FOCUS SPM Physics (2023)

Physics SPM Answers

Checkpoint 2.3 (b) Total momentum after collision (b) t = 0.2
mv – mu
= 1200 kg m s
–1
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 Magnitude of force = 1.5 × 10 N
2 200v = 600
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
lengthen the stopping time of
= (1500 × 15) + [1000 × (–20)]
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
v B = 0 m s –1 Q3 (a) Recoil velocity of the rifle = v reduce the impulsive force.
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
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
0 = 24 + (–10)t the car.
t = 2.4 s Q4 (a) When a large volume of water (b) The strong and rigid cell will
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
forces.
Q1 (a) Tin P opposite momentum is produced Q4 (a) The driver has inertia that keeps
causing the fireman to fall
(b) Tin P backward if not supported by 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 of matter in the object.
mass. Hence, its inertia is big. conservation of momentum Weight of an object is the
(b) The mass of a train is big. where the final total momentum gravitational force acting on the
Hence, its inertia is big. is still zero. object at a particular place.
(c) Steel tubes have big mass.
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
6
= 1200 kg m s –1 Magnitude of force = 1.5 × 10 N 6. A 7. C 8. B 9. C 10. B
476
ANSWER FOCUS PHYSICS.indd 476 10/03/2023 10:42 AM

Physics SPM Answers

11. C 12. C 13. D 14. D 15. B • The gradient of the graph (ii) Acceleration of the car, a
16. C 17. B 18. D 19. D 20. B gives the acceleration of the F
= —
21. B 22. D 23. D 24. C car. m
• The area under the graph 0.2
——
=
Subjective Questions gives the distance travelled by 2.0
Section A the car. = 0.1 m s –2
1. (a) Momentum is the product of (c) (i) • A short reaction time is Section C
mass and velocity. desirable.
(b) (i) 2.0 kg m s –1 So that, the car can reach 5. (a) (i) The mass of a body is the
(ii) 0.5 + 1.5 = 2.0 kg m s –1 the finishing line in a amount of matter in it.
(c) They are the same. shorter time. (ii) • An object at rest resists
(d) (i) The total momentum of the • A smaller mass is effort to move it.
balls before the collision and suitable. • An object in motion resists
after the collision are the same. So that, a bigger effort to stop it.
(ii) Principle of conservation of acceleration can be • The bigger the mass of an
momentum achieved. object, the more difficult to
(e) Inelastic collision • A bigger engine thrust is move it.
more desirable. • The bigger the mass of an
2. (a) Rate of change of velocity So that a bigger object, the more difficult to
(b) u = ——–—— acceleration can be stop it.

4.5
(5 × 0.02) achieved. • This is the property of all
= 45 cm s –1 • A smaller resistive force is objects with mass.
= —–——— more desirable. (iii) Inertia
22.5
v
(5 × 0.02) This is to produce a (b) • Hold the bottle upside down.
= 225 cm s –1 bigger resultant force • Give the bottle a downward
v – u forward. jerk.
a = ———
t (ii) Using the formula • The inertia of the sauce
225 – 45  100 m will cause it to continue its
= ————
2 × 0.1 t = Reaction time + ——– , movement downward and
F
= 900 cm s Car P:  100 × 1.6 hence, out of the bottle.
–2
= 9.0 m s –2 Time = 0.25 + —————
(10.5 – 3.8)
(c) Air resistance and friction = 5.14 s (c) (i) • The car stopped suddenly
between the ticker tape and the Car Q: causing a big impulsive

force.
100 × 2.0
ticker timer. Time = 0.45 + ————— • The front portion of the

10
3. (a) (i) a = —– = 4.88 s (12.8 – 2.6) car should be able to

5 Car R: crumple upon impact.

100 × 1.2
= 2 m s Time = 0.20 + ————— • This will increase the
–2
(ii) a = ———— (7.0 – 2.4) impact time and reduce
15 – 10
15 = 5.31 s the impulsive force acting
= 0.33 m s –2 Car S:  on the car.
100 × 1.8
(b) The beach has a larger friction Time = 0.55 + ————— (ii) • The driver flew out of the
(15.4 – 5.6)
against the motion of the bicycle = 4.84 s car because there are
than the road. Hence, the net Therefore, car S will win the no features in the car to
force acting on the bicycle is competition. secure him to the car.
reduced. • Safety belts should be fixed
(c) s = Area under the graph (d) (i) to the car to hold the driver.
= 25 + 187.5 • Air bag should be installed
= 212.5 m F F to absorb the impact.
2 1
Super • The air bag will cushion
Section B F 3 30° the driver from the
4. (a) A force can cause: steering wheel and the
• a stationary object to move. 20 N windscreen.
• a moving object to change its (iii) • Windscreen glass should
speed. Let F 1 , F 2 and F 3 be the be designed to fracture
• a moving object to change its engine thrust, resistive into rounded pieces
direction of motion. force and component of the upon impact instead of
• an object to change in size weight respectively. shattering.
and shape. F 1 = 12.8 N; F 2 = 2.6 N; • So that the driver is less
(Any one answer) F 3 = 20 sin 30° = 10 N likely to be cut by the
The resultant force up the shattered glass.
(b) v / m s –1 slope, F (iv) • The headlight of the car

= F 1 – (F 2 + F 3 ) should be bright and
= 12.8 – (2.6 + 10) powerful enough to shine
= 0.2 N over a long distance.
Therefore, the car can move • This is to enable the
up the slope. driver to see far ahead
0 t / s and have enough time to
avoid obstacles.
477
ANSWER FOCUS PHYSICS.indd 477 10/03/2023 10:42 AM

Physics SPM Answers

Natural satellite are any object in GM
Chapter (b) (i) g h = 2
3 Gravitation space orbiting the larger planets. (R + h)
The moon is a natural satellite
orbiting the Earth. GM
R
2
g
Checkpoint 3.1 Q2 v =  (ii) g h = GM
GM
Q1 F = G Mm r + h (R + h) 2 2
R 2 =  –11 24 g =  R  g h
–11
= 6.67 × 10 × 7 × 7 6.67 × 10 × 5.97 × 10 R + h
6.37 × 10 + 500 000
6
2
2
= 8.17 × 10 N  R + h  2
–10
= 7 613 m s = 7.61 km s –1 (iii) g h = R g
–1
1
Q2 (a) F ∝ 6 400 2
r 2 Q3 v =  2GM =  6 400 + 12 800 
(b) F R × 10 N kg –1
v Y : 1 791 =  2GM = 1.1 N kg –1
R
v X : v =  2G × 1.41 M W = mg = 81 × 1.1 = 89.1 N
0.919R Section C
v  2G × 1.41M R
= 4. (a) (i) A geostationary satellite is a
1 791 0.919R × 2GM satellite in the geostationary
v =  1.41 × 1 791 orbit round the Earth on an
r 0.919 equatorial plane. It moves
F
Q3 F = mg → g = m = 2 218 m s –1 in orbit round the Earth,
from east to west, and has
= 1 560 = 26 m s –2 3 a period of 24-hour or one-
60 SPM Practice
Q4 F because the force of gravity is the day.
same. Objective Questions (ii) 3
1. A 2. B 3. D 4. C 5. C (iii)
Q5 F c = T = F w
m × a c = M × g 6. C 7. B 8. D 9. B 10. C
0.05 × a c = 0.6 × 10 11. B 12. D 13. B 14. D 15. B
a c = 120 m s –2 16. C 17. C 18. D 19. B 20. C
21. D 22. B 23. D 24. C 25. A
a c = v 2
r Subjective Questions
v = 120 × 0.6 Section A
2
v = 8.5 m s –1
1. (a) • The force is proportional to the (b) (i) Mercury:
Checkpoint 3.2 product of the masses of the R = 1.85 × 10 km 3
3
23
two body. T = 7 744 day 2
2
Q1 T 2 = 1 2 • The force is inversely R 3
3
–2
19
r 3 (1.5 × 10 ) proportional to the distance T 2 = 2.52 × 10 km day
8 3
= 2.96 × 10 year km –3 between the two bodies.
2
–25
2 2 (b) (i) Same Mars:
Q2 T Earth = T Mercury R = 1.19 × 10 km 3
23
3
3
3
2
r Earth r Mercury (ii) Twice as big T = 471 969 day 2
2
3
1 2 = T Mercury 2. (a) (i) T = W = mg R 2 = 2.52 × 10 km day
19
–2
3
(1.50 × 10 ) (5.79 × 10 ) T
11 3
10 3
= 0.2 × 10 = 2.0 N
T Mercury = 0.575 (ii) F c = T = 2.0 N (ii) The value is about the
2
T Mercury = 0.24 year mv 2 same.
(b) F c = r (iii) Kepler III law states that the
Q3 r = GM T 2 2.0 = 1.0 × v 2 square of the orbital period
3
4p 2 1.0 of a planet is proportional

(6.67 × 10 × v = 20 = 4.47 m s –1 to the cube of the orbital
–11
5.98 × 10 ) (c) The orbital radius becomes
24
= (24 × 60 × 60) 2 radius.
4p 2 smaller. (iv) Can
(R + h) = 7.542 × 10 22 (d) The centripetal force acting on (c) • The force of gravity due to
3
R + h = 4.23 × 10 7 the stopper is perpendicular Earth’s gravitational field
h = 4.23 × 10 – 6.37 × 10 6 to the direction of motion of acted on the apple. The
7
the stopper. The stopper is not
= 3.59 × 10 m displaced in the direction of the weight of the apple causing
7
it to fall from its tree.
force, thus no work is done. • The gravittational force
Checkpoint 3.3
is a force that acts on
Q1 Man-made satellites are artificial 3. (a) (i) F = mg any pair of objects in the
satellite build and made by man. It is Mm universe. This gravitation
launched by a rocket into space and (ii) F = G R 2 force between the Earth
is placed in a specific orbit round the Mm M and the Moon produced the
earth for specific purposes. (iii) mg = G r 2 → g = G R 2 centripetal force that keep
The ISS is the largest man-made the Moon moving in circular
satellite orbiting the Earth. orbit round the Earth.
478
ANSWER FOCUS PHYSICS.indd 478 10/03/2023 10:42 AM

