Focus TG4 KSSM (Physics) Terbitan Penerbitan Pelangi Sdn Bhd

JAWAPAN




Chapter Checkpoint 1.2 Subjective Questions
1 Measurement Q1 (a) (i) An inference is an initial Section A
interpretation or explanation
concerning the observation. 1. (a) (i) s × m s = m
–2
2
(ii) A hypothesis is a statement (ii) Base quantity
Checkpoint 1.1
of an expected outcome (b) (i) 0.25 × 0.12 × 0.5 = 0.015 m 3
Q1 Physical Base or that usually states the (ii) 1.5 = 100 kg m –3
Description derived relationship between two 0.015
quantity or more variables intended
quantity 2. (a) m / kg T / s 2
2
Base to be given a direct
Total 55.2 g Mass experimental test. 2.0 0.8464
quantity (iii) A variable is a physical
3
150 cm of Volume Derived quantity that can be varied 3.0 1.2996
hot water quantity in an experiment. 4.0 1.7689
(b) (i) The extension of the spring
Base 5.0 2.1609
80°C Temperature depends on the weight of
quantity the object attached to it. 6.0 2.6569
Within 3 Time Base (ii) The bigger is the magnitude
minutes quantity of the weight, the greater is (b)
the extension of the spring.
9 kJ of Energy Derived (iii) • Manipulated variable: T /s 2
2
energy quantity
Weight of object attached 3.0
to the spring.
Q2 (a) (i) 7 500 000 m • Responding variable: 2.5
= 7 500 000 × 10 km Extension of the spring.
–3
= 7 500 km • Fixed variable: Type of 2.0
(ii) 7 500 000 m spring.
= 7 500 000 × 10 Mm Q2 (a) To investigate the relationship
–6
= 7.5 Mm between the distance of 1.5
extension of the elastic cord, x
7 853 m 1.0
(b) 7 853 m s = 1 s and the horizontal distance, d
–1
–3
7 853 × 10 km travelled by the ball. 0.5
= (b) The further the distance of
1
 60 × 60 h extension of the elastic cord, 0 m / kg
x, the further is the horizontal
distance, d travelled by the ball. 0 1.0 2.0 3.0 4.0 5.0 6.0
= 28 270.8 km h –1
(c) • Manipulated variable: The 2.65 – 1.30 2 –1
Q3 (a) Base quantity is quantity with distance of extension of the (c) m = 6.0 – 3.0 = 0.45 s kg
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 (ii) V is a responding variable
by the ball, d
Q4 Type of • Fixed variables: Type of because its value depends
Event Explanation on the other variables and
quantity elastic cord used, type of ball can only be determined
used and height of the table
1 Vector Magnitude (d) • Tabulation of data: from the experiment. l is
–1
700 km h and a manipulated variable
direction required x / cm because its value can be
to reach its d / cm fixed before the experiment
destination. and other variables depend
2 Scalar Only magnitude, • Analysis of data: on it and only can be
3 kg required. Plot a graph of d against x. determined from experiment.
If the gradient of the graph (b) (i) Graph of student A
3 Vector Magnitude of is positive, the hypothesis is • Weakness: Horizontal
force is 25 N accepted. axis, 6 divisions for 0.2
and the direction A is a difficult scale to
required to reach SPM Practice 1 determine the exact
the lift. position of value l.
Objective Questions • Weakness: Vertical axis,
4 Scalar Only magnitude
is involved, the 1. C 2. C 3. D 4. D 5. D 5 divisions for 0.4 V is a
temperature from 6. A 7. B 8. C 9. B 10. B difficult scale to determine
20 C to 100 C. 11. D 12. B 13. C 14. D the exact position of value
o
o
V.
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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, ticker tape = 1.1 m s –1
5 divisions for 0.2 A is an = 5 ticks × 0.02 s (c) Average velocity
easier scale to determine = 0.1 s
the exact position of value (a) Initial velocity, = Displacement
l. 3.0 Time taken
• Strength : Vertical axis, 5 u = 0.1 = 2.0
divisions for 0.5 V is an = 30.0 cm s –1 5.0
easier scale to determine = 0.4 m s –1
the exact position of value (b) Final velocity, v = 10.2 Q2 (a) s = Area under the graph from
0.1
V. 0 s to 14 s = 100 m
• Weakness : The graph = 102.0 cm s –1 (b) The bus stopped from t = 14 s
drawn is small. (c) Time taken between u and v, to 20 s. Therefore, it stopped for
(iii) Graph of student C t =  1 + 1 + 1 + 1  × 0.1 6 s.
• Weakness: Horizontal 2 2
axis, 5 divisions for 0.1 = 0.3 s (c) Gradient of graph
A is not suitable because Acceleration, a = v – u = 0 – 20
it does not cover the t 60 – 50
whole range of values 102.0 – 30.0 = –20
10
of l determined from the = 0.3 = –2.0 m s –2
experiment. = 240 cm s –2
• Weakness: Vertical axis, or 2.4 m s –2 (d) Distance between the two bus
5 divisions for 0.4 V is a stops = Distance travelled from
–1
–1
difficult scale to determine Q4 (a) u = 8 m s ; v = 4 m s ; s = 6 m 20 s to 60 s
the exact position of value s =  u + v  × t = Area under the graph from
V. 2 20 s to 60 s
• Weakness: The graph is t = 2s = 580 m
not complete because u + v Q3 (a) (i) Total distance travelled
some of the data from = 2 × 6 = 35 × 2
the experiment cannot be 4 + 8 = 70 m
plotted. = 1 s (ii) Average speed
(b) a = v – u 70
t = 55
Chapter = 4 – 8 = 1.27 m s (or 1.3 m s )
–1
–1
1
2 Force and Motion I = –4 m s –2 (b) t = 10 s to 20 s
(c)
(c) u = 4 m s ; v = 0 m s ; t = 3 s –1
–1
–1
Checkpoint 2.1 v / m s
s =  u + v  × t
Q1 (a) Total distance travelled 2 2
= 1.8 + 0.9 + 0.7 + 1.6 =  4 + 0  × 3 1
= 5.0 km 2
(b) Average speed = 5.0 km = 6 m 0 t / s
2 h 510152025303540455055
–1
–1
= 2.5 km h –1 Q5 u = 18 m s ; v = 20 m s ; s = 10 m
(a) Using the formula, -1
(c) Displacement = 1.2 km due 2 2
south of Farid’s house. v = u + 2as, -2
(d) Average velocity = 1.2 km 400 = 324 + 20 × a
76
2 h
= 0.6 km h –1 Therefore, a = 20 Q4 (a) (i) When t = 8 s to 13 s
Q2 (a) (i) Both tapes show uniform = 3.8 m s –2 (ii) The velocity of the lift was
zero during that time.
