CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Workbook answers
Chapter 1
Practice
Exercise 1.1
2
Focus
1
a
b
c
metre (m)
cubic metre (m3)
kilometre (km)
millimetre (mm)
100
1000
a
b
10 000 (100 × 100)
1 000 000 (1000 × 1000)
Challenge
3
a
3.50 m = 350 cm
Each face has an area of 350 × 350
There are six faces
Total surface area = 6 × (350 × 350)
= 735 000 cm2
b
3.5 m = 3500 mm
Volume of a cube = (length of side)3
(350 000)3 = 42 875 000 mm3
= 4.29 × 1010 mm3 (3 s.f)
Exercise 1.2
Focus
1
a
b
2
1
Material
State / type
Density / kg/m3
Density / g/cm3
water
liquid / non-metal
1 000
1.000
ethanol
liquid / non-metal
800
0.800
olive oil
liquid / non-metal
920
0.920
mercury
liquid / metal
13 500
13.500
ice
solid / non-metal
920
0.920
diamond
solid / non-metal
3 500
3.500
cork
solid / non-metal
250
0.250
chalk
solid / non-metal
2 700
2.700
iron
solid / metal
7 900
7.900
tungsten
solid / metal
19 300
19.300
aluminium
solid / metal
2 700
2.700
gold
solid / metal
19 300
19.300
Ice is less dense than water.
Since olive oil has a lower density than water the
olive oil will float on top of the water.
Cambridge IGCSE™ Physics Workbook – Hamilton © Cambridge University Press 2021
CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Practice
Exercise 1.3
3
Students should recognise that measuring 50
pulses is better than measuring 10 (provided that
the pulse rate is not changing). Also they should
appreciate that pulse rate can change, and that this
makes it less reliable than using a pendulum.
4
5
Disagree. Aluminium (metal) is less dense
than diamond (non-metal). But it is true that,
for the table, most metals are more dense than
most non-metals.
mass
​density = _______
​
​
volume
So mass = density × volume
= 19 300 × (0.20 × 0.15 × 0.1)
= 57.9 kg
mass
​density = ​ _______ ​​
volume
14
Volume in m3 = ______________
​​     ​​
100 × 100 × 100
Density
Self-assessment
Answers should include:
• mention of a form of exercise (for example,
walking/running upstairs)
• detail of that exercise (for example, 35 steps
rising to vertical height of 7 m, completed 20
times in 3 minutes)
• a statement of pulse measurement:
• whilst resting before exercise
• at regular intervals (30 s/1 min)
during exercise
• at regular interval after exercise
• the idea that the pendulum period will not
vary due to common external factors whilst
pulse rate can vary significantly.
0 . 270
= ​​ ________ ​​
14
________
​
​
1000000
= 19 300 kg/m3 (3 s.f )
This metal could possibly be tungsten.
Challenge
6
3
Measuring cylinder
• Note volume of water in
measuring cylinder.
• Immerse object in water and note
new reading.
• volume of object is the difference between
the readings.
Balance to measure mass
• Zero the balance.
• Place the object on the balance.
• Note the reading on the balance – this is
the mass of the object.
• Divide the mass by the volume to
calculate the density in g/cm3.
Practical points
• Place your eye level with the bottom
of the meniscus of the water in the
measuring cylinder when making
the measurements.
• Repeat and average values of volume.
a
Exercise 1.4
Focus
1
resultant
2N
4N
2
4N
a
resultant
2N
θ
b
To quote a vector, you MUST quote the
direction, so measure the angle θ. The
diagram above is another way to draw the
answer to 1a. In each case, the angle θ is
the angle between the horizontal and
the resultant.
10 N
–7 N
3 N = resultant
b
2
By calculation: 10 – 7 = 3 N to the right
Cambridge IGCSE™ Physics Workbook – Hamilton © Cambridge University Press 2021
CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Practice
4
a
Chapter 2
400 N
Exercise 2.1
200 N
300 N
1
resultant = 410 N
θ
b
___________
Resultant: √​​ ​100​​  2​ + ​400​​  2​ ​​ = 412 N,
5
a
SI unit
(name
and
symbol)
NonSI units
Measuring
instrument
distance
metre
(m)
mile,
etc.
tape measure,
rule
time
second
(s)
hour,
etc.
clock,
stopwatch
speed
metre
per
second
(m/s)
mile per light gate and
hour,
data logger,
etc.
ultrasound/
microwave
transducer
and
datalogger
Quantity
Vertical resultant by calculation:
300 – 200 = 100 N
100
​​ = 14°
θ = tan–1​____
​
400
Challenge
Focus
(20 – 10 – 5) N
θ
2
a
b
c
(12 + 7) N
resultant = 19.6 N
Distance travelled
distance
Speed = ________
​​
​​
time
2.8 m/s
Practice
3
a
b
c
Knowing the distance between the
detectors, calculate
distance
Speed = ________
​​
​​
time
24 m/s; this is within the speed limit
0.048 s
Challenge
4
b
If the time taken by a vehicle is equal to
or less than 0.048 s, the warning lights
are shown.
By calculation:
______________________
​​√
​(20 − 10 − 5)​​ 2​ + ​(12 + 7)​​  2​ ​​ = 19.6 N,
19
θ = tan–1​___
​   ​​ = 75°
5
3
Cambridge IGCSE™ Physics Workbook – Hamilton © Cambridge University Press 2021
CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Peer-assessment
Practice
Answers should include:
• a means of detecting the motion (for example,
light gates, radar or sonar transmitted
and received)
• how the speed will be calculated – for example:
• light gates: knowing the distance between
the light gates and the time measured for
the vehicle travelling between them, the
speed can be calculated
• radar/sonar: knowing the speed of light/
sound and the time between transmission
of the wave and the returning reflection,
the speed can be calculated from a pair
of measurements
• further detail on calculation:
• light gates: speed = distance/time
• radar/sonar: speed = distance/time,
where the distance travelled = speed
of light (or sound) × (time between
transmission and reflection return)/2, and
time is the time between measurements
3
making sign illuminate – if the measured
speed is greater than the pre-determined level,
a signal is sent to the sign to illuminate it.
Other relevant detail can be included, such as
reflections that scatter from the vehicle might
reduce the accuracy of the measurement, and light
gates are a potentially ugly permanent fixture and
easily damaged by weather.
•
4
Exercise 2.3
Focus
1
2
a
770 m
2
a
3.60 s
3
a
b
b
8.0 s
2.0 m
Their speeds might change during the race.
Exercise 2.4
Focus
1
Description of motion
Graph(s)
moving at a steady speed
B, D
stationary (not moving)
A
moving fastest
B
changing speed
C
4.2
23.8
Practice
green car
3.8
26.3
2
yellow car
4.7
21.3
1 200 000 m
80 min
4800 s
250 m/s
40 s
Challenge
red car
a
b
c
d
b
Practice
The green car should be circled as the fastest.
Time taken / s Speed / m/s
distance
distance
​​
​​ so time = _______
​​
Speed = _______
​​
speed
time
Distance = 2 × distance to object
200 × 2
Time = ​​ _______8 ​​= 1.3 × 10–6 seconds
3 × ​10​​  ​
Focus
Car
10 m/s
15 m/s
It is speeding up (accelerating).
15 m/s, because down is the positive
direction (or –15 m/s, because up is the
positive direction).
Challenge
Exercise 2.2
1
a
b
c
d
a
100
80
Distance
/m
60
40
20
0
4
0
2
4
6
8
Time / s
10
12
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
b
i
75.0 m
ii 6.5 s
iii 11.5 s
10.0 m/s
c
2
1.6 m/s2
30
3
Speed
/ m/s
20
B
10
A
Distance
0
0
0
20
40
60
Time / s
80
100
10
20
30
Time / s
40
Practice
3
Challenge
4
0
a
b
First section marked as faster
i
17.5 m/s
ii 10.0 m/s
Average speed = distance travelled/time taken
= 1000/100 = 10 m/s
Exercise 2.5
a
b
c
1
A = __
​​   ​​× 15 × 24 = 180 m
2
B = 25 × 24 = 600 m
Total distance = 780 m
Challenge
4
Speed
Focus
1
8 km/h
0
Practice
2
2.1 m/s2
3
34(.3) m/s
4
6.25 s
30
40
50
60
70
Focus
On Earth, time to stop = 10/9.8 = 1.02 seconds
On Pluto, time to stop = 10/0.62 = 16.13 seconds
Difference = 15.11 = 15 seconds to
2 significant figures.
1
a
b
c
d
e
Constant acceleration
Decreasing acceleration
Constant speed
(Increasing) deceleration
Constant speed
Practice
Focus
5
20
Exercise 2.7
Exercise 2.6
1
10
Time / s
Challenge
5
0
2
Description of motion
a
Graph(s)
moving at a steady speed
C
speeding up, then slowing
down
A
moving with constant
acceleration
D
accelerating to a steady speed
B
b
c
d
e
f
change in speed
​​
Acceleration = ______________
​​
time
6
= __
​​   ​​
3
= 2 m/s2
1
__
​​   ​​ = 0.25 m/s2
4
0 m/s2
6/3 = 2 m/s2
0 m/s2
The gradient of the tangent to the curve
CD at 14 seconds = –1.9 m/s2
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Challenge
3
a
b
4
As the car gets fasters, air resistance
increases, so the acceleration decreases,
making the gradient of the graph smaller.
The initial acceleration of car 2 is greater,
but for a shorter time.
Object 1’s acceleration decreases between
3 and 7 seconds, then travels at a constant
speed until 12 seconds. Object 2 travels
at a constant speed (higher) speed from 1
second to 15 seconds.
At the moment the skydiver jumps, they
accelerate at 9.8 m/s2. Since their speed
increases, air resistance increases. So the
acceleration decreases until the acceleration
falls to zero. This is terminal velocity.
The sky diver then opens their parachute
and air resistance increases, so the skydiver
decelerates. Air resistance decreases until the
deceleration falls to zero. The skydiver now
has a new, lower terminal velocity.
When they land, the force from the
ground provides a final deceleration to stop
the skydiver.
Forces and labels should be as follows:
Apple: (up) air resistance of air on apple;
(down) gravitational force of Earth on apple.
Car: (up) contact force of road on car; (down)
gravitational force of Earth on car; (back) air
resistance of air on car; (forwards) push of
engine on car.
Person on slide: (down) gravitational force of
Earth on person; (up slope) frictional force
of slide on person; (normal to slope) contact
force of slide on person.
Fish: (down) gravitational force of Earth on
fish; (up) upthrust of water on fish; (back)
drag of water on fish; (forwards) thrust caused
by fish’s movements, acting on fish.
Paperclip: (down) gravitational force of Earth
on clip; (up) magnetic force of magnet on clip.
Box: (down) gravitational force of Earth
on box; (up) contact force of floor on box;
(to right) push of person on box; (to left)
frictional force of ground on box.
Chapter 3
Exercise 3.1
Focus
1
size, shape (in either order)
Practice
2
Challenge
3
6
He appears to be floating in mid-air. Since he
has weight, he must either be supported, or be
accelerating towards the ground.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
4
See diagram in answer to Question 3.
Forces on object
Exercise 3.2
60 N
Focus
50 N
1
A
B
C
D
The van will accelerate/speed up.
The van will decelerate/slow down.
The tree will bend over to right.
The ball will accelerate downwards (but
follow a curved path).
40 N
Resultant force
20 N
100 N
40 N
100 N
100 N
30 N
Practice
2
a
40 N
Practice
b
Friction will make him go slower (better
answer: … reduce his acceleration).
Heating of both surfaces
c
2
The motion will change/the object
will accelerate.
3
The motion will remain constant/the object
will have a constant velocity.
4
Diagrams will vary; but must show a body
with four forces acting on it with resultant 4 N
acting vertically downwards.
Challenge
5
a
Challenge
3
The ground stops the phone in both cases.
This requires a force greater than the weight
of the phone.
Dropping from the window, it will be moving
faster than when I drop it from chair height
and so the force required to stop it over the
same distance is much greater.
1
Focus
Forces on object
80 N
Exercise 3.4
Focus
Exercise 3.3
1
b
32 + 42 = resultant2
______
Resultant= √​​ 9 + 16 ​​= 5 N
​3 ​​ = 63° to the direction of
At angle = tan–1 ​ __
2
the 2 N force
An arrow, drawn vertically down from
the middle of the rectangle, and
labelled ‘weight’.
Resultant force
45 N
20 N
20 N
c
2
20 N
35 N
The effect of gravity on an object
Weight = mass × gravitational field
strength, W = mg
Balance
g = 9.8 m/s2
Practice
3
40 N
a
b
Gravitational field strength falls with distance
from the Earth. The space station is very close
to the Earth and so gravity cannot have fallen
to zero at that height.
Therefore, the astronauts cannot be weightless.
20 N
7
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Peer-assessment
Exercise 3.5
An answer would include some of the following:
• mass is unchanged
• can only be weightless if the gravitational field
strength is zero
• the gravitational field strength depends on
how far from the planet you are when you
measure it
• it is the gravitational field strength that means
there is an atmosphere around Earth
• the Space Station is in the outer atmosphere
• therefore, there must still be a gravitational
field at this altitude
• the reason for feeling weightless is that both
the astronaut and the space station are falling
towards Earth at the same rate
• but they never hit the Earth because they are
moving in a circle around it at high speed.
Focus
1
The speed is increasing.
greater mass
with parachute
Challenge
8
4
The force depends on his mass, since the
force is required to move him horizontally.
