Focus SPM 2022 - physics

  Physics SPM  Chapter 5 Electronics

In bright condition In dark condition In bright condition In dark condition

Resistance of LDR: very Resistance of LDR: very Resistance of LDR: very Resistance of LDR: very
low. high. low. high.
Base voltage: Base voltage : high Base voltage: high Base voltage: too low
too low. Base current: high Base current: high Base current: too low to
Base current: too low to enough to switch on the enough to switch on the turn on the transistor.
turn on the transistor. transistor. transistor.
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No collector current and Collector current flows Collector current flows No collector current and
and LED lighted up. LED is off.
LED is off. and LED lighted up.

Conclusion: Conclusion:
The LED will turn on in the dark and turn off during The LED will turn on during bright day light and turn

bright day light. off in the dark.

Transistor as a Light-controlled Switch SPM Highlights

1. In the transistor circuit as a light-control Figure 5.51 shows a transistor circuit that acts as a
switch, the Light Dependent Resistor (LDR) light control switch.
is part of the potential divider circuit. With this,
the potential difference across the base changes R LED
depending on the bright or dark conditions. 1 k⍀ 6V

2. To act as a switch, the voltage across the base LDR
circuit must reach a minimum value depending
Form on the material of transistor. Thus, changing Figure 5.51
the resistance of the across the base circuit
will change the value of the base voltage Which condition can cause the bulb to light up?
based on the voltage divider and Ohm's law A Both the resistance of LDR and R are high.
(V ∝ R). The higher the resistance across the B The resistance of LDR high and the resistance of
base circuit, the higher the base voltage.
R low.
3. The Light Dependent Resistor (LDR) provides C The resistance of LDR low and resistance of R
a switching signal to the transistor to control a
load. LDR in the transistor circuit is commonly high.
D Both the resistance of LDR and R are low.
5 used in many applications such as controlling
road lighting systems automatically, controlling Examiner’s Tips
the power to LED or bulb directly, or turning Resistance of LDR must be high and resistance of
on a relay system. R must be low. With this, a high enough current can
flow through the base of transistor to turn it on.

Answer: B

420

Transistor as a Heat-controlled Switch Physics SPM  Chapter 5 Electronics  

1. In a transistor-based circuit that functions as a 6. As the temperature increases, the resistance
heat-control switch as shown in Figure 5.52, of the thermistor decreases considerably
a heat-sensitive resistor (thermistor) is used as compared to the resistance R. A large fraction
a heat or temperature detector is connected to of the 12 V voltage supply across R increases
the voltage-divider circuit. the base voltage. The transistor is switched
on and the collector current flows through the
relay coil and causes the switch to close.

7. An alarm is activated when the relay is turned
on.

8. Automatic heat-control switch is ideally used
in fire alarm systems.

9. The diode on the relay switch is to protect
the transistor from being damaged by large
induced current when a large e.m.f. is induced
in the relay coil due to the collector current
decreases to zero.

SPM Highlights

Figure 5.53 shows a circuit to turn on the Light
Emitting Diode (LED) during nighttime.
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switch
1 k⍀ 12 V
RB Alarm
10 k⍀
R

Figure 5.52 Heat-controlled switch

2. A thermistor is a type of resistor whose
resistance is dependent on temperature.

There are two types of thermistors:

PTC NTC

+t –t Resistor R1 Resistor R2 LED

• The PTC (Positive temperature coefficient) Light npn Cry cells
thermistors whose resistance increases dependent transistor
with increasing temperature.
resistor
• The NTC (Negative Temperature (LDR)
coefficient) thermistor whose resistance
decreases with increasing temperature. Figure 5.53

3. The potential difference across the base varies Which of the following actions can cause the LED
with the change in temperature of a fire.
To prevent a fire accident, the heat detector to light up at bright daytime?
generates a warning signal through the alarm.
A Replace the npn transistor with pnp transistor.
4. In the heat-control switch circuit as shown in RR12
Figure 5.52, the thermistor used is NTC types B Interchange the position of LDR with
whose resistance decreases with increasing C Interchange the position of LDR with
temperature.
D To increase the number of dry cells.
5. At room temperature, the thermistor has a high
resistance compared to resistor R. Therefore, Examiner’s Tips Form
the potential difference across the base of the
transistor is too low and not enough to turn on A transistor can only be turned on when there is
the transistor.
a large enough base current to pass through the

transistor. During nighttime, the resistance of LDR 5
is very high compare to R1. Therefore, the voltage
across the base circuit is large enough to switch on

the transistor. By interchange the position of LDR with
R1,
low during bright daytime, resistance of LDR is very
compare to R1. Therefore, the voltage across R1
is large to turn on the transistor circuit.

Answer: B

421

  Physics SPM  Chapter 5 Electronics

SPM Highlights (a) Label the terminal P, Q and R.
(b) Name the type of transistor shown on Figure
Figure 5.54 shows an NTC thermistor connected in
a transistor circuit. The circuit is used to control a 5.55.
240 V air conditioner.
S2 Figure 5.56 shows a simple transistor circuit.

24~0 V a.u M
T
Thermistor Relay
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R VC
VS
Resistor Air 9V
R condition

Resistor

Figure 5.56

Figure 5.54 (a) Name the type of transistor T.

Which of these situations can cause the air condition (b) On the figure above, label the base circuit and
to turn on?
A Low temperature thermistor and low base voltage. emitter circuit. V V
B Low temperature thermistor and high base s c
(c) Draw symbol dry cell at and so that the
voltage. bulb M can lights up.
C High temperature thermistor and high base (d) What is the function of resistor R?

voltage. S3 Azmin designed a transistor circuit to turn on a fan
D High temperature thermistor and low base in the classroom automatically. Figure 5.57 shows
the circuit for the electronic control system.
voltage.
X
Examiner’s Tips
a.c / a.u
When the temperature of the thermistor is high (hot P Relay
condition), the resistance of thermistor becomes
low compared to resistor R. Then the base voltage Y R1 Fan 9V
becomes high causing the current to flow through the 10 k⍀ Transistor
base of transistor to turn it on. The collector current
flows through the relay to turn on the air condition.

Answer: C

Checkpoint! 5 .2 Z

S1 Figure 5.55 shows the symbol of a transistor. Figure 5.57

P : ________________ (a) Name the component P.
(b) What is the function of P? Explain your
Form
R : ________________ answer.
(c) When the classroom becomes warm, the
5 Q : ________________
resistance of P decreases to 6 kΩ.
Figure 5.55 (i) State the potential difference across XZ.
(ii) Calculate the potential difference across

the themistor.

422

PenCONCEPT MAPerbita ELECTRONIC

n PelElectron
angiThermionic emission Semiconductor diode Transistor
Sproduces types

dnCathode rays The function of npn transistor pnp transistor
Bdeflected by semiconductor diode

effect

hdElectric field
Forward Reverse
biased biased
. Ato Features of
lldetermine Magnetic field transistor circuit

RSpeed of Used in
igelectron in
hts Resercathode ray tubeCharacteristics of Rectification of Base circuit Emitter circuit
cathode ray alternating current
Transistor Uses
– negatively charge as current
amplifier Transistor as
automatic
switch
– can be deflected Physics SPM  Chapter 5 Electronics  

5by electric field and One diode 4 diode

magnetic field423 Half Full
– produces fluorescence wave wave

effect

– can stopped by thin veCapacitor as current
d.smoother
sheet of metal

Form

  Physics SPM  Chapter 5 Electronics 5

SPM Practice

Objective Questions

1. Which of the followingPenerbitan Pelangi Sdn Bhd. All Rights Reserved.ElectronP 6. How much energy acquired
correctly explains thermionic beam Q by an electron in a vacuum
emission? tube if the electron is
A The process by which Figure 2 accelerated at a potential
electrons leave the hot difference of 2.0 kV. It is
filament. Which of the following given that the charge of one
B The process by which statements is not true? electron is 1.6 × 10-19 C.
ions leave the hot A The potential difference at A 1.6 × 10–19 J
filament. plate P is negative. B 1.6 × 10–16 J
C The process by which B The deflection would be C 3.2 × 10–19 J
molecules leave the hot greater if the potential D 3.2 × 10–16 J
filament. difference is greater.
D The process by which an C The deflection would be 7. Which of the following
electron leaves an atom. greater if the electrons is not true of n-type
are moving faster. semiconductors?
2. Which of the following is a D The electron beam will A It is formed by doping
cathode ray? return to straight line if trivalent atoms into pure
A Stream of fluorescent a suitable magnetic field silicon.
particles is applied between the B The majority charge
B Light rays from a screen plates. carriers are electrons.
C Beam of fast moving C The minority charge
particles 5. Which of the following figures carriers are holes.
D Light rays from hot shows the correct path of D It conducts electricity
filament the electron beam moving when connected to a
through two charged plates? circuit.
3. Figure 1 shows an image A
of a Maltese cross formed 8. Figure 3 shows a rectification
on the screen by a beam of ++++ circuit.
electrons.
P
U
T Electron
beam
––––

PR Input Q Output

B

Form Q Electron ++++ Figure 3
S beam ––––
Which of the following
Figure 1 C ++++ statements is true?
5 If two bar magnets are –––– A A rectifier changes d.c. to
placed near the tube, in Electron a.c.
beam ++++ B Device P allows current
which direction will the image –––– to flow in any direction.
D C Device Q acts as a
be shifted? C R rectifier.
A P D T Electron D The rectifier circuit would
B Q beam still work if device P is
reversed.
4. Figure 2 shows a beam
of electron beam being

deflected due to a potential
difference between plates P
and Q.

424

9. Figure 4 shows identical B Physics SPM  Chapter 5 Electronics  
bulbs labelled 12 V 6 W C C
connected in a circuit. Which D
bulb A, B, C or D is the D
brightest?
13. A C.R.O. is connectd to a
A circuit as shown in Figure 7.

B C.R.O
100 Ω
30 ΩPenerbitan Pelangi Sdn Bhd. All Rights Reserved.D

C

+–
12 v a.t.
Figure 4

10. Figure 5 shows the traces
displayed on a C.R.O.

100 Ω 200 Ω

PQ 12. A C.R.O. with the correct Figure 7
settings is connected across Which of the following
RS R as shown in Figure 6.
Figure 5 waveforms will be displayed
A.c. Input on the screen?
Which traces show the A
R
rectification of an a.c.? C.R.O B
A P and Q only
B P and R only Figure 6 C
C Q and S only
D P, Q, R and S Which of the following Form
waveforms will be displayed
11. Which of the following on the screen?
circuits will give a full-wave A
rectification of an a.c. supply?
A

D 5

B

425

  Physics SPM  Chapter 5 Electronics B Which component is used
to smoothen the rectification
14. Figure 8 shows a rectification –+ output?
–+ A P
circuit for an alternating –+ B Q
current. C R
D S
C.R.O
18. Which of the following is
Figure 8 the three terminals of a
Which waveform will be transistor?
A Base, Earth and Collector
displaced if the output B Emitter, Earth and
resistor is connected to the Collector
C.R.O.? C Base, Emitter and
A Collector
D Emitter, Earth and Base

19. Figure 11 shows the symbol
of a transistor.
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D


B

16. Figure 9 shows four diodes P
C connected to a dry cell and a Figure 11
resistor R.
D Name the type of the
I II
R transistor and terminal P.
15. Which of the following shows
IV III Type P
the correct connection of the Emitter
diodes in a bridge rectifier? Figure 9 A npn Collector
A Which diodes are in forward B pnp Emitter
C pnp Base
5 –+ bias? D npn
A I and II
Form B I and III 20. Figure 12 shows a transistor
C II and III circuit.
D II and IV
17. Figure 10 shows a full-wave X
rectification circuit.

SP Y
Figure 12
Q R C.R.O.

Figure 10

426

What are components X and B Physics SPM  Chapter 5 Electronics   
Y? C
R R
XY
C LDR
A Bulb Transistor
R D
B Light Bulb
D R
dependent

resistor
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C Variable Transistor
resistor

D Variable Light
resistor dependent
resistor
LDR

21. Figure 13 shows a transistor 24. Figure 14 shows a transistor
circuit. The milliammeter circuit. In dark conditions,
gives a reading when the the light dependent resistor
variable resistors, X and Y (LDR) is 1 MΩ. To turn on the
are adjusted to the same bulb, the minimum potential
value. difference across the LDR is
2 V.
mA
X
R R Bulb
1 kΩ T
23. An LDR has a resistance of 12 V
500 Ω when it is exposed to
Y light. Its resistance increases LDR
to 100 kΩ when dark. The
Figure 13 LDR is then connected to a Figure 14
transistor switching circuit.
Which of the following can be Which of the following circuits What is the maximum value
used to add a reading to the will switch on the bulb when of the resistor R to turn on
milliammeter? dark and switch off the bulb the bulb?
A Increase the value of R when bright? A 1 MΩ C 4 MΩ
B Increase the value of X A B 2 MΩ D 5 MΩ
C Increase the value of Y
D Increase the value of R 25. Figure 15 shows a transistor
voltage supply circuit acted as an amplifier.
LDR
22. Which of the following IC
transistor circuit can function? B Form
A Form
LDR
R IB RC 5
R
IE
VB VC

Figure 15
Which of the following is

true?
A IE > IC > IB C IC > IE > IB
B IE > IB > IC D IC > IB > IE

427

  Physics SPM  Chapter 5 Electronics P
Screen of C.R.O.
Soalan Subjektif
Output
Section A
1. Figure 1 shows a rectification circuit.

Input
signal
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Figure 1

(a) Name the type of input signal. [1 mark]
(b) Name the component P. [1 mark]
[1 mark]
(c) On the screen of C.R.O., draw the output signal. [1 mark]
(d) If component P is reverse biased, sketch the output signal.

2. (a) Figure 2.1 shows a simple cathode ray tube. Anode

Vacuum tube

Cathode

Heatng filament

– E.H.T + [1 mark]
[1 mark]
Figure 2.1 [1 mark]
(i) State the function of heating filament. [1 mark]
(ii) What is the function of cathode?
(iii) Name the process of emitting particles by the cathode in (a)(ii).
(iv) State a reason why extra high tension is used.

