FOCUS SPM Physics (2023)

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SPM
CC038543



FORM KSSM FOCUS
4∙5
Physics SPM


Physics SPM





FOCUS SPM KSSM Form 4 • 5 – a complete and precise series of reference books
with special features to enhance students’ learning as a whole. FORM
This series covers the latest Kurikulum Standard Sekolah Menengah (KSSM) and
integrates Sijil Pelajaran Malaysia (SPM) requirements. (Dual Language Programme) 4∙5 KSSM
A great resource for every student indeed!


REVISION REINFORCEMENT EXTRA Physics • Yew Kok Leh
› i-Study SPM & ASSESSMENT FEATURES • Chang See Leong
› Comprehensive Notes › SPM Practices › Examples • Abd Halim Bin Jama'in
› Concept Maps › SPM Model Paper › SPM Highlights
› Activities & Experiments › Checkpoint › Digital Resources QR Codes
› SPM Tips
› Complete Answers




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CC038543
ISBN: 978-629-7537-81-8
Based on the
LATEST SPM FORMAT
PELANGI

Format: 190mm X 260mm TPTV Focus SPM 2023 Phy BI version _pgi CRC












Physics SPM





FORM
4∙5
(Dual Language Programme) KSSM





• Yew Kok Leh
• Chang See Leong
• Abd Halim Bin Jama’in

















© Penerbitan Pelangi Sdn. Bhd. 2023
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ISBN: 978-629-7537-81-8
eISBN: 978-629-7537-82-5 (eBook)


First Published 2023




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Exclusive Features of This Book







Theme 4: Modern Physics Form 5
Chapter
7 Quantum Physics
Chapter Focus CHAPTER FOCUS Physics SPM Chapter 1 Measurement Bigger Standard Form SPM Tips
SPM Tips
Prefixes
lists the learning 7.1 Quantum Theory of 10 9 10 6 mega (M) 1. The distance of Pluto from the Earth is about points out
giga (G)
Light
7.2 Photoelectric Effect
6 000 000 000 000 m and the radius of a
7.3 Einstein’s Photoelectric
hydrogen atom is about 0.000 000 000 05 m.
objectives for an Theory FORM 10 3 kilo (k) small and a simpler way of expressing them important tips for
These quantities are either too large or too
scientific notation.
is by using standard form of representation or
overview of the 4 10 –3 10 –1 10 –2 deci (d) students to take
centi (c)
milli (m)
chapter. 10 –6 micro (μ) Electron note of.
10 –9
nano (n) Smaller
0.000 000 000 05 m
Proton
EXAMPLE 1.1
Hydrogen atom
Convert 2. In a standard form or scientific notation, a
(a) 0.0042 kg to g
(b) 5 800 g to kg numerical magnitude can be written as:
(c) 10 cm to m
Solution A × 10 n , where 1 < A  10 and n is an
integer
10 9
giga (G) Hence, the distance of Pluto from the Earth
10 6 can be written as 6 × 10 12 m and the radius of
FORM
mega (M) a hydrogen atom as 5 × 10 –11 m.
5
• What are light and quantum? (a) 10 3 10 3 kilo (k) EXAMPLE 1.2
• How is light associated with quantum matter?
• Why is quantum theory of light important in modern physics?
• What do quantum energy and photoelectric effects mean?
Access to
SPM
i-STUDI
i-STUDI
(c) 10 –2
10 –2 10 –1 deci (d) (b) 10 –3 For each of the following, express the magnitude
using scientific notation.
10 –3
centi (c)
milli (m) (a) The length of a virus = 0.000 000 08 m
10 –6 (b) The mass of a ship = 75 000 000 kg
micro (µ) Solution
SPM Highlights 454 (a) 0.0042 kg = 0.0042 × 10 3 g = 4.2 g (a) The length of a virus Example
10 –9

nano (n)
= 0.000 000 08 m

= 8 × 10 –8 m
provides exposure (b) 5 800 g = 5 800 × 10 –3 kg = 5.8 kg (b) The mass of a ship provides solution

(c) 10 cm = 10 × 10 –2 m = 0.1 m
= 75 000 000 kg

= 7.5 × 10 7 kg
to the frequently- Physics SPM Chapter 5 Electronics (a) Label the terminal P, Q and R. for example of
(b) Name the type of transistor shown on Figure
5.55.
Highlights
tested questions SPM a transistor circuit. The circuit is used to control a S2 Figure 5.56 shows a simple transistor circuit. 4
Figure 5.54 shows an NTC thermistor connected in
that appear in the 240 V air conditioner. Relay 240 V a.u R M T V C questions in the
~
V S
actual exam. Thermistor R R Resistor condition Air 9 V (a) Name the type of transistor T. Physics SPM Chapter 4 Heat subtopics.
Figure 5.56
Resistor
(c) Draw symbol dry cell at V s and V c so that the
Figure 5.54
emitter circuit.
Which of these situations can cause the air condition (b) On the figure above, label the base circuit and SPM Practice 4
bulb M can lights up.
(d) What is the function of resistor R?
A Low temperature thermistor and low base voltage.
to turn on? S3 Azmin designed a transistor circuit to turn on a fan Objective Questions A Specific heat capacity
B Low temperature thermistor and high base
1.
C High temperature thermistor and high base the circuit for the electronic control system. block from a cold beaker B Specific latent heat
voltage.
in the classroom automatically. Figure 5.57 shows A student transfers a metal
C Thermal equilibrium
D High temperature thermistor and low base X FORM temperature. 4. Alisa’s mother chose a
D Heat convection
voltage.
P into beaker Q at room
voltage. Relay a.c. Beaker P Beaker Q cooking pan as shown in
When the temperature of the thermistor is high (hot P Y R 1 Transistor Fan 4 9 V Water at 0°C Metal Water at 28°C about the characteristics of 200 ml of hot Figure 3 SPM Practice
Examiner’s Tips
Figure 2, and asked Alice
condition), the resistance of thermistor becomes
low compared to resistor R. Then the base voltage
water at 80°C
of the pan.
the material used as the base
becomes high causing the current to flow through the
block
base of transistor to turn it on. The collector current
10 k
A 20 o C
Figure 1
B 30 o C
Checkpoint flows through the relay to turn on the air condition. (a) Name the component P. Which of the following 7. A milk seller boils his milk provides ample
Answer: C
C 40 o C
D 60 o C
statements is true about a
Z
metal block?
5 .2
Figure 5.57
I
Checkpoint!
The temperature of metal
block will rise to 28 o C.
in a pot. The temperature of
provides questions S1 Figure 5.55 shows the symbol of a transistor. (b) What is the function of P? Explain your Figure 2 Base of pan of milk is 25 kg, what is the questions to test
the milk is raised from 10°C
II The temperature of metal
P : ________________
(c) When the classroom becomes warm, the
answer.
block increases but is
to 130° C with 12 000 kJ
lower than 28 o C.
(i) State the potential difference across XZ.
of heat energy. If the mass
resistance of P decreases to 6 kΩ. III The metal block releases
heat in beaker Q.
Which of the following is the
milk?
specific heat capacity of the
to test students’ FORM R : ________________ Q : ________________ (ii) Calculate the potential difference across most suitable characteristic of A 4 000 J kg –1 °C –1 students’ mastery
A I and II only
the themistor.
B II and III only
the material used as base of
the pan?
C I, II and III
B 8 000 J kg –1 °C –1
A Low melting point
C 12 000 J kg –1 °C –1
understanding of 5 Figure 5.55 2. An experiment is conducted B High density 8. Figure 4 shows a car of the chapter.
to calibrate a mercury
D 16 000 J kg –1 °C –1
C Low specific heat capacity
thermometer in an unmarked
D Low thermal conductivity
glass. The length of the
mercury column when
5. In an experiment, Ramly
radiator as a cooling agent.
found that 100 g of a radiator. Water is used in the
the subtopic and • put in boiling water is temperature from 30°C to
liquid requires 12 600 J of
• immersed in melting
ice is 4.5 cm,
thermal energy to raise the
22.0 cm,
60°C. The heat capacity of
reinforce learning. • placed in warm water B 420 J °C –1 .
100 g of liquid is
A 42 J °C –1 .
is 11.5 cm.
422 the warm water? C 4 200 J °C –1 . Figure 4
What is the temperature of
D 42 000 J °C –1 .
A 30 o C 6. Figure 3 shows a glass
B 40 o C
Water is used as cooling
agent because
containing 200 ml of hot
C 50 o C
D 55 o C water at 80 o C. A student A water evaporates easily.
filled in 200 ml of cold water
B water is not corrosive.
at 0 o C and stirred it with
3. Which of the following is the
a spoon. Assuming that C water has a high density.
concept used in measuring
D water can absorb a lot of
there is no heat loss to the
heat.
the temperature of a human
surrounding, what is the final
body using a thermometer?
temperature of the water in
the glass?
Physics SPM Chapter 1 Force and Motion II 134
Experiment 1.1
Aim: To find the relationship between the force and the extension of a spring
Inference: The force applied on a spring affects the extension of the spring.
Hypothesis: The larger the applied force, the larger the extension of the spring
(a) Manipulated variable: Force, F (slotted weight added)
Variable:
(b) Responding variable: Extension of the spring, x
(c) Constant variable: Length and type of spring
Steel spring (20 cm long), five 50 g slotted mass, slotted mass hanger, metre rule, retort stand with clamp.
Apparatus and Materials:
Metre rule
(a) The apparatus is set up as shown in Figure SPM Model Paper SPM Model
Procedure:
Activity/ (b) The mass hanger is fixed at the end of the Spring PAPER 1 Paper & Answers
1.42.
spring. The initial length of the spring, l o is
measured using a metre ruler.
and the new length, l of the spring is measured.
Experiment (c) A slotted mass, m = 50 g is slotted on the hanger of spring  o Extension x 1. Which of the following One hour and fiften minutes prepares students
The extension of the spring is calculated as,
Answer all questions.
(d) Step 3 is repeated by adding slotted mass of Original length
C
x = l – l o .
as impulse?
quantities has the same unit
Mass holder
Velocity
A Work
100 g, 150 g, 200 g and 250 g.
helps students (e) The weight of the slotted mass is calculated.  Weight Slotted weight D Kinetic energy 0 Velocity C for the SPM with
B Momentum
The weight represents the force, F act on the
C Weight
spring. [Using g = 10 N kg –1 ]
F = mg
to master hands- (f) The results are tabulated as below. Length of spring, Figure 1.42 x = (l – l o ) / cm 2. A car moves from a rest D Velocity 0 Time 0 0 A B D Time the actual exam
state with an acceleration of
Extension
1.5 m s –2 for 8 s. What is the
Initial length of spring, l o = 25 cm
Result:
velocity after 8 s?
A 6.0 m s –1 C 10.0 m s –1
on scientific Slotted mass added, Weight of slotted l / cm 2.5 3. Which of the following is a 0 0 Time 7. Figure 2 shows a bird in format.
Figure 1
B 8.0 m s –1 D 12.0 m s –1
mass, F / N
m / kg
27.5
flight. The bird is flying in a
vector quantity?
0.5
horizontal direction to the
A Speed
B Weight
knowledge and 0.05 0.10 1.0 1.5 30.0 33.0 5.0 8.0 10.0 4. An object initially moves with 5. A student jumps down from right. In which direction does
C Energy
air resistance act on the bird?
a height. He bends his legs
D Density
upon landing. Which of
correct?
the following statements is
A The student wanted to
a constant acceleration and
skills. FORM 5 0.15 0.20 2.0 2.5 35.0 38.0 13.0 Which of the following graphs B The student wanted to
then with constant velocity.
reduce the impulse on his
feet.
shows this motion?
A
0.25
reduce the impulsive
Velocity
A graph of extension, x against force F is plotted as shown. C The student wanted to Figure 2
force that acted on his
feet.
increase his velocity upon
0 landing.
0 D The student wanted to
Time 8. The escape velocity, v of an
B reduce his time of impact object on the surface of the
floor. A the mass of the object.
Velocity between his feet and the Earth does not depends on
266 6. A car is travelling along a B the mass of the Earth.
C the object’s distance from
straight horizontal road. The
the centre of the Earth.
speed-time graph is shown
0 in Figure 1. In the labelled D the Universal
0
Time part of the journey, which one Gravitational Constant.
indicates the resultant force 9. The value of gravitational
acting on the car is zero?
acceleration, g, on the
surface of planet A and
planet B is the same. Which
of the following quantities is
the same for planet A and
planet B?
1
ii
00 Prelims Physics 2023.indd 2 10/03/2023 10:35 AM