His weight has no effect. Since his mass is the
same in both situations, the force required is
the same.
5
Gravity on Jupiter is 25 m/s2, or 2.4 times that
on Earth.
Our visitor would therefore have to be 2.5
times as strong as an Earthling to do the
same everyday things (walking, jumping,
lifting, throwing).
So there is a good chance that they will be able
to jump much higher on Earth than someone
born on Earth and would probably be able to
beat world-class high-jumpers!
Of course, they might struggle with our
climate and our atmosphere.
Practice
2
a
b
See diagram – dots are at same heights.
The accelerations of the two objects
are equal.
Challenge
3
See diagram – crosses quickly become
equally spaced.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
4
a, b
B
Speed
air resistance
air resistance
A
B
weight
weight
A
Time
c
It would:
• rise more slowly (lower gradient)
• reach a lower maximum velocity.
Exercise 3.6
Focus
1
Exercise 3.7
Focus
1
It increases
At right angles
Practice
a
2
Quantity
Symbol
SI unit
force
F
newton (N)
mass
m
kilogram (kg)
acceleration
a
metre per second
squared (m/s2)
b
​m = __
​ F
a ​​
c
F
​a = ​ __
m ​​
2
14.4 N
3
3.5 m/s2
4
1667 kg (1670 kg)
Increase
Challenge
3
a
b
c
Practice
The force must increase. This is because
F = ma and since the mass has increased,
whilst the acceleration remains the same,
the resultant force must increase.
The force will decrease. This is because
the velocity is changing more slowly in the
larger circle, so the acceleration is smaller,
from F =ma.
The force will increase as the velocity is
changing more rapidly, due to the
higher speed of rotation, so there is a
larger acceleration.
Exercise 3.8
Challenge
Focus
5
1
a
b
c
2
Impulse = change in momentum
= 80 × 5 – 80 × 15
= –800 kg m/s
3
F = Δp/Δt or resultant force = rate of change
of momentum or resultant force = change in
momentum/time
a
2.4 N
8.0 N
b
9
a
b
7.0 m/s2
1200 kg m/s
15 000 kg m/s
0.030 kg m/s
Practice
4
a
b
c
d
9000 kg m/s
Its velocity is zero, so mv = 0.
3000 kg
9000 kg m/s
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
e
f
g
3.0 m/s
Impulse = change in momentum
= 2000 × 3 – 0
= 6000 kg m/s
Impulse = mv – mu
Ft = m(v – u)
m(v − u)
impulse ________
_______
​ = ​
​
​​
t
time
= ma
impulse
​
a = ___________
​
time × mass
6000
= __________
​
​
0 . 1 × 2000
= 30 ​m/s​​  2​​
Challenge
4
b
c
a
b
c
d
1
2400 N s (or 2400 kg m/s)
5000 kg m/s
400 kg m/s
14.8 m/s
The beam will have weight; there will be a
reaction at the pivot.
They both act through the pivot and so
have no moment about the pivot.
The moment of a force is the turning effect of
the force. It is calculated by multiplying the
size of the force by the perpendicular distance
of the line of action from the pivot.
Practice
2
a
b
c
x = 0.67 m
y = 0.71 m
z = 62.5 = 63 N to 2 significant figures
Challenge
3
Exercise 4.1
Focus
1
If each person on either end has the same
weight, it will balance.
2
a
b
Resultant force = 0
Resultant turning effect = 0
If the beam is uniform and pivoted in the
middle, then there will be no moment of the
weight about the pivot, so we can ignore it
here. (However, in order to fully consider
equilibrium, we would need to know the forces
acting at the pivot.)
Exercise 4.3
Focus
1
Practice
a
A
Focus
Chapter 4
3
B
Exercise 4.2
Challenge
5
a
force
a
b
Force 3 has the greatest moment about
point A.
Force 4 has no moment about point B.
Practice
2
b
10
Note: A vertical force at the end of the
handle is a satisfactory answer. Showing
the force arrow at 90° to the line joining
the handle to the wheel is better.
There is no resultant force, there is no net
moment – it is not accelerating.
a
b
3
Moment /
Nm
Clockwise or
anticlockwise?
A
6.0
anticlockwise
B
6.0
clockwise
C
8.0
clockwise
Force
Force C must be removed if the beam is to
be balanced.
F = 340 N
Challenge
4
a
b
distance x
This is the perpendicular distance from X
to the line of the force F.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Exercise 4.4
The horizontal force of the ground on the
base needs to be able to equal the opposite
horizontal force of the wind on the
turbine blades.
Focus
1
If the line of action of the weight falls outside
the base there will be a resultant moment, and
the object will topple.
Practice
2
a, b
Peer-assessment
Answer is entirely dependent on the descriptions
given by the students, but the expectation is that
they will learn from each other’s points and also
critically assess the validity of the points made by
others, in terms of the physics.
Chapter 5
A more stable object has a wider base and
a lower centre of mass. A less stable object
has a narrower base and a higher centre
of mass. Typical examples are shown.
3
Upward force: contact force; downward
force: weight
4
a
contact
force
Exercise 5.1
Focus
1
boss
contact
force
mass hanger
weight
b
clamp
ruler
weight
The object on the left will not topple over,
as its weight passes through its base and
so will cause it to tip to the left, returning
it to an upright position. The object on
the right will topple over, because its
weight is acting outside its base.
G clamp
Challenge
5
6
11
Car A will be able to corner at a higher speed.
It has a lower centre of mass, so it needs to
lean more before the line of action of the
weight falls outside the base. This means it can
travel faster.
It has a wider base. So, it needs to lean more
before the line of action of the weight falls
outside the base. This also means it can
travel faster.
The base needs to have a large weight – to
lower the centre of mass of the whole
wind turbine.
The base needs to be wide – to increase the
moment of the force the ground pushes
upwards on the base with.
Points to include:
• Equipment: clamp stand, 2 × boss, 2 ×
clamps, spring, mass carrier and 10 × 1 N
masses, metre ruler, optical pin (or other
fiducial marker), G clamp, set square.
• Use the G clamp to fix the clamp stand to
the desk, to avoid risk of injury through
the equipment toppling.
• Use the set square to ensure the 1 m ruler
is clamped vertically to the clamp stand
and is behind the optical pin, which is
attached to the top of the mass carrier.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
•
•
•
Read the position of the optical pin
against the ruler scale, keeping your eye
level with the pin, when only the mass
carrier is on the spring. Record this as the
unstretched length.
Add 1 N to the mass carrier, read the
position of the pin as before, and record
the reading; repeat six to nine times.
Remove each 1 N disc and record the
reading, unloading each time until only
the mass carrier remains. Calculate the
average extension for each value of
the force.
Plot a graph of extension against force.
c
50
45
40
Extension / mm
•
a
b
30
25
20
15
10
5
0
Practice
2
35
Load /
N
Length /
cm
Extension /
mm
3
a
0
1
25.0
0
1.0
25.4
4
2.0
25.8
8
3.0
26.2
12
Challenge
4.0
26.6
16
4
5.0
27.0
20
6.0
27.4
24
7.0
27.8
28
8.0
28.5
35
9.0
29.2
42
10.0
29.9
49
Force needed to produce an extension of
1 cm: approximately 2.5 N.
3
4
5 6 7
Load / N
8
9
10
4.0 mm/N. Indicates how easy it is the
stretch the spring.
Load at limit of proportionality:
approximately 7 N.
0
b
2
The cable stretches under its own weight.
The lift body will bounce on the end of a
very long cable (behaves like a spring) –
unsettling for miners, but unacceptable for
office workers!
The cable used needs to get stiffer as the length
needed gets longer – this make it heavier,
making the problem of stretching under its
own weight greater.
A very high building/very deep mine will take
a long time to get up and down from unless
the lift is very fast, but the forces involved in
accelerating the lift to these speeds presents
challenges for the cable too – increasing the
effect of the first two points.
The solution may lie in the use of carbon fibre
rather than steel – lighter cable, very strong.
For example, it is now being used in
bridge building.
Peer-assessment
Answer is entirely dependent on the descriptions
given by the students, but the expectation is that
they will learn from each other’s points and also
critically assess the validity of the points made by
others, in terms of the physics.
12
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Exercise 5.2
b
F = pA
F ​​
A = ​​ __
p
a
Standing on one foot, the downward
force (weight) acts on a smaller area, so
the pressure is greater and the ice is more
likely to break.
Need to spread weight over a larger area,
so lie down.
Focus
1
Stretched length − original length = extension
Stretched length = original length + extension
2
Practice
2
F = kx
= 12 × 0.1
= 1.2 N
3
​ = kx
F
10 = k × 0 . 03
b
Practice
10
​= 333 N/m​
k = _____
​
4
0 . 03
There are four wheels and so four springs.
Each spring bears the weight of one person,
on average.
The weight of one person,
W = mg = 90 × 9.8 = 882 N
F = kx, where x = 0.011 m
​882 = k × 0 . 011
882
k = ​ ______ ​
0 . 011
= 80 . 182 = 8kN/m to 2 s.f.​
Challenge
5
a
b
The spring constant would be half the
size of one spring’s k. This is because each
newton stretches each spring by the same
amount as it would stretch one spring on
its own, so each newton results in twice
the extension.
The spring constant would be twice the
spring constant of one spring alone.
This is because the load is borne by both
springs equally, so each spring supports
half the force. Each newton now results in
half the extension of one spring alone.
Exercise 5.3
Focus
1
13
a
Quantity
Symbol SI unit
pressure
p
pascal (Pa) or newton per
metre squared (N/m2)
force
F
newton (N)
area
A
metre squared (m2)
3
500 Pa
4
160 000 N (or 160 kN)
Challenge
5
There are tracks, rather than tyres – these have
a larger surface area, so reducing the pressure.
They give an even force distribution over
their area.
There are eight tracks. This reduces the load
carried by any one track and increases the
surface area overall.
Exercise 5.4
Focus
1
Pressure in a liquid = Δp = ρgΔh
Practice
2
24 500 Pa
3
a
b
8 km
It assumes a uniform density to the top of
the atmosphere, whereas density actually
decreases, so the Earth’s atmosphere will
extend further than 8 km.
Challenge
4
The pressure on the top of the block is
1000 × 9.8 × 5
The pressure on the bottom of the block is
1000 × 9.8 × 6
The pressure difference is then 1000 × 9.8 × 1
1000 × 9.8 × 1 × 2 = 19 600 N.
This is 19 600 N more on the bottom face
than on the top and so there is an upthrust of
19 600 N.
This is 490 N more on the bottom face than on
the top and so there is an upthrust of 490 N.
This is why an object in a liquid that is more
dense than air has a smaller weight than in air.
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5
Chapter 6
Exercise 6.1
Focus
1
Description
Store of energy
energy in the nucleus of a
uranium atom
nuclear
energy in diesel fuel
chemical
energy of a ball held above
your head
gravitational
potential
energy of a hot cup
of coffee
internal
2
Energy transfer:
chemical energy to …
sound
How we can tell
The rocket launch
is very noisy.
light
Bright flames
emerging from
rocket.
thermal energy
Flames are hot.
gravitational potential
energy
The rocket is rising.
kinetic energy
The rocket is
speeding up.
Practice
3
Form of
energy
Store or
transfer?
energy as visible
radiation
light
transfer
energy of a
stretched spring
strain
store
energy spreading
out from a hot
object
heat or
thermal or
infrared
transfer
energy in the
nucleus of a
uranium atom
nuclear
store
energy of a moving
car
kinetic
store
Description
In an energy flow diagram, the width of the
arrow represents the amount of energy. The
width of the initial arrow is equal to the sum
of the widths of the arrows it divides into, so
the total amount remains constant.
energy in diesel fuel chemical
store
energy of a ball
held above your
head
gravitational
potential
store
energy of a hot cup
of coffee
internal
store
energy carried by
an electric current
electrical
transfer
Challenge
6
The engine produces thermal energy that
is wasted, but less will be wasted in winter,
as some of it is used to heat the
passenger compartment.
7
a
chemical
(100 J)
power
station
national
grid
electricity
(60 J)
55 J
4
thermal
(40 J)
1200 J/s
kinetic energy
light
(47 J)
thermal
(8 J)
therma
(4.8 J)
47%
Exercise 6.2
1500 J/s
chemical energy
Focus
300 J/s
thermal energy
14
b
lamp
1
300 J
2
a
b
75%
The motor is not intended to
produce heat.
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Practice
3
a
b
4
The gas-fired power station is
more efficient.
Gas-fired station 45%; coal-fired
station 25%
30%
Challenge
5
a
800 J
reflected/thermal
1000 J
200 J
to inverter
1
tra 90 J
nsf to
orm
er
10 J thermal in inverter
9 J thermal
wasted
10 J thermal
wasted
1
to 81
ca J
bl
e
to er
1J m
17 sfor
n
tra
8 J thermal
wasted
162 J to consumer
b
​​3
16.2%
Self-assessment
Answer is entirely dependent on the descriptions
given by the students, but the expectation is that
they will learn from each other’s points and also
critically assess the validity of the points made by
others, in terms of the physics.