(b) Cathode ray tube is supplied with E.H.T. of 3 kV. Calculate the speed of electron moved from
cathode to anode. Given that the mass of electron, e = 9.1×10−31 kg and charge of one electron,
me = 1.6 × 10−19 C. [2 marks]

(c) Figure 2.2 shows an electron beam moving towards two charged plates with strong
Form
Form and uniform electric field. –

(i) In the figure, complete the path of cathode ray through the two plates.
[2 marks]
Give one reason for your answer.

5 3. Figure 3.1 shows a Maltese Cross tube which is used to study the characteristic of +
cathode rays.
Figure 2.2
Maltese
Filament Anode cross

6 V a.u Shadow

Cathode Screen

E.H.T
Figure 3.1

428

Physics SPM  Chapter 5 Electronics   

(a) Name the process of emitting electrons from the heated cathode. [1 mark]
(b) (i) Why the shadows is formed on the screen? [1 mark]
(ii) Why the green light region is formed on the screen? [1 mark]

(c) The extra high tension supplies 3000 V across cathode and anode. How much electrical potential energy
does an electron receive? [Charge of one electron, e = 1.6 × 10-19 C.] [2 marks]

(d) The electrical potential energy received by one electron is converted into the kinetic energy of the electron.
[1 mark]
What will happen to this kinetic energy when the electron hits the fluorescent screen?

(e) Two strong magnets are placed at the sides of the tube as shown in Figure 3.2 causing the shadow to
deflect.
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Filament Anode U
S

6 V a.c. Shadow
Screen
U
Cathode S

E.H.T [1 mark]
[1 mark]

Figure 3.2

(i) In the figure, draw an arrow to show the direction of the deflected shadow.
(ii) State the physics rule used to determine the direction of the shadow.

Section B

4. (a) Figure 4.1 shows a transistor circuit. Milliammeter S2

Microammeter R mA
1
μA R
1

T

Rheostat S
1

Figure 4.1

(i) Name the type of transistor T. [1 mark]

(ii) The transistor circuit in Figure 4.1 is used to amplify current.

State what happens to the readings on the microammeter and milliammeter when [1 mark]
SS11 SS22 [1 mark]
• is switched on and is switched off. Form
• is switched off and is switched on. Form

(iii) Explain how the circuit functions as a current amplifier when the rheostat is adjusted and both switches
[2 marks]
are switched on.

(iv) Based on Figure 4.1, the reading of microammeter is 10 μA and the reading of milliammeter is 1 mA. 5

Calculate [2 marks]
• the current amplification by using the formula [3 marks]
• emitter current, Ie.

(b) Figure 4.2 shows a circuit for an incomplete fire alarm system.

P Relay R
Q S


Figure 4.2

429

  Physics SPM  Chapter 5 Electronics

The table shows the four sets of electronic components used to complete the circuit in the figure.

Set of electronic Terminal P Terminal Q Terminal R Terminal S
components

W

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.LDRCapacitorAlarm Cell 1

X

Thermistor Resistor Alarm Cell 2

Y

Thermistor Capacitor LED Cell 2

Z

LDR Resistor LED Cell 1

You are required to determine the most suitable set of electronic components that can cause the fire

alarm system to function. Study the types of electronic components in the table. Explain the suitability

of each electronic component and determine the most appropriate set of electronic components. Give
[10 marks]
reasons for your choice.

Form
Form

5

430

JAWAPAN

FORM 4 Checkpoint 1.2 Subjective Questions

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

(ii) A hypothesis is a statement (ii) Base quantity
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. of an expected outcome
Checkpoint 1.1 that usually states the (b) (i) 0.25 × 0.12 × 0.5 = 0.015 m3
relationship between two
Q1 Base or or more variables intended (ii) 1.5 = 100 kg m–3
Description derived to be given a direct 0.015
Physical quantity experimental test.
quantity 2. (a) m / kg T2 / s2
(iii) A variable is a physical
Base quantity that can be varied 2.0 0.8464
quantity in an experiment.
Total 55.2 g Mass 3.0 1.2996
(b) (i) The extension of the spring
150 cm3 of Volume Derived depends on the weight of 4.0 1.7689
hot water quantity the object attached to it.
5.0 2.1609
(ii) The bigger is the magnitude
80°C Temperature Base of the weight, the greater is 6.0 2.6569
quantity the extension of the spring.

Within 3 Base (iii) • Manipulated variable: (b)
minutes quantity Weight of object attached
Time to the spring. T 2/s2

9 kJ of Energy Derived • Responding variable: 3.0
energy quantity Extension of the spring.

Q2 (a) (i) 7 500 000 m • Fixed variable: Type of 2.5
= 7 500 000 × 10–3 km spring.
= 7 500 km 2.0

(ii) 7 500 000 m Q2 (a) To investigate the relationship 1.5
= 7 500 000 × 10–6 Mm between the distance of
= 7.5 Mm extension of the elastic cord, x 1.0
and the horizontal distance, d
(b) 7 853 m s–1 = 7 853 m travelled by the ball.
1s
(b) The further the distance of 0.5
extension of the elastic cord,
= 7 853 × 10–3 km x, the further is the horizontal 0 m / kg
 1 h distance, d travelled by the ball. 0 1.0 2.0 3.0 4.0 5.0 6.0
60 × 60
(c) • Manipulated variable: The
= 28 270.8 km h–1 distance of extension of the (c) m = 2.65 – 1.30 = 0.45 s2 kg–1
elastic cord, x 6.0 – 3.0

Q3 (a) Base quantity is quantity with • Responding variable: The 3. (a) (i) Physical quantity is quantity
magnitude only. horizontal distance travelled that can be measured.
by the ball, d
(b) Vector quantity is quantity with (ii) V is a responding variable
magnitude and direction. • Fixed variables: Type of because its value depends
elastic cord used, type of ball on the other variables and
Q4 Event Type of Explanation used and height of the table can only be determined
quantity from the experiment. l is
(d) • Tabulation of data: a manipulated variable
1 Vector Magnitude because its value can be
700 km h–1 and x / cm fixed before the experiment
direction required and other variables depend
to reach its d / cm on it and only can be
destination. determined from experiment.
• Analysis of data:
2 Scalar Only magnitude, Plot a graph of d against x. (b) (i) Graph of student A
3 kg required. • Weakness: Horizontal
If the gradient of the graph
3 Vector Magnitude of is positive, the hypothesis is axis, 6 divisions for 0.2
force is 25 N accepted. A is a difficult scale to
and the direction determine the exact
required to reach SPM Practice 1 position of value l.
the lift. Objective Questions • Weakness: Vertical axis,
5 divisions for 0.4 V is a
4 Scalar Only magnitude 1. C 2. C 3. D 4. D 5. D difficult scale to determine
is involved, the 6. A 7. B 8. C 9. B 10. B the exact position of value
temperature from 11. D 12. B 13. C 14. D V.
20oC to 100oC.

487

  Physics SPM  Answers Q3 50 ticks per second. (b) Velocity = Gradient of graph
Therefore, 1 tick = 0.02 s
• Strength: The graph = 2.0 – 0.5
drawn is big. The time taken for each strip of 3.8 – 2.4

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

s = u + v2 × t = Area under the graph from
2 20 s to 60 s
= 580 m
t = u2+s v
Q3 (a) (i) Total distance travelled
= 2 × 6 = 35 × 2
4 + 8 = 70 m
=1s
(ii) Average speed
(b) a = v – u = 5750
t = 1.27 m s–1 (or 1.3 m s–1)
(b) t = 10 s to 20 s
2Chapter = 4 – 8 (c)
1 v / m s–1

Force and Motion I = –4 m s–2 2

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

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

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

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

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

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

was at rest.

488

Physics SPM  Answers  

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

u = 0 m s–1 momentum = 600(0) – 600(50)
(d) Total momentum after collision 0.2
s = ut + 1 at 2 = 100 × 6 + 200 × v
2 = 1200 kg m s–1 = –1.5 × 105 N
200v = 600
100 = 0+ 1 (10)t 2 v = 3 m s–1 Magnitude of force = 1.5 × 105 N
2
t = 20 (c) The time in (a) is 10 times
greater than in (b). The old tyres
= 2 5 s or 4.47 s Q2 (a) Total momentum before collision lengthen the stopping time of
= (1500 × 15) + [1000 × (–20)] the car and hence, reduce the
= 2500 kg m s–1 impulsive force acted on the car.

(d) afor=amFfix, eifdmvaislusemoafllF, .aWisithbigthgaetr,
the racing car can move faster.
Q2 (a) An object undergoes free fall ifPenerbitan Pelangi Sdn Bhd. All Rights Reserved.(b) Total momentum after collision
it is acted upon by gravitational = Total momentum before collision
force only. = 2500 kg m s–1

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

Q2 (a) A fully loaded lorry has a big Q1 (a) Mass of an object is the quantity
mass. Hence, its inertia is big.
of matter in the object.
(b) The mass of a train is big.
Hence, its inertia is big. Weight of an object is the

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

Checkpoint 2.6 (b) Weight Mass

Q1 m = 5.0 kg Vector quantity Scalar quantity

Q3 (a) The pencil box will continue to F = 12.0 N Varies with the Is the same
move and leave the toy car. F
a = m = 12.0 = 2.4 m s–2 value of g anywhere
(b) The inertia of the pencil box 5.0
causes it to remain in its original Measures Measured using
state of motion although the car Checkpoint 2.7 using spring inertia balance
has been stopped. balance
Q 1 (a) u = 50 m s–1, v = 0 m s–1,
(c) The pencil box will move with a (c) mg = 47 × 10
higher velocity and land further m = 600 kg = 470 N
from the obstacle. t =
F = 0m.0v2–, mu
0.02 SPM Practice 2
Objective Questions
Checkpoint 2.5 = 600(0) – 600(50)
0.02
Q1 (a) Total momentum before collision
= 100 × 2 + 200 × 5 = –1.5 × 106 N 1. B 2. C 3. A 4. B 5. C
= 1200 kg m s–1 Magnitude of force = 1.5 × 106 N 6. A 7. C 8. B 9. C 10. B

489

  Physics SPM  Answers

11. C 12. C 13. D 14. D 15. B (iii) Inertia [1] • The gradient of the graph
16. C 17. B 18. D 19. D 20. B
21. B 22. D 23. D 24. C (b) • Hold the bottle upside down. gives the acceleration of the

Subjective Questions [1] car. [1]

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

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

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

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

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

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

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

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

steering wheel and the

windscreen. [1]  CT iamreQ =: 0.45 + (—1120—.08—×–—22..—06)

(iii) • Windscreen glass should

be designed to fracture

into rounded pieces = 4.88 s
 Car R:
upon impact instead of —(170—.00—–× —21..4—2)
T ime = 0.20
shattering. [1] +

• So that the driver is less = 5.31 s

likely to be cut by the  Car S:

shattered glass. [1] T ime = 0.55 + (—1150—.04—×–—15..—86)

(iv) • The headlight of the car

should be bright and = 4.84 s [1]
Therefore, car S will win the
powerful enough to shine
competition. [1]
over a long distance. [1]

• This is to enable the 10

driver to see far ahead

and have enough time to (d) (i)

avoid obstacles. [1] F2 F1
F3
Max 10
Section C
30°

Section B 5. (a) A force can cause:

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

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

move it. [1] (b) v / m s–1

• The bigger the mass of an

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

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

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

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

5 0

490

Physics SPM  Answers  

(ii) Acceleration of the car, a Checkpoint 3.3 3. (a) (i) F = mg
= —mF (ii) F = GMRm2
Q1 Man-made satellites are artificial

= —02..—02 [1] satellite build and made by man. It is (iii) mg = GMrm2 → g = G M
R2
launched by a rocket into space and

= 0.1 m s–2 [1] is placed in a specific orbit round the (b) (i) gh = GM
(R + h)2
5 earth for specific purposes.

The ISS is the largest man-made GM

3Chapter satellite orbiting the Earth. g R2
gh GM
Natural satellite are any object in (ii) = (R + h)2

space orbiting the larger planets.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.Gravitation
The moon is a natural satellite
  R 2 gh
Checkpoint 3.1 orbiting the Earth. g = R+ h

Q1 F= G Mm  Q2 v = GM  (iii) gh = R+ h 2g
R2 r+h R
 =
7×7 6.67 × 10–11 × 5.97 × 1024   = 2
= 6.67 × 10–11 × 22 6.37 × 106 + 500 000 6 400
6 400 + 12 800
= 8.17 × 10–10 N
= 7 613 m s–1 = 7.61 km s–1 × 10 N kg–1

Q2 (a) F ∝ 1  Q3 v = 2GM = 1.1 N kg–1
r2 R W = mg = 81 × 1.1 = 89.1 N

(b) F  vY : 1 791 = 2GM Section C
R
 vX : v = 4. (a) (i) A geostationary satellite is a
v 2G × 1.41 M satellite in the geostationary
0.919R orbit round the Earth on an
 1 791 = equatorial plane. It moves
 2G × 1.41M × R in orbit round the Earth,
0.919R 2GM from east to west, and has
a period of 24-hour or one-
r v = 1.41 × 1 791 day.
0.919
F (ii) 3
Q3 F = mg → g = m = 2 218 m s–1 (iii)
= =
1 560 26 m s–2 SPM Practice 3
60 Objective Questions
Q4 F because the force of gravity is the

same.

Q5 Fc = T = Fw 1. A 2. B 3. D 4. C 5. C
m × ac = M × g 6. C 7. B 8. D 9. B 10. C
0.05 × ac = 0.6 × 10
ac = 120 m s–2 11. B 12. D 13. B 14. D 15. B
v2 16. C 17. C 18. D 19. B 20. C
r 21. D 22. B 23. D 24. C 25. A
ac = (b) (i) Mercury:
Subjective Questions
v2 = 120 × 0.6 R3 = 1.85 × 1023 km3
TRT223
v = 8.5 m s–1 Section A = 7 744 day2
  = 2.52 × 1019
1. (a) • The force is proportional to the
Checkpoint 3.2 product of the masses of the km3 day–2
two body.
Q1 T2 = 12 Mars:
r3 × 108)3 • The force is inversely R3 = 1.19 × 1023 km3
(1.5 proportional to the distance TRT223
between the two bodies. = 471 969 day2
= 2.96 × 10–25 year2 km–3   = 2.52 × 1019 km3
(b) (i) Same
T T2Earth 2Mercury (ii) Twice as big day–2
r = r3Earth 3Mercury
Q2

(1.50 12 1011)3 = T 2Mercury 2. (a) (i) T = W = mg (ii) The value is about the
× (5.79 × 1010)3 same.

