Cambridge O Level Physics
Answers to the Student Book
Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from
its past question papers which are contained in this publication. Exam-style questions and sample answers have been
written by the authors. In examinations, the way marks are awarded may be different. References to assessment and/or
assessment preparation are the publisher’s interpretation of the syllabus requirements and may not fully reflect the
approach of Cambridge Assessment International Education.
Section 1 Motion, forces and energy
1.1 Physical quantities, units and measurement
Test yourself questions
1
2
3
4
5
6
7
8
a 10
b 40
c 5
d 67
e 1000
a 3.00 b 5.50 c 8.70 d 0.43 e 0.1
a 1 × 105; 3.5 × 103; 4.28 × 108; 5.04 × 102; 2.7056 × 104
b 1000; 2 000 000; 69 200; 134; 1 000 000 000
a 1 × 10–3; 7 ×10–5; 1 × 10–7; 5 × 10–5
b 5 × 10–1; 8.4 × 10–2; 3.6 × 10–4; 1.04 × 10–3
Thickness of 100 pages = 100 × 0.10 mm = 10 mm
Thickness of two covers = 2 × 0.20 mm = 0.40 mm
Total thickness of book = 10 mm + 0.4 mm = 10.4 mm = 10 mm (correct to 2 s.f.)
a two
b three c four
d two
Volume of rectangular block = length × breadth × height
= 4.1 cm × 2.8 cm × 2.1 cm = 24 cm3
Stopwatch; at least 5
Now put this into practice questions
(Page 4)
1
Area of triangle =
× base × height =
× 8 cm × 12 cm = 48 cm2
2 Circumference 2πr = 2π × 6 cm = 38 cm
(Page 5)
Volume V = length × breadth × height = 30 cm × 25 cm × 15 cm = 11 250 cm3
= 11 250 × 10–6 m3 = 1.13 × 10–2 m3
2 Volume of cylinder V = πr2h = π × (5.0 cm)2 × 25cm = 2000 cm3 = 2 × 10–3 m3
(Page 9)
1
1
2
a
a
10
b
0.577 b
8.6
1.00
Cambridge O Level Physics
© Heather Kennett 2021
c
c
9.2
1.73
1
Cambridge O Level Physics answers
3
Let Fx = 5.0 and Fy = 7.0 then:
and
4
𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = �5.02 + 7.02 = √25 + 49 = √74 = 8.6 N
so 𝜃𝜃 = 54º
Resultant force is 8.6 N acting at 54º to the 5.0 N force
Let Fx = 6.0 and Fy = 8.0 then:
𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = �6.02 + 8.02 = √36 + 64 = √100 = 10 m/s
and
so 𝜃𝜃 = 53º
Resultant velocity is 10 m/s acting at 53º to the horizontal
Practical work questions
1
2
3
4
5
6
7
Students’ own responses based on their results
Students’ own responses based on their results
Students’ own responses based on their results
Students’ own plans
Record the time for at least 5 complete oscillations with a stopwatch; to determine the period
divide the time by the number of oscillations
BOAOB
Length of pendulum is the distance from the lower end of the metal plates to the centre of the
bob
Exam-style questions (Page 10)
1
a
b
c
2
a
b
3
a
b
4
a
b
c
Volume of chocolate bar, V1 = length × breadth × height
= 10 cm × 2 cm × 2 cm = 40 cm3;
Volume of chocolate bar, V2 = 2 cm × 2 cm × 2 cm = 8 cm3
Number of bars with same volume = V1 / V2 = 40 / 8 = 5
Time period = 8 s / 10 = 0.8 s
[1]
[2]
[1]
[2]
[2]
[Total: 8]
Average thickness = 6 mm / 60 = 0.1 mm
[2]
Number of blocks = (40 cm × 40 cm × 20 cm) / (10 cm × 10 cm × 4 cm) = 80
[5]
[Total: 7]
3
Volume of water = 6 cm × 6 cm × 7 cm = 252 cm
[3]
3
Volume of stone = 6 cm × 6 cm × (9 – 7) cm = 6 cm × 6 cm × 2 cm = 72 cm
[4]
[Total: 7]
Metre, second
[2]
3
[1]
2
2
i πr
ii 2πr
iii πr h
[4]
[Total: 7]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
2
Cambridge O Level Physics answers
5
a
b
2.31 mm
14.97 mm
6
a
b
C
13 N at 67º to 5 N force
[2]
[2]
[Total: 4]
[1]
[7]
[Total: 8]
1.2 Motion
Test yourself questions
1
2
3
4
5
6
7
8
a Average speed = distance moved / time taken = 400 m / 20 s = 20 m/s
b Distance moved / time taken = 1500 m / (4 × 60) s = 6.25 m/s
a Average speed = (10 m/s + 20 m/s) / 2 = 15 m/s
b Distance = v t = 15 m/s × 60 s = 900 m
a Acceleration = change of speed / time taken = 6 m/s / 3 s = 2 m/s2
b Acceleration = –6 m/s / 2 s = –3 m/s2
Time taken = change of speed / acceleration = 500 km/h / 10 km/h/s = 50 s
a Straight-line graph through the origin
b Positive, constant
c Acceleration = slope of graph = 16 m/s / 4 s = 4 m/s2
d Area of triangle = base × height / 2 = 4 s × 16 m/s / 2 = 32 m
e 32 m
a Straight line through the origin
b Speed = gradient of graph = constant
c Speed = distance/time = 18 m / 6 s = 3 m/s
At 1 s, gradient of tangent to curve = 20 m/s / 2.5 s = 8 m/s2; acceleration = 8 m/s2
Acceleration = change of speed / time taken, so
Time = change of speed / acceleration = (30 – 0) m/s / 10 m/s2 = 3 s
Distance = average speed × time = 30 m/s / 2 × 3 s = 45 m
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
3
Cambridge O Level Physics answers
a v = at = 9.8 m/s2 × 2 s = 20 m/s
b Distance = average velocity × time = (0 + 20) m/s / 2 × 2 s = 20 m
10 Let Fx = 12.0 and Fy = 5.0 then:
9
𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = �12.02 + 5.02 = √144 + 25 = √169 = 13 m/s
and
so 𝜃𝜃 = 23º
Resultant velocity is 13 m/s acting at 23º to the horizontal
Now put this into practice questions (Page 16)
1
2
v = u + at = 0 + 0.8 m/s2 × 4 s = 3.2 m/s
s = (u + v)t / 2 = (10 + 20) m/s × 5 / 2 = 75 m
3
s = ut +
at2 = 0 × 5 s +
(+2 m/s2) 52 s2 = 25 m
Practical work questions
1
2
3
4
The speed increases because the mass falls further in each 1/50 s
33 × 1/50 = 0.66 s
Reaction times will be longer than the small time interval to be measured. Also the stopwatch
would only give an average speed for the fall; changes in speed and acceleration could not then
be evaluated
No difference
Exam-style questions (Page 20)
1
a
b
c
2
Change in speed = at = 1 m/s2 × 15 s = 15 m/s;
final speed = (10 + 15) m/s = 25 m/s
a i 60 km
ii (6pm – 1pm) = 5 hours
3
Average speed = (0 + 8) / 2 = 4 m/s
s = v × t = 4 m/s × 4 s = 16 m
Acceleration = 2 m/s2
[Total: 4]
[1]
[1]
iii v = distance / time = 60 km / 5 h = 12 km/h
[1]
iv 2 (flat regions of graph)
[1]
v
[1]
(0.5 + 1.0) hours = 1.5 hours
vi v = distance / time = 60 km / (5 – 1.5) h = 60 km / 3.5 h = 17 km/h
4
[2]
[3]
[2]
[Total: 7]
b
Steepest line: EF
a
b
c
100 m
v = distance / time = 100 m / 5 s = 20 m/s
Slows down (slope of graph decreases)
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[2]
[2]
[Total: 9]
[1]
[1]
[2]
4
Cambridge O Level Physics answers
d
5
a
b
c
6
a
2
Acceleration = change in speed / time taken = (5 – 0) m/s / 4 s = 1.25 m/s
i Distance = average velocity × time = 5.0 m/s / 2 × 4 s = 10 m
(or area under slope = (base × height of triangle) / 2 =10 m)
ii Distance = average velocity × time = 5.0 m/s × 9 s = 45 m
(= area under horizontal line between 4 s and 13 s)
Remaining distance = (100 – 45 – 10) m = 45 m
Time taken for remaining distance = distance / velocity = 45 m / 5 m/s = 9 s
Total time for 100 m = (13 + 9) s = 22 s
Graph needs to be extended horizontally to 22 s
d
e
f
i Accelerating: OA, BC
ii Decelerating: DE
iii Constant speed: AB, CD
OA: a = change in speed / time = +80 km/h2
AB: v = 80 km/h
BC: a = (100 – 80) km/h / 0.5 h = +40 km/h2
CD: v = 100 km/h
DE: a = (0 – 100) km/h / 0.5 h = –200 km/h2
Distance = average speed × time
OA: distance = 80 km/h / 2 × 1 h = 40 km
AB: distance = 80 km/h × 2 h = 160 km
BC: distance = (80 + 100) km/h / 2 × 0.5 h = 90 km/h × 0.5 h = 45 km
CD: distance = 100 km/h ×1 h = 100 km
DE: distance = 100 km/h / 2 × 0.5 h = 25 km
(40 + 160 + 45 + 100 + 25) km = 370 km
Average speed = distance / time = 370 km / 5 h = 74 km/h
Zero and 5 hours
a
b
Constant speed
Distance = 600 m
b
c
7
[3]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[Total: 7]
[2]
[2]
[2]
[1]
[1]
[1]
[1]
[Total: 10]
[1]
[1]
[1]
[3]
[3]
[1]
[1]
[1]
[Total: 12]
[2]
[1]
5
Cambridge O Level Physics answers
8
c
Speed = distance / time = 600 m / 30 s = 20 m/s
a
i
ii
i
ii
b
v = at = 9.8 m/s2 × 1 s = 9.8 m/s
v = at = 9.8 m/s2 × 3 s = 29 m/s
Distance = average velocity × time = (0 + 9.8) / 2 m/s × 1 s = 4.9 m
Distance = average velocity × time = (0 + 29) / 2 m/s × 3 s = 44 m
[2]
[Total: 5]
[2]
[2]
[2]
[2]
[Total: 8]
1.3 Mass and weight
Test yourself questions
1
2
3
a
a
b
a
b
1N
b 50 N
c 0.50 N
Weight on Earth = mass × g = 12 kg × 9.8 m/s2 = 120 N
Weight on Moon = mass × gmoon = 12 kg × (10 / 6) m/s2 = 20 N
Weight on Moon = mg = 80 kg × 1.6 N/kg = 128 N
i 0 N/kg
ii 0 N
Exam-style questions (Page 25)
1
a
i
ii
Mass is a measure of the quantity of matter in an object (at rest relative
to an observer)
Weight is a gravitational force on an object that has mass
[1]
[2]
iii Using a balance
[1]
b
c
d
C, mass × g
B, mass
i N
ii
2
a
b
Gravitational field strength is the force per unit mass; g = W/m
i W = mg so m = W/g = 1850 N / 9.8 m/s2 = 189 kg
ii W = mg = 189 kg × 3.7 m/s2 = 700 N
3
a
*b
c
d
A gravitational field exerts a gravitational force on any mass in the field
Determines the weight of a mass.
Force per unit mass; g = W/m
W = mg = 200 kg × 8.8 N/kg = 1760 N
m/s2
iii N/kg
[1]
[1]
[3]
[Total: 9]
[2]
[2]
[2]
[Total: 6]
[2]
[1]
[2]
[2]
[Total: 7]
1.4 Density
Test yourself questions
1
a
b
i ρ = m/V = 100 g / 10 cm3 = 10 g/cm3
ii V = m/ρ = 38 g / 19 g/cm3 = 2.0 cm3
iii V = m/ρ = 95 g / 19 g/cm3 = 5.0 cm3
2
a
Volume of bar = 4 cm × 3 cm × 1 cm = 12 cm3
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
ii
ρ = m/V = 9 kg / 3 m3 = 3 kg/m3
6
Cambridge O Level Physics answers
3
b
a
ρ = m/V = 96 g /12 cm3 = 8.0 g/cm3
ρ = m/V = 96 × 10–3 kg / 12 × 10–6 m3 = 8.0 × 103 kg/m3
V = (60 – 50) cm3 = 10 cm3
b ρ = m/V = 60 g / 10 cm3 = 6 g/cm3
Now put this into practice questions (Page 27)
1
2
𝜌𝜌 =
𝑚𝑚
𝑉𝑉
= 2.7 g/cm3
a
m = V × ρ = 4 cm3 × 11 g/cm3 = 44 g
b
𝑉𝑉 =
𝑚𝑚
𝜌𝜌
=
55 g
11 g/cm3
= 5 cm3
Exam-style questions (Page 29)
1
a
b
A
i m = V × ρ = 5 m3 × 3000 kg/m3 = 15000 kg
ii m = V × ρ = 10 m × 5.0 m × 2.0 m × 1.3 kg/m3 = 130 kg
2
a
Measure the mass m1 of an empty beaker on a balance. Transfer a known
volume V of the liquid from a burette or a measuring cylinder into the beaker.
Measure the mass m2 of the beaker plus liquid. Mass of liquid m = (m2 – m1)
and so ρ = m/V can be calculated
Mass of liquid = (170 –130) g = 40 g
Density of liquid ρ = m/V = 40 g / 50 cm3 = 0.8 g/cm3
The density of ice is less than the density of water
The density of oil is less than the density of water
b
c
d
3
a
b
i V = 10 cm × 8 cm × 20 cm = 1600 cm3 = 1.6 × 10–3 m3
ii ρ = m/V = 1.2 kg / (1.6 × 10–3 m3) = 750 kg/m3
ρ = m/V = 33 g / 30 cm3 = 1100 kg/m3
[1]
[3]
[3]
[Total: 7]
[4]
[4]
[1]
[1]
[Total: 10]
[2]
[3]
[3]
[Total: 8]
1.5 Forces
Test yourself questions
1
2
OE
May cause an object to move / change speed / change direction; change the shape/size of an
object
3
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
7
Cambridge O Level Physics answers
Extension is proportional to stretching force over OE
4 k = F/x = 200 × 10–3 kg × 9.8 N/kg / (4 × 10–2) m = 49 N/m
5 The limit of proportionality can be defined as the point at which the load–extension graph
becomes non-linear because the extension is no longer proportional to the stretching force.
6 If the ring does not move there is no net force on the ring:
140 N = 100 N + FH, so FH = 140 N – 100 N = 40 N
7 Resulting forward force = 100 N – 50 N = 50 N
8 Total upward force = 2 F
Total downward force = 50 N
In equilibrium, TUF = TDF so 2 F = 50 N and F = 25 N
9 Resultant force is 50 N at an angle of 53o to the 30 N force
10 D; Total force = 20 N + 30 N = 50 N is larger than any other case
11 Since acceleration = 0, there is no resultant force and F = P = 20 N
12 a Resultant force = mass × acceleration = 1000 kg × 5 m/s2 = 5000 N
b Acceleration = resultant force / mass = 30 N / 2 kg = 15 m/s2
13 a Acceleration = change in speed / time = 8 m/s / 2 s = 4 m/s2
b F = ma = (500/1000) kg × 4 m/s2 = 2 N
14 a Friction occurs when there is motion between two surfaces
b impedes motion / reduces speed; causes heating
15 Air resistance between the air and the moving car acts to reduce speed; friction between the
tyres and the road acts to reduce the speed; the motor acts to increase the speed. When the
forces in each direction are equal and opposite, the resultant force is zero and the car maintains
a constant speed
16 The force is greater than the stalk can resist
17 Less
18 Yes, moments equal.
Clockwise moment = 40 N × 1 m = 40 N m; anticlockwise moment = 20 N × 2 m = 40 N m
19 Clockwise moment = W × (45 – 25) cm = W × 20 cm
Anticlockwise moment = 20 N × (25 – 10) cm = 300 N cm
Equating clockwise and anticlockwise moments gives
W × 20 cm = 300 N cm so W = 300 N cm / 20 cm = 15 N
20 C; moment of force = force × perpendicular distance from pivot
21 Yes; clockwise moment = 60 N × 0.5 m = 30 N m; anticlockwise moment = 20 N × 1.5 m =
30 N m so clockwise and anticlockwise moments equal
22 200 N; clockwise moment = W × 3 m;
Anticlockwise moment 160 N × 3 m + 120 N × 1 m = 480 N m + 120 N m = 600 N m
Equating moments gives W = 600 N m / 3 m = 200 N
23 a Centre of ruler
b Centre of sphere
24 a When the vertical line through its centre of gravity lies outside its base
b By lowering its centre of gravity and increasing the area of its base
25 i unstable equilibrium
ii stable equilibrium
iii neutral equilibrium
Cambridge O Level Physics 4th edition
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8
Cambridge O Level Physics answers
Now put this into practice questions
(Page 32)
1
k = F/x = 4 N / (2 × 10–3) m = 2000 N/m
2 F = kx so x = F/k = (2 N × 9.8 N/kg) / 250 N/m = 78 × 10–3 m = 7.8 mm
(Page 37)
1
2
a resultant force = 10 N – 8 N = 2 N.
b a = F/m = 2 N / 5 kg = 0.4 m/s2.