Challenge
4
Change in k.e. =
1
1
__
​​   ​ m​​v​  2​​​​  2​ − __
​   ​ m​​v​  1​​​​  2​
2
2
1
= ​ __ ​ × 600 × (​12​​  2​ − ​25​​  2​ )
2
= 144 300 J​
5
Step 1:
1
​k.e. = __
​   ​ m​v​​  2​
2
1
= __
​   ​ × 0 . 20 × ​8​​  2​ = 6 . 4 J​
2
Step 2:
Exercise 6.3
Focus
1
a
b
E
​m = _____
​ 2 ×2 ​​
​v​​  ​
ΔE
____
​m = ​   ​​
gΔh
Practice
2
15
1
k.e. = __
​​   ​​ mv2
2
​​  1 ​ × 600 × ​25​​  2​​
= __
2
= 187 500 J
ΔE = mgΔh
= 20 × 10 × 2500
= 500 000 J
6.4
​​ _________ ​ = 3.3 m​
0.2 × 9.8
6
By conservation of energy, g.p.e at top = k.e.
at bottom
Therefore, ​mgΔh = __
​ 1 ​ m​v​​  2​​
2
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Mass cancels on both sides, leaving
​gΔh = __
​ 1 ​ ​v​​  2​
2
2 × 10 = __
​ 1 ​ ​v​​  2​
2
2
​v​​  ​= 40
v = 6 . 3 m/s​
Chapter 7
Exercise 7.1
Focus
1
Description
Energy
resource
Renewable
or nonrenewable?
wood
biofuel
renewable
natural gas
fossil fuel
non-renewable
coal
fossil fuel
non-renewable
splitting of
uranium nuclei
nuclear
fission
non-renewable
hydrogen
nuclear
nuclei combine fusion
to release
energy
renewable
sunlight
captured
to make
electricity or
heat water
solar cell
(photocell)
renewable
hot rocks
underground
used to heat
water
geothermal
renewable
moving air
turns a turbine
wind power
renewable
water running
downhill turns
a turbine
hydroelectric
power
renewable
Practice
2
16
Each student’s diagram should show the
sun shining, the water cycle (evaporation,
convection, cloud formation, and rainfall on
mountains), a dammed river and a hydroelectric power station, with appropriate labels
and notes.
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Challenge
3
17
Resource
Renewable?
Cost per
MWh of
electricity
(indicative
figures)
Scale of
production
(LS = large
scale, >
1 GW; SS =
small scale)
Environmental
impact
(S = small;
L = large)
Reliability
nuclear
fission
no
129
LS
L in use if there is an reliable
accident/leak
in construction.
Concrete production
is a major source
of carbon dioxide
pollution and
nuclear power
stations use a great
deal of it.
solar
yes
54
LS
L once installed,
but extraction of
materials for the
construction of
the panels has
significant potential
for harm.
variable, but
varies across the
world
geothermal
yes
20
S
S, although some
greenhouse gases
are emitted.
reliable
hydroelectric yes
50
LS
S in use, although
reliable
methane gets
produced by
rotting vegetation.
Construction can
destroy habitats
and displace
populations.
Concrete production
is a major source
of carbon dioxide
pollution and
hydroelectric power
stations use a great
deal of it.
wind
yes
51
LS
L
low
wave
yes
390
SS
L
reliable
tidal
yes
200
SS
L in use
Construction can
destroy habitats.
reliable
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Exercise 7.2
Inability to store electrical energy efficiently,
so needs to be produced when it is to be used.
Physical size of a solar farm on a commercial
scale (much larger than a conventional
power station).
Environmental concerns in the mining of the
rare earth metals needed to make the
solar cells.
Focus
1
i
ii
iii
iv
v
TRUE
FALSE; 2008, not 2006
FALSE
TRUE
FALSE; more, not less
6
Practice
2
Suitable positions are: on a hilltop; in a wide
open space; on top of a tall building; on a
clifftop. These are places where the wind is
likely to be stronger.
Challenge
3
Wind speeds above 6 m/s and below 40 m/s
are needed; noise pollution; visual pollution;
cost of installation (especially off shore);
unreliability of the wind; potential threat to
birds; land area needed for a wind farm (large
in comparison to conventional power station)
Exercise 7.3
Focus
1
Solar cell/photovoltaic cell
2
Tick (originally from sun): fossil fuels, wind
power, hydro-electric power, wave, sunlight
Cross (not originally from sun): nuclear power,
tidal power, geothermal
Practice
3
Millions of years ago, plants grew using
the energy of sunlight. Plants died, became
buried, and gradually rotted to become coal.
4
Feature
Fission, fusion
or both?
large nuclei split into two
fission
two small nuclei join
together
fusion
energy is released
both
used in a uranium-fuelled
power station
fission
the energy source of the sun
fusion
helium can be a product
fusion
Challenge
5
18
Interesting facts that might come up:
How do they work?
• NIF: 192 lasers, fired simultaneously at
a pellet of fuel, causing massive pressure
and temperatures of 1–2 million °C.
• ITER: fuel is heated by being hit by high
energy negative ions (accelerated using
electric fields) and electric and magnetic
fields are used to heat it further to
150 million °C.
• Power output of NIF lasers is 5 × 1014 W.
This is only for 10–9 seconds.
Limitations
• ITER in South of France is designed
to produce ten times the energy that
it requires to make fusion happen.
However, this needs to be 15 times to
be commercially viable. NIF has not
approached this.
Why so hot?
• They need such high temperatures to
bring the nuclei close enough together to
fuse, but since they are positive, they repel
each other.
Prospects
• Both technologies claim to be the one
that will eventually provide commercially
viable fusion.
• Neither is likely to do so in the next 10
years. Fusion as a possibility was first
suggested in the 1930s.
Peer-assessment
Answer is entirely dependent on the descriptions
given by the students, but the expectation is that
they will learn from each other’s points and also
critically assess the validity of the points made
by others, in terms of the physics and the socioeconomic and environmental impact of
these approaches.
Unreliability of sunlight in most areas.
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4
Chapter 8
a
b
Exercise 8.1
Focus
1
Work done = energy transferred
Practice
2
Its weight (caused by gravity); increases;
kinetic; 2.0 J
3
a
b
c
4
The load is getting higher, so its g.p.e.
is increasing.
The girl provides the energy.
The upward pulling force of the rope does
work on the load.
c
Challenge
5
It is a bigger force; it moves further.
Challenge
5
It is friction that allows us to walk.
When the friction is lower, due to ice, we
cannot apply as large a force to the ground
before our feet slip.
Exercise 8.3
1
Focus
a
In the best-case scenario, the same amount of
work is done with and without the machine.
By conservation of energy, the amount of
work done between two points is the same, no
matter what path is taken.
If the force is smaller, then the distance moved
must be greater, so that force × distance is the
same.
Focus
Exercise 8.2
1
120 × 1.6 = 192 J
The friction with the slope – more friction
leads to more work.
The angle of the slope – the steeper the
angle the smaller the effect of friction
with the slope (as the angle increases, we
approach the vertical situation we
started with).
Although she will do more work, the
force she has to apply to do it is smaller,
which may make the task possible, or at
least easier.
A newton-meter (forcemeter,
spring balance)
b
a
b
c
60 J
60 × 60 = 3600 J (60 J/s for 60 s)
Most of the energy is transferred as heat,
not as light.
Practice
2
There are 60 × 60 × 24 seconds in a day, so the
rate of supply of energy is
energy supplied
10000000
______________
​     ​​ = 115.74
​​     ​ = ___________
time of supply
x
60 × 60 × 24
= 120 W to 2 significant figures
3
c
Work done = force × distance moved (in
the direction of the force)
d
Angle /
degrees
Force / N
contact force of road
a
forward force
of engine
air resistance
Distance
Work
moved / m done / J
weight
b
Practice
19
2
75 × 4.0 = 300 J
3
a
b
c
2500 × 6.0 = 15 000 J
By conservation of energy: 15 000 J
Gravitational potential energy (g.p.e.)
c
Work per second = force × distance per
second
= 1600 × 30
= 48 kJ
Work per second = power
= 48 kW
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Challenge
4
a
b
2
Fd = kv2
At maximum speed, the maximum force
from the engine = force due to
air resistance
force × distance
​
Since ​P = ______________
​
time
= force × velocity​
So max force from engine,
max power of engine
​F​ E​​ = ​  __________________
​
max velocity
P ​ = k​v​​  2​
​ __
v
P = k​v​​  3​
_1
317000 ​​  3​​
​
v=(
​​ _______
​
​​
0 . 34 )
v = 97.69 m/s
The difference between the theoretical and
actual top speed could be due to friction
between the road and the tyres, which has
not been considered in our solution.
melting
SOLID
LIQUID
freezing
3
a
b
boiling
GAS
condensing
0 °C
100 °C
Practice
4
a
Put salt solution in a plastic container
with space for expansion. Add a
temperature probe. Place in freezer with
the probe leads out of the door to a datalogger and computer (or temperature
display). Record data.
b
+20
Temperature /
°C
+10
freezing
point
0
Time
–10
Peer-assessment
Answer is entirely dependent on the content and
quality of the flashcards produced by the students,
but the expectation is that they will learn from
each other’s work, including points they missed,
improve their descriptions and definitions, and
also critically assess the usefulness of the cards
made by others.
Chapter 9
The freezing point is the temperature at
which the graph line becomes horizontal.
(The later horizontal section indicates
the lowest temperature achieved by the
freezer, ‒20 °C.)
Peer-assessment
Exercise 9.1
Focus
1
Description
State or states
occupies a fixed volume
solid, liquid
evaporates to become a gas liquid
20
–20
takes the shape of its
container
liquid, gas
has a fixed volume
solid
may become a liquid when
its temperature changes
solid, gas
Answer is entirely largely on the descriptions
given by the students, but the expectation is that
they will learn from each other’s points and also
critically assess the validity of the points made by
others, in terms of the completeness and empirical
validity of the method. If they are able to conduct
the experiment at home, a great deal more value
can be drawn from those accounts, potentially.
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Challenge
5
As ice: the molecules are in fixed in position/
bonded to each other in a regular, ordered
arrangement. They vibrate about this position.
As the ice is heated, the vibrations increase in
size/amplitude/speed.
As water: The bonds between the water
molecules are breaking. The molecules are
free to slide past each other. As the water
is heated, the molecules move faster. The
arrangement is less ordered.
As steam: The molecules are widely spaced
and there is no order to their arrangement.
They are free to move independently.
Exercise 9.2
Focus
1
State
solid
liquid
gas
How close are
particles to
their neighbours?
close
close
far apart
How do the particles
move?
vibrate about fixed
positions
move about within
liquid
move rapidly around,
bouncing off walls and
each other
Strength of forces
between molecules
(strong/weak/zero)
Strong
Weak
zero
2
a
b
c
Melting
Boiling
Condensation
Practice
3
It is called the ‘kinetic’ model because
particles are moving about, and this helps to
explain many phenomena.
Challenge
4
21
When the gas expands, it uses some of its
energy to do so (does work against atmospheric
pressure), so the amount of kinetic energy it
stores falls, and it cools down since temperature
is a measure of the average k.e. of the molecules.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Exercise 9.3
atmosphere, which hasn’t changed.
Focus
4
When the molecules collide with the walls of
the container, they rebound. So they have a
change of velocity, which means they have a
change of momentum (or an acceleration).
From F = ma, this results in a force on the
walls of the container. Pressure is force per
unit area.
5
The higher the temperature of the gas, the
higher the average kinetic energy store of
the molecules, so they are moving faster.
This means they have a greater change of
momentum when they hit the sides of the
container, and there will be more collisions
with the sides every second. These two mean
the rate of change of momentum is higher, so
there is a larger force and so a larger pressure.
1
light reflected
by smoke grains
light
Exercise 9.5
2
Smoke grains are too small to see with the
naked eye.
Practice
3
The observer sees bright specks of light
moving around in an irregular manner.
4
They are too small to see even with
the microscope.
In a gas such as air, the particles are moving
around quickly. When they collide with the
smoke grains, the grains are pushed around in
a random manner.
Focus
a
b
c
d
a
b
pV = constant or p1V1 = p2V2
Symbol
Name of
quantity
SI unit (name and
symbol)
p
pressure
pascal (Pa)
V
volume
metre cubed (m3)
It would increase.
Practice
3
5.0 × 105 Pa
4
a
Pressure of
gas / kPa
Exercise 9.4
1
1
2
Challenge
5
Focus
Particles bounce off walls; each collision
produces a tiny force; many collisions
result in pressure on walls.
Density in B is twice density in A.
Twice as many particles collide with walls
each second, so twice the pressure.
Increase their temperature.
Volume of
gas / cm3
Pressure ×
volume
100
88
8800
120
75
9000
140
63
8820
160
56
9000
180
50
9000
Practice
2
Increase its temperature; halve its volume
Challenge
3
22
It will be the same. The gas will expand until
the pressure equals that of the surrounding
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Practice
b 100
2
80
75
3
Volume /
cm3 60
56
a
b
a
b
40
20
100
120
150 160
Pressure / kPa
200
In the table, values assume that
pV = 9000 roughly.
Challenge
5
Greater, because at higher temperature,
molecules move faster, and so cause greater
pressure on walls of container.
Exercise 9.6
Focus
4
1
Convert temperatures to kelvin: 40 °C = 313 K
​P​  ​​ ​V​  ​​ ​P​ 2​​ ​V​ 2​​
_____
​
​​  1  ​1 = ​ _____
​T​ 1​​
​T​ 2​​
​T1​  ​​ = ​T2​  ​​
​P​  ​​ ​V​ 1​​
​V​ 2​​ = _____
​  1  ​
​P​ 2​​
200
× 100
= ​ _________
​​
500
= 40 cm3
Pressure /
kPa
Temperature / K
increases (it)
The aerosol canister might burst/explode.
increases
Since the temperature is unchanged the
(average) kinetic energy does not change
so the particles are moving at the same
(average) speed.
