Focus TG4 KSSM (Physics) Terbitan Penerbitan Pelangi Sdn Bhd

JAWAPAN

1Chapter Checkpoint 1.2 Subjective Questions

Measurement Q1 (a) (i) An inference is an initial Section A
interpretation or explanation
Checkpoint 1.1 concerning the observation. 1. (a) (i) s2 × m s–2 = m

(ii) A hypothesis is a statement (ii) Base quantity
of an expected outcome (b) (i) 0.25 × 0.12 × 0.5 = 0.015 m3
that usually states the
Q1 Physical Base or relationship between two (ii) 1.5 = 100 kg m–3
Description quantity derived or more variables intended 0.015
quantity to be given a direct
experimental test. 2. (a) m / kg T2 / s2
Base
Total 55.2 g Mass quantity (iii) A variable is a physical 2.0 0.8464
quantity that can be varied
150 cm3 of Volume Derived in an experiment. 3.0 1.2996
hot water quantity 4.0 1.7689
(b) (i) The extension of the spring 5.0 2.1609
80°C Temperature Base depends on the weight of 6.0 2.6569
quantity the object attached to it.

Within 3 Time Base (ii) The bigger is the magnitude (b)
minutes quantity of the weight, the greater is
the extension of the spring.
9 kJ of Energy Derived T 2/s2
energy quantity (iii) • Manipulated variable:
Weight of object attached 3.0
to the spring.
Q2 (a) (i) 7 500 000 m 2.5
= 7 500 000 × 10–3 km • Responding variable:
= 7 500 km Extension of the spring. 2.0

(ii) 7 500 000 m • Fixed variable: Type of
= 7 500 000 × 10–6 Mm spring.
= 7.5 Mm
Q2 (a) To investigate the relationship 1.5
between the distance of
(b) 7 853 m s–1 = 7 853 m extension of the elastic cord, x 1.0
1s and the horizontal distance, d
travelled by the ball.
= 7 853 × 10–3 km 0.5
(b) The further the distance of
 1 h extension of the elastic cord, 0 m / kg
60 × 60 x, the further is the horizontal 0 1.0 2.0 3.0 4.0 5.0 6.0
distance, d travelled by the ball.
= 28 270.8 km h–1 2.65 – 1.30
(c) • Manipulated variable: The 6.0 – 3.0
Q3 (a) Base quantity is quantity with distance of extension of the (c) m = = 0.45 s2 kg–1
magnitude only. elastic cord, x
3. (a) (i) Physical quantity is quantity
(b) Vector quantity is quantity with • Responding variable: The that can be measured.
magnitude and direction. horizontal distance travelled
by the ball, d (ii) V is a responding variable
Q4 Event Type of Explanation because its value depends
quantity • Fixed variables: Type of on the other variables and
elastic cord used, type of ball can only be determined
1 Vector Magnitude used and height of the table from the experiment. l is
700 km h–1 and a manipulated variable
direction required (d) • Tabulation of data: because its value can be
to reach its fixed before the experiment
destination. x / cm and other variables depend
on it and only can be
d / cm determined from experiment.

2 Scalar Only magnitude, • Analysis of data: (b) (i) Graph of student A
3 kg required. Plot a graph of d against x. • Weakness: Horizontal

3 Vector Magnitude of If the gradient of the graph axis, 6 divisions for 0.2
force is 25 N is positive, the hypothesis is A is a difficult scale to
and the direction accepted. determine the exact
required to reach position of value l.
the lift. SPM Practice 1 • Weakness: Vertical axis,
Objective Questions 5 divisions for 0.4 V is a
4 Scalar Only magnitude difficult scale to determine
is involved, the 1. C 2. C 3. D 4. D 5. D the exact position of value
temperature from 6. A 7. B 8. C 9. B 10. B V.
20oC to 100oC. 11. D 12. B 13. C 14. D

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Physics Form 4  Answers  

• Strength: The graph Q3 50 ticks per second. (b) Velocity = Gradient of graph
drawn is big. Therefore, 1 tick = 0.02 s
= 2.0 – 0.5
(ii) Graph of student B The time taken for each strip of 3.8 – 2.4
• Strength : Horizontal axis,
5 divisions for 0.2 A is an ticker tape = 1.1 m s–1
easier scale to determine = 5 ticks × 0.02 s
the exact position of value (c) Average velocity
l. = 0.1 s
• Strength : Vertical axis, 5 Displacement
divisions for 0.5 V is an (a) Initial velocity, = Time taken
easier scale to determine 3.0
the exact position of value u = 0.1 = 2.0
V. 5.0
• Weakness : The graph = 30.0 cm s–1
drawn is small. = 0.4 m s–1
10.2
(iii) Graph of student C (b) Final velocity, v = 0.1 Q2 (a) s = Area under the graph from
• Weakness: Horizontal 0 s to 14 s = 100 m
= 102.0 cm s–1
axis, 5 divisions for 0.1 (c) Time taken between u and v, (b) The bus stopped from t = 14 s
A is not suitable because to 20 s. Therefore, it stopped for
it does not cover the  2 t = 1 1 × 0.1 6 s.
whole range of values 2 +1+1+ 2
of l determined from the
experiment. = 0.3 s (c) Gradient of graph
• Weakness: Vertical axis,
5 divisions for 0.4 V is a Acceleration, a = v – u = 0 – 20
difficult scale to determine t 60 – 50
the exact position of value
V. = 102.0 – 30.0 = –20
• Weakness: The graph is 0.3 10
not complete because
some of the data from = 240 cm s–2 = –2.0 m s–2
the experiment cannot be
plotted.  or 2.4 m s–2 (d) Distance between the two bus
stops = Distance travelled from
Q4 (a) u = 8 m s–1; v = 4 m s–1; s = 6 m 20 s to 60 s

s = u + v2 × t = Area under the graph from
2 20 s to 60 s
= 580 m
t = u2+s v
Q3 (a) (i) Total distance travelled
= 2 × 6 = 35 × 2
4 + 8 = 70 m
=1s
(ii) Average speed
(b) a = v – u = 5705
t = 1.27 m s–1 (or 1.3 m s–1)
(b) t = 10 s to 20 s
2Chapter = 4 – 8 (c)
1 v / m s–1

Force and Motion I = –4 m s–2 2

Checkpoint 2.1 (c) u = 4 m s–1; v = 0 m s–1; t = 3 s

Q1 (a) Total distance travelled s = u + v2 × t
2
4 02
= 1.8 + 0.9 + 0.7 + 1.6 = + × 3 1
2
= 5.0 km
5.0 km
(b) Average speed = 2 h = 6 m 0 t/s
5 10 15 20 25 30 35 40 45 50 55
= 2.5 km h–1 Q5 u = 18 m s–1; v = 20 m s–1; s = 10 m
(a) Using the formula, -1
v2 = u2 + 2as,
(c) Displacement = 1.2 km due 400 = 324 + 20 × a

south of Farid’s house. -2
1.2 km
(d) Average velocity = 2 h Therefore, a = 76 Q4 (a) (i) When t = 8 s to 13 s
20
= 0.6 km h–1 (ii) The velocity of the lift was

Q2 (a) (i) Both tapes show uniform v u = 3.8 m s–2 zero during that time.
velocity. (b) (i) When t = 13 s to 20 s
(b) t = –
(ii) Tape P has a lower velocity a
compared with tape Q. (ii) The velocity of the lift during
= 20 – 18
(b) (i) Tape R: The separation 3.8 that time was negative,
between the dots is
increasing. Therefore, = 0.53 s indicating that the lift has
the trolley moved with
increasing velocity. The changed direction.
trolley was accelerating.
Checkpoint 2.2 (c) Total distance travelled
(ii) Tape S: The separation
between the dots is Q1 (a) • From t = 0 to 1 s, the crate = Total area under the graph
decreasing. Therefore,
the trolley moved with moved with uniform velocity. = 20 + 18
decreasing velocity. The • From t = 1 to 2.4 s, the crate
trolley was decelerating. = 38 m

was at rest. (d) Displacement = 20 – 18
• From t = 2.4 to 3.8 s, the
=2m

crate moved with uniform Q5 The area under the graph gives the
value for displacement of 100 m
velocity.
• From t = 3.8 to 5 s, the crate because this is a 100 m event

was at rest.