velocity. (b) t = v – u (b) (i) When t = 13 s to 20 s
(ii) Tape P has a lower velocity a (ii) The velocity of the lift during
compared with tape Q. = 20 – 18 that time was negative,
(b) (i) Tape R: The separation 3.8 indicating that the lift has
between the dots is = 0.53 s changed direction.
increasing. Therefore, Checkpoint (c) Total distance travelled
the trolley moved with 2.2 = Total area under the graph
increasing velocity. The Q1 (a) • From t = 0 to 1 s, the crate = 20 + 18
trolley was accelerating. moved with uniform velocity. = 38 m
(ii) Tape S: The separation • From t = 1 to 2.4 s, the crate (d) Displacement = 20 – 18
between the dots is was at rest. = 2 m
decreasing. Therefore, • From t = 2.4 to 3.8 s, the
the trolley moved with crate moved with uniform Q5 The area under the graph gives the
decreasing velocity. The velocity. value for displacement of 100 m
trolley was decelerating. • 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 = 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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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]
255
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Physics Form 4 Answers

(ii) Acceleration of the car, a Checkpoint 3.3 GM
F Q1 Man-made satellites are artificial (ii) g = R 2
= —
m
GM
0.2 [1] satellite build and made by man. It is g h (R + h) 2
——
=
2.0 launched by a rocket into space and R 2
= 0.1 m s –2 [1] is placed in a specific orbit round the g =  R + h  g h
5 earth for specific purposes.
The ISS is the largest man-made (iii) g h =  R + h  2 g
satellite orbiting the Earth. R
Chapter Natural satellite are any object in =  6 400  2
3 Gravitation space orbiting the larger planets. 6 400 + 12 800
The moon is a natural satellite × 10 N kg –1
orbiting the Earth. = 1.1 N kg –1
Checkpoint 3.1 W = mg = 81 × 1.1 = 89.1 N
GM
Q1 F = G Mm Q2 v =  r + h 3. (a) A geostationary satellite is a
R 2 satellite in the geostationary
–11
= 6.67 × 10 × 7 × 7 =  6.67 × 10 × 5.97 × 10 24 orbit round the Earth on an
–11
6.37 × 10 + 500 000
2
2
6
= 8.17 × 10 N equatorial plane. It moves in
–10
= 7 613 m s = 7.61 km s –1 orbit round the Earth, from east
–1
1
Q2 (a) F ∝ to west, and has a period of
r 2 Q3 v =  2GM 24-hour or one-day.
(b) F R (b) 3
v Y : 1 791 =  2GM (c)
R
v X : v =  2G × 1.41 M
0.919R
v  2G × 1.41M R
=
1 791 0.919R × 2GM
v =  1.41 × 1 791
r 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 = 0.2 × 10 = 2.0 N
Q4 F because the force of gravity is the Objective Questions (ii) F c = T = 2.0 N
same. (b) F c = mv 2
1. A 2. B 3. D 4. C 5. C r
Q5 F c = T = F w 6. C 7. B 8. D 9. B 10. C 1.0 × v 2
m × a c = M × g 2.0 = 1.0

0.05 × a c = 0.6 × 10 11. B 12. D 13. B 14. D 15. B v = 20 = 4.47 m s –1
a c = 120 m s –2 16. D 17. C 18. D 19. B 20. C (c) The orbital radius becomes
21. D 22. B 23. D 24. C 25. A
a c = v 2 smaller.
r Subjective Questions (d) The centripetal force acting on
v = 120 × 0.6 the stopper is perpendicular
2
v = 8.5 m s –1 Section A to the direction of motion of
1. (a) • The force is proportional to the the stopper. The stopper is not
Checkpoint 3.2 product of the masses of the displaced in the direction of the
two body.
Q1 T 2 = 1 2 • The force is inversely force, thus no work is done.
r 3 (1.5 × 10 ) 5. (a) Mercury:
8 3
= 2.96 × 10 year km –3 proportional to the distance R = 1.85 × 10 km 3
–25
2
23
3
between the two bodies.
2
2 2 T = 7 744 day 2
Q2 T Earth = T Mercury (b) (i) F = G Mm R 3
3
3
19
r Earth r Mercury R 2 T 2 = 2.52 × 10
2
1 2 = T Mercury = 6.67 × 10 ×
–11
10 3
(1.50 × 10 ) (5.79 × 10 ) (8.73 × 10 )(1.03 × 10 ) Saturn:
11 3
26
25
12 2 R = 1.26 × 10 km 3
3
24
2
T Mercury = 0.575 (1.63 × 10 ) 2 2
17
T Mercury = 0.24 year = 2.26 × 10 N T = 50 625 day
R
3
(ii) This is because the mass T 2 = 2.49 × 10
19
Q3 r = GM T 2 of the Sun is very large
3
4p 2 compared to the mass of Mars:
23
3
(6.67 × 10 × Uranus and Neptune. R = 1.19 × 10 km 3
–11
2
24
5.98 × 10 ) T = 471 969 day 2
= (24 × 60 × 60) 2 2. (a) (i) F = mg R 3
4p 2 Mm 2 = 2.52 × 10
19
(ii) F = G T
(R + h) = 7.542 × 10 22 R 2 (b) The value is about the same.
