Physics

PENERBIT
ILMU
BAKTI SDN. BHD.

Contents PENERBIT ILMU 1
BAKTI SDN. BHD.3
Chapter 1 Physical Quantities and Measurements 7
11
1.1 Dimensions of Physical Quantities
1.2 Scalars and Vectors 14
1.3 Significant Figures and Uncertainties Analysis 19
Summative Practice 1 22
27
Chapter 2 Kinematics of Linear Motion
30
2.1 Linear Motion 33
2.2 Uniformly Accelerated Motion 37
2.3 Projectile Motion 42
Summative Practice 2 46
53
Chapter 3 Dynamics of Linear Motion
55
3.1 Momentum and Impulse 60
3.2 Conservation of Linear Momentum 63
3.3 Basic of Forces and Free Body Diagram 65
3.4 Newton’s Laws of Motion
Summative Practice 3 68
Summative Assessment Test (UPS) 1 71
73
Chapter 4 Work, Energy and Power 76

4.1 Work 27/6/2022 7:29:28 PM
4.2 Energy and Conservation of Energy
4.3 Power
Summative Practice 4

Chapter 5 Circular Motion

5.1 Parameters in Circular Motion
5.2 Uniform Circular Motion
5.3 Centripetal Force
Summative Practice 5

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Chapter 6 Rotation of Rigid Body 78
82
6.1 Rotational Kinematics 85
6.2 Equilibrium of a Uniform Rigid Body 88
6.3 Rotational Dynamics 90
6.4 Conservation of Angular Momentum 93
Summative Practice 6
95
Summative Assessment Test (UPS) 2 102
PENERBIT ILMU 106
BAKTI SDN. BHD.Chapter 7 Oscillations and Waves 110
114
7.1 Kinematics of Simple Harmonic Motion 117
7.2 Graphs of Simple Harmonic Motion 122
7.3 Period of Simple Harmonic Motion 124
7.4 Properties of Waves
7.5 Superposition of Waves 131
7.6 Application of Standing Waves 134
7.7 Doppler Effect 137
Summative Practice 7 141
144
Chapter 8 Physics of Matters 150

8.1 Stress and Strain 153
8.2 Young’s Modulus 156
8.3 Heat Conduction 159
8.4 Thermal Expansion 161
Summative Practice 8 164
167
Summative Assessment Test (UPS) 3 172
175
Chapter 9 Kinetic Theory of Gases and Thermodynamics

9.1 Kinetic Theory of Gases
9.2 Molecular Kinetic Energy and Internal Energy
9.3 First Law of Thermodynamics
9.4 Thermodynamic Processes
9.5 Thermodynamic Work
Summative Practice 9

Semester Examination
Answers

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1CHAPTERPENERBIT ILMU
BAKTI SDN. BHD.
Physical Quantities and
Measurements

1.1 Dimensions of Physical Quantities

Learning Outcomes

● Define dimension.
● Determine the dimensions of derived quantities.
● Verify the homogeneity of equations using dimensional analysis.

1 Physical quantity is defined as anything that can be measured.
2 Dimension is the physical nature of a quantity. In other words, it indicates

how a physical quantity is related to its base quantities.
3 Uses of dimensional analysis:

(a) To determine the unit for a physical quantity
(b) To check the homogeneity/consistency/validity of an equation
(c) To create an equation
4 The symbols for base quantities are as follows:

Table 1.1 Symbols for base quantities

Base quantity Symbol Symbol of dimension

Length l L
Mass m M
Time t T

5 The dimensions of derived quantities can be determined by the following steps:
(a) Step 1: Write the equation of derived quantity in terms of base quantities.
(b) Step 2: Put the symbols for the base quantities of the physical quantity
in brackets [ ].

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 Physical Quantities and Measurements

(c) Step 3: Substitute the symbol of dimension of base quantity.
(d) Step 4: Simplify to the lowest terms.
6 Steps to determine the homogeneity of equations using dimensional analysis
are as follows:
(a) Step 1: Dimensional analysis of left-hand side (LHS) of the equation.

Write the physical quantities on the left-hand side in terms of symbols for
base quantities and put them in square brackets [ ]. Substitute the symbol
of dimension of base quantity. Then, simplify to the lowest terms.
(b) Step 2: Dimensional analysis of right-hand side (RHS) of the equation.
Write the physical quantities on the right-hand side in terms of symbols
for base quantities and put them in square brackets [ ]. Substitute the
symbol of dimension of base quantity. Then, simplify to the lowest terms.
(c) Step 3: Compare the results of the left-hand side (LHS) and the right-hand
side (RHS).
(d) Step 4: Make a conclusion. If both sides are the same, the equation
is homogeneous. If both sides are not the same, the equation is not
homogeneous.
Chapter 1 BAKTI SDN. BHD.