But the volume is smaller, so the particles
will collide more frequently with the walls
of the container (any mention of particles
colliding with each other is irrelevant and
should be treated as neutral). So a larger
force is exerted on each unit of) area
producing a larger pressure.
Credit will be awarded for a fuller
explanation of this larger force: Since the
particles collide with the walls and bounce
off there is a change of momentum Δ(mv)
because there are now more frequent
collisions. As FΔt = Δ(mv) then the force
is larger.
Challenge
5
a
Pressure / kPa
4.50
4.00
3.50
3.00
2.50
2.00
1.50
1.00
0.50
–200 –150 –100 –50
0
50
100
150
200
250
Temperature / °C
23
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b
Absolute zero = –273 °C from this graph.
It is the temperature at which a gas exerts
no pressure – so the atoms must have
stopped moving.
Chapter 10
Challenge
6
Work is being done on the gas, so energy is
being transferred to the gas. This raises the k.e
of the molecules. Temperature is a measure
of the average k.e. of the molecules, so the
temperature rises
7
Although the sparkler sparks have a very high
temperature, they do not possess a great deal
of internal energy, because they have a very
small mass.
The steel rod has a very much greater mass, so
even though it is at a lower temperature, it has
more internal energy and so will give a much
worse burn.
Exercise 10.1
Focus
1
The sum of the kinetic and potential energies
possessed by the molecules of a substance
2
Statement
Term or terms
increases when an object is
temperature,
supplied with thermal energy internal energy
a measure of the heat of an
object
temperature
energy moving from where
the temperature is higher to
where it is lower
thermal energy
a measure of the average
kinetic energy of the
particles of an object
temperature
tends to spread out from a
hot object
thermal energy
measured using a
thermometer
temperature
Steam has more internal energy than water at
the same temperature. Although the k.e. is the
same, there is more potential energy, as work
has been done to move the molecules
further apart.
1
Metal block; balance; immersion heater;
voltmeter and ammeter with a stop clock (or a
joulemeter); thermometer; insulation
Practice
Measure the mass of the block of metal, using
a balance. Measure the starting temperature
of the block, using a thermometer.
Supply a known amount of electrical energy
to the block, using an immersion heater,
power supply and either a voltmeter and
ammeter with a stop clock, or a joulemeter.
Voltmeter-ammeter method: measure the
current passing through the heater and
the p.d. across the heater for 5 minutes.
The energy supplied is given by: potential
difference × current × time in seconds.
Joulemeter method: record the reading on the
joulemeter connected to the heater once the
power supply has been turned on for 5 minutes.
Safety and accuracy:
•
Practice
24
Focus
2
the sum of all energies of the internal energy
particles of an object
3
Exercise 10.2
4
Supply thermal energy to increase
the temperature.
5
When the steam hits skin, the first thing it
does is to condense to water. In doing so,
it gives up the energy that it had, making it
steam. This causes burning. Then the boiling
water cools, giving up energy as it does so.
This also causes burning. So the steam has
more internal energy than water at the same
temperature and so does more damage.
•
•
•
Do not leave the heater out of the block
whilst turned on.
Do not touch the heater until it has been
given plenty of time to cool.
Wrap the block in insulation, to reduce
thermal energy loss.
Wait until the temperature of the
thermometer stops rising before taking
the final reading – some of the energy
supplied to the block in the time will
needs to be given time to affect the
thermometer reading, as it needs to
conduct across the block from the heater.
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•
Pre-heat the heater before use – to reduce
the uncertainty in the value of the energy
supplied to the block.
b
Challenge
3
Measure the mass of the block – cannot
calculate the specific heat capacity without
knowing this.
Heat using an immersion heater and power
supply – less thermal energy lost to the
surroundings. Not doing this will give a value
for the specific heat capacity that is too high.
Insulate the block – less thermal energy lost
to the surroundings. Not doing this will give a
value for the specific heat capacity that is
too high.
Measure the starting temperature and the
maximum temperature reached by the block
using a thermometer to calculate the rise
in temperature. Not doing this will give a
value for the specific heat capacity that is too
high, as not all the energy supplied will have
registered an effect on the thermometer.
Evaporation from the surface of the bottle
causes the temperature of the rest of the water
on the surface to fall.
This is because the average k.e. of the
molecules in the remaining water falls, due to
the loss of the molecules with the most k.e.
by evaporation.
The draught will increase the rate of
evaporation and so the rate of cooling of
the milk.
Self-assessment
Key words/terms: evaporation; average k.e.;
most k.e.; increase rate
Exercise 10.4
Exercise 10.3
Focus
Focus
1
1
2
a
b
0 °C
100 °C
Temperature of the liquid; temperature of the
surrounding atmosphere humidity; surface
area of the liquid; presence or absence of
draught over the surface of the liquid
3
a
b
4
It will cause the liquid to cool.
5
Boiling occurs when the average k.e. of the
all liquid molecules is enough to overcome
atmospheric pressure. It only occurs at a
specific temperature.
Evaporation happens because some of the
molecules, at the surface of the liquid, have
enough kinetic energy to escape against
atmospheric pressure. It can happen at
any temperature.
a
2
Aluminium
3
The temperature of the sea water will rise
more quickly, because its s.h.c. is lower, so
less energy is needed for each degree rise
in temperature.
4
Aluminium will cause a bigger rise in the
temperature of the water, because it requires
more energy to heat it, and so it will give out
more energy when it cools down, thus heating
the water more.
Melting
Boiling
Challenge
6
∆E = mc∆θ
Practice
Practice
25
7
arrangement. The density of the
substance increases.
The molecules will form strong bonds
between them and will become arranged
in a regular structure. The density of the
substance does not change very much
at all.
The molecules will come closer together
and weak bonds will form between
them. They will have a fairly random
5
Statement
True or false?
all metals have a lower s.h.c.
than all non-metals
false
metals generally have a lower true
s.h.c. than non-metals
the s.h.c. of water decreases
when it freezes
true
the s.h.c. of water decreases
when it boils
true
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6
154 kJ (154 000 J)
7
18 000 J
8
a
b
9
Water, because it has the highest specific heat
capacity, so will store the most energy per kg,
for the smallest rise in temperature.
155 J/(kg °C)
For example: some energy escapes; the
lead might be impure; the heater rating
might be incorrect; etc.
10 E = 1000 × 4200 × 80
= 3.36 × 108 J
They collide with the ions in the lattice,
passing on kinetic energy, so the ions vibrate
faster. This is shown by a rise in temperature,
since temperature is a measure of the average
kinetic energy of the ions or molecules in
the substance.
Challenge
5
Wax on the rod with low s.h.c. would melt first
because less energy would need to be supplied
to heat it up.
6
Most liquids and gases do not have bonds
between the atoms or molecules and so
vibrations cannot pass between them.
Most liquids and gases are covalent and
therefore do not have free electrons, so
conduction cannot occur by this mechanism.
Challenge
11 a
b
6.00 × 107 J
Answer should be a comparison, such as:
Although the water stored more energy,
much more water is needed to store the
same amount of energy as the concrete.
Cannot heat the water to as high a
temperature (without putting it under
high pressure, which is dangerous and
expensive).
Using a solid is more practical
(won’t spill).
A leak of water could be fatal, if it caused
a short circuit. Concrete is an electrical
insulator.
Exercise 11.2
Focus
1
The transfer of thermal energy through a
fluid (liquid or gas) by the movement of the
substance (convection current), carrying the
energy with it.
Practice
2
a
window
hot air
rises
Chapter 11
Exercise 11.1
cold air
sinks
Focus
1
a
b
c
2
Free (delocalised) electrons
Insulator
For example, brass, gold, diamond
For example, air, water, ice, plastics
Practice
26
3
a
b
4
All solids conduct through vibrations passing
through the lattice of ions or molecules. This
is slow. Conduction of heat in metals is also
by movement of free electrons and these are
also what is required to conduct electricity.
Thickness, length
The best conductor is the one where the
wax melts first. The worst conductor is the
one where the wax melts last.
3
heater
b
If the heater was high up on the wall,
close to the ceiling, the convection current
would remain close to the ceiling, and
heat would not be distributed lower down
in the room.
a
b
c
d
e
Increases
Stays the same
Decreases
Increases
Increases
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Challenge
4
9
reflector
The flame heats the air. The air becomes less
dense, and floats upwards. It is replaced by
colder air, which is then heated and rises.
Smoke is solid particles floating in air, and is
carried upwards by the rising air.
Exercise 11.3
Focus
1
2
Electromagnetic radiation, with a wavelength
longer than visible light, which is absorbed by
substances, causing an increase in
internal energy.
The hotter cup will cool faster initially
because the temperature difference between it
and the environment is higher.
Practice
3
Radiation can travel through the vacuum of
space. Conduction and convection require a
material to travel through.
4
Matt black
5
The temperature rises and the internal
energy increases.
6
An object cooler than its surroundings will
absorb more infrared than it absorbs until
it reaches thermal equilibrium with
the surroundings.
An object hotter than its surroundings
will emit more infrared than it emits until
it reaches thermal equilibrium with
the surroundings.
7
Once the object is at the same temperature
as the surroundings, the rate of emission and
absorption become equal and the temperature
remains the same as the surroundings.
Challenge
8
a
b
c
27
There is very little material through
which heat can conduct (the frame of the
window, for example).
There are very few gas molecules in the
gap that can move as a convection current
(no vacuum is perfect).
Yes, energy can escape by radiation,
because radiation can pass through
a vacuum.
TV
remote control
plan of room
10 Assuming the milk is at the same temperature
in both cases and that you use the same
amount in both cases, add the milk BEFORE
having a conversation.
The amount of energy required to change
the temperature of the milk will reduce the
temperature of the coffee at the start, which
will then reduce the temperature difference
between the coffee and the surroundings,
reducing the rate of cooling, so the coffee
loses less infrared radiation during
the conversation.
Adding it afterwards means that the coffee will
have lost more infrared radiation before the
milk is added, which may then leave the coffee
colder than you want after the same length
of conversation.
11 Points could include:
• Large surface area – larger surface
area increases rate of loss of
infrared radiation.
• Large mass of aluminium – larger
mass will absorb more thermal energy
by conduction for the same rise in
temperature, keeping the processor cooler.
• Matt surface – matt surfaces are better
emitters of infrared radiation than
shiny surfaces.
•
Black surface – black surfaces are better
emitters of infrared radiation than white
or silver surfaces.
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Inner glass wall: reduces energy loss by
conduction as glass is an insulator. Reduces
losses by radiation, as most radiation is
reflected by the inner silvered surface.
Radiation which is absorbed and therefore
raises the temperature of the glass is poorly
conducted and then poorly emitted by the
second silvered side, as shiny silver is a poor
emitter of infrared radiation
Plastic spacer: reduces energy loss by
conduction as plastic is an insulator.
Sealed air space: air is a good insulator,
reducing energy loss by conduction.
12 Infrared radiation entering the atmosphere
can be absorbed or reflected back towards
pace. Of the radiation that is absorbed, some
will be emitted back into the atmosphere.
The radiation reflected and emitted can leave
the atmosphere or it can be absorbed by the
gases in the atmosphere. Carbon dioxide and
methane are especially good at absorbing
infrared and so trap it within the atmosphere,
causing a rise in temperature.
Exercise 11.4
Focus
1
a
b
c
d
Mass of water
External temperature
Graph line 1 shows beaker A.
With a lid, it cools more slowly.
3
Beaker B could also be losing energy
by evaporation.
Insulation of sides and base would mean that
almost all heat was being lost through top,
which is the area of interest.
Challenge
4
28
For exemplar answer, see Question 4.
Exercise 11.5
Focus
Practice
2
Peer-assessment
Plastic stopper: reduces energy loss by
conduction as plastic is an insulator. The
stopper also prevents loss by convection, as
it is air tight, so convection currents cannot
leave the flask whilst the stopper is in place.
Plastic outer sleeve: reduces energy loss by
conduction as plastic is an insulator.
Outer glass wall: reduces energy loss by
conduction as glass is an insulator. Reduces
losses by radiation, as most radiation is
reflected by the inner surface. Radiation
absorbed and raises the temperature of the
glass is poorly conducted and then poorly
emitted by the second silvered side, as shiny
silver is a poor emitter of infrared radiation.
Vacuum: reduced thermal energy loss by
convection, as there are very few particles of
medium for the heat to transfer through.
1
Radiation
2
The Earth’s temperature would decrease to the
temperature of space (−270 °C).
3
a
b
Visible light, infrared, ultraviolet
It rises
Practice
4
Energy is transferred by radiation into space.
5
Until the Sun reappears in the sky, energy is
still being transferred.
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Challenge
6
Day-time temperatures would be higher,
night-time temperatures would be lower. This
is because any point on the Earth would be
exposed to sunlight for twice as long during
the day, and it would transfer energy at night
for twice as long.
Temperature
Time
Chapter 12
Challenge
8
Exercise 12.1
Exercise 12.2
Focus
1
a
b
c
d
2
330–350 m/s
3
20–20 000 Hz
Vibrating or oscillating
Strings (and body, and air inside)
Air (air column)
An echo
29
1
a
b
c
Longitudinal
P waves
Oscillations in the same plane as the
transfer of energy
2
microphone 2
timer
4
14.9 kHz, 16.5 kHz
5
Use a signal generator and loudspeaker (or
other source of high-frequency sounds),
change the frequency, and ask who can hear
each sound.
