Page 54 - Focus TG4 KSSM (Physics) Terbitan Penerbitan Pelangi Sdn Bhd

Physics Form 4 Answers

11. C 12. C 13. D 14. D 15. B (iii) Inertia [1] • The gradient of the graph
16. C 17. B 18. D 19. D 20. B (b) • Hold the bottle upside down. gives the acceleration of the
21. B 22. D 23. D 24. C [1] car. [1]
• Give the bottle a downward • The area under the graph
Subjective Questions jerk. [1] gives the distance travelled by
• The inertia of the sauce the car. [1]
Section A will cause it to continue its 4
1. (a) Momentum is the product of movement downward and (c) (i) • A short reaction time is
mass and velocity. hence, out of the bottle. [1] desirable. [1]
(b) (i) 2.0 kg m s –1 3 So that, the car can reach
(ii) 0.5 + 1.5 = 2.0 kg m s –1 (c) (i) • The car stopped suddenly the finishing line in a
(c) They are the same. causing a big impulsive shorter time. [1]
(d) (i) The total momentum of the force. [1] • A smaller mass is
balls before the collision and • The front portion of the suitable. [1]
after the collision are the same. car should be able to So that, a bigger
(ii) Principle of conservation of crumple upon impact. [1] acceleration can be
momentum • This will increase the achieved. [1]
(e) No external forces acting on the impact time and reduce • A bigger engine thrust is
system. the impulsive force acting more desirable. [1]
(f) Inelastic collision on the car. [1] So that a bigger
(ii) • The driver flew out of the acceleration can be
2. (a) Rate of change of velocity car because there are achieved. [1]

4.5
(b) u = ——–—— • A smaller resistive force is
(5 × 0.02) no features in the car to more desirable. [1]
= 45 cm s –1 secure him to the car. [1] This is to produce a
22.5
= —–——— • Safety belts should be bigger resultant force
v
(5 × 0.02) fixed to the car to hold the forward. [1]
= 225 cm s –1 driver. [1] (ii) Using the formula
v – u • Air bag should be installed  100 m
a = ———
t to absorb the impact. [1] t = Reaction time + ——– ,
F
225 – 45 • The air bag will cushion Car P: 
= ————
100 × 1.6
2 × 0.1 the driver from the Time = 0.25 + —————
(10.5 – 3.8)
–2
= 900 cm s steering wheel and the = 5.14 s
= 9.0 m s –2 windscreen. [1] Car Q:

(c) Air resistance and friction (iii) • Windscreen glass should Time = 0.45 + —————
100 × 2.0
between the ticker tape and the be designed to fracture (12.8 – 2.6)
ticker timer. into rounded pieces = 4.88 s
10

3. (a) (i) a = —– upon impact instead of [1] Car R:  100 × 1.2
Time = 0.20 + —————
shattering.
5 (7.0 – 2.4)
= 2 m s • So that the driver is less = 5.31 s
–2
likely to be cut by the
15 – 10 Car S:
(ii) a = ———— shattered glass. [1] 
100 × 1.8
15 Time = 0.55 + —————
= 0.33 m s –2 (iv) • The headlight of the car (15.4 – 5.6)
should be bright and
(b) The beach has a larger friction powerful enough to shine = 4.84 s [1]
against the motion of the bicycle over a long distance. [1] Therefore, car S will win the
than the road. Hence, the net • This is to enable the competition. [1]
force acting on the bicycle is driver to see far ahead 10
reduced. and have enough time to (d) (i)
(c) s = Area under the graph avoid obstacles. [1]
= 25 + 187.5 Max 10 F 2 F 1
= 212.5 m Section C F 3 30°
Section B 5. (a) A force can cause:
4. (a) (i) The mass of a body is the • a stationary object to move. 20 N
• a moving object to change its
amount of matter in it. [1] speed. Let F 1 , F 2 and F 3 be the
(ii) • An object at rest resists • a moving object to change its engine thrust, resistive
effort to move it. [1] direction of motion. force and component of the
• An object in motion resists • an object to change in size weight respectively.
effort to stop it. [1] and shape. F 1 = 12.8 N; F 2 = 2.6 N;
• The bigger the mass of an (Any one answer) [1] F 3 = 20 sin 30° = 10 N [1]
object, the more difficult to The resultant force up the
move it. [1] (b) v / m s –1 slope, F
• The bigger the mass of an = F 1 – (F 2 + F 3 )
object, the more difficult to = 12.8 – (2.6 + 10)
stop it. [1] = 0.2 N [1]
• This is the property of all
objects with mass. [1] Therefore, the car can move
[1]
up the slope.
5 0 t / s [2]
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ANS FOC PHYSICF F4 1P.indd 255 29/01/2020 2:04 PM