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  Physics  Form 4  Answers

Checkpoint 2.3 (b) Total momentum after collision (b) t = 0m.2v – mu
= 1200 kg m s–1 F = 0.2
Q1 s = 100 m, a = g = 10 m s–2 (c) Principle of conservation of

u = 0 m s–1 momentum = 600(0) – 600(50)
(d) Total momentum after collision 0.2
s = ut + 1 at 2 = 100 × 6 + 200 × v
2 = 1200 kg m s–1 = –1.5 × 105 N
200v = 600
100 = 0 + 1 (10)t 2 v = 3 m s–1 Magnitude of force = 1.5 × 105 N
2
t = 20 (c) The time in (a) is 10 times
greater than in (b). The old tyres
= 2 5 s or 4.47 s Q2 (a) Total momentum before collision lengthen the stopping time of
= (1500 × 15) + [1000 × (–20)] the car and hence, reduce the
Q2 (a) An object undergoes free fall if = 2500 kg m s–1 impulsive force acted on the car.
it is acted upon by gravitational
force only. (b) Total momentum after collision (d) faor=amFfix, eifdmvaislusemoafllF, .aWisithbigthgaetr,
= Total momentum before collision the racing car can move faster.
(b) The rubber ball will reach the = 2500 kg m s–1
ground first.
(c) Principle of conservation of
(c) The rubber ball has a compact
shape and has less frictional momentum Q2 The action of moving the hand
force from the air compared with backward, the worker will lengthen
a piece of paper. (d) (mvan + mcar)v = (mvanuvan + mcarucar) the impact time between the
watermelon and the hand and hence
Q3 (a) vA = 24 m s–1 (1500 + 1000)v = 2500 reduce the impulsive force.
vB = 0 m s–1 v = 1 m s–1
v2 = u2 + 2as Q3 (a) The front and rear crumple
02 = v2A = 2(–10)(s) zones are designed to crumple
0 = 576 – 20s Q3 (a) Recoil velocity of the rifle = v upon impact in a collision. A
s = 28.8 m 0.012 × 360 = 6v longer impact time will reduce
v = 0.72 m s–1 the impulsive force exerted on
(b) v = u + at (b) Mass of the wooden block = m the car.
0 = 24 + (–10)t 0.012 × 360 = (0.012 + m) × 12
t = 2.4 s m = 0.348 kg or 348 g (b) The strong and rigid cell will
prevent the roof from collapsing
(c) There is no frictional force acting Q4 (a) When a large volume of water on the passengers in the event
on the ball. rushes out of the hose with a when the car overturns. The
very high speed, it has a very passengers will be protected
Checkpoint 2.4 big momentum. According to from direct impact with external
the principle of conservation forces.
Q1 (a) Tin P of momentum, an equal and
opposite momentum is produced Q4 (a) The driver has inertia that keeps
(b) Tin P causing the fireman to fall him moving forward even when
backward if not supported by the car stops suddenly.
(c) The mass of tin Q is bigger another fireman.
(b) The air bag absorbs the initial
and hence, there is a bigger (b) Initially, the twins are at rest impact and cushions the driver
resistance for tin Q to change and the total momentum is from hard objects like the
zero. When they push each steering wheel and windscreen.
its state of being stationary or other and release their hands,
both will acquire momentum of
being in motion. That is, object equal magnitude but in opposite
directions to each other in
with bigger mass has bigger accordance to the principle of
conservation of momentum
inertia. where the final total momentum Checkpoint 2.8
is still zero.
Q2 (a) A fully loaded lorry has a big Q1 (a) Mass of an object is the quantity
mass. Hence, its inertia is big.
of matter in the object.
(b) The mass of a train is big.
Hence, its inertia is big. Weight of an object is the

(c) Steel tubes have big mass. gravitational force acting on the
Hence, the oil platform is not
easily moved. object at a particular place.

Checkpoint 2.6 (b) Weight Mass

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
a = m = 12.0 = 2.4 m s–2 value of g anywhere
(b) The inertia of the pencil box 5.0
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. balance
Q1 (a) u = 50 m s–1, v = 0 m s–1,
(c) The pencil box will move with a (c) mg = 47 × 10
higher velocity and land further m = 600 kg = 470 N
from the obstacle. t =
F = 0m.0v2–, mu
0.02 SPM Practice 2
Objective Questions
Checkpoint 2.5 = 600(0) – 600(50)
0.02
Q1 (a) Total momentum before collision
= 100 × 2 + 200 × 5 = –1.5 × 106 N 1. B 2. C 3. A 4. B 5. C
= 1200 kg m s–1 Magnitude of force = 1.5 × 106 N 6. A 7. C 8. B 9. C 10. B

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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
21. B 22. D 23. D 24. C (b) • Hold the bottle upside down. gives the acceleration of the

Subjective Questions [1] car. [1]

Section A • Give the bottle a downward • The area under the graph

1. (a) Momentum is the product of jerk. [1] gives the distance travelled by
mass and velocity.
• The inertia of the sauce the car. [1]
(b) (i) 2.0 kg m s–1
(ii) 0.5 + 1.5 = 2.0 kg m s–1 will cause it to continue its 4

(c) They are the same. movement downward and (c) (i) • A short reaction time is
(d) (i) The total momentum of the hence, out of the bottle. [1] desirable. [1]

balls before the collision and 3 So that, the car can reach
after the collision are the same.
(c) (i) • The car stopped suddenly the finishing line in a
(ii) Principle of conservation of
momentum causing a big impulsive shorter time. [1]

(e) No external forces acting on the force. [1] • A smaller mass is
system.
• The front portion of the suitable. [1]
(f) Inelastic collision
car should be able to So that, a bigger
2. (a) Rate of change of velocity
(b) u = —(5—×4–.05—.0—2) crumple upon impact. [1] acceleration can be

= 45 cm s–1 • This will increase the achieved. [1]
v = —(5–2×—20.—5.0—2)
impact time and reduce • A bigger engine thrust is
= 225 cm s–1
a = —v —–t —u the impulsive force acting more desirable. [1]
= 2—22—5× –—0.4—15
= 900 cm s–2 on the car. [1] So that a bigger
= 9.0 m s–2
(c) Air resistance and friction (ii) • The driver flew out of the acceleration can be

between the ticker tape and the car because there are achieved. [1]
ticker timer.
no features in the car to • A smaller resistive force is
3. (a ) (i) a = —150–
= 2 m s–2 secure him to the car. [1] more desirable. [1]

