days) × (24 h day−1) × (3600 s h−1) = 2.36 × 106 s.
ω=
2𝜋
𝑇
=
2𝜋𝑟𝑎𝑑
2.36𝑥106 𝑠
= 2.66 × 10−6 rad s−1 or more than 13 per day.
Alternatively, if the moon completes a 360 circuit of the Earth in 27.3 days, it must move 360/27.3 = 13.2
per day.
5
For all solid rotating bodies, ʋ = rω. If ʋ is to be kept constant when r is halved, ω must be doubled.
(Incidentally, the ω of a CD has to change continuously as the disc rotates.)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
19 Motion in a circle
Answers to Test yourself questions
Page 345 Test yourself
6
7
Newton’s second law states that:
a)
The rate of change of momentum of a body is equal to the resultant force exerted on the body.
b)
Fres = dp = d (mv)
dt
dt
Here ʋ2/r  9.8 m s−2  r > (18 m s−1)2/9.8 m s−2 = 33 m, so the tightest corner it can turn at this speed has a
radius of about 33 m.
8
Newton’s third law states: When body A exerts a force on body B, body B exerts a force of the same size
but in the opposite direction on body A.
9
ω = 2π/T
10
% difference =
11
The experiment assumes that (i) the string to the bung is always horizontal and (ii) that there is zero friction
(0.51−0.49) 𝑁
0.50 𝑁
x 100% = 4%
between the string and the glass tube. In addition to these problems, the measurements made - 10T and r are difficult to measure precisely.
12
There are two forces: the (forward) pull of the towing rope on the water-skier and the (backward) push of
the water on the water-skier. (These forces are equal in size.)
13
a)
For the drum:
ω=
2𝜋𝑥790
60
rad s−1
Speed of rim:
ʋ = rω = 0.27 m ×
2𝜋𝑥790
60
rad s−1
= 22 m s−1 (to 2 SF)
b)
Acceleration of the sock
a=
c)
𝑣2
𝑟
≈
(22𝑚𝑠 −1 )2
0.27𝑚
= 1800 m s−1
The sock gets this centripetal acceleration from the force of the drum wall pushing it inwards
towards the centre of rotation.
14
The cyclist has a centripetal acceleration given by:
v 2 (20ms −1 ) 2
=
r
25m
15
= 16 m s−2
The force towards the centre provided by the horizontal component of the weight is
Fsin30 = 1800 N × sin30 = 900 N.
The required centripetal force is
mv 2 97kg  (20ms −1 ) 2
=
r
25m
= 1552 N
Therefore the force F is insufficient to provide the centripetal acceleration.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
19 Motion in a circle
Answers to Test yourself questions
Page 346 Test yourself
16
A body appears to be ‘weightless’ when the only force acting on it is the gravitational pull of a planet or
moon on the body.
17
After a brief moment of free fall she will open her parachute and ‘float’ down to Earth under the action of
three forces that are in equilibrium – her weight downwards and the parachute ‘drag’, plus a small
upthrust, upwards.
18
Because we are not aware of the pull of the Earth, we are not conscious of the gravitational force, on us, so
we ’appear’ to be weightless.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
19 Motion in a circle
Answers to Exam practice questions
Pages 348–350 Exam practice questions
1
Answer C
[Total 1 mark]
2
Answer A
[Total 1 mark]
3
Answer B
[Total 1 mark]
4
At an average angular velocity of 8.2 rad s−1, it would take t = 2π rad/8.2 rad s−1 = 0.77 s to complete one
revolution
[2]
 It takes 5 × 0.77 s = 3.8 s to complete 5 revolutions
[1]
[Total 3 marks]
5
The centrifugal force is ‘the pull of the man on the post (which is part of the bus)’.
[2]
It is equal in size but in the opposite sense to ‘the pull of the post on the man’.
[1]
[Total 3 marks]
6
The acceleration a = rω2 = 0.12 m × (200 rad s−1)2
[1]
= 4800 m s−2
[1]
≈ 490g
[1]
[Total 3 marks]
7
Answer B
8
a)
[Total 1 mark]
[2]
b)
Ignoring forces in the direction of motion of the ‘car’
Newton’s second law gives:
⇒
F=
𝑚𝑣 2
𝑟
𝑚𝑣 2
𝑟
= F − mg
[1]
+ mg = 1800 kg × [(11 m s−1)2/8.5 m + 9.8 m s−2]
= 1800 kg × [14.2 + 9.8] m s−2 = 43 000 N or 43 kN
[1]
[1]
[Total 5 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
19 Motion in a circle
9
Answers to Exam practice questions
The centripetal force Fc is a component of the horizontal frictional force of the road on the car tyres.
[2]
This force Fc is towards the centre of the circle in which the car is turning. [1]
[Total 3 marks]
(The other component of the frictional force of the road on the car is F f . This force drives the car forward
around the bend – against air resistance etc.)
10
The bob of the pendulum, of mass m, is in equilibrium vertically.
Therefore applying Newton’s first law:
mg = Tcosθ
[1]
2
Horizontally the pendulum bob is accelerating towards the centre of the circle at ʋ /r.
[1]
Therefore applying Newton’s second law:
𝑚𝑣 2
= Tsinθ
𝑟
[1]
Dividing these equations yields:
𝑚𝑣 2
𝑚𝑔𝑟
⇒
=
𝑣2
𝑔𝑟
𝑇sin𝜃
[1]
𝑇cos𝜃
= tanθ
[1]
[Total 5 marks]
11
a)
2
A person at the centre of the ‘tyre’ experiences a centripetal acceleration rω .
For rω2 to equal g, ω2 must equal g/r
9.8 m s−2
ω=√
T=
2𝜋
𝜔
[1]
= 0.495 s−1
40 m
[1]
2π
=
[1]
0.495 s−1
= 12.9 s
 The space station must rotate once every 13 s!
b)
[1]
Some advantages might be:
• with careful design, eating and defecating might be made easier for the astronauts
• the need to exercise in order to reduce muscle and bone damage might be reduced, etc.
[2]
(There is no particular answer to this question – you are simply asked to ‘suggest’.)
[Total 6 marks]
12
a)
At the top of the circle in which the bucket of water is whirled, the water must not drop out
of the bucket. Therefore the pull of the Earth on the water must be too small to accelerate
the water downwards when the bucket of water is at the top of its circle, i.e. above the
teacher.
[1]
Therefore, at the top of the circle g must be less than ʋ2/r
[1]
If g <
𝑣2
𝑟
⇒
v > √𝑔𝑟
[1]
 ʋ must be greater than √(9.8 m s−2 ) × (1.2 m) = 3.4 m s−1
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
19 Motion in a circle
b)
Answers to Exam practice questions
One problem might be that it is difficult to whirl a heavy bucket of water at a constant speed in a
vertical circle.
[1]
[Total 5 marks]
13
Suppose the speed of the metal ball of mass m at the bottom of its swing is ʋ. By the principle
of conservation of energy ½mʋ2 = mgΔh, where Δh is the vertical distance the ball has
dropped. [1]
ʋ = √2𝑔𝛥ℎ = √2 × 9.8 m s−2 × 2.5 m = 7.0 m s−1
[1]
By Newton’s second law, at the bottom of the swing, the tension T in the cable is given by:
mʋ2/r = T − mg where r is the radius of the arc = 6 m
[1]
T = mʋ2/r + mg = m(ʋ2/r + g) = 2600 kg (7.02/6 + 9.8) m s−2
[1]
= 47 000 N (to 2 SF)
[1]
[Total 5 marks]
14
a)
The girl rotates once around the Earth’s axis each day, so T = 24 h × 3600 s h
−1
[1]
Her centripetal acceleration
a = rω2 = 4π2r/T2 (as ω = 2π/T)
=
4π2 × 6.4 × 106 m
(24 × 3600 s)2
[1]
= 0.034 m s−2
b)
[1]
To produce this acceleration for a girl of mass 56 kg requires a force F where
F = ma = 56 kg × 0.034 m s−2
[1]
= 1.9 N or about 2 N
[1]
[Total 5 marks]
15
‘Suggest’ means that you do not know the answer, but are being asked to think. Some of the points you
could have made might include:
•
The Earth is not a solid body.
•
Much of the Earth is a sticky liquid.
•
As the whole Earth rotates once every 24 hours (and used to rotate faster) greater centripetal
forces are required to pull in the outer parts of the Earth near the equator than at the poles.
16
•
Because of this the radius of the Earth near the equator is bigger than its radius near the poles.
•
Hence the Earth is oblate and hence flattened at the poles
[Total 3 marks]
The acceleration of the sock is rω2 i.e. just more than
0.25 m × (2π × 13 rev s−1)2 = 1700 m s−2 or about 170g.
[1]
It gets this acceleration from the push of the drum wall.
[1]
When the water is opposite one of holes in the drum, it cannot get this force.
[1]
So will try and continue in a straight line, according to Newton’s first law, and will pass through
the hole and out of the drum.
[1]
[Total 4 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
19 Motion in a circle
17
Answers to Exam practice questions
If the aeroplane banks to right or left, the upward push of the air on the wings can be resolved into vertical
and horizontal components.
[1]
The horizontal component provides the necessary centripetal force to enable the plane to turn in a circular
path.
[1]
For the plane to still fly horizontally, the vertical component must provides an upward lift equal to the pull
of the Earth on the plane (its weight), which continues to act vertically downwards. [2]
[Total 4 marks]
(In practice, as there is now only a component of the lift acting vertically upwards, the plane must increase
its speed in order to increase the lift so that the vertical component equals the weight of the aircraft).
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
Answers to Test yourself questions
Page 351 Test yourself on prior knowledge
1
W = mg = 42 kg × 9.8 N kg−1
= 410 N
2
½mʋ2 = mgΔh
 ʋ = √(2gΔh)
3
Because F = ma
4
Speed is a scalar and velocity is a vector
5
As KE = ½mʋ2 then, using m ≈ 80 kg and ʋ ≈ 7 m s−1, KE ≈ 2000 J
6
Chemical energy
7
ΔGPE = mgΔh = 14 kg × 9.8 N kg−1 × 0.65 m = 89 J
8
The velocity ʋ at t = 1.50 s is the gradient of a tangent to the graph line at t = 1.60 s.
v ≈ 18 m s−1.
Page 354 Test yourself
1
Baby’s mass = 35 N/9.8 N kg−1
= 3.6 kg
= 3.6 kg × 2.2 lb kg−1
= 7.9 lb
2
2.2 lb  1 kg which weighs 9.8 N
 1.0 lb will weigh 9.8 N/2.2 or about 4.45 N
3
The measured value of the free fall acceleration varies from 9.83 m s−2 at the poles to 9.78 m s−2 at the
equator.
The percentage difference =
4
(9.8 − 9.78)ms−2
09.805ms−2 100%
= 0.5%
a)
135 kg would weigh nearly 190 N (135 kg × 1.4 N kg−1) on the Moon’s surface.
b)
To lose 380 J would therefore involve a vertical move of 380J/190 N = 2.0 m
5
The feather will fall at 1.4 m s⌐2, because there is no air on the Moon’s surface.
6
The car and driver rise a vertical distance of Δh = 3 × 2.8 m = 8.4 m.
So their gain in GPE = mgΔh
= 1800 kg × 9.8 m s⌐2 × 8.4 m
= 1.48 × 105 J, i.e. about 0.15 MJ
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
Answers to Test yourself questions
Page 358 Test yourself
7
There are very high mountains on the Moon, and the Moon may not be exactly spherical. Both of these
would make for variations in the local value of g, possible up to 1%. Variations in the density of the
underlying rocks could also have an effect.
Centripetal acceleration is ʋ2/r. Here we have:
8
v=
2 πr
2 π  (384  10 6 m)
=
T
(27.3  24  36000 s)
= 1023 m s⌐1
 The centripetal acceleration is
v 2 (1023 m s −1 ) 2
=
r
384  10 6 m
= 2.73 × 10-3 m s-2
9
Newton’s law is F = GmEm/r2, so putting in the numbers gives
10
F=
(6.67  10 −11 Nm 2 kg −2 )(6.0  10 24 kg )(0.10kg )
(6.4  10 6 m) 2
= 9.8 N
11
The astronaut is moving very rapidly in a circle at a constant height above the Earth’s surface. He is pulled
towards the surface of the Earth all the time by a force that on Earth he would call his weight. (This is the
pull of gravity.) But he is moving so quickly that he keeps ‘falling over the horizon’, i.e. the rate at which he
falls is equal to the rate at which the Earth curves. So he is falling, but moving too quickly to ‘fall’ to Earth.
Page 362 Test yourself
12
a)
Make a list of values of r and g from the graph and calculate values of gr2.
b)
Ignoring the minus sign, values of r/106 m, g/N kg−1 and gr2/N m kg−1 are:
6
r/10 m
−1
g/N kg
2
−1
gr / N m kg
7.5
10.0
12.5
15.0
17.5
7.0
3.9
2.5
1.7
1.2
390
390
390
380
370
As gr2 is almost constant, the relationship g  1/r2 holds within the limits of reading from the graph.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
13
Answers to Test yourself questions
The relationship Vgrav = −GmE/rE means that the gravitational potential at the Earth’s surface Vgrav can be
calculated as −GmE/rE where G is the gravitational constant 6.67 × 10−11 N m2 kg−2 and the suffix E refers to
the Earth’s mass and radius. The minus sign results from the fact the zero of GPE is chosen to be zero at an
infinite distance from the Earth.
14
a)
At r = rE we have:
Vgrav = −GmE/r =
𝐺𝑚𝐸
𝑟
r/106 m
Vgrav/107 J kg−1
=
−4.0 ×1014 𝑁𝑚2 𝑘𝑔−1
6.4×106 𝑚
= − 6.25 × 107 etc…
6.4
8.0
10.0
15.0
20.0
−6.25
−5.00
−4.00
−2.67
−2.00
b)
15
The GPE needed is
Δ(GPE) = −GmEm(1/r − 1/rE) = − (4.0 × 1014 N m2 kg−1) × 11000 kg × (1/r − 1/rE)
Here we have
1
1
𝑟
𝑟𝐸
( −

)=
1
42 𝑥 106
−
1
6.4 𝑥 106
= −1.3 × 10−7 m−1
ΔGPE = − (4.0 × 1014 N m2 kg−1 × 11 000 kg) × − (1.3 × 10−7 m−1)
= 5.8 × 1011 J or 5800 GJ
16
Because the gravitational potential at a common point, at infinity, is the same for all planets or stars of any
mass m. If this were not the case, Δ(GPE) would be difficult to calculate, except in the region of one star or
planet.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
Answers to Exam practice questions
Pages 364–366 Exam practice questions
1
Answer C
2
Answer D
3
[Total 1 mark]
[Total 1 mark]
(GPE) = mgh = 450 kg × 9.8 N kg × 30 m
−1
a)
[1]
= 130 kJ
[1]
(GP) = mgh/m = gh = 9.8 N kg−1 × 30 m
b)
[1]
−1
= 290 J kg
[1]
[Total 4 marks]
4
Units of Gρr = (N m2 kg−2) × (kg m−3) × m
[2]
= N kg−1 or m s−2, either of which is the unit of g.
[1]
[Total 3 marks]
5
At the Earth’s surface:
gr =
𝐺𝑚E
𝑟E 2
=
𝐺𝑚E
(6.37× 106 m)2
[1]
At a height of 100 km above the Earth’s surface:
gr + 100 km =
𝐺𝑚E
(6.47× 106 m)2
[1]
𝑔𝑟
6.47 2
=(
) = 1.03
𝑔𝑟+100 km
6.37
So the Earth’s gravitational field reduces by only 3% in the first 100 km.
[1]
[Total 4 marks]
6
a)
−1 2
KE of stone at ground level = ½(3.0 kg)(50 m s ) = 3750 J
[1]
KE of stone at height h = ½(3.0 kg)(30 m s−1)2 = 1350 J
[1]
So the stone loses 2400 J in rising a height h in a uniform g-field
 3.0 kg × 9.8 N kg−1 × h = 2400 J
[1]
 h = 82 m
b)
[1]
No, the angle of projection does not enter the calculation.
[1]
This is because energy is a scalar quantity and has no direction associated with it.
[1]
[Total 6 marks]
7
a)
On Saturn’s equator, ω =
2π rad
3.7 × 104 s
= 1.7 × 10−4 rad s−1
 the centripetal acceleration = rω2 = (60 × 106 m)( 1.7 × 10−4 rad s−1)2
= 1.7 m s−2  1.7 N kg−1
b)
[1]
[1]
[1]
The gravitational field on the equator is 10.6 N kg−1
So the net force on a mass of 100 kg = 100 kg × (10.6 − 1.7) N kg−1
= 890 N
[1]
[1]
On Earth, a mass of 100 kg would experience a force of 980 N
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
Answers to Exam practice questions
890 N
 On Saturn the scales would read
980 N
× 100 kg = 91 kg
[1]
[Total 6 marks]
8
4
The volume of Venus = πr3 = 4/3π(6.0 × 106 m)3
[1]
3
= 9.05 × 1020 m3
[1]
 Mass of Venus mV = 9.05 × 10 m × 5200 kg m = 4.7 × 10 kg
20
At its surface g =
=
3
−3
24
[1]
𝐺𝑚V
𝑟2
(6.7 × 10−11 N m2 kg−2 )(4.7 × 1024 kg)
(6.0 × 106 m)
[1]
2
= 8.7 N kg−1
[1]
[Total 5 marks]
9
4
3𝑔
3
4𝜋𝜌𝑟
The ‘formula’ is g = πGρr, so Newton’s value for G would be
[1]
Assuming that he knew that the Earth’s radius was 6.4 × 10 6 m, he would have calculated a value for G of
G=
3 × (9.8 N kg−1 )
4π × (5000 kg m−3 ) × (6.4 × 106 m)
[2]
= 7.3 × 10−11 N m2 kg−2 (Not bad, though he would not have used these units!)
[1]
[Total 4 marks]
10
The centripetal acceleration of the bullet a =
But a is also equal to
𝐹
𝑚
=
=
𝐺𝑚M 𝑚
𝑟M 2 𝑚
=
𝑣2
𝑟
and the speed ʋ =
2𝜋𝑟
𝑇
𝐺𝑚M
[1]
𝑟M 2
(6.7 × 10−11 N m2 kg−2 ) × (7.3 × 1022 kg)
(1.7 × 106 m)
2
a = 1.7 N kg−1 or 1.7 m s−2
From a =
and T =
𝑣2
𝑟
2π𝑟
𝑣
[1]
 ʋ = √𝑎𝑟 = √(1.7 m s−2 ) × (1.7 × 106 m) = 1.7 × 103 m s−1
=
[1]
2π × 1.7 × 106 m
[1]
[1]
1.7 × 103 m s−1
= 6300 s = 105 minutes
[1]
[Total 6 marks]
11
a)
Applying
mass × centripetal acceleration = resultant force

b)
𝑚𝑣 2
𝐺𝑚𝑚E
=
(𝑟E +ℎ) (𝑟E +ℎ)2
[2]
From the above
υ2 =
𝐺𝑚E
(𝑟E +ℎ)
[1]
So as h increases, (rE + h) increases and υ decreases.
c)
[1]
[1]
The mass m of the satellite cancels out in step b), so it does not affect υ.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
Answers to Exam practice questions
All satellites will therefore have the same speed and the same centripetal acceleration at this
height h.
[2]
[Total 7 marks]
12
a)
As the net pull is zero, the gravitational fields of the Earth and the Moon must be of equal
magnitudes but act in opposite directions on the mass m, i.e.
𝑔𝐸 = 𝑔𝑀
[1]
so that
𝐺𝑚E
𝑅2
=
𝐺𝑚M
[1]
𝑟2
 R2 =
 R=(
𝑚E
𝑚M
× r2
[1]
𝑚E ½
) r
𝑚M
which is the required expression.
b)
[1]
From the above
R =√
𝑚E
𝑚E
×r=√
598 × 1022 kg
7.3 × 1022 kg
× r = 9.05r
[2]
But we are told that
R + r = 3.8 × 108 m
Substituting R = 9.05r
13
9.05r + r = 3.8 × 108 m  r = 0.38 × 108 m
[1]
 R = 3.42 × 108 m = 3.4 × 108 m
[1]
Infinity is a sensible choice for the zero of gravitational potential because all stars (or planets)
will then have the same value V = 0 at r = ∞.
[2]
Any other value, such as the surface of the Earth, will make it very complicated to calculate
ΔV between two places in our galaxy or in the solar system.
[1]
[Total 3 marks]
14
a)
i) Vg = −
𝐺𝑚E
𝑟
=−
(6.67 × 10−11 N m2 kg−2 ) × (5.98 × 1024 kg)
7500 × 103 m
= − 5.32 × 107 J
ii) Vg = −
𝐺𝑚E
𝑟
=−
[1]
[1]
(6.67 × 10−11 N m2 kg−2 ) × (5.98 × 1024 kg)
8000 × 103 m
[1]
= − 4.99 × 107 J
b)
ΔVg = (5.32 − 4.99) × 107 J kg−1
[1]
Δ(GPE) = ΔVg multiplied by the mass of the satellite (11 tonnes)
= (5.32 − 4.99) × 107 J kg−1 × 11 × 103 kg
10
= 3.63 × 10 J or about 36 GJ!
[1]
[1]
[Total 6 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
15
Answers to Exam practice questions
Suppose the mass of the object is m.
Its gravitational potential energy on the surface of the Moon is
GPE = −
𝐺𝑚𝑚M
[1]
𝑟M
In order to escape from the Moon its GPE must rise to zero, i.e. r  ∞
If its initial kinetic energy is ½mʋe2, then
𝐺𝑚𝑚M
½mʋe2 =
[1]
𝑟M
2𝐺𝑚M √2 × (6.67 × 10
=
𝑟M
 ʋe = √
−11
N m2 kg−2 ) × (7.34 × 1022 kg)
[1]
1.64 × 106 m
= 2.4 km s−1
[1]
[Total 4 marks]
16
The values of ln(T/days) and ln(r/103 km) are (to 3 SF):
a)
ln(T/days)
ln(r/103 km)
Io
0.57
6.05
Europa
1.30
6.51
Ganymede
1.97
6.98
Callisto
2.82
7.54
[1]
Your graph should look like the following figure.
b)
[4]
The gradient of this graph is
3.00
7.64 −5.68
= 1.53 ≈ 3/2
[2]
The linear graph suggests that T = krn, where n = 3/2
Therefore the data agree closely with the relationship T2  r3.
[1]
[Total 8 marks]
17
Our equation is
Vg = −
𝐺𝑚
𝑟
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
20 Universal gravitation
Answers to Exam practice questions
Differentiating both sides gives
d𝑉g
d𝑟
=
𝐺𝑚
[1]
𝑟2
This means that the gradient of the gravitational potential is equal to the gravitational field strength:
d𝑉g
d𝑟
=g
[1]
[1]
[Total 3 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Test yourself questions
Page 367 Test yourself on prior knowledge
1
640 N
2
ρ = Δp/gΔh
3
When a car uses its brakes
4
0.36 C
5
650 million joules
6
1.0 × 1011 electrons
7
The battery gives each coulomb of charge 6 joules of energy.
8
The ‘milli’ is wrong: the answer should be 400 V.
Page 370 Activity 21.2
1
Avoid contact with the copper sulphate solution and wash your hands if you touch it.
2
3
The gradient of the graph = 8.0 V/10 cm = 8.0 V/0.10 m
= 80 V m−1
4
The electric field between the plates: E = 80 N C−1.
Page 370 Test yourself
1
a)
Electron: mass 9.11 × 10−31 kg, charge minus 1.60 × 10−19 C
b)
Proton: mass 1.67 × 10−27 kg, charge plus 1.60 × 10−19 C
2
Magnetic, gravitational, electric.
3
Gravitational force on an electron in the Earth’s field:
FG = (9.11 × 10−31 kg) × 9.8 N kg−1 = 8.9 × 10−30 N downwards
Electric force on an electron in the Earth’s field:
FE = (1.60 × 10−19 C) × 120 N kg−1 = 1.9 × 10−17 N upwards
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
4
Answers to Test yourself questions
R = V/I
= 5000 V/0.005 A
= 1 000 000 Ω or 1 MΩ
5
N m/A s  J/C  V, so the unit N m/A s is equivalent to a volt.
6
E = 3000 N C−1 is the same as an electric field of 3000 V m−1.
From E = V/d  V = Ed =3000 V m−1 × 20 m = 60 000 V
 The p.d between places a distance 20 m apart on the lightning strike is 60 kV.
7
a)
Moving // to the electric field lines, the work done
= (3.6 × 10−3 C) × (300 N C−1) × 40 m
= 43 N m = 43 J
b)
Moving ⊥ to the electric field lines, i.e. along equipotentials,
the work done = zero.
8
Rubbing two insulating surfaces together can separate positive and negative charges, as can connecting
two conducting surfaces to a cell or battery of cells.
Page 374 Test yourself
9
Zero, because the two forces are equal in size but opposite in direction.
10
The dipole moment = Qdsinθ, where θ is the angle between the electric field and the dipole and d is the
distance between the dipole charges.
11
The leaves on whatever crop is being sprayed become positively charged by induction, as the plant
conducts electrons away from its leaves and stalks into the ground, leaving them positively charged. As
unlike charges attract, the leaves, top and bottom, attract the negatively charged spray.
12
When the balloon is rubbed on a nylon shirt, the balloon becomes negatively charged. The wall is a (very)
poor conductor, but when the charged balloon is placed on it, electrons in the wall migrate away from the
balloon (like charges repel), leaving the wall positively charged. As unlike charges attract, the balloon sticks
to the wall. (It probably won’t stick to the wall for long, as the charges tend gradually to re-combine!)
13
Q
2
r
As E = (constant) ×
, the constant, sometimes written as k and sometimes as
1
4 0 has units
14
NC−1  m2
C
 N m2 C−2.
The electric field between the plates is given by:
E = V/d
= 200 V/8.0 × 10−3 m
= 25 000 V m−1 (or 25 000 N C−1) downwards.
The gravitational force on the oil drop = (8.0 × 10−16 kg) × 9.8 N kg−1
= 7.8 × 10−15 N.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Test yourself questions
This must be balanced by an equal upward electric force, which means that the oil drop must have a
negative charge −Q.
Equating the upward electrical force to the downward gravitational force:
(25 000 N C−1) × −Q = 7.8 × 10−15 N  Q = −3.1 × 10−19 C
This charge is very close to −3.2 × 10−19 C which is the charge on two electrons, 2e.
Page 380 Test yourself
15
The unit of k or 1/4π0 is N m2 C−2. In base SI units this is
a)
(kg m s−2) × m2 × (A s)−2 = kg m s−2 m2 A−2 s−2 = kg m3 s−4 A−2
The unit of 0 is the inverse of the unit for k, as in the relationship k = 1/4π0 neither 4 nor π
b)
have SI units.
 The unit of 0 is (N m2 C−2)−1 = C2 m−2 N−1 = (A s)2 × m−2 × (kg m s−2)−1 = kg−1 m−3 s4 A2
16
Using Coulomb’s law, the force is kQ1Q2/r2 or Q1Q2/4π0r2
F=
(9.0  10 N m2 C−2 )(8.0−12 C)(12  10−12 C)
(0.010 m)2
= 8.6 × 10−9 N
17
The dry toast contains relatively few water molecules. It is these molecules that absorb energy from the
microwaves and would increase the internal energy of the toast and hence raise its temperature. In their
absence the toast does not warm up noticeably.
18
−19
kQ (9.0  109 N m 2 C−2 )(1.6 10 C)
V= r =
(1.0  10 −10 m)
= 14 J C−1 or 14 V to 2 SF
19
The difference in potential between a distance r1 and r2 from a charge Q is
ΔV = -kQ(1/r2 – 1/r1)
Here r1 is 1.0 × 10−10 m and r2 is , so kQ/r2 is zero.
ΔV = +14 J C−1 (from Question 18)
And the work done QΔV = (1.6 × 10−19 C) × (14 J C−1)
= 2.2 × 10−18 J to 2 SF
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Exam practice questions
Pages 382–385 Exam practice questions
1
Answer B
[Total 1 mark]
2
Answer B
[Total 1 mark]
3
Answer D
[Total 1 mark]
4
F = mg means ‘there is a gravitational force F on a mass m placed in a gravitational field of strength g’. The
force F is parallel to the field as m is a scalar quantity.
[3]
F = QE means ‘there is an electrical force F on a charge Q placed in an electrical field of strength E’. The force
F is parallel to the field as Q is a scalar quantity.
[2]
[Total 5 marks]
5
a)
Both F/Q and V/d represent the size of an electric field.
b)
F/Q has units N/C = N C
−1
[1]
[1]
V/d has units V/m  (J C −1) m−1
[1]
 N m C−1 m−1 = N C−1
[1]
So they are equivalent
A field of 3.0 × 106 N C−1  3.0 × 106 V m−1
c)
= 3000 V mm−1
[1]
[1]
[Total 6 marks]
6
Coulomb’s law gives us
F=
𝑄1 𝑄2
4𝜋𝜀0 𝑟 2
=
𝑘𝑄1 𝑄2
[1]
𝑟2
For a hydrogen atom Q1 = Q2 = e = 1.6 × 10−19 C
So F =
(9.0 × 109 N m2 C−2 ) × (1.6 × 10−19 C)
(5.3 × 10−11 m)
[1]
2
[1]
2
= 8.2 × 10−8 N
[1]
[Total 4 marks]
7
Force on molecule = QE = (1.6 × 10−19 C) × (360 N C−1)
= 5.76 × 10