T 2Mercury = 0.575 = 0.2 × 10 = 2.0 N (iii) Kepler III law states that the
T Mercury = 0.24 year (ii) Fc = T = 2.0 N square of the orbital period
mv2 of a planet is proportional
r3 GM T (b) Fc = r   to the cube of the orbital
4p2 radius.
Q3 = 2 2v.=0 =210.01=.×04v.42 7 m s–1
(iv) Can
(6.67 × 10–11 × (c) • The force of gravity due to

= 5.98 × 1024) (24 × 60 × 60)2 Earth’s gravitational field
4p2 (c) The orbital radius becomes acted on the apple. The
weight of the apple causing
(R + h)3 = 7.542 × 1022 smaller. it to fall from its tree.
R + h = 4.23 × 107 • The gravittational force
h = 4.23 × 107 – 6.37 × 106 (d) The centripetal force acting on is a force that acts on
any pair of objects in the
= 3.59 × 10­7 m the stopper is perpendicular universe. This gravitation

to the direction of motion of

the stopper. The stopper is not

displaced in the direction of the

force, thus no work is done.

491

  Physics SPM  Answers Q2 (a) Material T: The largest mass fast, random and continuous motion.
The gas molecules are always in
force between the Earth has the highest heat capacity. collision with each other and with
and the Moon produced the Q the wall of the container. When a
centripetal force that keep (b) Specific heat capacity, c = mDq , gas molecule collides with the wall
the Moon moving in circular and bounces back, the molecule
orbit round the Earth. means smaller mDq, the larger experiences a change in momentum.
• Without centripetal force, The rate of change of momentum
the Moon will continue specific heat capacity. Material from the rate of collisions with the
to move in a straight line P has the largest specific heat container wall produces a force. The
tangen to its orbit. Due to capacity due to the smallest mq force per unit area hit on the wall
attrative force of gravity produces gas pressure.
from the Earth, the Moon value.
always falling towards the
Earth. Q3 Q = CDq
• Since the linear speed
of the Moon is always
constant, then it can only
fall back into its own orbit,
Thus, the moon cannot
reach the Earth.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. = 1 400 × (25 – 5)

= 28 000 J
Q
Q4 c = mDq Q2 According to Charles’ Law, the
volume of gas increases with
= 54 000 J temperature. This is because when
0.6 kg × 200°C a fixed mass of gas is heated, the
gas molecules acquired energy and
= 450 J kg–1 °C­–1 increase their kinetic energy. Gas
molecules move faster and collide
Q5 t = mcDq with the wall of the container more
P frequent and this increases the
pressure of the gas. In order to keep
= 2 × 4 200 × 50 = 140 s the pressure constant, the excess
3 000 pressure of the gas will produce a
4Chapter force pushing the piston upward to
Heat Q6 (a) Q1 = mcDq increase the volume of the gas.
= 0.6 × 4 200 × (100 – 27)
= 183 960 J

Checkpoint 4.1 (b) Q2 = mcDq
= 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
contact until both are at the
same final temperature and Q1 The meaning is that to melt or Q4 This is because at 0°C, the gas
there is no net heat transfer freeze 1 kg of ice, the amount heat molecules are still in a state of
between the two objects. absorbed or released is 3.34 × 105J. motion and therefore the pressure
still exist.
(b) Hot coffee is cooled when ice Q2 (a) Q – mL = 0.2 × 3.34 × 105
cubes are added to it. Q5 P1 = 1.5 × 105 Pa; V1 = 50 cm3 ;
= 66 800 J V2 = 30 cm3 ; P2 = ?
Q2 • Net heat flow between P and Q (b) Q – mL = 0.4 × 2.26 × 106
is zero. P1V1
= 904 000 J P1V1 = P2V2 → P2 = V2  
• Temperature of P and Q is lower
than 70°C but higher than 20°C. Q3 (a) Melting point = 75°C = 1.5 × 105 × 50 
(b) t = 6 minute = 360 s 30
Q3 (a) Liquid suitable for X is mercury. m = 100 g = 0.1 kg
(b) Melting for ice is 0oC; boiling Q = Pt = 100 × 360 = 2.5 × 105 Pa
point is 100oC.
Q = 36 000 J Q6 P1 = 100 kPa; T1 = 270 K ;
m
Q4 100°C → (16 + 9) = 25 cm L=   = 36 000   T2 = 324 K ; P2 = ?
0.1
PT11   = P2 → P2 = 100
Therefore, 1265  × 100°C = 64°C = 3.6 × 105 J kg–1 T2 270   × 324

Q5 (a) 24°C Q4 (a) Q = mcDq + mL = 120 kPa
= 1.5 × 4 200 × 30 + 1.5 ×
(b) 17.2 cm Q7 V1 = 80 cm3 ; T1 = 273 K ;
3.34 × 105
(c) 2 cm → 0°C = 189 000 + 501 000 T2 = 338 K ; V2 = ?

and 16 cm → 68°C = 690 000 J VT11   = V2 → V2 = 80   × 338
(b) Q = mL + mcDq V2 273
Therefore, 14 cm : 68°C
= 0.2 × 2.26 × 106 + 0.2 × 4 200
For increase of temperature of × (100 – 60) = 99.0 cm3
= 452 000 + 33 600
20°C,

l = 20   × 14 = 4.12 cm = 485 600 J SPM Practice
68 Objective Questions
4
Q5 Q = mcDq(ice) + mL(ice) +
Checkpoint 4.2 mcDq(ice) + mL(steam)

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
the object, the larger the heat 105 + 3.2 × 4 200 × 100 + 3.2 6. C 7. A 8. D 9. A 10. D
capacity. × 2.26 × 106
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. 21. D 22. B 23. C 24. D 25. A
Checkpoint 4.4
(c) Types of material – different
types of materials have different Q1 Based on the kinetic energy of the
heat capacity. gas, the gas molecules are all in

492

Physics SPM  Answers  

Subjective Questions • Suggestion: The cup cover (b) Q has the same natural
should also be made of good frequency as the vibrating tuning
Section A heat insulator such as plastic. fork. Hence, Q will receive the
biggest amount of energy from
1. (a) (i) Thermal equilibrium Explanation: Good insulator the vibrating tuning fork.
can prevent heat loss to the
(ii) The change of oil surrounding. (c) Resonance

temperature is greater than • Suggestion: The cup material Q2 The hand moves with the same
should also have a high natural frequency as that of the
water. melting point. loaded spring. Resonance occurs
and the spring will receive maximum
(iii) The amount of heat energy Explanation: So that the cup energy.
does not easily change its
absorbed by oil and water is state or melt when filled with
liquid of high temperature.
the same.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. • Suggestion: The inside and
(iv) The specific heat capacity of outside of the cup are coated
with shiny metal.
oil is smaller than water. Checkpoint 5.3
(v) PT = mcDq Explanation: Shiny materials
500 × (2 × 60) = 0.5 × c × can prevent heat loss through Q1 Let the distance of the wall from the
radiation. student be d m
(97 – 25)
• Suggestion: The covered cup The distance travelled by sound to
c = 500 × 120 can be stored in a pocket and fro the wall = 2d m
0.5 × 72 made of bamboo cloth. 2d = v × t
= 340 × 2.0
= 1 667 J kg–1 °C–1 Explanation: So that there is = 680 m
a layer of air trapped inside it Therefore,
(b) (i) degree of hotness can reduce heat loss. h = 340 m

(ii) Volume of gas decreases. Q2 (a) v = 1500 m s–1, t = 0.12 s
Distance travelled by the pulse
(iii) Absolute zero temperature = v × t = 1500 × 0.12 = 180 m

(0 K or – 273°C)

(iv) The freezing gas molecules

do not move.

Section 2

2. (a) Heat is a form of energy that (b) Distance of the shoal of fish
can flow from area of high 180
temperatures to area of low below the boat = 2   = 90 m
temperatures.
5Chapter (c) Ultrasonic waves can transfer more
(b) (i) • The mass of water in
kettle P is smaller than Waves energy than audible sound
the mass of water in
kettle Q. waves.

• The heat energy supplied Checkpoint 5.1 Checkpoint 5.4
to the kettle P and the Q1 (a)
kettle Q are the same. Q1 (a) A wave is a travelling
disturbance from a vibrating or
• The temperature of water oscillating source.
rise in kettle P is higher
than the temperature of (b) Transverse waves
water rise in kettle Q.
(c)
(ii) The smaller the mass
of water, the higher the Deep Shallow
temperature rise.
Q2 (a) (i) A wave in which the (b) Refraction of waves
(iii) The heat supplied must be particles of the medium
constant. move in the direction Q2 (a) f = 16 Hz; λd = 3 ; vd = 6 cm s–1
perpendicular to the λs 2
(c) The initial temperature of the direction in which the wave
orange juice is higher than the propagates. λd = λλds
temperature of the ice cubes. vs
Heat from the orange juice flows (ii) Water waves λs
to the ice cubes to melt it. The (b) (i) A wave in which the Therefore, vs = λd vd
heat is continuously absorbed
by the cold melting water from particles of the medium = 2 ×6
the ice. The temperature of the move in the direction 3
orange juice decreased and parallel to the direction in
the temperature of the cold which the wave propagates. = 4 cm s–1
water from the ice increased. (ii) Sound waves
Therefore, orange juice is (b) λs = λs = 4 = 0.25 cm
cooled by adding ice cubes. Q3 (a) Period of oscillation, T = 0.5 s f 16

(d) • Suggestion: The cup should (b) Frequency, f = 1   = 1   Q3 (a), (b), (c)
be made of good heat T 0.5
insulator such as polystyrene Shallow Direction of
material. = 2.0 Hz area propagation

Explanation: Good heat (c) Wavelength, λ = 5 cm
insulation prevents heat from (d) Using v = f λ ,
hot drink being escape by v = 2.0 x 5 = 10 cm s–1
conduction.
Normal

Checkpoint 5.2 Deep
Q1 (a) Q area

493

  Physics SPM  Answers Q4 (a) The sound waves spread out Checkpoint 5.7
beyond the edge of the window
Q4 after passing through the Q1 1. They all transfer energy from
window.
Hotter air one place to another.
Speed of sound higher (b) Diffraction of sound waves
2. They are all transverse waves.

Refraction of sound waves occur 3. They can all travel through a
Listener vacuum.
Source Checkpoint 5.6
of s 4. They all travel with a speed of
Q1 (a) The principle of superposition 3 × 108 m s–1 in vacuum.
sound states that at any time, the
combined wave forms of two or Q2 (a) P : Microwaves; Q : Visible light;
Colder air more interfering waves is given R : X-rays
Speed of sound lower by the sum of the displacements
of the individual waves at each (b) P has higher frequency than
point of the medium.
The diagram shows the temperaturePenerbitan Pelangi Sdn Bhd. All Rights Reserved. radio waves.
of air at night. The layer of air
above the surface of the Earth is (c) Photography / Photosynthesis
colder compared the layer further
away from the surface of the Earth. (b) (i) by plants / Enables human
Since the speed of sound waves beings and animals to see. (Any
is higher at higher temperature of a 2a one answer)
air, refraction occurs and the sound (d) R has very high penetrating
waves are refracted towards the
surface of the Earth. This results in power and can cause cancer
the listener be able to hear clearer
sound at night. (ii) and genetic defects to living
a
cells.

Q3 v = s  , therefore,
t
s
a t = v  

Checkpoint 5.5 = 320 000 × 1000  
3 × 108
Q1 (a)
Q2 (a) • A: An up-and-down movement = 1.07 s

with a large amplitude Q4 P : Microwaves; Q : X-rays; R :
• B: No motion/stationary Visible light; S : Ultraviolet rays
(b) a = 3.0 cm; D = 30.0 cm;
x = 15.0 cm
ax SPM Practice 5
λ = D Objective Questions
(b)
(c) = 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 × 10–9 m 21. B 22. D 23. A 24. D 25. C
a = 0.4 mm = 0.4 × 10–3 m 26. D 27. D
D = 3.0 m
Using λ = ax ,
D
Subjective Questions
x = λaD = 540 × 10–9 × 3.0
0.4 × 10–3 Section A
1. (a)
= 4.05 × 10–3 m
Direction of
Q2 (a) Remains unchanged = 4.1 mm movement of hand
(b) Remains unchanged
(c) Remains unchanged (b) • x = λaD

Q3 (a) • Hence, if λ is bigger, x will

also be bigger. Therefore,

the distance between two (Accept single arrow)

adjacent bright fringes of light

will be bigger. v
f
Q4 (a) f = 600 Hz; v = 330 m s–1 (b) λ =
Using v = f λ,
(b) The leaves which act as λ = —vf = —6303—00 = 03.9.00
obstacles have small width.
Therefore, the effect of = 0.3 m
diffraction is greater. Hence the
waves join back after a short = 0.55 m (c)
distance from the leaves. (b) a = 1.5 m; D = 3.0 m;
x = λaD
(c) The log which is a larger 0.55 × 3.0
obstacle causes less diffraction. = 1.5
The waves join back after a
futher distance from the log. = 1.1 m

494

Physics SPM  Answers  

2. (a) Diffraction of waves Section B a particle of the medium
oscillates parallel with the
(b) • The amplitude of the waves 6. (a) A system that undergoes a direction of the waves. [1]
before passing through the • This shows that in
slit is higher than that the periodic to-and-fro movement. [1] the propagation of a
amplitude after passing (b) • Spring B carries a bigger longitudinal wave, the
through the slit. particle of the medium
mass than that of spring A. [1] vibrates/oscillates parallel
• This is because the waves • Spring B oscillates with a with the direction of the
spread over a larger area and wave. [1]
energy per unit area of the bigger period than spring A. [1]
diffracted waves is less. • Spring B oscillates with a Max 4

(c) lower frequency compared to
spring A. [1]
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• The bigger is the mass

attached to the spring, the (iii) • Work done by the hand in

lower is the frequency of its moving the spring

oscillation. [1] to-and-fro causes the

4 energy to be transferred

(c) (i) Oscillating with the in the form of waves

amplitude decreasing with along the spring. [1]

time. [1] • The string tied to the

(ii) • There is resistance spring which represents

3. (a) d = v × t = 1500 × 0.18 between the loaded spring a particle of the medium

and the oil. [1] vibrates/oscillates but

= 270 m • As a result, energy is lost does not move along with

(b) The depth of the sunken ship from the system as heat the wave. [1]

= d = 270 = 135 m energy. [1] 2
2 2
• If no external force is applied,

(c) Ultrasonic waves have more the spring will eventually 7. (a) • The convex-shaped peak of
the wave is like a convex lens.
energy than audio waves. stop completely. [1] [1]

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

(c) For taking the image of the • Move the hand
patient’s bones.
to-and-fro, perpendicular

5. (a) (i) High frequency sound waves. to the spring. [1]

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

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

bat and the bird. particle of the medium • When the angle of
incidence is not zero,
• With that, the distance vibrates/oscillates the waves change their

between the bat and the perpendicularly with the direction when moving from
deep to shallow areas. [1]
bird can be estimated direction of the wave. [1]
5
through the period of Max 4
(ii)
time transmission and

detection of the ultrasonic

waves.