( ii) a = —15—1–—5 1—0 • Safety belts should be This is to produce a
= 0.33 m s–2
fixed to the car to hold the bigger resultant force
(b) The beach has a larger friction
against the motion of the bicycle driver. [1] forward. [1]
than the road. Hence, the net
force acting on the bicycle is • Air bag should be installed  ( ii) UtCT i=samirnRePg e =:athc0et.i2of5onr+mti mu(—11ela00—+.05 —1—×–0—F0—13..—m6–8 ,)
reduced.
to absorb the impact. [1] = 5.14 s
(c) s = Area under the graph
= 25 + 187.5 • The air bag will cushion
= 212.5 m
the driver from the

steering wheel and the

windscreen. [1]  Car Q:

(iii) • Windscreen glass should T ime = 0.45 + (—1120—.08—×–—22..—06)

be designed to fracture

into rounded pieces = 4.88 s
 CT iamreR =: 0.20 + —(170—.00—–×—21..4—2)
upon impact instead of

shattering. [1]

• So that the driver is less = 5.31 s

likely to be cut by the  CT iamreS =: 0.55 + (—1150—.04—×–—15..—86)

shattered glass. [1]

(iv) • The headlight of the car

should be bright and = 4.84 s [1]
Therefore, car S will win the
powerful enough to shine

over a long distance. [1] competition. [1]

• This is to enable the 10

driver to see far ahead

and have enough time to (d) (i)

avoid obstacles. [1] F2 F1
F3
Max 10
Section C
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

amount of matter in it. [1] • a moving object to change its Let F1, F2 and F3 be the
engine thrust, resistive
(ii) • An object at rest resists speed. force and component of the
weight respectively.
effort to move it. [1] • a moving object to change its
F1 = 12.8 N; F2 = 2.6 N;
• An object in motion resists direction of motion. F3 = 20 sin 30° = 10 N [1]

effort to stop it. [1] • an object to change in size The resultant force up the
slope, F
• The bigger the mass of an and shape.
= F1 – (F2 + F3)
object, the more difficult to (Any one answer) [1]

move it. [1] (b) v / m s–1

• The bigger the mass of an

object, the more difficult to = 12.8 – (2.6 + 10)

stop it. [1] = 0.2 N [1]

• This is the property of all Therefore, the car can move

objects with mass. [1] t / s [2] up the slope. [1]

5 0

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  Physics  Form 4  Answers

(ii) Acceleration of the car, a Checkpoint 3.3 GM
= —mF R2
Q1 Man-made satellites are artificial (ii) g = GM
gh (R + h)2
= —02..—20 [1] satellite build and made by man. It is

launched by a rocket into space and   R 2 gh
g = R+
= 0.1 m s–2 [1] is placed in a specific orbit round the h

5 earth for specific purposes. R+ h
R
The ISS is the largest man-made  (iii) gh = 2g

3Chapter satellite orbiting the Earth.

  = 2
Natural satellite are any object in 6 400
6 400 + 12 800
Gravitation space orbiting the larger planets.

The moon is a natural satellite × 10 N kg–1

Checkpoint 3.1 orbiting the Earth. = 1.1 N kg–1
W = mg = 81 × 1.1 = 89.1 N
 Q2 v = GM
Q1 F=G Mm r+h 3. (a) A geostationary satellite is a
R2  = 6.67 × 10–11 × 5.97 × 1024 satellite in the geostationary
7×7 orbit round the Earth on an
= 6.67 × 10–11 × 22 6.37 × 106 + 500 000 equatorial plane. It moves in
orbit round the Earth, from east
= 8.17 × 10–10 N = 7 613 m s–1 = 7.61 km s–1 to west, and has a period of

Q2 (a) F ∝ 1  Q3 v = 2GM 24-hour or one-day.
r2 R (b) 3
(c)
(b) F  vY : 1 791 = 2GM
R
 vX : v =
v 2G × 1.41 M
0.919R
 1 791 =
 2G × 1.41M R
r v = 0.919R × 2GM

1.41 × 1 791
0.919
F
Q3 F = mg → g = m = 2 218 m s–1
= =
1 560 26 m s–2 SPM Practice 3 4. (a) (i) T = W = mg
60 Objective Questions = 0.2 × 10 = 2.0 N
(ii) Fc = T = 2.0 N
Q4 F because the force of gravity is the mv2
r
same. (b) Fc =  

Q5 Fc = T = Fw 1. A 2. B 3. D 4. C 5. C v2.=0 =210.01=.×04v.42 7 m s–1
m × ac = M × g 6. C 7. B 8. D 9. B 10. C
0.05 × ac = 0.6 × 10
ac = 120 m s–2 11. B 12. D 13. B 14. D 15. B
v2 16. D 17. C 18. D 19. B 20. C (c) The orbital radius becomes
r 21. D 22. B 23. D 24. C 25. A
ac = smaller.
Subjective Questions
v2 = 120 × 0.6 (d) The centripetal force acting on

v = 8.5 m s–1 Section A the stopper is perpendicular

to the direction of motion of
1. (a) • The force is proportional to the
Checkpoint 3.2 the stopper. The stopper is not
product of the masses of the
displaced in the direction of the
T2 two body.
Q1 r3 = 12 force, thus no work is done.
× 108)3 • The force is inversely
(1.5 5. (a) Mercury:
proportional to the distance
= 2.96 × 10–25 year2 km–3 R3 = 1.85 × 1023 km3
between the two bodies.
T T2Earth 2Mercury Mm T2 = 7 744 day2
Q2 r = r3Earth 3Mercury (b) (i) F = G R2 R3   = 2.52 × 1019
T2
12 T 2Mercury = 6.67 × 10–11 ×
(1.50 × 1011)3 = (5.79 × 1010)3 (8.73 × 1025)(1.03 × 1026) Saturn:
(1.63 × 1012)2
T 2Mercury = 0.575 R3 = 1.26 × 1024 km3
T Mercury = 0.24 year
= 2.26 × 1017 N T2 = 50 625 day2
R3   = 2.49 × 1019
GM (ii) This is because the mass T2
4p2
Q3 r3 = T 2 of the Sun is very large

compared to the mass of Mars:
R3 = 1.19 × 1023 km3
(6.67 × 10–11 × Uranus and Neptune. T2
R3
= 5.98 × 1024) (24 × 60 × 60)2 2. (a) (i) F = mg T2 = 471 969 day2
4p2 (ii) F = GMRm2   = 2.52 × 1019

(R + h)3 = 7.542 × 1022 (iii) mg = GMrm2 → g = G M (b) The value is about the same.
R + h = 4.23 × 107 R2 (c) Kepler III law states that the
h = 4.23 × 107 – 6.37 × 106
(b) (i) gh = GM square of the orbital period of
= 3.59 × 10­7 m (R + h)2 a planet is proportional to the
cube of the orbital radius.