Physics SPM Answers

• Without centripetal force, specific heat capacity. Material and bounces back, the molecule
the Moon will continue P has the largest specific heat experiences a change in momentum.
to move in a straight line capacity due to the smallest mq The rate of change of momentum
tangen to its orbit. Due to value. from the rate of collisions with the
attrative force of gravity Q3 Q = CDq container wall produces a force. The
from the Earth, the Moon = 1 400 × (25 – 5) force per unit area hit on the wall
always falling towards the = 28 000 J produces gas pressure.
Earth. Q Q2 According to Charles’ Law, the
• Since the linear speed Q4 c = mDq volume of gas increases with
of the Moon is always temperature. This is because when
54 000 J
constant, then it can only = 0.6 kg × 200°C a fixed mass of gas is heated, the
fall back into its own orbit,
–1
Thus, the moon cannot = 450 J kg °C –1 gas molecules acquired energy and
reach the Earth. Q5 t = mcDq increase their kinetic energy. Gas
P molecules move faster and collide
with the wall of the container more
= 2 × 4 200 × 50 = 140 s frequent and this increases the
3 000
Chapter pressure of the gas. In order to keep
4 Heat Q6 (a) Q 1 = mcDq the pressure constant, the excess
= 0.6 × 4 200 × (100 – 27)



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

68
= 485 600 J SPM Practice 4
Q5 Q = mcDq(ice) + mL(ice) +
Checkpoint 4.2 mcDq(ice) + mL(steam) Objective Questions
Q1 (a) Mass – the larger the mass of = 3.2 × 2 100 × 5 + 3.2 × 3.34 × 1. B 2. B 3. C 4. C 5. B
5
the object, the larger the heat 10 + 3.2 × 4 200 × 100 + 3.2 6. C 7. A 8. D 9. A 10. D
6
capacity. × 2.26 × 10 11. B 12. B 13. C 14. B 15. B
(b) Shape – does not affect the = 9 678 400 J 16. C 17. B 18. B 19. A 20. B
heat capacity. Checkpoint 21. D 22. B 23. C 24. D 25. A
(c) Types of material – different 4.4
types of materials have different Q1 Based on the kinetic energy of the Subjective Questions
heat capacity. gas, the gas molecules are all in Section A
Q2 (a) Material T: The largest mass fast, random and continuous motion. 1. (a) (i) Thermal equilibrium
has the highest heat capacity. The gas molecules are always in (ii) The change of oil
(b) Specific heat capacity, c = Q , collision with each other and with temperature is greater than
mDq the wall of the container. When a water.
means smaller mDq, the larger gas molecule collides with the wall
479
ANSWER FOCUS PHYSICS.indd 479 10/03/2023 10:42 AM

Physics SPM Answers

(iii) The amount of heat energy Explanation: So that the cup Checkpoint 5.3
absorbed by oil and water is does not easily change its
the same. state or melt when filled with Q1 Let the distance of the wall from the
(iv) The specific heat capacity of liquid of high temperature. student be d m
oil is smaller than water. • Suggestion: The inside and The distance travelled by sound to
(v) PT = mcDq outside of the cup are coated and fro the wall = 2d m
500 × (2 × 60) = 0.5 × c × with shiny metal. 2d = v × t
(97 – 25) Explanation: Shiny materials = 340 × 2.0
c = 500 × 120 can prevent heat loss through = 680 m
0.5 × 72 radiation. Therefore,
= 1 667 J kg °C • Suggestion: The covered cup h = 340 m
–1
–1
(b) (i) degree of hotness can be stored in a pocket Q2 (a) v = 1500 m s , t = 0.12 s
–1
(ii) Volume of gas decreases. made of bamboo cloth. Distance travelled by the pulse
(iii) Absolute zero temperature Explanation: So that there is = v × t = 1500 × 0.12 = 180 m
(0 K or – 273°C) a layer of air trapped inside it (b) Distance of the shoal of fish
can reduce heat loss. 180
Section C below the boat = 2 = 90 m
2. (a) Heat is a form of energy that (c) Ultrasonic waves can transfer more
can flow from area of high energy than audible sound
temperatures to area of low Chapter waves.
5
temperatures. Waves
(b) (i) • The mass of water in Checkpoint 5.4
kettle P is smaller than the Checkpoint 5.1
mass of water in kettle Q. Q1 (a)
• The heat energy supplied Q1 (a) A wave is a travelling
to the kettle P and the disturbance from a vibrating or
kettle Q are the same. oscillating source.
• The temperature of water (b) Transverse waves
rise in kettle P is higher (c)
than the temperature of Deep Shallow
water rise in kettle Q.
(ii) The smaller the mass (b) Refraction of waves
of water, the higher the Q2 (a) (i) A wave in which the
3
temperature rise. particles of the medium Q2 (a) f = 16 Hz; λ d = ; v d = 6 cm s –1
(iii) The heat supplied must be move in the direction λ s 2
constant. perpendicular to the λ d = λ d
(c) The initial temperature of the direction in which the wave v s λ s
orange juice is higher than the propagates. λ s
temperature of the ice cubes. (ii) Water waves Therefore, v s = λ d v d
Heat from the orange juice flows (b) (i) A wave in which the 2
to the ice cubes to melt it. The particles of the medium = 3 × 6
heat is continuously absorbed move in the direction = 4 cm s –1
by the cold melting water from parallel to the direction in
the ice. The temperature of the which the wave propagates. (b) λ s = λ s = 4 = 0.25 cm
orange juice decreased and (ii) Sound waves f 16
the temperature of the cold Q3 (a) Period of oscillation, T = 0.5 s
water from the ice increased. Q3 (a), (b), (c)
1
1
Therefore, orange juice is (b) Frequency, f = = 0.5 Direction of
T
cooled by adding ice cubes. = 2.0 Hz Shallow propagation
area
(d) • Suggestion: The cup should
be made of good heat (c) Wavelength, λ = 5 cm
insulator such as polystyrene (d) Using v = f λ , –1 Normal
material. v = 2.0 x 5 = 10 cm s Deep
Explanation: Good heat Checkpoint 5.2 area
insulation prevents heat from
hot drink being escape by Q1 (a) Q
conduction. (b) Q has the same natural
• Suggestion: The cup cover frequency as the vibrating tuning Q4
should also be made of good fork. Hence, Q will receive the Hotter air
heat insulator such as plastic. biggest amount of energy from Speed of sound higher
Explanation: Good insulator the vibrating tuning fork.
can prevent heat loss to the (c) Resonance
surrounding. Q2 The hand moves with the same Refraction of sound waves occur
Listener
• Suggestion: The cup material natural frequency as that of the Source s
should also have a high loaded spring. Resonance occurs of
sound
melting point. and the spring will receive maximum
energy. Colder air
Speed of sound lower
480




ANSWER FOCUS PHYSICS.indd 480 10/03/2023 10:42 AM

Physics SPM Answers

The diagram shows the temperature by the sum of the displacements (c) Photography / Photosynthesis
of air at night. The layer of air of the individual waves at each by plants / Enables human
above the surface of the Earth is point of the medium. beings and animals to see. (Any
colder compared the layer further (b) (i) one answer)
away from the surface of the Earth. (d) R has very high penetrating
Since the speed of sound waves a 2a power and can cause cancer
is higher at higher temperature of and genetic defects to living
air, refraction occurs and the sound cells.
waves are refracted towards the (ii) s
surface of the Earth. This results in Q3 v = t , therefore,
the listener be able to hear clearer a t = s
sound at night. v 320 000 × 1000
Checkpoint 5.5 a = 3 × 10 8
= 1.07 s
Q1 (a)
Q2 (a) • A: An up-and-down movement Q4 P : Microwaves; Q : X-rays; R :
with a large amplitude Visible light; S : Ultraviolet rays
• B: No motion/stationary
(b) a = 3.0 cm; D = 30.0 cm;
x = 15.0 cm SPM Practice 5
ax
(b) λ = D Objective Questions
3.0 × 15.0 1. D 2. B 3. D 4. C 5. C
= 30.0 6. B 7. C 8. D 9. B 10. B
= 1.5 cm 11. A 12. C 13. C 14. C 15. B
16. D 17. A 18. B 19. A 20. C
Q3 (a) λ = 540 nm = 540 × 10 m
–9
a = 0.4 mm = 0.4 × 10 m 21. B 22. D 23. A 24. D 25. C
–3
(c) D = 3.0 m 26. D 27. D
Using λ = ax , Subjective Questions
D
λD 540 × 10 × 3.0 Section A
–9
x = =
a 0.4 × 10 –3 1. (a)
= 4.05 × 10 m Direction of
–3
Q2 (a) Remains unchanged = 4.1 mm movement of hand
(b) Remains unchanged (b) • x = λD
(c) Remains unchanged a
• 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
will be bigger. (b) λ = f
= 0.90
Q4 (a) f = 600 Hz; v = 330 m s –1 3.0
Using v = fλ, = 0.3 m
v 330
(b) The leaves which act as λ = — = —— (c)
obstacles have small width. f 600
Therefore, the effect of = 0.55 m
diffraction is greater. Hence the (b) a = 1.5 m; D = 3.0 m;
waves join back after a short x = λD
distance from the leaves. a
(c) The log which is a larger = 0.55 × 3.0
obstacle causes less diffraction. 1.5
The waves join back after a = 1.1 m
futher distance from the log. 2. (a) Diffraction of waves
Q4 (a) The sound waves spread out Checkpoint 5.7 (b) • The amplitude of the waves
beyond the edge of the window Q1 1. They all transfer energy from before passing through the
after passing through the one place to another. slit is higher than that the
window. 2. They are all transverse waves. amplitude after passing
(b) Diffraction of sound waves 3. They can all travel through a through the slit.
vacuum. • This is because the waves
Checkpoint 5.6 4. They all travel with a speed of spread over a larger area and
8
–1
Q1 (a) The principle of superposition 3 × 10 m s in vacuum. energy per unit area of the
diffracted waves is less.
states that at any time, the Q2 (a) P : Microwaves; Q : Visible light;
combined wave forms of two or R : X-rays
more interfering waves is given (b) P has higher frequency than
radio waves.
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ANSWER FOCUS PHYSICS.indd 481 10/03/2023 10:42 AM