F = ma = 7 kg × 0.5 m/s2 = 3.5 N
Practical work questions
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A straight line
Extension proportional to stretching force
Use a set square to ensure the ruler is vertical; avoid parallax errors in readings; add pointer at
lower edge of spring to ensure readings taken from same place each time; repeat readings
Calculate force/extension for each pair of readings to test if a constant value is obtained or plot
a graph of stretching force against extension.
Two springs, 2 hooks to support springs, 1 kg weight, string, paper, board, pins, protractor,
ruler
Mass or angle POQ
Students’ own responses based on their results
Students’ own responses based on their results
Students’ own responses based on their results
Force and mass
Test independent variables separately: (a) calculate F/a for each pair of readings to see if a
constant value is obtained. (b) calculate ma for each pair of readings to see if a constant value is
obtained
Distances d1 and d2
a Moment = 5 N × 10 cm = 50 N cm
b Moment = 5 N × 15 cm = 75 N cm
c Moment = 5 N × 30 cm = 150 N cm
a Tie a piece of string onto a weight
b used to find a vertical line; the string indicates the vertical if the plumb line is allowed to
hang freely.
Vertically below the point of suspension
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
9
Cambridge O Level Physics answers
Exam-style questions (Page 51)
1
a
b
2
3
a
b
a
b
and
Hang a spring from a clamp and support a ruler vertically behind it. Mark the
position of the lower end of the spring. Hang different weights from the bottom of
the spring and record the scale readings in a table. Calculate the extension and plot
a load–extension graph. Precautions: ensure ruler is vertical and no parallax occurs
in readings, repeat readings.
[6]
The extension of the spring is proportional to the stretching force.
Mass/g
Stretching force/N
Scale reading/mm
Extension/mm
0
0
20.0
0
100
0.98
20.2
0.2
200
1.96
20.4
0.4
300
2.94
20.6
0.6
400
3.92
20.8
0.8
500
4.90
21.0
1.0
k = F/x1 = 4 N / (22 – 10) cm = 0.33 N/cm;
i x2 = F/k = 6 N / (4/12 N/cm) = 18 cm;
ii Total length of spring = original length + extension
= 10 cm + 18 cm = 28 cm
i
ii
12 N   5 N or 5 N   12N
Resultant force = 12 N – 5 N = 7 N;
7 N  or  7N
[4]
[Total:10]
[3]
[2]
[1]
[Total: 6]
[2]
[2]
𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = √52 + 122 = √25 + 144 = √169 = 13 N
so 𝜃𝜃 = 67º
Resultant force = 13 N at 67º to 5 N force
4
a
b
c
d
5
a
b
[6]
[Total: 10]
2
a = change in velocity / time = 5 m/s / 10 s = 0.5 m/s
[2]
Average speed = (0 + 5) m/s / 2 = 2.5 m/s
[2]
Distance = average speed × time = 2.5 m/s × 10 s = 25 m
[2]
Backward frictional forces (due to air resistance and friction between road and
tyres) are equal to the forward force exerted on the bicycle
[2]
[Total: 8]
Rearranging F = ma gives a = F/m. If F is large and m is small (due to the
use of lightweight material) the acceleration will be large, as is required for
a racing car
[3]
a = F/m; if the car has a small engine, F is small but m is also small. Low F
reduces a, but a low m will increase a. If m is low enough, the acceleration
can still be large.
[3]
[Total: 6]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
10
Cambridge O Level Physics answers
6
a
b
Weight = mg = 500 kg × 9.8 N/kg = 4.9 × 103 N
i Resultant force = 25 000 N – 4900 N = 20 100 N;
*ii From F = ma, initial acceleration a = F/m = 20 100 N / 500 kg = 40 m/s2
7
a
b
c
Friction between tyres and road
i larger
ii smaller
iii larger
Slicks allow greater speed in dry conditions but in wet conditions treads
provide frictional force to prevent skidding
8
9
[1]
[2]
[3]
[Total: 6]
[2]
[3]
[2]
[Total: 7]
Moment of force = force × perpendicular distance from pivot
a A: perpendicular distance = 0; moment = 0 Nm
b B: Taking crank to be at 45º to vertical, perpendicular distance = 0.2 cos 45° m
(or 0.2 sin 45° m) = 0.1414…m
Moment = 0.1414…m × 25 N = 3.5 Nm
c C: perpendicular distance = 0.2 m; moment = 0.2 m × 25 N = 5 Nm
Taking moments about pivot:
a clockwise moment = 10 N × (50 – 40) cm = 10 N × 10 cm = 100 N cm
b anticlockwise moment = 3 N × (40 –10) cm = 3 N × 30 cm = 90 N cm
c clockwise moment is greater than anticlockwise moment so beam tips to right
[2]
[2]
[2]
[Total: 6]
[3]
[3]
[2]
[Total: 8]
10 a
Suspend lamina so it can swing freely from a nail clamped in a stand. When
the lamina comes to rest, locate and draw the vertical line from the point of
suspension with a plumb line; repeat process using a different suspension point.
Centre of gravity located where the two lines cross.
b i Stability increases because centre of gravity is lowered
11 a
b
ii
Stability increases because the centre of gravity is lowered and the area of
the base is increased
i
ii
Weight
Air resistance
[5]
[2]
[2]
[Total: 9]
[1]
[1]
[2]
[Total: 4]
12 a Distance = vt = 20 m/s × 0.7 s = 14 m
[2]
b 50 m – 14 m = 36 m
[1]
c i Choose two from: tiredness, alcohol, drugs, eyesight, visibility of hazard
[2]
ii Choose two from: wet road, efficiency of brakes, depth of tyre tread
[2]
[Total: 7]
Alternative to practical (Page 53)
13 a
Falls at constant velocity (terminal velocity)
Support a spring in a clamp and fix ruler vertically behind it.
Draw up a table to record stretching force, scale reading and extension.
Record scale reading opposite the lower end of the unweighted spring.
Increase the load on the spring in 100g stages and record the scale reading at
lower end of spring for each load.
Calculate the extension for each load and plot a load–extension graph.
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[1]
[1]
[1]
[1]
[1]
11
Cambridge O Level Physics answers
b
Use set square to ensure the ruler is vertical; avoid parallax errors in readings;
add pointer at lower edge of spring to ensure readings taken from same place
each time; repeat readings.
14 a
i
[4]
[2]
[1]
[2]
[1]
[Total: 10]
[1]
[2]
[2]
[2]
ii OL
iii L
For example: gradient = 3 N / 6 mm = 0.5 N/mm
0.5 N/mm
b
c
15 a
[3]
[Total: 8]
i
ii
iii
iv
Mass/g
Force/N
Ruler
reading/cm
d/cm
F × d/N cm
50
0.49
5
20
9.8
A
50
0.49
10
15
7.4
B
50
0.49
15
10
4.9
C
50
0.49
20
5
2.5
D
100
0.98
30
5
4.9
E
100
0.98
35
10
9.8
F
100
0.98
40
15
14.7
G
150
1.47
20
5
7.4
H
150
1.47
35
10
14.7
I
b
AF, BH, CE
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[3]
[Total: 10]
12
Cambridge O Level Physics answers
1.6 Momentum
Test yourself questions
1
2
3
4
5
6
Momentum p = mv
a p = 10 kg × 5 m/s = 50 kg m/s
b p = 10 kg × 20 × 10–2 m/s = 2 kg m/s
c p = 10 kg × 36 × 103 m/s / (60 × 60) = 100 kg m/s
Momentum p = mv
Total momentum before collision is (1 kg × 2 m/s) + (1 kg × 0 m/s) = 2 kg m/s
Total momentum after collision is (1 kg × 0 m/s) + (1 kg × v) = 1 kg × v
If momentum is conserved: (1 kg × v) = 2 kg m/s, so v = 2 m/s
Momentum p = mv
Total momentum before boy jumps on trolley is (50 kg × 5 m/s) + (20 kg × 1.5 m/s)
= (250 + 30) kg m/s = 280 kg m/s
Total momentum after collision is (50 + 20) kg × v = 70 kg × v
If momentum is conserved: 70 kg × v = 280 kg m/s, so v = 280/70 = 4 m/s
Forward momentum of girl = mv = 50 kg × 3 m/s = 150 kg m/s
Backward momentum of boat = mv = 300 kg × v m/s
If momentum is conserved: 300 kg × v = 150 kg m/s, so v = 150/300 = 0.5 m/s
a Impulse = F Δt = 5 N × 0.02 s = 0.1 N s b Δp = F Δt = 0.1 N s
a Δp = m Δv =1000 kg × 24 m/s = 2.4 × 104 kg m/s
b F = Δp/Δt = 2.4 × 104 kg m/s / 1.2 s = 2.0 ×104 N
Now put this into practice questions (Page 55)
1
2
Total momentum before collision is (3 kg × 5 m/s) + (2 kg × 0 m/s) = 15 kg m/s
Total momentum after collision is (3 kg + 2 kg) × v = 5 kg × v
If momentum is conserved: 5 kg × v = 15 kg m/s, so v = 3 m/s
Total momentum before collision is (5 kg × 5 m/s) + (2 kg × 0 m/s) = 25 kg m/s
Total momentum after collision is (5 kg × 0 m/s) + 2 kg × v = 2 kg × v
If momentum is conserved: 2 kg × v = 25 kg m/s, so v = 12.5 m/s
Practical work questions
1
Momentum is conserved in a collision
2 To compensate for energy loss in friction between the trolley and the track
3
a
b
i
iii
i
iii
mv = 2 kg × 0.2 m/s = 0.4 kg m/s
mv = 2 kg × 5 × 10–2 m/s = 0.1 kg m/s
mv = 0.2 kg × 3 m/s = 0.6 kg m/s
mv = 1 kg × 3 m/s = 3 kg m/s
ii
mv = 2 kg × 0.8 m/s = 1.6 kg m/s
ii mv = 0.5 kg × 3 m/s = 1.5 kg m/s
Exam-style questions (Page 58)
1
a
b
c
Momentum p = mv
Momentum of truck A before the collision is (500 kg × 4 m/s) = 2000 kg m/s
Momentum of truck B before the collision is (1500 kg × 2 m/s) = 3000 kg m/s
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[1]
[2]
[2]
13
Cambridge O Level Physics answers
d
2
3
4
Total momentum after collision is (500 + 1500) kg × v = 2000 kg × v
If momentum is conserved: 2000 kg × v = (2000 + 3000) kg m/s = 5000 kg m/s
so v = 5000 kg m/s / 2000 kg = 2.5 m/s
[4]
[Total: 9]
Momentum p = mv
a Initial momentum (10 kg × 4 m/s) = 40 kg m/s
[2]
b Final momentum (10 kg × 8 m/s) = 80 kg m/s
[2]
c Total momentum gained in 2 s = (80 – 40) kg m/s = 40 kg m/s
[2]
2
d F = rate of change of momentum = 40 kg m/s / 2s = 20 kg m/s
[2]
e Impulse = FΔt = 20 N × 2 s = 40 N s
[2]
[Total: 10]
a Momentum = mass × velocity
[1]
b Backward momentum of ejected gas = 5 kg × 5000 m/s = 25 000 kg m/s
[2]
c Momentum is conserved in a collision if no external forces act
[2]
d If momentum is conserved forward momentum
= 25 000 kg m/s = (10 000 – 5) kg × v
so v = 25 000 kg m/s / 9995 kg = 2.5 m/s
[3]
[Total: 8]
2
a F = ma so a = F/m = 50 N / 0.03 kg = 1670 m/s
[2]
b Impulse = F Δt = 50 N × 0.001 s = 0.05 N s
[2]
c F Δt = Δ (mv) = 0.03 kg × v = 0.05 N s, so v = 0.05 N s / 0.03 kg = 1.7 m/s
[2]
d Apply force for a longer time; increase the size of the force
[2]
[Total: 8]
1.7 Energy, work and power
Test yourself questions
1
2
3
4
5
6
7
8
a
b
c
a
c
a
b
c
electric current
mechanical working
heating
chemical energy
b internal (thermal) energy
kinetic energy
d elastic (strain) energy
2
Ek = mv /2 = 1 kg × 2 m/s × 2 m/s / 2 = 2 J
Ek = mv2/2 = 0.002 kg × 400 m/s × 400 m/s / 2 = 160 J
Ek = mv2/2 = 500 kg × (72000 / (60 × 60)) m/s × (72000 / (60 × 60)) m/s / 2
= 100 000 = 105 J
a Rearranging Ek = mv2/2, gives v = √(2 × Ek/m) = √(2 × 200 J / 1 kg) = 20 m/s
b i ∆Ep = mgΔh = 5 kg × 9.8 N/kg × 3 m = 147 J
ii ∆Ep = mgΔh = 5 kg × 9.8 N/kg × 6 m = 294 J
∆Ep of water flowing over falls/s = mgΔh/t = 7 × 106 kg/s × 9.8 N/kg × 50 m
= 3.4 × 109 W = 3400 MW
W = F d = (3 × 9.8) N × 6 m
=176 J
W = F d = (51 × 9.8) N × 300 m = 1.5 × 105 J
W = ∆Ep = mgΔh, so mg = W/Δh = 80 J /5 m = 16 N
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
14
Cambridge O Level Physics answers
9
Renewable, non-polluting (i.e. no CO2, SO2 or dangerous waste), low initial building cost of
station to house energy converters, low running costs, high energy density, reliable, allows
output to be readily adjusted to varying energy demands
10 a Heating, lighting, refrigeration, television, computers, sound systems...
b i Install roof insulation, wall insulation, double-glazed windows, solar hot water system,
solar generating panels, turn off lights and electrical devices when not in use, wear a
jumper in cold weather so that home heating levels can be lowered, wait until washing
machine and dishwater are full before switching on, collect materials for recycling...