3
R + h = 4.23 × 10 7 (iii) mg = G Mm → g = G M 2 (c) Kepler III law states that the
R
r
2
7
h = 4.23 × 10 – 6.37 × 10 6 square of the orbital period of
7
= 3.59 × 10 m (b) (i) g h = GM a planet is proportional to the
(R + h) 2
cube of the orbital radius.
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Physics Form 4 Answers

(d) Can 3 3 Q3 Q = CDq force per unit area hit on the wall
(e) (i) R 2 = 386 000 = 1 400 × (25 – 5) produces gas pressure.
T 27 2 = 28 000 J
= 7.89 × 10 km day Q4 c = Q Q2 According to Charles’ Law, the
13
3
–2
volume of gas increases with
3
(ii) R 2 = J 3 2 = 7.89 × 10 13 mDq temperature. This is because when
T 1 54 000 J
J = 42 890 km = 0.6 kg × 200°C a fixed mass of gas is heated, the
gas molecules acquired energy and
–1
(iii) v = 2pJ = 2p(42 890) = 450 J kg °C –1 increase their kinetic energy. Gas
T 24 × 60 × 60 mcDq molecules move faster and collide
= 3.12 km s Q5 t = P with the wall of the container more
–1
= 2 × 4 200 × 50 = 140 s frequent and this increases the
pressure of the gas. In order to keep
3 000
Chapter the pressure constant, the excess
4 Heat Q6 (a) Q 1 = mcDq pressure of the gas will produce a



= 0.6 × 4 200 × (100 – 27)
= 183 960 J force pushing the piston upward to
increase the volume of the gas.
Checkpoint 4.1 (b) Q 2 = mcDq Q3 The mass and temperature of the
= 0.4 × 900 × (100 – 27)
Q1 (a) Thermal equilibrium is the = 26 280 J gas are kept constant.
process of transferring heat Q4 This is because at 0°C, the gas
between two objects in thermal Checkpoint 4.3 molecules are still in a state of
contact until both are at the motion and therefore the pressure
same final temperature and Q1 The meaning is that to melt or still exist.
there is no net heat transfer freeze 1 kg of ice, the amount heat Q5 P 1 = 1.5 × 10 Pa; V 1 = 50 cm ;
5
3
5
between the two objects. absorbed or released is 3.34 × 10 J. V 2 = 30 cm ; P 2 = ?
3
(b) Hot coffee is cooled when ice Q2 (a) Q – mL = 0.2 × 3.34 × 10 5
cubes are added to it. = 66 800 J P 1 V 1 = P 2 V 2 → P 2 = P 1 V 1
5
Q2 • Net heat flow between P and Q (b) Q – mL = 0.4 × 2.26 × 10 6 = 1.5 × 10 × 50 V 2
is zero. = 904 000 J 30
• Temperature of P and Q is lower Q3 (a) Melting point = 75°C = 2.5 × 10 Pa
5
than 70°C but higher than 20°C. (b) t = 6 minute = 360 s
Q3 (a) Liquid suitable for X is mercury. m = 100 g = 0.1 kg Q6 P 1 = 100 kPa; T 1 = 270 K ;
(b) Melting for ice is 0 C; boiling Q = Pt = 100 × 360 T 2 = 324 K ; P 2 = ?
o
point is 100 C. = 36 000 J P 1 = P 2 → P 2 = 100 × 324
o
270
Q4 100°C → (16 + 9) = 25 cm L = Q = 36 000 T 1 T 2 = 120 kPa
m
0.1
Therefore, 16 × 100°C = 64°C = 3.6 × 10 J kg
–1
5
25 Q7 V 1 = 80 cm ; T 1 = 273 K ;
3
Q5 (a) 24°C Q4 (a) Q = mcDq + mL T 2 = 338 K ; V 2 = ?
(b) 17.2 cm = 1.5 × 4 200 × 30 + 1.5 × 80
(c) 2 cm → 0°C 3.34 × 10 5 V 1 = V 2 → V 2 = 273 × 338
and 16 cm → 68°C = 189 000 + 501 000 T 1 V 2 = 99.0 cm 3
Therefore, 14 cm : 68°C = 690 000 J
For increase of temperature of (b) Q = mL + mcDq
6
20°C, = 0.2 × 2.26 × 10 + 0.2 × 4 200 SPM Practice 4
l = 20 × 14 = 4.12 cm × (100 – 60) Objective Questions
= 452 000 + 33 600

68
= 485 600 J 1. B 2. B 3. C 4. C 5. B
Q5 Q = mcDq(ice) + mL(ice) + 6. C 7. A 8. D 9. A 10. D
Checkpoint 4.2 mcDq(ice) + mL(steam) 11. B 12. B 13. C 14. B 15. B
= 3.2 × 2 100 × 5 + 3.2 × 3.34 ×
Q1 (a) Mass – the larger the mass of 10 + 3.2 × 4 200 × 100 + 3.2 16. C 17. B 18. B 19. A 20. B
5
the object, the larger the heat × 2.26 × 10 6 21. D 22. B 23. C 24. D 25. A
capacity. = 9 678 400 J
(b) Shape – does not affect the Subjective Questions
heat capacity. Checkpoint 4.4
(c) Types of material – different Section A
types of materials have different Q1 Based on the kinetic energy of the 1. (a) (i) Thermal equilibrium
heat capacity. gas, the gas molecules are all in (ii) The change of oil
Q2 (a) Material T: The largest mass fast, random and continuous motion. temperature is greater than
The gas molecules are always in
water.