Physics SPM Answers

(c) (c) (i) Oscillating with the does not move along with
amplitude decreasing with the wave.
time.
(ii) • There is resistance 7. (a) • The convex-shaped peak of
between the loaded spring the wave is like a convex lens.
and the oil. • The light rays from the lamp
• As a result, energy is lost passing through the peak will
from the system as heat be converged on the screen to
energy. produce a bright band.
• If no external force is applied, • The trough is like a concave
3. (a) d = v × t = 1500 × 0.18 the spring will eventually lens.
= 270 m stop completely. • The light rays from the lamp
(b) The depth of the sunken ship • Damping. passing through the trough will
= d 2 = 270 = 135 m (d) (i) be diverged causing a dark
band.
2
(c) Ultrasonic waves have more (b) (i) • The frequencies of the
energy than audio waves. waves in both the deep
4. (a) P = X-rays and shallow areas are the
same.
Q = Infrared • The wavelength of the
6
(b) (i) 100.6 × 10 Hz or • Move the hand waves in deep area is
1.006 × 10 Hz to-and-fro, perpendicular longer than that in shallow
8
(ii) Wavelength to the spring. area.
= 3.0 × 10 8 • The waves move forward • The speed of the waves
1.006 × 10 8 along the spring. in deep area is higher
= 2.98 m. • The string tied to the than that in shallow area.
spring which represents
(c) For taking the image of the a particle of the medium (ii) • When the angle of
patient’s bones. oscillates perpendicularly incidence is zero, the
direction of waves
to the direction of the
5. (a) • Ultrasonic waves transmitted waves. moving from deep to
by the bat is reflected by the • This shows that in shallow areas remains
body of the bird. the propagation of a unchanged.
• The reflected waves is transverse wave, the • When the angle of
detected by the bat. particle of the medium incidence is not zero,
• The time between the vibrates/oscillates the waves change their
transmission and detection of perpendicularly with the direction when moving from
the signal by the bat is equal direction of the wave. deep to shallow areas.
to twice the distance between (ii) (c) Refraction of waves
the bat and the bird. (d) (i) • The resort is to be built
• With that, the distance near the bay
between the bat and the bird • The waves at the bay are
can be estimated through the calmer than at the cape
period of time transmission • Move the hand to-and-fro, • due to the divergence of
and detection of the ultrasonic parallel with the spring. the waves’ energy from
waves. • The waves move forward the bay
(b) (i) Distance = d m along the spring. • and the convergence of
2d = 1 450 × 120 × 10 . • The string tied to the the waves at the cape
-3
Therefore d = 87 m. spring which represents (ii) • To reduce erosion,
(ii) v = fλ a particle of the medium retaining walls are built
3
1 450 = 45 × 10 × λ oscillates parallel with the • to reflect the waves from
1450 direction of the waves. the shore
Therefore, λ = • This shows that in • and to reduce direct
45 × 10 3
= 3.2 × 10 m the propagation of a impact of the waves on
-2
longitudinal wave, the the shore
Section C particle of the medium (iii) • Concrete structures with
6. (a) A system that undergoes a vibrates/oscillates parallel a gap in between are built
periodic to-and-fro movement. with the direction of the at the designated area for
(b) • Spring B carries a bigger wave. children
mass than that of spring A. (iii) • Work done by the hand in • Waves passing through
• Spring B oscillates with a moving the spring the gap will be diffracted
bigger period than spring A. to-and-fro causes the in the children’s area
• Spring B oscillates with a energy to be transferred • The smaller amplitude
lower frequency compared to in the form of waves of the diffracted waves
spring A. along the spring. causes the sea to be
• The bigger is the mass • The string tied to the calmer.
attached to the spring, the spring which represents 8. (a) Refraction of waves is a
lower is the frequency of its a particle of the medium phenomenon that occurs when
oscillation. vibrates/oscillates but there is a change of direction
482





ANSWER FOCUS PHYSICS.indd 482 10/03/2023 10:42 AM

Physics SPM Answers

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



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ANSWER FOCUS PHYSICS.indd 483 10/03/2023 10:42 AM

Physics SPM Answers

(b) The size of the image does not Q4 Checkpoint 6.6
change but the brightness of
image reduced by half. v/cm Q1 (a)
Q3 (a) Convex lens 60 P
v
(b) M = = 30 = 0.5 u = v Q
60
u
(c) 50
42 F
40
R
F 30
u = 60 cm f = 20 cm 20
v = 30 cm (b) The image is real, inverted and
10 magnified.
(c) The image is nearer to the
Focal length, f = 20 cm
0 u/cm mirror and becomes smaller in
10 18 20 30 40 50 60
Checkpoint 6.4 42 7 size.
v
1 1 1 (a) m = = 18 = = 2.33 Q2 (a)
3
Q1 (a) + = u
u v f (b) u = v = 25 cm
1 1 1 1 1 1
= – = – = 25
v f u 20 30 60 f = 2
v = 60 cm = 12.5
v 60 25°
m = = 30 = 2 Checkpoint 6.5 F C
u
Therefore, the image is real,
inverted, 60 cm from the lens, Q1 (a) For distant objects, the image
located opposite to the object must be formed at the focal
point of the lens. Therefore, the
and enlarged 2 times. distance between the lens and (b) The image is at F.
1 1 1 the film is 50 mm. (c) The image is virtual, upright and
(b) + =
u v f (b) The camera lens should be diminished.
1 1 1 1 1 1 adjusted outward toward the
= – = – = Q3 • Focal length of mirror P = 12 cm.
v f u 20 15 60 object. As the object distance
v = –60 cm is reduced, the image distance The image from distant object is
formed at the focal point of the
increases. With this, the
v 60 distance between the lens and mirror.
m = = = –4
u 15 the film is slightly longer than • Focal length of mirror Q = 10 cm.
The image is virtual, upright, at 50 mm. Object is located at radius of
a distance of 60 cm from the Q2 (a) Astronomical telescopes curvature, image distance = 2f.
lens on the same side as the (b) Objective lens: lens P, Eyepiece: • Therefore, the mirror Q has a
object and is magnified 4 times. lens Q shorter focal length.
v 15 (c) The separation distance of the
Q2 m = = h i = 10 = 1.5 lenses = 40 + 5 6
u
h o
Therefore, v = 1.5u = 45 cm SPM Practice
= 1.5 × 2.0 m = 3.0 m (d) 5 cm Objective Questions
1 1 1 40 cm
Using + = Parallel rays 1. B 2. C 3. B 4. C 5. B
u v f from distant 6. C 7. C 8. C 9. D 10. D
1 1 1 5 object 11. C 12. A 13. B 14. B 15. C
3
f
2 + = = 6 16. A 17. C 18. A 19. A 20. C
Therefore, focal length f = 1.2 m Lens P Lens Q 21. C 22. D 23. D 24. B 25. C
1 1 1
Q3 Using u + = Subjective Questions
f
v
1 1 1 1 1
v = – = – 10 – 30 Q3 Similarity: Section A
f
u
4 • Consists of two convex lenses 1. (a)
= – • The eyepiece function as a
30 magnifying glass
v = –7.5 m Differences:
30 • For microscope in normal
v 4 1
m = = = adjustment, the final image is at
u 30 4 near point. While for telescope in (b) (i) f = 10 cm, u = 12 cm
The image is virtual, upright, normal adjustment, the final image 1 1 1 1 1 1
at a distance of 7.5 cm from is at infinity. f = + → 10 = 12 + v
u
v
the lens on the same side • The distance between the 1 1 1 6 – 5 1
as the object and has linear objective lens and the eyepiece of v = 10 – 12 = 60 = 60
1
magnification of .
4 the microscope is L O > f O + f e . The v = 60 cm
distance between the objective Therefore, distance between
lens and the eyepiece of the the screen and the object
telescope is L O ≤ f O + f e . = u + v = 12 + 60 = 72 cm
484
ANSWER FOCUS PHYSICS.indd 484 10/03/2023 10:42 AM

Physics SPM Answers

v 60 (d) (i)
(ii) m = u = 12 = 5 FORM 5
(c) If the distance between the lens Right-angle Chapter
1
and the image = 10 cm = f, then prism Force and Motion II
object is at the focal point and
image formed is virtual and at Checkpoint 1.1
infinity. Therefore, image cannot Distant object
be seen on the screen. Cylindrical Q1 (a) 17 N, in the same direction as
tube the direction of the acting force
Image at infinity (b) 7 N, in the same direction as the
direction of 12 N
(c) F = 5 + 12 = 13 N
2
 2
F F –1 12
q = tan 5
Upright image = 67.4° with the horizontal
Section C Right-angle force of 5 N
2. (a) (i) The critical angle of glass is prism Q2 (a) Scale 1 cm : 2 N
42° which means that if the • A periscope is build using B
angle of incidence in glass two right-angle prisms
(denser medium) is 42°, arrange in an cylindrical
then the angle of refraction tube as shown in the 4.6 cm
in the air is 90°. diagram above. 3 cm
(ii) Total internal reflection • The prism acts like a
1 perfect mirror when light 27° 45°
(iii) sin c = O
n ray hits the inner surface 2 cm A
1
n = of the prism at an angle
sin c greater than 42°. The first Resultant force OB = 4.6 cm × 2 N
1
= sin 42 prism turns the image = 9.2 N
= 1.49 from a distant object 90°, Direction of the resultant force =
27° with the force of 4 N.
and then the second
prism turns the image
(b) back to upright. (b) Scale 1 cm : 2 N
• Object lens is placed in B
front of the first prism on
top to find a distant object.
45° 45° 4.4 cm
45° 45° The eyepiece is used to 3 cm
observe the image formed
by the second prism.
(ii) 37° 120°
Right-angle prism O A
45° 5 cm
45° Resultant force OB = 4.4 cm × 2 N
Distant
45° = 8.8 N
object at
the back Direction of the resultant force =
45° 45° 37° with the force of 10 N
45° Cylindrical
tube (c) Scale 1 cm : 2 N
O 3 cm A
(c) (i)
88° 120°
Glass
An inverted
Air image Right-angle prism
Total internal 5.2 cm 6 cm
Light ray reflection occurs here When the prism at the
top periscope is laterally
inverted, it can be used to
A light ray is transmitted observe the object behind
through a curved fibre optic the observer over an
that occurs a series of total obstacle. In this case, the B
internal reflections on the image seen is inverted
inner surface. (iii) • A right-angle prism uses Resultant force OB = 5.2 cm × 2 N
(ii) • Fiber optic cables are total internal reflection = 10.4 N
lighter and thinner. to reflect light. The light Direction of the resultant force =
• Transmission of signal reflected by the prism 88° with the force of 6 N
with almost no energy surface is 100%.
loss along the optical • Image formed by right- Q3 (a) Magnitude of the resistance
fiber. angle prism is sharper. = 20 N