ii Develop energy efficient cars, buildings, power stations and power transmission
systems, use waste energy from power stations (e.g. to heat local homes), encourage
people to walk, cycle or use trains and buses to travel to work, turn off heating in
offices at weekends and lights and computers at night, recycle materials such as metals,
paper, glass
11 Efficiency =
useful power output
× 100%
total power input
= (9 J/s / 20 J/s) × 100% = 45%
12 P = W/t = Fd/t = 600N × 10 m / 12 s = 500 W
13 P = ΔE/t = 2400 J / 60 s = 40 W
14 Work done = F × d = (60 × 70 × 9.8) N × 5 m = 206 000 J
P = work done / time taken = 206 000 J / 60 s = 3430 W = 3.4 kW
Now put this into practice questions
(Page 63)
1
Ek =
mv2 =
× 0.4 kg × (80 m/s)2 = 0.2 × 6400 kg m2/s2 = 1280 N m = 1280 J
2
Ek =
mv2 =
× (50 × 10–3) kg × (40 m/s)2 = 0.025 × 1600 kg m2/s2 = 40 N m = 40 J
(Page 63)
1 ΔEp = mgΔh = 0.2 kg × 9.8 N/kg × 2 m = 4 N m = 4 J
2 ΔEp = mgΔh = 0.4kg × 9.8 N/kg × 3 m = 12 N m = 12 J
(Page 64)
1
2
a Ek = mv2/2 = 2 kg × (10 m/s)2 / 2 = 100 kg m2/s2 = 100 N m = 100 J
b ∆Ep = Ek = 100 J
c ∆Ep = mgΔh so Δh = ∆Ep/mg = 100 J / 2 kg × 9.8 m/s2 = 5 m
∆Ep = mgΔh = 0.4 kg × 9.8 m/s2. × 30 m = 120 J. By conservation of energy
∆Ep = Ek =
mv2 = 120 J and so v2 = 2 × 120 J / 0.4 kg = 600 m2/s2; v = 24.5 m/s
(Page 73)
1
W = Fd = 500 N × 12 m = 6000 J;
efficiency = energy output / energy input × 100% = (6000 J / 8000 J) × 100% = 75%
2
Efficiency =
useful power output
× 100%
total power input
= (560 – 170) J/s / 560 J/s × 100% = (390 / 560) × 100% = 70 %
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
15
Cambridge O Level Physics answers
Practical questions
1
2
3
Students’ own responses based on their results
a To reduce the effect of frictional losses
b There is no longer a resultant force acting on the trolley
ΔEp = mgΔh = 0.3kg × 10 N/kg × 0.80 m = 2.4 N m = 2.4 J
4
Ek =
5
6
Chemical energy stored in your muscles is transferred to gravitational potential energy
Heat, sound, frictional losses
mv2 =
× 0.3 kg × (4.0 m/s)2 = 2.4 J
Exam-style questions (Page 76)
1
a
b
c
2
a
b
c
d
e
3
a
b
c
4
a
b
c
5
a
b
c
d
e
6
a
b
7
a
Electricity transferred to kinetic energy and thermal energy
Electricity transferred to heat
Electricity transferred to sound
[2]
[2]
[2]
[Total: 6]
∆Ep = mgΔh = 100/1000 kg × 9.8 N/kg × 1.8 m = 1.8 J
[2]
∆Ep transfers to Ek = 1.8 J
[1]
2
Rearranging Ek = mv /2, gives v = √(2 × Ek/m) = √( 2 × 1.8 J / 0.1 kg) = 6 m/s
[3]
∆Ep = mgΔh at high point of rebound = 0.1 kg × 9.8 N/kg × 1.25 m = 1.23 J
so Ek of rebound = 1.23 J
[2]
v = √(2 × Ek/m) = √(2 × 1.23 J / 0.1 kg) = 5 m/s
[3]
[Total: 11]
2
Rearranging Ek = mv /2, gives v = √(2 × Ek/m) = √( 2 × 100 J / 0.5 kg) = 20 m/s
[3]
Ek = ∆Ep = 100 J
[1]
2
∆Ep = mgΔh, so Δh = ∆Ep/mg = 100 J / (0.5 kg × 9.8 m/s ) = 20 m
[3]
[Total: 7]
Work done = F d = 100 N × 1.5 m = 150 J
[2]
150 J
[1]
Power = work done / time taken = 4 × 150 J / 60 s = 10 W
[3]
[Total: 6]
2%
[1]
Hydroelectric
[1]
Cannot be used up
[1]
Solar energy, wind energy
[2]
All energy ends up as thermal energy which is difficult to use and there is only
a limited supply of non-renewable sources
[2]
[Total: 7]
i Reliable, readily available at all times, high energy density
[2]
ii Polluting, non-renewable
[2]
i Renewable, non-polluting
[2]
ii Only available when the sun shines, needs large areas of land
[2]
[Total: 8]
Efficiency = (useful energy output / total energy input) × 100%
= (300 MJ / 1000 MJ) × 100 % = 30%
[3]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
16
Cambridge O Level Physics answers
b
c
Energy lost = 1000 MJ – 300 MJ = 700 MJ; thermal energy
Warms surroundings; lost from cooling towers
[2]
[2]
[Total: 7]
1.8 Pressure
Test yourself questions
1
2
3
4
To raise the level of the water supply above that in the reservoir and provide water pressure to
the taps in the building
The pressure in a liquid increases with depth so the wall at the bottom of the dam must
withstand a greater pressure than it does at the top of the dam
Δp = ρgΔh = 1000 kg/m3 × 9.8 m/s2 × 2 m = 2 × 104 Pa
Δp = ρgΔh so Δh = Δp/ρg = 3.0 × 106 Pa / (1.02 × 103 kg/m3 × 9.8 m/s2) = 300 m
Now put this into practice questions
(Page 78)
i The area of the base = 2 m × 2 m = 4 m2
ii Area of a side = 2 m × 5 m = 10 m2
b i Pressure on base = force / area = 80 N / 4 m2 = 20 Pa
ii Pressure on side = force / area = 80 N / 10 m2 = 8 Pa
2 a i Pressure = force / area = 50 N / 2.0 m2 = 25 Pa
ii F/A = 50 N / 100 m2 = 0.50 Pa
iii F/A = 50 N / 0.50 m2 = 100 Pa
b F = pressure × area = 10 Pa × 3.0 m2 = 30 N
(Page 79)
1
a
1
2
3
= 300 N
2
f= F ×
0.1 m
a
= 70 N ×
=7N
1.0 m 2
A
Incompressibility
Exam-style questions (Page 84)
1
2
A true, B true, C true, D false, E true, F false
a i p = F/A = 2000 kN / 2 m2 = 1000 kN/m2
ii p = F/A = 200 kN / 0.2 m2 = 1000 kN/m2
iii p = F/A = 0.5 kN / 0.0002 m2 = 2500 kN/m2
b High heels; they produce a pressure greater than 2000 kN/m2 on the floor
3
a
b
c
d
Pressure = force / area = 20 N / 0.20 m2 = 100 Pa
Force = pressure × area = 100 Pa × 2.0 m2 = 200 N
A liquid is nearly incompressible
A liquid transfers the pressure applied to it
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[Total: 6]
[2]
[2]
[2]
[2]
[Total: 8]
[2]
[2]
[1]
[1]
[Total: 6]
17
Cambridge O Level Physics answers
4
5
6
a A true, B true, C false
[3]
3
2
6
[4]
*b Δp = ρgΔh = 1150 kg/m × 9.8 N/kg × 100 m = 1 127 000 N/m = 1.13 × 10 Pa
[Total: 7]
a Δp = ρgΔh
[2]
b Pascal (Pa)
[1]
c Rearrange Δp = ρgΔh to give
Δh = Δp/ρg = 7.5 × 106 Pa / (1.03 × 103 kg/m3 × 9.8 m/s2) = 740 m
[3]
[Total: 6]
a Vacuum
[1]
b Atmospheric pressure
[1]
c 740 mm Hg
[1]
d It becomes less; atmospheric pressure lower (decreases with altitude)
[2]
[Total: 5]
Section 2 Thermal physics
2.1 Kinetic particle model of matter
Test yourself questions
1
2
3
4
5
6
7
a Air is readily compressed
b Steel is not easily compressed
a melting
b evaporation and boiling
c solidification d condensation
a Solid – ordered and densely packed particles vibrate about fixed positions
b Liquid – less ordered, particles further apart than in a solid and can slide over each other
c Gas – particles are wide apart and are free to move quickly in all directions
It is the temperature at which all particle motion ceases
When the temperature rises, the average speed of the gas particles increases leading to more
frequent collisions with the surfaces of the container so the pressure increases if the volume is
kept constant.
If the volume is reduced there will be more frequent collisions of the particles with the surfaces
of the container so the pressure will increase if the temperature is kept constant.
a It is believed to be the lowest possible temperature b They are the same size
Now put this into practice questions
(Page 93)
1 p1V1 = p2V2 ; V2 = p1 × V1 / p2 = 1 × 105 Pa × 9 cm3 / 3 ×105 Pa = 3 cm3
2 p1V1 = p2V2 ; p2 = p1 × V1 / V2 = 2 × 105 Pa × 40 cm3 / 20 cm3 = 4 × 105 Pa
(Page 94)
1
2
T = 273 + θ = 273 + 80 = 353 K
θ = T – 273 = 100 – 273 = –173 ºC
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
18
Cambridge O Level Physics answers
(Page 96)
1
2
p1V1 / T1 = p2V2 / T2
so p2 = p1V1T2 / T1V2 = 1 × 105 Pa × 9 cm3 × 310 K / (300 K × 5 cm3) = 1.9 × 105 Pa
p1V1/T1 = p2V2 /T2
so T2 = p2V2T1 / p1V1 = 3.2 × 105 Pa × 30 cm3 × 300 K / (2 × 105 Pa × 40 cm3)
= 360 K = 87 ºC
Practical work questions
1
2
3
4
5
6
Smoke particles
Haphazardly
Collisions with large numbers of light, fast moving air molecules.
a Temperature
b Pressure
Only one variable should be changed in an experiment at a time.
When taking a measurement, stop heating, stir the water and allow the gauge reading to
become steady.
7 a Pressure
b Volume
8 Graph is a straight line through the origin
9 a Temperature
b Volume
10 Volume is proportional to temperature measured in kelvin.
Exam-style questions (Page 98)
1
2
B
a
[Total: 1]
[2]
b
[2]
c
[2]
[Total: 6]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
19
Cambridge O Level Physics answers
3
a
b
7
a
Gas
[1]
Particles strike the surfaces of the container in large numbers per second and
cause a pressure on the surfaces
[2]
c The pressure increases
[1]
When the temperature rises so does the average speed of the gas particles and so, if the
volume remains fixed, the pressure increases because the change of momentum per second
when the particles collide with the surfaces increases, leading to a larger force per unit
area.
[3]
[Total: 7]
4 a i True
[1]
ii True
[1]
iii False
[1]
b i Absolute zero where particle motion ceases
[2]
ii T = 273 + θ = 273 – 273 = 0 K.
[1]
iii T = 273 + θ = 273 – 200= 73 K
[1]
[Total: 7]
5 a p 1V1 = p2V2 ; so p2 = p1V1 /V2 = 1 × 105 Pa × 10 cm / 40 cm = 2.5 × 104 Pa
[3]
5
4
b p1V1 = p3V3 ; so p3 = p1V1 /V2 = 1 × 10 Pa × 10 cm / 50 cm = 2.0 × 10 Pa
[3]
[Total: 6]
5
3
5
3
6 a p1V1 = p2V2; V2 = p1 × V1 / p2 = 1 × 10 Pa × 30 cm / 2 × 10 Pa = 15 cm
(pressure doubled, volume halved)
[3]
5
3
5
3
b V2 = p1 × V1 / p2 = 1 × 10 Pa × 30 cm / 5 × 10 Pa = 6 cm
[3]
[Total: 6]
Alternative to practical question (Page 99)
b
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[3]
[1]
20
Cambridge O Level Physics answers
pressure/ 105 Pa
volume/ cm3
1/volume / cm–3
24
1.0
1.00
12
2.0
0.50
8
3.0
0.33
6
4.0
0.25
4
6.0
0.17
c
[3]
d
Yes; values of pV all equal to 24 ×105 Pa cm3
[2]
[Total: 9]
2.2 Thermal properties and temperature
Test yourself questions
1
a
b
c
2
a
b
The metal lid expands more than the glass jar when it is held in hot water so becomes
looser and easier to unscrew
Wood and metal contract when the temperature falls, leading to creaking of the furniture
Concrete expands when the temperature rises and would buckle if gaps are not left
between sections; the pitch is easily squeezed out when expansion occurs
Aluminium
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
21
Cambridge O Level Physics answers
3
4
5
6
7
8
9
10
11
12
13
The particles in a liquid are further apart and more mobile than in a solid so expansion is easier
for liquids than for solids. In gases, particles are further apart than in liquids and can move
about freely at high speeds; this means they are able to expand much more easily than liquids
a Water expands when it is cooled from 4 ºC to 0 ºC; most liquids contract when the
temperature decreases
b Water pipes burst when the water in them freezes; water-based liquids may burst their
containers when they freeze; fish can survive below the frozen surface of a pond
Liquid in a glass bulb expands up a capillary tube when the bulb is heated; the temperature is
marked in degrees on a scale next to the capillary tube
C
Heat needed = mass × temperature change × specific heat capacity
= mΔθc = 5 kg × 10 ºC × 300 J/(kg ºC) = 15 000 J
Assume heat supplied ΔE = heat gained by water
then ΔE = 3000 J/s × t = mcΔθ
and t = mcΔθ / 3000 J/s = 5 kg × 4200 J/(kg ºC) × (50 – 30 )ºC / 3000 J/s = 140 s
a 1530 ºC
b 19 ºC c 0 ºC
d 100 ºC
e 37 ºC
a Ice draws heat from the drink when it melts (it has a high specific latent heat of fusion)
b Steam releases much heat when it condenses (it has a high specific latent heat of
vaporisation)
a i Remain constant
ii Absorbed
iii Remain constant
b i Remain constant
ii Increase
a A few energetic particles close to the surface of a liquid escape and become gas particles
lowering the average kinetic energy of the remaining particles
b Decreases
The water cools when evaporation occurs.
Now put this into practice questions
(Page 106)
1
Using
c = 25 000 J / (2 kg × (35 − 10) ºC)
= 25 000 J / 50 kg ºC
= 500 J/(kg ºC)
2
Rearrange equation
to give the heat equation:
ΔE = mcΔθ = 3 kg × 500 J/kg ºC × 10 ºC = 15 000 J
(Page 108)
1
2
Assume heat supplied to water = heat gained by water
then 3000 J/s × t = mcΔθ
and t = mcΔθ / 3000 J/s = 1 kg × 4200 J/(kg ºC) × (100 – 30 )ºC / 3000 J/s = 98 s
Heat lost by sphere = mcΔθ = 0.1 kg × c × (100 – 25) ºC = c × 7.5 kg ºC
Heat gained by water = 0.2 kg × 4200 J/kg ºC × (25 – 20) ºC = 4200 J
Equating heat lost to heat gained gives c = 4200 J / 7.5 kg ºC = 560 J/kg ºC
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
22
Cambridge O Level Physics answers
Practical work questions
1
2
Heat used to raise temperature of aluminium pan as well as the water; heat is lost to the
surroundings
Heat received by water = Power × t = 40 W × 300 s = 12 000 J
So
3
4
Reduce heat loss to surroundings by insulating the container and block
Heat received by aluminium cylinder = Power × t = 40 W × 300 s = 12 000 J
So
5
6
7
8
9
10
11
= 12 000 J / (1 kg × 2.5 ºC) = 4800 J/(kg ºC)
= 12 000 J / (1 kg × 12.5 ºC) = 960 J/(kg ºC)
Students’ own graphs bases on their results
Temperature remains constant
The liquid is solidifying
Faster
Students’ own responses based on their results
Collect the ice melted before the heater is switched on for the same time
Wrap insulation around the funnel
Exam-style questions (Page 113)
1
2
3
4
a
The particles in a gas are further apart than in a liquid and move around at
higher speeds. They have less interaction with each other than the particles in a
liquid and so a gas can expand more easily
[3]
b When water freezes to ice at 0 ºC, it expands and becomes less dense. The
expansion causes metal pipes to burst
[2]
[Total: 5]
A correct [1]
B correct [1]
C correct [1]
[Total: 3]
Heat supplied = ΔE = m × Δθ × c; so c = ΔE / (m × Δθ)
A: cA = 2000 J / (1.0 kg × 1.0 ºC) = 2000 J/(kg ºC);
[3]
[3]
B: cB = 2000 J / (2.0 kg × 5.0 ºC) = 200 J/(kg ºC);
C: cC = 2000 J / (0.5 kg × 4.0 ºC) = 1000 J/(kg ºC)
[3]
[Total: 9]
a Specific heat capacity of jam is higher than that of pastry so it cools down
more slowly
[2]
b Rearrange equation
to give
Δθ = ΔE / mc = 15 000 J / (3 kg × 500 J/kg ºC ) = 10 ºC
5
a
Heat supplied = ΔE = mΔθc = 10 g × 30 ºC × 4.0 J/(g ºC) = 1200 J.
b
Heat is conducted from the milk to the water which cools when it evaporates
[3]
[Total: 5]
[3]
[3]
[Total: 6]
6
a
i The temperature at which a solid changes to a liquid
ii The temperature at which a liquid changes to a gas
iii The temperature at which a liquid changes to a solid
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[3]
23
Cambridge O Level Physics answers
b
c
7
i
i
0 °C
released
ii
ii
100 °C
released
A incorrect [1] B incorrect [1] C incorrect [1] D correct [1]
E incorrect [1]
[2]
[2]
[Total: 7]
[Total: 5]
Alternative to practical question (Page 114)
8
a
b
c
9
80 ºC; temperature constant
They move nearer together and become more ordered
[4]
[2]
[2]
[Total: 8]
Set a container on the balance and record its mass mc. Pour some hot water into the
container so that its base is completely covered. Record the mass of the container
and water (mc + mw). At the same time start the stopwatch and record the value
of (mc + mw) at 15 s intervals for 5 minutes.
[1]
Using the same mass of water (mw), repeat the experiment using:
• containers with different base areas
[1]
• the hairdryer to direct an air flow (draught) over the surface of the water while
measurements are being taken.
[1]
Key control variables: initial mass of water and initial water temperature
[1]
Table headings: (mc + mw)/g, mw/g, time/s
[1]
Conclusions: compare the times taken for the same decrease in the mass of the water
from its initial value to occur in the different experiments; the shorter the time the
larger the rate of evaporation.