Acceleration =
𝐹
𝑚
=
−17
N
[1]
[1]
5.76 × 10−17 N
[1]
4.0 × 10−26 kg
= 1.4 × 109 m s−2
[1]
[Total 4 marks]
8
The negatively charged rod repels negatively charged electrons from the can into the (conducting) ground.[1]
The can therefore lacks electrons and becomes positively charged.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Exam practice questions
The electric field between the rod and the can will be as shown above.
[3]
[Total 5 marks]
9
a)
If the graph follows an inverse square law, then E  1/r , and Er will be constant.
2
2
[1]
We therefore need to calculate some values for Er2:
11
E/10 N C
−1
[3]
2.0
4.0
6.0
8.0
r/10−11 m
8.5
6.0
4.9
4.3
Er2/10−9 N m2
C−1
1.45
1.44
1.44
1.48
10.0
3.8
1.44
As you can see, the values of Er2 are roughly constant: therefore E is related to r by an
inverse-square law.
b)
As E =
𝑄
4𝜋𝜀0
𝑟2
=
𝑘𝑄
𝑟2
[1]
Q=
𝐸𝑟 2
[1]
𝑘
From the table, average value of Er2 is 1.45 × 10−9 N m2 C−1.
 Q=
[1]
1.45 × 10−9 N m2 C−1
[1]
9.0 × 109 N m2 C−2
= 1.6 × 10−19 C
[1]
[Total 9 marks]
Tip: As the field is due to an isolated proton, you would expect to get a value for the charge to be
that of a proton, i.e.1.6 × 10−19 C. If you don’t get this value, you should go back and check your
maths!
10 a)
b)
Ionisation occurs when one or more electrons have been removed from an atom, leaving the
atom positively charged.
[2]
The spark will occur when the air ionizes.
[1]
From E = V/d  V = Ed
[1]
The minimum p.d. needed across the gap is thus
V = (3 × 106 V m−1) × (0.67 × 10−3 m)
= 2000 V or 2 kV
[1]
[1]
[Total 6 marks]
11 a)
The resultant force is
F = mg (down) – eE (up)
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Exam practice questions
= (2.0  10−18 kg)(9.8 N kg−1) − (1.6 × 10−19 C)(150 N C−1)
= 1.96 × 10
−17
N − 2.40 × 10
−17
[1]
N
= − 4.4 × 10−18 N
[1]
The negative sign means that the resultant force is upwards.
[1]
Tip: Remember that force is a vector quantity, so you must include its direction in your answer –
there will be a mark for this.
[Total 5 marks]
12 There are two ways of approaching this problem. In both, s represents the distance the proton will travel
before it comes momentarily to rest.
Method 1:
The charge on a proton is +e and its mass is mp.
 The force F slowing the proton down is F = eE
[1]
The deceleration a of the proton will be (by Newton’s second law)
a=
𝐹
𝑚
=
𝑒𝐸
[1]
𝑚p
Using ʋ2 = u2 + 2as and substituting ʋ = 0 gives
[1]
s = u2/2a (the minus signs cancel as a is a deceleration and so is negative)
Hence
s=
𝑢2
2𝑒𝐸/𝑚𝑝
=
𝑚p 𝑢2
[1]
2𝑒𝐸
Method 2:
The initial energy of the proton is ½mu2 and the final energy 0.
[1]
The charge on a proton is e, so the force F slowing the proton down is F = eE.
[1]
By conservation of energy
Work done slowing proton down = loss of KE of proton
[1]
𝑚 p 𝑢2
2𝑒𝐸
[1]
eE × s = ½mu2 − 0  s =
Thus both methods give the same formula for the distance and yield
s=
1.67 × 10−27 kg × (4.2 × 105 m s−1 )
2
[1]
2 × 1.6 × 10−19 C × 45 × 103 N C−1
= 0.020 m or 20 mm.
[1]
The writer prefers Method 2 as it ignores directions. Which do you prefer?
13 a)
[Total 6 marks]
The electric field between the plates is
E = V/x and is downwards.
[1]
 The drop, which has charge −Q and mass m, has an acceleration given by
a=
𝐹
𝑚
=
−𝑄 × 𝐸
=−
[1]
𝑚
𝑄𝑉
[1]
𝑚𝑥
The negative sign means that the acceleration will be upwards as the field is downwards and
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Exam practice questions
the droplet is negatively charged.
[1]
Tip: If you have time, it is worth checking the units when you get a formula such as this – try for
yourself!
b)
c)
A drop having zero charge (Q = 0) will cross the plates in a straight line.
[1]
This assumes that the downward gravitational force is negligible.
[1]
If different drops can be given different charges Q, then QV/mx will vary for different drops,
which will then emerge from the plates moving in slightly different directions. Between them,
two sets of plates at right angles to each other can steer different charged drops to different
places on the paper.
[3]
[Total 9 marks]
Tip: You need to look at how many marks are given for each part of a question such as this so
that you can allocate your time appropriately.
14 a)
E = 490 V/0.008 m
[1]
= 61 000 V m−1 or 61 000 N C−1.
[1]
b)
[2]
c)
Gravitational force:
mg = (4.0 × 10−15 kg) × 9.8 N kg−1= 3.9 × 10−14 N
[1]
Electric force:
QE = Q × 61 000 N C−1
[1]
If these two forces are equal, then
Q=
d)
3.9 × 10−14 N
61 000 N C−1
= 6.4 × 10−19 C
[1]
The charge Q is 4 × 1.6 × 10−19 C, i.e. is 4 × the electronic charge e.
[1]
This is in agreement with the concept that all charges consist of an integral number of electron
charges.
[1]
[Total 9 marks]
15 Method a) is preferable to method b) because
•
if a good straight line can drawn through the points
[1]
•
any ‘poor’ measurement can be ignored
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Exam practice questions
•
the remaining results can be ‘averaged’
•
it is more difficult, if using method b), to neglect a ‘poor’ measurement when averaging the values of
[1]
Er2.
[1]
16 From Coulomb’s law, the two forces on the charge QX at X are:
k × QX × (40 × 10−6 C)
(0.40 m)2
and
k × QX × Q
(0.20 m)2
to the right
[2]
to the left
[1]
The k and the QX both cancel and the other two parts of the forces must balance, so
Q
40 × 10−6 C
=
2
(0.20 m)
(0.40 m)2
[1]
Q = 10 × 10−6 C or 10 µC
[1]
[Total 5 marks]
17 a)
Quantised here means that all charges are multiples of a basic charge.
[1]
Millikan called the basic charge e, so here quantised means that all charges can be given a
value Ne, where N is a integer.
b)
[1]
That Millikan ignored over 20% of his results means either that the experiments leading to his
conclusions were very difficult to perform with negligible uncertainty, or that he was a charlatan
(a fraudster).
[2]
The former suggestion is the more likely, especially as subsequent experiments supported the
idea that charge is quantised.
[2]
[Total 6 marks]
18
a)
The two forces acting on the particle at P are:
•
an electric force Fe = QE acting horizontally to the right
•
a gravitational force Fg = mg acting vertically downwards
Fe =
𝑄𝑉
𝑑
=
[1]
(8.0 × 10−18 C) × 100 V
[1]
0.020 m
= 4.0 × 10−14 N
Fg = (6.0 × 10
−15
kg) × 9.8 N kg
[1]
−1
= 6.0 × 10−14 N
[1]
b) The resultant of these two forces, acting in a vacuum, will accelerate the tiny particle
•
to the right, downwards
[1]
•
at an angle of 56 (tan−1 6/4) to the horizontal
[2]
[Total 7 marks]
Tip: Always draw a diagram to help you work out vector problems such as this. Remember,
vectors have direction!
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
21 Electric fields
Answers to Exam practice questions
19 It looks as if father and son both made discoveries about electrons: one that electrons behaved
in a particle-like manner and one that electrons behaved in a wave-like manner.
[1]
GP Thomson’s discovery gave support to the hypothesis of wave-particle duality that had been
proposed by de Broglie.
[2]
Nowadays we believe in wave-particle duality: that electromagnetic waves consist of photons
(‘particles’) and that electrons can behave like ‘waves’, for example in an electron microscope.
[2]
[Total 5 marks]
(Apparently, father and son were never able to reconcile their differences of opinion!)
Tip: In questions such as this the Specification states that ‘marks will be awarded for your ability to
structure your answer logically showing how the points that you make are related or follow on from each
other where appropriate’.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Answers to Test yourself questions
Page 386 Test yourself on prior knowledge
1
6.25 × 1018 electrons
2
I is a scalar quantity because it has not got a direction
3
Joule per coulomb
4
11 V
5
1/R = 1/R1 + 1/R2
Page 390 Activity 22.2
a)
A graph of mgh (the gain of GPE in joules) against ½CV2 (the loss of EPE in joules), or simply a graph of h
against V2, could be plotted. If the graph is a straight line through the origin, then h  V2, or Δ(GPE) is
proportional to Δ(EPE).
b)
The values of mgh and ½CV2 will not be equal because the efficiency of the energy transfer is not 100%.
Some energy is inevitably lost to internal energy in heating the wires of the circuit and in working against
friction in the motor. Over a wide range of values, the efficiency may not be constant and so Δ(GPE) may
not be proportional to Δ(EPE).
Page 390 Test yourself
𝑄
= 1.7 × 10−6 F = 1.7 µF
𝑐=
2
The graph is a straight line through the origin with a positive slope.
3
The gradient (of the graph in the previous question) is the capacitance C of the capacitor, as Q  V i.e. Q =
𝑉
=
20×10−6
1
12𝑉
CV.
4
Q = 3.5 A × 600 s = 2100 C
5
Work or energy, as kg m2 s-2 = kg m s-2 × m = N m = J
6
EPE = ½CV2
= ½ × 47 × 10−6 F × (60 V)2
= 0.085 J or 85 mJ
1 2
𝑄
1
EPE =3
8
The series circuit will contain a 12 V battery, the capacitor C, a switch that can be closed and a resistor R of
𝑐
=× 2
(16×10−3 𝐶 )2
7
47×10−6 𝐹
= 2.7 J
large resistance.
Page 394 Test yourself
9
2200 µF 25 V, MADE IN UK, E 77 18 and the + and − signs
10
a)
Q = It = 30 × 10-3 A × 120 s = 3.6 C
b)
Energy given to charge by cell = QV = 3.6 C × 1.5 V = 5.4 J
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Answers to Test yourself questions
11
In a camera flash unit, under the keys of a computer keyboard.
12
From C =
13
Starting at t = 25 min and taking readings every 5 minutes, the ratios are:
𝑘𝐴
𝑑
Fk=
𝐶𝑑
𝐴
=
𝐹𝑚
𝑚2
= F m-1
19/10.5 = 1.8
10.5/5.8 = 1.8
5.8/3.0 = 1.9
3.0/1.5 = 2.0
So the graph is showing exponential growth within the accuracy of reading off from the graph.
14
n
1.00 10.0 2.00 2.72 7.39
100
log10 (n)
0
1.00 0.301 0.435 0.869 2.00
lne(n)
0
2.30 0.693 1.00 2.00
4.61
Page 399 Core Practical 11
1
During the first 6 seconds (approximately) the capacitor is being charged through the resistor and the p.d.
rises exponentially from zero to just under 6 V. The capacitor then discharges exponentially, becoming
almost completely discharged after about 10 seconds.
2
Charging: V = V0 (1 - e-tRC)
Discharging: V = V0 e-tRC
3
There are various ways of finding the time constant RC. We can use
Charging: CR = time (from 0) for V  0.632V0
Discharging: CR = time (after start of discharge) for V  0.368V0
Charging or discharging: CR = 0.693 t½
You should determine at least 2 values and take an average. You should then find that CR is approximately
0.72 ± 0.05 s, bearing in mind the difficulty in reading off values from the graphs.
For CR = 0.67 s  C = 0.67 s / R = 0.67 s / 6.9 × 103 Ω = 9.71 × 10-5 F
= 97 μF to 2 SF
For CR = 0.77 s  C = 0.77 s / R = 0.77 s / 6.9 × 10 Ω = 1.12 × 10-4 F
3
= 110 μF to 2 SF
The capacitance is therefore estimated to be in the range 97–110 μF
This suggests the capacitor is probably a 100-μF capacitor (this is a ‘preferred value’)
4
This value is only an estimate as it was difficult to read off the times with any accuracy from the graphs.
This could be improved by expanding, say, the charging part of the graph so that the time scale was larger,
thereby making the times easier to read off.
Page 400 Test yourself
15
Because at any moment the rate at which the charge discharges is proportional to the charge on the
capacitor, i.e. dQ/dt  Q – the current I  Q. In other words, ‘The more you have, the quicker you lose it!’
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Answers to Test yourself questions
16
2.00 000 to 6 SF
17
Method 1: look for equal half lives starting at a variety of places.Method 2: calculate ratios of the current at
equal time interval.
Method 3: produce values for t and logI (or ln I ) and check that the log (or ln) graph is a straight line.
18
a)
b)
I = 12 V/22 000 Ω = 0.55 mA
𝑡1
19
Time constant 𝑡 =
2
0.693
=
25 𝑚𝑠
0.693
= 0.036 s
From t = CR 
C=
t
0.036s
=
R 22 ´103 W
= 1.6 × 10-6 F or 1.6 µF
20
Area is about ½ × (120 × 10-6 C) × (100 × 10-3 s) ≈ 6 × 10-6 C s
The curve means it is actually slightly less than this, so we could make a guess of 5 × 10 -6 C s.
21
a)
By Ohm’s law, the p.d. across the resistor is proportional to the current in the resistor (V = IR).
Knowing the value of R enables the current to be determined from the p.d. The computer
software does can be programmed to do this and display the current. As the capacitor is in
series, this will also be the current in the capacitor.
b)
At t = 0, at the moment the switch is connected to contact A, there is no charge on the capacitor
and so no p.d. across it. The full supply voltage is across the resistor, so the current is a
maximum. As the capacitor begins to charge up, the p.d. across it increases and so the p.d. across
the resistor, and therefore the current, decrease. Both curves are exponential and ‘mirror’ each
other. When the capacitor is fully charged, the p.d. reaches a maximum and the current is zero –
no more charge is flowing.
After about 8 seconds, the switch is switched to contact B and the capacitor now starts to
discharge through the resistor. As the capacitor is discharging, the charge flow in the opposite
direction to that when charging and so the current appears negative on the graph. The current is
initially large when the capacitor is fully charged. Then, as the charge on the capacitor gets less,
so the p.d. across the capacitor falls exponentially, as does the current.
c)
i) At t = 0, the current is 3.2 mA and the p.d. across the capacitor is 0.
The p.d. across the resistor must therefore be the full supply voltage (nominally 3 V, actually
3.3 V from the graph). This gives:
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Answers to Test yourself questions
R = V/I = 3.3 V / 3.2 mA = 1.0 kΩ to 2 SF
ii) The charge on the capacitor is the area under the current-time graph.
Taking a triangle of best fit gives us:
Q ≈ ½ × 2.5 s × 3.2 mA = 4.0 mC
iii) From C = Q/V  C = 4.0 mC / 3.3 V = 1.2 mF or 1200 μF
Note that these are only estimates because the data cannot be read off accurately from the
scaled down graphs.
Check:
C can also be found by determining the time constant CR from the voltage charging graph:
CR = time for V to rise to 0.632 of the maximum voltage
= time to rise to 0.632 × 3.3 V = 2.1 V
≈ 1.2 s
As we have found R = 1.0 kΩ, this gives us C = 1200 μF as before.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Answers to Exam practice questions
Pages 403–406 Exam practice questions
1
Answer B
2
C=
𝑄
𝑉
=
[Total 1 mark]
1.3 × 10−6 C
[1]
60 V
= 2.2 × 10–8 F or 22 pF
[1]
[Total 2 marks]
3
Answer A
[Total 1 mark]
4
Answer D
[Total 1 mark]
5
a)
From W = ½CV2  V2 =
V =√
2𝑊
𝐶
2 × 4.0 J
2𝑊
=√
𝐶
220 × 10−3 F
[1]
= 6.0 V
b)
[1]
From Q = CV  Q = ± (220 × 10−3 F) × 6.0 V
[1]
= ± 1.3 C on each of the plates
[1]
(This means that one plate will have a charge of + 1.3 C and the other plate will have a charge of
– 1.3 C)
6
a)
[Total 4 marks]
VR = IR = (8.5 × 10 A) × (220 × 10 Ω)
−6
3
[1]
= 1.9 V
b)
[1]
VC = 6.0 V – VR = 6.0 V − 1.9 V
[1]
= 4.1 V
[1]
This assumes that the p.d. across the ammeter VµA is zero
c)
[1]
−6
[1]
= ± 1.9 × 10−3 C or ± 1900 μC
[1]
Q = CVC = ± (470 × 10 F) × 4.1 V
[Total 7 marks]
7
The statement is telling you that WP = WQ and that CP = 2CQ
Using W = ½CV2  Answer C
8
a)
[Total 1 mark]
The charge Q = (22 × 10−3 F) × 10 V = 0.22 C
[1]
The energy stored = ½QV = ½ × 0.22 C × 10 V
[1]
= 1.1 J
b)
[1]
i) When the second capacitor is connected the charge on each capacitor is 0.11 C
1
𝑄2
2
𝐶
The energy stored in each capacitor = ×
=
1
(0.11 C)2
× 22 × 10−3 F
2
= 0.275 J
For the two identical capacitors the total energy stored = 2 × 0.275 J = 0.55 J
[1]
[1]
[1]
[1]
ii) Half the original energy, ½ × 1.1 J = 0.55 J, has been dissipated as internal energy in the
leads conducting the charge from one capacitor to another when the two capacitors
were connected
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Answers to Exam practice questions
[Total 8 marks]
9
Answer B
10 From C =
[Total 1 mark]
𝑘𝐴
𝑑
d=
=
𝑘𝐴
𝐶
[1]
(2 × 10−11 F m−1 ) × (12 × 10−3 m)
2
[2]
4.7 × 10−12 F
= 6.1 × 10−4 m or about 0.6 mm
[1]
[Total 4 marks]
11 a)
b)
For a spring: F  x
[1]
For a capacitor V  Q
[1]
Energy stored in a capacitor = ½VQ
 Energy stored in a spring = ½Fx
[2]
[Total 4 marks]
Tip: When doing a question like this, it is worthwhile just checking the units for the proposed
formula. Does ½Fx have the units of energy, J?
12 Referring to the circuit diagram, Figure 22.17 on page 397, if R is doubled the initial current in the
resistor and microammeter will be halved; in this case it will be about 30 μA (half of 60 µA).
[1]
But the total charge that must flow from the capacitor as it discharges is unchanged.
[1]
This charge is represented by the area under the I–t graph, so in order to get the same area the
discharge must be slower and last longer.
[1]
This is shown in the figure below.
[2]
[Total 5 marks]
13 The total energy of the discharge, power × time, is about
(6 × 1016 W) × (~1 × 10−9 s) ≈ 6 × 107 J
[1]
The energy stored in a capacitor is ½CV2 so
½ × 0.26 F × V2 ≈ 6 × 107 J
[2]
 V ≈ 20 000 V or 20 kV
[1]
The answer is only an estimate as the time of discharge (‘about a nanosecond’) is not even given
to one proper significant figure.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Answers to Exam practice questions
[Total 5 marks]
14 a)
The p.d. across each lamp at the start of the capacitor discharge in each circuit is 3.0 V.
[1]
Hence the initial brightness and the length of the flash is the same for each lamp in circuit 1
and circuit 2.
b)
[1]
With a 9.0 V supply the discharge of the capacitor should be through three parallel sets of
three lamps in series, i.e. nine lamps in all, each lamp being exactly like those in the first two
circuits.
[3]
[Total 5 marks]
15 The concept of half-life is the same for charging as for discharging a capacitor.
[1]
 Using t½ = RC × ln2
[1]
34 s = (22 × 103 Ω) × C × 0.693
[1]
 C = 2.2(3) × 10−3 F or 2200 μF
[1]
[Total 4 marks]
16
a)
The potential difference across each of the capacitors is 12 V because they are connected in
parallel.
b)
[2]
From Q = CV the charge on the capacitors will be:
22 μF × 12 V = 264 μF on the 22 μF capacitor
47 μF × 12 V = 564 μF on the 47 μF capacitor
c)
[2]
So a single capacitor of capacitance C is needed where
C=
𝑄
𝑉
=
(264+564) μF
12 V
=
828 μF
12 V
= 69 μF
[1]
This is the sum of the two capacitors because, for two capacitors in parallel:
C = C1 + C2 = 22 μF + 47 μF = 69 μF
[1]
[Total 6 marks]
17 The following table shows successive readings at 5 hour intervals:
t/h
N/103
5
1.6
10
3.0
15
5.7
20
10.5
25
18.8
[2]
The ratios of successive values of N are:
1.6
3.0
5.7
10.5
3.0
5.7
10.5
19.0
[2]
These ratios: 0.53, 0.52, 0.54 and 0.55 are close enough to mean that the curve is showing exponential
growth.
[2]
(This is much easier than trying to draw a log-lin graph.)
18 The current I =
d𝑄
d𝑡
, so I is the gradient of the graph, i.e. a tangent to the graph.
Using a large triangle at t½ = 28 s:
Gradient = −
108 μC
66 s
= −1.6(4) μA
Using a large triangle at 2t½ = 56 s:
[Total 6 marks]
[1]
[1]
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
Gradient = −
78 μC
96 s
Answers to Exam practice questions
= − 0.81(3) μA
[1]
(The negative sign means that this is the discharge current).
The current,
d𝑄
d𝑡
, at 2t½ is about half that at t½, which is what we might expect because the p.d. across the
plates at 2t½ will be half that at t½.
[2]
[Total 7 marks]
19 Each 1-cm square is equivalent to a charge of 0.50 μA × 5 s, = 2.5 µC.
[1]
Therefore each 1-cm square is equivalent to a charge of 2.5 µC flowing off the capacitor.
[1]
There are approximately 9 such squares (adding bits together), giving
[1]
Total charge ≈ 9.0 × 2.5 μC = 22.5 µC
[1]
So the initial charge on the capacitor must have been between 22 and 23 µC.
[1]
[Total 5 marks]
20 A capacitor C can be connected to the 12 V car battery through a resistor R.
[1]
Opening the car door disconnects the battery and discharges the capacitor.
[1]
This sends a diminishing current through the lamp.
[1]
The circuit is shown in the figure below.
[2]
Values for R and C of R ≈ 220 kΩ and C ≈ 22 µF would give a half-life of just under five seconds, as RC = 4.8 s
with the above values.
[2]
[Total 7 marks]
(This is not the whole story as the car owner can sometimes vary the time constant, and there may be
problems about resetting the switch to its original position.)
21 a)
b)
Q = Q0e−t/RC so from t = 0 to t = RC, the charge diminishes from Q0 to Q0/e.
[1]
As e = 2.718 to 4 SF, then 1/e = 0.3679 to 4 SF = 0.37 to 2 SF
[1]
Every half-life diminishes the initial charge by a factor of e = 2.718.
[1]
To reduce charge to 1% or 1/100 of its initial value takes N half lives, where
1
 0.01
(2.718)N
[1]

N  4.60 i.e. 5 half-lives must elapse
[1]
Tip: You might use the x1/y facility on your calculator to get this result
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
22 Capacitance
c)
From the above, for
Answers to Exam practice questions
1
to equal 0.01 means that N = 4.60 (almost exactly)
(2.718)N
[1]
The time constant RC = 22 kΩ × 470 μF = 10.34 s.
[1]
Time to discharge to 1 % of initial charge = 4.60 × 10.34 s = 47.6 s
[1]
[Total 8 marks]
22 a)
Your graph should look the same as the figure below.
[3]
b)
As Q = Q0e−t/RC the gradient or slope
d𝑄
d𝑡
=−
1
𝑅𝐶
d𝑄
d𝑡
of the graph is given by
Q0 e−t/RC
[1]
At t = 0, e−t/RC = 1

d𝑄
d𝑡
=−
[1]
𝑄0
[1]
𝑅𝐶
Looking at the triangle in the figure above, the gradient is its height Q0 divided by the length of
its base. Thus the base must be of length RC.
[1]
[Total 7 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
Answers to Test yourself questions
Page 407 Test yourself on prior knowledge
1
W = mg =  m = W/g = 35.3 N/ 9.81 N kg-1 = 3.60 kg.
2
m = W/g.
3
The kinetic energy of the car is converted into internal energy of the brakes (which get hot!).
4
I = ΔQ/Δt = 0.48 × 10-3 C /0.40 s =1.2 mA.
5
A watt is a joule per second (W = J s-1)
6
(i) 3.0 mA means 3.0 × 10-3 amperes; (ii) 98 kJ means 98 × 103 joules.
7
…p.d involves converting energy from electrical energy while e.m.f. involves converting energy into
electrical energy.
8
A 6 V battery converts 6 J of energy to electrical energy for each coulomb of charge passing through it.
9
2.0 V/0.5 A should equal 4.0 Ω.
10
This is drawn on page XX of Book 1
11
The magnet turns the pin into a magnet with ‘unlike’ poles then attracting to pick up the pin.
Page 411 Activity 23.1
1
For example:
I/A
1.2
2.5
3.4
4.2
5.0
m/g
0.45
0.90
1.25
1.50
1.85
F/10-3 N
4.4
8.8
12.3
14.7
18.1
2
3
The slope of this graph =
𝐹
𝐼
=
21.5×10−3 𝑁
5.5𝐴
= 3.8 × 10−3 𝑁𝐴−1
Using F = B⊥Iℓ with ℓ = 0.045 m
B⊥ =
4
3.8×10−3 𝑁𝐴−1
0.045𝑚
= 0.09 N A-1 m-1 or 0.09 T
The region over which the magnetic field is uniform cannot be determined for certain – it gets less near
each end of the magnets - so the value of B⊥ can only be approximate, e.g. to ± 5 %, or here to 1 SF.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
Answers to Test yourself questions
Page 412 Test yourself
1
The diagram should look similar to the one below.
2
Our definition of current is
I =ΔQ/Δt  1 A  1 C/s,
 1 C  1 A × s, i.e. 1 C in base SI units is 1 A s
3
a)
Nearest to the poles, N or S, of both magnets as the field lines are closest together here.
b)
The B-field is uniform as the field lines are parallel and equally spaced.
4
The diagram should look similar to the one below.
5
In London BV = 50 µT × sin65° = 45 µT
Page 414 Test yourself
6
7
a)
In nAʋQ the symbol n represents the number of charged particles per unit volume.
b)
The units of nAʋQ are m-3 × m2 × m s-1 × C  C s-1 = A
We need to use:
Charge on α-particle = 2e = 2 × 1.6 × 10-19 C
Speed of light = 3.00 × 108 m s-1
Then F = Beʋ = (250 × 10-3 T)(3.2 × 10-19 C)(3.0 × 108 m s-1) = 2.4 × 10-11 N
8
The unit of B⊥Qʋ is (N A-1 m-1) × C × m s-1  N.
(As C s-1  A, the ‘A’s cancel, as do the ‘m’s)
9
a)
9.0 A h means 9.0 C s-1 for 3600 s, so the battery can deliver 32 400 C.
b)
In 1.0 s the bicycle and rider move 5.0 m along the road, and therefore rise 0.33 m.
The rate of rise is therefore 0.33 m/s
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
Answers to Test yourself questions
The work done per second mgΔh/Δt = 90 kg × 9.8 N kg−1 × 0.33 m s−1
= 294 N m s−1 ≈ 300 J s−1 (300 W).
c)
i) From P = VI, A 36 V battery working at 300 W will supply a current of
300𝑊
36𝑉
= 8.3 A or just over 8 A.
ii) The battery can deliver 32 400 C. This means it supplies a current of
8 A for 32 400 C/8 A = 4050 s, or about 68 minutes.
iii) In an hour the cyclist will have travelled 5 m s-1 × 68 × 60 s = 20 250 m
= 20.25 km
This about half the manufacturer’s quoted 40 km. However, the cyclist has been climbing a 1
in 15 hill for the whole of this time, which cannot be considered to be “average riding
conditions”.
10
The Hall effect predicts that there will be a p.d. between the sides of a current carrying wire when it is
placed across a magnetic field.
Page 417 Test yourself
11
The size of the force on a moving charged particle depends upon the value of the magnetic field that is
perpendicular to the particle’s velocity. In F = B⊥Qʋ both B and ʋ are vector quantities, therefore if B is
parallel to ʋ, the force F will be zero, and an electron of charge Q = -e will experience no force and so the
electron will move in a straight line.
12
T m2 s-1  N A-1 m-1 × m2 × s-1  N m/A s  J/C = V, the volt.
13
a)
F = B⊥Qʋ
= (2.0 N A-1 m-1)(3.2 × 10-19 C)(1.2 × 107 m s-1)
= 7.7 × 10-12 N
b)
This force will make each α-particle move in a circle as B is perpendicular to ʋ.
c)
Using Newton’s second law:
a = F/m
= 7.7 × 10-12 N/ 6.6 × 10-27 kg
= 1.2 × 1015 m s-2 towards the centre of the circle.
14
The unit of (energy) × (time) is J × s = N m s
The unit of (momentum) × (displacement) is kg m s−1 × m = N s × m = N m s
Therefore the two units are the same.
Page 422 Test yourself
15
The e.m.f. induced in the racket will depend on
e=
-df -AdB
=
dt
dt where A
is the area of the racket frame cutting
the field.
The induced current is given by
I=
e
R
, where R is the resistance of the metal frame.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
The induced current is therefore
I=
Answers to Test yourself questions
A dB
R
dt
(Despite saying that ‘no calculations are expected’, it does not mean that you should not give the
underlying physics principles in your answer.)
When serving, the racket moves in a vertical plane, whilst when playing a forehand drive, the racket moves
in a horizontal plane (more or less, depending whether any spin is put on the ball by moving the racket
head up on impact, imparting topspin, or cutting down on the ball – a slice). Therefore different amounts of
A dB/dt will result in different currents.
16
a)
The units of µ0nI are N A-2 × m-1 × A  N A-1 m-1  T the tesla.
b)
As B = µ0nI,
then 𝐼 =
𝐵
µ0𝑛
=
2.0𝑇
(4𝜋×10−7 𝑁𝐴−2 )(5×103 𝑚−1 )
= 320𝐴
Page 424 Test yourself
17
a)
V0 = √2 × 12 V = 17 V
b)
I0 = V0/R = 17 V/220 Ω
= 0.077 A or 77 mA
18
i)
The frequency f is the number of times per second (Hz) that the a.c. current waveform repeats,
i.e. the number of cycles per second.
ii)
The period T is the length if time (s) that each complete cycle occupies.
iii)
The peak value I0 of the a.c. is its maximum value (A) during each cycle.
iv)
The r.m.s. value I of the a.c, current is its effective value during each cycle and is equal to I0/2 .
a)
Two points where the current is instantaneously not changing are marked (a). There are, of
19
course, lots of others – at the top each peak and bottom of each trough.
b)
Two points where the current is zero are marked (b). Again, there are lots of others – wherever
the line cuts the axis.
c)
As the frequency is 50 Hz, the period T = 1/50 s = 0.02 s. Each cycle (wavelength) should
therefore occupy 0.02s.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
20
a)
Answers to Test yourself questions
In an ideal transformer ISVS = IPVP