6
a
b
3.0 s
1650 m
7
a
b
... air.
... the student bangs two wooden blocks
together.
... the time interval between the sound
being detected by the two microphones.
... the distance between the microphones.
distance
​​
speed = ________
​​
time
d
Focus
Practice
Practice
c
5000 m/s
microphone 1
D IG
IT IM
INS
ER
MA
on
T
off
SE
RE
0.1
ms
1.0
6V
1.01
?
1.001
2V
ING
TIM
off
?
???
?
???
0.5A
~
~
on
stop
t
t
star
star
stop
???
0.5A
?
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Digital signals are less susceptible to
interference.
Digital signals can be regenerated, where they
are amplified without amplifying interference.
3
Displacement
Time
Peer-assessment
4
a
b
The answer could include: analogue is more
expensive; processing the signal introduces delays;
the signal is lost suddenly when quality drops too
far; it relies more on technology for reproduction –
old media (floppy discs, for example) are obsolete.
Sound B
2.3 ms (0.0023 s)
Challenge
5
Wave property
Sound property
Amplitude
loudness
Frequency
pitch
Exercise 13.1
Focus
Exercise 12.3
Focus
1
Chapter 13
1
Virtual
2
The rays of light appear to spread out from a
point, rather than actually coming from
that point.
3
68° (if angle with mirror is 22°, then the angle
with the normal = angle of incidence is 68°.
The angle of reflection = angle of incidence)
4
a
Ultrasound
Practice
2
No
3
a
b
Microphone
Oscilloscope
image
Challenge
4
The molecule moves up and down, oscillating
about its mean position.
mirror
Exercise 12.4
Focus
1
Sound waves are longitudinal waves. They
consist of compressions/rarefactions, which
are high/low pressure areas and rarefactions/
compressions, which are low/high
pressure areas.
b
c
d
3.0 cm
Virtual
Light does not pass through the mirror. It
appears to come from behind the mirror.
Practice
2
Sonar; medical scanning (fetal scans); nondestructive testing of materials; treatment of
soft tissue injuries
3
The LP and cassette were analogue, and the
phone and streaming service are digital.
Challenge
4
30
More digital information can be stored in the
same space.
More digital signals can be transmitted at the
same time.
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Practice
5
Practice
a, b
C
3
a
material 1
air
material 2
glass
A
refracted ray
40°
normal
B
c
incident ray
Parallel
Challenge
b
6
θmax
c
d
A
The ray bends towards the normal as it
enters glass. The angle in glass is smaller.
30°
20°
Challenge
h
4
a
B
θmax
b
x
The length of the mirror will be least when
the angle qmax is a maximum. This is tan–1​__
​  B
x ​​.
But, by geometry, the angle between the
line joining the top of the head and the ray
entering the eye from the feet is also qmax
and tan–1​__
​  A
x ​​.
B
__
So __
​​  A
x ​​must be equal to ​​ x ​​, so A = B and since
h
A + B = h, then A = __
​​   ​​.
2
The air nearest the road is least dense, as
it is hottest. As the light passes through
this air of decreasing density, it is refracted
away from the normal. This repeated
refraction causes the light to appear to
reflect from the road, as if it was wet.
Total internal reflection is when light
strikes the boundary between more dense
and less dense media at greater than the
critical angle. The effect in part a is due
to successive refraction. Total internal
reflection is a single event.
Exercise 13.3
Focus
1
Slower
2
Away
Practice
Exercise 13.2
3
Focus
1
It speeds up – light travels slower in a more
dense medium.
2
It increases
a
air
Perspex
angle of incidence
angle of refraction
31
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b
c
4
The light refracts away from the normal on
leaving the water. The light appears to come
from a point higher up in the water, as a
result. The brain works out where the object
must be, knowing that light travels in straight
lines. So the pool does not appear to be as
deep as it actually is.
5
Preparation – for all three blocks:
• place block on a sheet of plain paper
• draw a pencil line as close to one side
as possible
• move the block so that it just covers the
line just drawn
• draw another line along the opposite side
of the block
• remove the block, in order to draw a
normal on the side of the block on which
the light will be incident.
Drawing the incident normal:
a, c Using a protractor, draw a line at 90° to
the line representing the side on which the
light will be incident. This line needs to
project beyond the line inside the outline
of the block.
b Measure the length of the flat side of
the block.
Find the middle: using a protractor, draw
a line at 90° to the flat side at the mid
point. This line needs to project beyond
the line inside the outline of the block.
Replace the block.
Taking readings – for all three blocks:
•
•
•
•
•
•
32
Measuring the angle:
a, b Measure the angle of refraction inside the
block, using a protractor.
c Increase the angle at which the ray
strikes the flat side of the block until the
refracted ray leaves along the flat side of
the block. Looking along the edge stop
increasing the angle of incidence when the
ray is about to disappear.
The angle of refraction at this point is the
critical angle.
d Adjust the incident ray of white light so
that it strikes the block near the apex.
The refracted ray, as it leaves the block
will split into seven colours (red, orange,
yellow, green, blue, indigo, violet), where
the red is refracted least and the violet is
refracted most.
2 × 108 m/s
19.5°
Shine a fine beam of light in to the
block so that it hits the block where the
normal and block meet, at an angle of
incidence of about 30°. In the case of
the semi circular block, this means that
the light enters the curved side along a
radius to the curse of the block.
Place pencil crosses in the middle of
the beam as it approaches and as it
leaves the block.
Remove the block.
Join the crosses on the incident ray.
Join the crosses on the ray leaving the
block (refracted ray).
Join the incident and refracted rays
with a straight line.
Challenge
6
a
b
The bigger the difference in density
between two materials, the more
refraction there will be.
Wearing goggles means there is a layer of
air in contact with the cornea and so the
eye works normally.
Peer-assessment
For exemplar answer, see Question 6.
Exercise 13.4
Focus
1
a
X
X
45°
b
45°
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Practice
2
Ray is at 90° to surface, so it does not bend
(angle of incidence = 0°).
3
They would be the same – the rays would leave
the pool without changing direction.
Challenge
4
a
rays from
object
observer
b
5
The upper ray (from the top of the object)
is above the lower ray after two reflections.
If the curve of the fibre is too tight, the angle
of incidence on the inside will be less than the
critical angle and light will leave the fibre.
Exercise 13.5
Focus
1
a
b
c
Converging
The lens symbol is for a converging lens.
Principal focus (focal point)
Practice
2
a
F
O
3
33
image
axis
F
b
c
d
Image is smaller than object
Object is further from lens
Inverted
a
b
17 cm
4 mm
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Challenge
4
a
O
F
b
c
d
5
F
I
Upright
Virtual
Image is magnified because it is bigger/
taller than the object.
Convex
Exercise 13.6
Focus
1
a
b
c
d
Violet
Violet
Blue
Red
Practice
2
a
b
c
d
e
f
Wave A
8
14
The waves have equal lengths in the
same time.
Wave B
Wave A
Challenge
3
In a darkened room, pass white light through
a (60°) glass or Perspex prism. Observe the
light as it leaves the block on a white screen
(at least 1 m away from the prism).
4
Only one wavelength (or frequency) of
light present.
Tip
Although the word literally means ‘one colour’,
this is not an adequate answer to the question.
34
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Chapter 14
Exercise 14.1
Focus
1
a
b
c
d
e
2
a, b
Distance travelled by the wave
Wavelength
λ, metre (m)
Amplitude
A
Amplitude
T
c
500 Hz
3
a
b
c
Transverse
Longitudinal
Transverse
4
a
Move your hand from side to side, at right
angles to the length of the spring.
Move your hand back and forth, along
the length of the spring.
b
Practice
5
a
Wavelength
Amplitude
Wavelength
Wavelength
b
Longitudinal wave
(sound)
wavelength
compression
35
rarefaction
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Self-assessment
Practice
Students’ own answers.
2
Quantity
Increases / decreases /
stays the same?
Louder sounds are made by waves with
larger amplitudes
wave speed
decreases
wavelength
decreases
If the speed of sound changed with the
frequency of the wave casing it, we would hear
different notes at different times, even when
they were made at the same time!
frequency
stays the same
Challenge
6
7
3
a, b
Exercise 14.2
Focus
1
Symbol Quantity
v
2
Unit (name and
symbol)
wave speed metre per second
(m/s)
f
frequency
hertz (Hz)
λ
Challenge
wavelength
metre (m)
4
a
b
c
100 waves
330 m
330 m/s
Practice
3
a
b
625 m
3750 km
Chapter 15
Challenge
4
5
The wave speed may have varied as the wave
passed through different materials within the
Earth. The wave may not have travelled in a
straight line.
a
b
The sound waves have a much larger
wavelength than light, which is similar to the
size of the gap between the door and the wall.
The sound will diffract, but the light will not.
So the sound appears to go around corners,
but the light does not.
4.3 × 1014 Hz
The infrared wavelength is greater than
the wavelength of red light.
Exercise 15.1
Focus
1
a
b
c
d
Gamma rays
Radio waves
Ultraviolet
Gamma rays
Exercise 14.3
Focus
1
36
Description
Name
bouncing off a surface
reflection
changing direction because
of a change of speed
refraction
spreading out after passing
through a gap
diffraction
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Practice
Exercise 15.2
2
Focus
gamma rays
eyesight
X-rays
transmitting TV
programmes
ultraviolet
airport baggage
scanners
visible light
grilling food
infrared
sterilising medical
equipment
microwaves
communicating with
spacecraft
radio waves
forgery detection
banknotes
1
a
b
c
Radio waves
Microwaves
Visible or infrared waves
Practice
2
a
b
X-rays will penetrate walls (even better
than microwaves).
X-rays are ionising; X-rays damage living
tissue; X-rays can cause cancer.
Tip
X-rays are dangerous is NOT enough.
Answer needs to say WHY they are dangerous.
3
Can penetrate walls. Need a short aerial for
transmission and detection.
Challenge
4
Challenge
3
a
b
c
37
How they are used: X-rays can penetrate
flesh and bone. They are absorbed more
by bone than flesh and so create ‘shadows’
on film or other detectors.
Properties: absorption by bone;
transmission by flesh; detection
by photographic film or by
electronic detector.
How they are used: remote control
device sends beam of infrared, which
is detected by sensor on front of TV or
other appliance. The beam is pulsed with
code, which identifies the appliance (TV,
recorder, etc.) and gives instruction (e.g.
change channel).
Properties: travel in straight lines; diffract
into a beam from the controller, so do not
need precise direction.
How they are used: mobile phone signals
are digital signals carried by microwaves.
These are transmitted in both directions
between mobile phone mast and phone.
Microwaves are also used to transmit
between mobile phone masts
Properties: work in line of sight only; can
pass into buildings; high frequency means
many messages can be carried.
analog
digital
•
•
•
•
Analogue signals are continuously
variable between two extremes – can take
any value between two extreme values.
Digital signals can only take certain
values/made up of a series of 1s and 0s.
Digital signals can carry more
information/ digital signals can have a
greater data transmission rate.
Digital signals can be regenerated
accurately after being transmitted a long
distance. Analogue signals tend to be
corrupted/lose quality.
Peer-assessment
These flash cards are an excellent revision resource
and the process of making them requires the
student to process the chapter contents in order
to select and summarise the key information.
The main value is in their repeated, frequent use,
however, so the student is to be encouraged to use
them for retrieval practice regularly and frequently
over time.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
The peer review of the work is also useful, in the
same ways, but also in terms of critically assessing
the work of others and acting on feedback
from others.
Practice
2
Soft iron. Steel would retain its magnetism
when the current is turned off.
3
•
•
Chapter 16
Exercise 16.1
•
Focus
•
1
A device that exerts a force on
magnetic materials.
2
a, b
S
3
N
N
S
•
N
N
S
(Note: all the poles could be reversed.)
4
a
b
If it is magnetised then it will repel a known
magnet. If it is simply magnetic, it will always
attract to a known magnet.
Exercise 16.2
Focus
1
Due to interactions between magnetic fields
Practice
2
In contact with a permanent magnet, the steel
will become slightly permanently magnetised.
The iron will only be magnetised when in
contact with a permanent magnet.
Challenge
3
Challenge
4
An electromagnet is used. Need to be able
to position the magnet before it attracts the
metal – to avoid further damage to the eye.
Need to be able to vary the strength of the
magnet – to ensure the metal comes out gently
and in a controlled way.
5
Methods will probably involve a top pan
balance with a piece of iron on the pan. This
allows a direct measurement of small forces,
due to the reduction of the reading on the scale.
Good answers will note that this reading is in
grams and so needs to be converted to newtons.
Comparison requires control of variables,
such as the height of the pole above the
iron and ensuring that there is no residual
magnetism in the iron after each test.
Factors affecting the strength are: current in
the solenoid; number of turns (per metre) in
the solenoid; presence of a soft iron core in
the solenoid
Current should be varied from 0 A to 8 A.
Care should be taken to turn the current off
between measurements to avoid overheating
of the solenoid – danger of melting insulation
and of burns. Use a push-to-make switch.
End A: S. Since A is attracted to the
North pole, it must have a South
pole induced.
End B: N
Practice
5
•
S
The geographic and magnetic poles of the
Earth are not in the same place, so following
a compass will take you to the magnetic pole,
not the geographic pole.