[1]
[Total: 6]
2.3 Transfer of thermal energy
Test yourself questions
1
2
3
In conduction thermal energy is transferred through matter from places of higher temperatures
to places of lower temperature without movement of the matter as a whole.
a If small amounts of hot water are to be drawn off frequently, it may not be necessary to
heat the whole tank
b If large amounts of hot water are needed, it will be necessary to heat the whole tank
Hot air is less dense than cool air
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Cambridge O Level Physics answers
4
Black surfaces absorb radiation better than white ones; the ice on the black sections of the
canopy melts faster than on the white sections
5 Infrared
6 Clouds reduce the amount of energy radiated into space, keeping the ground warmer
7 Metal is a better conductor of heat than rubber
8 a Conduction and radiation
b Radiation and convection
Practical questions (Page 121)
a
b i (80 – 42) ºC = 38 ºC
ii (42 – 28) ºC = 14 ºC
c 80 ºC to 42 ºC
d Higher temperatures
(Page 123)
1 The environment
2 Use a thicker layer of insulation/better insulating material/work in warmer room
Exam-style questions (Page 125)
1
2
3
4
5
Fix a match to a metal rod with a little wax;[1] repeat with rods of identical
dimensions but of different materials.[1] Support the rods on a tripod in a fan
arrangement and heat where the ends are closest together.[1] The match will fall
first from the best thermal conductor and last from the worst.[1]
Atoms in the hot regions vibrate strongly and pass on some of their energy to
colder neighbouring atoms through lattice vibrations.[2] Some materials such as
metals also have large numbers of free electrons which when they gain kinetic
energy in the hot regions can travel faster and further;[1] they can interact with
atoms in cooler parts and transfer energy through the material quickly.[1]
a Convection transfers thermal energy through a fluid by movement of the fluid
from places of higher temperature to places of lower temperature;[2]
this occurs because the density of the fluid is lower when it is hot than when it
is cold.[1]
b Drop a few crystals of potassium permanganate down a tube into the bottom
of a beaker containing water.[1] When the tube is removed and the water is
heated from below, purple streaks mark the motion of the water.[2]
A true [1] B true [1] C false [1] D true [1]
a Black surfaces are better emitters of radiation than white surfaces;
dull surfaces are better emitters of radiation than shiny ones
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[Total: 4]
[Total: 4]
[Total: 6]
[Total: 4]
[2]
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Cambridge O Level Physics answers
b
c
6
7
White surfaces are better reflectors of radiation than black surfaces;
shiny surfaces are better reflectors of radiation than dull ones
Black surfaces are better absorbers of radiation than white surfaces;
dull surfaces are better absorbers of radiation than white ones
a Hold the back of your hands on either side of a hot copper sheet which has
one side polished and the other side blackened;
your hand will feel warmer near the better emitter of radiation
b Attach a coin with wax to two different surfaces, one black and dull,
the other shiny
Place each surface the same distance away from a heater; the wax will melt
faster and the coin will fall from the surface which is the better absorber
a
b
c
8 a
b
c
d
Newspaper is a poor conductor of heat
The fur would trap more air, which is a good insulator, and so keep
the wearer warmer
The holes in a string vest trap air, which is a poor conductor of heat,
next to the skin.
Fibreglass traps air which is a poor conductor of heat; using it for roof
insulation reduces heat loss by conduction
Plastic foam-filled cavity walls trap air which is a good insulator; using it in
the cavity walls reduces heat loss by conduction and convection
Double glazed windows trap air, which is a poor conductor of heat, so reduce
the loss of heat through windows by conduction
Fibreglass and plastic foam are both good insulators; they trap air which is a
poor conductor of heat
Convection cannot occur in the plastic foam
Implosion of glass could occur if glass is not strong enough
[2]
[2]
[Total: 6]
[3]
[1]
[2]
[2]
[Total: 8]
[1]
[2]
[2]
[Total: 5]
[1]
[2]
[1]
[2]
[1]
[1]
[Total: 8]
Alternative to practical (Page 126)
9
Method: Wrap the glass beaker with one layer of insulation and make an insulating lid with a
hole in it for a thermometer. Collect 100 cm3 of hot water into a measuring cylinder and pour
into the beaker. Cover the beaker with the lid and insert the thermometer through the lid into
the hot water.
Record the temperature of the water every 30 s. Repeat the experiment with two layers of
insulation on the beaker and lid.
[2]
Control variables: volume of water, starting temperature of water, room temperature and
environmental conditions.
[1]
Table headings: layers of insulation, time/s, temperature/ºC.
[2]
Conclusion: compare the times for the temperature of the water to fall over the same
temperature interval – the longer the time, the more effective the insulation.
[1]
[Total: 6]
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Cambridge O Level Physics answers
Section 3 Waves
3.1 General properties of waves
Test yourself questions
1
2
3
4
5
a λ = 5 cm /5 = 1 cm
b f = 5 cycles / 5 seconds = 1 cycle/s = 1 Hz
c v = f λ = 1 Hz × 1 cm = 1 cm/s
a Speed of ripple depends on the depth of the water
b AB since ripples travel more slowly towards it so the water is shallower in this direction
35°
B
C
Now put this into practice questions (Page 130)
1
2
3
λ = v/f = 5 cm/s / 50 Hz = 0.1 cm
Rearrange v = f λ to give f = v/ λ = 10 m/s / 1 m = 10 Hz
Rearrange v = f λ to give f = v/ λ = 10 m/s / 25 cm = 1000 cm/s / 25 cm = 40 Hz
Practical questions
1
2
Equally spaced parallel lines
Distance between crests of ripples
Exam-style questions (Page xx)
1
a
b
c
Trough
i λ = 3.0 mm (distance between consecutive wave-crests)
ii v = d/t = 3.0 mm × 5 / 1s = 15 mm/s
iii f = 5 cycles/s = 5 Hz
Two small balls fitted to the bar of a ripple tank
2
a
b
C
B
3
Wavelength longer after waves pass through gap
[1]
[2]
[3]
[2]
[2]
[Total: 10]
[1]
[1]
[Total: 2]
[Total: 6]
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Cambridge O Level Physics answers
3.2 Light
Test yourself questions
1
2
3
Larger, less bright
a Four images
b Brighter but blurred
C
4
5
Before; sound travels slower than light
A
6
7
8
B; the image is the same distance behind the plane mirror as the object is in front.
D
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Cambridge O Level Physics answers
9
Spear should be aimed below apparent position of the fish
10 D; glass has a higher refractive index than water so the ray is refracted towards the normal
11
12 Parallel
13 Distance from lens:
a Beyond 2F
b 2F
c Between F and 2F
d Nearer than F
14 Towards
15 Magnification = image length / object length = 10 cm / 5 cm = 2
16 a See Figure 3.2.42a (page 153)
b Behind the retina
17 C
18 A; the ray is refracted towards the normal at the first surface and away from the normal at the
second surface.
Now put this into practice questions
(Page 145)
1
n
so sin r = sin i / n = sin 30° / 1.5 = 0.50 / 1.5 = 0.33
and r = 20°
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Cambridge O Level Physics answers
(Page 147)
1
2
sin 32° = 0.53 so n = 1 / 0.53 = 1.9
Rearrange equation n = 1/sin c to give sin c = 1/n = 1/1.7 = 0.59 so c = 36°
Practical questions
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
Smaller, inverted, laterally inverted
Larger image
a brighter
b less sharp c same size
The angle of incidence equals the angle of reflection
The reflected ray does not emerge from the front surface of the mirror at the same point that the
incident ray strikes it.
Same size
Same points are perpendicular to glass
Same distance
Answer: image orientation changes from being parallel to 90°
Reflected and refracted
Towards
Away
They are parallel
Ray is undeviated
The ray strikes surface normally
Approximately 42° for glass
Incident, reflected and refracted rays
Parallel light from a distant object converges at principal focus of lens
A diverging lens produces a virtual image which cannot be formed on a screen
a Enlarged b Upright
Ray diagrams drawn as in Figures 3.2.39a–d. Values should roughly agree with measured
values of image position
Exam-style questions (Page 156)
1
a
b
c
d
40°
40°, 50°, 50°
Parallel, but turned through 180 ° (antiparallel)
Cambridge O Level Physics 4th edition
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[2]
[3]
[3]
[2]
[Total: 10]
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Cambridge O Level Physics answers
2
a
b
B
Top half
c
She must stand 1m from the mirror if she is to be 2 m from her image (the
image in a plane mirror is the same distance behind the mirror as the object
is in front). She must walk 4 m towards mirror
3
a
b
c
d
e
Refraction
POQ
Towards
40°
90° – 65° = 25°
4
a
b
c
Angle of incidence = 0°,
Angle of incidence is greater than the critical angle
CB is refracted away from the normal
[1]
[2]
[4]
[Total: 7]
[1]
[1]
[1]
[1]
[1]
[Total: 5]
[1]
[2]
[2]
[Total: 5]
5
a
b
i
The ray passes into the air and is refracted away from the normal,
since the angle of incidence is less than the critical angle
ii Total internal reflection occurs in water, since the angle of incidence is
greater than the critical angle
n = 1/sin c, so sin c = 1/n = 3/4 = 0.75
so c = 49°
Cambridge O Level Physics 4th edition
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[1]
[2]
[3]
[3]
[1]
[Total: 10]
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Cambridge O Level Physics answers
6
a
b
Converging
[1]
[3]
c
correct ray diagram
Image is 9 cm from the lens
Image is 3 cm high
7
a
b
c
Image height / object height = 8 cm / 4 cm = 2
8
Lens A: Object is near principal focus, F, of converging lens, so f = 10 cm
Lens B: Object is at 2F of converging lens, so f = 5 cm
4 cm
Correct ray diagram
Image is 8 cm behind lens,
virtual,
larger than object
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[2]
[2]
[2]
[Total: 10]
[1]
[2]
[1]
[1]
[1]
[2]
[Total: 8]
[3]
[3]
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Cambridge O Level Physics answers
9
a
b
Dispersion
Refraction in prism drawn correctly
Refraction of emergent rays drawn correctly
Red and blue rays in correct positions
[Total: 6]
[1]
[2]
[2]
[1]
[Total: 6]
Alternative to practical (Page 158)
10 a
b
c
d
For example: Draw a straight line AOB on the paper and then a line OC
perpendicular to AOB. Draw a further line OD at an angle of say 40°, to the
normal OC; record the angle in a table.
[1]
Set the mirror vertically on the line AOB. Press two of the pins, at least 5 cm
apart along the line OD into the cork board through the paper and view their
reflections in the mirror.
[1]
Find the viewing position where the image of the pins appears to be in line
(one behind the other) and mark the line, OE, by pressing two further pins
into the board. Draw in line OE with a ruler and measure the angle of reflection.
[1]
Repeat with line OD at different angles to the normal OC.
[1]
The table should have headings ‘angle of incidence’ and ‘angle of reflection’
[2]
Check if the values of the angle of incidence and angle of reflection agree
within experimental error
[2]
Ensure pins are vertical, ensure mirror is vertical and placed accurately on
AOB, draw lines thinly
[2]
[Total: 10]
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Cambridge O Level Physics answers
3.3 Electromagnetic spectrum
Test yourself questions
1
a
b
2 a
b
3 a
b
c
d
e
f
4 a
b
c
d
5 a
b
c
d
0.7 μm
0.4 μm
B
D
Ultraviolet
Microwave
Gamma rays
Infrared
infrared/microwaves
X-rays
ultraviolet
microwave
infrared
X-ray or gamma ray
microwave
radio
light or infrared
microwave
Now put this into practice questions (Page 160)
1
2
Rearrange the equation v = f λ to give f = v/λ then f = 3 × 108 m/s / 6 × 10–7 m
= 5 × 1014 Hz
Rearrange the equation v = f λ to give λ = v/f then
λ = 3 × 108 m/s / 4.0 × 1014 Hz = 7.5 × 10–7 m.
Exam-style questions (Page 165)
1
a
i
ii
i
ii
microwave
ultraviolet
gamma rays
radio waves
[1]
[1]
b
[1]
[1]
[Total: 4]
8
6
2 a v = f λ so λ = v/f = 3 × 10 m/s / 100 × 10 Hz = 3 m
[4]
3
8
–4
b v = s/t so t = s/v = 60 × 10 m / (3 × 10 m/s) = 2 × 10 s
[4]
[Total: 8]
3 D
[Total: 1]
4 a glass
[1]
b infrared and visible light
[2]
glass or plastic is transparent to infrared and visible light
[1]
higher frequency than radio waves and so can transmit information at a higher rate [2]
c
as a digital signal
[1]
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Cambridge O Level Physics answers
d
data can be regenerated accurately; can be carried over long distances; is secure
from electrical interference; signals can be transmitted at higher data rates; cannot be
hacked easily are cheaper and easier to install
[3]
[Total: 10]
3.4 Sound
Test yourself questions
1
2
3
4
5
6
7
8
9
Produced by a vibrating source such as a guitar string or loudspeaker
Longitudinally; molecules vibrate about a fixed position in the direction in which the wave
propagates
One wavelength
256 Hz
0.8 m (it has the highest frequency)
B
a 20 Hz to 20 000 Hz
b 330 m/s to 350 m/s
v = s/t so s = v t = 330 m/s × 5 s = 1650 m (about 1 mile)
a Reflection, refraction, diffraction
b In a transverse wave, vibrations are perpendicular to rather than along the direction of
travel of the wave. Sound waves are longitudinal.
Now put this into practice questions
(Page 170)
1 Rearrange the equation v = f λ to give λ = v/f then λ = 340 m/s / 512 Hz = 0.66 m
2 Rearrange the equation v = f λ to give f = v/λ then f = 340 m/s / 1.0 m = 340 Hz
(Page 171)
1
Using 2d = v t then d =
= 1400 m/s × 0.5 s / 2 = 350 m
2
Using 2d = v t then d =
= 1400 m/s × 2 s / 2 = 1400 m
Practical questions
1
2
Increase the distance between the microphones
v = d/t = 1.2 m / (3.6 × 10–3) s = 330 m/s
Exam-style questions (Page 174)
1
a
b
Time for 1 clap, t = 60 s / 60 = 1 s,
distance travelled to wall and back
d = 2 × 160 m = 320 m
Then v = d/t = 320 m / 1 s = 320 m/s
t = 60 s / 80 = 0.75 s,
d = 2 × 120 m = 240 m
so v = d/t = 240 m / 0.75 s = 320 m/s
Cambridge O Level Physics 4th edition
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[1]
[1]
[1]
[1]
[1]
[1]
35
Cambridge O Level Physics answers
c
t = 60 s / 30 = 2 s,
d = v t = 320 m/s × 2 s = 640 m;
distance to wall = d/2 = 320 m
[1]
[2]
[1]
[Total: 10]
2 a
i
[2]
b
i
ii
v = f λ so λ = v/f = 340 m/s / 340 Hz = 1.0 m
λ = v/f = 340 m/s / 170 Hz = 2.0 m
3 a
i
4
5
ii [2]
[3]
[2]
[Total: 9]
Compressions occur where the molecules of the medium transmitting a longitudinal
[3]
sound wave are closer together than normal
ii Rarefactions occur where the molecules of the transmitting medium are
further apart than normal
[3]
b λ/2
[1]
c v = f λ so λ = v/f = 330 m/s / 220 Hz = 1.5 m
[3]
[Total: 10]
a i solid
[1]
ii gas
[1]
b Time, t, for a sound wave to be reflected from an underwater surface and
return to the transmitter is recorded.
[2]
Knowing the speed of sound in the water, v, the depth of the reflecting
object d can be calculated from the equation v = 2d/t
[2]
c Medical imaging; non-destructive testing of materials
[2]
[Total: 8]
a The number and strength of overtones (exact multiples of the fundamental frequency)
differ.
[1]
b Sketches similar to Figure 3.4.7 (page 170).
[2]
[Total: 6]
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Cambridge O Level Physics answers
Section 4 Electricity and Magnetism
4.1 Simple magnetism and magnetic fields
Test yourself questions
1
2
3
C
a
NSNS
b
SNNS
Weaker: the magnetic field lines are further apart
Practical work questions
1
2
3
4
5
Magnetic fields interact with each other
Figure 4.1.9a N–S opposite N–S; Figure 4.1.9b N–S opposite S–N
Current through the coil, number of turns on the coil
The compass needle will point in the direction of the field lines emerging from the North pole
of the current carrying coil
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Cambridge O Level Physics answers
Exam-style questions (Page 183)
1
a, b
c
2
Magnetic field line goes from the North to the South pole of the magnet.
[1] [2]
[1]
[Total: 4]
a
Place a bar magnet on a piece of paper and a plotting compass near the N pole.
Mark the position of the N and S poles of the compass on the paper;[1] then
move the compass so that the S pole is at the point where the N pole was
previously and mark the new position of the N pole.[1] Continue until
compass is near the S pole then join up the points to give a field line.[1] Plot
other field lines by repeating the process with the compass at different
starting points.[1]
b Electromagnets are used where the strength of the magnetic field needs to be
varied and turned on and off.[2] Permanent magnets do not require an
electricity supply and are used when the magnetic field does not need to be varied [2]
c Examples: compass, computer hard disk, electric motor or generator,
microphone, loudspeaker, credit and debit cards.