Ip =
I sVs 2.0 mA  5.0 V
=
Vp
230 V
= 0.043 mA or 43 µA.
b)
The power transferred to the mobile is ISVS or IPVP = 10 mW.
c)
In five hours the energy transferred = 10 mW × 5 × 3600 s = 180 J.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
Answers to Exam practice questions
Pages 426–429 Exam practice questions
1
BH = (66 × 10−6 T) sin 25o
a)
[1]
= 28 µT
[1]
−6
b)
o
BV = (66 × 10 T) cos 25
= 60 µT
[1]
[Total 3 marks]
2
Answer A
[Total 1 mark]
3
Answer B
[Total 1 mark]
4
Answer A
[Total 1 mark]
5
Answer B
[Total 1 mark]
6
The transformer formula is IPVP = ISVS, where IPVP = 60 mW and Vs = 12 V
[1]
Rearranging:
IP =
60 mW
[1]
12 V
= 5.0 mA
[1]
[Total 3 marks]
7
Answer D
[Total 1 mark]
8
No calculation is possible, because the angle between the Earth’s magnetic field and the
direction of the cable is not given, i.e. you need to know that in BIℓsinθ, a value for θ
is not given.
[2]
You need to know the magnitude of the vertical component of the Earth’s magnetic field, i.e. the angle that
the Earth’s field makes with the vertical.
[1]
[Total 3 marks]
9
If a charge Q is travelling with speed v at an angle θ to a magnetic field B, it experiences a force
F = B sinθ Qv
[1]
The force will therefore be zero when θ = 0.
[1]
This means the charge must be travelling parallel to the magnetic field.
[1]
[Total 3 marks]
10
There is no force on conductor OQ as it is parallel to the magnetic field.
[1]
Force on OP is
F = B⊥Iℓ = (0.25 × 10−3 T) × 2.0 A × 0.50 m = 2.5 × 10−4 N
[1]
In calculating the force on OR, we need consider the length of OR that is perpendicular to the field. From
the diagram in the question we can see that this is 0.50 cm. Therefore:
F = B⊥Iℓ = (0.25 × 10−3 T) × 2.0 A × 0.50 m = 2.5 × 10−4 N
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
Answers to Exam practice questions
(This is exactly the same force that is exerted on OP because the component of OR that is perpendicular to
the field is effectively OP.)
Using the left hand rule, the forces on OP and OR are both vertically upwards.
[1]
[Total 5 marks]
11
Answer C
[Total 1 mark]
12
We need Ф = BAsinθ and A = πr = π × (8.0 × 10 m) = 2.0 × 10 m , then:
2
−3
2
−4
2
[1]
Maximum value of Ф is when sinθ = 1
a)
Фmax = (2.0 × 10−4 m2) × 1.5 T = 3.0 × 10−4 Wb
[1]
Minimum value of Ф is when sinθ = 0
Фmin = 0
ε=
b)
Δ𝛷
Δ𝑡
[1]
-4
=
3.0 × 10 Wb
[2]
0.30 s
= 1.0 × 10−3 V or 1.0 mV
[1]
[Total 6 marks]
(It could be argued that the minimum value of sinθ is −1, so the flux changes from + 3.0 × 10−4
Wb to − 3.0 × 10−4 Wb, making ΔΦ = 6.0 × 10−4 Wb and ε = 2.0 mV. This would mean rotating the
ring through 180°, which would be an unlikely scenario inside an MRI scanner!)
13
As the centripetal acceleration is a = ʋ2/r, only the radius of its circular path is needed.
a)
[2]
(Note that this does not need any knowledge of magnetic fields!)
b)
As the acceleration is always at right angles to its direction of motion, its speed does not change.
(Its velocity changes as it is continually changing direction as it moves round the circle.
[1]
[Total 3 marks]
14
The number of turns has doubled, but the area is only one quarter of the original.
Answer D
15
[Total 1 mark]
𝑉s
a)
𝑉p
=
𝑁s
𝑁p
 Np = Ns ×
= 200 turns ×
ISVS = IPVP  Is =
b)
𝑉p
240 V
12 V
𝐼p 𝑉p
𝑉𝑠
[1]
𝑉s
=
= 4000 turns
1.5 × 10−3 A × 240 V
12 V
= 0.030 A or 30 mA
[1]
[1]
[1]
[Total 4 marks]
16
Suppose a length ℓ of wire levitates, then BIℓ = µℓg so the ℓ cancels.
B=
𝜇𝑔
𝐼
=
80 × 10−3 kg m-1 × 9.8 N kg−1
[1]
[2]
5.6 A
= 0.14 N A−1 m−1 = 0.14 T
[1]
[Total 4 marks]
17
The answer here is to reword the caption to Figure 623 –‘whatever you try to do produces a
result that tries to stop you’ - in your own words.
[Total 3 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
18
We need to use F = B⊥Qʋ = (Bsinθ)Qv
a)
θ = 90°  F = (0.15 T)(3.2 × 10
−19
Answers to Exam practice questions
[1]
6
−1
C)(5.0 × 10 m s )
[1]
= 2.4 × 10−13 N
[1]
b)
θ = 45°  F = sin 45° × 2.4 × 10−13 N = 1.7 × 10−13 N
[1]
c)
θ = 30°  F = sin 30° × 2.4 × 10
[1]
−13
N = 1.2 × 10
−13
N
[Total 5 marks]
19
a)
Your argument should include the following points:
• There will be a sudden rise in the current IL in the left circuit as the switch is closed. This
current will quickly settle to a steady value.
• As IL rises from zero, a rising magnetic field B will be produced in the left coil.
• This B-field will link with the coil in the right circuit, but only as this field changes will a
current IR be induced in the right coil.
• Thus there will be a sudden rise in the current in the right coil, but this induced current will
quickly fall back to zero as the linking magnetic field settles to a steady value.
[Total 4 marks]
(In a question such as this, marks are awarded for indicative content and for how the answer is
structured and shows lines of reasoning. Broadly speaking, there would be one mark for each
valid point provided there is a well-structured argument.)
Tip: Answers to questions of this kind are often best laid out as a series of bullet points.
Examiners like this layout, as it often matches the mark scheme from which they are working.
20
Bℓʋ = (45 × 10−6 T) × 36 m × 240 m s−1
a)
= 0.39 T m2 s−1
[1]
From F = BIℓ  T  N A−1 m−1
b)
[1]
[1]
T m2 s−1  N A−1 m−1 m2 s−1

Nm
As

J
[1]
C
V
[1]
[Total 5 marks]
21
From ʋ = u + at, the downward speed of the pole after 1.3 s is
ʋ = gt = 9.8 m s−2 × 1.3 s
= 12.7 m s−1
[1]
[1]
Using Bℓʋ
The p.d. = (18 × 10−6 T) × 4.0 m × 12.7 m s−1
[1]
= 9.2 × 10−4 V or 0.92 mV
[1]
[Total 4 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
22
Answers to Exam practice questions
If the supply is sinusoidal we can write:
V = V0 sin(2πft)
[1]
From Q = CV
Q = CV0 sin(2πft)
[1]
Then
I=
d𝑄
d𝑡
= 2πf CV0 cos(2πft)
The maximum current is therefore:
I0 = 2πfCV0
[2]
= 2π(50 s−1)(0.47 × 10−6 F)(17 V)
[1]
= 2.5 × 10−3 A or 2.5 mA
[1]
[Total 6 marks]
23
Some of the points you might make are:
•
There are no toxic fumes from an electric car as it uses electric batteries and not petrol or diesel as
its energy source.
•
It is for this reason that such cars are thought to ‘produce no pollution’.
•
However, the electrical energy used to propel an electric car has been produced in power stations
that are less than 50% efficient.
•
Also, the batteries that the car uses need to be manufactured, which in itself produces pollution, as
would the many recharging points that need to be built and placed around countries that use
electric cars.
•
The main advantage of electric cars is that it does not pollute the atmosphere where it is being
driven, for example in towns.
•
A further advantage is that it is relatively quiet.
[Total 6 marks]
Tip: There is no ‘right’ answer to questions of this kind. What the examiner wants is to see is an argument,
preferably one that does not come down wholly on one side or the other.
24
Your discussion should include the following points:
•
‘Hybrid’ vehicles use petrol or diesel supplemented by energy from electric batteries.
•
The braking action in these cars converts the car’s kinetic energy into electrical energy. which is
stored as chemical energy in the car’s battery.
•
The d.c. motor that ‘drives’ the car spins in reverse as the car slows down. According to Faraday’s
law of electromagnetic induction the d.c. motor that drove the car now acts as a dynamo – it
generates electricity.
•
The rate of change of magnetic flux through the coils of what was the motor produces an induced
e.m.f. and hence an induced current that charges the car’s batteries.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
23 Magnetic fields
•
Answers to Exam practice questions
This process is making use of the car’s kinetic energy to slow it down, and is called regenerative
braking.
[Total 5 marks]
(As in question 19, marks are awarded for indicative content and for how the answer is structured and
shows lines of reasoning. Broadly speaking, there would be one mark for each valid point provided there is a
well-structured argument.)
25
Rough values for the strengths the Earth’s fields England are:
g ≈ 10 N kg−1, E ≈ 200 N C−1 and B ≈ 50 × 10−6 N A−1 m−1
[1]
The relevant relationships are:
Fg = mg, FE = eE and FB = Beʋ
The mass of a proton is about 1.7 × 10
[1]
−27
kg, the charge on a proton is +1.6 × 10
−19
C and the speed of light
is 3.0 × 108 m s−1.
 mg  1.7 × 10−26 N,
[1]
eE  3.2 × 10−17 N
and
Beʋ  2.4 × 10−16 N
[3]
[Total 6 marks]
Tip: You are reminded that within the context of Working as a Physicist, the Specification states that
students ‘should be able to estimate values for physical quantities and use their estimate to solve
problems’.
26
Faraday’s law of electromagnetic induction states that ε = −
d𝛷
[1]
d𝑡
Integrating both sides gives:
𝑡
𝛷
2
∫0 𝜀 dt = − ∫𝛷 d𝛷 = Φ1 – Φ2
[2]
1
The left side of this equation is the area between the induced e.m.f. and the time axis, which therefore
equals the change of magnetic flux in the experiment.
[1]
[Total 4 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
24 Electrons and nuclei
Answers to Test yourself questions
Page 430 Test yourself on prior knowledge
1
Z = 1 and Z = 2
2
The coulomb, C
3
1.6 × 10−19 C
4
1 A  1 C per s, i.e. 1 A = 1 C s−1
5
Because both consist on two protons and two neutrons.
6
1 J  1 N m, so a joule measures force times distance (energy).
7
Force is a vector quantity because forces have direction.
Page 431 Test yourself
1
Δm = 1.67493  10−27 kg −1.67262  10−27 kg
= 0.00231 × 10−27 kg
= 23.1 × 10−31 kg.
Δm is about 2.5me.
2
The carbon nucleus contains 6 protons and 6 (12 – 6) neutrons.
3
The number in the N box is the number of neutrons in the nucleus: here N = A − Z = 12.
4
You could have said radio, infra-red, light or ultra-violet waves as they are all part of the e-m spectrum.
5
X = 2 and Y = 2.
Page 434 Test yourself
6
Density ρ = m/V
 m = ρV = (1000 kg m−3) × (1200 m3)
 = 1.2 × 106 kg or 1200 tonnes.
7
A gold atom is ‘like’ a planetary system: the centre is a very tiny positively charged nucleus that contains
almost all the mass of the atom (rather like our Sun contains almost all the mass of the solar system).
Around the nucleus are lots of negatively charged electrons (rather like the planets of our solar system), but
a gold atom (like the solar system) is largely empty space.
8
9
a)
0.01 %  1 in 10 000 (a hundredth of a hundred)
b)
90   π/2 radians (as 2π radians are equal to 360)
Place a source of only α-particles a centimetre or so from a GM tube (a Geiger counter). The counter will
register many arriving α-particles. Place a single sheet of paper between the source and the GM tube and
the counter will register only background radiation, if anything.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
24 Electrons and nuclei
Answers to Test yourself questions
10
11
a)
Electric field: N C−1 – no name (Although we use Vm−1, this is not the defining unit)
b)
N A−1 m−1 or tesla, T.
Page 438 Test yourself
12
c = 3.00 × 108 m s−1.
13
1 eV is the energy gained by a particle of charge 1.6 × 10−19 C accelerated through 1 V. As 1 V  1 J C−1 then
1 eV  1 J C−1 × (1.6 × 10−19 C) = 1.6 × 10−19 J.
14
Be/m has units (N A−1 m−1) × C kg-1
As 1 N  kg m s−2 and 1 C  1 A s, then the unit of
𝐵𝑒
𝑚
is
kg m A s A-1 m-1 kg-1 = s.
15
10 MeV is the maximum energy because beyond this the protons begin to gain significant mass, which then
fails to synchronise (2πf =
16
𝐵𝑒
𝑚
) with the high frequency alternating p.d. between the Dees.
In both cyclotrons and linacs (i) charged particles move with constant energy when inside the Dees and
tubes respectively, (ii) the charged particles gain energy when moving between the Dees and tubes
respectively and (iii) in both devices the particles are accelerated by an electric field.
Page 442 Test yourself
17
Sub-atomic particles are usually detected by the ionisation they produce as they pass through other matter,
e.g. liquid hydrogen or photographic emulsion.
18
Sub-atomic particles with zero charge, e.g. neutrons, are not detected because they do not cause ionisation
in matter.
19
First, the electricity is not ‘used’ up – it is transferred to other forms of energy, e.g. light. Second:
2500 kW h  (2500 × 1000 J s−1) × 3600 s = 9.0 × 109 J or 9 GJ.
(This sounds an awful lot of energy – 9 billion joules! But, of course, the joule is a pretty small amount of
energy – about the work you would do lifting your textbook 15 cm!)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
24 Electrons and nuclei
Answers to Test yourself questions
20
1 J  1 N m = 1 kg m2 s−2. From the left hand side of the equation
J s2 m-2 = kg m2 s−2 s2 m-2, leaving only kg.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
24 Electrons and nuclei
Answers to Exam practice questions
Pages 445–448 Exam practice questions
1
Answer C
[Total 1 mark]
2
Answer B
[Total 1 mark]
3
Answer B
[Total 1 mark]
4
Answer C
[Total 1 mark]
5
Answer D
[Total 1 mark]
6
Answer A
[Total 1 mark]
7
Answer D
[Total 1 mark]
8
Answer B
[Total 1 mark]
9
a)
Answer D
[1]
b)
Answer A
[1]
[Total 2 marks]
10
Your diagram should look like the figure below, with appropriate labelling.
[Total 4 marks]
11
a)
EPE =
(9.0 × 109 N m2 C−2 ) × (2 × 1.6 ×10−19 C) ×(79 × 1.6 × 10−19 C)
[1]
5.0 × 10−14 m
= 7.3 × 10−13 J
b)
EPE =
[1]
7.3 × 10−13 J
1.6 × 10−19 J eV-1
= 4.6 × 106 eV = 4.6 MeV
[1]
From the Principle of Conservation of Energy EPE + KE = constant (here 4.6 MeV).
[1]
As the alpha-particle approaches the gold nucleus, the electric field of the nucleus does work
on the alpha- particle and slows it down. The KE of the alpha-particle is stored as PE in the
field. At the point of closest approach, all the KE of the α-particle will be stored as EPE and so
the α-particle will momentarily come to rest.
[1]
As the ‘collision’ is elastic, no KE is lost and so all the EPE is converted back to KE and the αparticle rebounds with the same KE as it had before the ‘collision’.
[1]
[Total 6 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
24 Electrons and nuclei
12
𝑟atom
The fraction is (
𝑟nucleus
)
Answers to Exam practice questions
3
[1]
3
10−10 m
=(
)
10−14 m
[1]
= 1012
[1]
[Total 3 marks]
13
The (inward) centripetal force of the proton on the electron is
F=
𝑘𝑒e 𝑒p
= m e rω 2
𝑟2
[1]
As ee = ep = e we have
merω2 =
f2 =
𝑘𝑒 2
 ω2 = (2πf)2 =
𝑟2
𝑘𝑒 2
4𝜋 2 𝑚𝑒 𝑟 3
 f=√

𝑘𝑒 2
𝑚𝑒 𝑟 3
(9.0 × 109 N m2 C−2 ) × (1.6 × 10−19 C)2
4π2 × (9.1 × 10−31 kg) × (5.3 × 10−11 m)3
f = 6.6 × 1015 Hz
[2]
[1]
Such a frequency lies in the ultraviolet part of the spectrum.
[1]
[Total 5 marks]
14
For speeds that are not too high, KE = E k = ½mʋ 2
Given that r =
𝑝
𝐵𝑒
 p = Ber = mʋ
[1]
From E k = ½mʋ 2  2mE k = (mʋ) 2 = (Ber) 2
This gives E k =
[1]
[1]
𝐵2 𝑒 2 𝑟 2
[1]
2𝑚
 E k  r 2 provided B, e and m are constant.
[1]
[Total 5 marks]
15
The maximum field between the tubes is
E max =
𝑉max
𝑑
=
200 × 103 V
[2]
25 × 10−3 m
= 8.0 × 10 6 V m −1 or 8.0 MN C −1
[1]
 The mean value of the electric field = ½ E max = 4.0 MN C −1
[1]
[Total 4 marks]
16
The protons move in circles. When the radius is r, their acceleration is ʋ2/r.
By Newton’s second law:
Centripetal force = Beʋ =
𝑚p 𝑣 2
[2]
𝑟
𝑣=
𝐵𝑒𝑟
𝑚p
But 2πr/ 𝑣 is the time T to complete one revolution at a speed 𝑣 is
T=
2𝜋𝑟
𝑣
 f=
1
𝑇
=
𝑣
2𝜋𝑟
=
𝐵𝑒𝑟
2𝜋𝑟𝑚p
=
𝐵𝑒
2𝜋𝑚p
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
24 Electrons and nuclei
Answers to Exam practice questions
[Total 4 marks]
17
The number of ionisations is equal to the area under the graph line, as interpreted by the axes.
[1]
For a particle that travels 60 mm in air, as on the graph, this is about 11 large squares.
[1]
−1
One square represents 10 mm × 2000 ionisations mm = 20 000 ionisations.
Thus 11 squares represents 11 × 20 000 ionisations = 220 000 ionisations
[1]
As each ionisation requires 30 eV, the initial energy E0 of the particle is
E0 = 30 eV per ionisation × 220 000 ionisations
[1]
= 6.6 × 106 eV or 6.6 MeV
[1]
[Total 5 marks]
18
Each particle has a mass of 6.65 × 10−27 kg (2 protons plus 2 neutrons).
Resolving vertically, the two momenta are therefore:
(6.65 × 10−27 kg) × (1.23 × 107 m s−1) × sin 35 = 4.7 × 10−20 kg m s−1
[2]
Down (6.65 × 10−27 kg) × (0.86 × 107 m s−1) × sin 55 = 4.7 × 10−20 kg m s−1
[2]
The momenta are equal to when rounded to 2 SF.
[1]
We have to assume that speeds are such that the rest masses still apply.
[1]
Up
[Total 6 marks]
19
Their mass increase Δm is given by Einstein’s equation ΔE = c2Δm
For an extra energy of 20 MeV
ΔE = (20 × 106) eV × (1.60 × 10−19 J eV−1)
= 3.20 × 10−12 J
[1]
This represents a mass increase of
Δm =
Δ𝑚
𝑚0
=
Δ𝐸
𝑐2
=
3.20 × 10−12 J
= 3.56 × 10−29 kg
(3.00 × 108 m s−1 )2
[2]
3.56 × 10−29 kg
[1]
1.67 × 10−27 kg
= 0.021 or just over 2 %
[1]
[Total 5 marks]
20
a)
Your diagram should look like the figure below. Note that you must include the direction of
the field lines.
[2]
The electric fields between the drift tubes accelerate the electrons as they pass in a vacuum
between them.
b)
[1]
i) They are called drift tubes because the electrons do not feel an electric acceleration
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
24 Electrons and nuclei
Answers to Exam practice questions
force as they travel within each tube. Within each tube the velocity of the electrons
remains constant.
[2]
ii) Once the electrons have reached a speed  0.9c, where c is the speed of light,
each addition of a few keV of energy does not noticeably alter their speed.
[2]
Thus their time in each drift tube remains essentially the same and consequently the length
of successive drift tubes does not noticeably alter.
c)
[2]
Using ΔE = c2Δm
 Δm =
(8.4 × 109 V) × (1.6 × 10−19 ) C
(3.0 × 108 m s−1 )
[2]
2
= 1.5 × 10−26 kg
d)
[1]
This is more than 16 000 times the rest mass, 9.1 × 10−31 kg, of an electron.
[1]
A volt is a J C−1
[1]
Therefore the unit for Δm is
JC
−1
×C
m2 s−2
But a joule is a N m  kg m s−2 × m = kg m2 s−2
This gives units for Δm as
kg m2 s−2 C−1 × C
m2 s−2
[1]
= kg
[1]
[Total 16 marks]
21
a)
b)
Measured radius of electron’s path:
At A: 46 mm, at B: 31 mm, at C: 22 mm
[1]
As photograph is ⅔ scale, the actual radii are 3/2 × measured values
[1]
Actual radius of electron’s path at A: 69 mm, at B: 46.5 mm, at C: 33 mm
[1]
In the relationship p = Ber, the above numbers represent r
Here we have:
Be = (1.2 N A−1 m−1)(1.6 × 10−19 C) = 1.9 × 10−19 kg s−1
[2]
Therefore the momenta are:
pA = 1.3 × 10−20 kg m s−1, pB = 8.9 × 10−21 kg m s−1, pC = 6.3 × 10−21 kg m s−1
c)
[2]
The speed of the electron is 3.0 × 108 m s−1 at each of A, B and C and the mass m of the electron
is p/v in each case, therefore
[1]
mA = 4.4 × 10−29 kg, mB = 3.0 × 10−29 kg, mC = 2.1 × 10−29 kg
[2]
(You may get slightly different values due to rounding down.)
d)
Each of these masses is much greater than the rest mass, 9.1 × 10−31 kg, of the electron.
[1]
[Total 11 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
25 Particle physics
Answers to Test yourself questions
Page 449 Test yourself on prior knowledge
1
F = B⊥Iℓ
2
F = BIℓsinθ
3
As 1.0 MeV = 1.0 × 106 eV and 1.0 eV  1.6 × 10−19 J, then the phrase means that the charged particle has an
energy of 1.6 × 10−13 J
4
In E = m0c2 the m0 represents the rest mass of the particle; extra energy, ΔE is then represented by the Δm
in the equation ΔE = c2Δm.
5
The units of BQʋ are (N A−1 m−1) × C × (m s−1). As a C s−1 is an ampere, A, then only a newton, N, is left.
6
The force F in F = BQʋ will be zero when ʋ is parallel to B, i.e. when the charged particle is moving along (or
against) the magnetic field.
7
J s, i.e. the joule second.
8
The energy of this photon is E = hf =hc/λ, where h is the plank constant,
h = 6.63 × 10−34 J s and c is the speed of e-m radiation in a vacuum
c = 3.00 m s−1.
9
Using λ = h/p with h = 6.63 × 10−34 J s and p = 2.73 × 10−22 N s
 λ = 2.43 × 10−12 m or 2.43 pm – the wavelength of a hard γ-ray.
Page 452 Test yourself
1
Kinetic energy is not lost in an elastic collision but is lost in an inelastic collision.
2
6 GeV = 6 × 109 eV  (6 × 109 eV) × (1.6 × 10−19 J eV−1) = 9.6 × 10−10 J or 0.96 nJ
3
F = kQ1Q2/r2 or F = Q1Q2/4πε0r2
4
A neutron is udd.
5
A positron has charge +1.6 × 10−19 C.
6
T C m  (N A−1 m−1) × (A s) × m  N s
7
a)
1.5m0
b)
between 95% and 96% of the speed of light (3.00 × 108 m s−1),
i.e. between 2.85 × 108 m s−1 and 2.88 × 108 m s−1.
Page 455 Test yourself
8
1 J  1 N m  1 kg m s−2 m  1 kg m2 s−2
9
9.1 × 10−31 kg  (9.1 × 10−31 kg) × (5.6 × 1029 MeV/c2 per kg) = 0.51 MeV/c2
10
34 u  34 u × 1.66 × 10−27 kg per u = 5.6 × 10−26 kg
11
Photon energy E = hc/λ where hc = (6.63 × 10−34 J s) × (3.00 × 108 m s−1)
E =1.99 × 10−25 J m / λ
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
25 Particle physics
12
Answers to Test yourself questions
By placing a magnetic field ⊥ to the plane of the pair produced and showing that the tracks of the two
particles show equal but opposite curvature.
13
1.02 MeV  1.02 MeV × (1.2× 106 eV) (1.6 × 10−19 J eV−1)
= 1.63 × 10−13 J
14
Using Einstein’s equation ΔE = c2Δm when Δm is the mass of two protons, gives
ΔE = (2 × 1.66 × 10−27 kg) × (3.00 × 108 m s−1)2
= 3.0 × 10−10 J
Half of this energy is the energy of each photon
= 1.5 × 10−10 J
= hf = (6.63 × 10−34 J s)f
From E = hf
 f = E/h = 1.5 × 10−10 J / 6.63 × 10−34 J s
= 2.3 × 1023 s−1
Page 460 Test yourself
15
a)
uud
b)
udd
16
Baryons consist of three quarks, mesons consist of two quarks.
17
The sss baryon is called the omega minus particle, its charge is −e.
18
No, because there are two baryons on the left of the relationship but only one baryon on the right. Both
the π0 particles are mesons.
19
The law of conservation of lepton number is correct as it is zero on the left and is (+1) + (−1) on the right.
20
đs as the full quark relationship is: sss = ssd + đs, i.e. strangeness is conserved and a neutral meson has
been produced in the decay.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
25 Particle physics
Answers to Exam practice questions
Pages 464–466 Exam practice questions
1
Answer D
[Total 1 mark]
2
Answer B
[Total 1 mark]
3
Answer D
[Total 1 mark]
4
Answer C
[Total 1 mark]
5
Answer B
[Total 1 mark]
6
Answer D
[Total 1 mark]
7
Answer D
[Total 1 mark]
8
Answer A
[Total 1 mark]
9
Answer C
[Total 1 mark]
10
Kinetic energy of tennis ball is
E = ½mʋ2 =½(57.5 × 10−3 kg)(220 000 m/3600 s)2
[1]
= 107 J
[1]
Quantum energy is
E = hf =
 λ=
ℎ𝑐
𝜆
λ=
ℎ𝑐
[1]
𝐸
(6.63 × 10−34 J s) × (3.00 × 108 m s−1 )
[1]
107 J
= 1.85 × 10−27 m
[1]
This is so small as to make any quantum effect shown by the tennis ball negligible.
[1]
[Total 6 marks]
11
a)
A mass of 30 GeV/c2 
(30 × 109 eV) × (1.6 × 10−19 J eV−1 )
[2]
(3.0 × 108 m s−1 )2
= 5.3 × 10−26 kg
b)
[1]
In terms of the mass of the electron this is
5.3 × 10−26 kg
9.1 × 10−31 kg
= 59 000
[2]
This means the Stanford accelerator speeds up electrons until they have an energy that is
59 000 times their rest mass energy.
12
[Total 5 marks]
Using E = mc2 the total energy producing the γ-rays
= 2 × (1.67 × 10−27 kg)(3.00 × 108 m s−1)2
= 3.00 × 10
−10
J
[1]
[1]
The energy of each photon is
½E = ½hf =
λ=
ℎ𝑐
2𝜆
λ=
ℎ𝑐
[1]
2𝐸
(6.63 × 10−34 J s) × (3.00 × 108 m s−1 )
[1]
(2 × 3.00 × 10−10 ) J
= 3.3 × 10−16 m ≈ 3 × 10−16 m
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
25 Particle physics
Answers to Exam practice questions
Assumption: that the proton and antiproton are at rest when they interact
[1]
[Total 6 marks]
13
The unit of 2hc/mʋ2 is
J s × m s−1
Js
=
kg × (m s−1 )2 kg × m s−1
[2]
Substituting that J  N m and N  kg m s−2 gives us
[2]
Nms
kg × m s−1
=
kg m s−2 m s
kg × m s−1
=m
[1]
The unit of 2hc/mʋ2 is therefore the metre.
14
[Total 5 marks]
The graph plots m/m0 against ʋ in multiples of c
From the equation:
For ʋ = 0.50 c  m/m0 =
1
√1 – (0.5 𝑐)
[1]
2
𝑐
=
1
√0.75
= 1.15, which agrees with the value read off the graph
[1]
= 1.51
[1]
= 2.1
[1]
Similarly
For ʋ = 0.75 c  m/m0 =
1
√0.438
For ʋ = 0.875 c  m/m0 =
1
√0.234
In each case, the values plotted agree with these calculations.
[1]
[Total 5 marks]
15
For the decay K− → π0 + π−
u̅s→ dd̅+ du̅
[2]
In terms of charge this is
(−e) → 0 + (−e)
Which in terms of quarks is
[(−⅔) + (−⅓)]→ [(−⅓) + (+⅓)] + [(−⅓) + (−⅔)]
[3]
[Total 5 marks]
16
a)
Applying the left hand rule, the magnetic field is upwards, out of the paper.
[1]
b)
The π+ meson’s path decreases in radius because the meson is losing energy and thus its
momentum is getting smaller.
[2]
As p = BQr (see Section 7.4) then p  r.
[1]
[Total 4 marks]
Tip: In answering questions such as this, you should look and see how many marks are being awarded for
the answer. Here, for only 2 marks, it would not be necessary to quote the formula in answering b) or to
show that p  r, but as the question asks you to ‘explain’ and there are 3 marks, both would be needed.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
25 Particle physics
17
The reaction here is: K− +
Answers to Exam practice questions
p → Ω− + K0 + K+
In terms of quarks this can be written as
su̅ + uud → sss + ds̅ + us̅
[2]
The d quark ‘survives’ as does the s quark and one of the u quarks.
A uu̅ has disappeared from the left and four quarks have appeared on the right – two each of s and s̅ .
Perhaps the energy for their appearance came from the disappearance of the uu̅.
Charge is conserved as the left side is −e + e and the right side is −e + 0 + e
[2]
Baryon number is conserved as each side contains one (three quark) baryon
[2]
Strangeness number is conserved with +1 on the left and +3 − 2 on the right
[2]
[Total 8 marks]
Tip: If you are asked to ‘explain’ you need to both state (‘charge is conserved’) and then explain why (as
the left side is −e + e and the right side is −e + 0 + e) to get both marks.
18
Your answer should follow a logical argument and should include the following points:
•
According to de Broglie, an electron can show wavelike properties.
•
In order for a stationary or standing wave pattern to form a circle, the circle can only contain
an even number of wavelengths.
•
Figure 25.15 shows 6 half wavelengths.
•
Different energy levels might correspond to 2, 4, 6… half wavelengths in the standing wave
pattern.
•
This could be why hydrogen shows a number of discrete (or quantized) energy levels.
[Total 5 marks]
19
a)
b)
i) Particles dss and uds are baryons having charges −e and 0 respectively
[3]
Particle u̅s is a meson of charge −e.
[2]
ii) The magnetic field is into the page.
[1]
i) Particle P is the proton because it is curving in the opposite direction to the negative
meson in part a).
[2]
The other particle has a negative charge (as it curves in the opposite direction to the
proton).
[1]
ii) Both P and Q have curved paths in the magnetic field, but as p = Ber, the particle showing
a path with the greater radius r has the greater momentum.
Therefore P has the greater momentum.
c)
[2]
When the xi particle decayed:
• charge was conserved (−e) on the left, (0) + (−e) on the right
[2]
• the strangeness number was conserved (−2 on both sides)
[2]
• some kinetic energy became mass (u and u̅ created)
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
25 Particle physics
Answers to Exam practice questions
[Total 17 marks]
20
You will find that the Wikipedia page has more than enough about ‘The twin paradox’.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Test yourself questions
Page 467 Test yourself on prior knowledge
1
2
a)
Alpha particles are helium nuclei, having a mass of 4 u and a charge of +2e.
b)
Beta particles are electrons, having electron mass and charge –e (or +e if β+ particles).
c)
Gamma rays are high frequency electromagnetic waves.
a)
Energy = 360 keV = 360 × 103 eV × 1.6 × 10-19 J eV−1 = 5.76 × 10−14 J
b)
If half-life is 6 days, then 30 days =
30 𝑑𝑎𝑦𝑠
6 𝑑𝑎𝑦𝑠
= 5 half-lives
(½)5 = (0.5)5 = 0.03125 (by calculator) = 1/32 (by pressing 1/x button)
Percentage remaining =
1
32
× 100 % = 3 %
(or 0.03125 × 100 % = 3 %)
3
From 𝐸 = ℎ𝑓 =
ℎ𝑐
𝜆
⇒𝜆=
ℎ𝑐
𝐸
=
6.63 × 10−34 J s × 3 × 108 m s −1
(160 × 103 ) eV×1.6 × 10−19 J eV −1
λ = 7.8 × 10−12 m
4
a)
An atom of 136𝐶 has
i) 6 protons
ii) 7 neutrons
iii) 6 electrons
b)
Isotopes are different forms of the same element having the same proton number (6 in the case
of carbon) but different masses due to a different number of neutrons in the nucleus (6 in the
case of carbon-12 and 7 in the case of carbon-13).
Page 476 Test yourself
1
a)
Background radiation is the radiation all round us in the environment mainly due to natural
sources but also as a result of human activity.
b)
The main sources of background radiation are
• radon gas from the ground and buildings
• radioactive elements in the Earth’s crust
• cosmic rays
• naturally occurring radioactive isotopes present in food and drink and the air we breathe
• medical uses (radiography)
• nuclear processes (e.g. nuclear power stations).
2
a)
Ionising radiation is harmful to humans as it can excite, or even ionise, the molecules in our
bodies, which can lead to damage to our DNA and subsequently cancer.
b)
When working with radiation we should
• keep as far away as possible
• not touch radioactive materials
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Test yourself questions
• keep sources pointing away from the body
• minimise exposure time
• keep material in lead container when not in use
• wash hands when finished.
In addition, people likely to be exposed to larger doses should wear protective clothing, such as
lead aprons.
3
4
a)
α causes the most ionisation.
b)
γ causes the least ionisation.
c)
γ is the most penetrating.
d)
α is the most easily absorbed.
e)
α has a positive charge.
f)
β has a negative charge.
g)
β is most easily deflected in a magnetic field.
h)
γ cannot be deflected in an electric field.
a)
An ionised atom is one that has lost or gained one or more electrons. Ions can be deflected by
electric or magnetic fields as they have an electric charge.
b)
Isotopes are different forms of the same element that have the same proton number (10 in the
case of neon) but different masses due to a different number of neutrons in the nucleus (e.g.
neon-20 has 10 neutrons whilst neon-22 has 12 neutrons.
c)
d)
Isotope
Proton number
Nucleon number
Number of neutrons
10
20𝑁𝑒
10
20
10
10
22𝑁𝑒
10
22
12
Let there be a fraction x of neon-20. The fraction of neon-22 will then be (1 – x)
We then have:
20x + 22(1− x) = 20.18
Expanding, we get:
20x + 22 – 22x = 20.18
Re-arranging:
20x − 22x = 20.18 – 22.0
− 2x = − 1.82
x = 0.91
This means that there is 91% of neon-20 and 9% of neon-22. The neon-20 isotope is therefore
only a small percentage of the mixture.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Test yourself questions
Page 480 Test yourself
5
a) b)
The missing parts of the decay chain are shown in blue:
c)
238
92𝑈
d)
206
From 226
88𝑅𝑎 to 82𝑃𝑏
230
234
and 234
92𝑈 are a pair of isotopes, as are 90𝑇ℎ and 90𝑇ℎ .
mass goes down by (226 – 206) = 20
proton number goes down by (88 – 82) = 6
β—decay does not change the mass, so the change of 20 must be due to
5 α-decays as each α-particle has a mass of 4.
5 α-decays will reduce the proton number by 5 × 2 = 10. As the proton number has only gone
down by 6, there must be (10 – 6) = 4 β—decays.
6
ΔE = c2m = (3.00 × 108 m2) × 1.66 × 10−27 kg = 1.49 × 10−10 J
a)
1u =
1.49  10 −10 J
= 9.31 108 eV = 931M eV
1.60  10 −19 J eV −1
For electron ΔE = c2m = (3.00 × 108 m2) × 9.11 × 10−31 kg = 8.20 × 10−14 J
b)
E =
8.20  10 −14 J
= 5.12  105 eV  0.5 M eV
1.60  10 −19 J eV −1
Ek = ½mv2  2mEk = m2v2 (by multiplying both sides by 2m)
c)
Momentum 𝑝 = 𝑚𝑣 = √2𝑚𝐸𝑘 where Ek = 8.20 × 10−14 J from part b)
=
h
6.63  10 −34 J s
=
= 2.68  10 −13 m
mv
2  (9.11 10 −31 kg)  8.20  10 −14 J
(Don’t forget to take the square root of the denominator!)
Page 482 Activity 26.1
1
Your graph should show an exponential decay, with some scatter. You should draw a smooth curve that
best fits the plotted points.
2
It takes about ‘3.7’, or approximately 4, throws for half the initial 48 dice to ‘decay’.
3
The rate of decay is proportional to the number of radioactive nuclei, which gives us the defining equation
for radioactive decay, −
𝑑𝑁
𝑑𝑡
= 𝜆𝑁.
The decay constant λ is therefore the probability of decay, which is 1 in 6 for throwing a six in our model.
x1 =
2
ln 2