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

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

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

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

495

  Physics SPM  Answers

• The waves at the bay are • Transfer energy from one
place to another sin i
calmer than at the cape [1] 0.9
(Other characteristics
• due to the divergence of acceptable) [1]

the waves’ energy from 3
(ii) 1. Sound wave does not
the bay [1] travel in vacuum. [1]
2. Sound wave is a
• and the convergence of longitudinal wave. [1] 0.7

the waves at the cape [1] 2

(ii) • To reduce erosion,

retaining walls are built [1] 0.5

• to reflect the waves from
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
the shore [1]

• and to reduce direct (d) (i) • Example 3 [1] 0.3 sin r
0.2 0.3 0.4 0.5 0.6
impact of the waves on • The microwaves

the shore [1] transmitted will be

(iii) • Concrete structures with reflected by the plane and (a) Set of reading for angle of
incident = 35°
a gap in between are built return to the receiver [1]

at the designated area for (ii) • Example 1 [1] n = 0.454
0.6561
children [1] • The mountain is an (b)

• Waves passing through obstacle [1] = 1.45

the gap will be diffracted • Radio waves with longer

in the children’s area [1] wavelength will diffract (c) sin 450 = 0.7071
r = sin–1 0.4877 = 29.2°
• The smaller amplitude around the mountain

of the diffracted waves more compared with Q4 (a) The speed of light in glass block

causes the sea to be radio waves with shorter is slower than the speed of

calmer. [1] wavelength [1] light in water. This is because

10 (iii) • Example 2 [1] the refractive index of glass is

• There is a vacuum larger. Therefore, its density is

Section C between the Sun and the greater than that of water.

8. (a) Refraction of waves is a Earth [1] cair
cmedium
phenomenon that occurs when • The fact that ultraviolet (b) n =

there is a change of direction rays reach the cair
cwater
of the propagation of waves Earth shows that 1.33 =

travelling from one medium to electromagnetic waves cair
cglass
another due to a change of travel through vacuum [1] and 1.52 =

speed. [1] Max 7 cair

(b) (i) Sound waves travel faster in cglass = cwater 1.33 = 0.88
cwater cair 1.52
air when the temperature of

the air is higher. [1] 6Chapter cglass

(ii) • At daytime, when it is hot,

the surface of the Earth Light and Optics Checkpoint 6.2

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

layer. [1]

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

away from the Earth. [1] = 1.34

• On a cool night, the Q2 Da = 12 – 10 = 2 cm 1 1
sin c sin 42°
surface of the Earth cools Dr Dr Q2 n= = = 1.49
Da 2
down faster than the air. n = → 1.50 = sin P
sin 40°
[1] 1.49 =

• Hence, the air near the Dr = 1.5 × 2 = 3.0 cm sin P = 1.49 × sin 40°

surface of the Earth \The thickness of glass block is
  3.0 cm
becomes cooler than the = 0.9578
P = 73.3°
upper layer. [1] Q = 90° – 40° = 50°
R = Q = 50°
• This causes the sound Q3 i sin i r sin r S = 90° – 50° = 40°

waves to be refracted 22 0.3746 15 0.2588

towards the Earth. [1]

6 28 0.4695 19 0.3256

(c) (i) • Transverse waves [1] 33 0.5446 22 0.3746 1.49 sin 40° = 1 sin T
[1] T = sin–1 (1.49 sin 40°)
• Travel with a speed of 35 0.5736 26 0.4384 T = 73.3°
3 × 108 m s–1 in
41 0.6561 27 0.4540
vacuum

496

Physics SPM  Answers  

Q3 Checkpoint 6.4 (a) m= v = 42 = 7 = 2.33
u 18 3
Q1 (a) 1u 1 1 (b) u = v = 25 cm
+ v = f

1 = 1 – 1 = 1 – 1 = 1 f = 25
v f u 20 30 60 2

v = 60 cm = 12.5

m = v = 60 = 2 Checkpoint 6.5
u 30
Q1 (a) For distant objects, the image
• Periscope is used to see an Therefore, the image is real, must be formed at the focal
object behind the observer. inverted, 60 cm from the lens, point of the lens. Therefore, the
located opposite to the object distance between the lens and
• Image formed is inverted and and enlarged 2 times. the film is 50 mm.
same size as object.
(b) The camera lens should be
adjusted outward toward the
object. As the object distance
is reduced, the image distance
increases. With this, the
distance between the lens and
the film is slightly longer than

50 mm.
CheckpointPenerbitan Pelangi Sdn Bhd. All Rights Reserved.6.3 (b) 1 + 1 = 1
u v f

Q1 (a) The focal point is the point on 1 = 1 – 1 = 1 – 1 = 1
the principal axis where the light v f u 20 15 60
rays parallel to the principal axis
after passing through the lens v = –60 cm
will converge on it.
m = v = 60 = –4
(b) (i) Virtual image is image that u 15
cannot be displayed on the Q2 (a) Astronomical telescopes
screen. The image is virtual, upright, at (b) Objective lens: lens P, Eyepiece:
lens Q
(ii) a distance of 60 cm from the

lens on the same side as the (c) The separation distance of the

object and is magnified 4 times. lenses = 40 + 5
v hi
1 cm Q2 m = u = ho = 15 = 1.5 = 45 cm
10
Converging lens 1 cm (d)

Therefore, v = 1.5u 5 cm
40 cm
= 1.5 × 2.0 m = 3.0 m Parallel rays
from distant
Using 1 + 1 = 1 object
u v f
F Main axis
Image Object f = 2 cm 1 + 1 = 1 = 5 Lens P Lens Q
2 3 f 6

Therefore, focal length f = 1.2 m

Q3 Using u1 + 1 = 1 Q3 Similarity:
v f • Consists of two convex lenses
1 1 • The eyepiece function as a
Virtual image is image that 1 = 1 – 1 = – 10 – 30 magnifying glass
Q2 (a) cannot be displaced on the v f u Differences:
screen = 2 cm • For microscope in normal
= – 4 adjustment, the final image is at
30 near point. While for telescope in
v = –7.5 m normal adjustment, the final image
is at infinity.
• The distance between the
v 30 objective lens and the eyepiece of
u the microscope is LO > fO + fe. The
m = = 4 = 1 distance between the objective
30 4 lens and the eyepiece of the
telescope is LO ≤ fO + fe.
The image is virtual, upright,
at a distance of 7.5 cm from
Image the lens on the same side
F as the object and has linear

magnification of 1 .
4

Q4

(b) The size of the image does not v/cm Checkpoint 6.6
change but the brightness of Q1 (a)
image reduced by half. 60 P
50 u = v
Q3 (a) CMo=nvuvex=le36n00s= 42
(b) 40
0.5 Q
30
(c)
20
F
10 R
0 10 1820 30 40 50 60 u/cm
F

u = 60 cm vf = 20 cm
= 30 cm
(b) The image is real, inverted and
Focal length, f = 20 cm magnified.

(c) The image is nearer to the
mirror and becomes smaller in
size.

497

  Physics SPM  Answers Section C • A periscope is build using
Q2 (a) two right-angle prisms
arrange in an cylindrical
2. (a) (i) The critical angle of glass is tube as shown in the
42° which means that if the diagram above.

angle of incidence in glass • The prism acts like a
perfect mirror when light
(denser medium) is 42°, ray hits the inner surface
of the prism at an angle
25° then the angle of refraction greater than 42°. The first
FC prism turns the image
in the air is 90°. from a distant object 90°,
and then the second
(ii) Total internal reflection prism turns the image
back to upright.
(iii) sin c= 1
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. n • Object lens is placed in
front of the first prism on
(b) The image is at F. n = 1 c top to find a distant object.
(c) The image is virtual, upright and sin The eyepiece is used to
1 observe the image formed
diminished. = sin 42 by the second prism.

Q3 • Focal length of mirror P = 12 cm. = 1.49 (ii)
The image from distant object is
formed at the focal point of the (b) Right-angle prism
mirror.
45° 45° Distant
• Focal length of mirror Q = 10 cm. 45° 45° object at
Object is located at radius of the back

curvature, image distance = 2f. 45° Cylindrical
• Therefore, the mirror Q has a tube
45°
shorter focal length. 45°

SPM Practice 6 45° 45°
Objective Questions 45°

1. B 2. C 3. B 4. C 5. B (c) (i) An inverted
6. C 7. C 8. C 9. D 10. D
11. C 12. A 13. B 14. B 15. C
16. A 17. C 18. A 19. A 20. C
21. C 22. D 23. D 24. B 25. C

Subjective Questions Glass image Right-angle prism

Section A Air When the prism at the
1. (a) top periscope is laterally
Light ray Total internal inverted, it can be used to
reflection occurs here observe the object behind
the observer over an
(b) (i) f = 10 cm, u = 12 cm A light ray is transmitted obstacle. In this case, the
through a curved fibre optic image seen is inverted
1 = 1 + 1 → 1 = 1 + 1 that occurs a series of total
f u v 10 12 v internal reflections on the (iii) • A right-angle prism uses
inner surface. total internal reflection
1 = 1 – 1 = 6–5 = 1 to reflect light. The light
v 10 12 60 60 (ii) • Fiber optic cables are reflected by the prism
lighter and thinner. surface is 100%.
v = 60 cm
• Transmission of signal • Image formed by right-
Therefore, distance between with almost no energy angle prism is sharper.
loss along the optical
the screen and the object fiber.
= u + v = 12 + 60 = 72 cm
v (d) (i)
(ii) m = u = 60 = 5 FORM 5
12 Right-angle
prism 1Chapter
(c) The minimum height of the
screen = 24 × 5 Forces and Motion

= 120 cm Distant object

(d) If the distance between the lens Cylindrical Checkpoint 1.1
and the image = 10 cm = f, then tube
object is at the focal point and
image formed is virtual and at Q1 (a) 17 N, in the same direction as
infinity. Therefore, image cannot the direction of the acting force
be seen on the screen.
(b) 7 N, in the same direction as the
Image at infinity Upright image (c) dFir=ect5io2n+o1f2122=N13 N

Right-angle q = tan–1 12
prism 5

FF = 67.4° with the horizontal

force of 5 N

498

Physics SPM  Answers  
(ii) Free body diagram
Q2 (a) Scale 1 cm : 2 N Checkpoint 1.2

B Q1 (a) Resolution of a force means
dividing a single force into two
4.6 cm component forces perpendicular 120 N
to each other.
3 cm
(b) (i) Vertical component upward
O 27° 45° = 12 cos 55° FR
2 cm A 39.8°
Horizontal component
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. Resultant force OB = 4.6 cm × 2 N = 12 sin 55°
(ii) Vertical component
= 9.2 N 100 N
downward = 25 sin 45°
Direction of the resultant force = Horizontal component Direction of resultant force, q
27° with the force of 4 N. = 25 cos 45°
(iii) Component parallel to the 100
(b) Scale 1 cm : 2 N = tan–1 120 = 39.8°
inclined plane = 30 cos 60°
B Component perpendicular Therefore, a force of 156.2 N

to the inclined plane = needed to act in the
30 sin 60°
4.4 cm direction 129.8° (39.8°
S2 (a) Net force = 0 N as the car is at
3 cm rest + 90°) with the force of

(b) Frictional force = component 100 N for the box to be in
weight of the car down the slope
120° equilibrium.
A = 12 000 sin 20° = 4 104 N
O 37° 5 cm (c) The supporting force Q2 Magnitude of the force
= 12 000 cos 20° = 11 276 N = 400 cos 30° + 400 cos 30°
Resultant force OB = 4.4 cm × 2 N
Q3 (a) Free body diagram: = 692.8 N

= 8.8 N Tension, T Q3 (a) (i)

Direction of the resultant force = 10°
37° with the force of 10 N 40° T1

(c) Scale 1 cm : 2 N 150 N

O 3 cm A

88° 120° T2

Air resistance, R (ii) 30°
Weight, W 60° P
5.2 cm 6 cm

(b) Vertical component of T = T cos 10°

Horizontal component of Q = 100 N 70°
T = T sin 10°
B
(c) T cos 10° = 1000 × 10

T = 10 000 = 10154 N R
cos 10°
Resultant force OB = 5.2 cm × 2 N 50°
40°
= 10.4 N
Checkpoint 1.3
Direction of the resultant force = (iii)
88° with the force of 6 N S1 (a) The forces are in equilibrium if
the resultant force produced is
Q3 (a) Magnitude of the resistance zero in all directions. FC 100° FA