[2]
[Total: 10]
3 a Through the interaction of magnetic fields; a magnet will experience a force
in a magnetic field
[2]
b Near the north and south poles
[2]
c The magnetic field lines are
i closest together
ii furthest apart
[2]
[Total: 6]
4.2 Electrical quantities
Test yourself questions
1
2
3
4
5
6
7
8
D
Electrons are transferred from the cloth to the polythene
a attracted
b repelled
Radial field lines are perpendicular to the surface of the sphere; direction is towards the centre
of the sphere
a Ink-jet printer, photocopiers, paint and crop sprayers, flue ash precipitators…
b Damage to electronic equipment, explosion in presence of flammable vapour, lightning
strikes…
By the movement of free electrons
Place ammeter in series with its positive terminal connected to the positive terminal of the
supply
a I = Q / t = 10 C / 2 s = 5 A
b I = Q / t = 20 C / 40 s = 0.5 A
c I = Q / t = 240 C / 120 s = 2 A
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Cambridge O Level Physics answers
9 Rearranging Q = I t gives t = Q/I = 5 C / 2 A = 2.5 s
10 a
b
c
11 Time for one cycle = 1/1000 = 10–3 s
12 a Electromotive force (e.m.f.) is the electrical work done by a source in moving a unit
charge around a complete circuit
b Potential difference (p.d.) is the work done by a unit charge passing through a component
13 a W = Q V = 1 C × 12 V = 12 J
b W = Q V = 5 C × 12 V = 60 J
c W = Q V = I t V = 2 A × 10 s × 12 V = 240 J
14 R = V / I = 12 V / 4 A = 3 Ω
15 V = I R = 2 A × 10 Ω = 20 V
16 I = V / R = 6 V / 3 Ω = 2 A
17 I = V / R = 12 V / 4 Ω = 3 A; Q = I × t = 3 A × 1 s = 3 C
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Cambridge O Level Physics answers
18 a
b
19 a
Calculate the gradient of the graph; resistance = 1/gradient
b
20 a
b
c
21 a
b
Resistance of filament increases as it heats up
E = P t = 100 J/s × 1 s = 100 J
E = P t = 100 J/s × 5 s = 500 J
E = P t = 100 J/s × 60 s = 6000 J
P = IV = 2 A × 12 V = 24 W
P = IV = 0.5 A × 6 V = 3 J/s
Now put this into practice questions
(Page 190)
1 Q = I × t = 2 A × 20 s = 40 C
2 I = Q / t = 3 C / 7 s = 0.4 A
(Page 195)
1 p.d. across the lamp =W/Q =8 J / 4 C = 2 V
2 W = Q V = 2 C × 6 V = 12 J
3 Q = I t = 3A × 10 s = 30 C so W = Q V = 30 C × 6 V = 180 J
(Page 196)
1 a 0.2 V
(Page 198)
1
2
3
b
upper scale
c
parallax error introduced
R = 4.5 V / 0.3 A = 15 Ω
V = IR = 0.2 A × 10 Ω = 2 V
so I = 12.0 V / 24 Ω = 0.5 A
(Page 200)
1
2
R = 60 Ω × 2 = 120 Ω since the length of wire is doubled
A1 / A2 = (0.20 mm)2 / (0.40 mm)2 = 0.25
so R2 = R1 × A1 /A2 = 60 Ω × 0.25 = 15 Ω
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40
Cambridge O Level Physics answers
(Page 203)
1
a p.d. V = IR = 1.0 A × 12 Ω = 12 V
b P = IV = 1.0 A × 12 V = 12 W = 12 J/s
c P = E/t so E = Pt = 12 J/s × 10 s = 120 J
2 a P = IV = 0.3 A × 12 V = 3.6 W
b 3.6 J/s
c P = E/t so E = Pt = 3.6 J/s × 60 s = 216 J
(Page 205)
1
2
Electrical energy E = Pt = 6.4 kW × 2 h = 12.8 kWh
Cost of using the oven = 12.8 kWh × 10 cents = 128 cents
Electrical energy E = Pt = 0.150 kW × 12 h = 1.8 kWh
Cost of using the refrigerator = 1.8 kWh × 10 cents = 18 cents
Practical questions
1
2
3
4
5
6
7
8
9
10
11
Rub the polythene with a cloth
Draw the rod firmly across the edge of the metal cap of the electroscope
The charge on the leaf has the same sign as that on the metal plate so is repelled from the plate
By connecting the metal cap to earth
When the lamp lights
a one
b no
a two
b yes
Students’ own responses based on their results
Students’ own responses based on their results
4.5 V / 3 = 1.5 V
a Connect voltmeter in parallel with the device; select appropriate range for the
measurement
b 1.5 V
12 Students’ own responses based on their results
13 a
b
14
15
16
17
Coil of wire, ammeter, voltmeter, variable resistor/rheostat, battery/power supply, circuit board
R = V/I = 4.5 V / 0.15 A = 30 Ω
P = IV = 30 V × 0.5 A = 15 W
Po = mgh/t = 0.5 kg × 9.8 m/s2 × 0.8 m / 4 s = 0.98 W
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Cambridge O Level Physics answers
Exam-style questions (Page 206)
1
a
b
c
d
2
a
b
c
d
3
4
Rubbing with a cloth removes electrons from the cellulose acetate leaving it
positively charged.
repelled
attracted
2
Charge a gold-leaf electroscope and then touch the cap with the test material.
The gold leaf will fall quickly when the electroscope is discharged through a
good conductor and only slowly or not at all when discharged through a bad
conductor or insulator
Any metal or carbon
Plastics such as polythene and cellulose acetate, Perspex and nylon
Conductors have some free electrons which can move through the material;
the electrons in insulators are firmly bound to their atoms and are not free
to move.
a An electric field is a region in which an electric charge experiences a force
b
c
coulomb (C)
a
The direction of an electric field at a point is the direction of the force on a
positive charge
[3]
[1]
[1]
[1]
[Total: 6]
[4]
[2]
[1]
[3]
[Total: 10]
[2]
[4]
[1]
[Total: 7]
[3]
b
5
6
B
C
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[4]
[Total: 7]
[Total: 1]
[Total: 1]
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Cambridge O Level Physics answers
7
a
b
c
8
a
b
9
a
b
10 b
c
d
e
f
11 a
b
c
12 a
b
13 a
b
c
Electrons
[1]
In d.c. electrons flow in one direction only; in a.c. the direction of flow reverses
regularly
[2]
Connect the ammeter in series in the circuit with the + of the ammeter closest to
the + of the battery. For a digital ammeter, choose the d.c. setting. For either
analogue or digital ammeters select a suitable range for the size of current.
[3]
[Total: 6]
An electric current is the charge passing a point per unit time and is given by
the equation I = Q/t
[2]
i I = Q/t = 180 C / 60 s = 3 A [2]
ii Q = I × t = 2 A × 60 s = 120 C
[3]
[Total: 7]
i Q = I × t = 5 A × 10 s = 50 C
[2]
ii Q = It = 5 A × 5 × 60 s = 1500 C
[2]
Rearrange equation I = Q/t to give t = Q/I = 300 C / 5A = 60 s
[3]
[Total: 7]
Very bright
[1]
Normal brightness
[1]
No light
[1]
Brighter than normal
[1]
Normal brightness
[1]
[Total: 5]
V2 = V – V1 = (18 – 12) V = 6 V
[2]
W = QV1 = I × t ×V1 = 0.5 A × 60 s × 12 V = 360 J
[4]
ammeter + and voltmeters + should be connected to the point nearest
the + (left) of the battery
[4]
[Total: 10]
V = IR; graph is a straight line through the origin indicating V is proportional
to I
[4]
R = V / I = 6V / 3A = 2 Ω
[3]
[Total: 7]
The resistance R of a wire of a given material is directly proportional to its
length l (R ∝ l), and inversely proportional to its cross-sectional area A (R ∝ 1/A ).
Combining these two statements gives 𝑅𝑅 ∝
𝑙𝑙
𝐴𝐴
R2 = R1 × l2/l1 = 70 Ω × 0.2 m / 1.0 m = 14 Ω
A1/A2 = (1.0 mm)2 / (0.5 mm)2 = 4.0
R2 = R1 × A1/A2 = 70 Ω × 4 = 280 Ω
[3]
[3]
[4]
[Total: 10]
14
[Total: 9]
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Cambridge O Level Physics answers
15 a
b
16 a
b
E=Pt
= 6400 J/s × 30 × 60 s
= 1.15 × 107 J
80 minutes / 60 minutes = 4/3 hours;
cost = 3 kW × 1.33 hours × 10 cents = 40 cents
i 2 kW
ii 60 W
iii 850 W
P = I V so I = P / V = 920 W / 230 V = 4 A
[1]
[1]
[1]
[3]
[Total: 6]
[3]
[4]
[Total: 7]
Alternative to practical questions (Page 208)
17 a
b
R = V/I
[1]
Connect the wire to an ammeter, rheostat and battery in series; connect a
voltmeter across the wire. Measure the current and voltage for different settings
of the rheostat. Calculate R from the equation R = V/I or plot a graph of V
versus I and determine R from the gradient of the graph
[4]
c i Straight line graph
[3]
ii Gradient = 12 V / 0.24 A = 50 Ω = R
[2]
[Total: 10]
18 Diagram: battery, ammeter and test wire connected in series and voltmeter connected across
test wire, with correct symbols used.
[2]
Procedure: measure and record current and voltmeter reading for different lengths of wire.[1]
Control variables: wire material, diameter of wire, temperature.
[1]
Table headings: length of wire/cm, I/A, V/V, R/ohms.
[2]
[Total: 6]
4.3 Electric circuits
Test yourself questions
1
2
3
All read 0.25 A (I1 = I2 + I4 and I2 = I3 = I1 / 2)
p.d. = 3 × 2 V = 6 V
a W=Q×V=1C×2V=2J
b W=Q×V=1C×3×2V=6J
4 a
5 a
6
R = R1 + R2 + R3
or
b the same
b larger
a
Cambridge O Level Physics 4th edition
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Cambridge O Level Physics answers
b
7
8
9
From the potential divider equation V1 / V2 = R1 / R2 = 12 Ω / 36 Ω = 1/3
Ratio of voltages is 1 : 3
c Dividing the supply voltage in the ratio 1 : 3 gives
V1 = 1 × 20 V / 4 = 5 V and V2 = 3 × 20 V / 4 = 15 V
A thermistor B lamp
C LDR
D cell
E resistor
A LED
B semiconductor diode
C relay
D variable resistor
Now put this into practice questions
(Page 212)
1
2
R = R1 + R2 + R3 = 4 Ω + 6 Ω + 8 Ω = 18 Ω
a I = V/R = 4.5 V / (3 Ω + 6 Ω) = 4.5 V / 9 Ω = 0.5 A
b V1 = IR1 = 0.5 A × 3 Ω = 1.5 V, V2 = IR2 = 0.5 A × 6 Ω = 3.0 V
(Page 214)
1
a R = 1 Ω + 2 Ω +3 Ω = 6 Ω
b current I = V/R = 12 V / 6 Ω = 2 A (same current in each resistor)
c V1 = IR1 = 2 A × 1 Ω = 2 V, V2 = IR2 = 2 A × 2 Ω = 4 V, V3 = IR3 = 2 A × 3 Ω = 6 V
2 a 1/Ra = 1 / (2 Ω) + 1 / (3 Ω) = 5 / (6 Ω) so Ra = 6 / 5 Ω
b 12 V, 12 V
c i
I = V/R = 12 V / 2 Ω = 6 A
ii I = V/R = 12 V / 3 Ω = 4 A
(Page 216)
1
2
V1 / V2 = R1 / R2
a From the potential divider equation V1 / V2 = R1 / R2 = 9 Ω / 6 Ω = 3/2
Ratio of voltages is 3 : 2
b Dividing the supply voltage in the ratio 3 : 2 gives
V1 = 3 × 30 V / 5 = 18 V and V2 = 2 × 30 V / 5 = 12 V
Exam-style questions (Page 220)
1
x = V1 + V2 = 12 + 6 = 18
y = V – V1 = 6 – 4 = 2
z = V – V2 = 12 – 4 = 8
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[2]
[2]
[2]
[Total: 6]
45
Cambridge O Level Physics answers
2
Resistors in series: Rtotal = R1 + R2 + R3 + R4 = 4R1
I = V / R = 12 / (4 R1) = 3 / R1;
VA = IR1 = 3 × R1 / R1 = 3 V;
VB = IR2 = 3 × R2 / R1 = 3 V;
VC = I(R3 + R4) = 3 × 2 × R1 / R1 = 6 V
3
a
b
4
5
6
[6]
[Total: 10]
a i R = (6 + 7 + 8) Ω = 21 Ω
[2]
ii Increased
[1]
b The p.d. across each lamp is fixed (at the supply p.d.), so the lamp shines with
the same brightness irrespective of how many other lamps are switched on.
Each lamp can be turned on and off independently; if one lamp fails, the others
can still be operated.
[4]
c Less
[1]
[Total: 8]
[2]
a V 1 = V2 = 3 V
b I = V/R = 6 V / (10 + 50) kΩ = 0.1 mA
so V1 = IR1 = 0.1 mA × 10 kΩ = 1 V and V2 = IR2 = 0.1mA × 50 kΩ = 5 V
[4]
c I = V/R = 6 V / (20 + 10) kΩ = 0.2 mA
so V1 = IR1 = 0.2 mA × 20 kΩ = 4 V and V2 = IR2 = 0.2 mA × 10 kΩ = 2 V
[4]
[Total: 10]
a LDR and R in series with a battery
[2]
b
7
Resistors in parallel: 1/R = 1/R1 + 1/R2 = 1/4 + 1/4 = 1/2 Ω–1; so R = 2 Ω
For resistors in parallel: 1/R// = 1/R1 + 1/R2 = 1/6 + 1/2 = 4/6 Ω–1
so R// = 6/4 =1.5 Ω; then combined resistance = (1.5 + 6) Ω = 7.5 Ω
[1]
[1]
[2]
[2]
[2]
[Total: 8]
[4]
c
i I = V/Rtotal = 12 V / (20 + 28) Ω = 0.25 A
ii V = IR = 0.25 A × 20 Ω = 5.0 V
iii V = IR = 0.25 A × 28 Ω = 7.0 V
i decreases ii increases iii increases
b
c
d
L1 lights, L2 does not; no current flows through D2 as it is reverse biased
L1 and L2 light; D forward biased and current splits between L2 and D
L1 lights, L2 does not; D2 reverse biased
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[2]
[2]
[2]
[3]
[Total: 11]
[3]
[4]
[3]
[Total: 10]
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Cambridge O Level Physics answers
4.4 Practical electricity
Test yourself questions
1
2
3
C: P = IV = 5 A × 230 V = 1150 W; number of bulbs = 1150 W / 100 W = 11
P = IV = 13 A × 230 V = 2990 W = 2.99 kW
a The metal case of the appliance
b Double insulation; enclose all metal parts in an non-conducting plastic case
Now put this into practice questions (Page 227)
1
2
3
a
a
a
I = P/V = 1500W / 240 V = 6.3 A b
I = P/V = 100 W / 240 V = 0.42 A b
I = P/V = 6400 W / 240 V = 27 A b
13 A fuse should be used
3 A fuse should be used
30 A trip switch setting should be chosen
Exam style questions (Page 228)
1
a
b
c
d
2
3
Overheated cables, damaged insulation, overloading circuit by connecting
too many appliances
Damp conditions, faulty wiring, damaged insulation
Disconnect appliance from electricity supply; look for any signs of a short
circuit; check correct size of fuse is being used
To prevent the user touching metal parts which could become live if a fault
developed in the appliance
[2]
[2]
[3]
[3]
[Total: 10]
a
It protects the circuit by ensuring the current-carrying capacity of the wiring
is not exceeded.
[2]
b In a, the fuse is in the live wire, so when it blows and breaks the circuit, the
lamp is disconnected from the live wire. In b, the fuse is in the neutral wire and
when it breaks, the lamp is still connected to the live wire
[4]
[Total: 6]
a i I = V/R = 240 V / 800 Ω = 0.3 A = 300 mA
[2]
ii Yes
[1]
iii By having dry hands or wearing shoes with rubber (insulating) soles
[2]
b i P = IV so I = P/V = 150 W / 230 V = 0.65 A; use a 3A fuse
[2]
ii I = P/V = 900 W / 230 V = 3.9 A; use a 13 A fuse
[2]
iii I = P/V = 2000 W / 230 V = 8.7 A; use a 13 A fuse
[2]
[Total: 11]
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Cambridge O Level Physics answers
4.5 Electromagnetic effects
Test yourself questions
1
D; the effect is called electromagnetic induction
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
Slip rings (rotate with the coil)
a
b coil
i horizontal
ii vertical
a North b East
S
a Inside the solenoid
b The current; the number of turns on the solenoid
When a large current is to be controlled by a small current
The pull-on current I = V/R = 15 V / 300 Ω = 0.05 A = 50 mA
R = V/I = 12 V / 0.06 A = 200 Ω
Sound waves are produced by a paper cone attached to a coil vibrating with the a.c. signal
D (use Fleming’s left-hand rule)
Decrease, because force on the electrons is greater
a Increases
b Reverses direction
Clockwise (use Fleming’s left-hand rule)
They act to reverse the current through the coil every half turn so that the coil continues
rotating in the same direction
B
B: Ns/Np = Vs/Vp so Ns = Np×Vs/Vp = 1000 × 46 / 230 = 200
Now put this into practice questions
(Page 245)
1
a
b
Np/Ns = Vp/Vs = 240 V / 12 V = 20 / 1
Np/Ns = 20 so Np = 20 Ns = 20 × 80 = 1600 turns
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Cambridge O Level Physics answers
2
a
b
Turns ratio = Np/Ns = Vp/Vs = 240 V/ 960 V = 1 / 4
Np/Ns = 1/4 so Ns = 4 Np = 4 × 500 = 2000 turns
Practical work questions
1
2
3
4
5
6
When there is relative motion between the coil and the magnet
Related by the equation V = IR
Strength of magnet, number of turns on the coil, current through the coil
Direction of rotation reversed
No; there is no changing magnetic field to induce an e.m.f in the secondary coil
Decrease
Exam-style questions (Page 247)
1
a
b
2
3
Connect a coil to a sensitive centre-zero ammeter in a complete circuit. Move
a bar magnet towards and away from the coil; the meter shows a current is
induced. Alternatively move the coil instead of the magnet
Increase the strength of the magnet, the speed of the motion, the number of
turns on the coil
a
The galvanometer needle swings alternately in one direction and then the
other as the rod vibrates
b This is due to an e.m.f. being induced in the metal rod when it cuts the
magnetic field lines; current flows in alternate directions round the circuit as
the rod moves in and out of the magnetic field
a
b
A: slip rings,
B: brushes; slip rings connect the rotating coil to the brushes
Increase the number of turns on the coil, the strength of the magnet and the
speed of rotation of the coil.