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Physics SPM Answers

(b) Resultant force, F = 30 – 20 = 10 N (b) (i) Triangle of forces (iii)
F = ma → 10 = 4 × a
a = 2.5 m s –2 F R B F F
Q4 (a) Resultant force, F = 0 N; C 100° A
Tension, T = weight of the fish O 50° 30°
= 0.8 × 10 F = 10 kN
= 8 N B
(b) Resultant force, F = 0 N; 120 N (b) (i) Scale 1 cm : 20 N
Tension, T = weight of the fish = 8 N 100 N
(c) Resultant force, F = ma = 0.8 O
× 0.5 = 0.4 N; T – 8 = 0.4, A
therefore T = 8.4 N 40°
Q5 (a) F = ma → 2 × 10 = (6 + 2) × a Resultant force, 9.7 cm
a = 2.5 m s –2 F R = 100 + 120 = 156.2 N 7.5 cm T 1
2
 2
(b) T = m × a = 6 × 2.5 = 15 N (ii) Free body diagram W
Checkpoint 1.2 T
2 50°
Q1 (a) Resolution of a force means A B
dividing a single force into two 6.1 cm
component forces perpendicular 120 N T 1 = 9.7 cm × 20 N = 194 N
to each other. T 2 = 6.1 cm × 20 N = 122 N
(b) (i) Vertical component upward F R (ii) Scale 1 cm : 20 N
= 12 cos 55° 39.8°
Horizontal component P
= 12 sin 55° 60° 4.1 cm
(ii) Vertical component 100 N
downward = 25 sin 45° Q 70°
Horizontal component
= 25 cos 45° Direction of resultant force, q 5 cm
–1 100
(iii) Component parallel to the = tan = 39.8°
inclined plane = 30 cos 60° 120 R 4.6 cm
Component perpendicular Therefore, a force of 156.2 N 50°
to the inclined plane = needed to act in the
30 sin 60° direction 129.8° (39.8°
S2 (a) Net force = 0 N as the car is at + 90°) with the force of P = 4.1 cm × 20 N = 82 N
100 N for the box to be in
rest equilibrium. R = 4.6 cm × 20 N = 92 N
(b) Frictional force = component
weight of the car down the slope Q2 Magnitude of the force (iii) Scale 1 cm : 20 kN
= 12 000 sin 20° = 4 104 N = 400 cos 30° + 400 cos 30°
(c) The supporting force = 692.8 N
= 12 000 cos 20° = 11 276 N Q3 (a) (i)
Q3 (a) Free body diagram: F C F A
40° 2.6 cm 3.9 cm
Tension, T T 1
150 N
10° 50° 30°
F B 5 cm
T F A = 3.9 cm × 20 kN = 78 kN
2
Air resistance, R F C = 2.6 cm × 20 kN = 72 kN
(ii)
Weight, W Checkpoint 1.4
(b) Vertical component of T = T cos 10° 60° 30° S1 (a) Spring extension, x = 18 cm –
Horizontal component of P 15 cm = 3 cm = 0.03 m
T = T sin 10° Magnitude of force, F = kx =
(c) T cos 10° = 1000 × 10 70° 100 N m × 0.03 m = 3 N
–1
T = 10 000 = 10154 N Q = 100 N (b) Length of compression,
cos 10° F
R x = k
2 N
Checkpoint 1.3 50° = 100 N m –1
40°
S1 (a) The forces are in equilibrium if = 0.02 m
the resultant force produced is = 2 cm
zero in all directions. Therefore, length of spring
l = 15 cm – 2 cm = 13 cm




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Physics SPM Answers

Q2 Length of compression, 2. (a) Direction of motion 3. (a) Weight due to force of gravity
x = 20 cm – 15 cm act on the spring.
= 5 cm (b) Vector has magnitude and
= 0.05 m direction.
1 (c) (i) I = l O + x
Energy needed, E = kx 2
2 T
1 10° Length of spring
–1
= 2 × 200 N m
× 0.05 m 2 1 kg 140
2
= 0.25 J 120
Q3 (a) Spring constant,
k = gradient of graph 10 N 100
8 N 80
= Resolving vertically
16 cm I = 76 cm
= 0.5 N cm –1 T cos 10° = 10 N 60
10
(b) Potential elastic energy of spring T = cos 10° 40 I 0
= Area under the graph = 10.2 N 20
1 (b) F = ma 9N
= × 8 × 0.16 = 0.64 J 0
2 T sin 10° = 1 kg × a 2 4 6 8 10 12 14
Q4 (a) (i) Hooke’s law a = 10.2 sin 10° Force/N
(ii) Mass of load of 1 kg is = 1.77 m s (ii) l = l O + x
–2
equivalent to a force of 10 N (c) (i) 76 = 40 + x
Substitute into formula, x = 76 – 40 = 36 cm
F = kx → 10 N = k × 2 cm (d) Curve upwards after 10 N (refer
Therefore, spring constant, the above graph)
k = 5 N cm atau 500 N m
–1
–1
(b) – Increase the stiffness
of spring by replacing 4. Scale 1 cm : 500 N
the copper spring with a (ii) Direction of motion 5 cm
stronger steel spring. a = – 1.77 m s -1
– Replace spring with spring
made of thicker wire. F 1
10° 7 cm R
SPM Practice 1 7 cm
Objective Questions 10.5cm
1. B 2. A 3. D 4. A 5. C
6. B 7. C 8. A 9. D 10. D 1 kg 36°
11. A 12. C 13. A 14. B 15. B 60°
16. A 17. A 18. B 19. C 20. D Due to deceleration, the 5 cm F 2
21. D 22. C 23. C 24. B 25. C direction of action is in the
opposite direction. Resultant force, R = 10.5 cm ×
Subjective Questions 500 N = 5 250 N
Direction of resultant force is
Section A q = 36° with the force of 2 500 N
1. (a)
5. (a) The net force acting on the system = 30 N – 10 N
T T = 20 N
(b) 3.33 m s –2

T 2.0 kg T
45° 45° 1 2
10 N
T T
1 2
(b) T + T = 10 2 3.33 m s –2 3 kg 1 kg 3.33 m s –2
2
2
2T = 100
2

T = 50 = 7.07 N
30 N 10 N
F = ma
20 = (3 + 2 + 1)a
a = 20
6
= 3.33 m s –2


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Physics SPM Answers

(c) (b) P = hρg (b) If the hydraulic fluid (oil) in the
jack is replaced by ordinary
P oil = P water air, the jack will not work.
T 1 ρ oil h oil g = ρ water h water g This is because ordinary air is
1000 × 0.03 compressible.
3.3 m s –2 3 kg ρ oil =
0.05 Q2 (a) Pascal’s principle states that
= 600 kg m −3 pressure applied to an enclosed
fluid is transmitted uniformly to
Checkpoint 2.2 every of the container of the
fluid.
30 N Q1 (a) (i) The length of the mercury F 12
column is 75 cm. Therefore, (b) (i) Pressure, P = = 9
A
3.3 m s –2 the atmospheric pressure is 4
75 cm Hg. = 3 N cm –2
3
2 kg (ii) P = hρg = 0.75 × 13.6 × 10 (ii) The pressure is transmitted
× 10 = 102000 Pa
T 1 T 2 (b) The length of the mercury equally, therefore the
column is still 75 cm. This is pressure exerted by the oil
because atmospheric pressure on the large piston is also
can support a vertical height of 4 N cm –2
T 75 cm of mercury. 3
2 (iii) F 2 = 4 × 10 8
Q2 (a) h = 76 cm 3
(b) Pressure at the point X = 76 + = 144 N
1 kg 3.3 m s –2 10 = 86 cm Hg = 0.86 × 13.6 × (c) A small input force can yield a
10 × 10 = 116960 Pa large output force.
3
(c) The level of mercury inside
the tube is the same as in the Q3 (a) 1. A hydraulic pump is more
10 N container outside the tube. flexible. It can be placed
in between the small gap
30 − T 1 = 3 × 3.33 This is because Moon does not between the wall and the
T 1 = 30 − 9.99 have atmosphere. Hence the floor.
= 20.01 N atmospheric pressure is equal to 2. A hydraulic pump is safer
T 2 − 10 = 1 × 3.33 zero. to use. Heavy machineries
T 2 = 3.33 + 10 Checkpoint may cause further collapse
= 13.33 N 2.3 of concrete structures and
Alternative: Q1 (a) The pressure of the gas is risk harming the victim
T 1 − T 2 = 2 × 3.33 higher than atmospheric trapped underneath the
20.01 − T 2 = 6.66 pressure. The level of oil in collapsed wall.
T 2 = 20.01 − 6.66 the arm attached to the gas (b) Using the formula,
= 13.35 N is lower than the other arm F input F output
(d) Yes, the system is still moving of the manometer. Hence the A input = A output
in the same direction. The pressure of the gas is the sum F output
system will move at a smaller of atmospheric pressure and A output = F input × A input
acceleration as the masses the pressure due to the 50 cm 80
increases. column of the oil. = 3 × 15
(b) Difference in pressure DP = hρg = 400 cm 2
= 0.5 × 800 × 10 = 4000 Pa
Checkpoint 2.5
Q2 Pressure of gas, P + Pressure due
Chapter
2 Pressure to the 60 cm column of mercury = Q1 (a) 6.3−5.5=0.8 N
Atmospheric pressure (b) 0.8 N
P + hρg = 10 (c) 0.8 N
5
Checkpoint 2.1 P = 100000 − (0.6 × 13.6 ×
10 × 10) Q2 (a) 0.18 × 0.06 × 10 = 0.108 N
3
Q1 Pressure, P = hρg = 100 × 1050 × 10 = 18400 Pa (b) Buoyant force = Weight of the
= 1.05 × 10 N m −2 displaced air
6
Force, F = P × A = 1.05 × 10 N m Checkpoint 2.4 = 1.3 × 0.06 × 10
6
−2
= 0.78 N
× 200 m = 2.10 × 10 N (c) Weight of the balloon + W
8
2
Q2 (a) The deeper the water, the higher Q1 (a) F 1 = F 2 = Buoyant force
A 2
A 1
the pressure. Therefore, the F 2 0.108 + W = 0.78
bigger the force acting on the F 1 = A 2 × A 1 W = 0.672 N
wall. 2 Q3 (a) For a floating object, the weight
D 1
(b) Pressure, P = hρg = (300 – 60) = F 2 2 × p  of the object is equal to the
2
D 2
6 p 
× 1000 × 10 = 2.4 × 10 P a 2 buoyant force acting on it. Since
Q3 (a) Pressure at the same level is 2 1.25 2 the same boat is used for both
D 1
the same. P and Q are at the =   × F 2 =  25  × 6000 situations, the buoyant force
same level. Hence both have D 2 acting on the ship is the same.
the same pressure. = 15 N
488
ANSWER FOCUS PHYSICS.indd 488 10/03/2023 10:42 AM