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Physics Form 4  Answers  

(d) Can R3 386 0003 Q3 Q = CDq force per unit area hit on the wall
(e) (i) T2 272 = 1 400 × (25 – 5) produces gas pressure.
  =
= 28 000 J
= 7.89 × 1013 km3 day–2 Q Q2 According to Charles’ Law, the
R3 1J32J Q4 c = mDq volume of gas increases with
(ii) T2   = = 7.89 × 1013 temperature. This is because when
= 42 890 km 54 000 J a fixed mass of gas is heated, the
= 0.6 kg × 200°C gas molecules acquired energy and
increase their kinetic energy. Gas
(iii) v = 2pJ   = 2p(42 890) = 450 J kg–1 °C­–1 molecules move faster and collide
T 24 × 60 × 60 with the wall of the container more
Q5 t = mcDq frequent and this increases the
= 3.12 km s–1 P pressure of the gas. In order to keep
the pressure constant, the excess
= 2 × 4 200 × 50 = 140 s pressure of the gas will produce a
3 000 force pushing the piston upward to
4Chapter increase the volume of the gas.
Heat Q6 (a) Q1 = mcDq
= 0.6 × 4 200 × (100 – 27)
= 183 960 J

Checkpoint 4.1 (b) Q2 = mcDq Q3 The mass and temperature of the
= 0.4 × 900 × (100 – 27) gas are kept constant.
Q1 (a) Thermal equilibrium is the = 26 280 J
process of transferring heat Q4 This is because at 0°C, the gas
between two objects in thermal
contact until both are at the Checkpoint 4.3 molecules are still in a state of
same final temperature and
there is no net heat transfer Q1 The meaning is that to melt or motion and therefore the pressure
between the two objects. freeze 1 kg of ice, the amount heat
absorbed or released is 3.34 × 105J. still exist.
(b) Hot coffee is cooled when ice
cubes are added to it. Q2 (a) Q – mL = 0.2 × 3.34 × 105 Q5 P1 = 1.5 × 105 Pa; V1 = 50 cm3 ;
V2 = 30 cm3 ; P2 = ?
Q2 • Net heat flow between P and Q P1V1
is zero. = 66 800 J P1V1 = P2V2 → P2 = V2  

• Temperature of P and Q is lower (b) Q – mL = 0.4 × 2.26 × 106 = 1.5 × 105 × 50 
than 70°C but higher than 20°C. 30
= 904 000 J
Q3 (a) Liquid suitable for X is mercury.
(b) Melting for ice is 0oC; boiling Q3 (a) Melting point = 75°C = 2.5 × 105 Pa
point is 100oC. (b) t = 6 minute = 360 s
m = 100 g = 0.1 kg Q6 P1 = 100 kPa; T1 = 270 K ;
Q = Pt = 100 × 360
T2 = 324 K ; P2 = ?
PT11   = P2
Q = 36 000 J T2 → P2 = 100   × 324
m 270
Q4 100°C → (16 + 9) = 25 cm L=   = 36 000  
0.1 = 120 kPa
Therefore, 2165  × 100°C = 64°C
= 3.6 × 105 J kg–1 Q7 V1 = 80 cm3 ; T1 = 273 K ;

Q5 (a) 24°C Q4 (a) Q = mcDq + mL T2 = 338 K ; V2 = ?
= 1.5 × 4 200 × 30 + 1.5 × VT11   = V2
(b) 17.2 cm V2 → V2 = 80   × 338
(c) 2 cm → 0°C 3.34 × 105 273
and 16 cm → 68°C = 189 000 + 501 000
= 99.0 cm3
Therefore, 14 cm : 68°C = 690 000 J
(b) Q = mL + mcDq
For increase of temperature of SPM Practice 4
= 0.2 × 2.26 × 106 + 0.2 × 4 200 Objective Questions
20°C, × (100 – 60)
= 452 000 + 33 600
l= 20   × 14 = 4.12 cm
68
= 485 600 J
1. B 2. B 3. C 4. C 5. B
Checkpoint 4.2 Q5 Q = mcDq(ice) + mL(ice) + 6. C 7. A 8. D 9. A 10. D
mcDq(ice) + mL(steam)
Q1 (a) Mass – the larger the mass of 11. B 12. B 13. C 14. B 15. B
the object, the larger the heat = 3.2 × 2 100 × 5 + 3.2 × 3.34 × 16. C 17. B 18. B 19. A 20. B
105 + 3.2 × 4 200 × 100 + 3.2 21. D 22. B 23. C 24. D 25. A
× 2.26 × 106
capacity.
= 9 678 400 J Subjective Questions
(b) Shape – does not affect the

heat capacity. Checkpoint 4.4 Section A

(c) Types of material – different Q1 Based on the kinetic energy of the 1. (a) (i) Thermal equilibrium
gas, the gas molecules are all in (ii) The change of oil
types of materials have different fast, random and continuous motion. temperature is greater than
The gas molecules are always in water.
heat capacity. collision with each other and with (iii) The amount of heat energy
the wall of the container. When a absorbed by oil and water is
Q2 (a) Material T: The largest mass gas molecule collides with the wall the same.
and bounces back, the molecule (iv) The specific heat capacity of
has the highest heat capacity. experiences a change in momentum. oil is smaller than water.
Q The rate of change of momentum (v) PT = mcDq
(b) Specific heat capacity, c = mDq , from the rate of collisions with the 500 × (2 × 60) = 0.5 × c ×
means smaller mDq, the container wall produces a force. The (97 – 25)
larger

specific heat capacity. Material
P has the largest specific heat
capacity due to the smallest mq

value.

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  Physics  Form 4  Answers

c = 500 × 120 Explanation: Shiny materials 2d = v × t
0.5 × 72 can prevent heat loss through = 340 × 2.0
radiation. = 680 m
= 1 667 J kg–1 °C–1 Therefore,
• Suggestion: The covered cup h = 340 m
(b) (i) degree of hotness can be stored in a pocket
made of bamboo cloth. Q2 (a) v = 1500 m s–1, t = 0.12 s
(ii) Volume of gas decreases. Distance travelled by the pulse
Explanation: So that there is = v × t = 1500 × 0.12 = 180 m
(iii) Absolute zero temperature a layer of air trapped inside it
can reduce heat loss.
(0 K or – 273°C)

(iv) The freezing gas molecules

do not move. (b) Distance of the shoal of fish

Section B below the boat = 180   = 90 m
2
2. (a) Heat is a form of energy that
can flow from area of high 5Chapter (c) Ultrasonic waves can transfer more
temperatures to area of low
temperatures. energy than audible sound

(b) (i) • The mass of water in Waves waves.
kettle P is smaller than
the mass of water in Checkpoint 5.1 Checkpoint 5.4
kettle Q.
Q1 (a) A wave is a travelling Q1 (a)
• The heat energy supplied disturbance from a vibrating or
to the kettle P and the oscillating source.
kettle Q are the same.
(b) Transverse waves
• The temperature of water
rise in kettle P is higher (c)
than the temperature of
water rise in kettle Q. Deep Shallow