=
0.693
= 4.2 throws  4 throws
1 /(6 throws)
% difference between experimental and theoretical values
=
4.2 − 3.7
 100% = 12%
4.2
(Note that the theoretical value is used as the denominator)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Test yourself questions
4
The experiment is a good model of radioactive decay insomuch as whether or not a six is thrown is a
random event, and the probability of throwing a six is proportional to the number of dice that are thrown.
However, unlike radioactive decay, it is not a continuous process. Also, the number of dice is many, many
times less than the number of nuclei involved in radioactive decay, which makes the ‘scatter’ greater.
Despite this, the value obtained for the ‘half life’ was within about 12% of the theoretical value.
Page 485 Test yourself
7
a)
The half-life of a nuclide it the average time that it takes for a given number of nuclei of that
nuclide to halve in number.
b)
The equation will be
226
88𝑅𝑎
c)
→
222
86𝑅𝑎
+ 43𝐻𝑒 +γ
If a thin sheet of paper is placed between the source and detector, the count rate will decrease
appreciably, showing that α-particles had been emitted but had been absorbed by the paper. The
remaining count could be due to β- or γ-radiation, but if a sheet of aluminium a few millimetres
thick is placed in the path of the radiation, the fact that the count rate is hardly reduced confirms
that it must be γ- rather than β-radiation.
(Note that you must make it clear that you have eliminated the possibility of
β-radiation being present.)
d)
Although there would not be a detectable amount of radiation from any radium-226 that was a
billion (109) years old, the radium-226 is being continuously generated from the decay of
uranium, in a number of stages, so that a state of equilibrium has been reached. The uranium
decay series is shown in Test Yourself Question 5 on page 480.
t1 =
2
8
a)
b)
ln 2

 =
ln 2
0.693
=
= 2.11 10 −6 s −1
t1
(3.8  24  60  60) s
2
As this is radium-226, the atomic mass is 226 u. This means that in 226 g of the radium there will
be 6.02 × 1023 nuclei.
If 226 g  6.02 × 1023 nuclei
15 mg 
𝑑𝑁
15 𝑥 10−3 𝑔
226𝑔
× 6.02 × 1023 nuclei = 4.00 × 1019 nuclei
= λN = 2.11 × 10−6 s−1 × 4.00 × 1019 nuclei = 8.44 × 1013 Bq
c)
−
d)
If half life is 3.8 days, then 19 days =
𝑑𝑡
19 𝑑𝑎𝑦𝑠
3.8 𝑑𝑎𝑦𝑠
= 5 half lives
(½)5 = (0.5)5 = 0.03125 (by calculator) = 1/32 (by pressing 1/x button)
Percentage remaining =
1
32
× 100 % = 3 % (or 0.03125 × 100 % = 3 %)
Page 486 Test yourself
9
a)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Test yourself questions
b)
By plotting a graph of the count-rate on the y-axis against time on the x-axis, you should get a
characteristic exponential decay curve.
c)
An average value for the half-life can be found by taking the average of the time to decay from,
for example, 500 to 250 counts min-1 and 250 to 125 counts min-1. You should get an average
value of about 72-73 s.
Page 486 Test yourself
10
a)
b)
You should find that the gradient of a graph of ln (A/min-1) on the y-axis against t on the x-axis is
a straight line of negative slope.
The gradient should be – (9.50 ± 0.10) s
From the equation for radioactive decay, A = A0e−λt, taking logarithms to base ‘e’ on both sides of
the equation, we get:
ln A = −λt + ln A0
The gradient is therefore – λ
t1 =
2
ln 2

=
0.693
0.693
=
= 73  1 s
− gradient 9.50  10 −3 s −1
Page 487 Core practical 15
1
2
For safety you should:
•
wear protective gloves
•
always handle the source with tongs
•
keep as far away from the source as possible
•
keep exposure time to a minimum.
The average background count is (73 + 67)/10 = 14 counts per minute. This is a relatively small count for
what is a random quantity. Taking the count for several minutes and repeating is therefore essential.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Test yourself questions
3
An absorber thickness of 14.9 mm could be obtained by using a combination of the 3.3 mm, 5.0 mm and
6.6 mm discs.
4
5
With reference to the graph shown in Figure 9.22:
At x = 0, N = 360 min−1 and at x = 2 mm, N = 320 min-1 ⇒ ratio = 360/320 = 1.13
At x = 2 mm, N = 320 min-1 and at x = 4 mm, N = 285 min−1 ⇒ ratio = 320/285 = 1.12
At x = 4 mm, N = 285 min-1 and at x = 6 mm, N = 250 min−1 ⇒ ratio = 285/250 = 1.14
Allowing for scatter due to the random nature of radiation, these ratios are sufficiently close together to
justify the supposition that the absorption of the gamma radiation is an exponential function of the
thickness of absorber.
6
If N = N0 e−μx, then taking logs to base e gives us
ln N = −μx + ln N0
which is a straight line of negative slope (gradient −μ) and intercept ln N0.
7
8
Allowing for a bit of scatter, which is to be expected because of the random nature of radioactive emission,
the graph is a straight line of negative gradient, with intercept ln N0. This confirms that the absorption of
the gamma radiation varies exponentially with the thickness of the absorber as proposed.
9
The gradient of the graph, found by drawing a large triangle, is about – 0.059, giving a value for the
absorption coefficient μ of 0.059 mm−1.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Test yourself questions
The left hand side of the equation
ln N = −μx + ln N0
has no unit since the value of a logarithm is just a number. Therefore the right-hand side has no unit. This
means that the term μx has no unit. As the unit of x is mm, the unit of μ must be mm−1.
Page 491 Test yourself
11
a)
If the half-life is 8 days, one month (say 31 days) is almost 4 half lives.
(½)4 = ½ × ½ × ½ × ½ = 1/16 (or 0.0625 by calculator)
1/16 (or 0.0625) × 32 kBq = 2 kBq
b)
𝑡1=𝐼𝑛2⇒𝜆=𝐼𝑛2=
2
1
𝑡2
𝜆
𝑑𝑁
0.693
=1.00×10−6 𝑠 −1
(8.0×24×60×60)𝑠
= 3.2 × 1010 nuclei
d)
If 6.0 × 1023 nuclei have a mass of 131 g, then
𝜆
=
32×103 𝑠 −1
−
𝑑𝑡
= 𝜆𝑁 ⇒ 𝑁 =
−𝑑𝑁/𝑑𝑡
c)
1.00×10−6 𝑠 −1
3.2 × 1010 nuclei will have a mass of
131 g x
e)
3.2𝑥1010 𝑛𝑢𝑐𝑙𝑒𝑖
6.0𝑥1023 𝑛𝑢𝑐𝑙𝑒𝑖
= 5.3 𝑥 10−14 𝑔
This isotope is particularly useful as a tracer in medicine as it is a β—emitter. Beta particles can
travel from inside the body to the external detector without too much absorption, which means
they will not do much harm to the patient. Alpha particles would be absorbed by the body, which
could be harmful, and would not be detected outside the body.
In addition, a half-life of 8 days is long enough for the medical procedure to be undertaken but
short enough for the radioactive isotope not to remain in the body by any appreciable amount
for too long.
f)
The decay process is 131
53𝐼 →
131
54𝑋𝑒
+ −10𝛽 + 𝛾 (+𝑉𝑒 )
g)
i) γ-radiation differs from α- and β-radiation in that it is electromagnetic radiation, whereas αand β-radiation are both charged particles.
ii) A quantum of γ-radiation is emitted as well as the β-radiation in order to return the excited
nucleus to a more stable state.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Exam practice questions
Pages 493–498 Exam practice questions
1
X-rays are quanta of energy emitted when electrons fall to a lower energy level, and so do not emanate from
the nucleus – Answer D.
2
[Total 1 mark]
Alpha particles, being the most massive (and therefore slowest moving for a given energy) and charged,
cause the most ionisation in air – Answer A.
[Total 1 mark]
3
Alpha particles have a typical range in air of a few centimetres – Answer C.
[Total 1 mark]
4
An alpha particle is a helium nucleus, 42He. When an alpha particle is emitted, the mass number will decrease
by 4 and the atomic number will decrease by 2 – Answer D.
5
A beta particle is equivalent to an electron,
0
−1e.
When a beta particle is emitted, the mass number will stay
the same and the atomic number will increase by 1 – Answer B.
6
[Total 1 mark]
[Total 1 mark]
One day (24h) is equal to 10 half lives.
The fraction of bromine-83 remaining after 1 day will be:
(½)10 = 9.76 × 10−4 ≈ 1 × 10−3 ≈ 0.001
Answer is C.
7
a)
[Total 1 mark]
Background radiation is the radiation that is constantly present in our environment. Sources
include:
• radioactive elements in the Earth’s crust;
• radioactive gases (mainly radon), emitted by the Earth;
• cosmic rays from outer space;
• medical uses, e.g. X-rays;
• nuclear industry;
• trace amounts of radioactivity in our bodies.
b)
[3]
You would need a Geiger-Müller tube connected to a counter or rate-meter. First of all, the
background count should be taken.
Then the G-M tube would be held about 1 cm above the soil (not to the side of the beaker as the
glass would absorb any alpha particles) and the count-rate noted. A sheet of paper should now
be inserted between the soil and the G-M tube.
•
If the count is reduced to background, then only α is present.
•
If the count is reduced considerably, then α with some β and/or γ is present.
•
If the count is not reduced, then only β and/or γ is present.
[3]
[2]
To check further whether β and/or γ is present, a 5 mm sheet of aluminium should be inserted.
•
If the count remains virtually the same, then only γ is present.
•
If the count is reduced a bit, then β and γ are present.
•
If the count is reduced virtually to background, only β is present.
[2]
Tip: A careful, logical description is needed. This is helped by the use of bullet points.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
8
a)
Answers to Exam practice questions
Ionisation is when one or more electrons are removed from an atom, leaving the atom with a
positive charge.
b)
[2]
An alpha particle consists of two protons and two neutrons (the nucleus of a helium atom). It
is emitted from the nucleus of an unstable (radioactive) atom.
c)
[1]
i) As the alpha particle travels through the air, it ionises the air molecules. The energy
required to do this (i.e. to remove electrons from the air molecules) comes from the alpha
particle’s kinetic energy, and so it slows down.
ii) The alpha particle eventually picks up two electrons, which are always around in the air,
and becomes an atom of helium gas.
d)
[3]
i) Isotopes are different forms of the same element (and so have the same atomic/proton
number) having different numbers of neutrons, and therefore different mass/nucleon numbers.
ii) The symbol 241
95Am shows that the atomic/proton number is 95 and the mass/nucleon
number is 241, which means there are (241 – 95) = 146 neutrons.
iii) 241
95Am →
e)
237
93Np
4
2He
+
[5]
i) The half-life of a particular isotope is the average time for a given number of radioactive
nuclei of that isotope to decay to half that number.
ln2
ii) t½ =
λ =
λ
ln2
t½
=
0.693
460 × (365 × 24 × 60 × 60) s
= 4.78 × 10−11 s−1 ≈ 4.8 × 10−11 s−1
iii) Number of nuclei =
1.6 × 10−8 g
241 g
× 6.0 × 1023
= 4.0 × 1013 nuclei
iv) Activity = 4.0 × 1013 × (4.8 × 10−11) s−1 = 1.9 × 103 Bq
f)
[1]
[1]
[1]
[1]
[1]
[1]
Americium-241 is a suitable source because
• it is an alpha emitter, which means that it readily ionises the air and the alpha particles don’t
travel very far (making it safe);
• it has a long half-life, so won’t need replacing;
• its activity is such that only a very small mass is needed.
[3]
[Total 20 marks]
9
a)
i)
239
94Pu
→
ii)
60
27Co
→
iii)
23
12Mg
→
235
92U
60
28Ni
+ 42He
+
23
11Na
0
−1e
+
0
+1e
[2]
+ 𝜐̅𝑒
+ 𝜐𝑒
[3]
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Exam practice questions
b)
[3]
[Total 10 marks]
10 a)
The half-life of a particular isotope is the average time for a given number of radioactive
nuclei of that isotope to decay to half that number.
b)
[1]
With all radioactive sources well away from the area in which the experiment is to be conducted,
the background count should be taken. This should be done for sufficient time, say two lots of 5
minutes, to get an average value.
[2]
The G-M tube should be placed close to the layer of protactinium-234 and the count-rate
measured at time intervals of, say, 20 s for about 3 minutes. This will give 10 values (including
that at t = 0). The background count should then be subtracted from each reading.
[3]
A graph of either count-rate A against time t (exponential decay) or ln A against t (straight line of
negative slope) should be plotted.
[2]
If A against t is plotted, the half life can be found directly by determining how long it takes for a
particular value of A to decay to A/2. This should be done for three different values of A and an
average found. If ln A against t is plotted, the gradient is numerically equal to the decay constant
λ. The gradient should be found by drawing a large triangle and the half-life found from t½ =
ln2
λ
.
[2]
c)
If the half-life is 28 years, the change in activity over even a few days would be too small to
detect, particularly as the decay is random.
A small amount of strontium-90, with a half-life of 28 years, would probably mean that the
activity was very low and may not be significantly greater than the background count.
[2]
[Total 12 marks]
11 a)
i) As alpha particles are charged and massive, they readily ionise matter as they pass through
it. They will therefore ionise the chemicals making up human cells and cause terrible internal
cell damage.
[2]
ii) Alpha particles are difficult to detect outside the body as they will be absorbed by even a
very thin layer of tissue and so will not penetrate as far as the outside of the body.
b)
i) From N = N0 e−λt ⇒
As λ =
ln2
𝑡½
⇒ ln
𝑁
𝑁0
𝑁
𝑁0
= e−λt ⇒ ln
= − ln 2 ×
𝑡
𝑡½
𝑁
𝑁0
[1]
= −λt
= − ln 2 ×
= − 1.83
365 days
138 days
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
This gives
𝑁
𝑁0
Answers to Exam practice questions
= e−1.83 = ‘anti-ln’ of − 1.83 = 0.16
This means the % remaining after one year = 16 %
[1]
Your graph should look like the one below.
[3]
Tip: Even if you are asked to ‘sketch’ a graph, relevant data should be plotted as far as possible. For
example, in the figure above some points that you might include are plotted – the activity at 1, 2, 3 and 4
half-lives, the activity that has just been found for 1 year and the activity after 2 years. You would only be
expected to show 3 or 4 values and you would not be required to use graph paper for a ‘sketch’. A grid has
been included in the figure above merely for information.
ii) λ =
ln2
𝑡½
=
ln2
[1]
(138 × 24 × 60 × 60 ) s
= 5.8 × 10−8 s−1
[1]
Activity A = λN = 5.8 × 10−8 s−1 × 2.9 × 1015 = 1.7 × 108 Bq
c)
[1]
Energy generated by 1 μg = 1.7 × 108 s−1 × (5.3 × 106) eV × (1.6 × 10−19) J eV−1
[1]
−4
= 1.43× 10 J s
[1]
The power generated by 1 g would be (1.43 × 10−4 × 106) W ≈ 140 W
[1]
−1
This is equivalent to a very powerful (150 W) filament lamp. If 1 g were to be contained in a
small capsule, this magnitude of power would soon melt the capsule.
[1]
[Total 16 marks]
12 The equations are:
4
2He
+
4
2He
12
7N
→
12
6C
12
6C
→
4
2He
+
4
2He
+
0
+1e
+
4
2He
→ 126C
+ 𝜐e
+
4
2He
[2]
[2]
[2]
[Total 6 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
13
a) i) The equation −
d𝑁
d𝑡
Answers to Exam practice questions
= λN is based on the concept that the rate of decay is proportional to the
Number of radioactive nuclei, i.e. the more nuclei there are, the more likely it is that one
will decay.
[1]
ii) If there are very large numbers of nuclei, as there usually are in radioactivity, then it is
statistically more probable that the above equation, and other mathematical laws, will
make calculations valid.
iii) −
d𝑁
d𝑡

= λN
 ln
𝑁
𝑁0
𝑁 1
∫𝑁
0
𝑁
[1]
𝑡
d𝑁 = −λ ∫0 d𝑡
[1]
= − λt
 N = N0 e− λt
[1]
When t is equal to the half life t½ we have N =

N = N0 e− λt


1
2
𝑁0
2
𝑁0
2
= N0 𝑒 −𝜆𝑡½
= 𝑒 −𝜆𝑡½  2 = 𝑒 𝜆𝑡½
λt½ = ln 2
b) i) From t½ =
ln 2
𝜆
 t½ =
[1]
ln 2
[1]
𝜆
it follows that an isotope with a very short half life t½ will have a very large
decay constant λ and so even a small amount of the isotope will have a detectable activity.
Also, an isotope with a short half-life will not stay in the blood flow very long and so will be
less harmful to the patient.
ii) Fluorine-18
( 189F)
[2]
does not have enough neutrons in the nucleus to provide the necessary
strong force that balances the electromagnetic repulsion between the protons. If it
undergoes β+ decay by emitting a positron it decays into 199F, which is stable.
Effectively 11p → 10n + +10e (plus an electron neutrino).
[2]
iii) In PET, the positrons are annihilated when they encounter an electron. As body tissue
does not contain positrons, using electrons would not work, as they would not encounter
any positrons to annihilate.
iv) From t½ =
ln 2
𝜆
 λ =
[2]
ln 2
𝑡½
=
0.693
(2 × 60 × 60)s
= 9.6 × 10−5 s−1
[1]
[1]
From N = N0 e− λt in one hour:

𝑁
𝑁0
= e− λt = exp[−9.6 × 10−5 s−1 × (1 × 60 × 60 s)]
= e− 0.346

𝑁
𝑁0
[1]
= 0.71  N = 0.71 N0
If 71% of the nuclei remain, then 29 % will have decayed.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
Answers to Exam practice questions
v) By conservation of mass-energy ΔE = c2Δm
The electron and positron each has a mass of 9.11 × 10−31 kg, so:
ΔE = (3.00 × 108 m s−1)2 × 2 × (9.11 × 10−31 kg)
1.64 × 10
= 1.64 × 10−13 J =
−13
J
[1]
= 1.024 MeV
1.6 × 10−19 J eV−1
[1]
Assuming the positron has lost its kinetic energy by the time it encounters an electron, by the
conservation of energy this will be the combined energy of the two γ photons produced in
the annihilation. To conserve momentum, the photons move off in opposite directions, each
with equal and opposite energy. This means they must share the 1.024 MeV equally, i.e. 512
keV each.
[2]
[Total 20 marks]
14 a)
210
82Pb
→
210
83Bi
→
β−
b)
210
84Po
β−
→
206
82Pb
α
It was necessary to pump out the air because α-particles are absorbed by a few centimetres of
air and would otherwise not even reach the beryllium target.
c)
[1]
As neutrons are particles without any electric charge they cannot cause ionisation and so could
not be detected directly by the ionisation chamber.
d)
[3]
[2]
In order to conserve both momentum and, as it is an elastic collision, kinetic energy, the neutron
must give up all its energy to the proton. The neutron will be brought to rest and the proton will
be ejected with the momentum and kinetic energy of the neutron. This is shown in the figure
below.
e)
[2]
Eγ = √½𝐸p 𝑚p 𝑐 2
= [0.5 × (5.7 ×106 eV) × (1.6 × 10−19 J eV−1)
× (1.007267 ×1.66 × 10−27 kg) × (3.0 × 108 m s−1)2]½ J
= 8.28 × 10−12 J =
8.28 × 10−12 J
[1]
[1]
1.6 × 10−19 J eV−1
= 5.17 × 107 eV ≈ 50 MeV
[1]
This is an extremely high energy for a photon and very unlikely to be produced as the result of αparticles of energy 5.7 Mev bombarding the beryllium.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
26 Nuclear decay
f)
Answers to Exam practice questions
Binding energy is given by:
ΔE = c2Δm = (3.00 × 108 m s−1)2 × (1.008 665 – 1.007 267) × 1.66 × 10−27 kg
= 2.09 × 10−13 J =
2.09 × 10−13 J
1.6 × 10−19 J eV−1
[1]
= 1.30 × 106 eV
[1]
This is ‘about 1 to 2 × 106 electron volts’ as Chadwick had predicted.
g)
In neutron decay:
1
0n
→
1
1p
+
0
−1e
[1]
+ 𝜈̅e
The electron anti-neutrino is necessary to conserve lepton numbers and spin. Charge is
conserved (= 0) as is the number of baryons (1).
h)
[3]
In terms of quarks, a proton is uud and a neutron is udd so, in neutron decay, an up quark
changes into a down quark. This involves the weak force.
[2]
[Total 20 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
Answers to Test yourself questions
Page 499 Test yourself on prior knowledge
1
a)
3 kW = 3000 J s−1
In one hour ΔE = PΔt = 3000 J s−1 × (20 × 60) s
= 1.08 × 107 J = 10.8 MJ
b)
c)
2
3
P 3000 W
=
= 12.5A
V
240 V
V 240 V
R= =
= 19.2 
I 12.5 V
P = IV  I =
a)
P = IV = 2.5 A × 12 V = 30 W
b)
ΔE = PΔt = 30 J s−1 × (60 × 60) s = 3.6 × 104 J = 36 kJ
a)
ΔE = mgΔh
b)
P=
dE dm
=
´gDh
dt dt
From density
𝜌=
𝑚
𝑣
⇒𝑚=
𝑑𝑚
𝑑𝑡
=𝜌×
𝑑𝑉
𝑑𝑡
dV
P= 
 g  h = 1.00  103 kg m −3 x 2300 m s −1  9.8 N kg −1  51 m
dt
= 1.1 × 109 J s−1 (W)
4
c)
ΔEk = ½mv2 = 0.5 × 375 × 103 kg × (80 m s−1)2 = 1.2 ×109 J
a)
Your graph should look like the figure below.
Features to note are:
• the gradients should be the same for both bounces
• the speed v2 after it bounces should be half the speed v1 just before the bounce
• the time t2 to reach the top of the bounce is half that of the initial time t1 to fall
• it takes a further time t3 = t2 to drop down again for the second bounce.
b)
The gradient is the acceleration of the ball, which, as it is falling freely under gravity, is constant
(=g), assuming that we ignore air resistance.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
Answers to Test yourself questions
c)
The ball hits the ground at the points marked G on the graph.
d)
The distance the ball drops is given by the shaded area A1 and the distances it rises back up
again by the shaded area A2.
e)
Before the ball is dropped it has gravitational potential energy (GPE). As it falls, GPE is
transferred into kinetic energy (KE) as the ball gains speed. A small amount of energy will also be
transferred into thermal energy due to air resistance.
On hitting the ground, the KE is transferred into potential energy stored in the bonds of the
rubber as the ball is momentarily brought to rest. When the ball bounces up again, most of this
energy is transferred back into KE. Some, however, will be transferred into thermal energy due to
hysteresis in the rubber and some into sound energy.
As the ball rises, the KE is transferred back into GPE, with some being transferred into thermal
energy doing work against air resistance.
Page 504 Test yourself
1
2
a)
660 °C = (660 + 273) K = 933 K
b)
−39 °C = (−39 + 273) K = 234 K
c)
1358 K = (1358 −273) °C = 1085 °C
d)
54 K = (54 −273) °C = −219 °C
a)
ΔE = mcΔθ = 0.150 kg × 2000 J kg−1 K−1 × (30 – 20) K = 3.00 kJ
b)
θfinal = (20 + 50) °C 70 °C
c)
d)
3
a)
ΔE = mcΔθ = 250 kg × 4200 J kg−1 K−1 × (40 – 20) K = 21 MJ
b)
ΔE = PΔt = 10.5 × 103 J s−1 × (7 × 60) s = 4.4 MJ
c)
As the shower takes only about 20% of the energy needed to heat the bath water, it is much
more energy efficient (and economical) to take a shower rather than a bath (provided you don’t
4
spend too long in the shower!).
d
dE
P=
= (mc ) = dm  c  
dt
dt
dt
dm = 7.5kg = 0.125kgs −1
60s
dt
dE = dm  c  
dt dt
10.5 × 103 W = 0.125 kg s−1 × 4200 J kg−1 K−1 × Δθ
 =
10.5  103 J s −1
= 20K
0.125 kg s −1  4200J kg −1K −1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
Answers to Test yourself questions
Temperature of water at shower head = (20 + 20) °C = 40°C
Page 505 Activity 27.1
1
E = mc  c =
=
E
m
(3.1A  10.3V)  (3.00  60)s
0.993kg  (27.3 − 21.3)K
= 980Jkg −1K −1
2
The instrument having the least precision will be the mercury thermometer. Although the temperature has
been recorded to a precision of 0.1°C (which you should always attempt), there is a degree of uncertainty
about the value, much more than in reading the meters and balance. In addition, Δθ is fairly small (5.9°C),
so any uncertainty in its value will give rise to a significant percentage uncertainty.
Page 506 Activity 27.2
E = mc  c =
1
E
(1.65 A  9.59 V)  (12  60) s
=
m
1.00 kg  (37.0 − 25.5) K
= 990.6 J kg −1K −1  990Jkg −1K −1
2
( to 2 SF)
As some energy will be transferred to the surroundings, it actually needs less energy to raise the
temperature of the aluminium block than that in the calculation. The actual value for the specific heat
capacity is therefore less than that calculated, i.e. the value calculated is too large.
Page 508 Activity 27.3
1
Power developed in resistor: P = IV = 0.77 A × 11.4 V = 8.8 W
As this is less than 11 W, a resistor rated at 11W would be entirely suitable.
2
E = mc  c = E
m where ΔE = PΔt = IVΔt
c=
=
IVt
m
−1
(0.77A  11.4 V)  (5.00  60) s
0.120 kg  (26.3 − 21.2) K
= 4300 J kg −1K −1
3
The main sources of error are:
•
energy taken by the heater itself (making c too large);
•
energy transferred to the surroundings, mainly from the top of the cup (also making c too large);
•
thermometer error, especially as Δ is only about 5K (could make c too large or too small, depending
on whether Δ were too small or too large respectively);
•
meter errors (which could affect the value of c either way);
•
small amount of energy taken by the thermometer and cup (making c too large).
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
Answers to Test yourself questions
Page 508 Test yourself
5
a)
V 12V
=
= 0.80 A
R 15
P = IV = 0.80 A × 12 V = 9.6 W
I=
As this is less than 10 W, a rating of 10 W for the resistor is suitable (just!).
b)
Δθ = (28.4 – 18.0) K
From ρ =
𝑚
𝑉
 m = ρV = 920 kg m−3 × (150 × 10−6) m3 = 0.138 kg
E = mc  c =
=
E
m
9.6 J s −1  (5.0  60) s
0.138 kg  10.4 K
= 2.00  103 J kg −1 K −1
6
a)
An object is in thermal equilibrium if it is giving out energy at the same rate as it is receiving it,
which means that its temperature will remain constant.
b)
i) The thermometer should be left in the oil for a time so that it is at the same temperature as
(or in thermal equilibrium with) the oil.
ii) Heat capacity of thermometer is 15 J K−1 and Δθ = (28.4 – 18.0) K = 10.4 K
So energy taken by thermometer = 15 J K−1 × 10.4K = 156 J
iii) Total energy over 5 minutes = 9.6 J s−1 × (5 × 60) s = 2880 J
% taken by thermometer =
156𝐽
2880𝐽
× 100 % ≈ 5 %
iv) As the thermometer has taken some of the energy, it actually needs less energy to raise the
temperature of the oil than the 2880 J used in the calculation. The value for the specific heat
capacity is therefore less than that calculated, i.e. the value calculated is too large.
7
a)
i) P = IV  I =
At 230 V: I =
𝑃
𝑉
2755 𝑊
230 𝑉
= 12.0 A and at 240 V: I =
3000 𝑊
240 𝑉
= 12.5 A
A 13 A fuse is therefore appropriate and consistent with the data.
ii) R =
𝑉
𝐼
At 230 V, R =
At 240 V, R =
230𝑉
= 19.2 Ω
12.0𝑉
240𝑉
12.5𝑉
= 19.2 Ω
The data is therefore consistent with the resistance of the element being 19 Ω.
b)
E = mc  c =
=
E
m
3000 J s −1  (3.00  60) s
01.5 kg  (100 − 20) K
= 4500 J kg −1K −1
c)
i) Fill the kettle with cold water to cool the kettle. Tip the water out, pour in the 1.0 litre of cold
water and wait for a few minutes for it to reach room temperature (20°C).
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
ii) ΔEsupplied = ΔEwater + ΔEkettle
Answers to Test yourself questions
ΔEsupplied = mwcwΔθ + ΔEkettle
iii) ΔEsupplied for 3.00 minutes = 3000 J s−1 × 180 s = 540 kJ
ΔEsupplied for 2.04 minutes = 3000 J s−1 × 124 s = 372 kJ
540 kJ = 1.5 kg × cw × 80 K + ΔEkettle …..(1)
372 kJ = 1.0 kg × cw × 80 K + ΔEkettle …..(2)
Subtracting (2) from (1)
168 kJ = 0.50 kg × cw × 80 K [Note that ΔEkettle is the same in both cases]
cw =
168 000 J
0.50 kg × 80 K
= 4200 J kg−1K−1
Substituting for cw in equation (2):
372 kJ = (1.0 kg × 4200 J kg−1K−1 × 80 K) + ΔEkettle
372 kJ = 336 kJ + ΔEkettle
ΔEkettle = 36 kJ
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
Answers to Exam practice questions
Pages 511–515 Exam practice questions
1 The unit of specific heat capacity is J kg−1 K−1.
J ≡ N m, and N ≡ kg ms−2, so
J kg−1 K−1 ≡ kg m s−2 × m × kg−1 K−1
The kg cancel, and so the answer is B.
[Total 1 mark]
2 Rearranging ΔE = mcΔθ,
c=
∆𝐸
𝑚∆𝜃
=
11 W × 180 s
0.1393 kg × (29.3−21.4) K
= 1800 J kg−1 K−1
So the answer is D.
[Total 1 mark]
Tip: Don’t forget to convert minutes to seconds, and g to kg.
3 The time (3.0 minutes) and the power (11 W) are only given to 2 SF, so the value for the specific
heat capacity should be given to 2 SF – Answer B.
[Total 1 mark]
4 Energy transferred to the surroundings, or taken by the cup, would mean that more energy would be needed
to raise the temperature – so the value for the specific heat capacity would be too large.
If the thermometer were reading systematically low, both temperatures would be low by the same amount
and so the temperature difference would be unaffected.
The answer is therefore D.
5
[Total 1 mark]
a)
−259°C = (−259 + 273) K = 14 K
[1]
b)
1540°C = (1540 + 273) K = 1813 K
[1]
c)
63 K = (63 − 273)°C = −210°C
[1]
d)
388 K = (388 − 273)°C = 115°C
[1]
[Total 4 marks]
6 ΔE = mcΔθ
m = density × volume = 1.3 kg m−3 × 0.23 m3
[1]
ΔE = 1.3 kg m−3 × 0.23 m3 × 1.0 × 103 J kg−1 K−1 × (20 − (−18)) K
[1]
= 11 kJ
[1]
[Total 3 marks]
7
a)
ΔE = mcΔθ = 45 kg × 800 J kg−1 K−1 × (70 −10) K
6
= 2.16 × 10 J ≈ 2 MJ
2.16 MJ
b)
Cost =
c)
Average power P =
3.6 MJ
[1]
× 10 p = 6 p
∆𝐸
∆𝑡
=
[1]
2.16 × 106 J
5 × 60 × 60 s
= 120 W
[1]
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
d)
Answers to Exam practice questions
Initially, when the temperature of the concrete block is well above room temperature, it will
emit energy at a much greater rate than the average of 120 W. As the block loses energy and its
temperature falls, its temperature difference above the temperature of the room (which is
getting warmer) gets less and so the rate at which it gives out energy to the room will get
gradually less and less, eventually falling to well below the 120 W average.
[2]
[Total 7 marks]
8
a)
ΔE = mcΔθ = 0.25 × 103 kg × 4200 J kg−1 K−1 × (35 − 15) K
[1]
= 21 MJ
b)
[1]
ΔE = PΔt
 Δt =
∆𝐸
𝑃
=
21 × 106 J
[1]
3000 J s−1
= 7000 s ≈ 1.9 h ≈ 2 h
c)
[1]
In practice it would take longer than this, as some energy would be used to heat the element of
the immersion heater and the tank, and some energy would be transferred to the surroundings.
[2]
d)
A shower uses much less water than a bath (try filling your bath with a shower head and see how
long it takes!) and so will consume less energy.
[1]
[Total 7 marks]
9
a)
Drawing a large triangle gives
∆𝜃
∆𝑡
=
(80 − 20) K
[1]
(5.0 − 0.0) min
= 12 K min−1 = 0.20 K s−1
[1]
Note: Drawing a tangent to a curve is not an exact science and so you may get a slightly different
value.
b)
Dividing both sides of ΔE = mcΔθ by Δt gives
∆𝐸
∆𝑡
= mc
∆𝜃
∆𝑡
= 0.117 kg × 4200 J kg−1 K−1 × 0.20 K s−1
[1]
= 98 W
[1]
This is just less than the 100 W rating of the immersion heater, which suggests a small amount of
energy may have been taken by the cup or transferred to the surroundings.
c)
[1]
As the water heats up, its temperature above that of the surroundings increases and so the rate
at which it transfers energy to the surroundings increases. If the water is losing energy at an
increasing rate, the rate at which the temperature rises will slow down.
[2]
[Total 7 marks]
10
a)
Electrical energy  internal energy of water
[1]
PΔt = mcΔθ
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
b)
Δθ =
=
Answers to Exam practice questions
𝑃∆𝑡
𝑚𝑐
10.8 × 103 J s−1 × 60 s
[1]
14 kg × 4200 J kg−1 K−1
= 11 K
[1]
This is a temperature rise, so the temperature of the hot water will be
(16 + 11)°C = 27°C.
[1]
[Total 5 marks]
11
a)
ΔE = mcΔθ = PΔt
c=
𝑃∆𝑡
𝑚∆𝜃
=
11.9 V × 4.12 A × 240 s
[1]
0.250 kg × 10.2 K
= 4614 J kg−1 K−1 ≈ 4600 J kg−1 K−1
b)
i) % difference =
[1]
4600 −4200
4200
× 100 % = 9.5 %
[1]
Tip: Not that the accepted value of 4200 J kg−1 K−1 has been used as the denominator
ii) For water ΔE = mcΔθ = 0.25 kg × 4200 J kg−1 K−1 × 10.2 K
[1]
= 10.7 kJ
[1]
iii) For beaker ΔE = mcΔθ = 0.134 kg × 780 J kg−1 K−1 × 10.2 K
= 1.07 kJ
[1]
iv) The energy taken by the beaker is 10 % of the energy taken by the water, which would
account for the experimental value for the specific heat capacity of the water being 9.5 %
too low. The teacher’s suggestion is therefore probably right.
c)
[2]
An expanded polystyrene cup
• has very little mass and is a poor conductor, so it will not absorb very much energy;
• is a good insulator and so very little energy will be transferred to the surroundings.
[2]
[Total 10 marks]
12
a)
The law of conservation of energy states that energy cannot be created or destroyed but
merely transferred from one type of energy to another.
b)
[2]
i) In Joule’s law, the chemical energy in the cell is transferred into electrical energy. In the
resistor, the moving charges (current) have to do work against the electric field of the
lattice ions (the resistance) and so the electrical energy is transferred into thermal energy
(‘heat’ in Joule’s words).
[2]
ii) Joule hypothesised that the gravitational potential energy of the water at the top of the
waterfall would be transferred into extra kinetic energy as the speed of the water
increased as it fell. This extra kinetic energy would be transferred into thermal energy
when the water hit the bottom, thus causing an ‘elevation of temperature’.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
c)
Answers to Exam practice questions
From conservation of energy: mgΔh → (½mv2) → mcΔθ
mgΔh = mcΔθ  Δθ =
=
𝑔Δℎ
𝑐
9.8 N kg−1 × 270 m
[1]
4200 J kg−1 K−1
= 0.63 K
d)
[1]
[1]
Joule had a ‘long thermometer in his hand’ because the predicted temperature difference was
very small, needing a very sensitive thermometer. If a thermometer is long, it will be more
sensitive because the mercury will expand more for a given rise in temperature.
e)
[2]
At the time, measuring a temperature difference as small as 0.63 K to any degree of precision
was not easy, even under laboratory conditions. Trying to do this at the top and bottom of a
waterfall would be even more difficult as:
• the water is continuously in motion;
• there would be variations in the water temperature at the top;
• the water at the bottom whose temperature is taken may not be the same water as that at
the top unless, say, a rubber ball was used as a ‘marker’;
• the experiment would really need two thermometers, one at the top and one at the bottom,
which would have to be calibrated against each other.
Apart from this, the whole exercise would be extremely hazardous!
f)
[3]
i) As blue light refracts more than red light, the passage of blue light will be as shown in
the figure below.
Note that there is greater refraction for blue light both as the light enters and as it leaves
the droplet.
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
27 Specific heat capacity
Answers to Exam practice questions
ii) On entering the water droplet, the white light is refracted. Each wavelength (colour) has a
different refractive index (due to a different speed in water) and so is refracted by a
different amount (red least, blue most). At the back surface of the drop, the light strikes
the surface at an angle less than the critical angle and so only partial internal reflection
takes place.
When the light leaves the droplet, it is refracted again, which further increases the
deviation of the blue light compared with the red light.
This would suggest that each drop forms a white light spectrum but, unfortunately, it is not
as simple as this. The red light is most strongly observed when its deviation is about 42°
whilst the blue light is most strongly observed when its deviation is about 40°. This means
that each colour of the rainbow comes from different drops that are at the appropriate
angle to the observer for that particular colour. This is why the rainbow is a circular arc. The
arc subtends an angle of 40° (blue) to 42° (red) at our eye, as shown in the figure below.
[4]
[Total 20 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
Page 516 Test yourself on prior knowledge
1
2
a)
2.7 K = (2.7 −273)°C = −270.3°C
b)
−196°C = (−196 + 273) K = 77 K
c)
310 K = (310 – 273)°C = 37°C
d)
The human body temperature is 37°C
a)
i) Ek = ½mv2 = 0.5 × (70 + 30) kg × (8.0 m s−1)2 = 3200 J
ii) Work done = force × distance moved
Ek = Fx
F=
=
Ek
x
3200 J
10 m
= 320 N
b)
By the conservation of energy, the mechanical work done bringing the cycle to rest is transferred
into thermal energy of the brake blocks and wheel rims. This increases the internal energy of the
blocks and rims so that their temperature increases – they get hot.
72 kph =
3
a)
72000 m
= 20 m s −1
(60  60) s
Ek = ½mv2 = 0.5 × 1500 kg × (20 m s−1)2 = 300 kJ
b)
Average power
P=
E 300 kJ
=
= 60 kW
t
5.0 s
(Note that keeping the energy in kJ gives the power directly in kW - a good use of quantity
algebra)
c)
As the manufacturer’s claim is that the car has a power of 125 kW and we have calculated that it
takes only 60 W to accelerate the car from rest to 72 kph in 5 seconds, the manufacturer’s claim
would appear to be reasonable.
d)
W = Fx
W
F=
x
=
300 kJ
30 m
= 10 kN
e)
At 72 kph the car has 300 kJ of kinetic energy. As the car brakes, the mechanical work done by
the frictional force of the brake pads against the brake discs is transferred into thermal energy of
the pads and discs. This energy is eventually dissipated into the air as random kinetic energy of
the air molecules.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
Page 521 Test yourself
1
a)
Work done by motor = KE gained by cyclist (+bike)
½mv2 = 0.5 × (75 + 25) kg × (7.2 m s−1)2
= 2592 J
P=
b)
W
t
 t =
=
W
P
2592 J
200 J s −1
= 12.96 s  13 s
c)
P = IV  I =
=
P
V
200 W
36 V
= 5.56 A
d)
ΔQ = IΔt = 5.56 A × 13 s = 72 C
e)
Charge stored in battery = 10 A h = 10 A × (60 × 60) s = 36 000 C
Percentage used =
2
a)
386 kph = (
72 𝐶
36000 𝐶
x 100% = 0.2%
386
60
) km min-1 = 6.43 km min−1
Distance travelled by blade tip in 1 revolution:
x = 2πr = 2π × 35 × 10−3 m = 0.22 m
Number of revolutions in 1 minute =
b)
P = IV  I =
=
6.43 𝑥103 𝑚
0.22𝑚
= 29 000 rpm
P
V
840 W
240 V
= 3.5 A
This is safely less than 5 A, so a 5 A fuse would be suitable.
c)
ΔE = mcΔθ = 1.0 kg × 4000 J kg−1 K−1 × (80 – 20) K = 240 kJ
d)
In liquidiser: machine uses 840 W for 6 minutes
ΔE = 840 J s−1 × (6 × 60) s = 302 kJ
Efficiency =
240 𝑘𝐽
302𝑘𝐽
× 100 % = 79 %
In microwave: output is 600 W for 9 minutes
ΔE = 600 J s−1 × (9 × 60) s = 324 kJ
Efficiency =
240 𝑘𝐽
324𝑘𝐽
× 100 % = 74 %
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
On the face of it, the microwave would appear to be almost as efficient, but to give an output of
600 W, the microwave may typically use as much as 1100 W. Its actual efficiency will therefore
be nearer 40% – considerably less than the liquidiser.
e)
In the liquidiser, the metal blades do mechanical work (ΔE = FΔx) against the frictional force
caused by the viscosity of the soup. This energy is transferred into vibrational kinetic energy of
the molecules of the liquid and blades, thus raising the temperature.
In the microwave, the energy, hf, of the quanta of electromagnetic radiation is absorbed by the
molecules of the soup, increasing their vibrational kinetic energy and thus raising the
temperature of the soup.
Page 523 Activity 28.1
1
You would need to set up the apparatus as shown in the figure below.
2
The melting point temperature is given by the horizontal section of the graph.
3
At the start, the stearic acid is a liquid. As its temperature falls, the kinetic energy of its molecules gets less
(and so does the potential energy, but not by much). Once the stearic acid reaches its melting point, only its
potential energy is reduced, as it changes state from liquid to solid – the molecules in the solid state have a
lower potential energy. As there is no change in the kinetic energy, the temperature remains constant.
Once all the stearic acid has solidified, it starts to cool down again, with a reduction in the kinetic energy of
its molecules (and a slight reduction in potential energy).
Page 525 Test yourself
3
a)
i) The latent heat of fusion of ice is 330 kJ kg−1 means that for ice at its melting point, it would
need 330 kJ of energy per kilogram to change it into water, without change in temperature.
ii) Thermal equilibrium is attained when the rate at which the melted water is absorbing energy
from the room is equal to the rate at which it is emitting it back to the room – the water will
be at room temperature.
b)
As the ice warms up from −18°C to 0°C its molecules gain both kinetic energy and some potential
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
energy from the room. At 0°C, whilst the ice is melting, the kinetic energy of the molecules
remains constant (constant temperature) whilst the potential energy of the molecules increases.
Once all the ice has melted into water, both the potential energy and the kinetic energy of the
water molecules increase until the water reaches the room temperature of 18°C. The energy of
the water molecules then remains constant as there is no further change in temperature.
c)
Energy needed for ice to warm from −18°C to 0°C:
ΔE = mcΔθ = 10 × 10−3 kg × 2100 J kg−1 × 18 K = 378 J
Energy needed to change ice into water at 0°C:
ΔE = LΔm = 330 × 103 J kg−1 × 10 × 10−3 kg = 3300 J
Energy needed for this water to warm from 0°C to a room temperature of 18°C
ΔE = mcΔθ = 10 × 10−3 kg × 4200 J kg−1 × 18 K = 756 J
Total energy taken from room = (378 + 3300 + 756) J = 4434 J
= 4.4 kJ (to 2 SF)
4
Your graph should look similar to this one.
Points to note are:
•
your graph should indicate the relevant temperatures;
•
the gradient of the line from −18°C to 0°C should be twice as steep as that from 0°C to 18°C because
the specific heat capacity for ice is only half that for water or, looking at it from another angle, the
time taken to warm from 0°C to 18°C is twice that to warm from −18°C to 0°C as the specific heat
capacity for water is twice that for ice;
•
the time for the ice to melt should be about 9 times as long as the time to go from −18°C to 0°C as
the energy needed (3300 J) is about 9 times more (9 × 378 J = 3402 J).
Page 526 Core practical 13A
1
By conservation of energy:
ΔE to melt ice + ΔE to warm this water to the final temperature
= ΔE lost by water in cup
ΔmL + Δm × c × Δθi = m × c × Δθw
23.2 ×10−3 kg × L + 23.2 ×10−3 kg × 4200 J kg−1 K−1 × (15 – 0) K
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
= 0.200 kg × 4200 J kg−1 K−1 × (25 – 15) K
23.2 ×10−3 kg × L + 1462 J = 8400 J
(8400 − 1462) J
= 3.0  105 J kg −1
23.2  10 −3 kg
2
The advantages of using an expanded polystyrene cup are:
•
the cup is a very good insulator and so the heat lost/gained to or from the room is negligible;
•
as the cup is a poor conductor and has very little mass, the energy taken/given out by the cup is also
negligible.
3
The ice needs to be melting so that it is at 0°C and the pieces of ice need to be small so that all the ice is at
0°C. Large pieces of ice are likely to be at a temperature below 0°C at the centre and so would take more
energy than that calculated, giving a value for the specific latent heat that is too large. The ice needs to be
dry – if not, the surface water will no longer need any energy to melt it. This would give rise to a value for
the specific latent heat that is too low.
4
Assuming the ice is added at a steady rate, the energy lost to the atmosphere by the water when it is above
room temperature will, to a good approximation, be equal to the energy gained by the water whilst it is
below room temperature.
5
In Question 1, we calculated that the energy to warm the melted ice to the final temperature was 1462 J.
The final temperature was 15.0 ± 0.25°C.
The % uncertainty in this temperature is
0.25℃
15.0℃
x 100% = 1.7%
The uncertainty in the energy is therefore 1.7% of 1462 J = 24 J
The energy lost by the warm water in the cup was 8400 J. This was calculated from a temperature
difference of (25 ± 0.25 – 15 ± 0.25) K.
The % uncertainty in this temperature difference of 10 K is therefore
0.50℃
10.0℃
x 100% = 5%
This gives an uncertainty of 5% of 8400 J = 420 J in the energy.
In the calculation of the specific latent heat, the energy is therefore
(8400 ±420 – 1462 ± 24) J.
The % uncertainty in this energy is
(420+24)𝐽
(8400−1462)𝐽
x 100% = 6%
Assuming no other sources of uncertainty, this is therefore the % uncertainty in the value for the
specific latent heat.
6
Percentage difference between the experimental value and the accepted value is given by
(3.34 − 3.0)  105 J kh −1
 100% = 10%
% difference =
3.34  105 J kg −1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
This is somewhat larger than the % uncertainty due to the uncertainties in measuring the temperatures,
which suggests that there are other uncertainties.
Note that the accepted value is used as the denominator when a comparison with an accepted value is
being made.
7
If the ice is not completely dry, less energy will be needed to melt the calculated mass of ice added to the
warm water and so the specific latent heat will be less than it should be.
Page 527 Core practical 13B
1
For the first 2 minutes the water is being heated up to its boiling point and so its mass will remain constant
(a little water may evaporate, but this won’t be detected by the balance). After 2 minutes, the water begins
to boil, steam is given off and the mass of water left in the beaker gradually reduces. The graph therefore
curves downward. After about 4½ minutes, all the water has reached its boiling point and the rate at which
it is being boiled off remains constant – the graph is a straight line of negative slope equal to − dm/dt.
2
Choosing a large triangle, the gradient of the graph is given by
−
3
dm (436 − 473)  10 −3 kg
=
= −2.06  10 −4 kg s −1  −2  10 −4 kg s −1
dt
(8.0 − 5.0)  60 s
For the straight part of the graph, assuming no energy is transferred to the surroundings:
rate at which energy is being supplied = rate of energy needed to change state of boiling water
From
E = Lm  power of heater =
dE
dm
=L
dt
dt
500 J s−1 = L × 2.06 × 10−4 kg s−1
L=
4
500 J s −1
= 2.43  10 6 J kg −1 = 2.4  10 6 J kg −1
2.06  10 −4 kg s −1
( to 2 SF)
Percentage difference between this value and the accepted value of 2.34 × 106 J kg−1 is given by]
% difference =
(2.43 − 2.34)  10 6 J kg −1
 100% = 4%
2.34  10 6 J kg −1
Note that the accepted value is used as the denominator when a comparison with an accepted value is
being made.
5
This value is probably larger than the accepted value because, despite expanded polystyrene being a good
insulator, some of the energy from the heater will be transferred to the surroundings, particularly as the
temperature of the boiling water is likely to be around 80°C above room temperature.
Page 528 Test yourself
5
a)
Volume of ice = 3 × 15 cm3 = 45 cm3
Mass of ice = density × volume
= 920 kg m−3 × 45 × 10−6 m3
= 0.0414 kg
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
Energy to raise temperature from −18°C to 0°C:
ΔE = mcΔθ
= 0.0414 kg × 2100 J kg−1 K−1 × 18 K
= 1565 J
Energy needed to change ice into water at 0°C:
ΔE = LΔm
= 330 × 103 J kg−1 × 0.0414 kg
= 13662 J
Total energy to melt ice cubes = (1565 + 13662) J = 15227 J
Let the final temperature of the drink be θ°C, then:
ΔE to melt ice cubes + ΔE to raise this water to θ°C
= ΔE taken from drink to cool it from 25°C to θ°C
Mass of drink = density × volume
= 1.04 × 103 kg m−3 × 330 × 10−6 m3
= 0.3432 kg
15227 J + 0.0414 kg × 4200 J kg−1 K−1 × (θ – 0) K
= 0.3432 × 4200 J kg−1 K−1 × (25 – θ) K
15227 + 174 θ = 441(25 – θ)
 (36025 – 15227) = (174 + 1441) θ
 θ=
20798
1615
= 12.9 °C
The final temperature of the drink is therefore 13°C
Note: In a question like this
• take great care with units;
• ask yourself whether your final answer seems reasonable – is 13°C reasonable?
b)
We also have to assume that
• the beaker does not gain or lose energy from or to the room;
• the ice does not gain energy from the room whilst it is being transferred to the water.
6
a)
Your graph should look similar to this one.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Test yourself questions
b)
To heat the water from 20°C to 100°C we have
E = mc 
dE
d
(100 − 20) K
= mc
= m  4200 
= 1120 m J kg −1s −1
dt
dt
5  60 s
To boil off 0.5m of water we have:
E = Lm 
dE
 t = Lm  1120m  t = L  0.5m
dt
L × 0.5m = 1120m J kg−1 s−1 × 18 × 60 s = m (1.21 × 106) J kg−1
m(1.21  10 6 )J kg −1
= 2.4  10 6 J kg −1
0.5 m
L=
Page 530 Test yourself
7
a)
At absolute zero, ignoring any quantum effects, we consider that molecules of an ideal gas have
no energy, either potential or kinetic.
b)
When ice melts into water at 0°C, work has to be done against the intermolecular forces. This
causes an increase in the potential energy between the bonds. However, as the temperature
remains at 0°C during the melting process, there is no change in the kinetic energy of the water
molecules.
c)
When steam at 100°C condenses into water droplets, the molecules become much closer
together and so there is a large reduction in their potential energy. As the temperature remains
constant at 100°C whilst condensation is taking place, the kinetic energy of the molecules stays
the same.
8
As physicists consider temperature to be a measure of molecular kinetic energy, the temperatures should
be measured on the kelvin scale on which the molecular kinetic energy is zero at 0 K.
On the kelvin scale:
Temperature in Hong Kong = (36 + 273) K = 309 K
Temperature in London
= (18 + 273) K = 291 K
The temperature in Hong Kong is therefore
(309 − 291) K
x100% = 5.8%  6%
309 K
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
Pages 532–536 Exam practice questions
1
In an ideal gas we assume that there are no inter-molecular forces (except during a collision) and so an
ideal gas has no potential energy. The gas will have internal energy due to the random kinetic energy of the
molecules. The answer is therefore C.
2
[Total 1 mark]
When ice melts, it does so at a constant temperature (0°C) and so the kinetic energy of its molecules does
not change. The volume decreases on melting and the potential energy (and hence the internal energy)
increases. The answer is therefore D.
3
[Total 1 mark]
Heat and work are both forms of energy and so are measured in J, the unit of energy. Power is the rate of
doing work and is measured in J s−1 (= W). The answer is therefore C.
4
a)
−39°C = (−39 + 273) K = 234 K
b)
630 K = (630 − 273)°C = 357°C
c)
156 K = (156 − 273)°C = -117°C
d)
79°C = (79 + 273) K = 352 K
e)
54 K = (54 − 273)°C = −219°C
f)
−183°C = (−183 + 273) K = 90 K
g)
1083°C = (1083 + 273) K = 1356 K
h)
2853 K = (2853 − 273)°C = 2580°C
[Total 1 mark]
[Total 4 marks]
5
The internal energy of a body is the energy that a body has due to the energy of the atoms or molecules
that make up the body. These atoms or molecules have forces between them – bonds – which give them
potential energy, and they are in continuous motion, which gives them kinetic energy. The potential energy
of the bonds is like the energy stored in an elastic band when it is stretched. Any object in motion has
kinetic energy – for example, when you kick a football, the energy from your foot gives the football kinetic
energy.
[2]
Heating is when energy is transferred between two objects due to a temperature difference. For example,
if you put your cold hand on a warm radiator, energy is transferred from the radiator to your hand. This
increases the internal kinetic energy of the molecules of your hand and so your hand warms up.
[2]
If you rub your hands together you are doing work against the frictional forces between your hands. This
work is transferred to the internal energy of the molecules of your hands and so your hands get warm.
[2]
[Total 6 marks]
Tip: Note that each of the three terms in italics here has been explained individually, with an everyday
example given for each. This is essential if you want to get good marks in a question like this.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
6
a)
Force due to 5 kg mass = 5 kg × 9.8 N kg−1 = 49 N
Frictional force due to bands = (49 – 16) N = 33 N
b)
[2]
Work done against friction per revolution = force × distance moved
ΔW = FΔx = 33 N × 2π × (29 × 10−3) m = 3.01 J ≈ 3 J
c)
Work done after 400 revolutions = 400 × 3.01 J = 1204 J
ΔW = mcΔθ  c =
d)
[2]
𝛥𝑊
𝑚𝛥𝜃
=
1204 J
0.197 kg × 14.5 K
= 420 J kg−1 K−1
[3]
This is likely to be higher than the accepted value as some of the calculated 1200 J of energy will
be transferred to the bands and to the surrounding air. The actual energy received by the
cylinder will be less than this, so the specific heat capacity will really be less than that calculated.
In other words, the experimental value will be too large.
[2]
[Total 9 marks]
7
a)
When a tray of water at room temperature is put into a freezer, where the temperature is
typically −18°C, the temperature of the water will fall and the kinetic energy of its molecules will
decrease, with little change in their potential energy. This continues until the water reaches 0°C,
at which point the temperature, and hence the kinetic energy of the molecules, remains constant
while the water changes from liquid to solid. A solid is more stable than a liquid and so the
potential energy of the molecules has been reduced. Once the water has all changed into ice, the
potential energy remains more or less constant while the kinetic energy of the molecules gets
less and less as the ice cools from 0°C to −18°C.
b)
[4]
Energy removed to cool water from 20° to 0°C:
ΔE = mcΔθ = 0.200 kg × 4.2 × 103 J kg−1 K−1 × (20 – 0) K = 16 800 J
[1]
Energy removed to change this water into ice:
ΔE = LΔm = 3.3 × 105 J kg−1 × 0.200 kg = 66 000 J
[1]
Energy removed to cool this ice from 0°C to –18°C:
ΔE = mcΔθ = 0.200 kg × 2.1 × 103 J kg−1 K−1 × (0 – {− 18}) K = 7 560 J
[1]
Total energy removed = (16 800 + 66 000 + 7560) J = 90 kJ
[1]
[Total 8 marks]
8
a)
i) Energy needed to raise temperature of water from 20°C to boiling point, 100°C:
ΔE = mcΔθ = 1.0 kg × 4.2 × 103 J kg−1 K−1 × (100 – 20) K = 3.36 × 105 J
[1]
Energy needed to change this boiling water into steam:
ΔE = LΔm = 2.3 × 106 J kg−1 × 1.0 kg = 2.3 × 106 J
[1]
Total energy needed = (3.36 × 105 + 2.3 × 106) J = 2.636 × 106 J
[1]
Kettle is supplying energy at the rate of 2.4 kW = 2400 J s
So time to boil dry would be
2.636 × 106 J
2400 J s−1
= 1098 s = 18.3 minutes
[2]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
ii) Before drawing the graph we must do a calculation. From the above, the energy to heat
the water from 20°C to 100°C was 3.36 × 105 J.
This would take 33 600 J / 2400 J s−1 = 140 s = 2.3 minutes
[1]
We can now plot a graph like the one below.
The line is horizontal at a constant volume of 1.0 litres whilst the temperature of the water
rises from 20°C to 100°C in 2.3 minutes. The line then slopes downwards at a constant
rate until all the water has boiled away after 18.3 minutes.
b)
i) Your graph should look like the one below.
[2]
[2]
ii) We must first calculate the resistance that would give an output p.d. of 3.0 V when the
thermistor is at 0°C. From the graph, the resistance of the thermistor at 0°C is 32 kΩ.
We can use the formula 𝑉out = 𝑉in ×
𝑉in
𝑉out
=
𝑅1 + 𝑅2
𝑅2

9V
3V
=3=
𝑅1 + 𝑅2
𝑅2
𝑅1
and re-arrange to give
𝑅2 + 𝑅2
 3R2 = R1 + R2  2R2 = R1
As R1 = 32 kΩ  R2 = 16 kΩ
[2]
The most suitable resistor would therefore be the 15 kΩ resistor.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
This is rather a long-winded way of calculating the required resistance. As the numbers
are simple, you should be able to spot that if the output is 3 V, then 6 V must be dropped
across the thermistor. The ratio of the voltage across the output to that across the
thermistor is then 1:2. The output resistance must therefore be half that of the thermistor
resistance, giving an output resistance of ½ × 32 kΩ = 16 kΩ.
iii)
The purpose of the potentiometer is to ‘fine tune’ the output resistance to the exact value
required for the output voltage to be 3.0 V when the thermistor is at 0°C.
iv)
[1]
The circuit can be calibrated by putting the thermistor in a funnel of melting ice and
adjusting the output voltage to be exactly 3.0 V by varying the resistance of the
potentiometer.
[2]
[Total 16 marks]
9
a)
Absolute zero is the lowest temperature that can theoretically be reached. In terms of the
kinetic theory, it is the temperature at which the molecules of matter have their lowest
possible average kinetic energy. In a simplified model, the molecules are considered to have
no kinetic energy at absolute zero, in other words they have no random movement. In
practice, quantum mechanics requires that they have a minimum kinetic energy, called the
zero-point energy.
b)
[2]
Kammerlingh Onnes’ achievement of preparing liquid helium led initially to experimental work in
low temperature physics, in particular in superconductivity and superfluidity. This gave rise to
the development of quantum mechanics to explain these phenomena. In turn, quantum
mechanics has contributed to today’s technology, such as computers and MRI scanners.
c)
[2]
A superconductor is a material that at low temperatures has no electrical resistance. This means
that once a current is created in it, the charge will continue to flow, even if the source of the
current (e.g. battery) is removed.
d)
[2]
Practical use of superconductors is now being made in creating very powerful electromagnets.
These are used, for example, in particle physics, such as the Large Hadron Collider (LHC) at CERN,
in MRI scanners and magnetic levitating trains.
Research is being undertaken in the application of superconductors in the generation and
transmission of electricity and in high-powered computers.
[2]
[Total 8 marks]
10
a)
Efficiency =
𝑇1 −𝑇2
𝑇1
× 100 % where the temperatures are in kelvin, K.
T1 = 600°C = (600 + 273) K = 873 K
 46 % =