CONTENTS











FORM 4



Theme 1 Elementary Physics Theme 3 Heat


Chapter Chapter Heat 93
1 Measurement 1 4

1.1 Physical Quantities 2 4.1 Thermal Equilibrium 94
1.2 Scientific Investigation 6 4.2 Specific Heat Capacity 98
4.3 Specific Latent Heat 110
SPM Practice 1 15
4.4 Gas Laws 119
SPM Practice 4 134

Theme 2 Newtonian Mechanics
Theme 4 Waves, Light and Optics

Chapter Chapter
2 Force and Motion I 19 5 Waves 140

2.1 Linear Motion 20 5.1 Fundamentals of Waves 141
2.2 Linear Motion Graphs 28 5.2 Damping and Resonance 148
2.3 Free Fall Motion 33 5.3 Reflection of Waves 151
2.4 Inertia 36 5.4 Refraction of Waves 155
2.5 Momentum 41 5.5 Diffraction of Waves 160
2.6 Force 49 5.6 Interference of Waves 165
5.7 Electromagnetic Waves 174
2.7 Impulse and Impulsive Force 54 SPM Practice 5 179
2.8 Weight 58
SPM Practice 2 60 Chapter
6 Light and Optics 187

6.1 Refraction of Light 188
Chapter
3 Gravitation 67 6.2 Total Internal Reflection 200
6.3 Image Formation by Lenses 207

3.1 Newton’s Law of Universal Gravitation 68 6.4 Thin Lens Formula 212
3.2 Kepler’s Laws 80 6.5 Optical Instruments 217
3.3 Man-made Satellites 83 6.6 Image Formation by Spherical

SPM Practice 3 88 Mirrors 222
SPM Practice 6 230



iii





00 Prelims Physics 2023.indd 3 10/03/2023 10:35 AM

FORM 5

4.2 Electromagnetic Induction 375
Theme 1 Newtonian Mechanics
4.3 Transformer 382
Chapter SPM Practice 4 388
1 Force and Motion II 235


1.1 Resultant Force 236
1.2 Resolution of Forces 252 Theme 3 Applied Physics
1.3 Forces in Equilibrium 256 Chapter
1.4 Elasticity 264 5 Electronics 397
SPM Practice 1 274
5.1 Electron 398
5.2 Semiconductor Diode 404
Chapter
2 Pressure 281 5.3 Transistor 413
SPM Practice 5 424
2.1 Pressure in Liquids 281
2.2 Atmospheric Pressure 289
2.3 Gas Pressure 293
Theme 4 Modern Physics
2.4 Pascal’s Principle 295
2.5 Archimedes’ Principle 297 Chapter
2.6 Bernoulli’s Principle 304 6 Nuclear Physics 431
SPM Practice 2 310 6.1 Radioactive Decay 432
6.2 Nuclear Energy 436
Theme 2 Electricity and Electromagnetism
SPM Practice 6 444

Chapter
3 Electricity 317 Chapter
7 Quantum Physics 454

3.1 Current and Potential Difference 318
7.1 Quantum Theory of Light 455
3.2 Resistance 329 7.2 Photoelectric Effect 460
3.3 Electromotive Force (e.m.f) and
7.3 Einstein’s Photoelectric Theory 463
Internal Resistance 344
3.4 Electrical Energy and Power 353 SPM Practice 7 470
SPM Practice 3 360 Answers


Chapter
4 Electromagnetism 366 SPM Model Paper & Answers

4.1 Force on a Current-carrying https://plus.pelangibooks.com/Resources/FocusSPM/
Conductor in a Magnetic Field 367 Physics/ModelPaperandAnswer.pdf




iv





00 Prelims Physics 2023.indd 4 10/03/2023 10:35 AM

Important Formulae





 l Gm m
Period of oscillation of pendulum, T = 2π  g Gravitational force, F = r 1 2 2


Velocity, Displacement,
v = u + at s = 1 (u + v)t
v = u + 2as 2 1
2
2
s = ut + 2 at 2
GM mv 2
Gravitational acceleration, g = Centripetal force, F =
r 2 c r

Centripetal acceleration, Kepler’s III law:
v 2 4π 2
2
3
2
a = T =   r → T ∝ r 3
c r GM
T 2 r 3
T 1 2 = r 3 1
2 2
 1
GM
Orbital velocity, v =  r Kinetic energy, KE = 2 mv 2
2GM
Escape velocity, v =   Gravitational potential energy, U = – GMm
r
r
Q Q
Specific heat capacity, c = mDq Specific latent heat, l = m

1 Charles’ law: V ∝ T


Boyle’s law: P ∝ V
Speed of waves, v = fl Refraction of wave:
v v
1 = 2
l l
1 2
Snell’s law: Critical angle, c :
n sin q = n sin q 1
1 1 2 2 n = sin c
sin i
n =
sin r
Interference of waves: Lens formula,
ax 1 1 1
l = D f = + v
u







v





00 Prelims Physics 2023.indd 5 10/03/2023 10:35 AM

Hooke’s law, Elastic potential energy,
1
F = kx E = Fx
2
p
1
E = kx 2
p 2
Pressure in Liquids Pascal’s principle:
P = hρg Force multiplier,
F 1 F 2
A = A
1 2
Archimedes’s principle: Bernoulli’s principle:
Buoyant force, Lift force, F = (P – P )A
2
1
F = ρVg
B
Electric field strength, Electric current,
F Q
E = q I = t


Potential difference, Resistance,
W ρl
V = Q R = A


Ohm’s law, Total resistance in a circuit, R :
T
V series: R = R + R + R +...
R = I T 1 2 3
parallel: 1 = 1 + 1 + 1 + ...
R R R R
T 1 2 3
Electromotive force, Electrical power,
ε = V + Ir P = VI = I R = V 2
2
R
Efficiency of transformer, Kinetic energy of electron,
1
Output power E = eV = mv 2 max
k
η = Input power × 100% 2
V I v is the maximum velocity of an electron.
max
= s s × 100%
V I
p p
Nuclear energy, Quantum energy,

E = mc 2 E = hf

Properties of wave-particle duality, Einstein’s photoelectric theory,
1
Momentum, hf = W + mv 2
p = mv (Particle property) 2
h
p = (Wave property)
l


vi





00 Prelims Physics 2023.indd 6 10/03/2023 10:35 AM

Theme 1: Elementary Physics Form 4

Chapter
1 Measurement










CHAPTER FOCUS



1.1 Physical Quantities FORM
1.2 Scientific Investigation 4














































Have you heard of the famous scientist Isaac Newton who sat under an apple tree? As a result
of an apple falling from the tree, he investigated the characteristics of objects falling towards
the surface of the Earth and eventually succeeded in discovering the theory of gravitational
force. Scientists always carry various scientific investigations to confirm certain scientific
theories, laws or principles. What is meant by a scientific investigation?

Access to
i-STUDI
i-STUDI SPM



1





01 FOC PHYSICS F4.indd 1 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement
7. We can represent a physical quantity by the
1.1 Physical Quantities symbol of the quantity, the numerical value
of the magnitude of the quantity and the unit

1. Physics is a branch of science concerning the of measurement of the quantity. For example,
study of natural phenomena like properties Figure 1.1 shows a footballer scoring a goal.
of matter and energy. Examples of natural The ball was kicked a distance of 8 m.
phenomena are:
(a) formation of rainbow in the sky




FORM

4
(b) lightning and thunder

8 m
Figure 1.1

(c) eclipse of sun and moon Symbol

Distance travelled by the ball, l = 8 m Unit

Numerical

(d) earthquake 8. There are two types of physical quantities, that
is, base quantities and derived quantities.
9. Base quantities are physical quantities that
cannot be defined in terms of other quantities.
Table 1.1 shows seven base quantities and their
respective S.I. units.
2. The word physics comes from the Latin word Table 1.1
physica meaning the science of natural things.
Up to the nineteenth century, physics was Base quantity Symbol S.I. Unit Symbol of
S.I. unit
called natural philosophy.
Length l metre m
3. Physics is based on experimental observations
and quantitative measurements. Physicists Mass m kilogram kg
always try to find the simplest explanation for Time t second s
a complex phenomenon. Temperature T kelvin K
4. Physical quantities are quantities that can be Electric current I ampere A
measured.
Light intensity I v candela cd
5. To describe a physical quantity, we first define Quantity of
the unit in which the measurement is made. matter n mole mol
There are many systems of units but the most
common system of units used by scientists is 10. Derived quantities are physical quantities
based on the metric system. derived from combinations of base quantities
6. The modernised version of the metric system is through multiplication or division or both
called International System of Units, officially multiplication and division.
abbreviated as SI.



2





01 FOC PHYSICS F4.indd 2 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement
11. Table 1.2 shows some derived quantities and their respective derived units.

Table 1.2
Derived quantity Symbol Relationship with base quantities Derived unit Unit in S.I. base unit
Area A Length × Length m 2 m × m = m 2

Volume V Length × Length × Length m 3 m × m × m = m 3
Mass kg
Density ρ Length × Length × Length kg m –3 m = kg m –3
3
Displacement m FORM
Velocity v Time m s –1 s = m s –1 4

Velocity m s –1
Acceleration a Time m s –2 s = m s –2

Force F Mass × Acceleration kg m s –2 kg × m s = kg m s –2
–2



Imperial Unit Prefix
1. Imperial unit was first used in Britain in the 1. Prefixes are used to simplify the description
year 1824 but is seldom being used today. of physical quantities that are either very big
Examples of imperial units are feet, inches, or very small in S.I. units.
yards, miles, gallon and psi. 2. Table 1.3 lists some commonly used S.I.
2. Compared with the S.I. unit, these units are prefixes and their multiplication factors.
more difficult to use.
Table 1.3
3. The S.I. unit which is based on the multiples Prefix Symbol Value
of 10 is easier to use than the imperial unit.
pico p 10 –12
How many
inches are in nano n 10 –9
3.5 yards?
micro µ 10 –6
It would be
easier if madam mili m 10 –3
use the metre
and centimetre centi c 10 –2
units.
deci d 10 –1
deca da 10 1
hecto h 10 2

kilo k 10 3
mega M 10 6
giga G 10 9
S.I. Unit tera T 10 12
1 m = 100 cm
Therefore, 3.5 m = 3.5 × 100 cm = 350 cm
Imperial Unit
1 yard = 36 inches
Therefore, 3.5 yards = 3.5 × 36 inches = 126 inches



3





01 FOC PHYSICS F4.indd 3 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

SPM Tips Standard Form
1. The distance of Pluto from the Earth is about
Prefixes Bigger 6 000 000 000 000 m and the radius of a
10 9 giga (G) hydrogen atom is about 0.000 000 000 05 m.
These quantities are either too large or too

10 6 mega (M) small and a simpler way of expressing them
is by using standard form of representation or
scientific notation.
10 3 kilo (k)



FORM
10 –1 deci (d)
10 –2 centi (c)
4
10 –3 milli (m)
Electron
10 –6 micro (μ)
0.000 000 000 05 m
10 –9 nano (n) Smaller
Proton


Hydrogen atom
2. In a standard form or scientific notation, a
EXAMPLE 1.1
numerical magnitude can be written as:
Convert
(a) 0.0042 kg to g A × 10 , where 1 < A  10 and n is an
n
(b) 5 800 g to kg integer
(c) 10 cm to m
Hence, the distance of Pluto from the Earth
Solution
can be written as 6 × 10 m and the radius of
12
10 9 giga (G) a hydrogen atom as 5 × 10 –11 m.
10 6 mega (M)
EXAMPLE 1.2
10 3 kilo (k)
(a) 10 3 (b) 10 –3 For each of the following, express the magnitude
using scientific notation.
(c) 10 –2 10 –1 deci (d) (a) The length of a virus = 0.000 000 08 m
10 –2 centi (c) (b) The mass of a ship = 75 000 000 kg
10 –3 milli (m)
Solution
10 –6 micro (µ) (a) The length of a virus
= 0.000 000 08 m
10 –9 nano (n) = 8 × 10 m
–8
(b) The mass of a ship
(a) 0.0042 kg = 0.0042 × 10 g = 4.2 g = 75 000 000 kg
3
–3
(b) 5 800 g = 5 800 × 10 kg = 5.8 kg = 7.5 × 10 kg
7
–2
(c) 10 cm = 10 × 10 m = 0.1 m





4





01 FOC PHYSICS F4.indd 4 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

Conversion of Units Involving Derived Scalar and Vector Quantities
Quantities
N
1. When converting units of derived quantities, W E
each of its base units involved must be
converted. The following example illustrates S
the conversion of derived units. 50 km