has the highest heat capacity. collision with each other and with (iii) The amount of heat energy
Q
(b) Specific heat capacity, c = mDq , the wall of the container. When a absorbed by oil and water is
means smaller mDq, the larger gas molecule collides with the wall the same.
specific heat capacity. Material and bounces back, the molecule (iv) The specific heat capacity of
P has the largest specific heat experiences a change in momentum. oil is smaller than water.
capacity due to the smallest mq The rate of change of momentum (v) PT = mcDq
value. from the rate of collisions with the 500 × (2 × 60) = 0.5 × c ×
container wall produces a force. The (97 – 25)
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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
–1
–1
= 1 667 J kg °C radiation. = 680 m
(b) (i) degree of hotness • Suggestion: The covered cup Therefore,
(ii) Volume of gas decreases. can be stored in a pocket h = 340 m
(iii) Absolute zero temperature made of bamboo cloth. Q2 (a) v = 1500 m s , t = 0.12 s
–1
(0 K or – 273°C) Explanation: So that there is Distance travelled by the pulse
(iv) The freezing gas molecules a layer of air trapped inside it = v × t = 1500 × 0.12 = 180 m
do not move. can reduce heat loss. (b) Distance of the shoal of fish
Section B below the boat = 180 = 90 m
2. (a) Heat is a form of energy that 2
can flow from area of high Chapter (c) Ultrasonic waves can transfer more
temperatures to area of low 5 energy than audible sound
temperatures. Waves waves.
(b) (i) • The mass of water in
kettle P is smaller than Checkpoint 5.1 Checkpoint 5.4
the mass of water in
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 (c)
rise in kettle P is higher
than the temperature of Deep Shallow
water rise in kettle Q.
(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 λ d 3 –1
temperature rise. move in the direction Q2 (a) f = 16 Hz; λ s = ; v d = 6 cm s
2
(iii) The heat supplied must be perpendicular to the λ d λ d
constant. direction in which the wave v s = λ s
(c) The initial temperature of the propagates.
orange juice is higher than the (ii) Water waves Therefore, v s = λ s v d
temperature of the ice cubes. (b) (i) A wave in which the λ d
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 = 4 cm s –1
by the cold melting water from which the wave propagates. 4
the ice. The temperature of the (ii) Sound waves (b) λ s = λ s f = 16 = 0.25 cm
orange juice decreased and
the temperature of the cold Q3 (a) Period of oscillation, T = 0.5 s
1
water from the ice increased. (b) Frequency, f = = 1 Q3 (a), (b), (c)
Therefore, orange juice is T 0.5 Direction of
cooled by adding ice cubes. = 2.0 Hz Shallow propagation
(d) • Suggestion: The cup should (c) Wavelength, λ = 5 cm area
be made of good heat (d) Using v = f λ ,
insulator such as polystyrene v = 2.0 x 5 = 10 cm s –1 Normal
material. Deep
Explanation: Good heat Checkpoint 5.2 area
insulation prevents heat from Q1 (a) Q
hot drink being escape by (b) Q has the same natural
conduction. frequency as the vibrating tuning
• Suggestion: The cup cover fork. Hence, Q will receive the Q4
should also be made of good biggest amount of energy from Hotter air
heat insulator such as plastic. the vibrating tuning fork. Speed of sound higher
Explanation: Good insulator (c) Resonance
can prevent heat loss to the
surrounding. Q2 The hand moves with the same Refraction of sound waves occur
• Suggestion: The cup material natural frequency as that of the Source Listener
should also have a high loaded spring. Resonance occurs of s
melting point. and the spring will receive maximum sound
Explanation: So that the cup energy. Colder air
does not easily change its Speed of sound lower
state or melt when filled with Checkpoint 5.3 The diagram shows the temperature
liquid of high temperature. Q1 Let the distance of the wall from the of air at night. The layer of air
• Suggestion: The inside and student be d m above the surface of the Earth is
outside of the cup are coated The distance travelled by sound to colder compared the layer further
with shiny metal. and fro the wall = 2d m away from the surface of the Earth.



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Physics Form 4 Answers

Since the speed of sound waves (ii) Q3 v = s , therefore,
is higher at higher temperature of t
air, refraction occurs and the sound a t = s
waves are refracted towards the v
surface of the Earth. This results in = 320 000 × 1000
8
the listener be able to hear clearer a 3 × 10
sound at night. = 1.07 s
Checkpoint 5.5 Q2 (a) • A: An up-and-down movement Q4 P : Microwaves; Q : X-rays; R :
with a large amplitude Visible light; S : Ultraviolet rays
Q1 (a) • B: No motion/stationary
(b) a = 3.0 cm; D = 30.0 cm; SPM Practice 5
x = 15.0 cm
ax Objective Questions
λ =
D 1. D 2. B 3. D 4. C 5. C
3.0 × 15.0 6. B 7. C 8. D 9. B 10. B
= 30.0
(b) = 1.5 cm 11. A 12. C 13. C 14. C 15. B
16. D 17. A 18. B 19. A 20. C
–9
Q3 (a) λ = 540 nm = 540 × 10 m 21. B 22. D 23. A 24. D 25. C
–3
a = 0.4 mm = 0.4 × 10 m 26. D 27. D
D = 3.0 m
Using λ = ax , Subjective Questions
D
(c) λD 540 × 10 × 3.0
–9
x = a = 0.4 × 10 –3 Section A
= 4.05 × 10 m 1. (a)
–3
= 4.1 mm Direction of
λD movement of hand
(b) • x = a
Q2 (a) Remains unchanged • Hence, if λ is bigger, x will
(b) Remains unchanged also be bigger. Therefore, (Accept single arrow)
(c) Remains unchanged the distance between two
adjacent bright fringes of light
Q3 (a) will be bigger. (b) λ = v
f
Q4 (a) f = 600 Hz; v = 330 m s –1 = 0.90
Using v = fλ, 3.0
v 330
λ = — = —— = 0.3 m
f 600 (c)
(b) The leaves which act as = 0.55 m
obstacles have small width. (b) a = 1.5 m; D = 3.0 m;
Therefore, the effect of x = λD
diffraction is greater. Hence the a
waves join back after a short = 0.55 × 3.0
distance from the leaves. 1.5
(c) The log which is a larger = 1.1 m
obstacle causes less diffraction.