Place the electromagnet on a flat surface.
Pass an electric current through
the solenoid.
Place a compass at one end of the
electromagnet.
The needle will point away from a Northseeking pole.
Confirm this by placing the compass at
the other end of the solenoid.
The needle will point in the opposite
direction now (towards a Southseeking pole).
Exercise 16.3
Focus
1
38
Can be turned on/off; the strength can
be varied
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Exercise 16.4
•
Focus
1
neutral point between the pair of magnets
with N poles facing (does not have to be
central, as question does not specify that
the magnets are identical).
Practice
2
N
S
S
N
S
N
S
N
N
S
N
current in
S
The method should include at least of eight
these points and must include the final point
for full marks:
1 Place the permanent magnet on a piece
of paper.
2 Draw around it with a sharp pencil. Note
the position of the magnet, so that it can
be replaced exactly, if it is moved.
3 Place a plotting compass next to one pole
of the magnet.
4 Draw a pencil dot as close to the compass
as possible, showing the direction of the
compass needle.
5 Move the plotting compass so the end of
the needle nearest the magnet is on the
spot you just made.
6 Repeat previous two steps until you have
arrived at the other pole of the magnet.
7 Join the spots with a smooth line.
8 Place the plotting compass next to the
first pole again, but this time in a
different position.
9 Repeat steps 1–8 until at least ten field
lines have been drawn.
10 The direction of the field lines is marked on
each line with an arrow. This is the direction
that the North-seeking end of the needle
was pointing as each line was plotted.
current out
Key points for the diagram:
• field lines run N to S
• closer together near the poles/ends
of solenoid
• getting further apart as the distance from
the poles/ends of solenoid increases
• never touch
• never cross
• straight lines or smooth curves – no kinks
• arrows on lines to show direction
39
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
summarise the key information. The main value is in
their repeated, frequent use, however, so the student
is to be encouraged to use them for retrieval practice
regularly and frequently over time.
Challenge
3
N
The peer review of the work is also useful, in the same
ways, but also in terms of critically assessing the work
of others and acting on feedback from others.
S
N
N
S
S
Chapter 17
Exercise 17.1
S
Focus
4
N
1
a
b
Each pole has a magnetic field around it, and
these fields interact. The field lines cannot
cross, so they repel each other.
2
Neutral
Practice
3
5
S
Conductor
Any metal or graphite (not just ‘carbon’)
a
b
Electrons are transferred, because they
have a negative charge. (NEVER talk
about transfer of positive charge.)
Protons
4
a
b
c
d
Friction
Negative
Attract
Opposite charges attract
5
The rod is suspended by a thread so that it
is free to turn. The rod is rubbed with the
cloth. The cloth is then removed. When the
rod is stationary, the cloth is brought towards
one end of the rod. The rod turns towards
the cloth due to the attraction of
opposite charges.
N
Challenge
6
S
N
The diagram with the most field lines is
the strongest.
These flash cards are an excellent revision resource
and the process of making them requires the student
to process the chapter contents in order to select and
40
Exercise 17.2
Focus
1
Peer-assessment
Conductors have free (or delocalised)
electrons. These move slowly and randomly
within the lattice of the material. When
a voltage is applied across the conductor
an electric field is produced, which makes
all these free electrons move in the same
direction. Insulators do not have free
electrons, so cannot conduct.
You should find that some plastic materials
charge up more readily than others, and that
some types of cloth are better than others.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
You should describe testing this using very
small scraps of paper. How close to the scraps
do you have to get before attraction occurs?
Use a standard rubbing method.
b
+++
+
+
++ +
Practice
2
a
b
You should see the stream of water bend
towards the comb.
The reason is that the water molecules,
which are normally neutral, get polarised
by the negatively charged comb, causing
the stream of water to have a positive
charge nearest the comb. They attract.
c
Peer-assessment
Answers should include some or all of
the following:
• Friction between comb and hair causes
transfer of electrons from one to the other.
• The charge on the comb attracts the opposite
charge in water molecules, causing them
to polarise.
•
•
•
This then causes the water to be attracted to the
comb, despite not having gained or lost electrons.
This is charging by induction.
Removing the comb allows the water
molecules to un-polarise and the stream
returns to vertical.
Challenge
3
It causes sparks (that might cause
explosions/damage).
Practice
2
+
–
+
–
+
–
+
–
+
–
+
–
3
An electric field is used to direct the ink drop,
since electric fields apply forces to charges.
4
The toner (dry ink) has a charge applied to it.
It sticks to the drum in the shape of the image
to be copied. The paper has the opposite
charge applied to it. The toner then sticks
to the paper when the paper is put on to
the drum.
Chapter 18
Focus
1
–
Challenge
Friction between the carpet and our footwear
causes a transfer of electrons from one to the
other. This charge builds up on us. When we
touch the door handle, it discharges, giving
us a shock.
Exercise 17.3
+
Exercise 18.1
Focus
1
a
a
Material
steel
+
✓
plastic
✓
glass
✓
copper
✓
silver
✓
wood
41
Conductor? Insulator?
✓
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
b
Practice
2
BUT
There are now three times as many lamps
in series, so it is three times as hard for the
current to flow.
SO
2 × bigger and 3 × smaller ends up as __
​​ 2 ​​ or
3
0.67 × original current
Connect an ammeter and 12 V d.c. power
supply/battery in series. Place the items in
the table into the circuit, using crocodile
clips to make a good electrical connection
with the item. Switch on the power supply.
If a current is shown on the ammeter, the
item is an electrical conductor.
a
6
current
None. The p.d. is still 1.5 V, the resistance is
still the same, so the current is unchanged.
Exercise 18.2
Focus
1
electron flow
b
3
Arrow points in the opposite direction to
that in part a.
a, b
voltmeter
Practice
V
2
4
Battery
It doesn’t – the p.d. across one cell and two
identical cells in parallel is the same.
charge
coulomb (C)
I
current
ampere (A)
t
time
second (s)
a
b
2.4 C
72 C
4
12.0 A
5
a
b
60 s
8 hours
Challenge
6
The p.d. is doubled, so this would double
the current.
Q
3
This is a consequence of the law of conservation
of energy. Whichever path you take between two
points (in this case across the power supply), the
energy change must be the same. This applies in
every circumstance – electrical, mechanical, etc.
5
Unit (name
and symbol)
1 ampere = 1 coulomb/second (or
equivalent)
Tip
Challenge
42
Symbol for
Quantity
quantity
b
A ammeter
c
a
Q = 500 µC discharging in t = 0.1 seconds
gives a current,
Q
​I = __
​   ​
t
500 × ​10​​  −6​
= ​ __________
​
1 × ​10​​  −1​
= 0 . 005A​
This is a very small current compared to the
current that could flow from mains supply,
even at a much smaller potential difference, so
not lethal in comparison.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Exercise 18.3
Challenge
3
Focus
1
a
b
20 V
20 V
Practice
Doubling the diameter alone means the
resistance will be _​​ 14 ​​of the original value and
tripling the length alone makes the resistance
three times bigger. The total effect is then to
have ​​ _34 ​​or 0.75 of the original resistance.
Exercise 18.5
2
Current – increase
or decrease?
Change
more resistance in the decrease
circuit
less resistance in the
circuit
increase
increase the voltage
increase
use thinner wires
decrease
use longer wires
decrease
Focus
1
a
ammeter
A
power +
supply
−
b
8.0 Ω
4
a
2
P.d. V / V
V
A
3
Current I / A Resistance R / Ω
2.0
0.37
5.4
4.1
0.75
5.5
5.9
1.20
4.9
7.9
1.60
4.9
5.2 Ω
4
2.0
0.10
20.0
4.0
0.18
22.2
6.0
0.25
24.0
8.0
0.31
25.8
10.0
0.36
27.8
12.0
0.40
30.0
Resistance increases from 20 Ω to 30 Ω.
a
0.3
0.25
0.2
0.15
0.1
0.05
0
Doubling the length doubles the resistance.
Practice
43
—
0.4
Focus
2
0.0
0.35
Exercise 18.4
1
0.0
0.45
Current / A
b
P.d. V / V
Current I / A Resistance R / Ω
Practice
power supply
c
V voltmeter
Voltmeter is connected in parallel with
the lamp.
Challenge
3
lamp
Doubling the diameter makes the area of crosssection four times bigger, since area = _​​ 14 ​​(p.d.)2,
so the resistance will be ​​ _14 ​​of the original value.
b
c
5
0
2
4
6
8
10
Potential difference / V
12
14
0.22 A
7.7 V
Sketch should show the I–V graph as a
straight line through the origin.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Challenge
6
As the p.d. across the lamp increases, the current through the lamp increases. This means the rate of
flow of electrons increases.
Therefore there are more collisions between the ions in the lattice of the filament and the free electrons.
The ions gain kinetic energy, and thus vibrate faster; this makes it more likely that the electrons will
collide with them.
So the electrons lose more energy through increased collisions; so the resistance has increased; and so
1
​​ .
the gradient of the graph decreases, as gradient = _________
​​
resistance
Exercise 18.6
Focus
1
a
energy transformed
_________________
​​
Power =
​​
time
b
Power = current × p.d.
Practice
2
3.0 W
3
a
500 W
b
500 J
b
96 W
c
50 Hz indicates the frequency of the alternating current
d
4.5 A
Challenge
4
a
5
Potential
difference / V
Current / A
Power / W
Time used
for / min
Cost of use for
this time / pence
200
3
600
20
3.6
220
2.72
600
90
18
120
4
480
100
16
200 000
1.5
2 00 000
10
1000
220
0.45
100
240
8
0.40 A
Peer-assessment
The mind map might look something like this:
p.d. is the work
done per unit charge
between to points in a circuit,
V = E/Q volts = joules
per coulomb
a potential
difference
measured in
volts (V) between
two points on a
conductor causes
Power = V × I
cost of using an
appliance is cost =
power (kW) × time used
(hours) × cost per unit (p)
a current (I), measured
in amps (A). It is rate of
flow of charge (Q) so
amps = coulombs per
second
current
electricity
if resistance (ohms) is
constant then current is
proportional to potential
difference (V = IR)
resistance of a metal
increases with
temperature
cells and batteries
produce a direct current
(d.c.). Mains power is
usually alternating
current (a.c.)
this leads us to Q = It
A good mind map contains both text and diagrams, which serve as memory prompts (‘dual coding’).
It should clearly show the inter-relationships between concepts and quantities (e.g. charge – current –
potential difference – energy-power – resistance). The pictures/diagrams could be quite personal to the student.
The use of hyperlinks in an electronic version can be very powerful.
44
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4
Chapter 19
Exercise 19.1
Focus
1
lamp
resistor
switch
Since the p.d. across R1 is 8 V, the p.d across
R2 is (12 – 8) = 4 V. Using the potential divider
equation:
R
V
​​ ___1 ​​ = ___
​​  1 ​​
R2 V2
R
8
​​ ___1 ​​ = __
​​   ​​
R2 4
R
​​ ___1 ​​ = 2
R2
Exercise 19.2
Focus
LDR
thermistor
fuse
diode
cell
transformer
Practice
1
Practice
2
Description
Component
gives out heat and light
lamp
resistance changes as the
temperature changes
thermistor
2
It prevents a reverse current, which could
damage the device.
Challenge
3
a, b
provides the ‘push’ to make a cell
current flow
‘blows’ when the current is
too high
fuse
makes and breaks a circuit
switch
has less resistance on a sunny LDR
day
adjusted to change resistance potentiometer
in a circuit
Exercise 19.3
Challenge
1
480 Ω
2
In series
3
The circuit sets the p.d. to 0 V across the lamp
when the slider is at the top. It will set the p.d.
to the supply voltage (12 V) when the slider is
at the bottom.
potentiometer
lamp
Focus
Practice
3
a
b
70 Ω
The current is the same all the way round
the circuit.
4
a
b
In parallel
A The combined resistance of the two
resistors must be less than 10 Ω. ✓
Challenge
5
45
0.50 A
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Exercise 19.4
Exercise 19.5
Focus
Focus
1
E1 + E2 + E3
1
2
This arrangement allows a larger current for
the same p.d.
3
30 Ω
a
b
c
5
a
b
6
A
LDR
6V
6V
Practice
4
a
B
Resistor C has the greatest resistance.
5.3 Ω
1.5 V
30 Ω
Two devices that need the same p.d., but
different currents can be connected to the
same supply. They can be switched on and
off independently.
c 1.40 A. 1.80 A flows into the parallel
arrangement, so by conservation of
charge, 1.80 A flows out. That means
1.40 A must flow into B.
d ____
​​  12  ​ = 8 . 6​Ω
1.4
The resistance must rise (in proportion to
the p.d.).
lamp
relay
b
Practice
2
6.7 Ω
8
Using V = IR, we have:
E1 + E2 = 4R (equation 1)
E1 – E2 = 2R (equation 2)
E1 = 2R + E2 (equation 3)
The p.d. across resistor B will increase.
Challenge
3
a
b
c
Challenge
7
LDR resistance decreases when light falls
on it.
d
e
Relay
This makes a bigger current flow in the
relay coil, so that the switch closes and the
lamp comes on.
For example: as a burglar alarm;
detecting when lights have been switched
on at night.
Change the LDR for a thermistor
A and B
Exercise 19.6
Focus
1
Substituting this expression for E1 into
equation 1, we have
2R + E2 + E2 = 4R
Damp environments are hazardous when
combined with mains electricity.