Figure 1.2
EXAMPLE 1.3
1. Physical quantities can be grouped into scalar FORM
Convert each of the following from one particular quantities and vector quantities.
unit to another and represent the quantity in 4
standard form. 2. Figure 1.2 shows a truck travelling a distance
(a) Convert the area of a button from 1.2 cm of 50 km in the eastward direction. We
2
into m . describe the journey of the truck by stating
2
(b) Convert the volume of a water tank from the magnitude and direction of its travel:
2.5 m into cm . (a) The magnitude is 50 km.
3
3
(c) Convert the density of mercury from (b) The direction is East.
13.6 g cm into kg m . 3. Scalar quantities are physical quantities that
–3
–3
Solution have magnitude only.
(a) 1 cm = 1 cm × 1 cm 4. Vector quantities are physical quantities that
2
= 10 m × 10 m have magnitude and direction.
–2
–2
= 10 m 2
–4
–4
2
Therefore, 1.2 cm = 1.2 × 10 m 2 5. Some examples of scalar and vector quantities
(b) 1 m = 1 m × 1 m × 1 m are listed in Table 1.4
3
= 10 cm × 10 cm × 10 cm Table 1.4
2
2
2
= 10 cm 3 Scalar quantities Vector quantities
6
Therefore, 2.5 m = 2.5 × 10 cm 3
6
3
1 g Length Displacement
(c) 1 g cm =
–3
1 cm 3 Time Velocity
10 kg Temperature Acceleration
–3
= Mass Momentum
10 m 3 Speed Force
–6
= 10 kg m
–3
3
Therefore, 13.6 g cm = 13.6 × 10 kg m –3
–3
3
= 1.36 × 10 kg m –3 SPM Highlights
4

Which of the following quantities is a vector quantity?
SPM Highlights A Mass
B Speed
C Energy
Which of the following is a derived quantity? D Force
A Length C Mass
B Speed D Time
Examiner’s Tip
Force has both magnitude and direction. The rest of
Examiner’s Tip the quantities have magnitude only.
Speed is Distance . Hence it is a derived quantity. Answer: D
Time
Answer: B
5
01 FOC PHYSICS F4.indd 5 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

Checkpoint 1.1
1.2 Scientific Investigation
Q1 Read the label of the box in Figure 1.3 carefully.
3 packets inside: Total 55.2 g 1. Scientists use scientific investigations to find
SUPER SOUP For each packet, add 150 cm of solutions and make discoveries of various
3
hot water at 80°C. Stir until thicken.
All done in 3 minutes. theories and laws in science.
9 kJ of energy per serving 2. A scientific investigator needs to arrange and
Expiry date: 02/02/2024 record the results of the investigation in a
systematic and effective way that will facilitate
analysis to determine relationships, patterns
4
and arrive at a conclusion.
FORM
Figure 1.3 3. Observations in an investigation are usually
4
Identify all the physical quantities stated. Then, done directly through seeing, hearing,
classify the physical quantities into base and
derived quantities. smelling, touching and tasting or with the
help of scientific instruments.
Q2 Figure 1.4 shows a satellite orbiting Earth.









Figure 1.4
(a) The orbit radius of the satellite is 7 500 000 m.
What is its radius in
(i) km? (ii) Mm?
(b) The satellite travels at a speed of 7853 m s . Process in a Scientific Investigation
–1
What is its speed in km h ? 1. If a phenomenon observed needs further
–1
Q3 What is a explanation and understanding, the question
(a) scalar quantity?
(b) vector quantity? will be phrased in the form of a problem
Q4 Table 1.5 shows some events involving physical statement. A scientific investigation process
quantities starts with the identification of the problem in
the form that can be tested scientifically. Below
Table 1.5
are examples of problem statements:
Event Description (a) How does the period of oscillation of a
1 A plane flies at 700 km h simple pendulum depend on its length?
–1
from Senai Airport to Kuching
International Airport. String
2 Theva buys 3 kg of flour for his Cork
mother to bake cookies.
3 A porter pushes a trolley with a
force of 25 N towards a lift. θ < 10° θ
Ruler
4 Jamal heats some water from
20°C to 100°C to make coffee.
For each event, determine whether each of the A C
quantities involved is a scalar or vector quantity. Stopwatch
Explain your answer. Bob B




6





01 FOC PHYSICS F4.indd 6 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement
(b) What is the relationship between the depth 5. After that, a scientific investigation is planned
and pressure of a liquid? and carried out. The apparatus and materials
needed are prepared. An experiment will then
be carried out with the proper and careful
arrangement of the apparatus and the procedure.
6. Systematic collection of experimental
data, usually in the form of tables will be
done to facilitate the analysis process later.
7. After that, analysis and interpretation of data FORM
will be carried out using scientific reasoning with
(c) What is the relationship between the speed the help of graphs or other scientific methods. 4
of flow of a fluid and its pressure? 8. Following that, a decision is made. It usually
is related to the relationship between the
manipulated variable and the responding
variable. After that, a conclusion is arrived
at and this is about accepting or rejecting
the hypothesis. To communicate the results
of the experiment to others, a report from
the beginning till the conclusion is written.

2. After that, a hypothesis is made. A hypothesis Identify the problem that can be tested by
is an intellectual guess on the relationship scientific investigation.
between two or more variables.

3. A variable is a physical quantity that can be
varied in an experiment. There are three types State a hypothesis
of variables.
(a) A manipulated variable is a physical
quantity with values that are fixed by
the experimenter before carrying out the Design how the variable is manipulated and
how the data is collected.
experiment.
(b) A responding variable is a physical
quantity that changes its value in response
to the change in the manipulated variable. Plan and carry out the scientific investigation.
(c) A fixed variable is a physical quantity
that is set to remain constant throughout
the experiment. Present the collected data.

4. A hypothesis
(a) must be brief and clear.
(b) must state the relationship between the Interpret the data and results using scientific
manipulated variable and the responding reasoning.
variable.
(c) is still not known to be correct yet and
needs to be tested. Make conclusion and present in a report.
(d) must be able to be tested by conducting
experiments.






7





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Physics SPM Chapter 1 Measurement

Graphical Method
1. The graphical method is one of the most important methods for analysing and interpreting experimental
data.
2. When using the graphical method, the following steps are taken to make sure a more accurate result
is achieved.
(a) Information or data needs to be collected and arranged in table form.
(b) From the tabulated data, a graph is plotted.
(c) Analysis of the graph is carried out.

Interpreting the Shape of the Graph to Determine the Relationship Between Two
Physical Quantities
FORM
Students frequently need to analyse graphs to come out with certain conclusions especially in dealing with
4
the relationship between two variables. Table 1.6 shows some graphs that students will usually come across
in scientific investigations.
Table 1.6
Type of graph Explanation

y 1. Equation y = mx is a straight line that passes through the origin
(0, 0).
a
2. Gradient of the line, m =
a b
3. Relationship between y and x:
b (a) y is directly proportional to x.
0 y = mx x
(b) Therefore, y ∝ x
1. Equation y = mx + c is a straight line that passes through the y-axis
y at c.
2. y-intercept = c
a 3. Gradient of the line, m = a
b
c b 4. Relationship between y and x:
0 y = mx + c x (a) y and x have a positive linear relationship.
(b) Therefore, y increases as x increases.

y y
1. Non-linear relationship.
2. y increases when x increases.
0 x 0 x

y y
1. Non-linear relationship.
2. y decreases when x increases.

0 x 0 x

y k
1. Equation y = x is a curve, where k is a constant.
2. Relationship between y and x:
(a) y is inversely proportional to x.
0 x 1
y = k x (b) Therefore, y ∝ x



8





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Physics SPM Chapter 1 Measurement

Analysing Graph to Derive the Conclusion of an Investigation
Extension of
spring, x / cm
10.0


8.0

6.0
Spring FORM
4.0 4

2.0
Load
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Weight of load, W / N
Figure 1.5
Figure 1.5 shows a graph that is plotted from a scientific investigation to determine the relationship between
the extension of spring and the weight of the load. Based on the graph, the following analyses are carried out.
(a) Determine the gradient of the Extension of
graph spring, x / cm
1. Two points that cover more than 10.0
half of the straight line are chosen.
8.0
2. From the two points, a triangle is
drawn and the gradient is determined 6.0
by the method shown in Figure 1.6.
6.7 – 2.7
3. The unit for the gradient is based 4.0 = 4.0
on the the unit of the vertical axis Gradient
divided by the horizontal axis. In this 2.0 ___
4.0
case it is in cm/N or cm N . 0.5 – 0.2 = 0.3
–1
= 0.3
= 13.3 cm N –1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
(b) Determine the area under the Figure 1.6 Weight of load, W / N
graph Extension of
1. Figure 1.7 shows the method to determine spring, x / cm
the area under the graph for a certain 10.0
required range of values on the horizontal
axis. 8.0
2. The unit for the area is the product of
both the units of the axes. In the coming 6.0
chapters, you will learn various physical 4.0 8.0
quantities related to the area under the
graph. 2.0
Area under graph = Area of trapezium 1.3
1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
= (1.3 + 8.0) × 0.5 Weight of load, W / N
2
= 2.33 N cm 0.5
Figure 1.7
9





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Physics SPM Chapter 1 Measurement
(c) Determine a certain physical quantity using the interpolation method

1. Interpolation of graph is a Extension of
spring, x / cm
process of estimating a value
that falls within the known range 10.0
of the graph.

2. What is the extension of the 8.0
spring if the weight of the load Interpolation
method
is 0.35 N? From Figure 1.8, If x = 6.0 cm 6.0
the value 0.35 N falls within W = 0.45 N
the range of 0.1 N and 0.6 N 4.0
with the known values of the
FORM
extension of the spring. Through 2.0
4
interpolation, the extension of the
spring when the load is 0.35 N 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
can be determined. Weight of load, W / N
Interpolation
3. If the extension of the spring is 6.0 cm, the method
weight of the load can also be determined with the If W = 0.35 N,
interpolation method as shown in the diagram. x = 4.6 cm
Figure 1.8
(d) Making prediction using the extrapolation method
1. Figure 1.9 shows two examples of the extrapolation method.

Extension of
spring, x / cm
Extrapolation
method 10.0
shows if
x = 9 cm, 8.0
W = 0.67 N
6.0


4.0

2.0


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Weight of load, W / N
Extrapolation method
shows if
W = 0 N, x = 0 cm

Figure 1.9

2. If a physical quantity to be estimated is not within the range of the graph, the graph can be extended
until that value to enable the estimation to be done. This method is known as extrapolation of graph.
3. The accuracy of the estimated value using the extrapolation method depends on a few factors. The
further the value to be estimated from the known range, the more difficult it is for accuracy to be
achieved. Furthermore, the extrapolation method assumes that the pattern of the graph continues out
of the known range. If the pattern changes, the estimation will not be accurate.



10





01 FOC PHYSICS F4.indd 10 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

SPM Highlights  Observation

The table shows the results of an experiment to The apple takes a longer time to swing to and fro
investigate the relationship between time and when the string is longer
temperature of a liquid in a beaker.
Time, t / minutes 0 1.0 2.0 3.0
Temperature,  Inference
q / °C 80.0 65.0 55.0 47.0 The time taken for the apple to swing to and fro
What is the manipulated variable? depends on the length of the string.
A Time, t FORM
B Temperature, q
C Type of liquid
D The volume of the liquid, V  Question or problem statement 4
What is the relationship between the length of
Examiner’s Tip
It is the physical quantity in the table that can be the string and the time taken for the apple to
fixed before the experiment. swing to and fro?
Answer: A
 Variables
Example of Scientific Investigation and The variables involved are the length of the
Writing a Complete Report string, the time of the swing and the mass of
1. During a telematch event, Rahman was the apple.
supposed to bite an apple without touching it
with his hands. The apple was tied to a string
and was swinging to and fro. Rahman observed
that the apple took a longer time to swing to  Hypothesis
and fro if the string was longer. The longer the length of the string, the longer
the time taken for the apple to swing to and fro.




 Designing an experiment
The experiment designed must be able to be
carried out in an ordinary school laboratory.
Hence, the case of the apple swinging is equivalent
Figure 1.10 to the oscillation of a simple pendulum. The time
to be measured is the period of the pendulum.
2. The flow chart in Figure 1.11 will guide you
through the process of scientific investigation Figure 1.11
of this example.
SPM Tips




Increasing accuracy of measurement
• Period is the time taken to complete one oscillation. For example, it is the time A B A
taken for a pendulum to swing from position A to B and back to A.
• To increase the accuracy of the result, time taken for more than one oscillation
is recorded. For example, time for 10 oscillations is recorded.