(873 − 𝑇2 ) K
46 × 873
100
873 K
× 100 %
[1]
= (873 – T2)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
 873 – T2 = 402
b)
 T2 = (873 – 402) K = 471 K = 198°C ≈ 200°C
[2]
T1 = 700°C = (700 + 273) K = 973 K, and T2 = 471 K as found in part a).
[1]
Efficiency =
=
𝑇1 −𝑇2
𝑇1
(973 − 471) K
973 K
× 100 %
× 100 % = 52 %
[1]
So the manufacturer’s claim that an efficiency of 50 % will be achieved with a steam temperature
of 700°C is valid.
c)
[1]
The remaining 50 % or so of the energy could be used as a piped hot water supply to the district
surrounding the power station for use in houses and factories. This is called a combined heat and
power (CHP) system. In some power stations this energy is recycled to heat the steam before it
enters the turbine.
[2]
[Total 8 marks]
11
a) i)
The atoms in a metal are arranged in a regular pattern called a lattice. The atoms in the
lattice are held in position by the inter-atomic bonds, rather like springs (see Figure 28.1
on page 518 of your textbook). This enables the atoms to vibrate about their mean
positions. The energy of these lattice vibrations is dependent on the temperature.
[1]
In a metal, the outer electrons are only loosely bound to the atom and can move
randomly from atom to atom. At any one time there is, on average, about one electron
per atom that is not attached to a particular atom. As this is the mechanism by which
metals conduct electricity, these electrons are called conduction electrons.
ii)
[1]
When an electric field is applied to a metal, the conduction electrons experience a force,
which causes them to drift in the direction of the applied field. The moving electrons
experience an equal and opposite force due to the positive lattice ions – electrical
resistance – and so the electrons drift at a constant speed – electric current.
b)
[2]
In the equation
c = αT3 + γT
the units must be the same on both sides of the equation. As the units of c are J kg−1 K−1, the
units of αT3 must also be J kg−1 K−1. This means the units of α must be:
J kg−1 K−1
K3
c)
= J kg−1 K−4
[1]
[2]
The lattice vibrations and conduction electrons make an equal contribution to the specific heat
capacity when:
αT3 = γT
[1]
Dividing both sides by T gives us:
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
αT2 = γ  T =
−1 −2
−3
𝛾
5.8 × 10 J kg K
√𝛼 = √
−1 −4
−4
1.7 × 10
J kg
[2]
K
= 5.8 K ≈ 6 K
d)
[1]
At 5.8 K, as αT3 = γT, we can express c = αT3 + γT as:
c = γT + γT = 2γT
[1]
−3
−1
−2
= 2 × 5.8 × 10 J kg K × 5.8 K
[1]
= 0.067 J kg−1 K−1
[1]
[Total 14 marks]
12
a)
You should find values the c at temperatures T of 1 K, 2 K, 3 K…. 10 K.
[2]
Then plot a graph of c against T. Your graph should look like the figure below.
[3]
The energy needed to increase the temperature of 5 g of silver from 0 K to 10 K can then be
found determined by finding the area under the graph from 0 K to 10 K (in J kg −1) and multiplying
this by 5.0 × 10−3 kg. You should get a value of about 4 × 10−3.
b)
[3]
δE = mcδT = m(αT3 + γT)δT
𝛼𝑇 4
10 K
E = m∫0
(α𝑇 3 + 𝛾𝑇)𝑑𝑇 = m [
= 5.0 × 10−3 kg × [
4
+
𝛾𝑇 2
2
10 K
]
[2]
0
1.7 × 10−4 J kg−1 K−4 × (10 K)4
5.8 × 10−3 J kg−1 K−2 × (10 K)2
]+[
]
4
2
= 5.0 × 10−3 kg × (0.425 + 0.290) J kg−1
[1]
−3
= 3.6 × 10 J
[1]
[Total 12 marks]
13
a)
2
As temperature is related to the kinetic energy (½mv ) of the atoms, slowing down the atoms
will reduce their kinetic energy and therefore the temperature.
b)
E = hf =
ℎ𝑐
𝜆
=
6.63 × 10−34 J s × 3.00 × 108 m s−1
589.6 × 10−9 m
= 3.37 × 10−19 J
E =
3.37 × 10−19 J
1.6 × 10−19 J eV−1
[1]
[1]
[1]
= 2.109 eV ≈ 2.1 eV
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
c)
From de Broglie:
ℎ
λ =
 p =
𝑝
ℎ
𝜆
=
6.63 × 10−34 J s
[1]
589.6 × 10−9 m
= 1.12 × 10−27 kg m s−1
d)
[1]
The law of conservation of momentum states the total momentum of a system remains the same
provided that no resultant external force acts on the system.
[1]
Consider the momentum of the system shown in the figure below.
Before the sodium atom absorbs the photon of momentum p (Figure a above) the momentum of
the system is (Mv – p) to the right.
[1]
After absorption (Figure b above), the momentum is M(v – Δv) to the left.
[1]
By conservation of momentum M(v – Δv) = (Mv – p)  MΔv = p
[1]
This means that Δv is in the same direction as p (to the left) so the atom will slow down.
[1]
e)
From d) we have MΔv = p  Δv =
f)
Δv =
𝑝
𝑀
=
𝑝
[1]
𝑀
1.12 × 10−27 kg m s−1
[1]
(23 × 1.66 × 10−27 )kg
= 0.029 m s−1 ≈ 3 cm s−1
g)
[1]
−1
The sodium atom has an original speed of 570 m s . Each collision slows it down by
0.029 m s−1, so to bring it to rest the number of collisions needed will be:
Number of collisions =
570 m s−1
[1]
0.029 m s−1
= 19 655 collisions ≈ 20 000 collisions
h)
As there are 10 absorptions per second, the time required will be:
t =
20 000 collisions
1 × 107 collisions s−1
= 2 ms
[1]
This would suggest that an atom would be stopped in a matter of milliseconds.
i)
[1]
7
[1]
A sodium atom will selectively absorb photons that have exactly the right amount of energy, hf,
to excite an electron to a higher energy level (hf = E2 – E1). The frequency f of photons for which
this occurs is called the resonant frequency.
j)
[2]
The laser is tuned to a frequency slightly below that of the resonant frequency so that the
Doppler shift, due the speed of approach of the atom relative to the photon, brings the
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
28 Internal energy, absolute zero and change of
state
Answers to Exam practice questions
frequency to exactly the resonant frequency. This means that an atom will absorb photons
coming head-on towards it, but not photons coming from behind, which will have a Doppler shift
away from the resonant frequency.
k)
Doppler shift:
Δλ
λ
As c = fλ  f =
 Δf =
𝑐𝑣
𝜆𝑐
=
v
λ
=
Δf
f
=
𝑣
𝑐
 Δf =
[2]
𝑓𝑣
𝑐
c
λ
=
570 m s−1
[1]
589.6 × 10−9 m
= 9.67 × 108 Hz ≈ 0.97 GHz
[1]
[Total 24 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Test yourself questions
Page 537 Test yourself on prior knowledge
1
p=
=
F
A
50 kg  9.8 N kg −1
245  10 −4 m 3
= 20 k Pa
2
p=
=
F
A
0.5  1000 kg  9.8 N kg −1
0.07 m  0.70 m
= 100 k Pa
This is equivalent to the pressure of the atmosphere!
3
4
a)
θ = (234 – 273)°C = −39°C
b)
T = (357 + 273)°C = 630 K
c)
T = (−219 + 273)°C = 54K
d)
θ = (90 – 273)°C = 183°C
a)
m = 32 × (1.66 × 10−27 kg)
Ek = ½mv2
= 0.5 × (32 × 1.66 × 10−27) kg × (491 m s−1)2
= 6.40 × 10−21 J
6.40  10 −21 J =
b)
5
a)
6.40  10 −21 J
= 0.040 eV
1.6  10 −19 J eV −1
i) Momentum is a vector quantity as it has direction as well as magnitude.
ii) In an elastic collision, no kinetic energy is lost by the system in the collision.
b)
i) m = 28 × (1.66 × 10−27 kg)
p = mv = (28 × 1.66 × 10−27 ) kg × 491 m s−1
= 2.28 × 10−23 kg m s−1
ii) After an elastic collision with a wall, the molecule will rebound with the same kinetic energy,
so its speed will be the same. However, its direction is directly opposite to its original
direction. This means its momentum will be −2.28 × 10−23 kg m s−1. The minus sign indicates
its new direction is opposite to the initial direction.
iii) The change in momentum will be:
Δ(mv) = momentum after collision – momentum before collision
= (−mv) – (mv) = −2mv
Note: It is important to pay careful attention to the signs (i.e. the direction) when dealing
with vector quantities such as momentum.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Test yourself questions
Page 539 Test yourself
1
a)
p = ρgΔh
= 13.6 × 103 kg m−3 × 9.81 N kg−1 × 0.758 m
= 101 kPa
b)
In the equation p = ρgΔh, the Earth’s gravitational field constant g varies from place to place on
the Earth and so the local value must be known in order to get an accurate value for the
pressure. The density of mercury ρ depends on temperature – it gets less as the temperature
increases. Therefore the value needs to be adjusted according to the temperature.
2
a)
Points X and Y are at the same pressure as they are at the same level in the water.
b)
The pressure due to the column of oil above point X must be equal to the pressure of the column
of liquid above point Y or else the liquids would flow one way or the other due to the pressure
difference.
c)
ρoil g Δhoil = ρwater g Δhwater
 oil =
d)
 water  hwater
hoil
=
1000 kg m −3  158 mm
= 790 kg m −3
200 mm
The following techniques should be employed:
•
a setsquare should be used, with one edge held against the vertical scale and the other
edge level with the bottom of the liquid meniscus as shown in Figure A12.1
•
the levels at W, X, Y and Z should be recorded, then
hoil = W – X and hwater = Z – Y.
•
your eye should be level with the bottom of the meniscus to avoid parallax error
•
repeat readings should be taken and averaged.
Note the use of bullet points to set out the answer clearly.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Test yourself questions
Page 541 Core Practical 14
1
2
You should have plotted a graph of p against 1/V as we are changing the volume (the independent variable)
and measuring the consequent pressure (the dependent variable). A graph of 1/V against p, in this
particular case, would be acceptable.
3
The graph is a straight line through the origin, which shows that 1/V is proportional to p.
1
 p  pV = constant
V
This confirms that the air under test obeys Boyle’s law.
4
It is important to allow time for thermal equilibrium to be reached before taking readings to ensure that
the air remains at a constant temperature, which is a necessary condition of Boyle’s law.
Page 546 Test yourself
3
When the table tennis ball is put into the hot water, thermal energy is conducted through the thin plastic
casing to the air inside. The random kinetic energy of the air molecules inside the ball increases and so the
pressure inside the ball increases. As the pressure inside the ball is now greater than the pressure outside,
the excess pressure pushes the dent back out again.
4
a)
The student has forgotten that the pressure of a gas is proportional to its temperature in kelvin,
not in °C.
𝑝1 𝑉1
𝑇1
=
𝑝2 𝑉2
𝑇2
⇒
𝑝1
𝑇1
=
𝑝2
𝑇2
when V is constant
In other words, the pressure has gone up by 9%.
b)
As before, ⇒
𝑝2
𝑃1
=
𝑇2
𝑇1
⇒
𝑇2
𝑇1
=2
 T2 = 2T1 = 2 × 300 K = 600 K = 327°C
5
a)
i) ΔE = mcΔθ = 25 × 10−3 kg × 2000 J kg−1 K−1 × (45 −20) K
= 1250 J
ii) The mass of air is much less than the mass of the rubber and so it is reasonable to ignore the
energy needed to heat the air.
iii) When the temperature of the ball increases, the random kinetic energy of the air molecules
increases, so they will move around faster.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Test yourself questions
b)
p1v1 p2 v2
p
p
=
 1= 2
T1
T2
T1 T2
 p2 =
=
c)
when V is constant
p1T2 101kPa  (273 + 45) K
=
T1
(273 + 20) K
101kPa  318 K
= 110 kPa
293 K
i) Energy from the racket strings compresses the rubber ball and transfers energy to increase
the internal potential energy stored in the distorted ball. Some of this energy is recovered to
give the ball kinetic energy as it flies off the racket but, due to hysteresis, some of the energy
stored in the rubbers is transferred into thermal energy. This energy is conducted to the air
inside the ball and increases the random kinetic energy of its molecules, therefore raising the
temperature.
ii) A warm ball bounces better than a cold one because rubber is more elastic at a higher
temperature. This means more of the kinetic energy given to the ball is recovered when it
bounces. The increased air pressure inside the ball makes the ball ‘harder’ and further
increases its elasticity.
d)
The initial GPE of the ball is 2.5 m × ρ × g.
After bouncing, the GPE at the top of the bounce is 0.75 m × ρ × g.
Ratio of
energy after bouncing
0.75  ρg
=
= 0.30
energy before bouncing
2.5  ρg
In other words, 0.70 (70%) of the energy is dissipated when the ball bounces on the floor.
Page 549 Test yourself
6
a)
An ideal gas is a gas that obeys the equation pV = NkT under all conditions of pressure and
temperature.
b)
A real gas will behave like an ideal gas when:
• the pressure is not too large
• it is above its critical temperature.
c)
pV = NkT  N =
=
𝑝𝑉
𝑘𝑇
where V = 22.4 × 1000 cm3 = 22.4 × 1000 × 10−6 m3
101  103 Pa  (22.4  1000  10 -6 ) m3
1.38  10 -23 JK -1  273K
= 6.01 × 1023 molecules ≈ 6 × 1023 molecules
7
a)
p1V1 p2V2
=
T1
T2
where p is constant
V1 V2
V T 400 cm 3  (100 + 273) K
=  V2 = 1 2 =
= 501cm 3  500 cm 3
T1 T2
T1
(25 + 273) K
b)
p1V1 p2V2
=
T1
T2
where T is constant (as it is an isothermal)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Test yourself questions
p1V1 = p2V2  V2 =
c)
p1V1 p2V2
=
T1
T2
p1V1 100 kPa  500 cm
=
= 125 kPa
V2
400 cm 3
3
where V is constant
p1 p2
p T 100 kPa  (100 + 273) K
=
 T2 = 2 1 =
= 298 K = 25 C
T2 T2
p2
125 kPa
d)
This figure shows that:
• A→B: Volume increases from 400 cm3 to 500 cm3 at constant pressure.
• B→C: An isothermal (p-V) curve to a pressure of 125 Pa and a volume of 400 cm3.
• C→A: Cooling back to a pressure of 100 kPa at constant volume
8
a)
i) Air is mainly made up of separate oxygen and nitrogen molecules. These molecules are
completely independent of each other – they do not combine in any way to form an ‘air’
molecule.
ii) We can think of an ‘air’ molecule as having a mass equal to the average mass of the oxygen
and nitrogen molecules from which the air is composed. As air is approximately 20% oxygen
(32 u) and 80% nitrogen (28 u), the average is given by
0.2 × 32 + 0.8 × 28 = 6.4 + 22.4 = 28.8 ≈ 29
We can therefore think of an ‘air’ molecule as having a molecular mass of 29 u.
b)
From pV = NkT 
pV
101 103 Pa  500  10 −6 m 3
=
kT 1.38  10 −23 J K −1  (273 + 37)K
= 1.18 × 1022 molecules
c)
i) As 20% of the air is oxygen, the number of oxygen molecules is
0.2 × 1.18 × 1022 molecules = 2.36 × 1021 molecules
ii) Mass of oxygen = 2.36 × 1021 molecules × 32 × 1.66 × 10−27 kg molecule−1
= 1.25 × 10−4 kg = 125 mg
Page 555 Test yourself
9
a)
An ideal gas is one which obeys the ideal gas equation pV = NkT under all conditions of
temperature and pressure.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Test yourself questions
b)
We assume that
• an ideal gas is made up of a very large number of molecules having random motion;
• intermolecular forces are negligible except during collision;
• collisions of the molecules with each other and with the walls of any container are elastic;
• the time spent in collisions is negligible compared with the time spent between collisions;
• the molecules themselves have negligible volume compared with the volume they move
around in.
c)
m = mass of each molecule.
c2 = mean square speed, i.e. the average of the squares of all the individual speeds of the
molecules.
k = Boltzmann constant, a constant having value 1.38 × 10−23 J K−1.
T = temperature of gas, measured in kelvin.
10
d)
The term ½ mc2 represents the average kinetic energy of each of the gas molecules.
a)
N = total number of molecules of the gas;
k = Boltzmann constant, a constant having value 1.38 × 10−23 J K−1.
b)
Units of pV are Pa × m3 = N m−2 × m3 = N m = J = units of energy
c)
As pV = NkT, the units of NkT must also be J
N is just a number (the number of molecules) so units of kT = J
As the unit of T is K, then units of k are J K−1
d)
From pV = NkT  N = pVkT where V = 10 m × 8 m × 3 m = 240 m3
N=
101 103 Pa  240 m 3
= 6.0  10 27 molecules
1.38  10 −23 J K −1  (273 + 19)K
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Exam practice questions
Pages 557–560 Exam practice questions
1
a)
The equation for Boyle’s law is pV = constant. This means:
• pV will be the same for all V values (graph A);
• a graph of p against V will be a rectangular hyperbola (graph B);
• a graph of p against 1/V will be a straight line through the origin (graph D).
The only incorrect graph is therefore C, which would infer that p  V.
b)
[1]
The equation for an ideal gas is pV = NkT. This means:
• p  T (graph A);
• V  T (graph B);
• pV/T = Nk = constant for all T values (graph C);
Therefore graph D is incorrect.
[1]
[Total 2 marks]
2
a)
Pressure p =
𝐹
𝐴
=
57 kg × 9.81 N kg−1
0.80 × 10−4 m2
= 6.99 × 106 N m−2 = 7.0 MPa
The answer is C.
[1]
Tip: Don’t forget to multiply the mass by g and to convert the area to m2.
b)
Pressure p =
𝐹
𝐴
=
1600 kg × 9.81 N kg−1
4 × 56 × 10−4 m2
= 7.01 × 105 N m−2 = 0.70 MPa
The answer is B.
[1]
[Total 2 marks]
3
Using
𝑝1 𝑉1
𝑇1
=
𝑝2 𝑉2
𝑇2
where V1 = V2
p1 = 101 kPa
T1 = (17 + 273) K = 290 K
T2 = (100 + 273) K = 373 K
p2 =
𝑝1 𝑉2
𝑇1
=
101 kPa × 373 K
290 K
= 130 kPa
The answer is C.
[Total 1 mark]
Tip: Remember that T must be in kelvin.
4
a)
Units of Δp are N m−2 = kg m s−2 × m−2 = kg m−1 s−2
[1]
Units of gΔh are kg m−3 × m s−2 × m = kg m−1 s−2
[1]
∴ Δp = gΔh is homogenous with respect to units.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Exam practice questions
b)
Δp = gΔh = 1.29 kg m−3 × 9.81 m s−2 × 240 m
[1]
= 3.04 kPa
[1]
We have to assume that the density of air and that the value of g remain constant over
this height.
c)
[2]
If atmospheric pressure is 101 kPa, the pressure difference of 3 kPa is only about 3 %. This is well
within the day-to-day variations in atmospheric pressure and so is unlikely to be
noticed.
d)
[2]
At a height of 10 000 m, the density of the air is considerably less than that at ground level. Also,
g will be slightly less. So atmospheric pressure at a height of 10 000 m cannot be determined by
simply using Δp = gΔh.
[2]
[Total 10 marks]
5
a)
−2
From the data, 1.1 atmospheres ≡ 1100 g cm
 1 atmosphere ≡ 1000 g cm−2
[1]
Atmospheric pressure must be expressed as weight per unit area, not mass per unit area, so
atmospheric pressure = 1000 g cm−2 × g
= 1000 × 10−3 kg × 104 m−2 × 9.81 m s−2
[1]
5
≈ 10 Pa or 100 kPa
[1]
Tip: A unit check is a good idea in a question like this, which is easy to do if you use quantity
algebra. The units can be arranged as
kg m s−2 × m−2 = N m−2 = Pa
b)
Using
𝑝1 𝑉1
𝑇1
=
𝑝2 𝑉2
𝑇2
where V1 = V2
p1 = 0.75 atmospheres above atmospheric pressure = 1.75 atmospheres
T1 = (21 + 273) K = 294 K
T2 = (7 +273) K = 280 K
 p2 =
𝑝1 𝑇2
𝑇1
=
1.75 atmospheres × 280 K
294 K
[1]
= 1.67 atmospheres
[1]
= 0.67 atmospheres above atmospheric pressure
[1]
The pressure is therefore within the prescribed limits.
c)
Tip: Atmospheric pressure should be added to find the total pressure and then subtracted again
at the end.
We have had to assume that the volume of the ball, and therefore the volume of air inside, does
not change.
[1]
[Total 7 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Exam practice questions
6
a)
Using
𝑝1 𝑉1
=
𝑇1
𝑝2 𝑉2
𝑇2
where V1 = V2
p1 = 87 kPa
T1 = (17 +273) K = 290 K
T2 = (77 +273) K = 350 K
 p2 =
𝑝1 𝑇2
𝑇1
=
87 kPa × 350 K
[1]
290 K
= 105 kPa
b)
[1]
From pV = NkT,
N=
𝑝𝑉
𝑘𝑇
=
87× 103 N m−2 × 110 × 10−6 m3
[2]
1.38 × 10−23 J K−1 × 290 K
= 2.4 × 1021 molecules
c)
Mass of argon =
[1]
2.4 × 1021 molecules
6.0 × 1023 molecules
× 40 g
[1]
= 0.16 g
[1]
[Total 7 marks]
7
a)
Units of ½mc2 are kg (m s−1)2 = kg m2 s−2
3
Units of kT are J K−1 × K = J = N m
[1]
[1]
2
= kg m s−2 × m = kg m2 s−2
[1]
∴ The two expressions are homogenous with respect to units.
b)
The term c2 means the average of the squares of the speeds of all the molecules.
c)
c2 =
3002 + 4002 + 5002 +6002 +7002
5
= 2.7 × 105 m2 s−2
d)
m2 s−2
[2]
[2]
[1]
3
½ mc2 = kT
2
3𝑘𝑇
 c2 =
[1]
𝑚
i) For air:
c2 =
3 × 1.38 ×10−23 J K−1 × 293 K
4.8 × 10−26 kg
= 2.5 × 105 m2 s−2
[1]
[1]
ii) For helium:
c2 =
3 × 1.38 ×10−23 J K−1 × 293 K
6.7 × 10−27 kg
= 1.8 × 106 m2 s−2
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Exam practice questions
e)
For air:
√⟨𝑐 2 ⟩ = √2.5 × 105 m2 s−2
[1]
= 500 m s−1 ≈ 1.5 × speed of sound in air
For helium:
√⟨𝑐 2 ⟩ = √1.8 × 106 m2 s−2
= 1300 m s−1 ≈ 1.3 × speed of sound in helium
[1]
The speed of sound is therefore related to the average speed of the molecules.
[1]
[Total 15 marks]
Tip: √⟨𝑐 2 ⟩ is not the average speed of the molecules, but is mathematically related to, and fairly
close in value to, the average speed.
8
a)
b)
Boyle’s law states that the pressure of a fixed mass of gas is inversely proportional to its
volume provided that the temperature is kept constant.
[2]
At A: pV = 600 × 103 Pa × 2 × 10−4 m3 = 120 J
[1]
At B: pV = 300 × 103 Pa × 4 × 10−4 m3 = 120 J
At C: pV = 150 × 103 Pa × 8 × 10−4 m3 = 120J
As the product pV remains constant, the gas obeys Boyle’s law.
c)
𝑝1 𝑉1
i) At C:
𝑝2 𝑉2
=
𝑇1
[1]
𝑇2
where V1 = V2
p1 = 150 kPa
T1 = 300 K
T2 = 400 K
 p2 =
𝑝1 𝑇2
𝑇1
=
150 kPa × 400 K
300 K
= 200 kPa
[1]
[1]
ii) At T1 = 300 K, the graph gives V1 = 3 × 10−4 m3 at p1 = 400 kPa.
At T2 = 400 K,
V2 =
𝑉1 𝑇2
𝑇1
=
3 × 10−4 m3 × 400 K
300 K
= 4 × 10−4 m3
[1]
[1]
iii) At A:
V1 = 2 × 10−4 m3
p1 = 600 kPa
T1 = 300 K
When p2 = 640 kPa,
T2 = 400 K
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Exam practice questions
V2 = ?
𝑝1 𝑉1
Using
𝑇1
𝑝2 𝑉2
=
 V2 =
𝑇2
𝑝1 𝑉1 𝑇2
𝑝2 𝑇1
600 kPa × 2 × 10−4 m3 × 400 K
=
[1]
640 kPa × 300 K
= 2.5 × 10−4 m3
d)
[1]
We can now plot the points:
V = 8.0 × 10−4 m3, p= 200 kPa
V = 4.0 × 10−4 m3, p= 400 kPa
V = 2.5 × 10−4 m3, p= 640 kPa
[2]
And draw a smooth curve through them, as shown by the blue line the figure below.
[2]
[Total 14 marks]
9
a)
b)
Volume of air V = 10 m × 10 m × 3.0 m = 300 m3
pV = NkT  N =
=
[1]
𝑝𝑉
𝑘𝑇
102 × 103 Pa × 300 m3
[2]
1.38 × 10−23 J K-1 × (18 + 273) K
= 7.62 × 1027 molecules
c)
[1]
Mass of each air molecule = 29 u = 29 × 1.66 × 10−27 kg = 4.81 × 10−26 kg
27
Mass of air in laboratory = 7.62 × 10 × 4.81 × 10
−26
kg = 367 kg
[1]
[1]
(That’s well over one third of a tonne! – where did the expression ‘as light as air’ come from?)
d)
Density of air ρ =
mass
volume
=
367 kg
[1]
300 m3
= 1.22 kg m−3
e)
[1]
Mean kinetic energy of each molecule:
1
2
3
3
2
2
m<c2> = kT =
× 1.38 × 10−23 J K−1 × (18 + 273) K
= 6.02 × 10−21 J
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Exam practice questions
f)
1
2𝐸k
2
𝑚
From m<c2> = Ek  <c2> =
Root mean square speed = √
2𝐸k
𝑚
[1]
2 × 6.02 × 10−21 J
4.81 × 10−26 kg
[1]
=√
= 500.3 m s−1 ≈ 500 m s−1
g)
[1]
The speed, v, of sound in the air in the laboratory is given by the formula
1.40𝑝
1.40 × (102 × 103 Pa)
= √
𝜌
1.22 kg m−3
v =√
[1]
= 342 m s−1
h)
[1]
Sound is propagated by means of longitudinal compression waves and so cannot travel faster
than the speed of the molecules oscillating back and forth creating the wave. It might be
expected, however, that the speed of sound would be of the same order of magnitude as, but
less than, the root mean square speed of the air molecules – which it is.
[2]
[Total 17 marks]
10
distance
Time t =
b)
Number of crossings of the laboratory in 1 second for each molecule =
speed
Number =
c)
=
10 m
a)
1s
0.02 s
500 m s−1
= 0.02 s
[1]
time (i.e. 1 second)
time for each crossing
= 50 times per second
[1]
Total number of crossings made by all the molecules = 7.62 × 1027 molecules × 50 s−1
= 3.81 × 1029 crossings per second
[1]
As, on average, 1/6 of the molecules are travelling towards, or away from, each wall and the
ceiling, the number hitting one wall will be 1/6 of 3.81 × 1029 s−1 = 6.35 × 1028 s−1.
d)
[1]
Momentum of molecule before hitting wall = mv.
After an elastic collision with the wall, momentum = −mv
Change in momentum = (−mv) – (mv) = −2mv
= −2 × (4.81 × 10−26 kg × 500 m s−1)
= − 4.81 × 10
e)
−23
−1
kg m s
[1]
[1]
[1]
Tip: Be careful with signs! The minus sign indicates the direction of the change in momentum
since momentum is a vector quantity.
Force = rate of change of momentum
= rate of collisions × change of momentum on each collision
= 6.35 × 1028 s−1 × − 4.81 × 10−23 kg m s−1
[1]
= − 3.05 × 106 N
[1]
Note that this is the force of the wall on the molecules that changes their momentum. By
Newton’s third law, the molecules will exert an equal and opposite force on the wall.
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
29 Gas laws and kinetic theory
Answers to Exam practice questions
Pressure exerted by molecules on wall p =
f)
F
A
=
3.05 × 106 N
10 m × 3 m
= 102 kPa
[1]
The air pressure in the laboratory was stated to be 102 Pa (see Question 9), so applying the
kinetic theory for a perfect gas gives the same value. The assumptions that we have made:
• the air behaves like an ideal gas
• the collisions with the wall are elastic
• on average, 1/6 of the molecules are moving in each mutually perpendicular direction is valid
because of the statistically large number of molecules
• to a good approximation, the root mean square speed can be taken to be the same as the
average speed
would all appear to be reasonable.
[4]
[Total 15 marks]
Tip: Note that we have used the terms ‘root mean square speed’ and ‘average speed’ for the
molecules. The average velocity of the molecules would be zero because the total volume of air
in the laboratory is stationary.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
30 Astrophysics
Answers to Test yourself questions
Page 561 Test yourself on prior knowledge
1
Distance = (3.0 × 108 m s-1)(1.0 × 60 s) = 1.8 × 1010 m or 18 Gm
2
90° = 90° × 2π rad/360° = π/2 rad or 1.6 rad
3
λT = k  λ = k/T
4
The gravitational pull of the Earth on the Moon: F = GmMmE/r2
5
W  J s-1  N m s-1  kg m s-2 × m s-1 = kg m2 s-3
6
ΔE stands for the energy transferred from a mass loss Δm.
The conversion factor is c2, the speed of light squared.
Page 565 Test yourself
1
A galaxy is a cluster of stars, up to thousands of billions, held together by gravity.
2
300 light years  300 × (3.0 × 108 m s-1) × (365 × 24 × 3600 s)
= 2.8 × 1018 m = 2.8 × 1015 km
3
The ‘fixed background of stars’ means those stars that are so far away from Earth that they show no
parallax (wobble) as the Earth moves around the Sun.
4
Lʘ is a symbol meaning the total power output of our Sun. It is presently about 3.9 × 10 26 W.
5
(In this chapter) Luminosity is the total power output of a star – unit J s-1 or W. Radiation flux is the
measured electromagnetic energy per second per unit area reaching the Earth from a particular star – unit
W m-2.
6
A quick way is to measure values of the radiation flux F by means of a light dependent resistor at a variety
of distances r from the lamp and then calculate the product Fr2 for each position. If Fr2 is constant, or
approximately constant, then F  1/r2. A better way would be to plot F against 1/r2; a straight line through
the origin would indicate that F  1/r2.
Page 570 Test yourself
7
Henrietta Swan Leavitt and Jocelyn Bell Burnell.
8
Polaris, the pole star, is a Cepheid variable star, i.e. its luminosity varies regularly.
9
2.5 × 106 light years
= (2.5 × 106) × (3.0 × 108 m s-1) × (365 × 24 × 3600 s)
= 2.4 × 1022 m = 2.5 × 1019 km
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
30 Astrophysics
Answers to Test yourself questions
10
11
a)
Red giant stars are, on average, a bit brighter than 100 times as bright as the Sun.
b)
The temperature of the brightest main sequence star is somewhere between 30 000 K and
40 000 K (this is difficult to judge with the logarithmic scale).
12
Elements between iron and uranium are produced in stars greater than 3 times the mass of the Sun. They
are produced during supernova explosions in these stars
13
We now refer to ‘the 5% universe’ because what the other 95% consists of is not, in 2014, known to us. It
consists of Dark matter and Dark energy.
Page 574 Test yourself
14
It is assumed that you act like a ‘black body radiator’, i.e. the e-m waves you produce obey Wien’s law: λmT
= 2.9 × 10-3 m K.
15
The Stefan-Boltzmann law states that L = σAT4. For a sphere A = 4πr2 and σ is a constant. This means that
L  r2T4 for a star, which can be assumed to be spherical.
16
Substituting values for the Sun into L = 4πσr2T4 yields
𝑟2 =
(3.9×1026 𝑊)
4𝜋(5.7×108 𝑊𝑚−2 𝐾 −4 (5800𝐾)4
) = 4.8 × 1017 𝑚2
r = 7.0 × 108 m (or 700 000 km)
17
A ‘black body’ is an object that completely absorbs all the electromagnetic radiation incident upon it.
Radiation emitted by such a body is called black body radiation. ‘Black bodies’ (for example all stars) obey
both Wien’s law: λmT = 2.9 × 10-3 m K and the Stefan-Boltzmann law: L = σAT4.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
30 Astrophysics
Answers to Exam practice questions
Pages 576–579 Exam practice questions
1
Answer C
[Total 1 mark]
2
Answer D
[Total 1 mark]
3
Answer D
[Total 1 mark]
4
Answer A
[Total 1 mark]
5
To get degrees, divide 0.01 arcsec by 3600 arcsec per degree  2.8 × 10−6 degrees.
To get from degrees to radians, multiply 2.8 × 10−6 degrees ×
2π rad
360°
 4.8 × 10−8 rad.
[2]
[2]
[Total 4 marks]
1000 W m−2
6
Reduction factor =
7
a)
They are related by tan p =
b)
i) Rearranging the above formula gives
d=
1350 W m−2
𝑟
tan 𝑝
=
= 0.74 so the answer is C
𝑟
[Total 1 mark]
[1]
𝑑
1.5 × 1011 m
[1]
tan(4.5 × 10−5 degrees)
= 1.9 × 1017 m
[1]
ii) 240 light years = 240 × (3.0 × 108 m s−1) × (365 × 24 × 3600 s) = 2.27 × 1018 m
[1]
 The parallax angle for this star will be
p = (4.5 × 10−5 degrees) ×
1.91 × 1017 m
2.27 × 1018 m
= 3.8 × 10−6 degrees
[1]
−6
= 3.8 × 10 degrees × 3600 arcsec per degree = 0.014 arcsec
[1]
[Total 6 marks]
8
You need to remember that
W  J s−1
JNm
N  kg m s−2
[2]
Radiation flux is measured in W m−2
[1]
W m  J s m  N m s m  kg m s m s m
−2
−1
−2
−1
−2
−2
−1
−2
So the base SI unit for radiation flux is kg s−3
[1]
[Total 4 marks]
9
From the formula
Radiation flux =
Luminosity
4𝜋𝑑2
 4πd2 =
Luminosity
Radiation flux
[1]
where d is the distance from Earth to the star.
So, for the star Rigel
4πd2 =
3.8 × 1031 W
5.4 × 10−8 W m−2
= 7.03 × 1038 m2
7.03 × 1038 m2
 d= √
= 7.5 × 1018 m
4π
[2]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
30 Astrophysics
Answers to Exam practice questions
[Total 4 marks]
10
We can only estimate the luminosity because both axes are logarithmic.
[1]
Four days will not be far to the right of 3 days on the horizontal axis.
[1]
Producing a vertical line from 4 days to the red line, a horizontal line then hits the vertical axis a
few mm above 102 (100).
[1]
27
The value is therefore between 100 and 1000, probably in the order of 500 × 10 W.
[1]
[Total 4 marks]
11
A grain of rice is roughly cylindrical, about 1 mm in diameter and 4 mm long.
[1]
Its approximate volume is therefore
V = πr2h = π × (0.5 × 10−3 m)2 × (4 × 10−3 m)
−9
[1]
3
≈ 3 × 10 m
[1]
From the formula for density
ρ=
𝑚
𝑉
 m = ρV ≈ (4 × 1017 kg m−3) × (3 × 10−9 m3)
[1]
≈ 12 × 108 kg ~ 109 kg
[1]
So the mass of a ‘grain’ of a neutron star would be about 1 billion kilograms!
[Total 5 marks]
Tip: Note that the Specification requires candidates to ‘be able to estimate values for physical quantities
and use their estimate to solve problems’.
12
The Stefan–Boltzmann law states that:
L = 4πσr2T4
[1]
We have here that:
L = 0.002Lʘ and σ = 5.7 × 10−8 W m−2 K−4
Rearranging the formula gives us
r2 =
𝐿
4𝜋𝜎𝑇
4=
0.002 × 3.9 ×1026 W
[2]
4π × (5.7 × 10−8 W m−2 K−4 ) × (20 × 103 K)
= 6.8 × 1012 m2
 r = 2.6 × 106 m or 2600 kilometres.
[1]
[Total 4 marks]
13
Questions like this are best answered using a list of ‘bullet points’. Your answer should follow a logical
argument, for which marks are awarded, and should include the following points:
•
The Sun will remain near S for a further 5 million years during which time its temperature and
luminosity will be about 6000 K and 4 × 1026 W (Lʘ).
•
[2]
Moving away from S to X it will become a ‘red giant’ star with a surface temperature of about 4000 K
and a luminosity of about 10000 times Lʘ.
•
[2]
Moving away from X the star will lose about half its mass and shrink to become a white dwarf star. Its
temperature and luminosity will be about 14000 K and just over 10−4Lʘ, i.e. 1/10000 of Lʘ.
[2]
[Total 6 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
30 Astrophysics
14
a)
Answers to Exam practice questions
Wien’s law gives us
λmT = 2.90 × 10−3 m K
 T=
2.90 × 10-3 m K
λm
=
2.90 × 10-3 m K
[1]
440 × 10-9 m
= 6600 K
[1]
Referring to a Hertzsprung-Russell diagram (such as those in Figures 13.7 and 13.11), a main
sequence star with this temperature will have a luminosity of about 6 times the luminosity Lʘ of
the Sun.
[1]
 L = 6Lʘ = 6 × 3.9 × 1026 W = 2.34 × 1027 W.
[1]
Its distance d from Earth is given by
Radiation flux =
4πd2 =
Luminosity
4πd2
2.34 × 1027 W
5.8 × 10−10 W m−2
 4πd2 =
Luminosity
Radiation flux
= 4.0 × 1036 m2
[1]
4.0× 1036 m2
d= √
= 5.6 × 1017 m
[1]
4π
In light years
d=
3.0 × 10
8
5.6 × 1017 m
× (365 × 24 × 3600 ) s ly−1
[1]
m s−1
= 60 light years
b)
[1]
Reading off the logarithmic scales of an H-R diagram with its band of stars is by far the least
reliable step in this calculation. So much so that the distance could be anywhere between 50 and
70 light years.
[2]
[Total 10 marks]
15
a)
Assuming that the mass mG of the galaxy all contributes to the Newtonian gravitational
attraction, then the centripetal force acting on the star S of mass m is:
F=
𝐺𝑚𝑚G
𝑟2
This produces a centripetal acceleration of ʋ2/r
Applying Newton’s 2nd Law to the circular motion of S yields:
F=
𝑚𝑣 2
𝑟
 mG =
b)
=
𝐺𝑚𝑚G
[2]
𝑟2
𝑣2𝑟
[1]
𝐺
Astrophysicists estimate the mass ‘mG’ by studying the electromagnetic radiation reaching
Earth from the galaxy.
[1]
Their conclusion, for this star and others, is that:
m ≠ mG
In fact mG ≈ 3.6 × m or m ≈ 0.28 × mG
[1]
So astrophysicists conclude that only 28% of the gravitationally attracting matter in the universe
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
30 Astrophysics
Answers to Exam practice questions
is ‘visible’ to us. Of the remaining 72%, about 67% is known as ‘dark energy’, leaving only 5% of
the universe as ‘hadronic’ matter. See also pages 133 and 244.
[2]
[Total 7 marks]
You will find lots of information about dark energy by putting ‘Dark Matter’ into a search engine.
16
Atoms of every kind beyond hydrogen are ‘made’ in stars by fusion and subsequent explosive
processes.
[2]
Our bodies are made of atoms that result from these events, atoms that have joined with others
to form particles of different materials and this ‘dust’ is what the Earth and animals living on
Earth are ‘made’ from.
[2]
[Total 4 marks]
17
Try a search engine with the words ‘dark matter’. Within Wikipedia you will find references
to the ‘Milky Way’, ‘baryonic matter’, ‘mass-energy’ and the ‘standard model’, all terms you should be
familiar with. You will also find the American Space Agency (NASA) site a useful resource.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
Answers to Test yourself questions
Page 580 Test yourself on prior knowledge
1
An alpha-particle is made up of two positively charged protons and two neutral neutrons.
2
A helium nucleus.
3
9.11 × 10−31 kg.
4
Because an electron has an extra mass Δm that depends on its speed ʋ.
5
a)
+1
b)
-1
6
E = hf
= (6.6 × 10−34 J s) × (4.9 × 1019 s−1)
= 3.2 × 10−14 J.
7
The energy of this photon = (3.2 × 10−14 J)  (1.6 × 10−19 J eV−1)
= 2.0 × 105 eV or 200 keV
Page 584 Test yourself
1
No, each cubic meter might contain some hydrogen atoms, perhaps a few helium atoms and sometimes a
dust particle.
2
Unit of ΔE (energy) is J  N m  kg m s−2 m  kg m2 s−2 = kg × (m s−1)2 which is the unit of Δm × c2, where c is
the speed of light.
3
 represents a neutrino and γ represents a gamma photon
(Strictly  represents an electron neutrino e, but in this text we rarely differentiate between different types
of neutrino.)
4
The total mass loss Δm = 2me where me is 9.1 × 10−31 kg.
 using Einstein’s equation:
E = c2Δm = (3.0 × 108 m s−1)2 × 2(9.1 × 10−31 kg)
= 1.6 × 10−13 J
5
The photon travels at the speed of light, 3.0 × 108 m s−1
 ʋ = s/t gives t = s/ʋ
= 150 × 109 m/3.0 × 108 m s−1
= 500 s or just over 8 min
(mn − m p ) (1.0087u −1.0073u)
=
mp
1.0073u
6
= 1.4 10−3 or 0.0014
7
1 keV = 1000 eV = 1000 eV × (1.6 × 10−19 J eV−1) = 1.6 × 10−16 J
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
Answers to Test yourself questions
8
Page 588 Test yourself
9
The electrostatic force is zero, as the slow neutron carries no charge.
10
The number of chemical reactions equivalent to 200 MeV
6
= 200 10 eV = 13million
15eV
11
Balancing the nucleon numbers:
235
12
a)
U + 1n → 144Ba + 89Kr + X neutrons is a reaction in which X = 236 − 233 = 3
The proton numbers of uranium and zirconium are 92 and 40 respectively. As neutrons carry no
electric charge and thus carry a proton number 0, then the proton number of tellurium must be
Z, where 92 = 40 + Z. Therefore Z = 52.
b)
In this calculation it was not necessary to equate the number of neutrons because neutrons have
zero charge.
13
No chemical process will work, and all atoms of uranium of whatever isotope are electrically neutral; the
only property that can be used is therefore the difference in mass of different uranium isotopes. From
Table 31.1:
(238.0508 − 235.0430) u = 3.0078 u
Converting to kg:
3.0078 u × (1.66 × 10−27 kg u−1) = 5.0 × 10−27 kg.
The difference is very small but the isotopes can be separated in a diffusion process whereby the lighter
U-235 diffuses faster than the heavier U-239.
Page 594 Test yourself
14
F = kQ1Q2/r2
= (9 × 109 N m2 C−2) × (1.6 × 10−19 C)2/(1.0 × 10−6 m)2
= 2.3 × 10−16 N
This force is much greater (more than 1010 times!) than the weight of a proton on Earth.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
15
Answers to Test yourself questions
(4 × 1.0078 u) − 4.0026 u = 0.0286 u  0.0286 u × 930 MeV u−1
= 26.6 MeV.
16
An emission spectrum is a set of bright lines showing the photons (wavelengths) given out, emitted, by a
hot gaseous element. An absorption spectrum is set of dark lines showing where the atoms of an element
have sucked in, absorbed, the photons (wavelengths) that those atoms would normally emit.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
Answers to Exam practice questions
Pages 597–601 Exam practice questions
1
Answer A
[Total 1 mark]
2
Answer B
[Total 1 mark]
3
Answer C
[Total 1 mark]
4
Answer A
[Total 1 mark]
5
Answer C
[Total 1 mark]
6
a)
b)
In the three isotopes:
1
1H
contains 1 proton only
2
1H
is made up of 1 proton and 1 neutron
3
1H
is made up of 1 proton and 2 neutrons
[3]
The mass deficit is:
Δm = 2.01355 u + 2.01355 u – (3.01550 u + 1.00728 u)
[2]
= 0.00432 u
c)
[1]
ΔE = 0.00432 u × 930 MeV u
–1
[1]
= 4.018 MeV or just over 4 MeV
[1]
[Total 8 marks]
7
a)
The α-particles have the same energy because their tracks in the cloud chamber have the
same length.
[1]
+ 73Li →
b)
1
1H
c)
17 MeV 
4
2He
4
2He
+
17 MeV
930 MeV u–1
+ energy
= 0.0183 u.
[3]
[1]
 the mass of a lithium nucleus = 2 × 4.0015 u + 0.0183 u − 1.0073 u
= 7.014 u
[2]
[1]
[Total 8 marks]
8
a)
The galaxy is moving away from Earth at a speed
ʋ = zc = 0.2 × 3.0 × 108 m s−1
= 6.0 × 107 m s−1
[1]
The distance to this galaxy is therefore
d=
𝑣
𝐻0
=
6.0 × 107 m s–1
[2]
2.3 × 10–18 s–1
= 2.6 × 1025 m
b)
[1]
In light years
d=
3.0 × 10
8
2.6 × 1025 m
× (365 × 24 × 3600) s ly–1
m s–1
= 2.8 × 109 ly or about 3 billion light years
[1]
[1]
[Total 6 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
9
Answers to Exam practice questions
4
The initial volume of the uranium nucleus is πR3
3
4
The volume of the two fission fragments will be 2 × πr3
3
So we have
4
4
3
3
2 × πr3 = πR3
[1]
Re-arranging:
𝑟 3
𝑟 3 1