Example 1

What is the unit for velocity?

Mid Semester Test 1 Solution

Writing the equation of the derived quantity in terms of base quantities.
∆s
v = ∆t

Smart Tips! Then, putting the symbols for the base quantities of the derived quantity in

Make sure brackets [ ] are ILMUbrackets.
used for physical quantity
only not for symbol [ ][v] = ∆s = L = L T–1
of dimension of base ∆t T
quantity.
Converting the symbols for the base quantities into units.
⸫ Unit for v is m s–1.

AnswersPENERBIT Example 2

Determine whether v = u + at is homogeneous or not.

Solution

Dimensional analysis of v:

[ ][v] = ∆s = L = LT –1
∆t T

Dimensional analysis of u and at:

Common Error [ ][u] = ∆s = L = LT –1
∆t T
Students conduct
dimensional analysis for [a][t] = (LT–2)(T) = LT–1
u + at but forget to do it
for v. Comparing the results of LHS to RHS.
The dimensions of both sides are the same.
2
І The equation is homogeneous.

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PENERBIT ILMU Physical Quantities and Measurements  Chapter 1
BAKTI SDN. BHD.
1.3 SignFiigfuirce a1.5nTthuFmbigpouintrs etostheadinrecdtionUofnrecsueltarntt aveicntorties Mid Semester Test 1
Analysis
Answers
Learning Outcomes

● State the significant figures of a given number.
● Use the rules for stating the significant figures at the end of calculation

(addition, subtraction, multiplication or division).
● Determine the uncertainty for average value and derived quantities.
● Calculate basic combination (propagation) of uncertainties.

Rules for Significant Figures

1 All non-zero digits are significant.
For example, 45.6 has three significant figures because all of the digits present
are non-zero digits.

2 Zeros are significant if:
(a) zeros are between two non-zero digits
For example, 1 042 has four significant figures because the zero is between
digits 1 and 4.
(b) trailing zeros to the right of a decimal
For example, 65.00 has four significant figures because of the zeros
trailing to the right.
(c) trailing zeros in a whole number with the decimal shown
For example, 670. has three significant figures because of the method
of convention by placing a decimal at the end of a number. This decimal
indicates a significant zero. "670." indicates that the trailing zero is
significant.
(d) exact numbers have an infinite number of significant figures
For example, 2 metres = 2.00 metres = 2.0000 metres =
2.0000000000000000000 metres. The value in order for the trailing zeros
to be counted as significant, must be followed by a decimal.

3 In the form of scientific notation, for example, N × 10x, all digits comprising N
are significant by the first 5 rules above but "10" and "x" are not significant.
For example, 8.03 × 107 has three significant figures because only "8.03" is
significant.

4 Zeros are not significant if:
(a) leading zeros are not significant
For example, 0.21 has two significant figures. 0.0021 also has two
significant figures because all the zeros are leading.
(b) trailing zeros in a whole number with no decimal shown are not
significant
For example, 320 has two significant figures because zero is not
significant.

Rules for stating the significant figures at the end of calculation

1 For addition or subtraction, the final answer must be stated in smallest decimal
places of the values given.

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Summative Physical Quantities and Measurements 

Practice 1

PENERBIT ILMU Objective Questions 6 State the number of significant figures for 0.987.
BAKTI SDN. BHD. A4
1 What is a dimension? B3
A It is about unit conversion C2
B A derived quantity that can be measured D1
C An indicator of the magnitude of a physical
quantity 7 State the number of significant figures for 7 900. Chapter 1
D A derived quantity shown in terms of base A4
quantities B3
C2
2 Which statement is true about scalar quantity? D1
A Velocity and acceleration are examples of
scalar quantity 8 Which of the following combinations of
B Quantity that cannot be measured uncertainties is not correct?
C Quantity that has magnitude only
D Quantity in two dimensions only A v= s ∆v = ∆s + ∆t Mid Semester Test 1
t v s t
3 Force is a vector quantity. What is a vector
quantity? B F = 2x ∆F = 2∆x
A A quantity that can be measured in two
dimensions only C v = 2u ∆v = u2
B A quantity that has both magnitude and v ∆u
direction
C A quantity that cannot be derived D v2 = u2 + 2as 2 ∆v = 2∆uu + ∆a + ∆s
D A quantity that has no unit v a s