A B

Figure 1.12

11





01 FOC PHYSICS F4.indd 11 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

Experiment 1.1


Situation: Rahman noticed that an apple tied to a longer string takes a longer time to oscillate.
Aim: To determine how the period of a simple pendulum is dependent on its length.
Problem: How is the period of a simple pendulum dependent on its length?
Inference: The period of a simple pendulum is dependent on its length
Hypothesis: The longer the length of the pendulum, the longer the period of its oscillation.
Variables:
(a) Manipulated variable: Length of pendulum
(b) Responding variable: Period of oscillation of the pendulum
FORM
(c) Fixed variable: Mass of pendulum bob
4
Material: 100 cm thread, two small pieces of plywood.
Apparatus: Retort stand with clamp, stopwatch, protractor, brass bob, and metre rule.
Procedure:
1. One end of the thread is tied to a brass bob and the other end is clamped
to the retort clamp with the help of two pieces of plywood as shown
in Figure 1.12.
2. The thread is adjusted so that from the point where it is clamped to the Plywood
centre of the bob, its length l is 20.0 cm.
Thread
0
3. The pendulum is made to oscillate at a small angle of 10 .
4. The time, t , for 20 complete oscillations is measured. The reading is Brass bob
1
recorded.
5. The time, t , for another 20 complete oscillations is measured. The reading
2
is recorded. Figure 1.12
6. The average of t and t is determined and recorded as t. Subsequently,
2
1
the time taken for one complete oscillation which gives the value of period of oscillation,
T = t is determined.
20
7. Steps 3 to 6 are repeated for l = 30.0 cm, 40.0 cm, 50.0 cm, 60.0 cm, and 70.0 cm.
8. The data is recorded in the table below.
9. Based on the data, the graphs of T against l and T against l are plotted.
2
Results:
Table 1.6
Length of Time taken for 20 complete oscillations, t / s Period
pendulum, t
l / cm t 1 t 2 Average, t T = 20 / s T / s 2
2
20.0 17.6 17.5 17.6 0.88 0.77
30.0 22.1 22.2 22.2 1.11 1.23

40.0 25.0 25.0 25.0 1.25 1.56
50.0 28.0 27.9 28.0 1.40 1.96
60.0 30.8 30.8 30.8 1.54 2.39
70.0 33.1 33.2 33.2 1.66 2.76




12





01 FOC PHYSICS F4.indd 12 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

Data Analysis:
T / s








0 l / cm
The graph of period, T against length, l shows a curve with a positive gradient. This means that when FORM
l increases, T also increases. The hypothesis is accepted.
T / s 4
2







0 l / cm
2
2
The graph of T against length, l is a straight line passing through the origin. Therefore, T is directly
2
proportional to l or T ∝ l.
Conclusion:
The longer the length of the pendulum, the longer the period of its oscillation. Hence, Rahman’s
observation about the apple tied on a longer string took longer time to oscillate has been proven to
be true in this experiment.


Checkpoint 1.2
Q1 (a) Explain the meaning of the following items in a Q2 Figure 1.14 shows the setup of an experiment to
scientific investigation. investigate how the distance of the extension of an
(i) Inference elastic chord, x, affects the horizontal distance of
(ii) Hypothesis travel, d, of the ball.
(iii) Variable
(b) Figure 1.13 shows objects attached to three Elastic cord
identical springs.
Ball x



d

Figure 1.14
Based on the setup,
(a) state the aim of the experiment.
(b) state a suitable hypothesis for the experiment.
(c) list down the manipulated, responding and fixed
Figure 1.13 variables of the experiment.
Based on the observation, (d) explain how you would tabulate and analyse
(i) write an appropriate inference. your data.
(ii) state a hypothesis.
(iii) state the manipulated, responding and fixed
variables.




13





01 FOC PHYSICS F4.indd 13 10/03/2023 7:48 AM

Physics
Chapter 1 Measurement
SPM
Physics SPM Chapter 1 Measurement



Experiment Analyse results Make conclusion Write report




Scientific Investigation Results of


FORM
4





Graph Analysis investigation





ELEMENTARY PHYSICS Measurement Vector Quantity Interpreting the shape Relationship between two physical quantities







Derived Quantity Scalar Quantity








Physical Quantity Unit Imperial









CONCEPT MAP Base Quantity Symbol S.I.
















14
14




01 FOC PHYSICS F4.indd 14 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

SPM Practice 1



Objective Questions

1. Physical quantity is a quantity B addition or division or 7. Fugire 2 shows a graph of
that is both addition and division distance against time.
A small of base quantities. s / m
B big C multiplication and division
C measurable of base quantities. 80
D variable D multiplication or division FORM
or both multiplication
2. Which of the following is a and division of base 60
base quantity? quantities. 4
A Weight 40
B Area 4. What is the S.I. unit of a
C Electric current physical quantity derived 20
D Density from the division of mass by
3. Derived quantity is a physical volume? 0 10 20 30 t / s
3
quantity that is derived from A g cm C g cm –3 Figure 2
3
A addition and multiplication B kg m D kg m –3 What is the gradient of the
of base quantities. graph?
A 0.94 m s –1 C 2.53 m s –1
B 1.07 m s –1 D 2. 67 m s –1
5. Which pair of the quantities is correct?
Scalar quantity Vector quantity 8. A machine lifts bricks up a
HOTS building under construction.
A Has magnitude only Has direction only Figure 3 shows a graph
of the height of the bricks
B Has direction only Has magnitude only against the time which the
machine has been switched
C Has magnitude and direction Has magnitude only
on.
D Has magnitude only Has magnitude and direction Height, h / m

6. Figure 1 shows a graph that What is the best conclusion 30
HOTS is plotted from a scientific that can be derived from the
investigation related to the graph? 20
relationship between the A The depth is directly
depth of a pole forced into proportional to the weight 10
the ground with the weight of of the load.
the load added on it. B The depth increases 0 5 10 15
linearly with the weight of Time, t / seconds
Depth, h / cm
the load. Figure 3
40 C The depth is inversely
proportional to the weight Which of the following best
30 describes the graph?
of the load.
20 A The height of the bricks is
D The weight of the load
10 will increase if the depth directly proportional to the
is increased. time which the machine
0 0.2 0.4 0.6 0.8 1.0 1.2 has been switched on.
Weight of load,W / N B Throughout the time
Figure 1 the machine has been
switched on, the height of
the bricks increases with
a constant rate.



15





01 FOC PHYSICS F4.indd 15 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

C The bricks are only What method is used in this Responding
lifted 3 seconds after estimation? variable Relationship
the machine has been
switched on. Estimated Method A a a is directly
D The machine is switched value proportional
on for only 14 seconds. A 7.6 m Interpolation to m

9. Figure 4 shows a graph of v B 7.6 m Extrapolation B a a is inversely
against t. C 8.0 m Interpolation proportional
to m
v
D 8.0 m Extrapolation C m m is directly
proportional
11. Figure 6 shows a graph of P 2 to a
FORM
against d. D m m is inversely
4
P 2 proportional
1
to a
13. Which of the following is not
t a suitable hypothesis?
Figure 4 HOTS
A The higher the ball is
Which of the following best dropped, the longer the
describes the graph? d time of its fall to the floor.
A v is inversely proportional Figure 6 B The lower the
to t. temperature of a liquid,
B The gradient of the graph Which of the following the longer the time for the
is negative. statements concerning this salt to dissolve.
C v increases with a uniform graph is not true? C The higher is the
rate. A P increases when d acceleration of a car, the
D v decreases with a increases. more difficult the use of
2
uniform rate. B P increases when d normal instruments to
increases. measure its velocity.
10. Figure 5 shows how the C P is directly proportional D The higher the
2
radius of a circle changes to d. temperature of a gas
with time. From the graph, D P has a linear relationship in a ball, the bigger the
estimate the radius of the with d. volume of the ball.
circle when the time is t = 36 s.
12. Figure 7 shows a graph
r / m 14. A manipulated variable
of relationship between
acceleration, a, and mass, m. A must be a base quantity.
8 Which pair is correct? B must be a derived
a quantity.
6 C must remain constant
throughout the
experiment.
4
D can be determined
before carrying out an
2
experiment.
t / s
1
0 10 20 30 40 –
m
Figure 5
Figure 7








16





01 FOC PHYSICS F4.indd 16 10/03/2023 7:48 AM

Physics SPM Chapter 1 Measurement

Subjective Questions
Section A

T g
2
1. A particular physical quantity L, is given by L = 2 where p is a constant, T is the time and the S.I. unit
for g is m s . 4p
–2
(a) (i) Determine the S.I. unit for L. [2 marks]
(ii) What type of physical quantity is L. [1 mark]
(b) A scientist successfully created a synthetic material with a very low density. Figure 1 shows a cuboid
that is made of this material. (Each surface of the cuboid is a rectangle). The mass of the cuboid is
1500 g. FORM
25 cm

12 cm 4



50 cm


Figure 1
(i) What is the volume of the cuboid in S.I. unit? [1 mark]
(ii) If density is defined as mass per unit volume, determine the density of the cuboid in S.I. unit.
[3 marks]

2. Figure 2 shows a inertia balance that is oscillating. An experiment is carried out to determine the period of
oscillation of the balance with different load attached to it. Table 1 shows the result of the experiment.

G-Clamp Table 1
Mass Period (Period) 2
Weight m / kg T / s T / s 2
2
m kg
2.0 0.92
3.0 1.14
Table 4.0 1.33
5.0 1.47
Inertia balance
6.0 1.63
Figure 2
(a) Complete the table by filling up the values of T . [3 marks]
2
(b) Draw the graph of m against T . [3 marks]
2
(c) From the graph in 2(b), determine the gradient of the graph. [3 marks]

3. Three students each draw a graph based on Table 2 derived from an experiment to investigate how the
potential difference, V, of an electrical circuit depends on the current, I, that flows round the circuit.
Table 2
Current 0.2 0.4 0.6 0.8 1.0 1.2
I / A
Potential difference 0.5 1.2 1.8 2.3 2.9 3.6
V / V
The graphs drawn by the students are shown in Figure 3.





17





01 FOC PHYSICS F4.indd 17 10/03/2023 12:18 PM

Physics SPM Chapter 1 Measurement


Potential difference Potential difference
V / V V / V

4.0
3.6
3.2 4.0
2.8 3.5
2.4 3.0
2.0 2.5
1.6 2.0
FORM
1.2 1.5
4
0.8 1.0
0.4 0.5
Current Current
0 I / A I / A
0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
Student A Student B

Potential difference
V / V
2.4
2.2
2.0
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
Current
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 I / A
Student C
Figure 3

(a) In this experiment, V and I, are the two physical quantities involved as variables.
(i) What is meant by physical quantity? [1 mark]
(ii) State what type of variable V and I are respectively. [2 marks]
(b) By analysing the three graphs drawn by the students from the aspect of scale of the horizontal axis,
scale of the vertical axis, ease of plotting the points, the size of the graph drawn, the ease to determine
HOTS
the gradient of the graph or other factors, state two strengths or weaknesses of each graphs.
(i) Graph of student A
(ii) Graph of student B
(iii) Graph of student C [6 marks]










18





01 FOC PHYSICS F4.indd 18 10/03/2023 7:48 AM

Theme 4 : Modern Physics Form 5

Chapter
6 Nuclear Physics











CHAPTER FOCUS



6.1 Radioactive Decay
6.2 Nuclear Energy











































We believe that there were dinosaurs for centuries before. Did you know, by calculating the
half-life of a radioactive decay, it helps archaeologists to identify the age of the dinosaur
fossils! Albert Einstein was the scientist who discovered a formula for quantifying nuclear
energy. This enables engineers to build reactor that has capable of controlling nuclear energy
production. Therefore, nuclear energy can be leveraged by using it to generate electricity and
fuel to propel large-scale vessels. In the future, nuclear energy could
be used as a rocket fuel for space exploration!
Access to
i-STUDI
i-STUDI SPM



431





06 SPM PHSICS F5.indd 431 10/03/2023 10:10 AM

Physics SPM Chapter 6 Nuclear Physics


6.1 Radioactive Decay

Alpha particle α
• An element can be represented in the following
nuclear notation:
Nucleon number A X
Proton number Z Symbol of element
Beta particle β
• Nucleon number is the sum of neutron number and
proton number.
A = N + Z
Gamma rays γ
(N = neutron number)

Figure 6.1
1. Radioactivity is the spontaneous decay of an unstable nucleus by emitting alpha particle, beta particle
or gamma ray.
2. There are three types of decays, which are alpha, beta, and gamma decay.