2
a
b
Simplifying we have:
E2 = R
Practice
3
Putting this into equation 3 we have:
E1 = 3R
​E​  ​​
so the ratio ___
​​  1 ​ = 3​
​E​ 2​​
There are too many devices connected to one
socket. It may result in overheating.
4
a
b
c
A metal case or exposed metal parts
Metal water pipes or an earth rod outside
The wire inside melts if too large a current
flows. This breaks the circuit.
It prevents overheating and therefore the
device and the building from fire.
It could catch fire.
Challenge
5
46
If too large a current flows, an electromagnet
causes the contacts to be separated, breaking
the circuit. It can be reset (unlike a fuse)
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
6
A fault develops. The live wire touches
exposed metal parts. These are connected to
the earth wire. The earth wire has a very small
resistance; so a large current flows; causing the
fuse to melt/trip switch to trip.
7
The fuse may not melt fast enough to prevent
them being electrocuted.
8
The 40 W lamp will get brighter and the 100 W
will not light much.
The reason is that the 100 W lamp has a
smaller resistance and so will take a smaller
proportion of the supply p.d (which must be
24 V, if the two 12 V, 40 W lamps in series are
properly lit).
So the p.d. across the 100 W is much less than
the p.d. across the 40 W lamp, which is now
much greater than the 12 V it needs to light.
For the 100 W lamp to light normally:
P=V×I
100 = 12 × I
100
I = ____
​​   ​​
12
I = 8.3 A
So, since V = I × R, R = ____
​​  12  ​​ = 1.44 Ω
8.3
For the 40 W lamp to light normally:
P=V×I
40 = 12 × I
40
I = ​​ ___ ​​
12
I = 3.3 A
So, since V = I × R, R = ____
​​  12  ​​ = 3.6 Ω
3.3
So, the p.d. across the 100W lamp will be:
V = IR
24 = I × (3.3 + 1.44)
I = 5.1 A
so V = IR
= 5.1 × 1.44
= 7.3 V
Therefore the p.d. across the 40 W lamp
= 24 – 7.3 = 16.7 V
The power developed by each is then:
P=V×I
40 W lamp: 16.7 × 5.1 = 85 W
100 W lamp: 7.3 × 5.1 = 37 W
9
A potential divider circuit with an LDR and
a potentiometer. Also in the circuit is a diode
pointing clockwise. These are in series with a
power supply.
In parallel with the LDR (or the potentiometer)
is a lamp. If it is across the LDR, it will come
on in the dark. If it is across the potentiometer,
it will come on in the light.
Peer-assessment
Exemplar answers supplied for Questions 8 and 9.
Chapter 20
Exercise 20.1
Focus
1
a
b
2
a
The current
The lines of magnetic force
coil
b
switch
To use a small current to switch on and
off a large one. Uses are any that fit with
this definition, for example car ignition,
power station switching, a control circuit
(low current) to turn on/off lights/open or
close doors, windows, curtains, etc.
(high current).
Practice
3
A current flows from the 12 V supply through
the coil. The coil is magnetised and attracts the
switch, closing the second circuit. A current
flows through the lamp, which lights up.
4
Answer should include:
• piece of card/paper at right angles to the
wire, with wire through the centre
• pass current up to 8 A through the wire
• plot the field lines using plotting
compasses/place iron filings on the card
and tap the card gently
• take care to avoid overheating of wire/
burns from wire.
So, we’d expect that the 100 W lamp would be
quite dim, whilst the 40 W lamp may well break.
47
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Challenge
5
a
Challenge
magnetic field
3
The motor will spin in the same direction,
but faster.
4
a
b
It would oscillate at 50 Hz.
It would oscillate about the
vertical position.
5
a
1: magnet; 2: diaphragm (or paper cone);
3: coil of wire
b
(4)
(7)
Coil of wire
Coils carrying
electric current
Diaphragm
(or paper cone)
Magnet
(5)
S
N
S
(6)
b
Answer should include:
• pass current up to 8 A through
the solenoid
• plot the field lines using plotting
compasses, starting at one end and
plotting around to the other
• join these points together
• pick a different starting point
and repeat
• take care to avoid overheating of
wire/burns from wire.
6
The field around the solenoid is stronger.
7
a
b
c
The field around the solenoid will
be stronger
Reverse the direction of the current
through the solenoid.
Exercise 20.2
Focus
1
a
b
d
Coil
Commutator
Practice
2
48
Increase:
• the current through the coil
• the number of turns in the coil
• the strength of the magnetic field.
A current flows through the coil, which
produces a magnetic field around the coil.
This magnetic field interacts with the field
between the poles of the (permanent)
magnet.
This produces a force, the direction of which
is given by Fleming’s left-hand rule. On the
left, the force will be down the page. On the
right, it will also be down the page (because,
although the direction of the current flow is
reversed, so is the direction of the magnetic
field between the poles of the magnet).
Therefore the coil moves down.
The current through the coil is
alternating. As the current changes
direction, so does the force on each side
of the coil. This makes the coil vibrate.
This makes the diaphragm (attached to
the coil) vibrate. This makes a series of
compressions and rarefactions in the air.
Making a sound wave.
Similarities:
• both use a coil of wire in a magnetic
field
• both have a current passing through
the coil
• both are explained using Fleming’s
left-hand rule and the interaction of
magnetic fields
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
the same factors affect the size of the
force on the coil.
Differences:
• motor magnet has two poles,
loudspeaker has three
• motor uses d.c., loudspeaker uses a.c.
• motor coil rotates, loudspeaker
coil vibrates.
The stronger the magnet, the bigger the
force will be for any given value of the
current through the coil. This will make
the loudspeaker more responsive.
The more turns of wire in the coil, the
bigger the force on the coil will be. BUT
this will also make the coil more massive
(have more mass). This will mean a larger
force is required to supply the same
acceleration – so the mass of the coil
needs to be kept as low as possible.
The mass of the diaphragm is also crucial
for the same reason.
The human hearing range is 20 Hz to
20 kHz. As the frequency of the current
increases, the acceleration of the coil and
diaphragm also increase. In order to be
able to supply this acceleration, the mass
of the cone and diaphragm must be kept
as low as possible.
•
e
Practice
3
a
thumb = motion
second finger
= current
b
c
4
a
b
Reversing the current reverses the force,
so that it becomes horizontal, away from
the power supply. The copper rod will roll
away from the power supply.
Increase the current; use stronger
magnetic field
Magnetic field is out of the page.
−
Exercise 20.3
+
Focus
1
2
Reverse the direction of: the current; the
magnetic field.
Challenge
5
Apply magnetic field vertically downwards.
a
b
6
Protons have positive charge, but neutrons
have no charge. A beam of neutrons is not an
electric current.
7
Reverse the direction of the magnetic field.
Protons are positive and electrons are negative,
so the same field direction will result in forces
in opposite directions.
Increase the strength of the magnetic field.
Protons have a much greater mass than
electrons and so a larger force is required
to produce the same change in motion (see
Chapter 3 to revise this, if required).
8
They are caused by charged particles (cosmic
rays) coming in from space and following
the Earth’s magnetic field lines. They ionise
molecules in the atmosphere, which causes
these spectacular lights in the sky.
Conventional current is towards the left.
S
copper rod
aluminium support rods
magnets
N
current
+
49
first finger = field
–
power
supply
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Peer-assessment
Answer is entirely dependent on the content
and quality of the flash cards produced by the
students, but the expectation is that they will
learn from each other’s work, including points
they missed, improving their descriptions and
definitions, and also critically assess the usefulness
of the cards made by others.
Chapter 21
Current
induced?
Case
a wire is moved through the
field of a magnet
Yes
a magnet is held close to a wire
No
a magnet is moved into a coil
of wire
Yes
a magnet is moved out of a
coil of wire
Yes
a magnet rests in a coil of wire
No
Practice
3
iron core
secondary coil
Challenge
Focus
2
2
primary coil
Exercise 21.1
1
Practice
Increase the strength of the magnetic field
around the coil; more turns of wire in the coil;
turn the coil faster; larger area of cross section
of the coil.
It is reversed
3
a
b
c
Max positive current: coil horizontal
Zero current (1/2 cycle): coil vertical
having performed 180° rotation
Max negative current: coil horizontal,
having performed 270° turn
max positive current: coil horizontal
Time
Current
Exercise 21.3
Focus
1
Secondary
1 cycle
2
Challenge
4
Exercise 21.2
Focus
1
50
Time
They are all increasing the rate of change of
the magnetic field through the coil. Another
way to look at that is that they all increase
the rate of cutting of field lines by the coil.
These are exactly the same thing said in two
different ways.
Current
Practice
3
a
b
Step-down
100 turns
4
a
b
c
83 turns
45 kW (45 000 W)
108 A
Slip rings
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Challenge
5
• This increases the induced p.d.
Purpose of the step up transformer:
• By stepping up the p.d., the current is
stepped down (by the same proportion).
• This reduces the heating of the cable
(P = I 2R), thereby increasing the efficiency
of the power transmission.
Why we need step-down transformers:
• Very high potential differences between
the cable and earth are dangerous.
• Closing the gap between the cable and the
ground is likely to result in electric shock.
• This can happen by accident in built
up areas.
• So very high p.d.s are used for crosscountry transmission, but lower p.d.s in
industrial and residential areas.
• Also, in use, p.d.s are generally much
lower, typically ranging from 30 kV
(industry) to 110 V (domestic).
Power loss in the cable, P = I 2R
20 000V: from P = V × I,
100000000
​​ = 5000 A
I = ​​ __________
20000
So power loss = 50002 × 2.5 = 62 500 000 W
useful power
____________
​​
Efficiency =
​​
total power
100000000 − 62500000
= ​​ ____________________
​​ = 0.375
100000000
400 000 V: from P = V × I,
100000000
​​ = 250 A
I = __________
​​
40000
So power loss = 2502 × 2.5 = 156 250 W
useful power
____________
​​
Efficiency =
​​
total power
100000000 − 156250
= ​​ __________________
​​ = 0.998
100000000
This is an increase of 0.623
Exercise 21.4
5
Focus
1
Electromagnetic induction
2
Primary
Practice
3
Since power out cannot be greater than power
in, by the Principle of Conservation of Energy,
and since the p.d. across the secondary is
greater, the current through it must be smaller.
So Ip × Vp = Is × Vs
Challenge
4
51
There would be no need for transformers
to reduce energy losses in long-distance
transmission, as there would be no losses. This
would reduce the infrastructure and so the
cost of the network.
It is also the case that a generator with a coil
made from this material would be able to
generate a larger current with a much smaller
coil, increasing the efficiency.
Peer-assessment
See answer to Question 5 for exemplar answer.
Physics:
• The p.d. across the primary coil changes, so
the current through the primary coil changes.
This
makes the magnetic field around the
•
primary coil change.
• This field goes through the secondary coil.
• This changing magnetic field induces a
p.d. across the secondary coil.
• If there is a complete circuit with the
secondary coil, an induced current flows.
• The soft iron laminated core is also
magnetised by the changing magnetic
field, increasing the strength of the field
passing through the secondary coil.
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Challenge
Chapter 22
3
Exercise 22.1
Focus
1
a
electron
nucleus
The previous model (‘plum pudding’) was one
where the mass and the positive charge were
evenly distributed throughout the volume
of the atom. It said that the electrons were
distributed on the surface of this volume.
Rutherford said that most of the atom is
empty space. Almost all of the mas and all the
positive charge of the atom is concentrated
in one small space within the volume of the
atom. The electrons orbit this ‘nucleus’ at a
considerable distance
Exercise 22.3
Focus
1
b
c
Nucleus
Nucleus
Practice
2
a
b
c
3
16
8
Practice
2
a
b
It loses one electron
It gains two electrons
Z=6
A = 13
N=7
​ ​O
​
Challenge
Challenge
3
4
a
b
By gaining three electrons
Three more protons and one less neutron
Exercise 22.2
Focus
1
a
b
c
Alpha particles have a positive charge.
An alpha particle is smaller than a
gold atom.
The nucleus of a gold atom has a
positive charge.
Practice
2
a
b
c
d
e
52
Protons, neutrons
‘Back-scattered’ means deflected through
more than 90°, so that it comes back out
of the foil rather than passing through.
The alpha particle labelled (1) was backscattered because it collided with the gold
nucleus. Positive charges repel each other.
‘Undeflected’ means that the alpha
particle continued in a straight line.
The alpha particle labelled (2) was
undeflected because it was far from the
gold nucleus so the repulsive force was
very small.
All the positive charge in the atom is in
one small volume.
Description
Which particles?
these particles make up
the nucleus
protons + neutrons
these particles orbit the
nucleus
electrons
these particles have very
little mass
electrons
these particles have no
electric charge
neutrons
these charge on these
particles is equal and
opposite to the charge
on an electron
protons
Exercise 22.4
Focus
1
a
b
Number of protons in nucleus
(proton number)
Number of neutrons (neutron number)
Practice
2
10
5
​​  ​​
​​  Be
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Challenge
3
Nuclide
Proton
number Z
Neutron
number N
Nucleon
number A
Name of
element
Nu-1
4
5
9
beryllium
Nu-2
5
7
12
boron
Nu-3
4
4
8
beryllium
Nu-4
6
5
11
carbon
Nu-5
5
6
11
boron
Exercise 22.5
Chapter 23
Focus
Exercise 23.1
1
Focus
235
Practice
2
+92
3
23492U
1
b
c
d
Challenge
4
23592U
a
23492U + 10n
Peer-assessment
A good mind map contains both text and
diagrams, which serve as memory prompts
(‘dual coding’).