Physics SPM Answers

(b) Since the buoyant force is equal (b) (i) Pressure at A caused by the
to the weight of the displaced SPM Practice 2 column of water P A = h A ρ A g
water, the weight of the Objective Questions = 0.3 × 1000 × 10 = 3000 N m .
−2
displaced seawater is equal to (ii) Pressure of air in the tube,
the weight of the displaced river 1. D 2. A 3. B 4. A 5. D P air = 10 − 3000
5
water. 6. C 7. A 8. A 9. B 10. C = 97000 N m −2
(c) Since the weight of the 11. B 12. B 13. C 14. B 15. D (c) When the clip is removed, air
displaced seawater is equal to 16. C 17. A 18. A 19. C 20. A enters into the tube and the
the weight of the displaced river 21. C 22. B 23. A 24. D pressure inside and outside
water, the tube becomes the same
ρ sea V sea g = ρ river V river g Subjective Questions and is equal to the atmospheric
ρ sea V sea = ρ river V river Section A pressure. Both the water and
Since V sea < V river therefore, ρ sea 1. (a) Pressure at A = Pressure at B liquid X columns will fall until the
 ρ river . The density of seawaters same levels as outside of the
is higher than the density of h 1 ρ 1 g + pressure of air in the tube.
river water. tube = h 2 ρ 2 g + pressure of air in
Q4 (a) Weight of the boat the tube
Therefore,
= Weight of the displaced seawater h 1 ρ 1 g = h 2 ρ 2 g
= ρVg = 1020 × 1.5 × 9.8
= 14994 N ρ 2 = h 1 × ρ 1
(b) For maximum load, h 2
Weight of the boat + Additional = 30 × 1000
weight 22
= Weight of the displaced seawater = 1364 kg m –2
= ρVg =1020 × 4.5 × 9.8
= 44982 N
Therefore, the additional weight 2. (a) Piston 1 V 1 V 2 Piston 2
that can be added = 44982 – Down Open Closed Up
14994 = 29988 N Up Closed Open Stationary
Checkpoint 2.6
(b) Assume that the liquid is • The pressure of air above
Q1 (a) The speed at P is higher than incompressible. the aerofoil is higher than
the speed at Q. Using the formula that below the aerofoil.
(b) The pressure at P is lower than F input F output • Due to the difference in
the pressure at Q. A input = A output pressure, a downward
(c) A output 32 force is produced.
P F output = F input × A input = 5 × 4 (iv) • A bigger piece of spoiler has
= 40 N a bigger surface area, A.
Q (c) Pascal’s principle F
• Since P = —, F = P × A.
Sail (d) The system can convert a small A
input force into a bigger output • Hence a bigger downward
Direction force. force is produced by a
of wind (e) Turn the release valve to allow bigger spoiler.
the hydraulic fluid to flow from (b) (i) • Design Q is most suitable.
Q2 increases; low; Atmospheric cylinder 2 into the reservoir. • This tube will produce a
Q3 (a) • The hydrofoil of the boat fast stream of air at the
has a curved upper surface Section B end of the tube because
and flat lower surface. This 3. (a) (i) Action on an object that can the cross-sectional area of
causes the water to travel at result in changes like size, the tube becomes smaller.
a higher speed at the upper shape, speed and direction. (ii) • Design P is not efficient
surface than the lower (ii) Aerofoil R because the cross-
surface of the hydrofoil. sectional area increases
• The pressure of water at the and the speed of air
upper surface is lower than decreases.
the pressure at the lower • Design R will not produce
surface of the hydrofoil. The a stream of air because
difference in the pressures D its end is closed.
results in a net force acting (iii) • Design Z is most suitable.
upward on the hydrofoil. • Due to the hole in the
Thus lifting the boat up. (iii) • Lines above aerofoil
(b) This will reduce the area of almost straight. stopper, the pressure of
contact between the boat and • Lines below aerofoil air in the container is the
the sea, hence reducing the curved. same as the pressure
resistance of the boat when it is • Air flows at a lower speed outside the container (or
travelling. above the aerofoil than atmospheric pressure).
that below the aerofoil.



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Physics SPM Answers

(iv) • Designs X and Y are • A very big balloon is Q2 (a) F = 2.0 × 10 N; e = 1.6 × 10 C
–19
–15
not suitable because as required to displace a F
more water is drawn out, very big volume of air. Substitution, E = Q
the pressure of the air • This is to produce a large Electric field strength, E
in the containers will be buoyant force. = 2.0 × 10 –15
reduced. (ii) • Hot air is produced by 1.6 × 10 –10
• Then it will not be able to heating air by burning gas = 12500 N C −1
force the water out of the at the base of the balloon.
–1
container. (iii) • Hot air rises to fill the (b) E = 12500 N C ; d = 5 cm
(v) • When air is blown out of balloon and displaces the = 0.05 m
cool air.
V
the tube at high speed, • A buoyant force is Using, E = , rearranged, V = Fd
d
• its pressure reduces. generated because hot Therefore, potential difference, V
• The higher atmospheric air is less dense than the = 12500 N C × 0.05 m
–1
pressure of the air in the cool air it displaces. = 625 V
container will force the • When more hot air fills the
water out of the container balloon until the buoyant Q3 The amount of charges
through the tube attached force is greater than the flowing through the TV in 10
to the container. total weight of the balloon, minutes,
• The water will be carried the balloon rises into the Q = It
by the air stream and sky. = 0.5 × 30 × 60
sprayed outward. (iv) • To make the balloon fall = 900 C
back onto the surface of Q 30000
Section C the earth, reduce the rate Q4 (a) I = = 50 × 60
t
–1
4. (a) (i) Density is mass per unit or stop the heating of the = 10 C s (A)
air.
volume of the material. • As less hot air fills the Q 1
(ii) • The egg sinks more in balloon, its buoyant (b) Q = ne → n = = 1.6 × 10 –19
e
water than in brine. force reduces. When the = 6.25 × 10 18
• The volume of the water buoyant force is less than
displaced is more than the weight, the balloon will Q5 V = W = 240 = 12 V
that of the brine. fall towards the earth. Q 20
• The density of water is Q6 Charge flowing through the lamp in
less than brine. 10 s,
• The weight of the egg is Q = It
the same when it is in the Chapter = 2 × 10
water and in the brine. 3 Electricity = 20 C
• The egg floats because its Electrical energy changed by the charge,
weight is balanced by the Checkpoint 3.1 E = VQ
buoyant force. = 240 × 20
• More liquid will be Q1 (a) = 4800 J
displaced when its density
is lower. Checkpoint 3.2
(iii) Archimedes’ principle 1
Q1 (a) R e = 8 + = 8 + 2 = 10 Ω
(b) • A large part of the ship is + 1 + 1
hollow. 3 6
• It displaces a very large (b) V = IR = 2(10) = 20 V
volume of water. V 1 = 2 × 8 = 16 V
• Hence a big buoyant force V 2 = 20 V – 16 V = 4 V
is generated to balance the (c) I 1 = V = 4 = 1.33 A
weight of the ship. R 3
(c)
Buoyant force (b) Q2 (a) Length, l; cross sectional area,
A; resisitivity or type of material
+ + (b) (i) R X > R Y > R Z
(ii) The gradient of the graph
+ +
shows the resistance of the
+ + conductor, then, gradient of
X > gradient of X > gradient
+ + of Y > gradient of Z
Q3 Using formula,
+ + (2.8 × 10 )(1000)
–8
Weight R = ρl =
t 1.5 × 10 –4
= 0.19 Ω
(i) • Low density material used
to make the balloon to
reduce its weight.
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Physics SPM Answers

Q4 Diameter, D = 0.100 × 10 m Q4 11. B 12. D 13. A 14. D 15. B
–3
Cross sectional area, V/V 16. A 17. C 18. A 19. A 20. A
21. C 22. A 23. A 24. B 25. C
A = p  0.100 × 10 –3 2

2 3.0 Subjective Questions
−9
= 7.85 × 10 m 2 Section A
ρl 2.0
Using equation R = rearrange 1.7 1. (a) Bulbs P and Q
A (b) Bulbs P and Q are connected in
RA series, the effective resistance
l = 1.0
e is larger and so less current
passing through them.
Therefore, the length of tungsten 0 I / A (c) (i) and (ii)
filament, l = 0.200 × 7.85 × 10 –9 0.2 0.4 0.6 0.8
5.6 × 10 –8 0.74 V
= 0.028 = 28 mm ɛ = I(R + r) = V + Ir P Q
ρl ɛ – V 3.0 –1.7
Q5 R = ; R P = ρ l P ; R Q = ρ l Q r = =
A A P A Q I 0.74
l Q = A Q → l Q = 0.0032 = 0.7 Ω A
I P A P 0.004 Checkpoint 3.4 R
= 8 cm
Q1 (a) P = VI = 12.0 × 1.2 = 14.4 W S
Checkpoint 3.3 (b) E = Pt = (14.4 × 60) = 864 J
Q1 (a) Using ɛ = I(R + r) Q2 (a) P = V R 2 = 240 2 E output
60
→ I = ɛ = 1.5 = 960 W = 0.96 kW 2. (a) (i) η = E input × 100%
R + r 2.5 + 0.5
= 0.5 A (b) E = Pt = (960)(20 × 60) = 95000 × 100%
= 1.152 × 10 J