Alpha Decay, α
1. Alpha decay happens when an unstable nucleus decays by emitting a helium atom which is called
alpha particle.


+ + + +
+ +
+ + + + +
+ + +
+ + +
+
Parent nucleus Daughter nucleus Alpha particle
Figure 6.2 Alpha decay
2. The decay has caused the parent nucleus to lose four nucleon numbers and two proton numbers to
form daughter nucleus and alpha particle.
3. So, the equation for alpha decay can be represented as:

4
A–4
A X → Y + He
Z Z–2 2
4. Example:

Alpha
decay
+
FORM
α–particles
5
(2p 144n)
Uranium-238 Thorium-234
Figure 6.3 Uranium-238 decay
The uranium-238 decay produces torium-234 and alpha particles.

4
234
238 U → Ra + He
90 88 2


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Physics SPM Chapter 6 Nuclear Physics

Beta decay, β
1. Beta decay occurs when the parent nucleus decays and emits high-speed electrons called beta particles.

+ + + +
+ + + +
+ + + + + –
+ +
+ + +

Parent nucleus Daughter nucleus Beta particle
Figure 6.4 Beta decay
2. The equation for beta decay can be represented as:
A X → Y + e
A
0
Z Z+1 –1
3. Example:
Beta
decay
+


β-particles
Carbon-14 Nitrogen-14
(6p 8n) (7p 7n)
Figure 6.5 Carbon-14 decay
The equation of carbon-14 decay is shown as follow:
0
14
14 C → N + e
6 7 –1
Gamma Decay
1. Gamma decay occurs when an unstable nucleus releases electromagnetic waves called gamma ray.

+ + + +
+ +
+ +
+ + + +
+ +
+ + +
Parent nucleus Daughter nucleus Gamma ray
Figure 6.6 Gamma decay
2. The equation for gamma decay can be represented as:

A
0
A X → X + γ
Z Z 0 FORM
3. Example 5

+


Gamma rays
Thorium-234 Thorium-234
Figure 6.7 Thorium-234 decay
The equation of thorium-234 decay is shown as follow:

234
234 Th → Th + γ
90 90
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Physics SPM Chapter 6 Nuclear Physics
4. Gamma rays can also be emitted when a 2. For example, a radioactive sample contains
nucleus undergoes alpha or beta decays. fluorin-20 of 32 g takes 11 s to decay to half of
5. For example, during uranium-234 decay, it its original mass. So, it means that the half-life
emits alpha particles and gamma ray. The of fluorin-20 is 11 s and the mass after 11 s is
equation of the decay is: 16 g.
Initial mass: 32 g
234 U → Th + He + γ Half of Another
4
230
half is
2
92
90
the initial
mass undecayed:
Radioactive Decay Series has 1 1 16 g
decayed: 2 2
1. In many cases, the daughter nucleus of a 16 g
radioactive decay may still be unstable.
2. Therefore, the daughter will decay again to
produce another nucleus. Figure 6.8 Half-life of fluorin-20 is 11 s
3. The decay process will continue until a stable 3. The radioactive decay occurs continuously
nucleus is obtained. until the nucleus becomes stable.
4. For example, radon-219 decays through a 4. Example:
series of five decays to become a stable (a) The half-life of Cobalt-60 is 5.3 years.
nucleus, lead-207. The following equation (b) After the 5.3 years, 50% of the undecayed
shows the decay process. nucleus takes another 5.3 years for the
second half-life to be 25% still undecayed.
β, γ
α, γ
β, γ
α
α
207
219 Rn → Po → Pb → Bi → Tl → Pb (c) Then the decaying process will continue
211
207
211
215
86 84 82 83 81 82
and the decaying time can be measured by
counting how many half-lives occurred.
Half-life

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.
1. The original number can be measured in term
of mass, decay rate or decay percentage.
Co-60 remaining (%)
Cobalt-60 – Half-life 5.3 years
100
Years 5.3 10.6 15.9 21.2 10 g
Curies 100 50 25 12.5 6.2 0
75
50%
75% 93.8% 5 g
100% 100%
87.5% 50
50%
25% 2.5 g
12.5% 6.2%
Half-life 1 2 3 4 After many 25 1.25 g
half-lives 12.5 0.625 g
FORM
Parent Material Daughter Material
0
5
0 1 2 3 4 5
Number of half-lives
Figure 6.9 Decay series of Cobalt-60





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Physics SPM Chapter 6 Nuclear Physics
(d) So, the half-life in a radioactive decay series can be measured by using the following equation:

N N N
N →   →   →   → ……
o
o
o
o
T 1 2 T 1 4 T 1 8 T 1
2 2 2 2
(e) The number of undecayed nucleus, N is
1 n
N =   N o What is meant by the term
2
‘Half-life’?
where, N = Number of initial nucleus VIDEO
o
n = Number of half-life (always in positive)
= Half-life
T 1
2
Applications of Half-life
Ray
Radioisotope Half-life Application
produced
Chromium-51 α 27.7 days To label red blood cell in a blood test.
Sodium-24 β 15 hours To identify the leakage of underground pipe.
Hydrogen-3 β 12.3 years To identify the age of an underground water source.
Americium-241 α 432 years To detect smoke in a fire alarm system.
Caesium-137 β 30.2 years To measure and control paper thickness in a machine.
Cobalt-60 γ 5.27 years To sterilise medical tools.
Gadolinium-153 γ 240 days To measure bone fragility in osteoporosis inspection.
Solving Problems Involving Half-life
Step 2: Identify the equation to be used.
EXAMPLE 6.1 N N N
o
o
o
Figure 6.10 shows a decay curve of a radioactive N →   →   →   → ……
2
8
4
o
T 1
T 1
T 1
T 1
sample. Find the value of t. 2 2 2 2
Step 3: Substitute the numbers in the equation and
Activity / min –1
1600 solve it.
1400 1 600 ⎯→ 800 ⎯→ 400 ⎯→ 200
1200 20 days 20 days 20 days
1000 So,
800
600 t = 20 + 20 + 20
400 = 60 days
200
20 t Time / day FORM
Figure 6.10
Solution 5
EXAMPLE 6.2
Step 1: List the information from the graph.
T 1 = 20 days A radioactive sample with the mass of 12 g was
2
N = 1 600 min –1 decaying to become stable. Find the undecayed
mass after the fourth half-life.
o
N = 200 min –1

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Physics SPM Chapter 6 Nuclear Physics

Solution Based on the graph,
Step 1: List the information from the graph. Activity / count per minute
N = 12 g 2000
o
n = 4
Step 2: Identify the equation to be used. 1500
1 n
N =   N o 1000
2
Step 3: Substitute the numbers in the equation and
solve it. 500
1 4
N =   (12)
2
= 0.75 g 0 2 4 6 8 10 12 14
Decay time / s
The half-life of fluorine-22 is 4 s
1 n
(b) The undecayed activity, N =   N
2
o
1 3
Therefore, N =   (2 000)
EXAMPLE 6.3 2
= 250 counts per minute.
Figure 6.11 shows the decay rate of flourine-22.
Activity / count per minute Checkpoint! 6.1
2000
Q1 Figure 6.12 shows the decay rate of sodium-24.
1500 Activity / count per minute
2000
1000
1500
500
1000

0 2 4 6 8 10 12 14
Decay time / s 500
Figure 6.11
4
3
(a) What is the half-life of fluorine-22? 0 1 2 Decay time / s
(b) Find the undecayed activity left after third half-
life is completed. Figure 6.12
Solution (a) What is the half-life of sodium-24?
(b) Find the undecayed activity left after fourth
(a) When fluorine-22 completes a half-life, the half-life is completed.
activity becomes half from initial count.
FORM
Therefore, the activity = 2000 = 1000 counts
2
5
per minute
6.2 Nuclear Energy

When an unstable nucleus undergoes fission or
fusion, a large energy is produced. The large energy
is called nuclear energy.





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Physics SPM Chapter 6 Nuclear Physics

Nuclear Fission

Nuclear fission is the splitting of a heavy nucleus into two or more lighter nuclei while releasing a large amount of
energy.



A nucleus is The nucleus The nucleus Produces two Releases
bombarded by a becomes undergoes fission. different nuclei. energy and
neutron. unstable. three neutrons.


Kripton-92, Kr
92
+ 36
+ +
+ +
+
+ + +
+ + +
+ + + + ENERGY
1
Neutron n + + + + + + 3 neutrons, 3 n
1
0
0
+ + +
Uranium-235 235 92 U +
Uranium-236 236 92 U +
+ +
+ Barium-141, 141 Ba
56
1 n + U → Ba + Kr + 3 n + Energy
141
92
235
1
0 92 56 36 0
Figure 6.13 Nuclear fission of uranium-235
1. The uranium-235 is Nine U-235
bombarded with a neutron nuclei fissions
to become a highly unstable
uranium-236.
2. Subsequently, it splits into
two different nuclei and Three U-235
releases three neutrons and nuclei fissions
also large energy.
3. Then, the three neutrons U-235
will bombard other Neutron
uranium-235 and nuclear
fission is going to happen
again. This is called chain Three FORM
reaction. neutrons
released 5


Nine neutrons
released




Figure 6.14 Chain reaction



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Physics SPM Chapter 6 Nuclear Physics

Nuclear Fusion

Nuclear fusion is the combining of two lighter nuclei to form a heavier nucleus while releasing a large amount of
energy.



Two nuclei move Both nuclei collide Combine to form a Release energy and
with high speed. with each other. heavier nucleus. neutron.



Deuterium, H Helium, He
4
2
1 2
+ Fusion +
+
+
Energy
+
+


3
Tritium, H Neutron, n
1
1 0
3
1
2 H + H → He + n + Energy
3
1 1 2 0
Figure 6.15 Nuclear fusion between deuterium and tritrium
1. Deuterium and tritium move with high speed. Energy in a Nuclear Reaction
2. Then, they collide each other and combine to 1. Nuclear reactions cause the loss of mass when
form a heavier nucleus. a large amount of energy is released.
3. The combining of two nuclei releases a large 2 So, the mass of nucleus after reaction is less
amount of energy. than the mass before reaction.
3. The loss of mass is known as mass defect.
Atomic Mass Unit
1. The atomic mass unit (amu) is the unit of 4. The mass defect is changed to energy released
mass for atoms and subatomic particles such during the reaction.
as proton, neutron and electron. 5. The relationship between the mass defect and
2. Isotope carbon-12 is chosen as a standard the energy released is defined in the following
definition of atomic mass unit. Einstein’s equation:
1 E = mc 2
1 amu = 1 u = of the mass of carbon-12
12
Therefore, where,
FORM
1 u = 1.66 × 10 kg E = Energy released (J)
–27
m = Mass defect (kg)
5
3. Common atomic masses: c = Speed of light = 3.00 × 10 m s .
–1
8
Atom / subatom Mass (u)
Proton 1.007276 u 6. Nuclear energy also can be defined by using
electron-volt (eV) unit.
Electron 0.000549 u
Neutron 1.008665 u 1 eV= 1.6 × 10 –19 J
Hydrogen 1.007825 u


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Physics SPM Chapter 6 Nuclear Physics