The waves join back after a Checkpoint 5.7 2. (a) Diffraction of waves
futher distance from the log. Q1 1. They all transfer energy from (b) • The amplitude of the waves
Q4 (a) The sound waves spread out one place to another. before passing through the
beyond the edge of the window 2. They are all transverse waves. slit is higher than that the
after passing through the 3. They can all travel through a amplitude after passing
window. vacuum. through the slit.
(b) Diffraction of sound waves 4. They all travel with a speed of • This is because the waves
3 × 10 m s in vacuum. spread over a larger area and
8
–1
energy per unit area of the
Checkpoint 5.6 Q2 (a) P : Microwaves; Q : Visible light; diffracted waves is less.
Q1 (a) The principle of superposition R : X-rays (c)
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 (c) Photography / Photosynthesis
by the sum of the displacements by plants / Enables human
of the individual waves at each beings and animals to see. (Any
one answer)
point of the medium. (d) R has very high penetrating
(b) (i) power and can cause cancer
2a and genetic defects to living
a 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]
from the system as heat
2
2
2
(c) Ultrasonic waves have more energy. [1]
energy than audio waves. • If no external force is applied, 7. (a) • The convex-shaped peak of
the spring will eventually the wave is like a convex lens.
4. (a) P = X-rays stop completely. [1] [1]
Q = Infrared • Damping. [1] • The light rays from the lamp
(b) (i) 100.6 × 10 Hz or 4 passing through the peak will
6
1.006 × 10 Hz (d) (i) be converged on the screen to
8
produce a bright band.
(ii) Wavelength • The trough is like a concave [1]
= 3.0 × 10 8 lens. [1]
1.006 × 10 8
= 2.98 m. • The light rays from the lamp
passing through the trough will
(c) For taking the image of the [1] be diverged causing a dark
patient’s bones. • Move the hand band. [1]
to-and-fro, perpendicular 4
5. (a) (i) High frequency sound to the spring. [1]
waves. • The waves move forward (b) (i) • The frequencies of the
(ii) • Ultrasonic waves along the spring. [1] waves in both the deep
transmitted by the bat is • The string tied to the and shallow areas are the
reflected by the body of spring which represents same. [1]
the bird. a particle of the medium • The wavelength of the
• The reflected waves is oscillates perpendicularly waves in deep area is
detected by the bat. to the direction of the longer than that in shallow
• The time between waves. [1] area. [1]
the transmission and • This shows that in • The speed of the waves
detection of the signal by the propagation of a in deep area is higher
the bat is equal to twice transverse wave, the than that in shallow area.
the distance between the particle of the medium [1]
bat and the bird. vibrates/oscillates (ii) • When the angle of
• With that, the distance perpendicularly with the incidence is zero, the
between the bat and the direction of the wave. [1] direction of waves
bird can be estimated moving from deep to
through the period of Max 4 shallow areas remains
time transmission and (ii) unchanged. [1]
detection of the ultrasonic • When the angle of
waves. incidence is not zero,
(b) (i) Distance = d m [1] the waves change their
2d = 1 450 × 120 × 10 . direction when moving from
-3
Therefore d = 87 m. • Move the hand to-and-fro, deep to shallow areas. [1]
parallel with the spring. [1]
(ii) v = fλ • The waves move forward 5
1 450 = 45 × 10 × λ along the spring. [1] (c) Refraction of waves [1]
3
1450 • The string tied to the
Therefore, λ = (d) (i) • The resort is to be built
45 × 10 3 spring which represents near the bay [1]
= 3.2 × 10 m a particle of the medium • The waves at the bay are
-2
oscillates parallel with the
Section B direction of the waves. [1] calmer than at the cape [1]
6. (a) A system that undergoes a • This shows that in • due to the divergence of
periodic to-and-fro movement. [1] the propagation of a the waves’ energy from
(b) • Spring B carries a bigger longitudinal wave, the the bay [1]
mass than that of spring A. [1] particle of the medium • and the convergence of
• Spring B oscillates with a vibrates/oscillates parallel the waves at the cape [1]
bigger period than spring A. [1] with the direction of the (ii) • To reduce erosion,
• Spring B oscillates with a wave. [1] retaining walls are built [1]
lower frequency compared to Max 4 • to reflect the waves from
spring A. [1] 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
a gap in between are built
4 in the form of waves at the designated area for
(c) (i) Oscillating with the along the spring. [1] children [1]
amplitude decreasing with • The string tied to the • Waves passing through
time. [1] spring which represents
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Physics Form 4 Answers

the gap will be diffracted • Radio waves with longer (c) sin 45 = 0.7071
0
–1
in the children’s area [1] wavelength will diffract r = sin 0.4877 = 29.2°
• The smaller amplitude around the mountain Q4 (a) The speed of light in glass block
of the diffracted waves more compared with is slower than the speed of
causes the sea to be radio waves with shorter light in water. This is because
calmer. [1] wavelength [1] the refractive index of glass is
10 (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 c air
8. (a) Refraction of waves is a Earth [1] (b) n = c medium
phenomenon that occurs when • The fact that ultraviolet c air
there is a change of direction rays reach the 1.33 = c water
of the propagation of waves Earth shows that c air
travelling from one medium to electromagnetic waves and 1.52 = c glass
another due to a change of travel through vacuum [1] c air
speed. [1] Max 7 c glass c water 1.33
(b) (i) Sound waves travel faster in c water = c air 1.52 = 0.88
air when the temperature of c glass
the air is higher. [1]
(ii) • At daytime, when it is hot, Chapter Checkpoint 6.2
the surface of the Earth 6 Light and Optics
heats up faster than the Q1 (a) The angle of incidence ray BC
air. [1] Checkpoint is larger than the critical angle.