[4]
[3]
[Total: 7]
[2]
[4]
[Total: 6]
[1]
[2]
[3]
c
[4]
[Total: 10]
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Cambridge O Level Physics answers
4
5
b
The direction of the magnetic field reverses
a
b
To complete the circuits to the negative terminal of the battery
A contains the relay contacts and starter;
B contains the starter switch and relay coil
Wire A carries a much larger current to the starter motor than wire B
It allows the wires to the starter switch to be thin since they carry only the
small current needed to energise the relay
c
d
6
7
A small coil placed between the poles of a magnet receives a varying current;
the magnetic fields of the coil and the magnet interact causing the coil to vibrate
with the frequency of the a.c. signal
sound is produced by the vibration of a paper cone attached to the coil
a
b
8
9
[4]
a
[1]
[Total: 5]
[2]
[2]
[1]
[2]
[Total: 7]
[2]
[3]
[1]
[Total: 6]
Support a wire horizontally between the poles of a magnet so that the
direction of the wire is perpendicular to a horizontal magnetic field;
the wire moves up or down when a current is passed through the wire;
if the direction of the current or of the magnetic field is reversed, the wire
moves in the opposite direction
i
increases [1]
ii decreases [1]
iii increases [1]
[2]
[Total: 8]
[Total: 1]
[3]
A
a
b
[2]
[1]
The commutator rotates with the coil, so that the current through the coil
reverses direction every half-turn
The forces on the coil then always act in the same direction, producing
continuous rotation
If the current is as shown in the figure, Fleming’s left-hand rule gives an
upward force on ab and a downward force on cd, producing clockwise rotation
Cambridge O Level Physics 4th edition
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[1]
[1]
[1]
50
Cambridge O Level Physics answers
c
Reversal of the battery connections would give anticlockwise rotation of the coil
[1]
i The motor would rotate in the opposite direction
[1]
ii The motor would rotate in the opposite direction
[1]
iii No change in the direction of rotation
[1]
[Total: 10]
10 a
b
c
It is made up of two coils of wire, a primary and a secondary, wound on a
complete soft iron core.
It changes an a.c. voltage to an a.c. voltage of greater value
Np/Ns = Vp/Vs so Ns = Np Vs/ Vp = 120 × 720 / 240 = 360 turns
11 a Ns/Np = Vs/Vp so Ns = Np × Vs /Vp = 460 × 12 V / 230 V = 24
*b Ip × Vp = Is × Vs so Is = Ip × Vp /Vs = 0.10 A × 230 V / 12 V = 1.9 A
12 a
b
c
13 a
b
[4]
[2]
[4]
[Total: 10]
[4]
[4]
[Total: 8]
When the switch is first closed, current will flow in coil A and a magnetic
field will build up in A. Coil B will be cut by a changing magnetic field, a p.d.
will be induced and a current will flow through the galvanometer deflecting
the needle.
[2]
Once the current and magnetic field become constant in coil A there will be
no changing magnetic field linking it with coil B, no induced p.d. and the
galvanometer reading will return to zero; this happens almost immediately
after the switch is closed.
[2]
When the switch is opened again the magnetic field in coil A falls, there is
again a changing magnetic field linking it to coil B and a p.d. is induced in
the opposite direction to previously; the galvanometer needle will swing briefly
in the opposite direction before returning to zero again
[3]
The deflection on the galvanometer would increase; the soft iron wires will
become magnetised by the coil and will increase the magnetic field linking
coil A with coil B. When the switch is closed or opened the induced p.d. will
be larger
[3]
The deflection on the galvanometer would increase
[1]
[Total: 11]
Transformers step a.c. voltages up or down efficiently; p.d.s are stepped up at
the power station before transmission and stepped down at sub-stations for
local distribution.
[3]
Energy lost as heat in cables is reduced; smaller currents result which allow
thinner, cheaper wires to be used.
[2]
[Total: 5]
4.6 Uses of an oscilloscope
Test yourself questions
1
2
3
a Voltage = 3.0 div × 0.5 V/div = 1.5 V
b d.c. voltage = 4.2 div × 2.0 V/div = 8.4 V
Amplitude = (8.6 div × 1.0 V/div)/2 = 4.3 V
Time for one complete wave, T = 4.0 div × 0.001 s/div = 0.004 s;
Frequency = 1/T = 1 / 0.004 s = 250 Hz
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Cambridge O Level Physics answers
Exam-style question (Page 252)
1 a
b
c
i
Timebase – controls the time interval displayed on the horizontal axis of the
oscilloscope display.
[2]
ii Y-gain amplifies the input voltage to make it large enough to appear a suitable
size on the screen.
[2]
Timebase
[1]
0.28 V
[2]
[Total: 7]
Section 5 Nuclear physics
5.1 Nuclear physics
Test yourself questions
1
2
3
4
5
6
7
8
9
Positive charge and most of the mass are concentrated in a small dense nucleus; electrons with
remaining mass carry the negative charge and orbit around the nucleus
Removal of an orbital electron from the atom
B
a i 7
ii 3
iii (7 – 3) = 4
iv 3
b
a Z = 11, A = 23
b 11
+2
a 37
b mass of nucleus = total mass of the nucleons
a Very high temperature needed so that the nuclei collide at speeds high enough to overcome
the repulsive electrostatic forces between them
b Large amounts of energy are released by the conversion of mass into energy
a Fission
b Fusion
Now put this into practice questions (Page 258)
1
2
Carbon-12:
a Number of protons Z = 6
b number of nucleons A =12
c number of neutrons = A – Z = 6
Carbon-14:
a Number of protons Z = 6
b number of nucleons A =14
c number of neutrons = A – Z = 8
A different
B same
C different
Exam-style questions (Page 261)
1
a
b
c
i
Z is the number of protons in the nucleus of an atom
ii A is the total number of protons and neutrons in the nucleus
Isotopes of an element have the same number of protons in the nucleus but
different number of neutrons.
In
there is one less neutron in the nucleus than in
[2]
[2]
[2]
[Total: 6]
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Cambridge O Level Physics answers
2
3
a
b
c
A = 40, Z = 20
i 20
ii 40
Positively charged.
a
The large deflection of a few alpha particles showed that most of the mass of
the nucleus was concentrated at its centre and had a positive charge
Most of the alpha particles were undeflected showing that the central nucleus
was very small compared with the size of the whole atom
i +1
ii 0
iii +2
b
4
a
b
iii 20
[2]
[4]
[1]
[Total: 7]
iv 20
[3]
[2]
[3]
[Total: 8]
i Nuclear fission is the break-up of a nucleus into smaller fragments
[2]
ii In nuclear fusion light nuclei join up to form heavier nuclei
[2]
A chain reaction is set up by the fission of uranium-235. Uranium rods are inserted into the
graphite core (moderator) of the reactor. Control rods inserted into the core absorb neutrons
and control the rate of the chain reaction. A coolant is circulated in the core and transfers
thermal energy to water in a heat exchanger leading to the production of steam that is used
to drive a turbine.
[5]
[Total: 8]
5.2 Radioactivity
Test yourself questions
1
Any two from: cosmic rays, radon gas, rocks/buildings, food/drink, radioisotopes used in
medicine
2 Ionising effect
*3 a 300 counts / 60 s = 5 counts/s
b (190 – 5) counts/s = 185 counts/s
4 a α
b γ
c α
d β–
*5 a β
b γ
c α
d γ
6 a Corrected count rates for background radiation
Time/s
Corrected counts/s
0
30
60
90
120
150
180
160
107
72
48
32
22
15
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53
Cambridge O Level Physics answers
b
half-life = 52 s
7 D: 8 minutes = 4 half-lives so activity falls by 1/24 = 1/16
8 a Gamma rays b Alpha-particles
9
If a radioisotope which emits gamma rays is placed on one side of a moving sheet of
material and a GM tube on the other, the count rate increases if the thickness of the sheet
decreases.
Now put this into practice questions
(Page 268)
+

1
2

+
(Page 270)
1
a
b
2
3
60 minute / 15 minutes = 4; after 4 half-lives fraction left = 1/2 × 1/2 × 1/2 × 1/2 = 1/16
After 1 half-life count rate =140 / 2 = 70 counts/minute, after 2 half-lives count rate = 70 / 2 =
35 counts/minute so 60 minutes = 2 half-lives. Half-life of the material = 60 / 2 = 30 minutes
After 1 × 5700 years the activity will be 80 / 2 = 40 counts per minute;
after 2 × 5700 years the activity will be 40 / 2 = 20 counts per minute.
Estimated age of the canoe is 2 × 5700 = 11 400 years
4
The count rate drops by half between 15 and 20 minutes
Half-life = 17 min
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Cambridge O Level Physics answers
Exam-style questions (Page 275)
1
a
b
B
i C
ii A
iii B
[3]
[1]
[1]
[1]
iv A
[1]
[Total: 7]
2
a
b
3
4
a
The half-life of a radioisotope is the time taken for half the nuclei of that
isotope in a particular sample to decay
2 minutes; count rate falls by a half every 2 minutes
i
ii
iii
iv
b
After 1 × 1500 million years there will be N/2 atoms of argon left
Number of potassium atoms formed = N − (N/2) = N/2
Ar : K ratio = N/2 : N/2 = 1 : 1
After 2 × 1500 = 3000 million years, there would be
(N/2)/2 = N/4 argon atoms left
v and N − (N/4) = 3N/4 potassium atoms formed
vi Ar : K ratio of N/4 : 3N/4 = 1 : 3
Measured ratio is 1 : 3 so the rock must be about 3000 million years old
a
i
C
b
ii
i
220
86Rn
ii
5
a
B
234
90Th
4
→ 218
84Po + 2He
0
→ 234
91Pa + −1e
[2]
[4]
[Total: 6]
[1]
[1]
[1]
[1]
[1]
[1]
[1]
[Total: 7]
[1]
[3]
[1]
[3]
[Total: 8]
Can damage living cells and tissue leading to cell death, gene mutation,
cancer, eye problems, radiation burns and sickness
b i alpha
ii gamma
iii gamma
iv alpha
c Two from: reduce exposure time, increase distance between source and
person/handle source with forceps, keep away from eyes, shield source, store
source in a lead box when not in use
[2]
[4]
[2]
[Total: 8]
6
a
b
A small amount of an α-particle emitter, such as Americium-241, causes
ionisation of the air in an ionisation chamber and results in a current flow
between two metal electrodes. When smoke enters the detector, it impedes the
flow of ions and the current reduces. The fall in current is detected
electronically and an alarm activated
An α-particle emitter is chosen because α-particles have only a short range in
air; Americium-241 has a long half-life so its activity remains fairly constant
over time
A radioisotope is placed on one side of a moving sheet of material and a GM
tube on the opposite side; the count rate decreases when the thickness increases
β-emitters are suitable for thin sheets, but γ-rays are needed for thicker sheets
because of their penetrating power; a long half-life source is preferred so that
the activity of the source remains fairly constant over time
Cambridge O Level Physics 4th edition
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[3]
[1]
[2]
[1]
55
Cambridge O Level Physics answers
c
Gamma rays kill bacteria on the food without damaging the food itself
This leads to a longer shelf-life for the product
A γ-rays source is used for its high penetrating power; a long half-life is
preferred so that the activity stays fairly constant over time
[1]
[1]
[1]
[Total: 10]
Section 6 Space physics
6.1 Earth and Solar System
Test yourself questions
1
2
3
4
5
6
D
v = 2 π r / T = 2 π 380 000 km / (27 × 24 × 60 × 60) s = 1.02 km/s
Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto
C
A true B true C true
a Circumference = 2 π r = 2 π × 228 × 106 km = 1.43 × 109 km
b v = 2 π r / T = 1.43 × 109 km / (687 × 24 × 60 × 60) s = 24 km/s
7 a 23.1 N/kg
b weight = mgJ = 50 kg × 23.1 N/kg = 1155 N
8 Higher; Jupiter is nearer the Sun than Saturn
Now put this into practice questions
(Page 282)
1
2
t = distance / speed = 5.8 × 1010 m / 3 × 108 m/s = 193 s
t = distance / speed = 5.9 × 1012 m / 3 × 108 m/s = 2.0 × 104 s (over 5 hours)
Exam-style questions (Page 286)
1
2
3
D
Time = distance / speed of light = 228 × 109 m / 3 × 108 m/s = 760 s
a 2πR
= 2 π × 58 × 106 km
= 3.64 × 108 km
b T = 88 × 24 × 60 × 60
= 7.6 × 106 s
v = 2 π R / T = 48 km/s
4
a
b
i
i
ii
increase
ii increase
Greater as Venus is nearer the Sun
Less, as circumference of orbit is less and it travels faster
5
a i Lower; Jupiter is further from the Sun so receives less radiation/m2
ii Larger
iii Lower
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[Total: 1]
[Total: 3]
[3]
[4]
[Total: 7]
[3]
[2]
[3]
[Total: 8]
[2]
[1]
[1]
56
Cambridge O Level Physics answers
b
i
ii
6
a
b
c
v = distance / time = 2 π r / T so vM/vE = (2 π RM / TM) / (2 π RE / TE)
= (RM × TE) / (RE × TM) = 1.5 / 1.9 = 0.78
Slower
Circumference = 2 π r
Speed, v = distance / time so T = 2 π r / v
T = 2 π r / v = 2 × π × 6400 × 103 m / 8000 m/s = 5000 s (84 min)
[3]
[1]
[Total: 8]
[1]
[2]
[4]
[Total: 7]
6.2 Stars and the Universe
Test yourself questions
1
2
3
4
5
6
Infrared, visible light and ultraviolet
Nuclear fusion of hydrogen into helium
D
D
B
D
Exam-style questions (Page 294)
1
a
b
2
3
C
a
b
4
a
b
Light from glowing hydrogen and other gases in stars in distant galaxies has a
longer wavelength than it does on Earth – the light is ‘shifted’ towards the red
end of the spectrum
Distant galaxies are moving away from us; the further away they are the faster
the speed of recession
Clouds of hydrogen gas collapse due to gravitational attraction and form a
protostar
As the protostar grows in size it becomes hotter and when the temperature is
high enough in the core, nuclear fusion of hydrogen into helium starts;
large amounts of energy are released which sustain the fusion process. The
protostar has then become a star, powered by nuclear fusion
When most of the hydrogen in the core has been used up in the nuclear fusion
of hydrogen into helium
Gravity acts inwards; thermal pressure, due to the high temperature of the
star, acts outwards
Low mass star → red giant → planetary nebula → white dwarf → black dwarf
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[3]
[2]
[Total: 5]
[Total: 1]
[1]
[2]
[2]
[2]
[Total: 7]
[4]
[5]
[Total: 9]
57
Cambridge O Level Physics answers
Mathematics for physics
1
2
3
4
5
6
a
b
c
d
e
f
g
h
i
a
b
c
d
e
f
g
h
a
3
5
8/3
20
12
6
2
3
8
f = v/λ
λ = v/f
I = V/R
R = V/I
m=d×V
V = m/d
s = vt
t = s/v
I2 = P/R
b
c
d
e
f
g
h
a
b
c
d
e
I = √(P/R)
f
3 × 108
a
b
c
2.0 × 105
10
8
d
2.0 × 108
e
f
a
b
c
d
20
300
4
2
5
8
a = 2s/t2
t2 = 2s/a
t = √(2s/a)
v = √(2gh)
y = Dλ/a
ρ = AR/l
10
34
2/3
1/10
10
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Cambridge O Level Physics answers
e 2/3
f –3/4
g 13/6
h –16
i 1
7 a = (v – u)/t
a 5
b 60
c 75
8 a = (v2 – u2)/2s
9 a Students’ graphs of extension against mass
b Extension ∝ mass because the graph is a straight line through the origin
10 a Students’ graphs of m against v
b No: graph is a straight line but does not pass through the origin
c 32
11 a Graph is a curve
b
Graph is a straight line through the origin, therefore s ∝ t2 or s/t2 = a constant = 2
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Cambridge O Level Physics answers
Additional exam-style questions
Motion forces and energy
1
2
3
4
5
6
7
8
D
a
b
a
b
c
A
a
b
c
D
D
a
b
c
d
a
Average speed = s/t = 1600 m / (8 × 60) s = 3.3 m/s
Increases
mg = 90 kg × 9.8 m/s2 = 880 N
90 kg
mg = 90 kg × 9.8 m/s2 / 6 = 147 N
Yes, 1 mm = 0.001 m
kg/m3
Density = m/V = 120 g / 15 cm3 = 8.0 g/cm3 = 8000 kg/m3
W = F d, where d is the distance moved by the force F in the direction of the force
joules, J
W = F d = 3 N × 5 m = 15 N m = 15 J
15 J
9
i Electrical to heat and light
ii Chemical to electrical
iii Electrical to kinetic energy
iv Kinetic to electrical
b Efficiency is the ratio of the useful energy output to the total energy input expressed as a
percentage
10 D
11 a Ep = mgΔh = 2 kg × 9.8 m/s2 × 4 m = 78 J
b i
Ek = mv2/2; rearrange to v2 = 2Ek/m = 2 × 100 J / 2 kg = 100 m2/s2 and v = 10 m/s
ii Ek = Ep = mgΔh; rearrange to Δh = Ek/mg = 100 J / (2 kg × 9.8 m/s2) = 5.1 m
12 a W = F d = Ek; rearrange to give d = Ek /F = 10 J / 5 N = 2 m
b F = ma = 6 kg × 3 m/s2 = 18 N
c i Impulse = FΔt = 18 N × 3 s = 54 N s
ii Δp = FΔt = 54 Ns
Thermal physics
13 a
Consists of equal lengths of two different metals of different expansivity welded together.