6
V = IR 120000
= 0.5(2.5) Q3 P = V 2 → R = V 2 = 240 2 = 28.8 Ω = 79.17%
= 1.25 V R P 2000 (ii) P = 90000
ɛ P 2000 60
(b) ɛ = I(R + r) → r = – R P = VI → I = = = 8.3 A
I V 240 = 1500 W
9 2 2 (b) The electric kettle can boil the
= – 5 = 4 Ω Q4 (a) P 10Ω : P 25Ω = V : V
1 water first. This is because the
R 10Ω R 25Ω rate of thermal energy supplied
2
2
(c) V = IR → l = V = 5 = 12 : 12 to water is larger and therefore
25
10
R 15 rate of heat transferred to water
1 = 5 : 2
= A is faster.
3 V 2 V 2
ɛ = I(R + r) (b) P output = R 10Ω + R 25Ω = 20.2 W 3. (a) Increasing the length of wire
1 increases the resistance of wire.
= × (15 + 3) Increasing the diameter of wire
3 Q5 (a) P = VI → I = P
= 6.0 V 24 36 V decreases the resistance of
wire.
V 10 I = 12 + 12 = 5 A l kl
Q2 V = IR → R = = = 12.5 Ω (b) R ∝ → R =
I 0.8 V 2 V 2 A A
ɛ (b) P = R → R = P kl
ɛ = I(R + r) → r = – R R 1 = = 16 Ω
I R 1 = 12 2 = 6 Ω A
1
= 12 – 12.5 24 k   l
0.8 12 2 R 2 = 2 = kl
= 2.5 Ω R 2 = 36 = 4 Ω 2A 4A
kl
1
Q3 (a) 4.5 V (c) R = 6 Ω + 4 Ω = 10 Ω = 1   = (16 Ω) = 4 Ω
1 1 1 1 4 A 4
(b) = + = 2 → r = Ω V 12
r 1 1 2 I = R = 10 = 1.2 A (c) (i) Effective e.m.f. = 2E
ɛ
(c) ɛ = I(R + r) → l = Because the batteries are
R + r The bulbs become dimmer. connected in parallel.
4.5
1
= 5 + 0.5 (ii) 1 = 1 + = 5
= 0.82 A SPM Practice 3 R 16 4 16
16
(d) V = IR = 0.82(5) = 4.1 V Objective Questions R = 5 = 3.2 Ω
(e) V drop = Ir = 0.82 × 0.5 = 0.41 V (iii) I> I 2 > I 1
1. B 2. C 3. D 4. C 5. D
6. B 7. B 8. C 9. D 10. C
491
ANSWER FOCUS PHYSICS.indd 491 10/03/2023 10:42 AM

Physics SPM Answers

Section B in Figure 4.2, dry cells are (ii) Number of battery used: 3
4. (a) (i) Electromotive force of 1.5 V connected in parallel. Reason: Increase the
meant the amount of energy The voltmeter reading in Figure magnitude of current
supplied to a coulomb 4.2 is smaller. (iii) Shape of permanent
charge to pass through a The ammeter reading in Figure magnet: Semi circular-shape
complete circuit is 1.5 J. 4.2 is smaller. Reason: More cutting the
(ii) V = IR = 1.0 × 2 = 2 V The larger the voltage supplied, magnetic flux
(iii) e = I(R + r) the greater the energy used to (iv) Electric motor A
3.0 = (1.0)(2 + 2r) move the electrical charges in 3. (a) Catapult field is a resultant force
r = 0.5 Ω the circuit. cause a conductor wire moves.
(b) In Figure 4.1, the dry cells are The larger the current flow in (b)
connected in series. Whereas the circuit, the faster the energy
transferred. K
N S
(c) (i) and (ii)
L
Characteristic Reason
Low resistivity To reduce heat loss in the cables (c) • Interaction between two
magnetic fields
Low density The cables will become lighter • cause a region to have a
combination of magnetic
Low cost Cost of project will be lower field with opposite direction
between each other.
Low rate of thermal expansion The cables will not expand under hot weather • So, the strength of the
magnetic field in that region
Choose: Cable S becomes weak.
Reasons: Low resistivity, low density, medium cost, low rate of thermal (d) (i) Fleming’s left-hand rule and
expansion right-hand grip rule
(ii)

Checkpoint 4.3
Chapter
4 Electromagnetism N S
S1 η = P output × 100
P input
Checkpoint 4.1 V s I s
= × 1000
V p I p
S1 (a) The direction of force can be 6000 × 0.20 4. (a) Direct current motor
identified by using Fleming’s left- = 240 × 6.25 × 100 (b) (i) A-B-C-D
hand rule (ii) downward, upward
= 80%
(iii) clockwise
(c) Long lasting and no
B SPM Practice 4 maintenance required because
S
C no carbon brush used.
Objective Questions
5. (a) Voltage
N 1. B 2. C 3. D 4. A 5. C (b) • When a bar magnet passes
A 6. C 7. B 8. D 9. C 10. D through the solenoid, the
+ D 11. B 12. A 13. C 14. B 15. C magnetic flux is being cut.
– • So, induced e.m.f is occurred
Subjective Questions and the galvanometer pointer
(b) AB moves upward and CD Section A is deflected.
moves downward. (c) • The magnet is fall from higher
1. (a) • The conductor will displace. place.
• The direction follows by the
Checkpoint 4.2 • So, the magnet will passes
Fleming left-hand rule. through the solenoid with
S1 (a) When the bar magnet is pushed (b) Extended higher speed.
into the solenoid, induced (c) • Extended (d) • No pointer deflection occurs.
current is produced thus • Use more battery to increase • No magnetic flux is being cut.
deflected the galvanometer current.
pointer. (d) • Inverts current direction 6. (a) P: North; Q: North
(b) If the bar magnet is stay • Inverts magnetic field polarity (b) P: South; Q: South
stationary inside the solenoid, no 2. (a) (i) C (c) Lenz’s law
induced current is produced thus (ii) The conductor will move to
no deflection of galvanometer the direction of A. Section B
pointer. (b) (i) Number of turn of coil: 50 7. (a) National grid network is a
turns network of cables connects all
Reason : More current flow power stations in a country to
through the coil consumers.
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Physics SPM Answers

(b) X – Step-up transformer 9. (a) (i) Step-down transformer
Y – Step down transformer (ii) • Number of turns at input is larger than number
(c) • Power can be distributed based on location demand. of turns at output.
• If a power station is breakdown or under • Voltage input is higher than voltage output.
maintenance, other power station can continue to • The larger the number of turns, the higher the
supply electricity without disruption. voltage produced.
• Reduced cost and pollution (b) (i) P – Step-up
(d) (i) P =VI Q – Step-down
10 000 000 = 144 000 I R – Step-down
2
I = 69.44 A (ii) • P loss = I R
(ii) P = I R • If the current increases, the loss of power also
2
= 69.44 × 300 increases.
2
= 1.45 × 10 W • By the Ohm’s law, when the voltage increases,
6
the current will decreases, thus the loss of
(e) Aspect Reason power also decreases.
Soft iron core Easy to magnetised and (c) Aspect Reason
material demagnetised.
The number of turns of To make step-up
Made by laminated Reduced eddy current primary coil is less then transformer
core secondary coil
K-shaped Reduced the leakage of Copper wire Low resistance
magnetic flux Soft iron core Can be magnetised and
Ratio of 40 : 1 Able to reduce 240 V to 6 V demagnetised easily.
The best Because easy to magnetised Laminated iron core Less hysteresis
transformer is U and demagnetised, reduced The secondary coil is Less magnetic flux
eddy current, reduced the wounded on the primary leakage.
leakage of magnetic flux and coil.
able to reduce 240 V to 6 V.
Section C
8. (a) Induced current is the current produced from a Chapter
5
conductor cutting magnetic flux. Electronics
(b) (i) • Both have equal magnetic field strength
• The number of turns in Figure 8.2 is higher than Checkpoint 5.1
Figure 8.1.
• The degree of deflection in Figure 8.2 is higher Q1 (a) Thermionic emission is the process of releasing free
than Figure 8.1. electron from hot metal surface. Cathode ray is a fast-
(ii) • The higher the number of turns of coil, the moving electron beam.
higher the degree of galvanometer pointer (b) – negative charge,
deflection. – can be deflected by electric field and magnetic
• The higher the number of turns of coil, the field,
higher the rate of cutting magnetic flux. – can produce fluorescent effect,
• The higher the rate of cutting magnetic flux, the – can stopped by thin metal.
higher the magnitude of induced current.
(c) • When the coil rotates, it cuts the magnetic flux from Q2 (a)
the magnet.
• Induced current is produced. Cathode ray
• The direction of induced current can be identified by
using Fleming’s right-hand rule.