It should clearly show the inter-relationships
between concepts and quantities (for example,
unstable – decay – modes of decay – uses –
dangers – activity – half-life – contamination –
irradiation). The pictures/diagrams could be quite
personal to the student.
The use of hyperlinks in an electronic version can
be very powerful.
Nuclide
symbol ¡​AZ​​X
i
Alpha
ii Beta
iii Gamma
Gamma
Alpha
Alpha, beta and gamma
2
An ion is a particle that has become
charged because it has gained or lost one or
more electrons.
3
a
b
c
d
e
f
g
4
The fact that it is ionising.
Gamma
Alpha
Beta
Alpha
Gamma
Beta
Gamma
Practice
53
5
Excess neutrons in the nucleus; nucleus being
too heavy
6
Contains an inert gas at low pressure, inside
a tube which forms one electrode, and has
another electrode along the axis of this tube.
There is a p.d. of around 400 V between
the electrodes.
When ionising radiation enters the tube
through the thin window at the front,
electrons removed by ionisation are attracted
to the anode and the positive ion is attracted
to the cathode; this causes a pulse of current,
which is counted by the scaler.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
thin mica
window
+
metal tube
(cathode)
central wire
(anode)
a
b
8
9
Exercise 23.2
50. This is the difference between the
count rate either side of the paper barrier.
240 counts per second. Only background
remains after the 10 cm thick block
of lead.
Point a GM tube at the source and note
the vertical position that gives the
maximum reading.
Gamma rays will be unaffected and so will still
be detectable with the GM tube in the original
position, even with a strong electric field
around the beam of emitted radiation.
Alpha radiation will be attracted to the
negative electrode. Moving the detector
vertically towards this electrode will show that
some of the original beam is now deflected in
this direction.
Beta radiation will be deflected towards
the negative electrode. Moving the detector
vertically towards this electrode will show that
some of the original beam is now deflected in
this direction.
Since we detect no deflection and also have
two deflections in opposite directions, we
know that there are three types of ionising
radiations being emitted.
Alpha particles have much greater mass than
beta, so they have greater kinetic energy and
are therefore more ionising.
Alpha particles also have twice the charge of
a beta particle and so are more ionising
than beta
Gamma rays have neither charge nor (rest)
mass and so are not very ionising.
Focus
1
Particle
Symbol
Composition
alpha, α
​​​  42​​ ​  He
2 protons + 2 neutrons
beta, β
​​​  –10​​ ​  e
1 electron
2
a
b
c
d
Ra
Rn
Alpha
Protons: 88 = 86 + 2
3
a
b
Gamma emission
Beta emission; number of
protons increases
Practice
4
Nucleons: 15 = 15 + 0
Protons: 6 = 7 ‒ 1 or 6 = 7 + (−1)
5
6
207
4
​​ 211
84​ ​Po ​→ ​  82​ ​Pb ​+ ​  2​ ​  He + energy
a ​​  231
140​ ​Pa
227
4
b ​​  231
140​ ​Po ​→ ​  138​ ​Pb ​+ ​  2​ ​  He + energy
Challenge
7
Mass number = 234 (decay by alpha, 238 goes
to 234; decay by beta, no change; decay by
beta, 234 again)
Atomic number = 92 (decay by alpha, 92
becomes 90; each beta decay adds 1; making
the final atomic number 92)
This is uranium,​​ ​  234
​
92​​U
Exercise 23.3
Focus
1
54
–
scaler
low-pressure gas
(mainly argon)
Challenge
7
400 V
a
b
c
300
2100
Nuclear decay is a process that is random
in time and direction.
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Practice
2
a
b
6
172 800
3 counts/s
a
b
c
20 counts per minute
100 counts per minute
You can use the pattern of numbers,
100–50–25–12–6–3–1, showing 6 half-lives
in about 7 days, or refer to the graph.
Challenge
3
700 counts per min due to the source, of
which:
745 – 622 = 123 are due to alpha
622 – 600 = 22 due to beta
600 due to gamma
None of the background count was affected
by the material placed between the source and
the detector.
Exercise 23.4
140
120
Count
rate /
counts
per
minute
80
60
40
20
Focus
1
100
0
Decay is the loss of particles or energy from
an unstable nucleus.
Practice
2
0
1
2
3
4 5
6
Time / days
7
8
9
Half-life is approximately 1.2 days.
There are three modes of decay:
• Alpha: reduces the mass number of the
parent nucleus by four and the atomic
number by two.
• Beta: the mass number of the nucleus is
unchanged, the mass number increases
by one.
This is due to the decay of a neutron into a
proton and an electron:
neutron → proton + electron
• Gamma: atomic and mass numbers
are unchanged. It is the emission of
electromagnetic radiation from
the nucleus.
Challenge
Use of radioactivity
Code
number(s)
3
250
4
E
39 years
finding the age of an
ancient object
destroying cancerous tissue
A, C
sterilising medical equipment
C
7
Exercise 23.5
Focus
1
5
Activity / 500
counts 400
per second 300
250
200
controlling the thickness of paper
B, D
in a paper mill
100
0
After 32 hours:
• A has1 had 2 half-lives and so its activity is
now ​​ _4 ​​of the starting value = 32 counts/s
• B has1had 4 half-lives and so its activity is
now ​​ __
​​of the starting value = 8 counts/s
16
Total count rate = 40 counts/s
0
2
2.5h
4
6
8
detecting smoke in the air
B, D
tracing leaks from
underground pipes
A, D
irradiating food
A, C
10
Time / h
Half-life is approximately 2.5 h.
55
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
Practice
2
4
Reduce exposure time of living tissue to
radiation, to reduce the chances of damage to
living tissue.
Increase distance of living tissue from
source, to reduce the intensity of the
radiation received.
Shielding – absorbing radiation before it
reaches living tissue.
Challenge
3
a
b
A medical tracer
• Tc-99
• Gamma source – high penetration, so
will be detectable outside the body;
low ionisation ability, so minimum
risk of harm to patient.
• Half-life long enough for a medical
tracing procedure, but count rate
will fall rapidly, reducing danger to
patients and those around them.
To check the thickness of paper in a mill
Kr-85
Beta source – will be affected by
thickness of paper.
• Long half-life, so source does not
need to be replaced too frequently,
and equipment does not need to be
recalibrated too often due to falling
count rate.
In a domestic smoke detector
• Am-241
• Alpha source – will be absorbed
by smoke.
• Half-life long enough to allow
constant count rate over the life of
detector, but not so long as to pose a
very long-term hazard.
To treat cancer by radiotherapy
• Ba-133
• Gamma source source – high
penetration, so will be detectable
outside the body; low ionisation
ability, so minimum risk of harm
to patient.
• Long half-life, so source does not
need to be replaced too frequently,
and equipment does not need to be
recalibrated too often due to falling
count rate.
•
•
c
d
56
Students’ own answers. See Peer assessment
for feedback.
Peer-assesment
Answer is entirely dependent on the content
and quality of the flash cards produced by the
students, but the expectation is that they will
learn from each other’s work, including points
they missed, improving their descriptions and
definitions, and also critically assess the usefulness
of the cards made by others.
Chapter 24
Exercise 24.1
Focus
1
a
b
1 day, 24 hours
1 year, 365 days
Practice
2
The sun rises in the East and sets in the West
each day, giving us day and night.
Challenge
3
The sun is stationary relative to the Earth,
so if the appearance of the moon changes, it
must be due to the motion of the moon.
We see the moon by light from the sun
reflecting from it.
As the moon orbits the Earth, different
portions of it are illuminated by the light from
the sun, so it seems to change shape (these are
the phases).
4
1.5 × 108 km = 1.5 × 108 × 1000 m
365 days = 365 × 24 × 60 × 60 seconds
2 × 3 . 14 × 1 . 5 × ​10​​ 8​× 1000
_________________________
​​
​v = ​
365 × 24 × 60 × 60
v = 29 871 m/s
Exercise 24.2
Focus
1
There are three celestial bodies that orbit the
sun. These are planets, minor planets and
comets. Moons orbit planets.
2
Elliptical
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
km/h, it would take over 336.5 million years to
get there).
Practice
3
Mercury, Venus, Earth, Mars
4
a
b
The conclusion is not valid. Uranus is
much larger than Earth, but has weaker
gravity. Mercury is bigger than Mars, but
has the same strength of gravity in the
table. Saturn is much larger than Earth,
but has a weaker gravity.
Gravity on Venus is much greater than on
Mercury. This means it is likely to have
a more dense atmosphere. This is likely
to trap more of the sun’s thermal energy,
which makes the surface hotter, despite
being further away from the sun.
Challenge
5
6
The planets are believed to have formed from
a spinning disc of dust. The rocky planets
formed too close to the sun for gases on the
surface to condense to solid particles, but
rather to vapourise. Only matter with a very
high melting point could remain solid here.
a
b
c
The orbits are ellipses, so the distance to
the sun varies over the orbit.
As the planet gets closer to the sun, it
will travel faster. This is because the sun’s
gravity is stronger closer to the sun. So
there will be a larger force on the planet,
causing it to travel faster.
The total energy is constant. Further away
from the sun, the gravitational energy is
greater, so the kinetic energy is less. Closer
to the sun, there is less gravitational
energy and so the kinetic energy increases
– the planet moves faster.
Exercise 24.3
Chapter 25
Exercise 25.1
Focus
1
9.5 × 1015 m
2
a
b
3
cm wavelength waves detected at all points in
the universe
4
It suggests that the universe was much hotter
and smaller in the past, with the waves
redshifting as the universe expanded and
cooled.
Practice
5
4.03 × 1013 km = 4.03 × 1016 m
1 light-year = 9.5 × 1015 m
4 . 03 × ​10​​  16​
So Proxima Centauri is ___________
​​
​​
9 . 5 × ​10​​  15​
= 4.23 light-years away from Earth
6
v: the redshift of light from stars in the galaxy
d: from measuring the brightness of a
supernova in the galaxy
Challenge
7
Our galaxy, made up of billions of stars
2
The distance travelled by light in one year
Practice
3
The apparent increase in the wavelength of
electromagnetic radiation from distant stars
Challenge
4
57
Other galaxies are more than 10 000 lightyears away (Canis Major is 25 000 light-years
away, actually).
It is not possible for us to travel this far
because it would take too long (at 80 000
Rewriting the equation, we get:
d
d
__
​​  1  ​​, since __
​​  v ​​ = ___
​​  v ​​ = time
H0
This gives us the age of the universe.
8
It suggests that the universe started from a
single point, which is consistent with the Big
Bang Theory.
9
A cloud of dust and hydrogen collapses
under gravity, forming a protostar. Nuclear
fusion begins – the star is stable, now that the
outward force of fusion balances the inward
pull of gravity.
When most of the hydrogen has become
helium through nuclear fusion, the star
collapses, due to gravity.
More nuclear reactions occur due to higher
temperature and pressure from the collapse.
The star expands to become a red giant if it
has a mass less than eight times the mass of
the sun.
Focus
1
v = H0d
2.2 × 10–18 per second
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CAMBRIDGE IGCSE™ PHYSICS WORKBOOK ANSWERS
After being a red giant, the star becomes
a white dwarf at the centre of a
planetary nebula.
It becomes a red supergiant, and then a
supernova, if the star has a mass greater than
eight times the mass of the sun.
Elements heavier than iron are made in the
supernova. These may be part of the planetary
formation of another solar system.
After the supernova, either a neutron star or a
black hole is left behind.
10 Points about day and night:
• The Earth rotates points about an
imaginary line through the poles (its axis).
• This means that most parts of the world
rotate into and out of sunlight each day –
this makes day and night.
Points about seasons:
The Earth’s distance from the sun varies
over the year.
• The Earth is tilted on an imaginary line
through the poles (its axis), around which
it rotates.
• Because of this tilt, for part of the year,
the Northern hemisphere is tilted toward
the sun making the time it spends in
daylight longer – this makes the
day longer.
• At the same time, the Southern
hemisphere is tilted away from the sun;
this means it spends less time in sunlight –
making the days shorter.
• The Earth is closest to the sun in January,
but the tilt on the axis still means that
because the Northern hemisphere is tilted
away from the sun at this time, it is
winter here.
Points about the phases of the moon:
• We see the moon because it reflects the
light from the sun.
• The portion of the moon we see depends
on the amount of the moon we can see
that is illuminated.
• The moon rotates around the Earth every
27.3 days, during which time the position
of the sun relative to the Earth has not
changed much – so the portion of the
illuminated part of the moon changes.
•
58
A new moon occurs when the moon
is between the sun and the Earth – so
that none of the side facing the Earth is
illuminated by the sun.
• An eclipse occurs when the moon moves
into the Earth’s shadow – so that the
moon cannot be directly lit by the sun,
since the Earth is between the sun and
the moon.
Points about the origin of the sun (points in
brackets are supplementary content):
• A cloud of dust and gas (stellar nebula)
collapsed under its own gravity, making
it hot.
• It became a protostar, when it got
hot enough.
• Nuclear fusion began where hydrogen
nuclei fuse to make helium nuclei and
release energy.
• This is what the sun is doing now.
•
Self-assessment
See answer to Question 10 for exemplar answer.
Cambridge IGCSE™ Physics Workbook – Hamilton © Cambridge University Press 2021