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 a 10 b 40 c 5 d 67 e 1000
2 a 3.00 b 5.50 c 8.70 d 0.43 e 0.1
3 a 1 × 10 ; 3.5 × 10 ; 4.28 × 10 ; 5.04 × 10 ; 2.7056 × 104
5 3 8 2
b 1000; 2 000 000; 69 200; 134; 1 000 000 000
4 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
5 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.)
6 a two b three c four d two
7 Volume of rectangular block = length × breadth × height
= 4.1 cm × 2.8 cm × 2.1 cm = 24 cm3
8 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)
1 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 a 10 b 8.6 c 9.2
2 a 0.577 b 1.00 c 1.73
,3 Let Fx = 5.0 and Fy = 7.0 then:
𝐹 = �𝐹2 + 𝐹2 = �5.02 + 7.02 = √25 + 49 = √74 = 8.6 N
𝑋 F
and
so 𝜃𝜃 = 54º
Resultant force is 8.6 N acting at 54º to the 5.0 N force
4 Let Fx = 6.0 and Fy = 8.0 then:
𝐹 = �𝐹2 + 𝐹2 = �6.02 + 8.02 = √36 + 64 = √100 = 10 m/s
𝑋 F
and
so 𝜃𝜃 = 53º
Resultant velocity is 10 m/s acting at 53º to the horizontal
Practical work questions
1 Students’ own responses based on their results
2 Students’ own responses based on their results
3 Students’ own responses based on their results
4 Students’ own plans
5 Record the time for at least 5 complete oscillations with a stopwatch; to determine the period
divide the time by the number of oscillations
6 BOAOB
7 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 Volume of chocolate bar, V1 = length × breadth × height [1]
= 10 cm × 2 cm × 2 cm = 40 cm3; [2]
b Volume of chocolate bar, V2 = 2 cm × 2 cm × 2 cm = 8 cm3 [1]
Number of bars with same volume = V1 / V2 = = 5 [2]
c Time period = 8 s / 10 = 0.8 s [2]
[Total: 8]
2 a Average thickness = 6 mm / 60 = 0.1 mm [2]
b Number of blocks = (40 cm × 40 cm × 20 cm) / (10 cm × 10 cm × 4 cm) = 80 [5]
[Total: 7]
3 a Volume of water = 6 cm × 6 cm × 7 cm = 252 cm3 [3]
b Volume of stone = 6 cm × 6 cm × (9 – 7) cm = 6 cm × 6 cm × 2 cm = 72 cm3
[4]
[Total: 7]
4 a Metre, second [2]
b 3 [1]
c i πr 2
ii 2πr iii πr h
2
[4]
[Total: 7]
,5 a 2.31 mm [2]
b 14.97 mm [2]
[Total: 4]
6 a C [1]
b 13 N at 67º to 5 N force [7]
[Total: 8]
1.2 Motion
Test yourself questions
1 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
2 a Average speed = (10 m/s + 20 m/s) / 2 = 15 m/s
b Distance = v t = 15 m/s × 60 s = 900 m
3 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
4 Time taken = change of speed / acceleration = 500 km/h / 10 km/h/s = 50 s
5 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
6 a Straight line through the origin
b Speed = gradient of graph = constant
c Speed = distance/time = 18 m / 6 s = 3 m/s
7 At 1 s, gradient of tangent to curve = 20 m/s / 2.5 s = 8 m/s2; acceleration = 8 m/s2
8 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
, 9 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:
𝐹 = �𝐹2 + 𝐹2 = �12.02 + 5.02 = √144 + 25 = √169 = 13 m/s
𝑋 F
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 v = u + at = 0 + 0.8 m/s2 × 4 s = 3.2 m/s
2 s = (u + v)t / 2 = (10 + 20) m/s × = 75 m
3 s = ut + at2 = 0 × 5 s + (+2 m/s2) 52 s2 = 25 m
Practical work questions
1 The speed increases because the mass falls further in each 1/50 s
2 33 × 1/50 = 0.66 s
3 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
4 No difference
Exam-style questions (Page 20)
1 a Average speed = (0 + 8) / 2 = 4 m/s [2]
b s = v × t = 4 m/s × 4 s = 16 m [3]
c Acceleration = 2 m/s2 [2]
[Total: 7]
2 Change in speed = at = 1 m/s2 × 15 s = 15 m/s;
final speed = (10 + 15) m/s = 25 m/s [Total: 4]
3 a i 60 km [1]
ii (6pm – 1pm) = 5 hours [1]
iii v = distance / time = 60 km / 5 h = 12 km/h [1]
iv 2 (flat regions of graph) [1]
v (0.5 + 1.0) hours = 1.5 hours [1]
vi v = distance / time = 60 km / (5 – 1.5) h = 60 km / 3.5 h = 17 km/h [2]
b Steepest line: EF [2]
[Total: 9]
4 a 100 m [1]
b v = distance / time = 100 m / 5 s = 20 m/s [1]
c Slows down (slope of graph decreases) [2]
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 a 10 b 40 c 5 d 67 e 1000
2 a 3.00 b 5.50 c 8.70 d 0.43 e 0.1
3 a 1 × 10 ; 3.5 × 10 ; 4.28 × 10 ; 5.04 × 10 ; 2.7056 × 104
5 3 8 2
b 1000; 2 000 000; 69 200; 134; 1 000 000 000
4 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
5 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.)