= 20 N (b) (i) Triangle of forces
(b) Resultant force, F = 30 – 20 = 10 N
F = ma → 10 = 4 × a FR B 50° 30°
FB = 10 kN
a = 2.5 m s–2 O

Q4 (a) Resultant force, F = 0 N; (b) (i) Scale 1 cm : 20 N
Tension, T = weight of the fish O

= 0.8 × 10

=8N
(b) Resultant force, F = 0 N;
Tension, T = weight of the fish = 8 N 120 N 40°

(c) Resultant force, F = ma = 0.8 100 N 7.5 cm 9.7 cm
W T1
× 0.5 = 0.4 N; T – 8 = 0.4, A
therefore T = 8.4 N

Q5 (a) F = ma → 2 × 10 = (6 + 2) × a Resultant force, T2 50°
a = 2.5 m s–2 FR = 1002 + 1202 = 156.2 N A 6.1 cm B
T1 = 9.7 cm × 20 N = 194 N
(b) T = m × a = 6 × 2.5 = 15 N T2 = 6.1 cm × 20 N = 122 N



499

  Physics SPM  Answers

(ii) Scale 1 cm : 20 N (ii) Mass of load of 1 kg is (ii) Direction of motion
60° P 4.1 cm equivalent to a force of 10 N a = – 1.77 m s-1
10°
Q 70° Substitute into formula,
5 cm R 4.6 cm F = kx → 10 N = k × 2 cm
Therefore, spring constant,
50° k = 5 N cm–1 atau 500 N m–1
(b) – Increase the stiffness

of spring by replacing
the copper spring with a
stronger steel spring.
– Replace spring with spring
made of thicker wire.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. 1 kg
P = 4.1 cm × 20 N = 82 N SPM Practice 1
Objective Questions Due to deceleration, the
R = 4.6 cm × 20 N = 92 N direction of action is in the
opposite direction.
(iii) Scale 1 cm : 20 kN 1. B 2. A 3. D 4. A 5. C
6. B 7. C 8. A 9. D 10. D 3. (a) Weight due to force of gravity
FC FA 11. A 12. C 13. A 14. B 15. B act on the spring.
2.6 cm 3.9 cm 16. A 17. A 18. B 19. C 20. D
21. D 22. C 23. C 24. B 25. C (b) Vector has magnitude and
direction.
Subjective Questions
(c) (i) I = lO + x
50° 30° Section A T
FB 5 cm 1. (a) Panjang spring/cm
FA = 3.9 cm × 20 kN = 78 kN
T 140
FC = 2.6 cm × 20 kN = 72 kN
120
Checkpoint 1.4
100
S1 (a) Spring extension, x = 18 cm – 45° 45° 80
I = 76 cm
15 cm = 3 cm = 0.03 m 10 N 60
Magnitude of force, F = kx = 40 I0
(b) T2 + T2 = 102 20
100 N m–1 × 0.03 m = 3 N 9N
0
(b) Length of compression, 2T2 = 15000 2 4 6 8 10 12 14
F T = daya/N
x = k 2N = 7.07 N (ii) l = lO + x
= 100 N m–1 76 = 40 + x
2. (a) Direction of motion x = 76 – 40 = 36 cm
(d) Curve upwards after 10 N (refer
= 0.02 m the above graph)

= 2 cm Section B
4. (a) Scale 1 cm : 500 N
Therefore, length of spring T
l = 15 cm – 2 cm = 13 cm 10° 5 cm

Q2 Length of compression, 1 kg
x = 20 cm – 15 cm

= 5 cm

= 0.05 m 10 N F1
7 cm
Energy needed, E = 1 kx2 R
2 10.5cm

= 21 × 200 N m–1 Resolving vertically 7 cm
T cos 10° = 10 N
× 0.052 m2
T = 10
= 0.25 J cos 10°

Q3 (a) Spring constant, = 10.2 N 36°
k = gradient of graph 60°
(b) F = ma
8 N T sin 10° = 1 kg × a 5 cm F2
= 16 cm a = 10.2 sin 10°
Resultant force, R = 10.5 cm ×
= 0.5 N cm–1 = 1.77 m s–2 500 N = 5 250 N

(b) Potential elastic energy of spring (c) (i) Direction of resultant force is
q = 36° with the force of 2 500 N
= Area under the graph

= 1 × 8 × 0.16 = 0.64 J (b) (i) The net force acting on the
2
system = 30 N – 10 N
Q4 (a) (i) Hooke’s law
= 20 N

500

Physics SPM  Answers  

(ii) T1 3.33 m s–2 T2 (c) The level of mercury inside
2.0 kg the tube is the same as in the
T container outside the tube.
1 T 3.33 m s–2 This is because Moon does not
2 have atmosphere. Hence the
3.33 m s–2 3 kg 1 kg atmospheric pressure is equal to
zero.

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.30 N 10 N Checkpoint 2.3

F = ma Q1 (a) The pressure of the gas is
20 = (3 + 2 + 1)a higher than atmospheric
pressure. The level of oil in
a = 20 the arm attached to the gas
6 is lower than the other arm
of the manometer. Hence the
= 3.33 m s–2 pressure of the gas is the sum
of atmospheric pressure and
(iii) T1 2Chapter the pressure due to the 50 cm
column of the oil.
Pressure
(b) Difference in pressure DP = hρg
3.3 m s–2 3 kg Checkpoint 2.1 = 0.5 × 800 × 10 = 4000 Pa

Q1 Pressure, P = hρg = 100 × 1050 × 10 Q2 Pressure of gas, P + Pressure due
= 1.05 × 106 N m−2 to the 60 cm column of mercury =
Atmospheric pressure

Force, F = P × A = 1.05 × 106 N m−2 P + hρg = 105
× 200 m2 = 2.10 × 108 N P = 100000 − (0.6 × 13.6 ×

103 × 10)
= 18400 Pa

30 N Checkpoint 2.4

3.3 m s–2 Q2 (a) The deeper the water, the higher Q1 (a) F1 = F2
the pressure. Therefore, the A1 A2
2 kg bigger the force acting on the F2
TT wall. F1 = A2 × A1

12 (b) Pressure, P = hρg = (300 – 60)     F2 D1 2
× 1000 × 10 = 2.4 × 106 Pa 2
T2 = D2 2 × p
2
1 kg 3.3 m s–2 Q3 (a) Pressure at the same level is p
the same. P and Q are at the
10 N same level. Hence both have     D1 2 1.25 2
(iv) 30 − T1 = 3 × 3.33 the same pressure. D2 F2 = 25
T1 = 30 − 9.99 = × × 6000
= 20.01 N
T 2 − 10 = 1 × 3.33 (b) P = hρg = 15 N
T2 = 3.33 + 10
= 13.33 N Poil = Pwater (b) If the hydraulic fluid (oil) in the
Alternative: ρoilhoilg = ρwaterhwaterg
T1 − T2 = 2 × 3.33 jack is replaced by ordinary
20.01 − T2 = 6.66
T2 = 20.01 − 6.66 air, the jack will not work.
= 13.35 N
(v) Yes, the system is still ρoil = 1000 × 0.03 This is because ordinary air is
0.05
moving in the same compressible.
direction. The system
will move at a smaller = 600 kg m−3 Q2 (a) Pascal’s principle states that
acceleration as the masses
increases. pressure applied to an enclosed

Checkpoint 2.2 fluid is transmitted uniformly to

Q1 (a) (i) The length of the mercury every of the container of the
column is 75 cm. Therefore,
the atmospheric pressure is fluid. F
75 cm Hg. A
(b) (i) Pressure, P = = 12
(ii) P = hρg = 0.75 × 13.6 × 103 9
× 10 = 102000 Pa
= 4 N cm–2
(b) The length of the mercury 3
column is still 75 cm. This is
because atmospheric pressure (ii) The pressure is transmitted
can support a vertical height of
75 cm of mercury. equally, therefore the

Q2 (a) h = 76 cm pressure exerted by the oil
(b) Pressure at the point X = 76 +
10 = 86 cm Hg = 0.86 × 13.6 × on the large piston is also
103 × 10 = 116960 Pa
4 N cm–2
3
4
(iii) F2 = 3 × 108

= 144 N

(c) A small input force can yield a
large output force.

501

  Physics SPM  Answers

Q3 (a) 1. A hydraulic pump is more Therefore, the additional weight SPM Practice 2
that can be added = 44982 – Objective Questions
flexible. It can be placed 14994 = 29988 N

in between the small gap

between the wall and the Checkpoint 2.6 1. D 2. A 3. B 4. A 5. D
6. C 7. A 8. A 9. B 10. C
floor. Q1 (a) The speed at P is higher than 11. B 12. B 13. C 14. B 15. D
the speed at Q. 16. C 17. A 18. A 19. C 20. A
2. A hydraulic pump is safer 21. C 22. B 23. A 24. D
(b) The pressure at P is lower than
to use. Heavy machineries the pressure at Q.

may cause further collapse (c)
P
of concrete structures and

risk harming the victim Subjective Questions
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
trapped underneath the

collapsed wall. Section A

(b) Using the formula, 1. (a) Pressure at A = Pressure at B
F Finput Q
A = Ainput output Sail ℎ1ρ1g + pressure of air in the
output tube = ℎ2ρ2g + pressure of air in
Foutput
Aoutput = Finput × Ainput Direction the tube
of wind Therefore,
80 ℎ 1ρ1g = ℎ2ρ2g
= 3 × 15 Q2 increases; low; Atmospheric

= 400 cm2 Q3 (a) • The hydrofoil of the boat ρ2 = h1 × ρ1
has a curved upper surface h2
and flat lower surface. This
Checkpoint 2.5 causes the water to travel at = 30 × 1000
a higher speed at the upper 22
surface than the lower
Q1 (a) 6.3−5.5=0.8 N surface of the hydrofoil. = 1364 kg m–2
(b) 0.8 N (b) (i) Pressure at A is the same
• The pressure of water at the
(c) 0.8 N upper surface is lower than as atmospheric pressure
the pressure at the lower
Q2 (a) 0.18 × 0.06 × 10 = 0.108 N surface of the hydrofoil. The 105 N m–2.
difference in the pressures (ii) Pressure at A caused by the
(b) Buoyant force = Weight of the results in a net force acting
upward on the hydrofoil. column of water PA = ℎA ρAg =
displaced air Thus lifting the boat up. 0.3 × 1000 × 10 = 3000 N m−2.

= 1.3 × 0.06 × 10 (b) This will reduce the area of (iii) Pressure of air in the tube,
contact between the boat and Pair = 105 − 3000
= 0.78 N the sea, hence reducing the
(c) Weight of the balloon + W resistance of the boat when it is = 97000 N m−2
= Buoyant force travelling.
0.108 + W = 0.78 (c) When the clip is removed, air

W = 0.672 N enters into the tube and the

Q3 (a) For a floating object, the weight pressure inside and outside
of the object is equal to the
buoyant force acting on it. Since the tube becomes the same
the same boat is used for both
situations, the buoyant force and is equal to the atmospheric
acting on the ship is the same.
pressure. Both the water and
(b) Since the buoyant force is equal liquid X columns will fall until the
to the weight of the displaced
water, the weight of the same levels as outside of the
displaced seawater is equal to
the weight of the displaced river tube.
water.
2. (a) Piston 1 V1 V2 Piston 2
(c) Since the weight of the Down Open Closed Up
displaced seawater is equal to Up Closed Open
the weight of the displaced river Stationary
water,
(b) Using the formula Foutput (ii) Aerofoil R
ρseaVseag = ρriverVriverg Finput Aoutput
ρseaVsea = ρ Vriver river Ainput =
Since Vsea < Vriver therefore, ρsea
Foutput = Finput× AAoinuptpuutt = 5 × 32
 ρriver. The density of seawaters 4
is higher than the density of
river water. = 40 N

(c) Pascal’s principle D

(d) The system can convert a small (iii) • Lines above aerofoil
almost straight.
Q4 (a) Weight of the boat input force into a bigger output
= weight of the displaced seawater • Lines below aerofoil
= ρVg = 1020 × 1.5 × 9.8 force. curved.
= 14994 N
(e) Turn the release valve to allow • Air flows at a lower speed
(b) For maximum load, above the aerofoil than
Weight of the boat + Additional the hydraulic fluid to flow from that below the aerofoil.

weight cylinder 2 into the reservoir. • The pressure of air above
= Weight of the displaced seawater the aerofoil is higher than
= ρVg =1020 × 4.5 × 9.8 Section B that below the aerofoil.
= 44982 N
3. (a) (i) Action on an object that can
result in changes like size,
shape, speed and direction.

502

Physics SPM  Answers  

• Due to the difference in • The weight of the egg is 3Chapter Electricity
pressure, a downward the same when it is in the
force is produced. water and in the brine.

5 • The egg floats because its Checkpoint 3.1
(iv) • A bigger piece of spoiler has weight is balanced by the
a bigger surface area, A. buoyant force. Q1 (a)
• S ince P = AF—, F = P × A.
• More liquid will be
• Hence a bigger downward displaced when its density
force is produced by a is lower.
bigger spoiler. Maksimum 5
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
3 (iii) Archimedes’ principle +
(b) (i) • Design Q is most suitable. (b) • A large part of the ship is

• This tube will produce a hollow.
• It displaces a very large
fast stream of air at the
volume of water.
end of the tube because • Hence a big buoyant force

the cross-sectional area of is g­ enerated to balance the
weight of the ship.
the tube becomes smaller. (c) Buoyant force
(ii) • Design P is not efficient
Weight
because the cross- (b)
+
sectional area increases +
+
and the speed of air + +
+ +
decreases. +
• Design R will not produce +
+
a stream of air because

its end is closed.
(iii) • Design Z is most suitable.

• Due to the hole in the

stopper, the pressure of

air in the container is the

same as the pressure

outside the container (or (i) • Low density material used
to make the balloon to
atmospheric pressure). reduce its weight. Q2 (a) F = 2.0 × 10–15 N; e = 1.6 × 10–19 C
(iv) • Designs X and Y are
• A very big balloon is F
not suitable because as required to displace a Substitution, E = Q
very big volume of air.
more water is drawn out, Electric field strength, E
• This is to produce a large
the pressure of the air buoyant force. 2.0 × 10–15
1.6 × 10–10
in the containers will be (ii) • Hot air is produced by =
heating air by burning gas
reduced. at the base of the balloon.