4 What is the dot product? 9 Determine the combination of uncertainties for
A The product of two vectors that have different
directions s = ut.
B The product of two vectors that are parallel to
each other A ∆s = ∆ut
C The product of two vectors that have the same
magnitude B ∆s = ∆u + ∆t
D The product of two vectors that are
perpendicular to each other C ∆s = ∆u + ∆t Answers
s u t
5 What is the meaning of cross product? ∆s ∆u ∆t
A The product of two vectors that are D s = u × t
perpendicular to each other
B The product of two vectors that are parallel to 10 Which of the following combinations of
each other uncertainties is true?
C The product of two vectors that have opposite
directions A v= s ∆v = ∆s + ∆t
D The product of two vectors that have the same t
direction
B r= 1 d ∆r = 2∆d
2

C τ = 3θ ∆τ = ∆θ

D v2 = v12 ∆v2 = 2 ∆v1
v2 v1

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 Physical Quantities and Measurements

Structured Questions
1 Find the S.I. unit for acceleration using dimension.

2 Verify the homogeneity of the equation s = at + 1 ut.
2

3 Verify the consistency of the equation v 2 = u2 + 2as. EXAM CLONEBAKTI SDN. BHD.

4 Determine whether the equation v = f λ is dimensionally correct.

5 Determine the equation for the period of a simple pendulum where the mass is m, length is l
and acceleration due to gravity is g.

Chapter 1 6 A force, F1 of 50 N acts on a box at an angle of 60° to the floor. At the same time, the frictional
force of 5 N also acts on the box in the negative x-direction. Determine the resultant force

acting on the box. (Other forces are neglected.)

7 A vector is given as Q = i + 2j + k. Draw a diagram to show the x, y and z-components
of this vector.

8 Figure 1 shows a vector, v. y

Mid Semester Test 1 yA

v = 12.5 m s–1

θ1 Cx
θ2 0
35° x
0 B

Figure 1
ILMU
Determine the x-component and y-component of the above vector.

9 Three vectors are given as A = 15 m, B y= 20 m and C = 30 m as shown in Figure 2.

y E = 35 N C–1
30° 1

yA
AnswersPENERBIT
v = 12.5 m s–1 Ox

θ1 E20= 23 N C–1C x
θ2

35° x
0
B

Figure 2

If θ1 = 30° and θ2 = 45°, calculate the magnitude of the resultant vector.

10 The x-coymponent of the velocity of an object is 25 m s–1 in the positive x-direction and its
yo-bcjoemct.po3n0e°ntEi1s=3305 NmCs––11 in the negative y-direction. Determine the resultant velocity of the

Ox
E2 = 23 N C–1

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θ1 Cx
θ2 0
35° x
0 B

Physical Quantities and Measurements 

11 Figure 3 shows two electric fields of strengths, E 1 and E 2 acting on the origin, O.

y
30° E1 = 35 N C–1

O x

PENERBIT ILMU E = 23 N C–1
BAKTI SDN. BHD. 2

Figure 3

Determine the magnitude and direction of the resultant electric field strength, E .

12 Figure 4 shows two forces, F1 and F2 exerted on an object. Chapter 1

y y

10° x A = 2i + 4j
0 20° x

F1 = 11 N F2 = 15 N B = –1.5i
0

Figure 4

Calculate the magnitude of the resultant force exerted on the object.

13 VectorsyA and B are shown in Figure 5. y Mid Semester Test 1

y

10° x A = 40 N A = 2i + 4j
0 20°

F1 = 11 N F2 = 15 N B = –610.5°i y x x y
20° x 0
00 A = 2i + 4j
10° x

B = 3F0igNure05

Calculate the magnitude of vector CFi1f=C11 iNs the resultaFn2t=v1e5cNtor of A and BB. = –1.5i

14 ADy ebtoeArym=k4iin0cekNsitas ball with a velocity of 25 m s–1 at an angle of 40° above the positive x-axis.
x-component and y-component.