When heated, one of the metals expands more than the other and the strip bends. It can be
used in many applications from thermostats to fire alarms
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Cambridge O Level Physics answers
b
i pV = constant
ii p1V1 = p2V2 so p2 = p1V1 / V2 = p1 / 2; pressure halves
c When the temperature of a gas increases the average speed of its molecules increase. If the
volume of the gas remains constant, there will be more frequent collisions with the walls of
the container and the pressure of the gas will increase
14 a Wood is a less good conductor of heat than metal, which conducts heat away from the
hand
b When a fluid is heated it expands and becomes less dense. Parts that are warm will rise
above colder denser regions leading to a convection current being set up in the fluid
c Conduction and convection require a medium to transfer thermal energy.
Waves
15 a
b
16 a
b
c
17 C
18 a
b
c
19 a
b
20 a
b
i Longitudinal
ii Compression
iii Rarefaction
i Become circular
ii No change
iii No change
60°
30°
The image is:
as far behind the mirror as the object is in front with the line joining the
object and image being perpendicular to the mirror;
the same size as the object; virtual; laterally inverted
Dispersion
i Red
ii Violet
Violet, indigo, blue, green, yellow, orange, red
i The line through the centre of a lens at right angles to the lens
ii The point on the principal axis of the lens to which a parallel beam of light passing
through the lens converges
i Less than the focal length of the lens
ii Upright
iii Virtual
i Infrared
ii X-rays
i Radio
ii γ-rays
Transverse
Communications, microwave ovens for cooking
c
d
21 A
22 D
23 a Two from: electromagnetic (light, radio …), water, seismic S-waves,
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Cambridge O Level Physics answers
b Two from: sound, mechanical waves on a spring, seismic P-waves
c Three from: reflected, refracted, diffracted, obey the wave equation, carry energy from
place to place
Electricity and magnetism
24 C
25 a
b
c
26a
b
c
R = R1 + R2 = 2 Ω + 1 Ω = 3 Ω
V = IR so I = V/R = 6 V / 3 Ω = 2 A (same in each resistor)
V = IR: V1 = 2 A × 1 Ω = 2 V; V2 = 2 A × 2 Ω = 4 V
resistors in parallel: 1/R = 1/R1 + 1/R2 = 1/2 + 1/2 = 1 Ω–1; so R = 1 Ω
V = IR so I = V/R = 6 V / 2 Ω = 3 A (same in each resistor)
6V
27 a
b
i P = IV, rearrange to I = P/V = 3000 W / 230 V = 13 A
Number of kWh = 3 × (100 × 10–3) × 10 = 3
Total cost = 3 ×10 cents = 30 cents
ii
D
28 B
29 a
b
c
d
e
coulomb, C
ampere, A
volt, V
joule, J
watt, W
Nuclear physics
30 a
b
After 1 half-life count rate will be 50 counts/s; after 2 half-lives count rate will be 25
counts/s. Time for two half-lives = 2 × 1 hour = 2 hours
i alpha-particles < beta-particles < gamma rays
ii alpha-particles > beta-particles > gamma rays
Space physics
31 a
b
c
d
32 a
b
33 a
b
West
At noon, around June 21
On or near to Dec 22
March and September
distance = 2 π r = 2 × π × 385 000 km = 2.42 × 106 km
average speed = distance / time so, time = distance / speed = 2.42 × 106 km / 1 km/s
= 2.42 × 106 s. Time in days = 2.42 × 106 s / (60 × 60 × 24) = 28 days.
2.85 × 1014 km / (9.5 ×1012) km/ly = 30 light years
i 100 000 light years
ii 500 s
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Cambridge O Level Physics answers
35 a
b
Redshift is the increase in the observed wavelength of electromagnetic radiation emitted
from receding galaxies.
The further away a galaxy is, the greater the redshift, which indicates that the speed away
from the Earth is greater. This suggests that the Universe is expanding and originated in a
Big Bang.
Theory past paper questions
Motion, forces and energy
1
a
b
2
a
b
c
3
a
b
c
d
e
4
a
b
Measure distance between two points with a tape measure; time how long it takes
cyclist to travel between the two points with a stopwatch;
average speed = total distance travelled / total time taken
[3]
i Speed decreases
[1]
2
ii Δv = at =1.8 m/s × 2.5 s = 4.5 m/s
[2]
[Total: 6]
i 58 km
[1]
ii 48 minutes
[1]
The gradient is steeper
[1]
Average speed = total distance travelled/total time taken
= 58 km / 4.0 hours = 14.5 km/hour
(= 58 000 m / (4.0 × 60 × 60) s = 4.0 m/s)
[2]
[Total: 5]
The extension is no longer proportional to the load.
[2]
i Mass is a measure of the quantity of matter in an object at rest relative to the
observer
[1]
ii Weight is a gravitational force acting on an object that has mass
[2]
i Weight = mg = 0.70 kg × 10 N/kg = 7.0 N
[1]
ii Load / extension = 7.0 N / (0.97 – 0.80) m = 41 N/m (=k )
[1]
i Load = k × extension = 41 N/m × (1.70 – 0.80) m = 37 N
Mass in can = (37 – 7)N / 10 N/kg = 3.0 kg
[3]
ii Straight line drawn from 0.80 m on length axis to the point marked limit of
proportionality, then the gradient of the graph gradually increases
[3]
Strain energy or elastic potential energy  kinetic or gravitational potential energy [2]
[Total: 15]
i C
ii Each increase in load produces the same extension of the spring except for spring C
where the extension changes from (11.3 – 10.5) cm = 0.8 cm at low loads to
(16.9 −13.1) cm = 3.8 cm at high loads
[2]
iii Length of spring = (6.1 – 1.6) cm
= 4.5 cm (since extension/N = (7.7 – 6.1) = 1.6 cm)
[1]
Plot a weight–extension graph for spring A; measure extension of spring when rock
attached, and read off weight for extension produced by rock; divide weight by g to find
the mass of the rock
[2]
[Total: 5]
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Cambridge O Level Physics answers
5
a
b
6
a
b
7
a
b
c
8
a
b
c
d
e
9
a
b
i No resultant force
ii Box must move along the floor
iii Box moves at constant speed
Resultant force = 100 N – 85 N = 15 N;
acceleration a = resultant force / m = 15 N / 25 kg = 0.60 m/s2
[1]
[1]
[1]
[3]
[Total: 6]
[2]
F = ma = 35 kg × 2.6 m/s2 = 91 N
i Straight line sloping from origin to a point on t1 above time axis then
slope gradually decreases before becoming a horizontal line parallel to time axis. [2]
ii Calculate area under speed–time graph
[1]
[Total: 5]
2
i F = ma = 160 kg × 0.35 m/s = 56 N
[2]
2
[2]
ii v = at = 0.35 m/s × 1.2 s = 0.42 m/s
i Straight line from origin to a point above the time axis where t =1.2 s followed by a
straight horizontal line to t = 3.0 s.
[2]
ii Calculate the area between the line of the speed–time graph and the x-axis
[2]
i 1. Arrow towards centre of Earth
[1]
2. The gravitational attraction of the Earth
[2]
ii Change in displacement per unit time
[2]
iii 1. Changes; the direction changes
[1]
2. Unchanged; speed remains constant
[1]
[Total: 15]
Matter; motion / rest
[2]
i The region in which a mass experiences a force due to gravitational attraction [1]
ii Mass of water = ρV = 1000 kg/m3 × 2.4 × 10−2 m3 = 24 kg
Mass of water + bucket = 24 kg + 1.0 kg = 25 kg
Weight = 25 kg × 10 N/kg = 250 N
[3]
i Moment = Fd = 250 N × 0.12 m = 30 Nm
[2]
ii Force = 30 Nm / 0.40 m = 75 N
[1]
i Friction between axle and frame in windings; mass/weight of rope neglected;
rope winds unevenly on cylinder
[2]
ii Chemical energy  gravitational potential energy and thermal energy
[3]
Upward and downward forces are equal
[1]
[Total: 15]
W = mg = 26 kg × 10 N/kg = 260 N
[1]
i When an object is in equilibrium the sum of the clockwise moments about any point
is equal to the sum of the anticlockwise moments about the same point. There is no
resultant moment on an object that is in equilibrium.
[2]
ii Taking moments about the hinge:
clockwise moment 260 N × 0.35 m = 91 N m
anticlockwise moment = F × 0.65 m
Equating moments gives total upward force F = 91 N m/0.65 m = 140 N
[3]
[Total: 6]
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64
Cambridge O Level Physics answers
10 a
b
c
Work done = force × by the distance moved in the direction of the force
ΔEp = mgΔh = 0.35 kg × 10 N/kg × 6.0 m = 21 J
Friction between rope and pole opposes the motion; weight of rope neglected,
kinetic energy is gained by flag
11 Use a stopwatch to time 10 complete oscillations and divide by 10 to obtain an
average value.
b i ΔE = mgΔh so m = ΔE/gΔh = 240 J /10 N/kg × 0.60 m = 40 kg
ii 1. Friction at supports
2. Air resistance
12 a
b
13 a
b
[2]
[2]
[2]
[Total: 6]
[2]
[2]
[2]
[Total: 6]
7
i E = Pt = 75 J/s × (63 × 60 × 60) s = 1.7 × 10 J
[3]
7
6
[1]
ii Light energy = 12 × 1.7 × 10 / 100 = 2.0 × 10 J
i Non-renewable; not replaceable
[1]
ii Greenhouse gas emission leading to global warming / air pollution resulting in acid
rain that can affect health of people and trees / oil spills that damage sea-life
[2]
[Total: 7]
5
−4
2
i F = pA = 3.8 × 10 Pa × 6.1 × 10 m = 230 N
[2]
ii There is a force on the piston due to air / atmospheric pressure; the pressure
of trapped air decreases as valve opens and air enters the tyre
[1]
Molecules have more frequent collisions with the walls of the container
[3]
[Total: 6]
Thermal physics
1 a Less regular arrangement than in a solid.
[1]
b Solid – vibrate
Liquid – slide past each other
Gas – move quickly in all directions
[3]
c Particles far apart in a gas so attractive forces between them small; particles move
more quickly in a gas.
[2]
[Total: 6]
2 a Two from: evaporation occurs at any temperature; bubbles do not form in the
liquid; the remaining liquid cools
[2]
b More energetic molecules escape from the surface of the liquid; the average kinetic
energy of the remaining liquid falls so its temperature decreases
[2]
c i Rate of evaporation increases with surface area
[1]
ii Rate of evaporation is greater the higher the temperature of the liquid
[1]
[Total: 6]
3 a i Boiling occurs at a specific temperature while evaporation can occur at any
temperature
The temperature of the liquid remains constant during boiling but decreases in
evaporation
Bubbles form in the liquid in boiling, while evaporation occurs only at the
surface of the liquid
A source of thermal energy is needed for boiling but not for evaporation. [4]
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Cambridge O Level Physics answers
b
4 a
b
5 a
b
c
6 a
b
ii Thermal energy is needed to overcome the attractive forces between molecules
and allow them to escape to the gas phase
[2]
3
−3
3
i Mass = ρ × V = 1000 kg/m × 4.6 × 10 m / 60 s = 0.077 kg
[2]
ii ΔE = mcΔθ = 0.077 kg × 4200 J/(kg ℃) × (37 – 16)℃ = 6800 J
[3]
iii 1. Energy supplied to shower
2. Efficiency = (useful energy output) / (total energy input)
[2]
iv To avoid electric shock if the heater develops a fault and a live wire
touches the pipe
[2]
[Total: 15]
Particles in a liquid are much closer together than in a gas so the attractive forces
between them are large; liquid particles move more slowly than gas particles [2]
i p1V1 = p2V2
[1]
ii V2 = 100 cm3 – (5 × 8) cm3 = 60 cm3
p2 = p1V1/V2= 1.2 × 105 Pa × 100 cm3/60 cm3 = 2 × 105 Pa
[3]
iii When the volume decreases the molecules have more frequent collisions with
sides of container so the pressure increases
[1]
[Total: 7]
i I = P/V = 2800 W / 230 V = 12 A
[3]
ii Between 13 A and 20 A
[1]
5
i ΔE = mcΔθ = 6.3 kg × 4200 J/(kg ℃) × (49 – 23)℃ = 6.9 ×10 J
[2]
ii Thermal energy is lost to the environment at a higher rate when the temperature
of the water is higher
[1]
i Convection – hot water is less dense than cold water so the water near the
heater rises and cooler water falls to take its place; convection currents are set
up which transfer the thermal energy
[3]
ii Conduction – thermal energy is transferred by free electrons moving through
the steel; particles in the hot regions vibrate strongly and pass on thermal
energy to their neighbours through lattice vibrations
[3]
iii The particles move further apart, so the volume increases and the density
decreases
[2]
[Total: 15]
Coal, oil
[1]
i Water becomes heated to a very high temperature and pressure underground;
as it rises to the surface its pressure falls and it turns into steam which is used to
drive the turbine
[2]
7
ii ΔE = mcΔθ = 90 kg × 4200 J/(kg ℃) × (160 – 30)℃ = 4.9 ×10 J
[2]
[Total: 5]
Waves
1 a The minimum angle of incidence at which total internal reflection occurs
[2]
b Ray at 90 º normal to water / air boundary passes straight through undeviated
Ray at 60º to horizontal passes into air but is refracted away from the normal
Ray at 30º to horizontal undergoes total internal reflection.
[3]
[Total: 5]
2 a Direction of vibration is parallel to the direction of energy transfer
[1]
b i f = v / λ = 0.17 m/s / 0.019 m = 8.9 Hz
[2]
ii Unchanged
[1]
iii Decreases
[2]
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Cambridge O Level Physics answers
c
3 a
b
4 a
b
5 a
b
c
6 a
b
7 a
b
c
i Electromagnetic, transverse
[1]
ii sin r = sin i / n
= sin 60º / 1.6
= 0.5413 so r = 33º
[2]
iii 1. sin c = 1 / n
= 1 / 1.6
= 0.625, so c = 39º
[2]
2. 90º – 33º = 67º; this is greater than c so the light undergoes total internal
reflection
[2]
3. Reflected at 67º to normal at P; at 30º to right hand side vertical
[2]
[Total: 15]
i Long sightedness
[1]
ii Image formed behind the retina
[2]
i Converging lens drawn between rays and eye; rays meet on retina
[2]
ii Converging lens
[1]
[Total: 6]
i Red
[1]
ii Blue
[1]
X-rays – imaging of broken bones, cancer diagnosis/treatment, security
scanners, flaw detection in metals
microwaves – satellite television, mobile phones, radar, Bluetooth, microwave
ovens
gamma rays – cancer diagnosis/treatment, sterilisation, flaw detection
in metals
[4]
[Total: 6]
Reflection of sound waves
[1]
i Reduced
[1]
ii Unchanged
[1]
λ=v/f
= 330 m/s / 3700 Hz
= 0.089 m = 8.9 cm
[2]
[Total: 5]
Longitudinal wave – direction of vibration is parallel to direction of energy transfer
Transverse wave – direction of vibration is at right-angles to the direction
of energy transfer.