Solution
EXAMPLE 6.4
(a) Write the equation for the fission of
Calculate the energy released during the decay of a uranium-235.
nucleus of radium-226. Step 1: List the information related to the equation.
222
226 Ra → Rn + He
4
88 86 2 Elements before fission: U, n.
235
1
[Given that the mass of radium-226 = 226.025 u, 92 0
94
1
the mass of radon-222 = 222.018 u, the mass of Elements after fission: 140 Xe, Sr, and some n
0
54
38
–27
alpha particle = 4.003 u, 1 u = 1.66 × 10 kg,
8
c = 3.00 × 10 m s , and 1 eV = 1.6 × 10 J] Step 2: Write the equation by labelling the number
–1
-19
of neutrons with = N
Solution
94
1
235 U + n → Xe + Sr + N n
1
140
Step 1: Identify the masses of each element in the 92 0 54 38 0
equation. Step 3: Get the value of N by balancing the
226 Ra → Rn + He nucleon number before and after the
4
222
88
86
2
226.025 → 222.018 + 4.003 fission.
Step 2: Calculate the mass difference between 235 + 1 = 140 + 94 + N (1)
before and after decay to get the value of N = 2
mass defect, m.
Total of mass before decay = 226.025 Step 4: Re-write the equation with the value of N
Total of mass after decay = 222.018 + 4.003 235 U + n → Xe + Sr + 2 n
140
1
94
1
= 226.021 92 0 54 38 0
m = total of mass before decay – total of mass after (b) Calculate the energy released during fission
decay
= 226.025 – 226.021 Step 1: Identify the masses of each element in the
= 0.004 u equation.
= 0.004 × 1.66 × 10 kg
–27
1
140
94
1
= 6.64 × 10 kg 235 U + n → Xe + Sr + 2 n
–30
92
0
54
38
0
Step 3: Calculate the energy released by using 235.043925 + 1.008665 →
Einstein’s equation. 139.921636 + 93.915360 + 2 (1.008665)
E = mc 2
= (6.64 × 10 ) × (3.00 × 10 ) J Step 2: Calculate the mass different between
–30
8 2
= 5.98 × 10 –13 J before and after decay to get the value of
5.98 × 10 –13
= 1.6 × 10 –19 eV mass defect, m.
= 3.74 × 10 eV m = (235.043925 + 1.008665)
6
– [139.921636 + 93.915360 + 2 (1.008665)]
= 0.198264 u
= 0.198264 × 1.66 × 10 kg
–27
EXAMPLE 6.5
= 3.291 × 10 kg FORM
–28
The fission of uranium-235 ( U) is due to the
235
92
bombardment by a neutron and produces xenon-140 Step 3: Calculate the energy released by using
( Xe), strontium-94 ( Sr) and some neutrons. Einstein’s equation. 5
140
94
54
38
(a) Write an equation for the fission. E = mc 2
–28
8 2
(b) Calculate the energy released. = (3.291 × 10 ) × (3.00 × 10 ) J
= 2.96 × 10 –11 J
[ U = 235.043925 u, Xe = 139.921636 u, = 2.96 × 10 –11
140
235
92
54
1
94 Sr = 93.915360, n = 1.008665 u, 1 u = 1.66 × 1.6 × 10 –19 eV
38 0
8
10 kg, c = 3.00 × 10 m s , 1 eV= 1.6 × 10 –19 J] = 1.85 × 10 eV
–1
8
–27
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Physics SPM Chapter 6 Nuclear Physics

Step 2: Calculate the mass different between
EXAMPLE 6.5 before and after decay to get the value of
mass defect, m.
A nucleus reaction is represented by the following
equation: m = (2.01410 + 2.01410) – (3.01603 + 1.00867)
2
3
1
2 H + H → He + n + Energy = 0.0035 u
1 1 2 0 = 0.0035 × 1.66 × 10 kg
–27
Calculate the energy released. = 5.81 × 10 kg
–30
[ H = 2.01410 u, He = 3.01603 u, n = 1.00867 u,
1
2
3
0
2
1
1 u = 1.66 × 10 kg, c = 3.00 × 10 m s , Step 3: Calculate the energy released by using
8
–27
–1
1 eV= 1.6 × 10 J] Einstein’s equation.
–19
Solution E = mc 2
Step 1: Identify the masses of each element in the = (5.81 × 10 ) × (3.00 × 10 ) J
–30
8 2
equation. = 5.229 × 10 –13 J
5.229 × 10 –13
1
3
2
2 H + H → He + n = –19 eV
1 1 2 0 1.6 × 10
2.01410 + 2.01410 → 3.01603 + 1.00867 = 3.27 × 10 eV
6
Generation of Electrical Energy in a Nuclear Reactor
Figure 6.16 Nuclear power station
1. Nowadays, many developed countries use nuclear energy to generate their electricity.
2. A nuclear reactor is a structure that conducts a nuclear reaction in controlled and safe manner with the
consequent release of energy.

3. The energy released from nuclear fission can be used to generate electricity.



Boron control rods
- To control the rate of nuclear
reaction by absorbing excess
neutrons.
Hot gas outlet
Graphite moderator
FORM
- To reduce the speed of neutrons.
Slower neutrons are more readily
5
captured by the uranium nuclei.
Uranium fuel rod
- To perform nuclear fusion reactions Cold gas
and release a large amount of energy.
The energy released heats up the cold Cold gas inlet
gas that passes through the reactor core.


Figure 6.17 Nuclear reactor

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Physics SPM Chapter 6 Nuclear Physics

4. Figure 6.18 shows the structure of a nuclear reactor and the process of electrical energy generation at
a nuclear power station.

2
Steam generator
Steam
Concrete shield
4
Reactor
Uranium rod
Dynamo
1
Pump
Water Turbine
Hot gases Hot water Cooling
tower

Gas pump Cold water
Seawater
Cold gases
Chiller
3 Condenser
Figure 6.18 Nuclear power station

1 • Nuclear fission of uranium-235 occurs in the reactor and produces

energy.
• The energy produced is used to heat up the gas.
• Hot gases flow to the steam generator. The heat is transferred to
the water.


2 • In the steam generator, heat energy from the hot gas


boils the water into high pressure steam.
• Then, steam is channeled to the turbine.


3
• High acceleration of steam produces a force to rotate the turbine.
• Then, the steam is cooled by the chiller. The steam is condensed
back into water.
• The water is pumped into steam generator again. FORM



4 • As the turbine rotates, the dynamo will generate 5

electricity.
• Electricity is delivered to the consumers.


Figure 6.19 The generation of electricity from the nuclear energy






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Physics SPM Chapter 6 Nuclear Physics

Component Function
Reactor The location where the nuclear fission occurs.
• Uranium fuel To release heat energy through the nuclear fission to heat up the gas that flows through the
rod reactor core.
• Concrete shield To prevents radiation leakage from the reactor core.
• Gas pump To pump cold gases into the reactor.
Steam generator The location where the water is heated up by hot gases and then change to steam.
• Water pump To pump water from the condenser into the steam generator.
Condenser To cool down the steam after turning the turbine.
• Turbine High acceleration of steam produces a force to rotate the turbine that connects directly to
the axis of a dynamo.
• Dynamo Contains turning coil and magnet.
When the turbine axis rotates, magnetic flux is being cut and thus electricity is generated.
Chiller A different water looping system which is used to absorb heat in the condenser and then
released into the atmosphere.
• Seawater A large amount of water reservoir to lower steam temperature in the condenser.
• Cooling tower Heat in water is released into the atmosphere.




1 The cost to operate a power 2 The usage of fossil fuels is

reduced.
plant is cheaper.

The advantages No air pollution with no sulfur
6 provided for power plant power to generate 3 dioxide and carbon dioxide is
of using nuclear
Many jobs opportunities are
operation.
released into the atmosphere.
electricity
5 Small amount of fuel is needed 4 Safer as very few accidents were
reported if compared to the fossil
to produce high electrical energy.
fuel power plant.


1 With the high safety measurement,
the cost to build a power plant is
too expensive.


The disadvantages
FORM
4 High risk of disaster. power to generate 2 Produce radioactive waste
of using nuclear
5
that is dangerous to the
environment and the residents.
electricity
3 The cost of disposing of
radioactive waste is expensive.





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Physics SPM Chapter 6 Nuclear Physics


EXAMPLE 6.7
The following table shows an equation of a reaction and total atomic mass before and after a nuclear fission.

Nuclear fission
Before reaction After reaction
1
93
Equation 239 Pu + n 145 Ba + Sr + 2 n + Energy
1
0
94
38
0
56
Total atomic mass 240.0608 u 239.8583 u
Calculate
(a) mass defect in kg.
(b) the energy released in the reaction.
[1 u = 1.66 × 10 kg; c = 3.0 × 10 m s ]
8
–27
–1
Solution
(a) Mass defect, m = 240.0608 u – 239.8583 u
= 0.2025 u
= 0.2025 × 1.66 × 10 –27
= 3.36 × 10 kg
–28
(b) The energy released, E = mc 2
8 2
–28
= (3.36 × 10 ) (3.0 × 10 )
= 3.02 × 10 –11 J


Checkpoint! 6.2
Q1 The following table shows an equation of a reaction and total atomic mass before and after a nuclear fusion.
Nuclear fusion
Before reaction After reaction
Equation 2 1 H + H 4 2 He + n + Energy
1
3
0
1
Total atomic mass 5.0104 u 5.0301 u
Calculate
(a) mass defect in kg.
(b) the energy released in the reaction.
[1 u = 1.66 × 10 kg; c = 3.0 × 10 m s ] FORM
–1
–27
8


5













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Physics SPM Chapter 6 Nuclear Physics


CONCEPT MAP



NUCLEAR PHYSICS



Radioactive Nuclear
Decay Energy


Decay Half-life Nuclear Mass Generation of
reaction defect electricity
α
Nuclear Release of Nuclear
β fission energy reactor
γ
Nuclear
fusion






SPM Practice 6



Objective Questions

1. The nucleus of an atom 3. Figure 1 shows an 5. The equation below shows a
consists of incomplete decay equation. radioactive decay series.
A proton only. 192 0 211 X 211 Y 207 Z 207
B neutron only. 77 Ir → + e + γ 82 Pb → 83 Bi → 81 Po → 83 Bi
–1
C electron only. Figure 1 What is the radioactive ray
D proton dan neutron only. emitted at X, Y and Z?
Which of the following
2. The following equation elements should be filled to X Y Z
represents an alpha decay. complete the equation? A Alpha Beta Alpha
A 192 Pt
X → Y + α particle 78 particle particle particle
B 192 Ir
Beta
Alpha Gamma
77
Which of the following is C 188 Re B particle particle ray
correct about element Y? 75
D 226 Os C Alpha Beta Gamma
Proton Nucleon 76 particle particle ray
Number Number 4. An atom 230 Y has decayed D Beta Alpha Beta
FORM
91
A No change Increases and produced a nucleus A, particle particle particle
5
by 1 an alpha particle and a beta
B Increases No change particle. What is A? 6. The decay activity of sample
by 1 A 226 A X becomes 6.25% from the
88
C Decreases Decreases B 226 A initial value. What is the
89
by 2 by 4 C 226 A number of half-lives that have
90
D Decreases Decreases 226 occurred on samplel X?
A 3
C 5
by 4 by 2 D 91 A B 4 D 6
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Physics SPM Chapter 6 Nuclear Physics

7. A total of 25% of Galium-65 10. Figure 4 shows a decay 14. Figure 5 shows a decay
have left undecayed after 15 curve of a radioactive curve of a radioisotope.
minutes. What is the half-life material. Activity / %
of Gallium-65?
A 7.5 minutes Activity / counts per minute 100
B 15.0 minutes 800
C 23.4 minutes
D 30.0 minutes 400
50
8. Figure 2 shows a decay
curve for a radioactive 25
material.
0 15 Time / minute
Activity / counts per minute
4000 Figure 4 0 5 10 Time / years
What is the activity left after Figure 5
3000 1 hour? Based on the graph, suggest
A 50 C 200
B 100 D 400 the suitable usage for this
2000 radioisotope.
11. The activity for a fossil is A Used to detect the
24 counts per minute for 1 g position of blood clots in
1000
of carbon-14. Calculate the veins
undecayed activities after B Used for food sterilisation
Time / fourth half-life.
0 1.0 2.0 3.0 4.0 year C Used to test the leakage
A 0.75 C 3.0 of underground pipes
Figure 2 B 1.5 D 48.0 D Used to measure the
What is the half-life of the 12. The mass of two radioactive age of archaeological
radioactive material? materials, K and L are specimens
A 0.5 year 200 g and X g respectively. 15. Which radioisotope is
B 1.0 year The half-life of K and L is usually used to measure the
C 2.0 year 6 hours and 12 hours. After absorption of fertiliser by
D 4.0 year 24 hours, both have the plants?
same mass. What is the A Carbon-14
9. Figure 3 shows a decay value of X?
curve for sample Z. B Nitrogen-15
A 12.5 C Iodine-131
Activity / counts per second B 50.0 D Cobalt-60
C 100.0
160 D 400.0 16. Figure 6 shows a series of
radioactive decays for the
13. The ratemeter on a G-M nucleus of uranium-238 to
tube shows a background that of radium-226.
80 radiation of 50 counts per
minute. When a radioactive Nucleon number
material is placed in front of 238 U
the G-M tube, the ratemeter Th Pa
15 Time / day 234 U
records 170 counts per FORM
Figure 3 minute. After 8 hours, the 230 Th
reading of the ratemeter
If the initial activity for the becomes 7.5 counts per 226 5
sample Z is 160 counts per minute. Determine the 222
second, what is the activity 87 88 89 90 91 92
after 75 days? half-life of the radioactive Proton number
material.
A 2.5 A 1.0 hour Figure 6
B 5.0 B 1.5 hours What is the number of the
C 10.0 C 2.0 hours alpha particles and beta
D 20.0 particles emitted during this
D 2.5 hours
process?