(ii) The smaller the mass Q2 (a) (i) A wave in which the (b) Refraction of waves
of water, the higher the particles of the medium
temperature rise. move in the direction Q2 (a) f = 16 Hz; λd = 3 ; vd = 6 cm s–1
perpendicular to the λs 2
(iii) The heat supplied must be direction in which the wave
constant. propagates. λd = λλds
vs
(c) The initial temperature of the (ii) Water waves λs
orange juice is higher than the (b) (i) A wave in which the Therefore, vs = λd vd
temperature of the ice cubes.
Heat from the orange juice flows particles of the medium = 2 × 6
to the ice cubes to melt it. The move in the direction 3
heat is continuously absorbed parallel to the direction in
by the cold melting water from which the wave propagates. = 4 cm s–1
the ice. The temperature of the (ii) Sound waves
orange juice decreased and (b) λs = λs = 4 = 0.25 cm
the temperature of the cold f 16
water from the ice increased.
Therefore, orange juice is Q3 (a) Period of oscillation, T = 0.5 s Q3 (a), (b), (c)
cooled by adding ice cubes.
(b) Frequency, f = 1   = 1   Shallow Direction of
(d) • Suggestion: The cup should T 0.5 area propagation
be made of good heat
insulator such as polystyrene = 2.0 Hz
material.
(c) Wavelength, λ = 5 cm Normal
Explanation: Good heat (d) Using v = f λ ,
insulation prevents heat from v = 2.0 x 5 = 10 cm s–1
hot drink being escape by
conduction. Checkpoint 5.2 Deep
area
• Suggestion: The cup cover
should also be made of good Q1 (a) Q
heat insulator such as plastic. (b) Q has the same natural

Explanation: Good insulator frequency as the vibrating tuning Q4
can prevent heat loss to the fork. Hence, Q will receive the
surrounding. Hotter air
biggest amount of energy from Speed of sound higher
• Suggestion: The cup material
should also have a high the vibrating tuning fork.
melting point.
(c) Resonance
Explanation: So that the cup
does not easily change its Q2 The hand moves with the same Refraction of sound waves occur
state or melt when filled with natural frequency as that of the Listener
liquid of high temperature. loaded spring. Resonance occurs Source
and the spring will receive maximum of s
• Suggestion: The inside and energy.
outside of the cup are coated sound
with shiny metal.
Colder air
Speed of sound lower

Checkpoint 5.3 The diagram shows the temperature
of air at night. The layer of air
Q1 Let the distance of the wall from the above the surface of the Earth is
student be d m colder compared the layer further
away from the surface of the Earth.
The distance travelled by sound to
and fro the wall = 2d m

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Since the speed of sound waves (ii) Q3 v = s  , therefore,
is higher at higher temperature of a t
air, refraction occurs and the sound s
waves are refracted towards the t = v  
surface of the Earth. This results in
the listener be able to hear clearer = 320 000 × 1000  
sound at night. 3 × 108
a

= 1.07 s

Checkpoint 5.5 Q2 (a) • A: An up-and-down movement Q4 P : Microwaves; Q : X-rays; R :
Visible light; S : Ultraviolet rays
with a large amplitude
• B: No motion/stationary
Q1 (a) (b) a = 3.0 cm; D = 30.0 cm; SPM Practice
Objective Questions 5
x = 15.0 cm
ax
λ = D 1. D 2. B 3. D 4. C 5. C
6. B 7. C 8. D 9. B 10. B
= 3.0 × 15.0 11. A 12. C 13. C 14. C 15. B
30.0 16. D 17. A 18. B 19. A 20. C
(b) = 1.5 cm 21. B 22. D 23. A 24. D 25. C
(c) 26. D 27. D
Q3 (a) λ = 540 nm = 540 × 10–9 m
a = 0.4 mm = 0.4 × 10–3 m
D = 3.0 m
Using λ = ax , Subjective Questions
D
x = λaD =
540 × 10–9 × 3.0 Section A
0.4 × 10–3 1. (a)

= 4.05 × 10–3 m Direction of
movement of hand
= 4.1 mm
Q2 (a) Remains unchanged (b) • x = λaD
(b) Remains unchanged
(c) Remains unchanged • Hence, if λ is bigger, x will

Q3 (a) also be bigger. Therefore, (Accept single arrow)

the distance between two

adjacent bright fringes of light v
f
will be bigger. (b) λ =

Q4 (a) f = 600 Hz; v = 330 m s–1 = 03.9.00
Using v = f λ,
λ = —vf = —6303—00 = 0.3 m

= 0.55 m (c)
(b) a = 1.5 m; D = 3.0 m;
(b) The leaves which act as x = λaD
obstacles have small width. 0.55 × 3.0
Therefore, the effect of = 1.5
diffraction is greater. Hence the
waves join back after a short = 1.1 m
distance from the leaves.
Checkpoint 5.7 2. (a) Diffraction of waves
(c) The log which is a larger
obstacle causes less diffraction. Q1 1. They all transfer energy from (b) • The amplitude of the waves
The waves join back after a one place to another. before passing through the
futher distance from the log. slit is higher than that the
2. They are all transverse waves. amplitude after passing
Q4 (a) The sound waves spread out through the slit.
beyond the edge of the window 3. They can all travel through a
after passing through the vacuum. • This is because the waves
window. spread over a larger area and
4. They all travel with a speed of energy per unit area of the
(b) Diffraction of sound waves 3 × 108 m s–1 in vacuum. diffracted waves is less.

Checkpoint 5.6 Q2 (a) P : Microwaves; Q : Visible light; (c)
R : X-rays
Q1 (a) The principle of superposition
states that at any time, the (b) P has higher frequency than
combined wave forms of two or radio waves.
more interfering waves is given
by the sum of the displacements (c) Photography / Photosynthesis
of the individual waves at each by plants / Enables human
point of the medium. beings and animals to see. (Any
one answer)
(b) (i)
(d) R has very high penetrating
a 2a power and can cause cancer
and genetic defects to living
cells.

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  Physics  Form 4  Answers

3. (a) d = v × t = 1500 × 0.18 (ii) • There is resistance a particle of the medium

= 270 m between the loaded spring vibrates/oscillates but

(b) The depth of the sunken ship and the oil. [1] does not move along with

= d = 270 = 135 m • As a result, energy is lost the wave. [1]
2 2
from the system as heat 2

(c) Ultrasonic waves have more energy. [1] 7. (a) • The convex-shaped peak of
the wave is like a convex lens.
energy than audio waves. • If no external force is applied, [1]

4. (a) P = X-rays the spring will eventually • The light rays from the lamp
Q = Infrared passing through the peak will
stop completely. [1] be converged on the screen to
produce a bright band. [1]
• Damping. [1]
• The trough is like a concave
(b) (i) 100.6 × 106 Hz or 4 lens. [1]
1.006 × 108 Hz
(d) (i) • The light rays from the lamp
(ii) Wavelength passing through the trough will
be diverged causing a dark
= 3.0 × 108   band. [1]
1.006 × 108
4
= 2.98 m.

