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
Science in context guidance
•
To calculate speed you are going to need to
know both displacement and time.
•
To measure displacement you would need
to include an object of known length (for
example, a meter ruler) in the photograph.
This would allow you to convert a
displacement measured on the photograph
into a real-life distance.
•
The stroboscope would provide you with the
time period between photographs.
•
However, calculating the speed is not as simple
as using the equation speed= distance
. Think
time
about the vertical motion of the ball; once the
ball has left the hand it starts to decelerate
(due to gravity). As it reaches the top of its
motion it is momentarily stationary, before
accelerating downwards again. The speed
is never constant. Realistically the best you
could measure would be the average speed
between photographs.
Self-assessment questions
1
average speed =
2
a
mm s−1
b
km h-1
c
km s−1
d
10000
1625.17
= 6.15 m s−1
6
or example, attach a card to a weight and
F
drop it through a light gate. Alternatively,
attach ticker-tape to the falling mass.
7
a
Displacement
8
b
Speed
c
Velocity
d
Distance
distance s = v × t = 1500 × 0.2 = 300 m
(Remember: the 0.4 s total time is that taken
for the sound waves to travel out and be
reflected back from the surface of the water.)
9
t ime taken for orbit is one year = 1 × 365.25 ×
24 × 60 × 60 = 31 557 600 s.
distance travelled = circumference of orbit =
2 × π × 1.5 × 1011 = 9.425 × 1011 m
so, the Earth’s speed = 29.9 km s−1 ≈ 30 km s−1
As the Earth orbits the Sun, its direction of
motion keeps changing. Hence, its velocity
keeps changing. In the course of one year, its
displacement is zero, so its average velocity is
zero.
10 Sloping sections: bus moving; horizontal
sections: bus stationary (e.g., at bus stops)
11 O
A: constant speed; AB: stationary; BC:
reduced constant speed; CD: running back
to gate
s
C
−1
m s
3
distance = 12 cm = 120 mm
4
so, average speed = 120
60
0.05 m
average speed = 0.40 s
A
B
= 2.0 mm s
−1`
= 0.0125 m s−1 ≈
0.013 m s−1
5
1
a
Constant speed
b
Increasing speed (accelerating)
0
D
0
t
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
12 a
85 m s−1
b
13 a
Graph
is a straight line through the
origin, with gradient = 85 m s−1
b
17.3 m s−1 ≈ 17 m s−1
c
43.9° ≈ 44° to the vertical
18 a
10 m s−1 North
Graph is a straight line for the first 3 h;
then less steep for the last hour
b
0 m s−1
b
car’s speed in first three hours = 23 km h−1
c
7.1 m s−1 045° or N45°E
c
84
4
d
7.1 m s−1 315° or N45°W
car’s average speed in first four hours =
Exam-style questions
= 21 km h−1
14 atotal distance travelled = 3.0 + 4.0 =
7.0 km
b, c T
he two parts of the journey are at 90° to
each other, so the resultant displacement
is given by Pythagoras’ theorem.
1
A[1]
2
C[1]
3
a
( )
4.0
3.0
b
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]
c
istance travelled in 1 minute =
d
0.5 × circumference but, displacement =
diameter of track[1]
circumference
=
π
4000 m
=
= 1270 m[1]
π
= 53° E of N or 37° N
15 a, b8.5 km; 48° W of S or a bearing of 228°
A
48° 45°
8.5 km
8.0 km
π
W
π
SE
12.0 km
4
b
magnitude2 = 2.02 + 0.82 = 4.64 so magnitude =
4.64 = 2.154 ≈ 2.2 m s−1
( )
vertical
18 ms–1
angle at B = tan−1
[1]
( 800
600 )
displacement = 1000 m at an angle 53° W
of N or a bearing of 307°[1]
direction = tan−1 02.8 ≈ 22° to the direct route
(68° to the river bank)
17 a
resultant
25 ms–1
a
By Pythagoras’ theorem, distance2
= 6002 + 8002 m2[1]
= 1000 000 = 1000 m [1]
16 S
wimmer aims directly across river; river flows
at right angles to where she aims. So, resultant
velocity is given by geometry:
44°
[1]
= 4.0 km[1]
displacement2 = 3.02 + 4.02 = 25.0, so
displacement = 5.0 km
angle = tan−1
of E
distance = speed × time
120 × 2.0
=
60 c
5
velocity =
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]
total distance = 2.2 + 15 = 17.2 km[1]
horizontal
17 ms–1
2
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
b
By Pythagoras’ theorem, displacement =
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]
( )
c
b
t ime for 2.2 km at 2.0 m s =
= 1100 s[1]
d
average speed = distance
time
17200
=
[1]
2000
= 8.6 m s−1[1]
e
average velocity =
displacement
time
15200
=
[1]
2000
= 7.6 m s−1[1]
resultant velocity = 1.02 + 2.402
24
speed = distance
= 0.1
[1]
time
= 240 cm s−1[1]
c
2200
2
−1
total time = 1100 + 900 = 2000 s[1]
6
At least two examples: 108 − 84 = 24,
84 − 60 = 24, 60 − 36 = 24 cm[1]
108 + 2 × 24[1]
= 156 cm[1]
d
distance = 240 × 0.001 = 0.24 cm[1]
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]
100 km h–1
= 2.6 m s [1]
at an angle of tan−1 12..04 [1]
−1
( )
7
N
= 23° E of N or a bearing of 023°[1]
a
Distance in a (particular) direction[1]
b
resultant
hen athlete returns to his original
W
position or the start[1]
500 km h–1
(direct) distance from original position
zero[1]
8
boy
s / m 40
38
36 35
girl
30
25
20
15
b
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]
10
c
5
11 a
0
9
3
0.25 × 510 = 128 ≈ 130 km 11° W of N[1]
velocity of aircr
B
0
1
2
3
4
5
6
7
8
9 10 11 12
t/s
7.5 m s–1
a
Straight line from t = 0, s = 0 to t = 12,
s = 36[1]
b
Correct vectors drawn and labelled[1]
traight line from t = 0, s = 0 to t = 5,
S
s = 10[1]
15 m s–1
A
Straight line from t = 5, s = 10 to t = 12,
s = 38[1]
Correct vector diagram[1]
Velocity of aircraft in still air in easterly
direction or calculation[1]
c
b
10 s where the graphs cross[1]
a
Each second, it travels a constant
distance.[1]
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 =
10000
703.7
= 14.2 m s−1[1]
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020