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 4
5
Self-assessment questions
1
a
direction
of
travel
2
3
force of parachute
= 2000 N
60°
30°
weight of parachutist
= 1000 N
500 N
b
force upwards = 2000 − 1000 = 1000 N
upwards
b
c omponent of weight down slope = 500
sin 30° = 250 N
c
he will accelerate upwards (i.e.,
S
decelerate).
c
he contact force of the slope is a normal
T
reaction, so it is at 90° to the slope.
d
Friction; up the slope
a
Yes, the ship is in equilibrium, because
it travels at a constant velocity (not
accelerating, so no resultant force acting
on it).
b
pthrust is equal and opposite to weight
U
of boat, as it is floating, so = 1000 kN
c
ecause the velocity is constant, we know
B
that the drag is equal and opposite to the
force of the engines, so = 50 kN
6
7
No, there is a net force acting upon it.
ith rope horizontal, the force pulling the
W
box is F. With the rope at an angle θ to the
horizontal, the horizontal component
(= F cos θ) is less, since cos θ is less than 1.
et force down slope = 40 × 9.81 ×
n
sin 25° − 80 = 85.8
acceleration = 8540.8 = 2.1 m s−2
a
sum of clockwise moments = sum of
anticlockwise moments
400 × 0.20 = F × 1.20
so, force required is F = 4001.×200.20 = 67 N
a
vertical component of force = weight −
upthrust = 2.5 − 0.5 = 2.0 N downwards
b
a
component of acceleration parallel to
slope = 9.81 × sin 25° = 4.1 m s−2
b
b
horizontal component of force = 1.5 N
so, resultant force is obtained from R2 =
(2.0)2 + (1.5)2 = 6.25
so, R = 2.5 N
angle = tan−1 12..50 = 37° to vertical
4
a contact force
s um of clockwise moments = sum of
anticlockwise moments
400 × 0.20 = F × 0.50
so, force on legs of wheelbarrow is
F = 4000×.500.20 = 160 N
8
a
Remember that weight = mg, and that
the acceleration g is the same for all the
masses; in our moments equation, g
cancels out from both sides.
sum of clockwise moments = sum of
anticlockwise moments
(100 × 30) + (10 × 45) = M × 20
so, mass M = 300020+ 450 = 172.5 ≈ 173 g
1
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
b
y this method, weighing could be carried
B
out with a limited selection of relatively
small masses.
c
pwards force at pivot = sum of the
U
weights of M, the 100 g and the 10 g mass
= (0.1725 + 0.110) × 9.81
c
s um of clockwise moments = 2.5 + 2.5 =
5 N m = sum of anticlockwise moments
so, yes, the moments are balanced
10 torque = force × radius, so
force = torque
= 0137
= 761 N ≈ 760 N
radius
.18
= 2.77 N
9
a, bF1 = 0 N m
F2 = 10 × 0.25 = 2.5 N m clockwise
F3 = 10 sin 30° × 0.50 = 2.5 N m clockwise
F4 = 5 × 1.0 = 5 N m anticlockwise
2
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020