(c) For taking the image of the [1]
patient’s bones.
• Move the hand

5. (a) (i) High frequency sound to-and-fro, perpendicular

waves. to the spring. [1] (b) (i) • The frequencies of the
waves in both the deep
(ii) • Ultrasonic waves • The waves move forward and shallow areas are the
same. [1]
transmitted by the bat is along the spring. [1]
• The wavelength of the
reflected by the body of • The string tied to the waves in deep area is
longer than that in shallow
the bird. spring which represents area. [1]

• The reflected waves is a particle of the medium • The speed of the waves
in deep area is higher
detected by the bat. oscillates perpendicularly than that in shallow area.
[1]
• The time between to the direction of the
(ii) • When the angle of
the transmission and waves. [1] incidence is zero, the
direction of waves
detection of the signal by • This shows that in moving from deep to
shallow areas remains
the bat is equal to twice the propagation of a unchanged. [1]

the distance between the transverse wave, the • When the angle of
incidence is not zero,
bat and the bird. particle of the medium the waves change their

• With that, the distance vibrates/oscillates direction when moving from
deep to shallow areas. [1]
between the bat and the perpendicularly with the
5
bird can be estimated direction of the wave. [1]

through the period of Max 4
(ii)
time transmission and

detection of the ultrasonic

waves.

(b) (i) Distance = d m [1]
2d = 1 450 × 120 × 10-3.
Therefore d = 87 m. • Move the hand to-and-fro,

(ii) v = fλ parallel with the spring. [1]
1 450 = 45 × 103 × λ
• The waves move forward

1450 along the spring. [1] (c) Refraction of waves [1]
45 × 103
Therefore, λ = • The string tied to the (d) (i) • The resort is to be built

= 3.2 × 10-2 m spring which represents near the bay [1]

a particle of the medium • The waves at the bay are

Section B oscillates parallel with the calmer than at the cape [1]

6. (a) A system that undergoes a direction of the waves. [1] • due to the divergence of

periodic to-and-fro movement. [1] • This shows that in the waves’ energy from
(b) • Spring B carries a bigger
the propagation of a the bay [1]
mass than that of spring A. [1]
• Spring B oscillates with a longitudinal wave, the • and the convergence of

bigger period than spring A. [1] particle of the medium the waves at the cape [1]
• Spring B oscillates with a
vibrates/oscillates parallel (ii) • To reduce erosion,

with the direction of the retaining walls are built [1]

lower frequency compared to wave. [1] • to reflect the waves from
spring A. [1]
Max 4 the shore [1]

• The bigger is the mass (iii) • Work done by the hand in • and to reduce direct

attached to the spring, the moving the spring impact of the waves on

lower is the frequency of its to-and-fro causes the the shore [1]

oscillation. [1] energy to be transferred (iii) • Concrete structures with

4 in the form of waves a gap in between are built
(c) (i) Oscillating with the
amplitude decreasing with along the spring. [1] at the designated area for
time. [1]
• The string tied to the children [1]

spring which represents • Waves passing through

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Physics Form 4  Answers  

the gap will be diffracted • Radio waves with longer (c) sin 450 = 0.7071
in the children’s area [1] r = sin–1 0.4877 = 29.2°
• The smaller amplitude wavelength will diffract
of the diffracted waves
causes the sea to be around the mountain Q4 (a) The speed of light in glass block
calmer. [1]
more compared with is slower than the speed of
10
radio waves with shorter light in water. This is because

wavelength [1] the refractive index of glass is

(iii) • Example 2 [1] larger. Therefore, its density is

• There is a vacuum greater than that of water.

Section C between the Sun and the n = cair
cmedium
8. (a) Refraction of waves is a Earth [1] (b)

phenomenon that occurs when • The fact that ultraviolet cair
cwater
there is a change of direction rays reach the 1.33 =

of the propagation of waves Earth shows that cair
cglass
travelling from one medium to electromagnetic waves and 1.52 =

another due to a change of travel through vacuum [1] cair

speed. [1] Max 7 cglass cwater 1.33
cwater cair 1.52
(b) (i) Sound waves travel faster in = = 0.88

air when the temperature of cglass

the air is higher. [1] 6Chapter

(ii) • At daytime, when it is hot, Checkpoint 6.2

the surface of the Earth Light and Optics Q1 (a) The angle of incidence ray BC
is larger than the critical angle.
heats up faster than the Thus, total internal reflection
occurs.
air. [1] Checkpoint 6.1
(b) Light ray DE refract away
• Hence, the hot surface of Q1 i = 140° – 90° = 50° from normal because it leaves
from glass prism to air, from a
the Earth causes the layer r = 125° – 90° = 35° denser medium to a less dense
medium.
of air near the surface to sin i
sin r
be warmer than the upper n =

layer. [1]

• This causes the sound = sin 50°
sin 35°
waves to be refracted

away from the Earth. [1] = 1.34 n= 1 1
sin c sin 42°
• On a cool night, the Q2 Da = 12 – 10 = 2 cm Q2 = = 1.49

surface of the Earth cools Dr Dr sin P
Da 2 sin 40°
down faster than the air. n = → 1.50 = 1.49 =

[1] sin P = 1.49 × sin 40°

• Hence, the air near the Dr = 1.5 × 2 = 3.0 cm

surface of the Earth \The thickness of glass block is = 0.9578
P = 73.3°
becomes cooler than the   3.0 cm Q = 90° – 40° = 50°
R = Q = 50°
upper layer. [1] S = 90° – 50° = 40°

• This causes the sound Q3 i sin i r sin r

waves to be refracted 22 0.3746 15 0.2588

towards the Earth. [1]

6 28 0.4695 19 0.3256 1.49 sin 40° = 1 sin T
T = sin–1 (1.49 sin 40°)
(c) (i) • Transverse waves [1] 33 0.5446 22 0.3746 T = 73.3°

• Travel with a speed of 35 0.5736 26 0.4384

3 × 108 m s–1 in 41 0.6561 27 0.4540 Q3

vacuum [1]

• Transfer energy from one
sin i
place to another

(Other characteristics 0.9

acceptable) [1]

3 0.7 • Periscope is used to see an
(ii) 1. Sound wave does not object behind the observer.
travel in vacuum. [1] 0.5
2. Sound wave is a • Image formed is inverted and
longitudinal wave. [1] same size as object.

2

(d) (i) • Example 3 [1] 0.3 sin r Checkpoint 6.3
0.2 0.3 0.4 0.5 0.6
• The microwaves

transmitted will be Q1 (a) The focal point is the point on
the principal axis where the light
reflected by the plane and (a) Set of reading for angle of rays parallel to the principal axis
incident = 35° after passing through the lens
return to the receiver [1] will converge on it.

(ii) • Example 1 [1] n = 0.454
0.6561
• The mountain is an (b)

obstacle [1] = 1.45

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(b) (i) Virtual image is image that Q2 m = v = hi = 15 = 1.5 (c) The separation distance of the
u ho 10
cannot be displayed on the lenses

screen. Therefore, v = 1.5u = 40 + 5

(ii) 1 cm = 1.5 × 2.0 m = 3.0 m = 45 cm

Converging lens Using 1 + 1 = 1 (d)
u v f
1 cm Parallel rays 5 cm
40 cm
1 1 1 5
2 + 3 = f = 6 from distant

Therefore, focal length f = 1.2 m object

Image Object F Main axis Q3 Using 1u + 1 = 1 Lens P Lens Q
f = 2 cm v f
1 1
1 = 1 – 1 = – 10 – 30
v f u
Q3 Similarity:
= – 4 • Consists of two convex lenses
30 • The eyepiece function as a
v = –7.5 m magnifying glass
Virtual image is image that Differences:
Q2 (a) cannot be displaced on the • For microscope in normal
screen = 2 cm v 30 adjustment, the final image is at
u near point. While for telescope in
m = = 4 = 1 normal adjustment, the final image
30 4 is at infinity.
• The distance between the
The image is virtual, upright, objective lens and the eyepiece of
the microscope is LO > fO + fe. The
at a distance of 7.5 cm from distance between the objective
lens and the eyepiece of the
the lens on the same side telescope is LO ≤ fO + fe.