(d) Aspect Reason (b) The cathode ray is a negatively charged electron
beam, so it is attracted to a positively charged plate
Number of turns must be More magnetic fluxes can and forms a parabolic path in the electric field between
higher be cut. the two plates.
Use higher magnetic Produce higher strength Q3 (a) The kinetic energy acquired by electron is equal to the
strength of magnetic field. electric potential energy
Use round magnet shape Less leakage of cutting = 4 keV
−19
the magnetic flux. = (4000)(1.6 × 10 )
= 6.4 × 10 J
−16
Brushless motor • Long lasting. 1
2
• No maintenance. (b) E = mv max = eV
2
The magnet is inside and The cutting of magnetic 1 × 9.1 × 10 × v max = 6.4 × 10
–31
–16
2
is surrounded by coil. flux is optimum. 2 
2 × 6.4 × 10
–16
v max = 
9.1 × 10 –31
= 3.75 × 10 m s –1
7
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Physics SPM Answers

Checkpoint 5.2 Subjective Questions (e) (i)
Q1 (a) The diode only allows current to Section A Filaments Anode S N
pass through it in one direction
only. 1. (a) Alternating current 6 V a.c. Shadow
(b) (i) Bulb Q is not lights up (b) Diode N
because diode D 2 is (c) Cathode S Screen
connected in reversed bias.
(ii) Terminal Y is positive. E.H.T
(iii) Both diodes are used to
indicate the polarity of (ii) Fleming’s left-hand rule
battery.
Q2 (a) Rectifier is a device that can Section B
convert AC voltage or current to 4. (a) (i) Transistor npn
DC voltage or current. (ii) • Both ammeters do not
(b) (i) Full wave rectification circuit (d) show any reading.
(ii) Bridge rectifier • Microammeter shows
(iii) To smoothen the direct reading and milliammeter
current output. does not.
(iv) (b) When the rheostat is adjusted
and both switches are switched
on, a base current is generated
to activate the transistor. This
causes a larger collector current
to flow through the circuit.
2. (a) (i) To heat up the cathode (c) • the current amplification by
Checkpoint 5.3 using the formula
(ii) To emit electron when
Q1 (a) P : Collector; Q : Emitter; heated by heating filament. b = I c
R : Base (iii) Thermionic emission I b –3
(b) pnp transistor (iv) To supply high potential = 1 × 10 –6 = 100
Q2 (a) npn transistor difference to accelerate the 10 × 10
(b) and (c) beam of electron to high • emitter current Ie.
speed. I c = I b + I e
1
(b) eV = m e V 2 I e = I c – I b
2
M = 1 × 10 – 10 × 10 –6
–3
1.6 × 10 × 3000 = = 9.9 × 10 A
–19
–4
1
–31
× 9.1 × 10 × v 2 (d) • To detect fire, a thermistor is
R 2
V C
T 2 15 used to detect heat and needs
v = 1.055 × 10 to be connected to terminal
v = 3.25 × 10 m s –1
7
V B P as a potential difference
(c) (i) divider.
Base circuit Emitter circuit • A resistor must be connected
(d) To limit the base current flow – to terminal Q to control the
into transistor. Parabola size of the current flowing into
the base of transistor.
Q3 (a) Thermistor Straight line • At the collector terminal of
(b) Function of P is to detect transistor, which is terminal R,
temperature change. The + should be connected with an
resistance of P decreases with alarm to alert the public when
temperature. activated.
(c) (i) 9 V (ii) The negatively charged • Since the transistor used in
electron is attracted by the
6
(ii) V XY =  6 + 10  (9) positively charged plate. the circuit is npn type, cell
2 is used and connected to
= 3.375 V terminal S.
3. (a) Thermionic emission • The most suitable set is set
(b) (i) The light from the heated X. Set X uses thermistor as a
SPM Practice 5 filament is blocked by the detector, a resistor to control
Maltese cross.
transistor base current, an
Objective Questions (ii) The electrons strike the alarm to alert the public when
fluorescent screen. fire occurs and cell 2 is give
1. A 2. C 3. A 4. C 5. D (c) Electrical potential energy = eV correct connection to the
6. D 7. A 8. D 9. D 10. A = 1.6 × 10 × 3000 transistor.
–19
11. B 12. C 13. C 14. B 15. D = 4.8 × 10 J
–16
16. D 17. B 18. C 19. C 20. D (d) The kinetic energy is converted
21. C 22. A 23. A 24. D 25. A to heat and light energy.
494
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Physics SPM Answers

(ii) 5730 × 3 = 17190 years (b) Sodium-24
Chapter
4
2
6 Nuclear Physics 5. (a) The half-life of a radioactive (c) 8 → → → 1
sample is the time taken for the 8 8 8 8
number of undeca yed nuclei in \ n = 3
Checkpoint 6.1 the sample to be reduced to half t = n T 1

Q1 (a) When sodium-24 completes a of its original number. 2 = 3(132) = 396 days
half-life, the activity becomes (b) 6000 counts per second 9. (a) Nuclear fusion is the combining
half from initial count. (c) of two lighter nuclei to form a
2000
Therefore, the activity = 2 Activity / count per second heavier nucleus while releasing
= 1000 count per minute. a large amount of energy.
Based on the graph, the half-life 6000 (b) (i) m = (2.014012 u + 3.016029 u)
of sodium-24 is 1 s. 5000 − (4.0022603 u +
(b) The undecayed activity, 4000 1.008665 u)
1
1
N =   n N o =   4 (2000) 3000 = 0.018863 u −27
= 0.018863 × 1.66 × 10
2
2
2000
= 3.13 × 10 kg
= 125 count per minute. 1000 (ii) E = mc 2 −29
0 t / minutes = (3.13 × 10 )(3.0 × 10 )
8 2
−29
Checkpoint 6.2 2 4 6 8
= 2.82 × 10 J
−12
Q1 (a) Mass defect, m 2 minutes (c) Sun
= 5.0301 − 5.0104 (d) Not suitable because the half-life 10. (a) Reactor
= 0.0197 u × 1.66 × 10 −27 of material X is too short. (b) (i) Control the rate of fission
= 3.27 × 10 kg 6. (a) Nuclear fission is the splitting of reaction by absorbing
−29
(b) Energy released, a heavy nucleus into two or more excess neutrons.
E = mc = (3.27 × 10 )(3.0 × 10 ) lighter nuclei while releasing a (ii) Reduce the speed of
−29
2
8 2
= 2.94 × 10 J neutrons so that slower
−12
large amount of energy.
(b) (i) 92 U + 0 n → 56 Ba + 36 Kr + N 0 n neutrons are more readily
236
92
1
1
141
SPM Practice 6 (ii) 236 92 U + 0 n → 56 Ba + 36 Kr + N 0 n captured by the uranium
141
1
1
92
nuclei
Objective Questions Balance the nucleon (c) – Energy released by nucleus
numbers before and after fission will heat up gas.
1. D 2. C 3. A 4. C 5. D – The gas will heat up water
6. B 7. A 8. B 9. B 10. A reaction, thus changes to a moving
11. B 12. B 13. C 14. B 15. B 236 + 1 = 141 + 92 + N(1) steam with high acceleration.
16. B 17. C 18. D 19. C 20. C N = 4 −27 – The steam will rotate turbine.
21. B 22. A 23. D 24. C 25. D (iii) m = 0.18606 × 1.66 × 10 – The connected turbine
−28
26. C 27. A 28. C 29. C 30. C = 3.09 × 10 kg
E = mc 2 and dynamo will generate
−30
8 2
Subjective Questions = (3.09 × 10 )(3.0 × 10 ) electricity.
= 2.78 × 10 J (d) – No air pollution.
−11
Section A (2.78 × 10 ) – Fuel cost cheaper.
–11
1. (a) Alpha decay = 1.6 × 10 –19 – Provide many jobs opportunity.
(b) 222 86 Rn
= 1.74 × 10 eV

(c) 4; 2 (c) Generate electricity 8 Section B
2. (a) (i) 239 11. (a) Radioactive decay series occurs
(ii) 91 7. (a) Chain reactions are ongoing because a decayed nucleus is
still in an unstable state.
(b) Nucleus is not stable. reactions due to the result of (b) 1 because Uranium-234 lost 4
(c) 239 91 Pa → 239 92 U + –1 e the reaction that could carry a nucleon number and 2 proton
0
similar reaction.
3. (a) Gamma decay is a decay which (b) 222 − 86 = 136 number to form a helium.
releases an electromagnetic 222 218 4 (c) (i) 234 92 U → 90 Th + 2 He + energy
4
230
wave. (c) 86 Ra → 84 Ps + 2 He 230 230 0
(b) w = b particle, x = γ ray (d) – There is a 3 times decrease (ii) 90 Th → 91 Pa + –1 e + energy
in the nucleon number. So, (d) (i) Mass defect,
(c) (i) Gamma ray, because it has
high ionising power. the number of a produced m = 238.029 − (232.038 +
(ii) Kills microorganisms without = 3 4.003)
damaging equipment – There is a 2 times increase = 1.988 u −27
structure. in the proton number. So, = 1.988 × 1.66 × 10
the number of b produced = 3.300 × 10 kg
−27
4. (a) – It is easily found from the = 2 2
surronding. (ii) E = mc −27 8 2
– It has a very long half-life. 8. (a) (i) Short; Water is not exposed = 3.300 × 10 × (3.0 × 10 )
= 2.970 × 10 J
−10
to radiation for a long time.
(b) (i) (ii) Beta; High penetrating 2.97 × 10 –10
100% ⎯→ 50% ⎯→ 25% ⎯→ 12.5% power but limit with pipe = 1.6 × 10 –29
19
T 1 T 2 T 3 wall. = 1.86 × 10 eV
So, the fossil has undergone (iii) Liquid; Easy to dilute with
3 times of the half-life. water.
495
ANSWER FOCUS PHYSICS.indd 495 10/03/2023 10:42 AM