6 a two b three c four d two
7 Volume of rectangular block = length × breadth × height
= 4.1 cm × 2.8 cm × 2.1 cm = 24 cm3
8 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)
1 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 a 10 b 8.6 c 9.2
2 a 0.577 b 1.00 c 1.73
,3 Let Fx = 5.0 and Fy = 7.0 then:
𝐹 = �𝐹2 + 𝐹2 = �5.02 + 7.02 = √25 + 49 = √74 = 8.6 N
𝑋 F
and
so 𝜃𝜃 = 54º
Resultant force is 8.6 N acting at 54º to the 5.0 N force
4 Let Fx = 6.0 and Fy = 8.0 then:
𝐹 = �𝐹2 + 𝐹2 = �6.02 + 8.02 = √36 + 64 = √100 = 10 m/s
𝑋 F
and
so 𝜃𝜃 = 53º
Resultant velocity is 10 m/s acting at 53º to the horizontal
Practical work questions
1 Students’ own responses based on their results
2 Students’ own responses based on their results
3 Students’ own responses based on their results
4 Students’ own plans
5 Record the time for at least 5 complete oscillations with a stopwatch; to determine the period
divide the time by the number of oscillations
6 BOAOB
7 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 Volume of chocolate bar, V1 = length × breadth × height [1]
= 10 cm × 2 cm × 2 cm = 40 cm3; [2]
b Volume of chocolate bar, V2 = 2 cm × 2 cm × 2 cm = 8 cm3 [1]
Number of bars with same volume = V1 / V2 = = 5 [2]
c Time period = 8 s / 10 = 0.8 s [2]
[Total: 8]
2 a Average thickness = 6 mm / 60 = 0.1 mm [2]
b Number of blocks = (40 cm × 40 cm × 20 cm) / (10 cm × 10 cm × 4 cm) = 80 [5]
[Total: 7]
3 a Volume of water = 6 cm × 6 cm × 7 cm = 252 cm3 [3]
b Volume of stone = 6 cm × 6 cm × (9 – 7) cm = 6 cm × 6 cm × 2 cm = 72 cm3
[4]
[Total: 7]
4 a Metre, second [2]
b 3 [1]
c i πr 2
ii 2πr iii πr h
2
[4]
[Total: 7]
,5 a 2.31 mm [2]
b 14.97 mm [2]
[Total: 4]
6 a C [1]
b 13 N at 67º to 5 N force [7]
[Total: 8]
1.2 Motion
Test yourself questions
1 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
2 a Average speed = (10 m/s + 20 m/s) / 2 = 15 m/s
b Distance = v t = 15 m/s × 60 s = 900 m
3 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
4 Time taken = change of speed / acceleration = 500 km/h / 10 km/h/s = 50 s
5 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
6 a Straight line through the origin
b Speed = gradient of graph = constant
c Speed = distance/time = 18 m / 6 s = 3 m/s
7 At 1 s, gradient of tangent to curve = 20 m/s / 2.5 s = 8 m/s2; acceleration = 8 m/s2
8 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
, 9 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:
𝐹 = �𝐹2 + 𝐹2 = �12.02 + 5.02 = √144 + 25 = √169 = 13 m/s
𝑋 F
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 v = u + at = 0 + 0.8 m/s2 × 4 s = 3.2 m/s
2 s = (u + v)t / 2 = (10 + 20) m/s × = 75 m
3 s = ut + at2 = 0 × 5 s + (+2 m/s2) 52 s2 = 25 m
Practical work questions
1 The speed increases because the mass falls further in each 1/50 s
2 33 × 1/50 = 0.66 s
3 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
4 No difference
Exam-style questions (Page 20)
1 a Average speed = (0 + 8) / 2 = 4 m/s [2]
b s = v × t = 4 m/s × 4 s = 16 m [3]
c Acceleration = 2 m/s2 [2]
[Total: 7]
2 Change in speed = at = 1 m/s2 × 15 s = 15 m/s;
final speed = (10 + 15) m/s = 25 m/s [Total: 4]
3 a i 60 km [1]
ii (6pm – 1pm) = 5 hours [1]
iii v = distance / time = 60 km / 5 h = 12 km/h [1]
iv 2 (flat regions of graph) [1]
v (0.5 + 1.0) hours = 1.5 hours [1]
vi v = distance / time = 60 km / (5 – 1.5) h = 60 km / 3.5 h = 17 km/h [2]
b Steepest line: EF [2]
[Total: 9]
4 a 100 m [1]
b v = distance / time = 100 m / 5 s = 20 m/s [1]
c Slows down (slope of graph decreases) [2]
