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Physics Form 4 Answers

Checkpoint 2.3 (b) Total momentum after collision (b) t = 0.2
= 1200 kg m s –1 mv – mu
Q1 s = 100 m, a = g = 10 m s –2 (c) Principle of conservation of F = 0.2
u = 0 m s –1 momentum 600(0) – 600(50)
1 (d) Total momentum after collision = 0.2
s = ut + at 2 = 100 × 6 + 200 × v = –1.5 × 10 N
2
5
1 = 1200 kg m s –1 5
100 = 0 + (10)t 2 200v = 600 Magnitude of force = 1.5 × 10 N
2
t = 20 v = 3 m s –1 (c) The time in (a) is 10 times
= 2 5 s or 4.47 s Q2 (a) Total momentum before collision greater than in (b). The old tyres
= (1500 × 15) + [1000 × (–20)] lengthen the stopping time of
Q2 (a) An object undergoes free fall if = 2500 kg m s –1 the car and hence, reduce the
it is acted upon by gravitational impulsive force acted on the car.
force only. (b) Total momentum after collision (d) a = F , if m is small, a is bigger

= Total momentum before collision
m
(b) The rubber ball will reach the = 2500 kg m s –1 for a fixed value of F. With that,
ground first. (c) Principle of conservation of the racing car can move faster.
(c) The rubber ball has a compact momentum
shape and has less frictional Q2 The action of moving the hand
force from the air compared with (d) (m van + m car )v = (m van u van + m car u car ) backward, the worker will lengthen
a piece of paper. (1500 + 1000)v = 2500 the impact time between the
Q3 (a) v A = 24 m s –1 v = 1 m s –1 watermelon and the hand and hence
reduce the impulsive force.
v B = 0 m s –1 Q3 (a) Recoil velocity of the rifle = v
2
v = u + 2as 0.012 × 360 = 6v Q3 (a) The front and rear crumple
2
2
0 = v A = 2(–10)(s) v = 0.72 m s –1 zones are designed to crumple
2
0 = 576 – 20s (b) Mass of the wooden block = m upon impact in a collision. A
s = 28.8 m 0.012 × 360 = (0.012 + m) × 12 longer impact time will reduce
(b) v = u + at m = 0.348 kg or 348 g the impulsive force exerted on
the car.
0 = 24 + (–10)t (b) The strong and rigid cell will
t = 2.4 s Q4 (a) When a large volume of water prevent the roof from collapsing
rushes out of the hose with a
(c) There is no frictional force acting very high speed, it has a very on the passengers in the event
on the ball. big momentum. According to when the car overturns. The
the principle of conservation passengers will be protected
Checkpoint 2.4 of momentum, an equal and from direct impact with external
Q1 (a) Tin P opposite momentum is produced forces.
causing the fireman to fall
(b) Tin P backward if not supported by Q4 (a) The driver has inertia that keeps
him moving forward even when
(c) The mass of tin Q is bigger another fireman. the car stops suddenly.
and hence, there is a bigger (b) Initially, the twins are at rest (b) The air bag absorbs the initial
resistance for tin Q to change and the total momentum is impact and cushions the driver
its state of being stationary or zero. When they push each from hard objects like the
being in motion. That is, object other and release their hands, steering wheel and windscreen.
with bigger mass has bigger both will acquire momentum of
inertia. equal magnitude but in opposite Checkpoint 2.8
directions to each other in Q1 (a) Mass of an object is the quantity
Q2 (a) A fully loaded lorry has a big accordance to the principle of
mass. Hence, its inertia is big. conservation of momentum of matter in the object.
(b) The mass of a train is big. where the final total momentum Weight of an object is the
Hence, its inertia is big. is still zero. gravitational force acting on the
(c) Steel tubes have big mass. object at a particular place.
Hence, the oil platform is not Checkpoint 2.6 (b) Weight Mass
easily moved.
Q1 m = 5.0 kg Vector quantity Scalar quantity
Q3 (a) The pencil box will continue to F = 12.0 N Varies with the Is the same
move and leave the toy car. F 12.0 value of g anywhere
(b) The inertia of the pencil box a = m = 5.0 = 2.4 m s –2
causes it to remain in its original Measures Measured using
state of motion although the car Checkpoint 2.7 using spring inertia balance
has been stopped. –1 –1 balance
(c) The pencil box will move with a Q1 (a) u = 50 m s , v = 0 m s , (c) mg = 47 × 10
m = 600 kg

higher velocity and land further t = 0.02, = 470 N
from the obstacle. mv – mu
F = 0.02
Checkpoint 2.5 = 600(0) – 600(50) SPM Practice 2
Q1 (a) Total momentum before collision 0.02 Objective Questions
= 100 × 2 + 200 × 5 = –1.5 × 10 N 1. B 2. C 3. A 4. B 5. C
6
= 1200 kg m s –1 Magnitude of force = 1.5 × 10 N 6. A 7. C 8. B 9. C 10. B
6
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ANS FOC PHYSICF F4 1P.indd 254 29/01/2020 2:04 PM