= √0.5
𝑅
2
𝑅
( ) =
[1]
= 0.79 ≈ 0.8
[1]
[Total 3 marks]
10
Your answer should include the following points:
•
The moving source S produces a series of waves 1,2,3, etc. at a regular frequency f.
•
The source S is moving forward at a steady speed ʋS,
•
The ‘ripples’ or wavefronts of these waves seem to be closer together in the forward direction and
further apart in the backward direction.
•
An observer or detector placed at O1 will judge that the waves are arriving with a lower wavelength,
and thus a higher frequency, than that produced by the source S.
•
At O2 the opposite effect will be observed, i.e. f1  f and f2  f.
[Total 6 marks]
Tip: Questions like this are best answered using a list of ‘bullet points’. Your answer should follow a logical
argument, for which marks are awarded.
11
With a radius of r = 0.70 × 109 m and a period of rotation T = 2.2 × 106 s, the speed at which the ‘edge’ of
the Sun is moving away from us is given by
ʋ=
2𝜋𝑟
𝑇
=
2π × 0.70 × 109 m
[1]
2.2 × 106 s
= 2000 m s−1
[1]
The change in wavelength of light from the edge of the Sun that is moving away from us is an increase (red
shift) given by
Δ𝜆
𝜆
=
𝑣
𝑐
 Δλ =
=
𝜆𝑣
[1]
𝑐
520 nm × 2000 m s-1
[1]
3.00 × 108 m s-1
= 3.5 × 10−3 nm = 0.0035 nm.
[1]
[Total 5 marks]
(The difference in the wavelengths of light at about 520 nm from opposite sides of the Sun would therefore
be twice this, i.e. 0.007 nm.)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
Answers to Exam practice questions
Tip: Note that it is important to state that the change in wavelength is an increase. The answer also shows
the value of quantity algebra. If the wavelength is left in nm, the quantity algebra shows that the change in
wavelength is also in nm.
12
a)
The variation in λ is because the star S moves both towards the Earth and then away from
Earth as S orbits its star M.
[1]
The variation is
Δλ = (656.323 − 656.241) nm = 0.082 nm
[1]
Applying the Doppler equation:
Δ𝜆
2𝑣
=
𝜆
𝑐Δ𝜆
 𝑣=
𝑐
2𝜆
(The factor 2 occurs because the shift in wavelength is due to the difference between the
movement towards and away from the Earth.)
=
3.00 × 108 m s–1 × 0.082 nm
[1]
[1]
2 × 656 nm
= 19 000 m s−1
So the speed of S around M ≈ 20 km s–1.
b)
As ʋ =
2𝜋𝑟
𝑇
r=
𝑣𝑇
2𝜋
=
19 000 m s–1
[1]
8
× 2.7 × 10 s
2π
= 8.2 × 1011 m
[1]
[1]
The acceleration of the star S towards the star M is
𝑣2
𝑟
𝐺𝑚M
=
𝑟2
 mM =
𝑣2𝑟
[1]
𝐺
2
=
(19 000 ms–1 ) × 8.2 × 1011 m
[1]
6.7 × 10–11 N m2 kg–2
= 4.4 × 1030 kg, or 4.3 x 1030 kg if data is kept in calculator.
[1]
[Total 10 marks]
13
a)
The scale of Figure 14.20 is that 5.0 nm cover 74 mm.
 The red shift in M is 0.38 mm ×
and the red shift in S is 0.67 mm ×
5.0 mm
74 mm
5.0 mm
74 mm
= 0.0257 nm
[1]
= 0.0453 nm
[1]
The mean wavelength is 427.5 nm, so
[1]
Applying the Doppler equation:
Δ𝜆
𝜆
=
𝑣
𝑐
 𝑣=
𝑐Δ𝜆
𝜆
In March Arcturus was receding from the Earth at
ʋM =
3.00 × 108 m s–1 × 0.0257 nm
427.5 nm
= 18 000 m s−1 or 18 km s–1
[1]
In September Arcturus was receding from the Earth at
ʋS =
3.00 × 108 m s–1 × 0.0453 nm
427.5 nm
= 32 000 m s−1 or 32 km s–1
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
b)
Answers to Exam practice questions
This variation suggests that Arcturus is receding from Earth at 25 km s –1, but that the Earth’s
motion round the Sun subtracts and adds to this number by 7 km s –1 every six months.
[3]
[Total 8 marks]
14
a)
ρ0 =
3𝐻0 2
8𝜋𝐺
=
2
3 × (2.3 × 10–18 s–1 )
[1]
8π × 6.7 × 10–11 N m2 kg–2
= 9.4 × 10−27 kg m−3
[1]
The value of H0 is known only to ±5 %.
[1]
Here it is squared, so the figure for the average density of matter is at best only calculated
to ±10 %.
[1]
This gives us a value of about (10 ±1) × 10−27 kg m–3.
[1]
Putting in that N  kg m s−2 gives us that the unit is
[1]
s–2
kg m s–2
b)
× m2 × kg–2
= kg m−3 as expected for density.
[1]
The mass of a hydrogen atom 11H is about 1.7 ×10−27 kg, therefore the number of hydrogen atoms
in a cubic meter of space ≈
9.4 × 10–27 kg
[1]
1.7 × 10–27 kg
= 5.5, i.e. about 5 or 6.
c)
[1]
If the Hubble constant was known more precisely, the distance between the solar system and
very distant galaxies could be found more easily, as ʋ = H0d, and ʋ can be found from a galaxy’s
red shift.
[3]
[Total 12 marks]
15
Astrophysicists use supernovae type Ia as standard candles, as the maximum luminosity L of such
explosions is known.
[1]
This enables them to calculate a value for the Hubble constant H0.
[1]
They then calculate z by finding Δλ/λ.
[1]
Knowing an approximate value for H0 and that c is known to be 3.0 × 108 m s–1, a value for d can be found as
d = zc/H0.
[2]
[Total 5 marks]
16
The Sun is rotating, so light from the edge moving away from Earth and the light from the edge moving
towards Earth each show a Doppler shift.
[1]
This breaks the absorption line into two adjacent lines.
[1]
The value of 2Δλ is very small, but means that the lines in the Sun’s absorption spectrum appear as two
lines very close together.
[1]
[Total 3 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
31 Cosmology
17
a)
Answers to Exam practice questions
The Sun’s luminosity is
Lʘ = 3.9 × 1026 W
Multiplying this by 1.7 × 109 suggests that the maximum luminosity of this supernova
explosion was
L = (1.7 × 109) × (3.9 × 1026 W)
[1]
35
= 6.6 × 10 W
[1]
Given the uncertainties in this calculation, we can say that this is about 7 × 10 35 W.
b)
[1]
This was a type Ia supernova, but there are other types of supernovae explosions that
cosmologists cannot use in measuring the universe.
[1]
This is because they do not believe that these explosions have the same maximum power output,
i.e. cannot be used as standard candles.
[2]
[Total 6 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Test yourself questions
Page 602 Test yourself on prior knowledge
1
2
31.25𝑠
a)
𝑇=
b)
𝑓=
a)
The amplitude i) is the distance from the mid-point of the oscillation to the point of maximum
25
1
𝑇
=
= 1.25𝑠
1
1.25𝑠
= 0.80Hz
displacement. The wavelength ii) is the distance between adjacent points that are in phase.
These are shown in the figure below.
b)
i) The time period T is the time to complete one oscillation, or one wavelength, for example
from A to E. This is 20 ms.
ii) Frequency is 𝑓 =
1
𝑇
=
1
20×10−3 𝑠
= 50Hz
c)
Points A and E and B and F are in phase with each other.
d)
i) Points A and B have a phase difference of ¼ wavelength (or cycle) or π/2 (or 90°).
ii) Points B and D have a phase difference of ½ wavelength (or cycle) or π (or 180°)
Page 605 Activity 32.1
1
2
A graph of F against x should be a straight line through the origin and very close to the point (90 mm, 2.40
N).
3
Gradient =
(2.40−0.00) 𝑁
(90−0)×10−3 𝑚
= 27 𝑁𝑚−1
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Test yourself questions
Page 606 Activity 32.2
1
2
A graph of T2 against m should be a straight line through the origin and close to the point (0.35 kg, 0.525 s2).
3
Gradient =
4
From 𝑇 = 2𝜋√ ⇒ 𝑇 2 = 4𝜋 2
𝑘=
5
4𝜋2
gradient
(0.525−0.000) 𝑠 2
(0.350−0.000) kg
=
= 1.50(𝑠 2 kg −1 )
𝑚
𝑚
𝑘
𝑘
4𝜋2
1.50 𝑠 2 kg −1
⇒
𝑇2
𝑚
×
4𝜋2
𝑘
= gradient ×
4𝜋2
𝑘
= 26 𝑁kg −1
Percentage difference between the two values of 𝑘 = 1
(27−26) 𝑁kg −1
⁄2(27+26) 𝑁kg −1
× 100 % = 4 %
(Note that the average of the two values should be used as the denominator).
We need to compare the likely uncertainty in Δx with that in T2. If we take mid-values:
Δx = 50 ± 2 mm and 40T = 20.0 ± 0.2 s
% uncertainty in Δx =
2 mm
50 mm
% uncertainty in T2 = 2 ×
× 100% = 4%
0.2 𝑠
20 𝑠
× 100% = 2%
(Note: that you must take the uncertainty in the actual timings, and then T will have the same %
uncertainty. The factor of 2 is because we have T squared).
This would suggest that the timing method is probably the more reliable, although both methods have an
uncertainty of less than 5%.
Tip: Note that a question like this should be answered quantitatively. You must back-up your argument
with some data or else you won’t get many marks.
Page 607 Core Practical 16a
1
𝑚
From 𝑇 = 2𝜋√ , by squaring both sides of the equation we get
𝑘
T2 =
4 2 m
4 2
T2 =
m
k
k
This means that a graph of T2 against m should give a straight line through the origin. Interpolating from a
straight line rather than from a curve, which a graph of T against m would give, is more accurate.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
2
Answers to Test yourself questions
Your graph of T2 against m should be a straight line through the origin and close to the point (0.35 kg, 0.525
s2).
3
The period 𝑇𝑤 =
11.63+11.69
40
𝑠 = 0.583 𝑠 ⇒ 𝑇 2 = 0.340 𝑠 2
From your graph you should find that when T2 = 0.340 s2, mw ≈ 0.225 kg
4
If you have vernier or digital callipers that can measure up to 75 mm then these could be used. If not, you
should attempt to make measurements to a precision of a mm (or better) using a metre rule.
Whatever instrument you use, where you take the measurements is critical. You should take two
measurements, in different places, across each of the six faces of the block, and then average the values for
opposite faces. This will give you average values for the length, width and height of the block from which
the volume can be calculated.
5
Volume V = (75 × 10−3) m × (75 × 10−3) m × (72 × 10−3) m = 4.05 × 10−4 m3
6
Density of wood =
𝑚
𝑉
=
0.225 kg
4.05×10−4 𝑚3
= 560 kg𝑚−3
Note that the answer is given to 2 SF, which is the precision of the measurements of the dimensions of the
block.
7
The % uncertainty in the mass is determined by the precision of the timing. Although the timings are
recorded to the resolution of the stopwatch (0.01 s), the uncertainty will be determined by the reaction
time of the observer. Realistically this is 0.1 s. As the measured times are in the order of 10 s, the %
uncertainty is about
0.1 𝑆
× 100 %. = 1 %. The % uncertainty in T2 will therefore be twice this, i.e. about 2 %.
10 𝑆
For the volume, the % uncertainty will be:
1mm
1 mm
1 mm
 100 % +
 100 % +
 100 % = 1.33 % + 1.33 % + 1.39 % = 4 %
75mm
75 mm
72 mm
The % uncertainty in the density will therefore be 2 % + 4 % = 6 %
The uncertainty will be
6
100
× 560 kg m−3 = 0.34 kg m−3
The density should strictly be expressed as (0.56 ± 0.03) × 103 kg m−3 to show 2 SF precision.
In practice we would probably say it was 560 ± 30 kg m−3.
Tip: Remember that the total percentage uncertainty is found by adding together the percentage
uncertainty for each measurement if the measurements are multiplied together or divided.
Page 608 Test yourself
1
a)
𝑇=
8.8𝑠
5
= 1.76 𝑠
𝑙
4𝜋2 𝑙
𝑔
𝑇2
2𝜋√ ⇒ 𝑔 =
=
4𝜋2 ×0.75 𝒎
(1.76 𝒔)2
b)
Percentage difference =
c)
Realistic uncertainties:
±0.1 in 5𝑇 =
0.1𝒔
8.8𝒔
= 9.6 𝑁 kg −1
(9.8−9.6)𝑁 kg −1
9.8 𝑁 kg −1
× 100 % = 2 %
× 100% = 1.1% = uncertainty in 𝑇
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
 2 mm in l =
Answers to Test yourself questions
2  10−3 m
 100% = 0.3%
0.75m
% uncertainty in g = 2 × % uncertainty in T (as we have T2) + % uncertainty in l
= 2 × 1.1% + 0.3% = 2.5%
d)
The length measured should be from the point of suspension to the centre (of mass) of the bob.
e)
The student only counted 5 oscillations and did not repeat the timing. Had 10 oscillations been
taken, and repeated, the % uncertainty in the value for T would be at least halved and so the %
uncertainty in T2 would be 4 times less – about 0.5%. This would reduce the overall percentage
uncertainty to less than 1% - a significant improvement. This shows the importance of timing at
least 10 oscillations (if possible) and repeating. If the oscillations are heavily damped so that only
5 oscillations can be timed, then the timing should be repeated to give at least 4 × 5 oscillations.
The length was presumably measured to the nearest mm, so it should have been recorded as
75.0 cm (and not just 75 cm) to reflect this precision.
One value for the length is not adequate – T should be found for at least six values of l over a
wide range and then g found from a graph of T2 against l.
2
a)
The length of the pendulum is l = H – h
𝑙
(𝐻−ℎ)
𝑔
𝑔
𝑇 = 2𝜋√ ⇒ 𝑇 = 2𝜋√
(𝐻−ℎ)
4𝜋2 𝐻
4𝜋2 ℎ
b)
Squaring both sides: 𝑇 2 = 4𝜋 2 ×
c)
T should be found for a wide range of values of h by shortening the length of string each time.
𝑔
=
𝑔
−
𝑔
Twenty oscillations should be timed, repeated and averaged, for each value of h. Then a graph of
T2 against h should be plotted.
4 p 2H
You should get a straight line of negative slope. The intercept on the T2–axis will be g from
4p 2
which H can be found, and the gradient will be − g from which g can be found.
Page 617 Test yourself
3
a)
amax = −ω2A = − (2πf)2A
T=
14.1 s
1
1
= 0.705 s  f = =
= 1.42 Hz
20
T 0.705 s
amax = (2π × 1.42 s−1)2 × 50 × 10−3 m = 3.97 m s−2
=
3.97 𝑚𝑠 −2
9.8 𝑚𝑠 −2
≈ 0.41𝑔 (Answer C)
b)
The maximum acceleration occurs at the top and bottom of each oscillation. (Answer D)
c)
The maximum velocity is as the mass passes through the middle. (Answer A)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
d)
𝑚
4𝜋2 𝑚
𝑘
𝑇2
𝑇 = 2𝜋√ ⇒ 𝑘 =
=
4
Answers to Test yourself questions
4𝜋2 ×0.200kg
(0.705𝒔)2
= 15.9 𝑁𝑚−1 ≈ 16 𝑁𝑚−1 (Answer C)
vmax = ± ωA = 2πfA where A = ½ × 16 mm (16 mm = top to bottom) = 8 mm
f = 720 min −1 =
720 min −1
= 12 s −1
60 s min −1
vmax = 2πfA = 2π × 12 s−1 × 8 × 10−3 m = 0.60 m s−1
5
a)
l=
b)
𝑙
𝑇 2𝑔
𝑔
4𝜋2
𝑇 = 2𝜋√ ⇒ 𝑙 =
(9.5 s) 2  9.8 m s −2
 22.4 m
4 2
At maximum displacement of 1.2 m:
sin  
1.2 m
 0.0536    3
22.4 m
This is a small enough angle for the approximation that, in radians, sin θ = θ on which assumption
the pendulum formula is derived.
Check: In radians θ = 0.0536 rad (to 3 s.f.)
c)
amax = −ω2A = − (2πf)2A
T=
95 s
= 9.5 s
10
T = 9.5 s  f =
1
1
=
= 0.105 Hz
T 9.5 s
amax = (2π × 0.105 s−1)2 × 1.2 m = 0.52 m s−2
The maximum acceleration will occur at the end of each swing.
d)
The general equation for SHM is x = A cos ωt = A cos (2πf)t
In this case, A = −1.2 m, f = 0.105 Hz and t = 3.0 s
What does the minus sign for A mean? Imagine starting the timing when the pendulum is at the
end of its swing, to your left. Its displacement back from the centre is then – 1.20 m.
x = −1.2 m × cos (2π × 0.105 s−1 × 3.0 s) = −1.2 m × cos (1.979)
= −1.2 m × (− 0.397) = + 0.48 m
(remember to put calculator in ‘rad’ mode!)
What does the plus sign mean? When the displacement is + 0.48 m it means that the pendulum
is 0.48 m from the centre away from you. The pendulum will therefore be (1.20 + 0.48) m = 1.68
m from the end at which the timing was started.
(If you had started with A = + 1.2 m, the displacement after 3.0 s would come out at – 0.48 m and
the converse argument would be true.)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Test yourself questions
Page 622 Test yourself
6
a)
b)
Within the experimental error associated with such an experiment, the ratios are equal. This
suggests that the damping is exponential.
Page 625 Activity 32.6
1
For lightly damped oscillations:
31¾ oscillations in 100 s  T =
100𝑆
31.75
= 3.15 s
For heavier damping:
12½ oscillations in 40 s  T =
30𝑆
12.5
= 3.12 s
The difference between the two values is not significant as the values agree within the uncertainty of
reading off the data from the graph.
2
a)
From a = a0e−kt, if we take logarithms to base e on both sides of the equation we get:
ln (a/mm) = ln a0 + ln (e−kt)  ln (a/mm) = − kt + ln(a0 /mm)
b)
c)
N
0
1
2
3
4
5
6
7
a/mm
25.0
21.5
19.0
17.0
15.0
12.5
11.0
9.5
ln(a/mm)
3.22
3.07
2.94
2.83
2.71
2.53
2.40
2.25
If you plot a graph of ln(a/mm) against the number of oscillations, N, you should get a straight
line of negative slope. Note that the ln(a/mm) axis should not start at the origin as the scale will
then be too small. You should start the scale at, say, 2.0.
d)
As the line, within the uncertainty of reading off the data, is straight, with a negative slope, it
shows that the amplitude decreases exponentially with the number of oscillations. In the
equation
ln (a/mm) = − kt + ln (a0/mm)
the time t = NT where T is the period, so the gradient of the graph is − kT
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Test yourself questions
Page 626 Activity 32.7
1
For example:
2
For example:
3
For example:
Page 629 Test yourself
7
a)
If the father’s leg has a period of 1.0 s, he will take TWO strides each second (try this for yourself
– remember one oscillation is there and back again!). He will therefore walk a distance of 2 ×
0.70 m = 1.4 m in one second. His speed will then be 1.4 m s−1.
Speed = 1.4 m s−1 × (60 × 60) s h−1 = 5040 m h−1 ≈ 5 km h−1
b)
i) Given that T2  length of leg, if the girl’s leg is only half as long as that of her father, her
period of oscillation will be a factor of √2 smaller than her father’s, i.e. 0.71 s.
ii) Frequency 𝑓 =
1
𝑇
=
1
0.71 𝑠
= 1.4 Hz
iii) If her leg is only half as long, her stride will only be half as long as her father’s:
girl’s stride = ½ × 0.70 m = 0.35 m
iv) Her natural speed of walking will be:
speed =
c)
distance 2  0.35 m
=
= 1.0 m s −1 (= 3.6 km h −1 )
time
0.17 s
i) When we walk at a pace when our legs are moving at their natural frequency, there is
maximum energy transfer from our body to our legs, i.e. a resonance effect. This means we
are walking with maximum efficiency.
ii) In order to keep up, the little girl has to walk at a rate of 1.4 times (or 40%) faster than her
natural frequency, which means that the energy transfer is not very efficient. Thus she
struggles to keep up.
(1.4 times because ratio of her natural speed to that of her father is 1.0 m s−1: 1.4 m s−1 =
1:1.4)
8
a)
The amplitude of an oscillation is the distance from the centre of the oscillation to the maximum
displacement from the centre.
b)
When the girl is at one end of the swing, she has maximum gravitational potential energy and no
kinetic energy. As she swings down, some of this GPE is converted into KE due to her speed and a
little is transferred into random molecular kinetic energy, due to air resistance and friction at the
point of suspension (and also, probably, some sound). At the bottom of her swing, she has
minimum GPE and maximum KE. As she rises again, she slows down and her KE is transferred
back into GPE. When she reaches the end of the swing, she momentarily comes to rest, all the
energy becoming GPE. Each time her father pushes her, the work he does is converted into extra
KE.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
c)
Answers to Test yourself questions
As her father is pushing her at the end of each swing, the frequency of him doing work on her is
the same as the frequency of her swinging. Therefore there is resonance, with maximum transfer
of energy and, after a few swings, she will have a large amplitude.
9
a)
For a spring:
T = 2
f =
b)
m
20 kg
= 2
= 0.497 s
k
3200 N m −1
1
1
=
= 2.01 Hz  2 Hz
T 0.497 s
For a spring:
F = kx
If the spring constant is 3200 N m−1 and the girl’s mass is 20 kg:
F = kx  x =
F 20 kg  9.8 N kg −1
=
= 0.061 m = 6.1 cm
k
3200 N m −1
Although she will be able to increase this displacement by pushing on the ground, the motion is
likely to be heavily damped and so her amplitude will soon become very small and she will stop
oscillating unless her father lends a helping hand!
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Exam practice questions
Pages 632–636 Exam practice questions
1
The acceleration, energy and velocity in simple harmonic motion will all increase if the amplitude is larger.
The frequency does not depend on the amplitude – answer C.
2
The acceleration is a maximum at each end of the swing and the total energy and the frequency remain the
same. The velocity is a maximum at the centre – answer D.
3
[Total 1 mark]
Following on from question 2, at the end of the swing the pendulum momentarily comes to rest and so has
zero velocity – answer D.
4
[Total 1 mark]
f=
20 oscillations
16 s
[Total 1 mark]
= 1.25 s–1
ω = 2πf = 2π × 1.25 s–1 = 2.50π s–1
The answer is D.
[Total 1 mark]
5
The peak power is twice the average power – answer C
[Total 1 mark]
6
a)
A 12 V alternating current means that the peak voltage is √2 × 12 V = 17 V.
[1]
A suitable setting for the input voltage might be 5 V per division as the trace would have an
amplitude of ±3.4 divisions.
[1]
If the frequency is 50 Hz, the time period will be 1/50 s = 20 ms.
[1]
A suitable time base setting might be 5 ms per division, then a wavelength would be 4 divisions.
[1]
b)
Your sketch should be a sine wave that reflects your data. A sketch for the data above is shown in
the figure below.
[2]
[Total 6 marks]
7
a)
The particular frequency at which a given part of a car vibrates is called its natural frequency
of vibration.
b)
[1]
Resonance occurs when an object is forced to vibrate and the frequency of the driving force is
equal to the natural frequency of vibration of the object. This results in a maximum transfer of
c)
energy to the object, which vibrates with large amplitude.
[2]
Your graph should look like the figure below.
[3]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Exam practice questions
d)
This graph is also shown in the figure below.
[2]
e)
If the suspension of a car is designed such that its natural frequency of vibration is nowhere near