• Then it will not be able to (iii) • Hot air rises to fill the = 12500 N C−1
balloon and displaces the
force the water out of the cool air. (b) E = 12500 N C–1 ; d = 5 cm

container. • A buoyant force is
generated because hot
(v) • When air is blown out of air is less dense than the = 0.05 m pVdo,treenatiraral ndgiffeedr,eVnc=e,FVd
cool air it displaces.
the tube at high speed, Using, E =
• When more hot air fills the Therefore,
• its pressure reduces. balloon until the buoyant
force is greater than the
• The higher atmospheric total weight of the balloon, = 12500 N C–1 × 0.05 m
the balloon rises into the
pressure of the air in the sky. = 625 V

container will force the (iv) • To make the balloon fall
back onto the surface of
water out of the container the earth, reduce the rate Q3 The amount of charges
or stop the heating of the flowing through the TV in 10
through the tube attached air. minutes,

to the container. • As less hot air fills the Q = It
balloon, its buoyant = 0.5 × 30 × 60
• The water will be carried force reduces. When the = 900 C
buoyant force is less than
by the air stream and the weight, the balloon will
fall towards the earth.
sprayed outward. [1]

Max 10

Section C Q4 (a) I= Qt = 30000
50 × 60
3. (a) (i) Density is mass per unit = 10 C s–1 (A)
volume of the material.
(b) Q = ne → n = Qe = 1
(ii) • The egg sinks more in 1.6 × 10–19
water than in brine. = 6.25 × 1018

• The volume of the water
displaced is more than Q5 V = W = 240 = 12 V
that of the brine. Q 20

• The density of water is
less than brine.

503

  Physics SPM  Answers

Q6 Charge flowing through the lamp in (b) ɛ = I(R + r) → r = ɛ – R Q4 (a) P10Ω : P25Ω = V2 : V2
10 s, I R10Ω R25Ω

Q = It = 9 – 5=4Ω = 122 : 122
= 2 × 10 1 10 25
= 20 C
(c) V = IR → l = V = 5 =5:2
Electrical energy changed by the charge, R 15
E = VQ (b) Poutput = V2 + V2 = 20.2 W
= 240 × 20 1 R10Ω R25Ω
= 4800 J = 3 A

ɛ = I(R + r) Q5 (a) P = VI → I = P
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. V
Checkpoint 3.2 = 1 × (15 + 3) 24 36
3 I = 12 + 12 =5A

Q1 (a) Re = 8 + 1 = 8 + 2 = 10 Ω = 6.0 V V2 V2
R P
1 + 1 Q2 V = IR → R = V = 10 = 12.5 Ω (b) P = → R =
3 6 I 0.8
122
(b) V = IR = 2(10) = 20 V ɛ R1 = 24 = 6 Ω
V1 = 2 × 8 = 16 V I
V2 = ɛ = I(R + r) → r = –R R2 = 122
I1 = 36
20 V – 16 V = 4 V 12 = 4 Ω
(c) V 4 = 1.33 A 0.8
R = = – 12.5 (c) R = 6 Ω + 4 Ω = 10 Ω
3

= 2.5 Ω I= V = 12 = 1.2 A
R 10
Q2 (a) Length, l; cross sectional area, Q3 (a) 4.5 V
A; resisitivity or type of material
(b) 1 = 1 + 1 =2→r= 1 Ω The bulbs become dimmer.
(b) (i) RX > RY > RZ r 1 1 2
(ii) The gradient of the graph ɛ
shows the resistance of the (c) ɛ= I(R + r) → l = R+r SPM Practice 3
conductor, then, gradient of Objective Questions
X > gradient of X > gradient = 5 4.5
of Y > gradient of Z + 0.5

= 0.82 A 1. B 2. C 3. D 4. C 5. D
(d) V = IR = 0.82(5) = 4.1 V 6. B 7. B 8. C 9. D 10. C
Q3 Using formula, (e) Vdrop = Ir = 0.82 × 0.5 = 0.41 V 11. B 12. D 13. A 14. D 15. B
16. A 17. C 18. A 19. A 20. A
R = ρl = (2.8 × 10–8)(1000) Q4 21. C 22. A 23. A 24. B 25. C
t 1.5 × 10–4

= 0.19 Ω V/V

Q4 Diameter, D = 0.100 × 10–3 m 3.0 Subjective Questions

Cross sectional area,

0.100 × 10–3 2 Section A
2
  A = p 2.0 1. (a) Bulbs P and Q
1.7 (b) Bulbs P and Q are connected in
= 7.85 × 10−9 m2 series, the effective resistance
1.0 is larger and so less current
Using equation R = ρl rearrange passing through them.
A (c) (i) and (ii)
RA
l = e 0 0.2 0.4 0.6 0.8 I/A
0.74
Therefore, the length of tungsten V
0.200 × 7.85 × 10–9 ɛ = I(R + r) = V + Ir PQ
filament, l = 5.6 × 10–8
ɛ–V 3.0 –1.7
= 0.028 = 28 mm r = I = 0.74

Q5 R = ρl RP lP RQ = lQ = 0.7 Ω AR
A AP AQ S
; = ρ ; ρ Checkpoint 3.4

lQ = AQ → lQ = 0.0032 Q1 (a) P = VI = 12.0 × 1.2 = 14.4 W
IP AP 0.004 (b) E = Pt = (14.4 × 60) = 864 J

= 8 cm V2
R
Checkpoint 3.3 Q2 (a) P = = 264002 2. (a) (i) η = Eoutput × 100%
Einput
Q1 (a) Using ɛ = I(R + r) = 960 W = 0.96 kW

ɛ 1.5 (b) E = Pt = (960)(20 × 60) = 95000 × 100%
R+r 2.5 + 0.5 120000
→ I = = = 1.152 × 106 J

= 0.5 A Q3 P = V2 → R = V2 = 2402 = 28.8 Ω = 79.17%
R P 2000 90000
V = IR P (ii) P= 60
V 2000
= 0.5(2.5) P = VI → I = = 240 = 8.3 A = 1500 W

= 1.25 V

504

Physics SPM  Answers  

(b) The electric kettle can boil the Section B SPM Practice 4
water first. This is because the Objective Questions
rate of thermal energy supplied 4. (a) (i) Electromotive force of 1.5 V
to water is larger and therefore meant the amount of energy 1. B 2. C 3. D 4. A 5. C
rate of heat transferred to water supplied to a coulomb 6. C 7. B 8. D 9. C 10. D
is faster. charge to pass through a 11. B 12. A 13. C 14. B 15. C
complete circuit is 1.5 J.
3. (a) Increasing the length of wire Subjective Questions
increases the resistance of wire. (ii) V = IR = 1.0 × 2 = 2 V
(iii) e = I(R + r)
Increasing the diameter of wire 3.0 = (1.0)(2 + 2r)
r = 0.5 Ω
decreases the resistance of (b) In Figure 4.1, the dry cells are Section A
connected in series. Whereas
in Figure 4.2, dry cells are 1. (a) • The conductor will displace.
connected in parallel. • The direction follows by the
The voltmeter reading in Figure Fleming left-hand rule.
4.2 is smaller.
The ammeter reading in Figure (b) Extended
4.2 is smaller. (c) • Extended
The larger the voltage supplied, • Use more battery to increase
the greater the energy used to
move the electrical charges in current.
the circuit. (d) • Inverts current direction
The larger the current flow in • Inverts magnetic field polarity
the circuit, the faster the energy
transferred. 2. (a) (i) Fleming left-hand rule
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.wire. (ii) C
(iii) The conductor will move to
(b) R∝ l → R= kl the direction of A.
A A (iv) • increase the magnitude of
kl current
R1 = A = 16 Ω • increase the strength of
magnetic field
  k 1 l kl
2 4A (b) (i) Number of turn of coil: 50
R2 = 2A = turns

  = 1 kl = 1 (16 Ω) = 4 Ω Reason : More current flow
4 A 4 through the coil

(c) (i) Effective e.m.f. = 2E (ii) Number of battery used: 3
1 1 1 5 Reason: Increase the
(ii) R = 16 + 4 = 16
magnitude of current
R = 16 = 3.2 Ω (iii) Shape of permanent
5
(iii) I> I2 > I1 magnet: Semi circular-shape
Reason: More cutting the
(c) (i) and (ii)
magnetic flux
Characteristic Reason (iv) Electric motor A
Low resistivity To reduce heat loss in the cables
3. (a) Catapult field is a resultant force
Low density The cables will become lighter cause a conductor wire moves.

Low cost Cost of project will be lower (b)

Low rate of thermal expansion The cables will not expand under hot weather K

Choose: Cable S
Reasons: Low resistivity, low density, medium cost, low rate of thermal

expansion

4Chapter Electromagnetism Checkpoint 4.2 NS
L
S1 (a) When the bar magnet is pushed
Checkpoint 4.1 into the solenoid, induced (c) • Interaction between two
current is produced thus magnetic fields
S1 (a) The direction of force can be deflected the galvanometer
identified by using Fleming’s left- pointer. • cause a region to have a
hand rule combination of magnetic
(b) If the bar magnet is stay field with opposite direction
S B stationary inside the solenoid, no between each other.
D induced current is produced thus
A no deflection of galvanometer • So, the strength of the
pointer. magnetic field in that region
+– becomes weak.
C Checkpoint 4.3
(d) (i) Fleming’s left-hand rule
U S1 η = Poutput × 100
Pinput (ii)

= VsIs × 1000
VpIp
US
= 6000 × 0.20 × 100
240 × 6.25

(b) AB moves upward and CD = 80%
moves downward.

505

  Physics SPM  Answers

4. (a) Direct current motor • The higher the rate of cutting magnetic flux, the
(b) (i) A-B-C-D higher the magnitude of induced current.
(ii) downward, upward
(iii) clockwise (c) • When the coil rotates, it cuts the magnetic flux from
(c) • Less friction during rotation. the magnet.
• Long lasting and no maintenance required because
no carbon brush used. • Induced current is produced.
• The direction of induced current can be identified by
5. (a) Voltage
(b) • When a bar magnet passes through the solenoid, using Fleming’s right-hand rule.
the magnetic flux is being cut.
• So, induced e.m.f is occurred and the galvanometer (d) Aspek Sebab
pointer is deflected.
(c) • The magnet is fall from higher place. 1. Number of turns More magnetic fluxes can
• So, the magnet will passes through the solenoid with must be higher be cut.
higher speed.
(d) • No pointer deflection occurs.
• No magnetic flux is being cut.

6. (a) P: North; Q: North
(b) P: South; Q: South
(c) Lenz’s law
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. 2. Use higher magnetic Produce higher strength

strength of magnetic field.

3. Use round magnet Less leakage of cutting
shape the magnetic flux.

4. Brushless motor • Long lasting.
• No maintenance.

5. The magnet is inside The cutting of magnetic
and is surrounded by flux is optimum.
coil.

Section B

7. (a) National grid network is a network of cables connects 9. (a) (i) Step-down transformer
(ii) • Number of turns at input is larger than number
all power stations in a country to consumers. of turns at output.
(b) X – Step-up transformer • Voltage input is higher than voltage output.
Y – Step down transformer • The larger the number of turns, the higher the
voltage produced.
(c) • Power can be distributed based on location demand.
(b) (i) P – Step-up
• If a power station is breakdown or under Q – Step-down
R – Step-down
maintenance, other power station can continue to
(ii) • Ploss = I2R
supply electricity without disruption. • If the current increases, the loss of power also

• Reduced cost and pollution increases.
P =VI • By the Ohm’s law, when the voltage increases,
(d) (i)
the current will decreases, thus the loss of
10 000 000 = 144 000 I power also decreases.
I = 69.44 A

(ii) P = I2R

= 69.442 × 300

= 1.45 × 106 W

(e) Aspect Reason (c) Aspect Reason
To make step-up
Soft iron core Easy to magnetised and The number of turns of transformer
material demagnetised. primary coil is less then
secondary coil Low resistance
Made by laminated Reduced eddy current Can be magnetised and
core Copper wire demagnetised easily.
Less hysteresis
K-shaped Reduced the leakage of Soft iron core Less magnetic flux
magnetic flux leakage.
Laminated iron core
Ratio of 40 : 1 Able to reduce 240 V to 6 V The secondary coil is
wounded on the primary
The best Because easy to magnetised coil.
transformer is U and demagnetised, reduced
eddy current, reduced the Electronics
leakage of magnetic flux and 5Chapter
able to reduce 240 V to 6 V.

Section C Checkpoint 5.1

8. (a) Induced current is the current produced from a Q1 (a) Thermionic emission is the process of releasing free
conductor cutting magnetic flux. electron from hot metal surface. Cathode ray is a fast-
moving electron beam.
(b) (i) • Both have equal magnetic field strength
• The number of turns in Figure 8.2 is higher than (b) – negative charge,
Figure 8.1. – can be deflected by electric field and magnetic
• The degree of deflection in Figure 8.2 is higher
than Figure 8.1. field,
(ii) • The higher the number of turns of coil, the – can produce fluorescent effect,
higher the degree of galvanometer pointer – can stopped by thin metal.
deflection.
• The higher the number of turns of coil, the
higher the rate of cutting magnetic flux.

506

Physics SPM  Answers  

Q2 (a) (d) To limit the base current flow –
Cathode ray into transistor.
Parabola
(b) The cathode ray is a negatively Q3 (a) Thermistor Straight line
charged electron beam, so it is (b) Function of P is to detect
attracted to a positively charged temperature change. The +
plate and forms a parabolic path resistance of P decreases with
in the electric field between the temperature.
two plates. (c) (i) 9 V

 (ii) VXY =
6 (9) 3. (a) Thermionic emission
6 + 10
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. (b) (i) The light from the heated

= 3.375 V filament is blocked by the

Maltese cross.