• Hence, the hot surface of 6.1 Thus, total internal reflection
the Earth causes the layer Q1 i = 140° – 90° = 50° occurs.
of air near the surface to r = 125° – 90° = 35° (b) Light ray DE refract away
be warmer than the upper sin i from normal because it leaves
layer. [1] n = sin r from glass prism to air, from a
• This causes the sound sin 50° denser medium to a less dense
waves to be refracted = sin 35° medium.
away from the Earth. [1] = 1.34 1 1
• On a cool night, the Q2 n = sin c = sin 42° = 1.49
surface of the Earth cools Q2 D a = 12 – 10 = 2 cm
down faster than the air. D r D r 1.49 = sin P
[1] n = D a → 1.50 = 2 sin 40°
• Hence, the air near the D r = 1.5 × 2 = 3.0 cm sin P = 1.49 × sin 40°
surface of the Earth = 0.9578
becomes cooler than the \The thickness of glass block is P = 73.3°
upper layer. [1] 3.0 cm Q = 90° – 40° = 50°
• This causes the sound Q3 i sin i r sin r
waves to be refracted R = Q = 50°
towards the Earth. [1] 22 0.3746 15 0.2588 S = 90° – 50° = 40°
6 28 0.4695 19 0.3256 1.49 sin 40° = 1 sin T
–1
(c) (i) • Transverse waves [1] 33 0.5446 22 0.3746 T = sin (1.49 sin 40°)
• Travel with a speed of 35 0.5736 26 0.4384 T = 73.3°
3 × 10 m s in
8
–1
vacuum [1] 41 0.6561 27 0.4540 Q3
• Transfer energy from one
place to another sin i
(Other characteristics
acceptable) [1] 0.9
3
(ii) 1. Sound wave does not
travel in vacuum. [1] 0.7 • Periscope is used to see an
2. Sound wave is a object behind the observer.
longitudinal wave. [1] • Image formed is inverted and
2 0.5 same size as object.
(d) (i) • Example 3 [1]
• The microwaves 0.3 sin r Checkpoint 6.3
transmitted will be 0.2 0.3 0.4 0.5 0.6 Q1 (a) The focal point is the point on
reflected by the plane and (a) Set of reading for angle of the principal axis where the light
return to the receiver [1] incident = 35° rays parallel to the principal axis
(ii) • Example 1 [1] after passing through the lens
• The mountain is an (b) n = 0.454 will converge on it.
obstacle [1] 0.6561
= 1.45


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Physics Form 4 Answers

(b) (i) Virtual image is image that Q2 m = = h i = 15 = 1.5 (c) The separation distance of the
v
cannot be displayed on the u h o 10 lenses
screen. Therefore, v = 1.5u = 40 + 5
(ii) = 1.5 × 2.0 m = 3.0 m = 45 cm
1 cm (d)
1
1
1
Converging lens Using + =
1 cm u v f 5 cm
1 1 1 5 Parallel rays 40 cm
f
2 + = = 6 from distant
3
Therefore, focal length f = 1.2 m object
1
1
F Main axis Q3 Using 1 + =
Image Object f = 2 cm u v f 1 Lens P Lens Q
1
1
1 = – = – 1 –
v f u 10 30
= – 4 Q3 Similarity:
30 • Consists of two convex lenses
Virtual image is image that v = –7.5 m • The eyepiece function as a
cannot be displaced on the 30 magnifying glass
screen = 2 cm v 4 1 Differences:
m = = = • For microscope in normal
Q2 (a) u 30 4 adjustment, the final image is at
The image is virtual, upright,
at a distance of 7.5 cm from near point. While for telescope in
the lens on the same side normal adjustment, the final image
as the object and has linear is at infinity.
Image 1 • The distance between the
magnification of .
F 4 objective lens and the eyepiece of
Q4 the microscope is L O > f O + f e . The
distance between the objective
v/cm lens and the eyepiece of the
telescope is L O ≤ f O + f e .
(b) The size of the image does not 60
change but the brightness of
image reduced by half. 50 u = v Checkpoint 6.6
Q3 (a) Convex lens 42 Q1 (a)
v
(b) M = = 30 = 0.5 40 P
u
60
(c) 30
Q
20
F F
10 R
u = 60 cm f = 20 cm
v = 30 cm 0 u/cm
10 18 20 30 40 50 60
Focal length, f = 20 cm (b) The image is real, inverted and
v 42 7
(a) m = = = = 2.33 magnified.
Checkpoint 6.4 u 18 3 (c) The image is nearer to the
1 1 1 (b) u = v = 25 cm mirror and becomes smaller in
Q1 (a) + =
u v f f = 25 size.
1 1 1 1 1 1 2
v = – = 20 – 30 = 60 = 12.5 Q2 (a)
f
u
v = 60 cm
v 60 Checkpoint 6.5
m = = = 2
u 30
Therefore, the image is real, Q1 (a) For distant objects, the image
must be formed at the focal
inverted, 60 cm from the lens, point of the lens. Therefore, the 25°
located opposite to the object distance between the lens and F C
and enlarged 2 times. the film is 50 mm.
1 1 1
(b) + = (b) The camera lens should be
u v f adjusted outward toward the
1 1 1 1 1 1
v = – = 20 – 15 = 60 object. As the object distance
f
u
v = –60 cm is reduced, the image distance (b) The image is at F.
increases. With this, the
v 60 distance between the lens and (c) The image is virtual, upright and
m = = 15 = –4 the film is slightly longer than diminished.
u
The image is virtual, upright, at 50 mm. Q3 • Focal length of mirror P = 12 cm.
a distance of 60 cm from the Q2 (a) Astronomical telescopes The image from distant object is
lens on the same side as the (b) Objective lens: lens P, Eyepiece: formed at the focal point of the
object and is magnified 4 times. lens Q mirror.