Image as the object and has linear
F magnification of 41.

Q4

v/cm

(b) The size of the image does not 60 Checkpoint 6.6
change but the brightness of 50 u = v Q1 (a)
image reduced by half. 42 P
40 Q
Q3 (a) CMo=nvuvex=le63n00s= F
(b) R
0.5

(c) 30

F 20

u = 60 cm 10

vf = 20 cm 0 10 1820 30 40 50 60 u/cm
= 30 cm

Focal length, f = 20 cm v (b) The image is real, inverted and
u magnified.
Checkpoint 6.4 (a) m = = 42 = 7 = 2.33
18 3 (c) The image is nearer to the
(b) u = v = 25 cm mirror and becomes smaller in
Q1 (a) 1u + 1 = 1 size.
v f f = 25
2 Q2 (a)
1 = 1 – 1 = 1 – 1 = 1
v f u 20 30 60 = 12.5

v = 60 cm
v
m = u = 60 = 2 Checkpoint 6.5
30
Q1 (a) For distant objects, the image
Therefore, the image is real, must be formed at the focal 25°
inverted, 60 cm from the lens, point of the lens. Therefore, the FC
located opposite to the object distance between the lens and
and enlarged 2 times. the film is 50 mm. (b) The image is at F.
(c) The image is virtual, upright and
(b) 1 + 1 = 1 (b) The camera lens should be
u v f adjusted outward toward the diminished.
object. As the object distance Q3 • Focal length of mirror P = 12 cm.
1 = 1 – 1 = 1 – 1 = 1 is reduced, the image distance
v f u 20 15 60 increases. With this, the The image from distant object is
distance between the lens and formed at the focal point of the
v = –60 cm the film is slightly longer than mirror.

m = v = 60 = –4 50 mm.
u 15

The image is virtual, upright, at Q2 (a) Astronomical telescopes
a distance of 60 cm from the (b) Objective lens: lens P, Eyepiece:
lens on the same side as the lens Q
object and is magnified 4 times.

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Physics Form 4  Answers  

• Focal length of mirror Q = 10 cm. (b) The eyepiece is used to
Object is located at radius of observe the image formed
45° 45° by the second prism.
curvature, image distance = 2f. 45° 45° (ii)
• Therefore, the mirror Q has a Right-angle prism

shorter focal length. 45° Distant
object at
SPM Practice 6 45° the back
Objective Questions 45°
Cylindrical
1. B 2. C 3. B 4. C 5. B tube
6. C 7. C 8. C 9. D 10. D
11. C 12. A 13. B 14. B 15. C 45° 45°
16. A 17. C 18. A 19. A 20. C 45°
21. C 22. D 23. D 24. B 25. C

Subjective Questions An inverted

image Right-angle prism

Section A (c) (i) When the prism at the
1. (a) top periscope is laterally
Glass inverted, it can be used to
observe the object behind
Air the observer over an
obstacle. In this case, the
Light ray Total internal image seen is inverted
reflection occurs here
(iii) • A right-angle prism uses
(b) (i) f = 10 cm, u = 12 cm A light ray is transmitted total internal reflection
through a curved fibre optic to reflect light. The light
1 = 1 + 1 → 1 = 1 + 1 that occurs a series of total reflected by the prism
f u v 10 12 v internal reflections on the surface is 100%.
inner surface.
1 = 1 – 1 = 6–5 = 1 • Image formed by right-
v 10 12 60 60 (ii) • Fiber optic cables are angle prism is sharper.
lighter and thinner.
v = 60 cm
• Transmission of signal
Therefore, distance between with almost no energy
loss along the optical
the screen and the object fiber. PRE-SPM MODEL PAPER
= u + v = 12 + 60 = 72 cm
v (d) (i) Objective Questions
(ii) m = u = 60 = 5
12 Right-angle 1. B 2. B 3. C 4. A 5. D
prism 6. B 7. C 8. A 9. B 10. C
(c) The minimum height of the 11. A 12. A 13. B 14. A 15. A
screen = 24 × 5 16. B 17. C 18. A 19. D 20. C
21. B 22. A 23. C 24. A 25. B
= 120 cm 26. C 27. D 28. A 29. D 30. A
31. C 32. C 33. A 34. D 35. C
(d) If the distance between the lens Distant object Cylindrical 36. D 37. A 38. D 39. D 40. D
and the image = 10 cm = f, then tube 41. C 42. B 43. A 44. B 45. B
object is at the focal point and 46. B 47. D 48. C 49. C 50. C
image formed is virtual and at
infinity. Therefore, image cannot Subjective Questions
be seen on the screen.
Section A
Image at infinity Upright image
1. (a) • time / length
FF Right-angle • velocity / force
prism (b) (i) acceleration
Section B (ii) the original length of the
• A periscope is build using spring
2. (a) (i) The critical angle of glass is two right-angle prisms
42° which means that if the arrange in an cylindrical 2. (a) Initially, there is no air resistance
tube as shown in the acted on the skydiver. Therefore,
angle of incidence in glass diagram above. the skydiver experience a free
fall due to gravitational force.
(denser medium) is 42°, • The prism acts like a
perfect mirror when light (b) When the parachute is opened,
then the angle of refraction ray hits the inner surface the air resistance acting on the
of the prism at an angle parachute is greater than the
in the air is 90°. greater than 42°. The first weight of the skydiver. The net
prism turns the image force acting on the parachute
(ii) Total internal reflection from a distant object 90°, upward causes upward
and then the second acceleration.
(iii) sin c = 1 prism turns the image
n back to upright.

n = 1 c • Object lens is placed in
sin front of the first prism on
1 top to find a distant object.
= sin 42

= 1.49

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  Physics  Form 4  Answers

(c) the air resistance is balanced by (d) Size of image formed does not • the increase in air
the weight of the skydiver. change, but the brightness of temperature in Diagram
the image formed is less. 9.1 is smaller.
3. (a) • The force holding the satellites
(ii) The temperature increases
and the Moon in their orbit is 7. (a) Heat is a formed of energy. as the heat supplied
(b) (i) The thermometer reading in increases. As the air
the gravitional force. temperature increases, the
T2 air pressure in the flask
• The ratio R­3 for all satellites is Diagram 7.1 is higher. increases.

a constant. (ii) The volume of water in (c) According to kinetic theory, the
kinetic energy of gas molecules
RT2­3 = (386 000)3 Diagram 7.2 is larger. increases as heat is absorbed.
(27 × 24)­2 The air molecules in the flask
(b) (iii) The time taken to heat the move faster and collide more
frequently with the flask wall and
water is the same, which is thus increases the air pressure
in the flask.
=1.37 × 1011 km3 jam–2 5 minutes.
(d) (i) The sheath mass should be
(c) (2R4E)­23 = 1.37× 1011 (c) (i) The smaller the volume of small and light so that the
R3E = 1.37 × 1011 × 242 aneroid barometer is easy
RE = 42 892 km water, the larger the reading to carry.

of the thermometer. (ii) The chamber should be
made of light metal box
(ii) The smaller the mass of enclosed with corrugated
edges so that it is flexible
water, the higher the rise in and can move up or down.