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Physics SPM Chapter 6 Nuclear Physics

The The 20. The mass defect in a fission 25. In a nuclear reaction, a total
–3
–12
number number reaction is 3 × 10 kg. What of 4.30 × 10 J of energy is
of alpha of beta is the energy produced? released. What is the mass
–1
8
particles particles [c = 3.0 × 10 m s ] defect?
14
8
–1
A 2 3 A 1.8 × 10 J [c = 3.00 × 10 m s ]
14
B 2.7 × 10 J
–08
B 3 2 C 1.8 × 10 J A 1.43 × 10 kg
15
–17
B 4.78 × 10 kg
C 4 1 D 2.7 × 10 J C 1.43 × 10 kg
15
–20
D 1 1 21. The mass defect in a nuclear D 4.78 × 10 kg
–29
reaction is 0.05 u. What is
17. A nucleus of uranium-235 will the energy released? 26. In a nuclear reactor, the
–1
8
split when it captures [c = 3.00 × 10 m s , excess neutrons are
A an alpha particle. 1 u = 1.66 × 10 –27 kg] absorbed by
–20
B a gamma ray. A 2.49 × 10 J A uranium rods.
–12
C a neutron. B 7.47 × 10 J B graphite moderator.
–12
D an electron. C 7.65 × 10 J C boron control rod.
D 8.97 × 10 J
–11
D concrete wall.
18. Which of the following 22. Thorium-228 decays by
nuclear reactions is an emitting an α-particle. 27. What is the correct energy
example of fusion? change in a nuclear power
A 14 C → N + e + energy 228 Th → 224 Ra + α
0
14
90
88
6 7 –1 station?
B 21 Na → 121 Mg + β + energy What is the mass defect of A Heat energy → Potential
11
12
the decay?
C 226 Ra → 222 Rn + α + [Mass of 228 Th = 227.97929 u, energy → Kinetic energy
86
88
90
energy mass of Ra + α = 227.97340 u, → Electrical energy
224
88
–27
D 2 1 H + H → He + e + 1 u = 1.66 × 10 kg B Nuclear energy →
0
2
3
Potential energy→ Heat
2
1
–1
–30
energy A 9.78 × 10 kg energy → Electrical
B 9.78 × 10 kg energy
–29
–29
19. Figure 7 shows the beginning C 3.44 × 10 kg C Heat energy → Nuclear
–27
of a chain reaction. D 3.44 × 10 kg energy → Electrical
23. The mass defect from the
88 energy → Kinetic energy
38 Sr decay of radium-226 is D Potential energy → Heat
8.68 × 10 kg. Calculate the energy → Kinetic energy
–30
energy released.
Released [c = 3.00 × 10 m s , → Electrical energy
–1
8
Neutron neutrons –19
238 1 eV = 1.60 × 10 J] 28. Which of the following is the
92 U A 1.36 × 10 eV
–23
B 7.81 × 10 eV advantage of using nuclear
–13
6
131 Xe C 1.57 × 10 eV fission to generate electricity?
54 D 4.71 × 10 eV I Able to generate more
14
power than other types of
Figure 7 24. A nuclear reactor has power station.
The chain reaction is undergone a fission reaction. II Reduce greenhouse
occurred because After several days, it is effect.
A the energy produced is found that 0.09% from 1 g of III The cost to build the
too large. uranium-235 was changed to power station is cheap.
FORM
B it releases neutron. energy. Calculate the energy IV Provide many job
5
C free neutron bombarded released.
–1
8
other uranium nucleus. [c = 3.00 × 10 m s ] opportunities.
2
D Sr and Xe are A 2.7 × 10 J A I, II and III
5
radioactives. B 2.7 × 10 J B I, III and IV
C 8.1 × 10 J C I, II and IV
13
15
D 8.1 × 10 J D I, II, III and IV
446
06 SPM PHSICS F5.indd 446 10/03/2023 10:10 AM

Physics SPM Chapter 6 Nuclear Physics

29. Figure 8 shows the steam into the steam generator 30. Which of the following
rotating a turbine in a nuclear again. Which of the following statements is not true about
power plant. statement is the possible way the advantages of using
in order to cold down the nuclear fusion rather than
Hot gas Dynamo steam? nuclear fission in order to
Steam Turbine
A Channel the steam out so generate electricity?
that heat can release to A Fuel is everywhere.
surrounding. B No radioactive waste.
B Use hot gases rather than C No greenhouse effect.
steam as a material to D No risk of disaster to
rotate the turbine. environment.
Pump C Transfer the heat energy
Water to another separate
Pump
water channel so that the
Figure 8 heat can be released to
surrounding.
After rotating the turbine,
it is found that the steam D Increase the pressure of
suction pump so that the
is still hot and need to be
cooled before it is pumped steam flows with higher
speed.

Subjective Questions

Section A

1. Figure 1 shows an incomplete equation to represent the decay process of radium.

226 Ra → Rn + He
4
88 2
Figure 1
(a) Name the type of decay. [1 mark]
(b) Complete the equation above. [2 marks]
(c) Based on your answer in 1(a) and 1(b), complete the following statement.
When radium-226 decays, nucleon number is reduced by , and proton number also reduced by
. [2 marks]

2. Figure 2 shows the beta decay of protactinium-239.


Beta
decay
+

β–Particles FORM
Pa-239 U-239
(239n 92p) 5
Figure 2
(a) Complete the following statement.
(i) The nucleon number of protactinium-239 before the decay is .
(ii) The proton number of protactinium-239 before the decay is .
[2 marks]
(b) Why does the radioactive decay occur? [1 mark]
(c) Write a complete equation for the decay. [2 marks]



447





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Physics SPM Chapter 6 Nuclear Physics

3. The following equation shows the decay of a nuclide.
192 Ir → 192 Pt + w + x
77 78
(a) State the meaning of gamma decay. [1 mark]
(b) Identify w and x. [2 marks]
(c) Figure 3 shows syringes that need to be sterilised before being marketed as
medical equipment.
(i) Choose the best radioactive ray to sterilise the syringes. Give a reason
for your answer. [2 marks]
(ii) State one advantage of using radioactive ray to sterilise the medical
equipment. [1 mark]
Figure 3
4. Figure 4 shows a new found fossil by an archaeologist.
(a) State two characteristics of carbon-14 that allow it to be a reference to
determine the life span of a fossil. [2 marks]
(b) The archaeologist is trying to determine the life span of the fossil.
(i) Specify the number of times of half-life of the fossil if the total amount of
carbon-14 which has yet to decay in the fossil is 12.5%. [2 marks]
(ii) Determine the life span of the fossil.
[Half-life of carbon-14: 5730 years] [1 mark]


5. Figure 5 shows a decay curve for sample X. Figure 4
Activity / counts per second
6000
5000
4000

3000
2000

1000
t / minute
0 2 4 6 8
Figure 5
(a) What is half-life? [1 mark]
(b) Determine the original activity value of the sample X. [1 mark]
(c) Specify the half-life of the sample X. Show on the graph how you get it. [2 marks]
(d) Fire alarm system requires a suitable radioactive material for smoke detection. Is the sample X suitable
for that purpose? Give a reason for your answer. [2 marks]

6. Figure 6 shows the nuclear fission for uranium-235. Neutron
(a) What is the meaning of nuclear fission?
FORM
[1 mark]
5
(b) A nucleus of uranium-235 ( U) is bombarded Kripton
235
92
by a neutron and produces barium-141
92
( 141 Ba), kripton-92 ( Kr) and some neutrons. Neutron Neutron
56
36
(i) Write the equation of the fission by
labelling the number of neutron as N. Barium
[1 mark] Uranium-235
Neutron
Figure 6
448





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Physics SPM Chapter 6 Nuclear Physics

(ii) Specify the value of N. [1 mark]
(ii) Calculate the energy released in electron-volt unit if the mass defect is 0.18606 u.
[1 u = 1.66 × 10 kg, c = 3.00 × 108 m s , 1 eV= 1.6 × 10 J]
–1
–19
–27
[2 marks]
(c) State one application of nuclear fission. [1 mark]
7. Figure 7 shows a chain reaction of radon-222.
Nucleon number


222 Rn

Po
218
214 Pb Bi Po
Pb
210

206

Proton number
81 82 83 84 85 86 87
Figure 7
(a) What is the meaning of chain reaction? [1 mark]
(b) What is the neutron number for radon-222? [1 mark]
(c) Write the equation to show the decay of Rn-222 to Po-218. [1 mark]
(d) Determine the number of α and β particles produced during the decay of radon-222 to lead-210.
[1 mark]
8. A technician tries to detect a leak in the underground water pipe.




Ratemeter Detector





Water flowing


Figure 8 FORM
A small amount of a radioisotope is put into the water retention tank. There are four radioisotopes that had
been suggested to be used in the detection of pipe leakage. 5
Radioisotope Half-life Radiation type State of matter
Gadolinium-153 240 days Gamma Solid
Iodine-131 8 days Beta Liquid
Amerecium-241 432 years Alfa Solid
Sodium-24 15 days Beta Liquid




449





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Physics SPM Chapter 6 Nuclear Physics

Geiger-Muller tube is moved on above ground. The ratemeter records high radiation levels if there is a pipe
leakage.
(a) Based on the table, state the characteristics of the radioactive which is used to detect the pipe leakage.
Give reason for each of the following characteristic.
(i) Half-life [2 mark]
(ii) Radiation type [2 mark]
(iii) State of matter [2 mark]
(b) Based on your answer in 8(a), which of the following radioisotopes is suitable to be used in the detection
of pipe leakage? [1 mark]
1
(c) The half-life of polonium-210 is 132 days. How long does it take to reduce the activity to from its initial
8
state? [2 marks]
9. A nuclear fusion reactor is suggested to generate electricity.
Fusion reactions Hot fluid
occur in hot
plasma fuel
Turbine
generator
Heat
Plasma exchanger

Fast neutrons Output
from fusion power
reactions heat
the moderator
Cold fluid
Figure 9
(a) What is the meaning of nucleus fusion? [1 mark]
(b) A fusion reaction is shown as follows.
2 H + H → He + n + energy
3
0
4
1 1 2 1
[ H = 2.014012 u, H = 3.016029 u, He = 4.002603 u, n = 1.008665 u, u = 1.66 × 10 kg]
0
4
–27
3
2
1 1 2 1
(i) Calculate the mass defect, m. [2 marks]
(ii) Find the amount of energy released. [2 marks]
(c) Give another example of where the nuclear fusion is possibly found. [1 mark]
10. Figure 10 shows a nuclear power station.
Concrete wall
Boron control rods
Steam
Turbine
FORM
X
5
Pump
Cold water

Graphite moderator Uranium rod
Figure 10





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Physics SPM Chapter 6 Nuclear Physics

(a) Name the part where the process of nucleus fission occurs. [1 mark]
(b) Explain the functions of each of the following structures.
(i) Boron control rods [1 mark]
(ii) Graphite moderator [1 mark]
(c) Explain how nuclear energy can generate electricity. [4 marks]
(d) State two advantages of generating electricity by using nuclear power. [2 marks]


Section B
11. Figure 11 shows the decay of uranium-234 to radium-226 by emitting α-particle.
234 U → 230 Th → 230 Pa
92 90 91
Figure 11
(a) Explain why the radioactive decay series occurs. [1 mark]
(b) How many α-particles are emitted? Explain your answer. [2 marks]
(c) Complete the equations below.
(i) 234 U → 230 Th + …… + energy
92
90
(ii) 230 Th → 230 Pa + …… + energy
90 91
[2 marks]
(d) The following table shows the mass of three elements involved
in the decay of uranium-234. Element Atomic mass unit (u)
(i) Calculate the mass defect in unit kg. Uranium-234 238.029
[1 u = 1.66 × 10 kg] [3 marks]
–27
(ii) What is the amount of energy released? Thorium-230 232.038
[ c = 3.00 × 10 m s ] [2 marks] α-particle 4.003
8
–1
(e) The following table shows the characteristics of four
radioisotopes.
HOTS
Radioisotope Half-life Type of radiation State of matter Detector
J 29.4 years Beta Solid Spark counter
K 6.34 years Gamma Solid GM-tube
L 15.1 days Beta Liquid Spark counter
M 16 hours Gamma Liquid GM-tube

Based on the table above, identify the suitable characteristics of a radioisotope to estimate the capacity
of underground mineral water reservoir. Give reasons for each of the following characteristic.
(i) Half-life
(ii) Type of radiation
(iii) State of matter
(iv) Detector
After that, choose the most suitable radioisotope to be used to estimate the capacity of underground FORM
mineral water reservoir. [10 marks]

12. (a) A nuclear power station will be built in your residence area to generate electricity. The following table 5
shows four different types of nuclear power stations and their specifications.
HOTS
Nuclear power Type of reaction Fuel resource Operating life Water reservoir
station time near with
P Fission Uranium 60 years Sea
Q Fusion Hydrogen 58 years Sea
R Fission Plutonium 50 years Lake
S Fusion Hydrogen 45 years Lake

451





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Physics SPM Chapter 6 Nuclear Physics

Explain the suitability of each feature of a nuclear power station to make it safer to use. Identify the safest
nuclear power station to generate electricity. Give reasons for your answer. [10 marks]
(b) Figure 12 shows a nuclear fusion reaction with their masses in atomic mass unit (amu).
2 H + H → He + n + Energy
4
1
2
1 1 2 0
2.01410 3.01605 4.00260 1.00867
(i) What is the mass defect of the reaction in kg?
[1 u = 1.66 × 10 kg] [3 marks]
–27
(ii) What is the amount of heat released in eV unit?
[ c = 3.0 × 10 m s , 1 eV= 1.6 × 10 J] [3 marks]
–19
8
–1
(c) State two disadvantages of nuclear power station and suggest ways to overcome them. [4 marks]

Section C
13. Figure 13 shows radioactive decay graphs for source A and source B.