Q3 (a) The kinetic energy acquired by SPM Practice 5 (ii) The electrons strike the
electron is equal to the electric Objective Questions
fluorescent screen.
potential energy (c) Electrical potential energy = eV

= 4 keV = 1.6 × 10–19 × 3000

= (4000)(1.6 × 10−19) 1. A 2. C 3. A 4. C 5. D = 4.8 × 10–16 J
6. D 7. A 8. D 9. D 10. A
= 6.4 × 10−16 J 11. B 12. C 13. C 14. B 15. D (d) The kinetic energy is converted
16. D 17. B 18. C 19. C 20. D
1 21. C 22. A 23. A 24. D 25. A to heat and light energy.
2
(b) E= mv2max = eV (e) (i)

× 9.11 v2 × Filaments Anode U
max S
2
× 10–31 × = 6.4 10–16 Subjective Questions

vmax = 2 × 6.4 × 10–16 6 V a.c. Shadow
9.1 × Screen
10–31 Section A U
Cathode S
= 3.75 × 107 m s–1 1. (a) Alternating current
(b) Diode VLT
Checkpoint 5.2 (c)

Q1 (a) The diode only allows current to (ii) Fleming’s left-hand rule
pass through it in one direction
only. Section B

(b) (i) Bulb Q is not lights up 4. (a) (i) Transistor npn
because diode D2 is
connected in reversed bias. (ii) • Both ammeters do not

(ii) Terminal Y is positive. show any reading.
(iii) Both diodes are used to
• Microammeter shows
indicate the polarity of
battery. reading and milliammeter

Q2 (a) Rectifier is a device that can (d) does not.
convert AC voltage or current to
DC voltage or current. (b) When the rheostat is adjusted

(b) (i) Full wave rectification circuit and both switches are switched
(ii) Bridge rectifier
(iii) To smoothen the direct on, a base current is generated
current output.
(iv) to activate the transistor. This

causes a larger collector current

to flow through the circuit.

(c) • the current amplification by

using the formula
Ic
2. (a) (i) To heat up the cathode b = Ib

(ii) To emit electron when = 1 × 10–3 = 100
10 × 10–6
heated by heating filament.

(iii) Thermionic emission • emitter current Ie.

(iv) To supply high potential Ic = Ib + Ie

Checkpoint 5.3 difference to accelerate the Ie = Ic – Ib
= 1 × 10–3 – 10 × 10–6
Q1 (a) P : Collector; Q : Emitter; beam of electron to high = 9.9 × 10–4 A
R : Base
speed.
(b) pnp transistor 1
(b) eV = 2 meV2 (d) • To detect fire, a thermistor is
Q2 (a) npn transistor used to detect heat and needs
(b) and (c) 1.6 × 10–19 × 3000 = to be connected to terminal
P as a potential difference
1 × 9.1 × 10–31 × v2 divider.
2
v2 = 1.055 × 1015 • A resistor must be connected
M v = 3.25 × 107 m s–1 to terminal Q to control the
size of the current flowing into
VC (c) (i) The negatively charged the base of transistor.
T
R electron is attracted by the • At the collector terminal of
VB transistor, which is terminal R,
positively charged plate. should be connected with an

Base circuit Emitter circuit

507

  Physics SPM  Answers (b) w = b particle, x = γ ray – There is a 2 times increase

alarm to alert the public when (c) (i) Gamma ray, because it has in the proton number. So, the
activated. high ionising power. number of b produced = 2
• Since the transistor used in
the circuit is npn type, cell (ii) Kills microorganisms without 8. (a) (i) Short; Water is not exposed
2 is used and connected to damaging equipment
terminal S. structure. to radiation for a long time.
• The most suitable set is set
X. Set X uses thermistor as a 4. (a) – It is easily found from the (ii) Beta; High penetrating power
detector, a resistor to control surronding.
transistor base current, an but limit with pipe wall.
alarm to alert the public when – It has a very long half-life.
fire occurs and cell 2 is give (iii) Liquid; Easy to dilute with
correct connection to the (b) (i)
transistor. 100% ⎯T1→ 50% ⎯T2→ 25% ⎯T3→ 12.5% water.

So, the fossil has undergone (b) Sodium-24
3 times of the half-life.

(ii) 5730 × 3 = 17190 years

5. (a) The half-life of a radioactive
sample is the time taken for the
number of undecayed nuclei in
the sample to be reduced to half
of its original number.

(b) 6000 counts per second
(c)
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. (c) 8 → 4 → 2 → 1
8 8 8 8
\n=3

6Chapter   t = n T1 = 3(132) = 396 days
2

Nuclear Physics 9. (a) Nuclear fusion is the combining
of two lighter nuclei to form a
Checkpoint 6.1
heavier nucleus while releasing
Q1 (a) When sodium-24 completes a Activity / count per second
half-life, the activity becomes a large amount of energy.
(b) (i) m = (2.014012 u + 3.016029 u)
half from initial count. 6000
T=h1e0r0e0forceo,utnhtepaecrtimviitnyu=te.20200 5000 − (4.0022603 u +
Based on the graph, the half-life 4000
of sodium-24 is 1 s. 1.008665 u)
3000
(b) The undecayed activity, 2000 = 0.018863 u
1000 = 0.018863 × 1.66 × 10−27
= 3.13 × 10−29 kg
0
(ii) E = mc2
= (3.13 × 10−29)(3.0 × 108)2­
= 2.82 × 10−12 J

(c) Sun

    N= 1 n No = 1 4 10. (a) Reactor
2 2 (b) (i) Control the rate of fission
(2000) reaction by absorbing
excess neutrons.
= 125 count per minute. t / minutes (ii) Reduce the speed of
2468 neutrons so that slower
neutrons are more readily
Checkpoint 6.2 (d) 2 minutes captured by the uranium
(e) Not suitable because the half-life nuclei
Q1 (a) Mass defect, m (c) – Energy released by nucleus
of material X is too short. fission will heat up gas.
= 5.0301 − 5.0104 – The gas will heat up water
6. (a) Nuclear fission is the splitting of thus changes to a moving
= 0.0197 u × 1.66 × 10−27 steam with high acceleration.
a heavy nucleus into two or more – The steam will rotate turbine.
= 3.27 × 10−29 kg – The connected turbine
lighter nuclei while releasing a and dynamo will generate
(b) Energy released, electricity.
E = mc2 = (3.27 × 10−29)(3.0 × 108)2 large amount of energy. (d) – No air pollution.
– Fuel cost cheaper.
= 2.94 × 10−12 J (b) (i) ­23962U + 10n → 14516Ba + 9326Kr + N10n – Provide many jobs opportunity.

SPM Practice (ii) U236 + 10n →1 4516Ba + 9326Kr + N10n
Objective Questions 92

6 Balance the nucleon

numbers before and after

1. D 2. C 3. A 4. C 5. D reaction,
6. B 7. A 8. B 9. B 10. A 236 + 1 = 141 + 92 + N(1)
11. B 12. B 13. C 14. B 15. B N = 4
16. B 17. C 18. D 19. C 20. C (iii) m = 0.18606 × 1.66 × 10−27
21. B 22. A 23. D 24. C 25. D
26. C 27. A 28. C 29. C 30. C = 3.09 × 10−28 kg
E = mc2

= (3.09 × 10−30)(3.0 × 108)2 Section B

= 2.78 × 10−11 J 11. (a) Radioactive decay series occurs
Subjective Questions (2.78 × 10–11)
= 1.6 × 10–19 because a decayed nucleus is

Section A = 1.74 × 108 eV still in an unstable state.

1. (a) Alpha decay (c) Generate electricity (b) 1 because Uranium-234 lost 4
(b) 22826Rn
(c) 4; 2 nucleon number and 2 proton

7. (a) Chain reactions are ongoing number to form a helium.
reactions due to the result of
the reaction that could carry a (c) (i) U234 → 23900Th + 42He + energy
similar reaction. 92

2. (a) (i) 239 (b) 222 − 86 = 136 (ii) 23900Th → 23901Pa +–01e + energy
(ii) 91 (c) 22826Ra → 21884Ps + 42He
(d) – There is a 3 times decrease in (d) (i) Mass defect,
(b) Nucleus is not stable. m = 238.029 − (232.038 +
23991Pa U239 + –01e the nucleon number. So, the
(c) → 92 number of a produced = 3 4.003)

3. (a) Gamma decay is a decay which = 1.988 u
releases an electromagnetic = 1.988 × 1.66 × 10−27
= 3.300 × 10−27 kg
wave.

508

Physics SPM  Answers  

(ii) E = mc2 (e) Aspect Reason

= 3.300 × 10−27 × (3.0 × 108)2 Short half-life Unstable element will not stay
long in plant.
= 2.970 × 10−10 J

= 2.97 × 10–10 Gamma ray High penetration power
1.6 × 10–29
= 1.86 × 1019 eV
Low radiation rate No side effect to human body

(e) Aspect Reason High boiling point The material maintains in liquid

Short half-life Radioactive material activity is Energy released Not damaging the tree and
rapidly weakening
is lower human body.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
Type of radiation High penetrating power 14. (a) Activity represent decay rate.
is gamma ray (b) (i) Half-life of P is shorter than Q.
(ii) The gradient of decay curve of P is higher than Q.
State of matter is Easy to dilute in water (iii) The rate of decay of P is faster than Q.
liquid (c) – The shorter the half-life, the higher the gradient of
decay curve.
Type of detector is High sensitivity – The higher the gradient of decay curve, the higher
GM-tube the rate of decay.
Because it lowers the activity (d) – After 5 days, bismuth-210 undergoes first half-life.
Radioisotope M faster, has high penetrating – The activity drops to 400 counts per minutes.
power, easy to dilute in water – Bismut-210 undergoes second half-life.
and high sensitivity. – After 10 days, the activity is reduced half to become
200 counts per minutes.
12. (a) Aspect Reason
– Able to produce higher
Type of reaction is (e) Aspect Reason
fusion energy.
– Can be found anywhere Unstable element • Produced radiation that can
Fuel resource is be detected.
Hydrogen – Low maintenance
– Provide large quantity of Beta particle • High ionisation power.
Long lifetime
water in order to cold down Long half-life • Provide enough time to detect
Near with sea the steam. liquid waste flow.

(b) (i) m = (2.01410 + 3.01605) − (4.00260 + 1.00867) Low penetrating • Able to control the release of
power the ray.

= 0.01888 u No radioactive • Safe to environment.
= 0.01888 × 1.66 × 10−29
= 3.13 × 10−29 kg waste is produced
(ii) E = mc2 = 3.13 × 10−29 × (3.0 × 108)2
= 2.82 × 10−12 J 15. (a) Mass defect is the loss of mass after a nuclear
reaction.
= 2.82 × 10–12
1.6 × 10–29 (b) – The proton number of radium-226 is larger than
radon-222.
= 1.76 ×107 eV
– The nucleon number of radium-226 is larger than
(c) Disadvantage Way to overcome radon-222.

The waste is hot and Bury at abandon – The mass of radium-226 is larger than radon-222.
radioactive. location. – The larger the number of nucleons, the larger the

Risk of destruction of Do geological research mass of the element.
nuclear power stations on location selection (c) m = 226.025 – (222.018 + 4.003) = 0.004 u
due to natural disaster before building a nuclear (d) – A neutron is bombarded to a nucleus and it
power system.
becomes unstable.
Section C – Nuclear fission is occurred.
- The mass is decreased after the reaction and
13. (a) The half-life of a radioactive sample is the time taken
for the number of undecayed nuclei in the sample to changed to energy form.
be reduced to half of its original number.
(e) Aspect Reason
(b) (i) – Initial decay activity of source A is the same
with source B. Concrete wall Prevent radiation escapes
to surrounding.
– Half-life of source A is longer than source B.
– Rate of decay for source A is lower than source Use boron control rod Absorb excess neutrons.

B. High specific heat More heat can be
(ii) The higher the half-life, the lower the rate of decay. capacity transferred at one time.
(c) Source A = 30 years,; Source B = 3 years,
(d) (i) Source A, because it has longer half-life. Heat transfer from Turbine is not exposed
(ii) Alpha gas to water to rotate to extremely high
the turbine temperature.

Low gas density Easy to channel

509

  Physics SPM  Answers

7Chapter Quantum Physics (iii) The intensity of light does not
affect the number of electrons
= Checkpoint 7.1 emitted. If the incident light has
a low intensity, the metal surface
h h must be exposed continuously
l l to obtain enough energy to emit
electrons

Q1 Using p = mv ; p= → mv = Checkpoint 7.3

l = h 6.63 × 10–34 Q1 (a) Yes. The number of electrons
mv = 9.1 × 10–31 × 5.0 × 106 released depends on the
number of photons incident on
it.

(b) Not necessarily. The energy
(not the number) of the
emitted electrons depends on
the frequency of the incident
photons. For example, bright
blue light (high intensity) can
emit more electrons at lower
energies than low light intensity
light.
= 1.46 × 10−10 mPenerbitan Pelangi Sdn Bhd. All Rights Reserved.

Q2 (a) E = hf = 6.63 × 10−34 × 6.2 × 1014 = 4.1 × 10−19 J

E= hc →l= hc = 6.63 × 1034 × 3.0 × 108
l E 4.1 × 10–19

= 4.8 × 10–7 m

Alternative: can use l = c = 3.0 × 108
f 6.2 × 1014

= 4.8 × 10−7 m

(b) Using f = c = 3.0 × 108 Q2 (a) The three properties are:
l 400 × 10–9

= 7.5 × 1014 Hz • Photons are either quanta or

Energy, E = hf = 6.63 × 10–34 × 7.5 × 1014 discrete energy carriers.

= 4.97 × 10–19 J • Photon energy is proportional

nhc El to frequency.
l hc
Q3 (a) E = → n = • Photons transfer all of their

1.0 × 103 × 10.0 × 10–6 energy to the electrons they
6.63 × 10–34 × 3.0 × 108
= interact with.