15 Calculate the resultant vector of A and B . Answers
60°
0 x y
A = 40 N
B = 30 N

60° x
0

B = 30 N

Figure 6

16 State the number of significant figures of the following numbers:

(a) 23.03 (b) 1.26 (c) 0.005

17 State the final answer in standard form of rules of significant figures for the following operations:

(a) 34.1 + 20.456 = (b) 23.91 ÷ 2.7 =

18 Write an expression to determine the uncertainty for power, P = Fts.

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5 The motion of the object can be in one dimension (only x-direction or Kinematics of Linear Motion 
y-direction) or two dimensions (x-direction and y-direction).
Smart Tips!
Example 1
Before using the
An object moves with a constant acceleration from rest to 15 m s–1 in 10 s. equations, make a list of
Calculate the acceleration and distance travelled by the object. the data given:
PENERBIT ILMU u = 0, v = 15 m s‒1,
BAKTI SDN. BHD.Solution t = 10 s

Use the appropriate linear motion equation to find a, Common Error

v = u + at Students do not realise
that u = 0. Object at rest
15 = 0 + a(10) means static, u = 0.

a = 1.5 m s–2 Alternative Chapter 2
Method
Use the appropriate linear motion equation to find s,
1 Other linear motion
s = ut + 2 at2 equations can also be
s = used to find the distance
(0)(10) + 1 (1.5)(10)2 travelled by the object.
2
s = 75 m v 2 = u2 + 2as
152 = 02 + 2(1.5)s
Example 2
s = 75 m
A car moves with a constant velocity of 40 km h–1 for 1 minute. Then, it Mid Semester Test 1
accelerates until its velocity reaches 90 km h–1 in 5 minutes. Calculate the

displacement of the car.

Solution

For first 1 minute, constant v, a = 0
1
s = ut + 2 at2

40 × 103
60 × 60
( )s1 = (60) + 1 (0)(60)2
2

= 6.67 × 102 m

For the next 5 minutes,

v = u + at
( ) ( )90 × 103
60 × 60
= 40 × 103 + a(5 × 60) Answers
60 × 60

25 = 11.11 + 300a

a = 0.05 m s–2
1
І s2 = ut2 + 2 at 2
2

40 × 103
60 × 60
( )s2 = (5 × 60) + 1 (0.05)(5 × 60)2
2
= 5.58 × 103 m

Total displacement

= s1 + s2
= (6.67 × 102) + (5.58 × 103)

= 6.25 × 103 m

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 Kinematics of Linear Motion

Example 2

A ball is kicked with an initial velocity of 40 m s–1 at an angle of 35° above the
positive x-axis. How far will it travel in the x-direction?
Solution
BAKTI SDN. BHD.
Smart Tips! 40 m s–1

• For such type of 35°
questions, always
sketch the diagram of sx
the situation.
Chapter 2 For horizontal motion, sx = ux t
• The total distance
can be determined by sx = (40 cos 35°)t … 1
first finding the total
travelling time of the For the whole journey, sy = 0
ball. 1
sy = uy t – 2 gt 2
0 = (40
sin 35°)t – 1 (9.81)t 2
2

t = 4.68 s, t = 0 s (neglected because t = 0 is at the launch point)

І Total travelling time, t = 4.68 s

Mid Semester Test 1 Substitute the value of t into equation 1 .

Total distance travelled, sx = (40 cos 35°)(4.68)
sx = 153.35 m

ILMU
Quick Check 2.3
AnswersPENERBIT
1 A ball is thrown vertically upwards with a velocity of 25 m s–1. Calculate the
maximum height reached by the ball.

2 An archer shoots an arrow from the top of a 15-m high building with a
velocity of 20 m s–1 at an angle of 30° to the horizontal. Calculate the velocity
of the arrow just before it strikes the ground.

3 An object is dropped from a height of 50 m. Calculate its velocity just before
it hits the ground.

4 A ball is thrown vertically upwards with a velocity of 25 m s–1. Calculate the
time taken for the ball to reach the maximum height.

5 An arrow is launched with a velocity of 23 m s–1 at an angle of 30° to the
horizontal from the roof of a building of height 40 m. Determine the time
taken for it to hit the ground.

6 A ball is launched horizontally off the edge of a table with a velocity of
20 m s–1. If the time taken by the ball to hit the ground is 2.0 s, calculate the
magnitude of its velocity just before it hits the ground.

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Dynamics of Linear Motion 

Quick Check 3.1

1 A 30-g ball is thrown horizontally at a speed of 10 m s–1. The ball strikes a
wall and rebounds at the same speed. Calculate
(a) the initial momentum of the ball,
(b) the final momentum of the ball,
(c) the change in momentum of the ball,
(d) the impulse of the ball on the wall,
(e) the impulse of the wall on the ball,
(f) the magnitude of impulsive force if the contact time is 0.01 s.