[2]
i λ=v/f
= 330 m/s / 3800 Hz
= 0.087 m = 8.7 cm
[2]
ii Not audible; humans cannot hear frequencies below 20 Hz
[1]
[Total: 5]
Number of vibrations per second
[1]
i λ=v/f
= 330 m/s / 2200 Hz
= 0.15 m = 15 cm
[2]
ii 1. No change
2. Increases
[1]
i 1. Electrical signal causes the diaphragm of loudspeaker to vibrate and
energy is transferred to the air molecules
[1]
2. In a longitudinal wave, by compressions and rarefactions in the air
[2]
ii There are fewer particles to transfer the energy.
[1]
[Total: 8]
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67
Cambridge O Level Physics answers
Electricity and magnetism
1 a There is the same number of positive and negative charges.
[1]
b i Negatively charged electrons transferred from the cloth to the rod
[2]
ii Bring rod close to a gold leaf electroscope – leaf will rise/when rod is
brought close to small pieces of aluminium foil they will be picked up/
suspend rod and bring another charged rod close to it – suspended rod
will be repelled
[2]
[Total: 5]
2 a Change the value of the variable resistor
[1]
b i The current in a metallic conductor is directly proportional to the potential
difference (p.d.) across its ends if the temperature and other conditions are
constant.
[2]
ii R1 = V / I
= 7.6 V / 0.320 A
= 23.8 Ω = 24 Ω (to 2 significant figures)
R2 = V / I
= 5.2 V / 0.220 A
= 23.6 Ω = 24 Ω
R3 = V / I
= 2.4 V / 0.100 A
= 24.0 Ω = 24 Ω
The values for R all agree so Ohm’s law is obeyed
[1]
iii For 220 mA and 320 mA readings use 0–10 A setting;
for 100 mA use 0–200 mA setting
[1]
[Total: 5]
3 a i 1 / R = 1 / R1 + 1 / R2
= 1 / 1800 + 1 / 9000
= 6 / 9000 Ω−1
so total resistance R = 9000 / 6 =1500 Ω
[2]
ii I = V / R
= 4.5 V / 1500 Ω = 3 mA
[2]
b i Increases; resistance of LDR decreases so current I = 4.5 / R LDR increases [1]
ii Does not change; resistance and e.m.f. do not change
[1]
[Total: 6]
4 a i 1 / R// = 1 / R1 + 1 / R2
= 1 / 3.6 + 1 / 1.8
= 3 / 3.6 Ω−1 so R// = 3.6 / 3 =1.2 Ω
Rtotal = (2.8 + 1.2) Ω= 4.0 Ω
[3]
ii I = V / R
= 6.0 V / 4.0 Ω
= 1.5 A
[2]
b i Work done by unit charge passing through the component
[1]
ii
iii Current decreases; R// increases, so p.d. across R// increases
and p.d. across 2.8 Ω resistor decreases.
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[1]
[2]
[Total: 9]
68
Cambridge O Level Physics answers
5 a p.d. = 2.0 div × 2.0 V/div = 4.0 V
[1]
b I=V/R
= 4.0 V / 700 Ω;
R LDR = VLDR / I
= 8 V × 700 Ω / 4.0 V
= 1400 Ω
[3]
c Trace moves up. When light intensity increases, the resistance of the LDR
decreases so the p.d. across it decreases and the p.d across the 700 Ω resistor
increases
[3]
[Total: 7]
6 a Fuse melts and breaks circuit to hair dryer if the current in the wires becomes too
high
[1]
b i I=P/V
= 1500 W / 240 V
= 6.25 A
[2]
ii 13 A
[1]
c Double insulation
[1]
[Total: 5]
7 a i The coil is cut by a changing magnetic field when the magnet moves.
[2]
ii Increase the speed at which the torch is shaken / use a coil with more turns /
use a stronger magnet
[1]
b i
ii Ratio of power output to power input is higher
8 a
b
c
d
9 a
b
10 a
b
Coil A; p.d. must be stepped down before entering the house house
To increase the number of field lines cutting the secondary coil
A changing magnetic field in coil A cuts coil B
Advantage: less easily damaged / less intrusive
Disadvantages: harder to find to repair / may be damaged accidently by
diggers / more expensive to install
[1]
[1]
[Total: 5]
[1]
[1]
[1]
[2]
[Total: 5]
i Current carrying wire perpendicular to a magnetic field experiences a force;
direction of force given by Fleming’s left-hand rule
[1]
ii Reverses current in coil every half revolution
[2]
i P = IV
= 2.0 A × 12 V
= 24 J/s
Energy supplied = 24 J/s × 8 s = 192 J
[2]
ii Efficiency = useful power output / total power input
= 140 J / 192 J = 0.73 or 73%
[1]
[Total: 6]
i X: N; Y: S; Z: N
[2]
ii They are attracted to each other because they have opposite polarity
[1]
Starter motor in a car – large current is needed for the motor that is too large
to go through the switch on the control panel
Electric bell / mains supply for appliances – a relay allows a small current to
control a large current
[2]
[Total: 5]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
69
Cambridge O Level Physics answers
Nuclear Physics
1 a Same number of protons; different number of neutrons
[2]
b i 218 – 84 = 134
[2]
ii 84 + 2 = 86
[2]
iii A neutral helium atom has two electrons / an alpha particle has two
protons and two neutrons only
[1]
10
c i 2.1 × 10
[1]
ii 2.8 × 1010 – 2.1 × 1010 = 0.7 × 1010 = 7 × 109
[1]
iii After 2 half-lives count = 2.8 × 1010 / 4 = 7 × 109
so half-life of radon = 7.6 days / 2 = 3.8 days
[3]
d Alpha particles do not penetrate matter very far
[1]
e Cause ionisation in lung tissue that can damage human cells and cause cancer [2]
[Total: 15]
2 a i Same number of protons / electrons
[1]
ii Different number of neutrons
[1]
b i
;
[3]
5
4
ii 1.2 × 10 / 2.4 × 10 = 5 half-lives
Count rate = 6400 / 25
= 6400 / 32
= 200 counts/s
[3]
iii 1. Curved path downwards
[2]
2. Direction of force found from Fleming’s left-hand rule
[1]
c i Breaking up of a nucleus into smaller parts
[1]
ii Struck by a neutron
[1]
iii 1. Greenhouse gases not emitted
[1]
2. Power can be generated continuously
[1]
[Total: 15]
3 a i 11 protons, 13 neutrons, 11 electrons; nucleus contains protons and neutrons;
electrons orbit nucleus
[3]
ii Sodium-24 has one more neutron
[1]
b i Electron
[1]
ii
[3]
c High-energy electromagnetic waves
[2]
d i Upward curve
[1]
ii Continue in a straight line
[1]
iii Negatively charged beta-particles are attracted to the positively
charged plate; gamma-rays are unaffected by an electric field
[1]
e Allows sufficient time for tracing; activity of source falls quickly enough not to
cause ongoing radiation threat
[2]
[Total: 15]
4 a Small nuclei combine together to form larger nuclei releasing energy in
the process; hydrogen is converted into helium
[3]
b It travels through space in the form of electromagnetic radiation
[2]
[Total: 5]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
70
Cambridge O Level Physics answers
Practical Test past paper questions
1 Values of t1 and t2 will differ if springs with spring constants other than 25 N/m are
used
a i t1 = 49.0 ± 2.0 s
[1]
ii Decreases
[1]
b i t2 = 49.5 ± 2.0 s
[1]
ii Amplitude decreases faster with card
[1]
iii Air resistance higher
[1]
[Total: 5]
2 a Line TU and angle r drawn correctly
[2]
b For glass of refractive index1.5,
i = 55º ± 1º. r = 33º ± 3º
[2]
c n = sin i / sin r
= sin 55º / sin 33º
= 1.5
[1]
[Total: 5]
3 a i 15º ± 3º
[1]
ii Angle θ increasing from 15º to 45º
[2]
b Difficulties: movement of protractor / beam; parallax error due to separation of
beam and protractor
Improvements: tape protractor to the beam / reduce separation between
protractor and beam
[1]
c Each end of beam at same distance above bench; appeared level with beam
support / horizontal edge of window frame
[1]
[Total: 5]
4 a Water is hot so danger of scalding / spilling water / breaking thermometer
[1]
b Temperature falls quickly at first and then less quickly
[1]
c Two readings correctly averaged to give value for TA
[1]
[1]
d Two readings correctly averaged to give value for TF
e If TF > TA, the results suggest that the thermometer cools more slowly when air is
trapped around it
[1]
[Total: 5]
5 a i Value of V stated
[1]
ii Five readings with increasing trend
[2]
b
Use of any V / 47 to calculate a current; yes and use of another value of V / 47 with a
correct comparison
[2]
[Total: 5]
6 Values given for a lens with f = 15 cm and lamp 110 cm above floor
a i u = 18 cm
[1]
ii v = 92 cm
[1]
iii Difficulty in determining position of filament and centre of lens / reducing
parallax error by having eye perpendicular to the scale marking on the
ruler / keeping ruler vertical
[1]
b f = uv / (u + v)
= 18 × 92 / (18 + 92)
= 15 cm
[2]
[Total: 5]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
71
Cambridge O Level Physics answers
7 a VS = 4.5 ± 0.5 V
[1]
b i 1. Temperature > 70℃ ; 2. VAB in range 0.5 V to 2.5 V recorded in table [2]
ii A minimum of seven readings of temperature and VAB recorded as
temperature falls
[1]
c Table completed with values for headings and units, Vt/V, I/A and Rt/Ω
[4]
d Graph: axes correctly oriented and labelled with quantity and unit; easy to
read / plot scales chosen; points correctly plotted to 1/2 small square; good
best fit curve drawn with a thin continuous line.
[4]
e Tangent drawn at 65℃ and gradient calculated to 2 or 3 significant figures by
triangle method
[3]
[Total: 15]
8 a Repeat readings and take average; t1 = 4 ± 1 s
[2]
b t2 < t1
[1]
c t1/ t2 > 1 (in range 1.8 to 2.2)
[2]
[Total: 5]
9 (30 cm length of SWG 30 nichrome wire wound with turns not touching is suitable
for R.)
[1]
a Room temperature θr given, with units in ℃
b i Current I less than 1 A with units given in A
[1]
[1]
ii P = I2R = 4 I2 (W)
10 a Used a set square against each face of ruler and bench.
[1]
b i x in range 15.0 to 35.0 cm
ii h measured from same part of ball to ramp;
iii Ball lifted vertically from sand; experiment repeated and
iv Average value of x calculated
[4]
[Total: 5]
11 a x0 in range 15.0 cm to 25.0 cm (values will differ if springs with spring constants)
other than 25 N/m are used)
[1]
c Check that the clay cylinder can oscillate freely in the measuring cylinder
[1]
d s in range 6.0 to 10.0 cm
[1]
e s larger than in part d, and x less than x0
[2]
f Headings: s/cm, x/cm
Table completed for a range of at least 5 values of x with s increasing as
x decreases
[4]
g Graph: axes correctly oriented and labelled with quantity and unit; easy to
read / plot scales chosen: points correctly plotted to 1/2 small square; narrow
line of best fit drawn
[4]
h G calculated correctly to 2 or 3 significant figures by triangle method
[2]
[Total: 15]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
72
Cambridge O Level Physics answers
Alternative to Practical Test past
paper questions
1
a
b
Screen moved slowly towards and away from the lens
i u = 6.0 cm v = 4.3 cm
ii f = 2.5 cm
iii Repeat measurements and calculate average values
2
a
b
Ammeter in series, voltmeter in parallel with lamp
i Changes value of variable resistor
ii 1.20 A, 1.65 A
iii 4.0 V / 1.20 A = 3.3 Ω and 8.0 V / 1.65 A = 4.8 Ω
1. Becomes hotter
2. Becomes brighter
c
3
a
b
c
d
4
a
b
5
a
b
c
d
6
a
[1]
[1]
[1]
[1]
[Total: 4]
[2]
[1]
[1]
[1]
[2]
[Total: 7]
Time = 5 divisions × 0.5 ms/div = 2.5 ms
[1]
Ruler worn at ends and zero position unclear
[1]
v = distance / time = 82.0 × 10 −2m / (2.5 × 10−3) s = 328 m/s
[1]
Use a longer distance between microphones / repeat measurement and calculate
an average value
[1]
[Total: 4]
i Micrometer screw gauge
[1]
ii 1. 3.4 cm
[1]
2. Set bead between two parallel vertical surfaces and measure the perpendicular
distance between the surfaces with a ruler
[1]
i Measure the volume of water in a measuring cylinder V1; add beads and measure
the new volume V2; volume of beads = V2 – V1.
[3]
ii Ensure that the beads are completely submerged / take measurement of volume with
line of sight perpendicular to scale reading to avoid parallax error / take volume
readings from bottom of the meniscus
[1]
iii ρ = 523 g / 196 cm3
[1]
= 2.67 g/cm3
[Total: 8]
Light source and narrow slit, ruler, pencil and paper
[1]
Draw outline of block and normal at centre of side AB; shine a light beam at an
angle of incidence (i) to the normal and mark its position on the paper. Mark the line
of the exit path of the ray. Remove the glass block and join up the path of the ray
through the block. Repeat for different angles of incidence
[3]
i = 50º ± 1º, r = 31º ± 1º
[2]
Table headings: i / º, r / º, sin i, sin r, n;
Results: use a minimum of four values of i between 10º and 80º
[2]
[Total: 8]
i Distance from bottom of metal ball to bench top
[1]
ii Distance between bottom of glass ball in its initial and final positions on bench [1]
iii Metre ruler
[1]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
73
Cambridge O Level Physics answers
7
8
iv (23.5 + 20.8 + 22.3) cm/3 = 22.2 cm
[1]
v mgh = 0.0081 kg × 10 N/kg × 0.120 m = 0.009 72 J
[2]
b i E / J = 0.009 72, dav /cm = 22.2
ii Graph: axes correctly oriented and labelled with quantity and unit; easy to
read / plot scales chosen; points correctly plotted to 1/2 small square; good
best fit curve drawn with a thin continuous line
[4]
c To reduce distance travelled by the glass ball by increasing the friction between
the ball and the surface
[1]
[Total: 11]
a Micrometer screw gauge
[1]
b i So the wire does not heat up and its resistance change
[1]
ii 0.34 A, 2.8 V
[2]
c i
[3]
a
diameter/mm
voltage/V
current/A
resistance/Ω
0.15
2.8
0.34
8.2
0.24
2.8
0.85
3.3
ii
Resistance is reduced when the diameter of the wire becomes larger
i
ii
Deceleration
Distance between the dots is smaller so car travels a shorter distance in 0.020
b
c
9
d/cm
t/s
0
1.000
6
1.020
4
1.040
2
1.060
[1]
[Total: 8]
[2]
[4]
Average speed = total distance/total time
= 12 cm / 0.060 s
= 200 cm/s
= 2 m/s
[1]
[Total: 7]
a i Lines correctly drawn
`
[1]
ii Straight line drawn through P1 and P2 to AB and through P4 and P3 to CD; connect
points where each meet the mirrors with a straight line.
[2]
iii Ruler and sharp pencil used
[1]
b i Q labelled correctly
ii Normal drawn at 90º to AB
iii i = 44º ± 1º
iv r = 47º ± 1º
[3]
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
74
Cambridge O Level Physics answers
c
Increase the distance between P4 and P3 / use base of pins to align / ensure pins
upright
10 a
b
c
[1]
[Total: 8]
[1]
[1]
[1]
[2]
Ball drawn with its lower edge opposite 60 mark on ruler
To estimate where to look for the bounce height
i Difficult to measure bounce height more accurately
ii hav= 40 + 39 + 40 + 42 + 40) cm /5 = 40 cm (to 2 significant figures)
d i 40 cm entered in table
ii Another person / steps needed to drop the ball; measure height with a tape
measure
[2]
iii Graph: axes correctly oriented and labelled with quantity and unit; easy to read / plot
scales chosen; points correctly plotted to 1/2 small square; good best fit curve drawn
with a thin continuous line.
[4]
e i Bounce height too small to view easily without parallax error
[1]
ii Correct value read from graph
[1]
[Total: 13]
11 a
[1]
c
d
e
2.8 V
Insert into some melting ice cubes
Rroom temperature = (5000 /2 .8) – 1000 = 1786 – 1000 = 790 Ω
R0℃ = (5000 / 1.8) – 1000 = 2778 – 1000 = 1800 Ω
Resistance increases when temperature decreases (or vice versa)
Cambridge O Level Physics 4th edition
© Heather Kennett 2021
[1]
[1]
[2]
[1]
[Total: 6]
75