4. (a) Water vapour temperature. The air in the chamber
is released into a partial
(b) The heat energy supplied (d) (i) Amount of heat supplied, vacuum, so that the space
Q = 16 × (1 × 60 × 60) can respond to changes
is used to release the water in the outside air pressure
= 57 600 J more effectively.
molecules into the air. The (ii) Q = mcDq
57 600 = 0.8 × 4 200 × Dq (iii) Spring must be strong
kinetic energy of the molecules and easily extend in order
to respond well to the
does not change. 57 600 movement of the chamber.
0.8 × 4 200
(c) Heat used to change 20 g of DT = (iv) The lever should be light
and easy to move in order
water to steam, = 17.1°C to simplify the movement of
Q = 52 000 – 6 000 the chamber and pointers.

= 46 000 J 8. (a) (i) Velocity is the rate of 10. (a) A sound wave is a kind of
change of displacement longitudinal wave, produced
Mass of water change to steam, from vibrating object in which
m = 20 g = 20 × 10–3 kg or distance move in one the particles in the medium
move forward and backward
Specific latent heat of second in a specific in the direction of the wave
propagation.
Lva p=ormQisation, direction.
(b) (i) The distance x in Diagram
(ii) Velocity change because its 10.1 is smaller than the
distance x in Diagram 10.2.
= 46 000 direction change. The distance a in Diagram
20 × 10–3 10.1 is smaller than the
(iii) Force of gravity towards distance a in Diagram 10.2.
The larger the value of a,
= 2.3 × 106 J kg–1 the centre of Earth. This the smaller value of x.

(d) When water evaporates, force produces centripetal (ii) The distance D, between
the water molecules acquire the speaker and the
energy and change into vapour force perpendicularly to the interference pattern, must
and leaves the less energy be constant. The physical
molecules behind. This reduces direction of motion. phenomenon is interference
the average kinetic energy of of sound wave.
the molecules and thus the (b) (i) Resultant force
water is cooled down. (c) (i) Sound waves with
= Upthrust – rocket weight frequencies exceeding
5. (a) transverse wave
(b) • same frequency Upthrust 20 kHz are called ultrasonic
• ld > lc = 50 000 N + 40 000 × 10 N waves.
(c) • ld = 1.5 cm
• lc = 1.0 cm = 450 000 N (ii) • Ultrasonic signals are
(d) v = fl = 5.0 × 1.0 = 5.0 cm s–1 (ii) Resultant force, F = ma transmitted into the sea
(e) 50 000 = 40 000 × a until reaches the sea floor.

Acceleration, a = 50 000
40 000

= 1.25 m s–2

(c) (i)

Velocity

Deep Shallow
region region

6. (a) Refraction of light 0 2 4 6 8 10 12
(b) (i) Thickness of lens P < Time after launching / min
thickness of lens Q
(ii) f1 > f2 (ii) Area under the graph
(iii) h1 > h2
(c) (i) The thicker the lens, the Section B
shorter the focal length.
(ii) The longer the focal length, 9. (a) To measure gas presure
the smaller the size of (b) (i) • the reading of Bourdon
image formed. gauge in Diagram 9.1 is
smaller.
• the heat supplied to
Diagram 9.1 is smaller.

264

ANS FOC PHYSICF F4 1P.indd 264 29/01/2020 2:04 PM

Physics Form 4  Answers  

• The time interval for the (iii) The cartwheel must be large Mass, m = 200 g = 0.2 kg
signal or echo to arrive is Using, Q = mcDq
recorded. in diameter. So that cart

• From ultrasonic velocity, can be easily move and Therefore, specific heat
uvs=in2gtdt,hdeefpotrhmiuslacadlc=ulav2tte.d capacity of, X,
the pushing force can be
(d) (i) The speakers should be
arranged in such a way that reduced. c = Q 6 000
no sound wave interference mDq 0.2 × 40
occurs so that the same (iv) The cartwheel should made =
loudness can be heard in all
directions. of hard rubber. So that the

(ii) The speaker must be cart can be easily move and = 750 J kg–1 °C
powerful enough to
transmit sound waves in not easily damaged when (ii) Time taken, t
all directions so that sound =8–4=4
can be heard clearly and carrying heavy goods. = 4 × 60
in all directions with equal The L trolley is the perfect = 240 s
loudness.
choice because it has the right Heat supply, Q
(iii) A good amplifier along with = 25 × 240
a good microphone is used height of handle, light and strong = 6 000 J
so that a person›s voice
can be clearly detected and base, large wheel diameter and Mass, m = 200 g = 0.2 kg
amplified. Using, Q = mL
made of hard rubber
(iv) The walls of the concert hall
should be fitted with small (d) (i) Change in velocity
hollow soft boards so that
sound waves can be fully = –22 + –25
absorbed without reflection.
= –47 m s–1
The floor should be
carpeted to avoid sound (ii) Average acceleration Therefore, specific latent
wave reflection. heat of fusion of X,
a = Change in velocity
Section C Time taken
Q
11. (a) Impulse is change in momentum = 47 L = m = 6 000 J
(b) When the player hits a tennis 0.0013 0.2 kg
ball, he or she can increase the
contact time with the racket. = 3.6 × 104 = 30 000 J kg–1
This will add impulse to the ball
and increase the momentum (iii) Average force, (c) • The cover should be made
change on the ball. Therefore, F = m × a of hollow plastic. Air trapped
the speed of the tennis ball in hollow plastic is a good
increases. = 0.16 × 3.6 × 104 thermal insulator and can
prevent heat lost to the
(c) (i) The height of the handle = 5 760 N surrounding.
should be approximately the
height of the shoulder of the 12. (a) (i) The specific latent heat of • Space P should be a vacuum.
pusher. So that the cart is vaporisation of water is the The vacuum can prevent heat
easy to push. Lower handle amount of heat required to lost through conduction and
needs more force to push. convert 1 kg of water into convection.
steam without temperature
(ii) The base of the cart should change. • Two-layer wall must be made
be light and strong. So that of glass. Glass is a good
the cart is easily push and (ii) Water has a large specific insulator of heat and has a
the hard material is not latent heat of vaporisation. greater specific heat capacity.
easily damaged and can When steam condenses it This feature can reduce heat
last longer. releases a large amount transferred to the surrounding.
of heat. Foods like fish in
a plate can absorb a lot of • The wall should be coated
heat from hot steam. The with shinny paint. The shiny
food becomes cooked when surface is a good heat
the food reaches thermal reflector and can reduce heat
equilibrium with the steam loss.
after some time.
• The thermos flask S is the
(b) (i) Time taken, t = 4 min best choice because it uses
= 4 × 60 s hollow plastic as cover, the
= 240 s space P is vacuum, the two-
layer wall made of glass and
Change in temperature, ΔT coated with shiny paint.
= 60° – 20°

= 40°C
Heat supplied, Q
= P × t
= 25 × 240
= 6 000 J

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NOTE

ANS FOC PHYSICF F4 1P.indd 266 29/01/2020 2:04 PM