the typical engine frequency, and the suspension is also damped, the transfer of energy to the
suspension will be minimised and unwanted resonance will be avoided.
[2]
[Total 10 marks]
8
a)
The relevant equation is a = − ω x. As ω is a constant, a graph of acceleration ω against
2
displacement x is a straight line through the origin of negative gradient (= − ω2) – answer C.
b)
[1]
The potential energy has its maximum value at each end of the oscillation and is zero at the
centre – think of a pendulum – so the answer is D.
[1]
[Total 2 marks]
9
a)
Your graphs should look like those in the figure below.
[2] and [3]
b)
For SHM, the acceleration must be proportional to the displacement and towards the equilibrium
position. For the bouncing ball, as it descends, the gradient, and therefore the acceleration, is
constant (if air resistance is ignored) and equal to the acceleration due to gravity, g, while the
ball is in the air. At the point of impact, this downward acceleration decreases and then suddenly
becomes a large upward acceleration (very steep gradient) due to the contact force of the hard
surface. Once the ball has left the surface, the acceleration is once more g downwards. The
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Exam practice questions
acceleration of the ball is therefore not proportional to the displacement and so the ball does not
have SHM.
[3]
[Total 8 marks]
Note: The question also arises as to where the equilibrium position of the bouncing ball is –
about which point does it oscillate? If this is taken to be the mid-point of its motion, then the
acceleration is not always towards this point, again indicating that the motion is not SHM.
10
a)
If the stroke (top to bottom) is 80 mm, the amplitude A is 40 mm = 4.0 × 10–2 m.
[1]
–1
At 6000 rpm, the frequency f = 6000/(60 s) = 100 s .
[1]
The SHM equation x = A cos(2πf)t is therefore
x = 4.0 × 10–2 cos(2π × 100)t
[1]
= 4.0 × 10–2 cos628t
b)
i) From a = – (2πf)2x, maximum acceleration = (2πf)2A
 amax = (628 s–1)2 × 4.0 × 10–2 m
[1]
= 1.6 × 104 m s–2
[1]
ii) From v = – (2πf)A sin(2πf), maximum speed = (2πf)A
 vmax = (628 s–1) × 4.0 × 10–2 m
[1]
= 25 m s–1
c)
[1]
Your graphs should look like those in the figure below
[5]
[Total 12 marks]
Tip: To get full marks you must put values on the axes, if the values are known. To get the final
mark, your velocity graph must be negative with respect to the displacement graph
11
a)
The amplitude, A, is the distance mass is pulled down from its equilibrium position,
i.e. 60 mm.
b)
T=
11.4 s
20
ω = 2πf =
[1]
= 0.57 s
2𝜋
𝑇
=
2π
0.57 s
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
Answers to Exam practice questions
= 11.02 s−1 ≈ 11 s−1
c)
[1]
i) amax = − ω2A = (11 s−1)2 × 60 ×10−3 m
[1]
= − 7.3 m s−2
[1]
ii) vmax = ± ωA = ± 11 s × 60 ×10 m
−1
−3
[1]
= ± 0.66 m s−1
d)
[1]
2
−1
Maximum KE = ½mvmax = ½ × 0.400 kg × (± 0.66 m s )
[1]
= 0.087 J ≈ 0.09 J
e)
[1]
Maximum F = mamax = 0.400 kg × 7.3 m s–2
[1]
= 2.9 N ≈ 3 N
f)
[1]
Spring constant (stiffness) is given by F = kx.
k=
𝐹
𝑥
=
2.9 N
[1]
60 × 10–3 m
= 48 N m–1 (or 49 N m–1 if all data is kept in the calculator)
[1]
𝑚
𝑘
Tip: You might like to check that the equation T = 2π√ gives the same answer for k.
g)
Maximum Ep = ½kA2 = ½× 49 N m–1 × (60 × 10–3 m)2 = 0.087 J
[1]
This is equal to the maximum kinetic energy found in part d), which it should be, because the
total energy is conserved.
h)
[1]
The mass is released at t = 0, at which point its KE is zero. As it passes through the mid-point of
the oscillation for the first time (after ¼T) the mass reaches its maximum velocity and so has
maximum KE of 0.09 J. When it reaches the top of the oscillation (after ½T) it momentarily comes
to rest, so its KE is zero. As it goes back down, its KE at the mid-point (after ¾T) is again 0.09 J,
becoming zero as it reaches the bottom (after time T).
[1]
In between these values the KE actually follows a (sin)2 curve as v = ± ωA sin ωt, making the KE =
½mω2A2 sin2ωt. Because it is a (sin)2 curve it is positive for each half cycle and the curves are
symmetrical about their mid-point. This is shown in blue in the figure below.
[1]
Likewise, Ep = ½kx = ½mω A cos ωt. This is plotted in red. Note that it is the mirror image of the
2
2 2
2
KE curve.
[1]
The total energy is the sum of the KE and PE and, by conservation of energy, must be constant.
This is therefore a horizontal straight line of value 0.09 J, as shown in purple.
[1]
[Total 20 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
12
a)
Answers to Exam practice questions
For the motion of the particles to be simple harmonic, the particles must have an acceleration
that is proportional to their displacement and which is directed towards their equilibrium
position.
b)
[2]
As the total energy must – by the conservation of energy – remain constant, the total energy line
must be a horizontal line at 2.0 × 10−6 J. The kinetic energy must therefore be a mirror image of
the potential energy curve so that Ek + Ep = ET = 2.0 × 10−6 J throughout the motion. This is shown
in the figure below.
c)
[2]
Drawing an analogy with a spring, the potential energy stored in the bonds is given by
Ep = ½kx2
[1]
From the graph, maximum Ep = 2.0 × 10−6 J when x = 10 × 10−2 m
[1]
k=
2𝐸p
𝑥2
=
2 × 2.0 × 10–6 J
(10 × 10–2 m)
[1]
2
= 4.0 × 10−4 N m−1
d)
[1]
Buildings in earthquake zones can be designed to be safer by employing:
• ductile construction materials;
[1]
• different types of damping mechanisms.
[1]
The ductile materials absorb the energy as they are compressed and then relax again
(hysteresis), and the damping mechanisms reduce the energy transfer from the earthquake to
the building.
[2]
[Total 12 marks]
Tip: The question asks you to ‘explain’, so merely stating two ways (as in the bullet points)
would only get half the marks.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
13
a)
Answers to Exam practice questions
At the mid-point the student has no acceleration, so the resultant force on her must be zero.
– the push of the seat must equal her weight. At the end of the oscillation, she will experience a
resultant force towards the mid-point. This is provided by a component of the push of the swing.
Your free-body diagram should therefore look like the figure below.
[4]
Tips: Remember that:
•
free-body force diagrams should only show the forces acting on the body. If you put in any
additional forces you will be penalised
•
b)
‘gravity’ is not a force. You must state ‘weight’ or ‘mg’ to get the mark.
i) The motion may not be simple harmonic because
•
it is fairly heavily damped;
•
the girl may move her centre of gravity, and therefore change the resultant force, during
the oscillation.
[2]
ii) The velocity at the mid-point of the first oscillation will be given by the gradient where the
curve crosses the axis for the first time.
[1]
If we draw a tangent to the curve at this point and extend it to the top and bottom edge of
the grid we get:
v=
2.2 m
[1]
0.9 s
= 2.4 m s−1
c)
[1]
At the mid-point, the velocity has its maximum value when sin (2πft) = 1
vmax = 2πfA = 2π × 0.333 s−1 × 1.2 m
[1]
= 2.5 m s−1
d)
[1]
If the girl on the swing were to be given a push each time she reached the end of each oscillation,
the driving force (the push) would have the same frequency as her frequency of oscillation. She
would therefore absorb this energy and oscillate with large amplitude – this is resonance.
[3]
[Total 14 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
14
a)
b)
Answers to Exam practice questions
For SHM, the body must have an acceleration that is proportional to its displacement from a
fixed point and which is always directed back towards the fixed point.
[2]
The graph of acceleration as a function of displacement is shown in the figure below.
[2]
As the graph is a straight line through the origin it shows that the acceleration is proportional to
the displacement. The negative slope indicates that the acceleration is always in the opposite
direction to the displacement, i.e. back towards the centre of the oscillation.
[2]
c)
From x = A cos(2πf)t
d𝑥
d𝑡
d2 𝑥
d𝑡 2
From x = A sin(2πf)t
= − (2πf) A sin(2πf)t
d𝑥
d𝑡
= − (2πf)2 A cos(2πf)t
d2 𝑥
d𝑡 2
a = − (2πf)2x
= (2πf)A cos(2πf)t
= − (2πf)2 A sin(2πf)t
a = − (2πf)2x
[3]
[3]
The solution depends on the value taken for x when t = 0. For the sine wave, x = 0 when t = 0 and
for the cosine wave, x = A when t = 0. There are an infinite number of solutions between these
two as x can have any value between 0 and A.
d)
[2]
From x = A cos(2πf)t
v=
d𝑥
d𝑡
= − (2πf)A sin(2πf)t
[1]
The maximum value that sin(2πf)t can have is ±1
[1]
vmax = ± (2πf)A
[Total 16 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
15
a)
Answers to Exam practice questions
For the mass to oscillate with SHM, the spring must obey Hooke’s law, i.e. its extension must
be proportional to the resultant force acting on it.
b)
𝑚
𝑘
0.400 kg
25 N m–1
For the spring: T = 2π√ = 2𝜋√
[2]
[1]
= 0.795 s
f=
1
𝑇
=
1
0.795 s
= 1.26 Hz
[1]
Maximum KE = ½mvmax2 = ½ × m × (2πfA)2
c)
= ½ × 0.400 kg × (2π × 1.26 s−1 × 0.10 m)2
[1]
= 0.125 J
[1]
At the centre of an oscillation, the extension of the spring is just the extension produced by the
mass of 0.400 kg.
F = kx  x =
𝐹
𝑘
=
0.400 N × 10 N kg–1
[1]
25 N m–1
= 0.16 m
[1]
Elastic PEcentre= ½kx2 = ½ × 25 N m−1 × (0.16 m)2
= 0.320 J
[1]
[1]
At bottom of oscillation spring has been pulled down a further 10 cm, so:
Total extension of spring = 0.16 m + 0.10 m = 0.26 m
−1
2
Elastic PEbottom = ½ × 25 N m × (0.26 m) = 0.845 J
[1]
[1]
At top of oscillation spring goes up 10 cm above the equilibrium position, so:
d)
Extension of spring = 0.16 m − 0.10 m = 0.06 m
[1]
Elastic PEtop = ½ × 25 N m−1 × (0.06 m)2 = 0.045 J
[1]
At the bottom of an oscillation, the mass is 5 cm above the bench, so:
GPE of mass at bottom = mgΔh = 0.400 kg × 10 N kg−1 × 0.05 m
= 0.200 J
[1]
[1]
At the top of an oscillation, the mass is 5 cm + 10 cm + 10 cm = 25 cm above bench, so:
GPE of mass at top = 0.400 kg × 10 N kg−1 × 0.25 m
= 1.000 J
[1]
[1]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
32 Oscillations
e)
Answers to Exam practice questions
The graphs are shown in the figure below.
Points to note are:
•
The gravitational potential energy of the mass will vary linearly with the displacement (from
ΔE = mgΔh) – draw this first, from 0.200 J at the bottom to 1.00 J at the top.
•
The kinetic energy of the mass will be a maximum of 0.125 J at the midpoint and zero at the
top and bottom of the oscillation, following an x2 curve between these points.
•
[1]
Next plot the elastic potential energy at the bottom (0.845 J), middle (0.320 J) and top
(0.045 J) and sketch a curve between the points
•
[1]
[2]
At the top and bottom, when the kinetic energy of the mass is zero, the total energy of the
system will be the sum of the potential energies (0.200 + 0.845 = 1.045 J at the bottom and
1.000 + 0.045 = 1.045 J at the top). They must, of course, be the same because the total
energy of the system is conserved. In other words the total energy is constant.
[2]
[Total 24 marks]
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
Page 637 Test yourself
1
2
3
4
5
6
7
8
a)
8.3 × 10−2
b)
3.6 × 10−3
c)
6.1 × 10−1
d)
1.7 × 102
e)
8.2 × 102
a)
20
b)
23.9
c)
2.0
a)
1012
b)
108
c)
10−9
d)
103
e)
109
f)
102
a)
1
b)
0.01
c)
0.1
d)
1 × 109
e)
10 000
a)
1.45 × 103
b)
1.21 × 10−4
c)
0.20
d)
742
a)
74
b)
32
c)
2.0 × 104
d)
100
e)
9
f)
12.7
a)
100%
b)
200%
a)
100
b)
222
c)
0.18
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
9
Tolerance: 3.14–3.26 mm; 2nd and 8th batch out of tolerance
10
a)
400 V
b)
3.0 × 108 m s−1
c)
0.10 s
Page 639 Test yourself
1
2
3
a)
2.3 × 10−2 m
b)
1.8 × 10−3 m
c)
4.5 × 10−7 m
d)
1.2 × 103 m
e)
3.8 × 10−2 kg
f)
4.2 × 10−1 kg
g)
1.2 × 10−5 kg
h)
1.8 × 103 s
i)
2.0 × 10−8 s
j)
4.3 × 104 s
k)
3.0 × 10−1 A
l)
1.5 × 10−3 m3
a)
2.3 × 10−4 m
b)
8.4 × 10−6 m2
c)
2.3 × 10−4 m2
d)
3.3 × 10−6 m3
e)
9.1 × 10−8 m3
a)
1.2 m2
b)
0.18 m2
c)
8.4 × 10−2 m2
d)
1.2 × 10−4 m2
4
1.52 × 105 N
5
3.7 × 10−4
6
a)
𝑣=
b)
𝑟=√
c)
ℎ=
d)
𝑡=
e)
𝑠
𝑡
𝐴
𝑎=
𝑙
𝑉
𝜋𝑟 2
𝑣−𝑢
𝑎
2(𝑠 − 𝑢𝑡)
𝑡2
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
𝑇1
f)
𝑇2 =
g)
𝐼=√
1−𝜂
𝑃
𝑅
𝜀−𝑉
𝑟=√
h)
𝑇 2𝑔
𝑙=
i)
𝐼
4𝜋2
14
7
5.02 × 10 Hz
8
8.5 × 10−2 m
9
4.2 mm
10
5060 s
11
N = kg m s−2
12
Pa = N m−2 = kg m s−2 m−2 = kg m−1 s−2
13
J = kg m s−2 m = kg m2 s−2
14
W = kg m2 s−2 s−1 = kg m2 s−3
15
C=As
16
V = kg m2 s−2 (A s)−1 = kg m2 A−1 s−3
17
Hz = s−1
18
𝐶=
19
a)
45 m s−1
b)
5 × 10−2 J
c)
25 Pa
d)
64 kg
a)
J = kg m2 s−2
b)
m−3 = kg m2 s−2 m−3 = kg m−1 s−2; Pa = kg m s−2 m−2 = kg m−1 s−2; ε has no unit
c)
Pa = kg m s−2 m−2 = kg m−1 s−2; ρgh =(kg m−3)(m s−2)(m) = kg m−1 s−2
d)
√m s−2 = s
e)
J = kg m2 s−2; (m s−1)2(kg) = kg m2 s−
20
𝑄
𝑉
As
⇒𝐹=
kg m2 A−1 s
= A2 kg −1 m−2
mgh = (kg)(m s−2)(m) = kg m2 s−2
m
Page 640 Test yourself
1
2
a)
x is greater than 12 cm
b)
m is greater than 10 kg and equal to, or less than, 20 kg
c)
V is proportional to r cubed
d)
the change in x is 25 mm
a)
sin 𝜃𝑚 ∝
b)
𝑝∝
𝜆
𝑑
𝑚⟨𝑐 2 ⟩
𝑉
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
3
4
5
6
7
8
9
a)
True
b)
True
c)
True
d)
False
e)
True
f)
False
g)
True
h)
True
a)
ABC
b)
B
a)
50 m s−1
b)
2.5 m s−2
c)
2.5 Ω
d)
25 µF
e)
4.0 × 1014 m3 s−2
f)
3.8 × 10−15 V s
a)
160 m
b)
140 m
c)
3.0 × 10−2 J
a)
36–38 J
b)
3300–3400 µC
a)
Plot Ek against ʋ 2; gradient = ½m
b)
Plot λ against ; gradient = c
c)
Plot F against
d)
Plot f2 against l; gradient =
e)
Plot T2 against m; gradient =
f)
Plot Vs against f; gradient = h/e
1
𝑓
1
𝑟2
; gradient = G m1 m2
1
4π2 𝑔
4π2
𝑘
ℎ
𝑒
a)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
10
b)
No
c)
−0.73 Ω
d)
m = −0.73 Ω, c = 1.52 V
a)
l1
b)
𝑥2
c)
P = 4π × gradient = 4π × 0.48 W = 6.0 W
Page 643 Test yourself
1
2
3
a)
12°
b)
115°
c)
90°
a)
0.436 rad
b)
0.698 rad
c)
1.13 rad
π
a)
2 rad
b)
2π rad
π
c)
4
5
6 rad
a)
60
b)
3600
a)
1 radian is the angle subtended by an arc of a circle equal to the radius.
b)
Draw a circle with 2 radii at an angle of about 114°
6
60 mm
7
a)
0.36 rad
b)
21°
a)
12.6 m
b)
Assume that the pole is at right angles to the ground.
8
9
tan
1°
3600
=
1.5 × 1011 m
1 pc
; 1 pc = 3.09 × 1016 m
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
Page 644 Test yourself
1
2
a)
0.597
b)
0.815
c)
0.711
a)
i) 0
ii) 1
b)
i) 1
ii) 0
c)
i) 1/√2
ii) 1/√2
3
4
5
6
7
8
9
10
a)
0.594
b)
36.4°
a)
0.808, 36.1°
b)
0.735, 36.3°
a)
10.4 cm
b)
4.20 cm
a)
x = 24 m s−1cos 30 ° = 21 m s−1; y = 24 m s−1cos 60 ° = 12 m s−1
b)
x = 9.8 m s−2cos 63 ° = 4.4 m s−2; y = 9.8 m s−2cos 27 ° = 8.7 m s−2
c)
x = 0.15 T cos 40 ° = 0.11 T; y = 0.15 T cos 50 ° = 0.10 T
a)
yes
b)
no
c)
yes
d)
yes
e)
no
f)
yes
g)
yes
a)
5.8 cm
b)
5.6 cm
c)
0.754 cm
a)
23 N
b)
31 °
a)
T cos θ = 10 N
b)
T sin θ = 8.0 N
c)
tan θ = 0.8; θ = 39 °
d)
x = 50 cm sin 39 ° = 31 cm
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
Page 646 Test yourself
1
a)
sin θ = 0; cos θ = 1
b)
sin θ = 1; cos θ = 0
2
There is a phase difference of π/2 radians (a quarter of a cycle)
3
0 ° and 180 °
4
0 ° and 180 °
5
a)
radians per second
b)
radians
a)
6π × 10−3 m s−1 (0.0188 m s−1)
b)
60π2 m s−2 (590 m s−1)
a)
i) v = 0
6
7
ii) v = 2πs02
b)
a = −4π2s02s
8
x = −1.62 m
9
x = −1.18 m
10
ω = 100π = 314 rad s−1
11
a)
f = 2.3 × 10−5 Hz
b)
vmax = 5.1 × 10−3 m s−1
12
Page 647 Test yourself
1
2
a)
1.48
b)
2.48
c)
3.48
a)
3.40
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
b)
5.70
c)
8.01
3
The first figure represents the power of ten.
4
a)
i) 1.00
ii) 2.00
iii) 3.00
b)
i) 1.00
ii) 2.00
iii) 3.00
5
Logs are numbers with no physical quantity.
6
a)
5.47
b)
5.47
a)
0.183
b)
0.183
7
8
3.12 × 10−53
9
Fixed percentage chance of growth until saturation, i.e.
10
11
a)
when people become immune to the virus and,
b)
when sufficient people own an i-pad so that demand begins to fall.
a)
24 month−1
b)
80 month−1
a)
18.0 µC
b)
13.5 µC
12
5730 years
13
a)
14
15
b)
Half-life = 30 s
a)
seconds
b)
0.231 s
a)
2800 s−1
b)
7 half-lives (
1
128
)
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
16
a)
k h must be unit-less
b)
i) 54 kPa
ii) 29 kPa
17
a)
b)
The straight line graph shows that it is exponential.
c)
The value of N increases by equal proportions for equal time intervals.
d)
Method c)
Page 648 Test yourself
1
a)
50π (= 157) m s−1
b)
π2 × 104 (= 9.87 × 104) m s−2
c)
Speed is a scalar and acceleration is a vector quantity.
2
dQ/dt = −kQ0 sin kt; this is the current
3
720 µC
4
Area = distance travelled
5
gradient = resistance
6
a)
ii)
b)
7
d𝐹
i) gradient = 0
d𝐹
d𝑡
d𝑡
=0
= 6.0 × 103 N s−1
Area ≈ 25 N s
dEk
a)
d𝑣
b)
= mv
momentum
d(𝑚𝜐)
is more general; F = 𝑚
dυ
8
F=
9
a)
Minus sign means that N is decreasing with time
b)
i) The rate of change in N is proportional to N
d𝑡
d𝑡
or F = ma only applies to objects of fixed mass.
ii) ln N = ln N0 – λt
10
a)
Not proportional – the line does not pass through the origin
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
33 Maths in physics
Answers to Test yourself questions
b)
11
12
c)
Graph of V against t will show exponential decay.
a)
The values of N cover an extremely large range.
b)
Graph of ln N against t is a straight line through the origin.
a)
i) −3/r4
ii) −2/r3
iii) ln r
b)
i) −1/2r2
ii) −1/r
iii) ln r
13
a)
i) m = −0.45 day−1
ii) c = 1.8
b)
ln p = −0.45t + 1.8
c)
Excess pressures decays exponentially with time.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
34 Practical skills
Answers to Exam practice questions
Page 659 Exam practice questions
1
We had that
T2 = k2L3 where the units for k were s m–3/2 .
From E =
⟹ k2 =
16 𝜋 2 𝑀𝐿3
𝑤𝑡 3 𝑇 2
⟹ T2 =
16 𝜋2 𝑀𝐿3
𝑤𝑡 3 𝐸
⟹ T2 =
16 𝜋 2 𝑀
𝑤𝑡 3 𝐸
× L3
16 𝜋2 𝑀
𝑤𝑡 3 𝐸
Substituting units: k2 =
kg
m × m3 × N m−2
=
kg
m2 × kg m s−2
= s2 m–3
Units of k = √ s2 m–3 = s m–3/2
This is consistent with the units of k found from T2 = k2L3
2
a)
i) When the frequency of the signal generator, that is the frequency of vibration of the
loudspeaker, is equal to the natural frequency of vibration of the air inside the flask,
resonance occurs.
Energy is transferred with maximum efficiency from the loudspeaker to the air in the flask
and so this air vibrates very loudly.
ii) The frequency should gradually be increased until resonance occurs. This frequency
should be recorded. The frequency should then be increased beyond the resonant frequency
and gradually reduced until once again resonance is detected. This frequency should also
be recorded and the average of the two values should be taken as the resonant frequency.
Tip: Note that the technique should be described in detail – an answer such as ‘The reading
was repeated and the average taken’ is not sufficient.
iii) A measuring cylinder is required, then
• the measuring cylinder is used to fill the flask to the top with water so that the volume Vf
of the flask can be determined
• the flask is then emptied
• a known volume v of water is now added to the flask
• the volume of air will therefore be V = Vf – v
• further volumes of water are added to give different volumes V of air in the flask.
iv) The proposed equation is f  Vn, or f = kVn. If we take logarithms on both sides of the
equation we get
lnf = n lnV + lnk
which is of the form
y = mx + c.
So if a graph of lnf against lnV is plotted, this should be a straight line with gradient equal
to the constant n and intercept on the y-axis equal to lnk.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
34 Practical skills
Answers to Exam practice questions
Tip: Note that you could take natural logs (ln button)) or logs to base 10 (log button). It is a good
idea to stick to natural logs as this works for exponential functions as well as for power
equations.
b)
i)
V/cm3
f/Hz
In(V/cm)
In(f/Hz)
554
219
6.32
5.39
454
242
6.12
5.49
354
274
5.87
5.61
254
324
5.54
5.78
204
361
5.32
5.89
154
415
5.04
6.03
ii) Your graph should be a straight line of negative slope.
Note that:
• the points should occupy at least half the graph paper in each direction, which means
that the scales should not start at the origin;
• the axes should be labelled exactly as shown: ln(f/Hz) on the y-axis and ln(V/cm) on the
x-axis
• the points should be plotted to a precision of at least half a square;
• a thin, straight line of best fit should then be drawn.
iii) The constant n is the gradient of the graph. Taking a large triangle,
gradient = n =
6.05 − 5.35
5.00 − 6.40
=
+0.70
−1.40
= − 0.50
Don’t forget the minus sign!
iv) A negative value of n means that as the volume of air in the flask gets less, the frequency
of vibration (pitch) of the air gets higher, which is consistent with the student’s initial
observation.
A value of − 0.50 suggests that n = − ½, meaning that
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019
34 Practical skills
Answers to Exam practice questions
f  𝑉 −½ , or f 
1
√𝑉
In other words, the frequency of vibration is inversely proportional to the square root of
the volume of air.
© Nick England, Jeremy Pollard, Nicky Thomas & Carol Davenport 2019