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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
curvature, image distance = 2f. by the second prism.
• Therefore, the mirror Q has a 45° 45° (ii)
shorter focal length. 45° 45° Right-angle prism
SPM Practice 6 Distant
object at
Objective Questions the back
1. B 2. C 3. B 4. C 5. B 45° Cylindrical
6. C 7. C 8. C 9. D 10. D 45° tube
11. C 12. A 13. B 14. B 15. C 45°
16. A 17. C 18. A 19. A 20. C
21. C 22. D 23. D 24. B 25. C
45° 45° An inverted
Subjective Questions 45° image Right-angle prism
Section A (c) (i) When the prism at the
Glass top periscope is laterally
1. (a) inverted, it can be used to
Air observe the object behind
Total internal the observer over an
Light ray reflection occurs here obstacle. In this case, the
image seen is inverted
A light ray is transmitted (iii) • A right-angle prism uses
(b) (i) f = 10 cm, u = 12 cm through a curved fibre optic total internal reflection
1 1 1 1 1 1 that occurs a series of total to reflect light. The light
f = + → 10 = 12 + v internal reflections on the reflected by the prism
u
v
1 1 1 6 – 5 1 inner surface. surface is 100%.
v = 10 – 12 = 60 = 60 (ii) • Fiber optic cables are • Image formed by right-
v = 60 cm lighter and thinner. angle prism is sharper.
Therefore, distance between • Transmission of signal
the screen and the object with almost no energy
= u + v = 12 + 60 = 72 cm loss along the optical PRE-SP M MODEL P APER
PRE-SPM MODEL PAPER
v 60 fiber.
(ii) m = = = 5 (d) (i)
u 12 Objective Questions
(c) The minimum height of the Right-angle
screen = 24 × 5 prism 1. B 2. B 3. C 4. A 5. D
= 120 cm 6. B 7. C 8. A 9. B 10. C
(d) If the distance between the lens 11. A 12. A 13. B 14. A 15. A
and the image = 10 cm = f, then Distant object Cylindrical 16. B 17. C 18. A 19. D 20. C
object is at the focal point and tube 21. B 22. A 23. C 24. A 25. B
image formed is virtual and at 26. C 27. D 28. A 29. D 30. A
infinity. Therefore, image cannot 31. C 32. C 33. A 34. D 35. C
be seen on the screen. 36. D 37. A 38. D 39. D 40. D
41. C 42. B 43. A 44. B 45. B
Image at infinity 46. B 47. D 48. C 49. C 50. C
Upright image
Right-angle Subjective Questions
prism
F F Section A
• A periscope is build using 1. (a) • time / length
two right-angle prisms • velocity / force
arrange in an cylindrical (b) (i) acceleration
Section B tube as shown in the (ii) the original length of the
2. (a) (i) The critical angle of glass is diagram above. spring
42° which means that if the • The prism acts like a
angle of incidence in glass perfect mirror when light 2. (a) Initially, there is no air resistance
(denser medium) is 42°, ray hits the inner surface acted on the skydiver. Therefore,
then the angle of refraction of the prism at an angle the skydiver experience a free
in the air is 90°. greater than 42°. The first fall due to gravitational force.
(ii) Total internal reflection prism turns the image (b) When the parachute is opened,
the air resistance acting on the
1 from a distant object 90°,
(iii) sin c = n and then the second parachute is greater than the
1 prism turns the image weight of the skydiver. The net
n = force acting on the parachute
sin c back to upright. upward causes upward
1
= sin 42 • Object lens is placed in acceleration.
front of the first prism on
= 1.49 top to find a distant object.
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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
3. (a) • The force holding the satellites the image formed is less. 9.1 is smaller.
and the Moon in their orbit is 7. (a) Heat is a formed of energy. (ii) The temperature increases
the gravitional force. (b) (i) The thermometer reading in as the heat supplied
T 2 increases. As the air
• The ratio R 3 for all satellites is Diagram 7.1 is higher. temperature increases, the
a constant. (ii) The volume of water in air pressure in the flask
Diagram 7.2 is larger.
R 3 (386 000) 3 (iii) The time taken to heat the increases.
(b) 2 = (c) According to kinetic theory, the
T (27 × 24) 2 water is the same, which is kinetic energy of gas molecules
=1.37 × 10 km jam –2 5 minutes. increases as heat is absorbed.
11
3
(c) (i) The smaller the volume of
(R E ) 3 The air molecules in the flask
(c) 2 = 1.37× 10 11 water, the larger the reading move faster and collide more
24 of the thermometer.
R E = 1.37 × 10 × 24 2 (ii) The smaller the mass of frequently with the flask wall and
11
3
thus increases the air pressure
R E = 42 892 km water, the higher the rise in in the flask.
4. (a) Water vapour temperature. (d) (i) The sheath mass should be
(b) The heat energy supplied (d) (i) Amount of heat supplied, small and light so that the
is used to release the water Q = 16 × (1 × 60 × 60) aneroid barometer is easy
molecules into the air. The = 57 600 J to carry.
kinetic energy of the molecules (ii) Q = mcDq (ii) The chamber should be
does not change. 57 600 = 0.8 × 4 200 × Dq made of light metal box
(c) Heat used to change 20 g of DT = 57 600 enclosed with corrugated
water to steam, 0.8 × 4 200 edges so that it is flexible
Q = 52 000 – 6 000 = 17.1°C and can move up or down.
= 46 000 J The air in the chamber
Mass of water change to steam, 8. (a) (i) Velocity is the rate of is released into a partial
m = 20 g = 20 × 10 kg change of displacement vacuum, so that the space
–3
Specific latent heat of or distance move in one can respond to changes
vaporisation, second in a specific in the outside air pressure
direction.
more effectively.
