CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Coursebook answers
Chapter 1
Exam-style questions
By Pythagoras’ theorem, displacement =
1
A[1]
b
2
C[1]
3
a
2.22 + 152
= 15 200 m[1]
at an angle = tan−1 215.2 [1]
= 8° E of N or a bearing of 008°[1]
distance = speed × time
120 × 2.0
=
60 ( )
[1]
= 4.0 km[1]
b
c
4
c
he car’s direction of motion keeps
T
changing. Hence, its velocity keeps
changing. In the course of one lap,
its displacement is zero, so its average
velocity is zero.[1]
istance travelled in 1 minute =
d
0.5 × circumference but, displacement =
diameter of track[1]
circumference
=
π
4000 m
=
= 1270 m[1]
π
total time = 1100 + 900 = 2000 s[1]
6
a
By Pythagoras’ theorem, distance
= 6002 + 8002 m2[1]
[1]
( 800
600 )
5
velocity =
7
total distance = 2.2 + 15 = 17.2 km[1]
average velocity =
displacement
time
15200
[1]
2000
= 7.6 m s−1[1]
resultant velocity = 1.02 + 2.402
= 23° E of N or a bearing of 023°[1]
a
Distance in a (particular) direction[1]
b
hen athlete returns to his original
W
position or the start[1]
(direct) distance from original position
zero[1]
1000
60
= 16.7 m s−1[1]
at an angle 53° W of N[1]
a
distance in car = 0.25 × 60 = 15 km[1]
e
( )
displacement = 1000 m at an angle 53° W
of N or a bearing of 307°[1]
c
average speed = distance
time
17200
=
[1]
2000
= 8.6 m s−1[1]
= 2.6 m s−1[1]
at an angle of tan−1 12..04 [1]
= 1000 000 = 1000 m [1]
angle at B = tan−1
d
=
2
b
t ime for 2.2 km at 2.0 m s−1 = 2200
2
= 1100 s[1]
8
boy
s / m 40
38
36 35
girl
30
25
20
15
10
5
0
1
0
1
2
3
4
5
6
7
8
9 10 11 12
t/s
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
a
Straight line from t = 0, s = 0 to t = 12,
s = 36[1]
b
100 km h–1
traight line from t = 0, s = 0 to t = 5,
S
s = 10[1]
N
Straight line from t = 5, s = 10 to t = 12,
s = 38[1]
c
9
resultant
500 km h–1
10 s where the graphs cross[1]
a
Each second, it travels a constant
distance.[1]
At least two examples: 108 − 84 = 24,
84 − 60 = 24, 60 − 36 = 24 cm[1]
b
d
Correct vectors drawn and labelled[1]
108 + 2 × 24[1]
Scale stated and diagram of
sufficient size[1]
Resultant velocity 510 (±10) km h−1[1]
11° W of N or a bearing of 349° (±3°)[1]
= 156 cm[1]
c
distance = 240 × 0.001 = 0.24 cm[1]
11 a
24
speed = distance
= 0.1
[1]
time
= 240 cm s−1[1]
c
b
The smallest scale division on the ruler is
2 cm and so each dot is blurred by about
1/10th of a scale division. This might just
be observable but difficult to see[1]
10 a
Vector quantities have direction, and
scalar quantities do not.[1]
One example of a vector, e.g., velocity,
acceleration, displacement, force[1]
One example of a vector, e.g., speed, time,
mass, pressure[1]
0.25 × 510 = 128 ≈ 130 km 11° W of N[1]
velocity of aircr
B
7.5 m s–1
15 m s–1
A
Correct vector diagram[1]
Velocity of aircraft in still air in easterly
direction or calculation[1]
b
t =
5000
15
= 333 s and
5000
13.5
= 370 s[1]
total time = 703 or 704 s or 703.7 s[1]
average speed =
2
10000
703.7
= 14.2 m s−1[1]
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020