80 80
Source A Source B
Activity / count per second 60 Activity / count per second 60




40
40

20 20

Decay time / Decay time /
0 Year 0
10 20 30 1 2 3 Year
Figure 13
(a) What is the meaning of half-life? [1 mark]
(b) Based on the graphs above,
(i) Compare the initial decay activity, half-life and the rate of radioactive decay. [2 marks]
(ii) State the relationship between half-life and the rate of radioactive decay. [2 marks]
(c) Identify the time taken for both sources to complete their third half-life. [2 marks]
(d) Between source A and source B,
(i) Which is suitable to detect smoke? Give reason for your answer. [2 marks]
(ii) What is the most suitable radioactive rays used to detect the smoke? [1 mark]
(e) You are required to find ways to kill pests in a farm. State the characteristics of suitable radioactive
HOTS materials used for this purpose based on the following aspects:
(i) Half-life
FORM
(ii) Emitted rays
(iii) The rate of radiation
5
(iv) The boiling point of the material
(v) Energy released
[10 marks]









452





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Physics SPM Chapter 6 Nuclear Physics

14. Figure 14 shows the decay curve of two Activity (counts per minute)
radioactive nuclides, P and Q.
(a) What is represented by the activity in the N
graph? [1 mark]
(b) Based on the graph, compare N
(i) Half-life [1 mark] 2
(ii) The gradient of decay curve [1 mark] N
(iii) The decay rate [1 mark] 4 Q
P
(c) Therefore, state the relationship between half- 0 T T 2T 2T Time /
life and the gradient of decay curve in order p Q p Q days
to deduce the relationship between the gradient Figure 14
of decay curve and the decay rate. [2 marks]
(d) Bismut-210 has a half-life of 5 days and the initial activity is 800 counts per minute. Explain how the
nucleus decay occurs until the activity decreases to 200 counts per minute. [4 marks]
(e) You are required to identify the characteristics of the elements that are suitable to study the underground
liquid waste flow. Explain the stability of the element, the radiation released, its half-life and the harmless
HOTS
properties that the element must have. [10 marks]
15. The equation below represents a spontaneous radioactive decay.
4
226 Ra → 222 Rn + He
88 86 2
The table below shows the mass of the above three elements.
Element Mass (u)
226 Ra 226.025
88
222 Rn 222.018
86
4 He 4.003
2
(a) What is the meaning of mass defect? [1 mark]
(b) Based on the information above, compare radium-226 with radon-222 in term of proton number, nucleon
number and the mass. Then, state the relationship between nucleon number and the mass of the
element. [4 marks]
(c) Calculate the mass defect in unit u. [2 marks]
(d) Explain how the nuclear fission produces a large amount of energy. [3 marks]
(e) Figure 15 shows nuclear reactor that is built to meet the growing demand for domestic electricity.
HOTS









Cold gas FORM
5


Figure 15
Study the characteristics of a safe nuclear reactor in term of the following:
(i) Material to build a wall shield
(ii) How to control the fission reaction
(iii) The characteristics of the gas for transferring heat to the turbine. [10 marks]


453





06 SPM PHSICS F5.indd 453 10/03/2023 10:10 AM

ANSWERS






FORM 4 Checkpoint 1.2 Subjective Questions
Chapter Q1 (a) (i) An inference is an initial Section A
1 Measurement interpretation or explanation 1. (a) (i) s × m s = m
2
–2
concerning the observation. (ii) Base quantity
Checkpoint 1.1 (ii) A hypothesis is a statement (b) (i) 0.25 × 0.12 × 0.5 = 0.015 m 3
of an expected outcome 1.5 –3
Q1 Physical Base or that usually states the (ii) 0.015 = 100 kg m
Description derived relationship between two
quantity 2. (a) m / kg T / s 2
2
quantity or more variables intended
Base to be given a direct 2.0 0.8464
Total 55.2 g Mass experimental test.
quantity 3.0 1.2996
(iii) A variable is a physical
150 cm of Derived quantity that can be varied 4.0 1.7689
3
hot water Volume quantity in an experiment. 5.0 2.1609
Base (b) (i) The extension of the spring
80°C Temperature depends on the weight of 6.0 2.6569
quantity the object attached to it.
Within 3 Base (ii) The bigger is the magnitude (b)
minutes Time quantity of the weight, the greater is T /s 2
2
9 kJ of Derived the extension of the spring.
energy Energy quantity (iii) • Manipulated variable: 3.0
Weight of object attached
to the spring. 2.5
Q2 (a) (i) 7 500 000 m • Responding variable:
= 7 500 000 × 10 km Extension of the spring.
–3
= 7 500 km • Fixed variable: Type of 2.0
(ii) 7 500 000 m spring. 1.5
–6
= 7 500 000 × 10 Mm Q2 (a) To investigate the relationship
= 7.5 Mm between the distance of 1.0
7 853 m extension of the elastic cord, x
(b) 7 853 m s = 1 s and the horizontal distance, d 0.5
–1
7 853 × 10 km travelled by the ball.
–3
= (b) The further the distance of
1
 60 × 60 h extension of the elastic cord, 0 0 1.0 2.0 3.0 4.0 5.0 6.0 m / kg
x, the further is the horizontal
= 28 270.8 km h –1 distance, d travelled by the ball. (c) m = 2.65 – 1.30 = 0.45 s kg
2
–1
(c) • Manipulated variable: The 6.0 – 3.0
Q3 (a) Base quantity is quantity with distance of extension of the
magnitude only. elastic cord, x 3. (a) (i) Physical quantity is quantity
that can be measured.
(b) Vector quantity is quantity with • Responding variable: The (ii) V is a responding variable
magnitude and direction. horizontal distance travelled because its value depends
by the ball, d
Q4 Type of • Fixed variables: Type of on the other variables and
Event Explanation can only be determined
quantity elastic cord used, type of ball from the experiment. l is
1 Vector Magnitude used and height of the table a manipulated variable
700 km h and (d) • Tabulation of data: because its value can be
–1
direction required x / cm fixed before the experiment
to reach its and other variables depend
destination. d / cm on it and only can be
2 Scalar Only magnitude, • Analysis of data: determined from experiment.
3 kg required. Plot a graph of d against x. (b) (i) Graph of student A
If the gradient of the graph • Weakness: Horizontal
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 position of value l.
the lift. SPM Practice 1 • Weakness: Vertical axis,
4 Scalar Only magnitude Objective Questions 5 divisions for 0.4 V is a
is involved, the difficult scale to determine
temperature from 1. C 2. C 3. D 4. D 5. D the exact position of value
o
20 C to 100 C. 6. A 7. B 8. C 9. B 10. B V.
o
11. D 12. B 13. C 14. D
474
ANSWER FOCUS PHYSICS.indd 474 10/03/2023 10:42 AM

Physics SPM Answers

• Strength: The graph Q3 50 ticks per second. (b) Velocity = Gradient of graph
drawn is big. Therefore, 1 tick = 0.02 s 2.0 – 0.5
(ii) Graph of student B The time taken for each strip of = 3.8 – 2.4
• Strength : Horizontal axis, ticker tape = 1.1 m s –1
5 divisions for 0.2 A is an = 5 ticks × 0.02 s (c) Average velocity
easier scale to determine = 0.1 s
the exact position of value (a) Initial velocity, = Displacement
l. 3.0 Time taken
• Strength : Vertical axis, 5 u = 0.1 = 2.0
divisions for 0.5 V is an = 30.0 cm s –1 5.0
easier scale to determine = 0.4 m s –1
the exact position of value (b) Final velocity, v = 10.2 Q2 (a) s = Area under the graph from
0.1
V. 0 s to 14 s = 100 m
• Weakness : The graph = 102.0 cm s –1 (b) The bus stopped from t = 14 s
drawn is small. (c) Time taken between u and v, to 20 s. Therefore, it stopped for
(iii) Graph of student C t =  1 + 1 + 1 + 1  × 0.1 6 s.
• Weakness: Horizontal 2 2
axis, 5 divisions for 0.1 = 0.3 s (c) Gradient of graph
A is not suitable because v – u = 0 – 20
it does not cover the Acceleration, a = t 60 – 50
whole range of values 102.0 – 30.0 = –20
of l determined from the = 0.3 10 –2
experiment. = 240 cm s –2 = –2.0 m s
• Weakness: Vertical axis, or 2.4 m s –2 (d) Distance between the two bus
5 divisions for 0.4 V is a stops = Distance travelled from
–1
–1
difficult scale to determine Q4 (a) u = 8 m s ; v = 4 m s ; s = 6 m 20 s to 60 s
the exact position of value s =  u + v  × t = Area under the graph from
V. 2 20 s to 60 s
• Weakness: The graph is 2s = 580 m
not complete because t = u + v
some of the data from = 2 × 6 Q3 (a) (i) Total distance travelled
the experiment cannot be 4 + 8 = 35 × 2
= 70 m
plotted. = 1 s (ii) Average speed
(b) a = v – u 70
t = 55
Chapter = 4 – 8 = 1.27 m s (or 1.3 m s )
–1
–1
2 Force and Motion I 1
= –4 m s –2 (b) t = 10 s to 20 s
(c)
–1
–1
(c) u = 4 m s ; v = 0 m s ; t = 3 s
Checkpoint 2.1 v / m s –1
s =  u + v  × t
Q1 (a) Total distance travelled 2 2
= 1.8 + 0.9 + 0.7 + 1.6 =  4 + 0  × 3 1
= 5.0 km 2
(b) Average speed = 5.0 km = 6 m 0 t / s
2 h
= 2.5 km h –1 Q5 u = 18 m s ; v = 20 m s ; s = 10 m -1 5 10 15 20 25 30 35 40 45 50 55
–1
–1
(c) Displacement = 1.2 km due (a) Using the formula,
2
2
south of Farid’s house. v = u + 2as, -2
(d) Average velocity = 1.2 km 400 = 324 + 20 × a
76
2 h
= 0.6 km h –1 Therefore, a = 20 Q4 (a) (i) When t = 8 s to 13 s
Q2 (a) (i) Both tapes show uniform = 3.8 m s –2 (ii) The velocity of the lift was
velocity. (b) t = v – u zero during that time.
(ii) Tape P has a lower velocity a (b) (i) When t = 13 s to 20 s
compared with tape Q. = 20 – 18 (ii) The velocity of the lift during
(b) (i) Tape R: The separation 3.8 that time was negative,
between the dots is = 0.53 s indicating that the lift has
increasing. Therefore, changed direction.
the trolley moved with Checkpoint 2.2 (c) Total distance travelled
increasing velocity. The Q1 (a) • From t = 0 to 1 s, the crate = Total area under the graph
trolley was accelerating. moved with uniform velocity. = 20 + 18
(ii) Tape S: The separation • From t = 1 to 2.4 s, the crate = 38 m
between the dots is was at rest. (d) Displacement = 20 – 18
decreasing. Therefore, • From t = 2.4 to 3.8 s, the = 2 m
the trolley moved with crate moved with uniform Q5 The area under the graph gives the
decreasing velocity. The velocity. value for displacement of 100 m
trolley was decelerating. • From t = 3.8 to 5 s, the crate because this is a 100 m event
was at rest.
475
ANSWER FOCUS PHYSICS.indd 475 10/03/2023 10:42 AM