(b) Einstein’s Photoelectric equation
1
= 5.0 × 1022 2 mv2­max = hf – W

(b) E = nhc = 1.5 × 1013 × 6.63 × 10–34 × 3.0 × 108 Graph of K.E.max against frequency
l 3.0 × 10–12

= 0.9945 J

Q4 P = nhf – For a given metal, the 1 mv2max
n = minimum frequency of the 2
Ptt incident light below which
hf = 6.63 × 20 × 1 × 1014 no photoelectron emission
10–34 × 5 occurs. This frequency is
known as the threshold
= 6.0 × 1019 frequency.

Q5 (a) P= h = 6.63 × 10–34 – Above the threshold
l 650 × 10–9 frequency, the maximum
kinetic energy of the emitted
= 1.02 × 10−27 kg m s−1 photon does not depend on
the intensity of the light but
(b) E = hf = hc = pc depends on the frequency or f0 f
l wavelength of the light.
hc
= 1.02 × 10−27 × 3.0 × 108 Q2 The above observations cannot be Q3 (a) E = l = 6.63 × 10−34 × 3.0 × 108
= 3.06 × 10−19 J explained by wave theory because: 420 × 10–9

Checkpoint 7.2 According to wave theory: = 4.74 × 10−19 J
(i) Light waves of any frequency
Q1 (a) When light with a suitable = 4.74 × 10−19 = 2.96 eV
frequency is incident on a should emit electrons, provided 1.6 × 10–19
metal surface, electrons are that a certain intensity is hc
emitted from the metal surface. maintained. (b) K.E.max = l – W
This phenomenon is called (ii) The kinetic energy of
photoelectric effect. photoelectrons increases with = 2.96 − 2.71 = 0.25 eV
light intensity, light with high
(b) Three features of photoelectric intensity will emit electrons Q4 (a) W = hfo = hc
effects that cannot be explained with high kinetic energy. The lmax
by wave theory of light are: frequency of light does not hc
– For a given metal and the affect the kinetic energy of the lmax = W
frequency of incident light, electrons emitted.
the number of photoelectrons = 6.63 × 10–34 × 3 × 108
emitted at any given time is 3 × 1.6 × 10–19
proportional to the intensity of
the incident light. = 4.14 × 10−7

= 414 nm

510

Physics SPM  Answers  

(b) K.E.max = hc – W (c) (b) (i)
l Photoelectric current
1 Graph of K.E against frequency
2 max

= 6.63 × 10–34 × 3 × 108 mv2
350 × 10–9 max

– 3 × 1.6 × 10–19 Saturation current

= 81.8.62 × 10–20 = 0.55 eV Stopping
× 10–19 potential
f0 Frequency / f
1 mv2 × 10–19 W0
 (c) 2 = 0.88

Penerbitan Pelangi Sdn Bhd. All Rights Reserved. v =2 × 0.88 × 10–19 Vs 0 V
9.11 × 10–31

= 4.4 × 105 m s–1 3. (a) From de Broglie relation, (ii) At stopping potential, the
photocurrent just equal
SPM Practice 7 ve = vp = v to zero. The electrons
Objective Questions with the maximum kinetic
l = h energy just fail to reach the
mv anode. This shows that the
maximum kinetic energy
1. A 2. C 3. C 4. C 5. B The de Broglie wavelength is of the photoelectrons is
6. A 7. A 8. D 9. C 10. A just equal to the electric
11. A 12. B 13. D 14. C 15. D inversely proportional to the potential energy of the
16. C 17. C 18. C 19. B 20. B photoelectrons. Hence, the
21. B 22. B 23. B 24. A 25. A mass of the particle. maximum kinetic energy of
a photoelectron is equal to
Since the mass of electron is eVs.

smaller than the mass of proton, (iii) The number of electrons
emitted per second from the
therefore, the electron has a cathode is a constant. When
the total number of electrons
Subjective Questions longer de Broglie wavelength. emitted per second equal
(b) ve = vp = v to the total number of
electrons reaching the
Section A Therefore, the ratio of the de anode per second, the
current cannot increase
Broglie wavelength is further even though the
  h potential difference applied
1. (a) Photon mev is increased. Hence, the
(b) The brighter light beam will eject le = h current reaches a saturated
a greater number of electrons. lp  mpv value.
This is because the number of
photoelectrons ejected depends = mp = 1.67 × 10–27 = 1833
on the number of incident me 9.11 × 10–31
photons.
(c) Not necessarily. The energy (not Section C
the number) of ejected electrons
depends on the frequency of the 4. (a) (i) Photoelectric effect
incident photons.
(ii) The minimum amount of

energy required to emit an

2. (a) The three characteristics are: electron from the surface of

• Photon carries quanta or the metal.

discrete energy. (iii) Photon energy, SPM MODEL PAPER

• Photon energy is hc 6.63 × 10–34 ×
l =
proportional to frequency. E = hf = 3 × 108
330 × 10–9
• Photon transfers all its PAPER 1
1. B 2. D 3. B 4. A 5. B
energy to the electron which = 6.03 × 10–19 J 6. C 7. D 8. A 9. D 10. D
11. A 12. C 13. D 14. B 15. C
1 interact with it. = 6.03 × 10–19 16. D 17. A 18. B 19. D 20. A
2 1.602 × 10–19 21. C 22. D 23. C 24. A 25. C
(b) mvmax2 = hf – W0 26. D 27. A 28. C 29. C 30. B
31. B 32. A 33. A 34. B 35. C
= 3.76 eV 36. A 37. B 38. D 39. C 40. D
e21nemrgvmyaxo2f (iv) K.E. = hf − WO
is the maximum kinetic PAPER 2
the emitted electrons, = 6.03 × 10−19 – 2.2 ×
Section A
where m is the mass of electron 1.602 × 10−19
1. (a) Resultant force
and vmax is the maximum velocity = 2.51 × 10−19 J = 100 kN cos 20o + 100 kN cos 20o

of the ejected electrons. (v) Assume that all the = 188 kN

hf is the photon energy where electrical energy is The direction of the resultant
h is the Planck’s constant force is 20° from the 100 kN
and f is the frequency of the converted into photon force follow the direction of
motion of the ship.
energy. N(thf) = N
E t (b) The ship is moving in a straight
electromagnetic waves. P = t = (hf) line at constant speed because

W0 is the work function, the = nhf
minimum energy to eject an
Therefore, number of

photons per second,
electron from the metal. P 60
n= hf = 6.03 × 10–19

= 9.95 × 1019 photon s–1.

511

  Physics SPM  Answers

the resultant force acting on it is (ii) ε = IR = 0.24 × (40 + 10) = 12 V suitable because the scale is
zero. The resultant force acting (b) Voltmeter 0 – 2 V is not suitable too large, and 10 V is difficult to
on a ship is balanced by the read.
drag (friction). because the scale is too small Voltmeter 0 – 20 V is suitable
and not enough to measure 10 V. and 10 V is in the middle of the
2. (a) The specific latent heat is Voltmeter 0 – 200 V is not scale.
defined as the amount of heat
energy required to change an (c) Quantity Increase, decrease Brief explanation
object from solid state into a or stay the same
liquid state without change in after resistor R is The resistance is
temperature. smaller.
connected
(b) (i) Energy needed, The potential difference
Q = mL increases as the current
= (5.0 × 10−3 kg)(3.3 × increases.

105 J kg−1) The potential difference
= 1650 J across the the resistor
of 10 Ω increases
(ii) Ice acquires heat energy
from the air through
conduction through glass
funnel.

3. (a)
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. The current in the resistor Increase
of 10 Ω

The potential difference Increase
across the resistor of
10 Ω

The potential difference Decrease
across the resistor of
40 Ω

Object Lens

6. (a) • Located at Geostationary (c) (i) V = IR = 0.045 mA × 1.2 Ω
= 0.045 × 10−3 A × 1.2 Ω
P Image Orbit. = 5.4 × 10−5 V = 0.054 mV

• Orbit round the Earth in the (ii) Q = It = (0.045 × 10−3)(0.14)
= 6.3 × 10−6 C
1 cm direction same as the direction
1 cm
of rotation of the Earth.

• Always located above the 8. (a) (i) 2

(b) Real / Inverted same geographical position on (ii) Neutron

(c) Linear magnification the Earth’s surface. (iii) 2
v hi
= u= ho = 3.4 = 1.7 • The orbital period is the same (iv) 4
2
as the period of rotation of the (b) Dm = E = 2.82 × 10–12
c2 (3.0 × 108)2
(d) Projector Earth, which is 24 hours.

a = v – u 25 – 0 (any two answers) = 3.13 × 10−29 kg
t 14
4. (a) (i) = (b) (i) Force of gravity (c) Nuclei repel each other and

= 1.79 m s−1 (ii) The satellite will move in required high kinetic energy

(ii) straight line tangent to its to overcome the repulsive

orbit. force. Therefore, very high

60 (c) g = k → k = gR2 temperature is needed for nuclei
Speed / m s–1 R2
gbRb2 = gkRk2 to acquired high kinetic energy.
40
Rb2 (d) (i) Half-life is the time required
Rk2
gk = × gb = 6400 × 10 for a number of active
19200
20 nucleus reduce to half of its

= 3.33 N kg−1 initial value.

Fck = Wk = mk gk (ii)
= 900 × 3.33
00 20 40 60 80 16 000 ⎯→ 8 000 ⎯→ 4 000 ⎯→ 2000
Time / s = 2997 N 12 yrs 12 yrs 12 yrs

(b) When the brake is applied, the (d) (i) The orbital period of satellite ⎯→ 1000 The number of atoms
K < the orbital period of 12 yrs present is 1 000.
satellite L

(ii) The Kepler’s third Law
car slows down but the driver’s

body still moves forward due

to its inertia. The seat belt Section B

produces the force to resist 7. (a) (i) There are no free electrons 9. (a) (i) 0.84 N
in the plastic. (ii)
the driver moving forward. The
(ii) Aluminium is not a magnetic Extension / cm
inertial of the bag will cause it material. Iron is a soft
magnetic material and
to slide forward as the frictional cannot make a permanent
magnet.
force on the seat is not strong 12
(b) When magnet is moving, the
enough from preventing it to magnetic field lines are cut by 10
solenoid wire. The changing 8X
move forward. magnetic fields causing induced
e.m.f. in the solenoid and thus 6
5. (a) (i) Current, I = V induced current flows in the 4.8
R solenoid.
4
= 9.6
40 2

= 0.24 A 0
0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
Load / N

512

(b) (i) P = hρg = 0.035 × 1000 × 10 (ii) K.E.after : K.E.before = K.E.after Physics SPM  Answers  
K.E.before
= 350 Pa (ii) • The waves pattern before
passing through the gap
(ii) F = PA = 350(0.04 × 0.04) 1 (mx + my) × v2 is parallel.
2
= 0.56 N 1 • Waves after passing
2 (mx) × (5v)2 through the narrow gap
(iii) W = 0.84 + 0.56 = 1.4 N are circular.

(c) (i) The atmospheric pressure = mx + 4mx × v2 • Waves after passing
exerts a downward force mx 25v2 through the wide gap are
cancels out the extra parallel but curve at two
upward force. = 5 × 1 = 1 = 0.2 sides.
25 5
(ii) Vector has direction and • The wavelength before
magnitude and scalar has and after passing through
magnitude only. the gaps is the same.

(d) (i) The total force acted on the • Waves that pass through
plastic block is a constant the gap spread out at the
and is given by edge of the gap causing
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. (iii) For a perfectly elastic the waves to diffract. The
Tension of spring, FE + buoyant force due collision, there is no lost in physic concept involved is
to water, FB = weight,W kinetic energy. diffraction.

Therefore, as more and Therefore, the ratio = 1 (iii) The wave phenomenon is
more plastic block immersed (iv) From t = 0 to t = 20 ms, the diffraction of waves
in water, the buoyant force
increases and the tension in resultant force acting on (b) (i) To build retaining walls, the
spring decreases. block X is zero because the concrete structures must
block moving at constant be solid, rigid and can
(ii) speed. withstand the impact of the
From t = 20 ms to t = 40 ms, waves. The wall must be
Extension, e / cm the momentum of X strong and not easily crack.
decreases uniformly
showing that the magnitude The wall must have many
of resultant force acting narrow gaps to reduce the
on it is a constant and the force from the incoming
direction is opposing the waves.
momentum.
From t = 40 ms to t = 60 ms, The narrow gaps allow
the resultant force acting on waves to diffract and spread
block X is zero because the out to reduce the amplitude
block moving at constant of the waves causes the
speed. rough sea to calm down.
(v)
The wall must be high to
Momentum reduce overspill of the
waves.
d
Length of block, d / cm (ii) The new jetty should be
built in the region of the bay.
(iii) Archimedes’ principle.
The bay is deeper and
10. (a) (i) The weight of the steel ball waves in the bay are usually
calmer.
is very large compared to
(iii) The wave energy diverges
the force of air resistance at the bay and spread out to
larger area.
acting on the ball. 00 10 20 30 40 50 6t 0/ ms
1 Section C The amplitude and size of
(ii) s = ut + 2 at2 11. (a) (i) The water waves formed act the waves are smaller at the
bay.
(0.280) = 0(t) + 1 (9.8)t2 like lenses that converge
2 and diverge the parallel The wave energy converges
t = 0.24 s light rays from the lamp as at the cape.
(iii) s = 0.280 + 0.080 = 0.360 m shown in the figure.
1 Light from lamp 513
0.360 = 0(t) + 2 (9.8)(t2)
t = 0.27 s Water wave
Dt = 0.27 − 0.24
Light converge Light diverge to
= 0.03 s to from bright from dark bands
bands
(iv) The resultant force acting

on the light ball is less

as compare to steel ball.

Therefore, the velocity of

the ball at beam of light is

less and so the time interval

is longer.

(b) (i) mx × 5v = (mx + my) × v
5mx = mx + my
4mmyx my
mx = 4
=

SPM CC038542 FOCUS SPM
FORM
4∙5
KSSM
PHYSICS Penerbitan Pelangi Sdn Bhd. All Rights Reserved.

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üSPM Tips
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REINFORCEMENT Sijil Pelajaran Malaysia (SPM) requirements.
& ASSESSMENT
A great resource for every student indeed!
üSPM Practices üSPM Model Paper
üCheckpoint üComplete Answers REVISION
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EXTRA FEATURES
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