Physics SPM Answers Physics SPM Answers

(e) Aspect Reason 14. (a) Activity represent decay rate.
(b) (i) Half-life of P is shorter than Q.
Short half-life Radioactive material activity is (ii) The gradient of decay curve of P is higher than Q.
rapidly weakening (iii) The rate of decay of P is faster than Q.
Type of radiation High penetrating power (c) – The shorter the half-life, the higher the gradient of
is gamma ray decay curve.
State of matter is Easy to dilute in water – The higher the gradient of decay curve, the higher
the rate of decay.
liquid (d) – After 5 days, bismuth-210 undergoes first half-life.
Type of detector High sensitivity – The activity drops to 400 counts per minutes.
is GM-tube – Bismut-210 undergoes second half-life.
Radioisotope M Because it lowers the activity – After 10 days, the activity is reduced half to become
faster, has high penetrating 200 counts per minutes.
power, easy to dilute in water (e) Aspect Reason
and high sensitivity.
Unstable element • Produced radiation that can
12. (a) Aspect Reason be detected.
Type of reaction – Able to produce higher energy. Beta particle • High ionisation power.
is fusion Long half-life • Provide enough time to detect
Fuel resource is – Can be found anywhere liquid waste flow.
Hydrogen Low penetrating • Able to control the release of
Long lifetime – Low maintenance power the ray.
Near with sea – Provide large quantity of water No radioactive • Safe to environment.
in order to cold down the steam. waste is produced
Nuclear power station Q 15. (a) Mass defect is the loss of mass after a nuclear reaction.
(b) (i) m = (2.01410 + 3.01605) − (4.00260 + 1.00867) (b) – The proton number of radium-226 is larger than
= 0.01888 u radon-222.
= 0.01888 × 1.66 × 10 −29 – The nucleon number of radium-226 is larger than
= 3.13 × 10 kg radon-222.
−29
(ii) E = mc = 3.13 × 10 × (3.0 × 10 ) – The mass of radium-226 is larger than radon-222.
−29
8 2
2
= 2.82 × 10 J – The larger the number of nucleons, the larger the
−12
2.82 × 10 –12 mass of the element.
= (c) m = 226.025 – (222.018 + 4.003) = 0.004 u
1.6 × 10 –29 (d) – A neutron is bombarded to a nucleus and it
= 1.76 ×10 eV becomes unstable.
7
(c) Disadvantage Way to overcome – Nuclear fission is occurred.
- The mass is decreased after the reaction and
The waste is hot and Bury at abandon location. changed to energy form.
radioactive.
Risk of destruction of Do geological research on (e) Aspect Reason
nuclear power stations location selection before Concrete wall Prevent radiation escapes
due to natural disaster building a nuclear power to surrounding.
system.
Use boron control rod Absorb excess neutrons.
Section C High specific heat More heat can be
13. (a) The half-life of a radioactive sample is the time taken capacity transferred at one time.
for the number of undecayed nuclei in the sample to Heat transfer from Turbine is not exposed
be reduced to half of its original number. gas to water to rotate to extremely high
(b) (i) – Initial decay activity of source A is the same the turbine temperature.
with source B.
– Half-life of source A is longer than source B. Low gas density Easy to channel
– Rate of decay for source A is lower than source
B.
(ii) The higher the half-life, the lower the rate of decay.
(c) Source A = 30 years,; Source B = 3 years,
(d) (i) Source A, because it has longer half-life. Chapter
7
(ii) Alpha Quantum Physics
(e) Aspect Reason
Short half-life Unstable element will not stay Checkpoint 7.1
long in plant. h h
Q1 Using p = mv ; p = → mv =
λ
Gamma ray High penetration power h 6.63 × 10 –34 λ
Low radiation rate No side effect to human body λ = mv = 9.1 × 10 × 5.0 × 10 6
–31
High boiling point The material maintains in liquid = 1.46 × 10 m
−10
Energy released Not damaging the tree and
is lower human body.
496



ANSWER FOCUS PHYSICS.indd 496 10/03/2023 10:42 AM

Physics SPM Answers

Q2 (a) E = hf = 6.63 × 10 × 6.2 × 10 = 4.1 × 10 J • Photon energy is proportional
14
−19
−34
hc hc 6.63 × 10 × 3.0 × 10 8 to frequency.
34
E = → λ = = • Photons transfer all of their
λ E 4.1 × 10 –19
= 4.8 × 10 m energy to the electrons they
–7
interact with.
c 3.0 × 10 8 (b) Einstein’s Photoelectric equation
Alternative: can use λ = = 1
f 6.2 × 10 14 mv max = hf – W
2
= 4.8 × 10 m 2
−7
c 3.0 × 10 8 Graph of K.E. against frequency
(b) Using f = = max
λ 400 × 10 –9
14
= 7.5 × 10 Hz 1
Energy, E = hf = 6.63 × 10 × 7.5 × 10 14 2 mv 2 max
–34
= 4.97 × 10 J
–19
nhc E λ
Q3 (a) E = → n =
λ hc
3
= 1.0 × 10 × 10.0 × 10 –6
6.63 × 10 × 3.0 × 10 8
–34
= 5.0 × 10 22
nhc 1.5 × 10 × 6.63 × 10 × 3.0 × 10 8
–34
13
(b) E = = f 0 f
λ 3.0 × 10 –12
= 0.9945 J
hc 6.63 × 10 × 3.0 × 10 8
−34
Q3 (a) E = = –9
nhf depends on the frequency or λ 420 × 10
Q4 P = t wavelength of the light. = 4.74 × 10 J
−19
Pt 20 × 1 4.74 × 10 −19
n = = Q2 The above observations cannot be = = 2.96 eV
hf 6.63 × 10 × 5 × 10 14 explained by wave theory because: 1.6 × 10 –19
–34
= 6.0 × 10 19 According to wave theory: hc
h 6.63 × 10 –34 (i) Light waves of any frequency (b) K.E. max = – W
Q5 (a) P = = λ
λ 650 × 10 –9 should emit electrons, provided = 2.96 − 2.71 = 0.25 eV
= 1.02 × 10 kg m s −1 that a certain intensity is
−27
hc maintained.
(b) E = hf = = pc (ii) The kinetic energy of hc
λ photoelectrons increases with Q4 (a) W = hf o =
= 1.02 × 10 × 3.0 × 10 8 light intensity, light with high hc λ max
−27
= 3.06 × 10 J intensity will emit electrons λ max = W
−19
–34
with high kinetic energy. The 6.63 × 10 × 3 × 10 8
Checkpoint 7.2 =
frequency of light does not 3 × 1.6 × 10 –19
Q1 (a) When light with a suitable affect the kinetic energy of the = 4.14 × 10 −7
frequency is incident on a electrons emitted. = 414 nm
metal surface, electrons are (iii) The intensity of light does not hc
emitted from the metal surface. affect the number of electrons (b) K.E. max = λ – W
This phenomenon is called emitted. If the incident light has –34 8
photoelectric effect. a low intensity, the metal surface = 6.63 × 10 × 3 × 10
350 × 10
–9
(b) Three features of photoelectric must be exposed continuously – 3 × 1.6 × 10 –19
to obtain enough energy to emit
effects that cannot be explained electrons 8.82 × 10 –20
by wave theory of light are: = –19 = 0.55 eV
– For a given metal and the Checkpoint 7.3 1.6 × 10
frequency of incident light, 1
2
the number of photoelectrons Q1 (a) Yes. The number of electrons (c) 2 mv = 0.88 × 10 –19
emitted at any given time is released depends on the v =  
2 × 0.88 × 10
–19
proportional to the intensity of number of photons incident on 9.11 × 10 –31
the incident light. it. 5 –1
– For a given metal, the (b) Not necessarily. The energy = 4.4 × 10 m s
minimum frequency of the (not the number) of the
incident light below which emitted electrons depends on 7
no photoelectron emission the frequency of the incident SPM Practice
occurs. This frequency is photons. For example, bright Objective Questions
known as the threshold blue light (high intensity) can
frequency. emit more electrons at lower 1. A 2. C 3. C 4. C 5. B
– Above the threshold energies than low light intensity 6. A 7. A 8. D 9. C 10. A
frequency, the maximum light. 11. A 12. B 13. D 14. C 15. D
kinetic energy of the emitted Q2 (a) The three properties are: 16. C 17. C 18. C 19. B 20. B
photon does not depend on • Photons are either quanta or 21. B 22. B 23. B 24. A 25. A
the intensity of the light but
discrete energy carriers.
497
ANSWER FOCUS PHYSICS.indd 497 10/03/2023 10:42 AM

Physics SPM Answers

Subjective Questions 3. (a) From de Broglie relation, 5. (a)
v e = v p = v Photoelectric current
Section A h
1. (a) Photon λ = mv
(b) The brighter light beam will eject The de Broglie wavelength is
a greater number of electrons. inversely proportional to the Saturation current
This is because the number of mass of the particle. Stopping
photoelectrons ejected depends Since the mass of electron is potential
on the number of incident smaller than the mass of proton,
photons. therefore, the electron has a
(c) Not necessarily. The energy (not longer de Broglie wavelength. V s 0 V
the number) of ejected electrons (b) v e = v p = v (b) At stopping potential, the
depends on the frequency of the Therefore, the ratio of the de photocurrent just equal to zero.
incident photons. Broglie wavelength is The electrons with the maximum
h
2. (a) The three characteristics are:  m e v  kinetic energy just fail to reach
• Photon carries quanta or λ e = the anode. This shows that the
h
discrete energy. λ p  m p v  maximum kinetic energy of the
• Photon energy is –27 photoelectrons is just equal to
proportional to frequency. = m p = 1.67 × 10 = 1833 the electric potential energy of
• Photon transfers all its m e 9.11 × 10 –31 the photoelectrons. Hence, the
maximum kinetic energy of a
energy to the electron which photoelectron is equal to eV s .
interact with it. 4. (a) Photoelectric effect (c) The number of electrons emitted
1 (b) The minimum amount of energy
(b) mv max = hf – W 0 required to emit an electron from per second from the cathode
2
2 is a constant. When the total
1 the surface of the metal. number of electrons emitted
2 mv max is the maximum kinetic (c) Photon energy, per second equal to the total
2
–34
energy of the emitted electrons, 6.63 × 10 × number of electrons reaching
where m is the mass of electron E = hf = hc = 3 × 10 8 the anode per second, the
and v max is the maximum velocity λ 330 × 10 –9 current cannot increase further
–19
of the ejected electrons. = 6.03 × 10 J even though the potential
hf is the photon energy where = 6.03 × 10 –19 difference applied is increased.
h is the Planck’s constant 1.602 × 10 –19 Hence, the current reaches a
and f is the frequency of the = 3.76 eV saturated value.
electromagnetic waves. (d) K max = hf − W O
−19
W 0 is the work function, the = 6.03 × 10 – 2.2 ×
1.602 × 10
−19
minimum energy to eject an = 2.51 × 10 J
−19
electron from the metal. (e) Assume that all the electrical
(c) energy is converted into photon
Graph of K max against frequency energy.
1 mv 2 E N(hf) N
2 max P = t = t = t (hf)
= nhf
Therefore, number of
photons per second,
f 0 Frequency / f P 60
n = hf = 6.03 × 10 –19
= 9.95 × 10 photon s .
–1
19
W 0















498





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