L = Q (ii) Velocity change because its (iii) Spring must be strong
m
= 46 000 –3 direction change. and easily extend in order
20 × 10 (iii) Force of gravity towards to respond well to the
–1
= 2.3 × 10 J kg the centre of Earth. This movement of the chamber.
6
force produces centripetal
(d) When water evaporates, force perpendicularly to the (iv) The lever should be light
the water molecules acquire direction of motion. and easy to move in order
energy and change into vapour (b) (i) Resultant force to simplify the movement of
and leaves the less energy = Upthrust – rocket weight the chamber and pointers.
molecules behind. This reduces Upthrust 10. (a) A sound wave is a kind of
the average kinetic energy of = 50 000 N + 40 000 × 10 N longitudinal wave, produced
the molecules and thus the = 450 000 N from vibrating object in which
water is cooled down. (ii) Resultant force, F = ma the particles in the medium
5. (a) transverse wave 50 000 = 40 000 × a move forward and backward
(b) • same frequency 50 000 in the direction of the wave
• λ d > λ c Acceleration, a = 40 000 propagation.
(c) • λ d = 1.5 cm = 1.25 m s –2 (b) (i) The distance x in Diagram
• λ c = 1.0 cm (c) (i) 10.1 is smaller than the
(d) v = fλ = 5.0 × 1.0 = 5.0 cm s –1 Velocity distance x in Diagram 10.2.
(e) The distance a in Diagram
10.1 is smaller than the
distance a in Diagram 10.2.
The larger the value of a,
Deep Shallow the smaller value of x.
region region (ii) The distance D, between
0 2 4 6 8 10 12 the speaker and the
Time after launching / min interference pattern, must
6. (a) Refraction of light be constant. The physical
(b) (i) Thickness of lens P < (ii) Area under the graph phenomenon is interference
thickness of lens Q Section B of sound wave.
(ii) f 1 > f 2 9. (a) To measure gas presure (c) (i) Sound waves with
(iii) h 1 > h 2 (b) (i) • the reading of Bourdon frequencies exceeding
(c) (i) The thicker the lens, the gauge in Diagram 9.1 is 20 kHz are called ultrasonic
shorter the focal length. smaller. waves.
(ii) The longer the focal length, • the heat supplied to (ii) • Ultrasonic signals are
the smaller the size of Diagram 9.1 is smaller. transmitted into the sea
image formed. until reaches the sea floor.
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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 in diameter. So that cart Using, Q = mcDq
recorded. can be easily move and Therefore, specific heat
• From ultrasonic velocity, the pushing force can be capacity of, X,
2d reduced.
v = , depth is calculated Q 6 000
t vt (iv) The cartwheel should made c = =
using the formula d = . of hard rubber. So that the mDq 0.2 × 40
2
–1
(d) (i) The speakers should be cart can be easily move and = 750 J kg °C
not easily damaged when
arranged in such a way that carrying heavy goods. (ii) Time taken, t
no sound wave interference The L trolley is the perfect = 8 – 4 = 4
occurs so that the same choice because it has the right = 4 × 60
loudness can be heard in all height of handle, light and strong = 240 s
directions. base, large wheel diameter and
(ii) The speaker must be made of hard rubber Heat supply, Q
powerful enough to (d) (i) Change in velocity = 25 × 240
transmit sound waves in = –22 + –25 = 6 000 J
all directions so that sound = –47 m s –1 Mass, m = 200 g = 0.2 kg
can be heard clearly and (ii) Average acceleration Using, Q = mL
in all directions with equal Change in velocity Therefore, specific latent
loudness. a = Time taken heat of fusion of X,
(iii) A good amplifier along with 47 Q 6 000 J
a good microphone is used = 0.0013 L = m = 0.2 kg
so that a person›s voice = 3.6 × 10 4 = 30 000 J kg –1
can be clearly detected and (iii) Average force,
amplified. F = m × a (c) • The cover should be made
(iv) The walls of the concert hall = 0.16 × 3.6 × 10 4 of hollow plastic. Air trapped
should be fitted with small = 5 760 N in hollow plastic is a good
hollow soft boards so that thermal insulator and can
sound waves can be fully 12. (a) (i) The specific latent heat of prevent heat lost to the
absorbed without reflection. vaporisation of water is the surrounding.
The floor should be amount of heat required to • Space P should be a vacuum.
carpeted to avoid sound convert 1 kg of water into The vacuum can prevent heat
wave reflection. steam without temperature lost through conduction and
change. convection.
Section C (ii) Water has a large specific • Two-layer wall must be made
latent heat of vaporisation. of glass. Glass is a good
11. (a) Impulse is change in momentum When steam condenses it
(b) When the player hits a tennis releases a large amount insulator of heat and has a
ball, he or she can increase the of heat. Foods like fish in greater specific heat capacity.
contact time with the racket. a plate can absorb a lot of This feature can reduce heat
This will add impulse to the ball heat from hot steam. The transferred to the surrounding.
and increase the momentum food becomes cooked when • The wall should be coated
change on the ball. Therefore, the food reaches thermal with shinny paint. The shiny
the speed of the tennis ball equilibrium with the steam surface is a good heat
increases. after some time. reflector and can reduce heat
(c) (i) The height of the handle (b) (i) Time taken, t = 4 min loss.
should be approximately the = 4 × 60 s • The thermos flask S is the
height of the shoulder of the = 240 s best choice because it uses
pusher. So that the cart is hollow plastic as cover, the
easy to push. Lower handle Change in temperature, DT space P is vacuum, the two-
needs more force to push. = 60° – 20° layer wall made of glass and
(ii) The base of the cart should = 40°C coated with shiny paint.
be light and strong. So that Heat supplied, Q
the cart is easily push and = P × t
the hard material is not = 25 × 240
easily damaged and can = 6 000 J
last longer.















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