2 A ball of mass 0.3 kg moving to the right strikes a wall with a velocity of
20 m s–1 and rebounds at the same speed. Calculate the magnitude of the
impulse of the ball.

3 A car of mass 1 200 kg moves with an initial velocity of 60 km h–1. Calculate
its initial momentum in S.I. unit.

4 A golf club strikes a golf ball of mass 45 g, causing the ball to fly with a
velocity of 35 m s–1. Calculate the force exerted on the ball by the club if the
contact time is 2 ms.

5 A ball of mass 30 g is struck by a racket. The related force-time graph is
shown below.

F (N)
PENERBIT ILMU Chapter 3
BAKTI SDN. BHD.
20 Mid Semester Test 1

0 1.5 2.4 t (ms)

Calculate the impulse.

6 A ball of mass 0.2 g initially at rest is struck by a wooden bat with a force of
18 N in 1.5 ms. Calculate the speed of the ball after it is struck by the bat.

3.2 Conservation of Linear Momentum Answers

Learning Outcomes 33
● State the principle of conservation of linear momentum.
● Apply the principle of conservation of momentum in elastic and inelastic 27/6/2022 7:12:17 PM

collisions in two-dimensional collisions.
● Differentiate elastic and inelastic collisions.

1 In a closed system (Fexternal = 0), the total momentum is conserved. This is called
the principle of conservation of linear momentum.
∑ pi = ∑ pf

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Dynamics of Linear Motion 

4 In perfectly elastic collisions, the objects separate after the collision.
5 In perfectly inelastic collisions, the objects stick and move together after the

collision.

6 Thus, the motion of objects can be separated or the objects may stick and
move together after collision. This, however, is not an indicator of the type of
collision.

7 The only way to determine the type of collision is by comparing ∑Ki and Kf.
If ∑ Ki is equal to ∑ Kf , there is an elastic collision. If ∑ Ki is not equal to ∑ Kf ,
there is an inelastic collision.
PENERBIT ILMU
BAKTI SDN. BHD.Example 1

Ball A of mass 3 kg moving to the right with an initial velocity of 5 m s–1 Chapter 3

collides with a stationary ball B of mass 2 kg.
(a) If the velocity of ball B is 2 m s–1 to the right after the collision, calculate

the final velocity of ball A.

(b) What is the type of the collision?

Solution uB = 0 m s–1 vA vB = 2 m s–1 Smart Tips! Mid Semester Test 1
(a) B AB
Always sketch diagrams
uA = 5 m s–1 to illustrate the situation.
A
Smart Tips!
mA = 3 kg mB = 2 kg
Remember:
Before collision After collision • Elastic collision:

In a closed system (Fexternal = 0), ∑Ki = ∑Kf
∑ pi = ∑ pf • Inelastic collision:

mAuA + mBuB = mAvA + mBvB ∑Ki ≠ ∑Kf
(3)(5) + (2)(0) = (3)vA + (2)(2)
Common Error
vA = +3.67 m s–1 (to the right)
• Students assume that
(b) Before collision, ∑ Ki = KiA + KiB since A and B do not
1 1 stick together after the
∑ Ki = 2 mAuA2 + 2 mBuB2 collision, it must be an
elastic collision.
∑ Ki = 1 (3)(5)2 + 1 (2)(0)2 Answers
2 2 • Students state the type
of collision as either
= 37.5 J ‘separate’ or ‘stick
together’.
After collision, ∑ Kf = KfA + Kf B

∑ Kf = 1 mAvA2 + 1 mBvB2
2 2
1 1
∑ Kf = 2 (3)(3.67)2 + 2 (2)(2)2

= 24.20 J

Comparing between ∑ K f and ∑ Ki,
So, ∑ Ki ≠ ∑ K f
∑ K is not conserved.

This collision is an inelastic collision.

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Summative Assessment Test (UPS) 1 

Summative Assessment Test (UPS) 1

Time: 30 minutes
[13 marks]

Answers all questions.
PENERBIT ILMU
1 Which of the following statements is not trueBAKTI SDN. BHD.6 A ball moves with constant velocity. Which UPS 1
about the dimension of physical quantities? statement is true about the motion of the ball?
A A method by which physical quantity can be A The ball does not accelerate Answers
expressed in the form of combinations of basic B The ball has non-zero constant acceleration
quantities C The ball has non-zero constant deceleration
B The principle of homogeneity can be used to D The ball accelerates with a value greater than 1
determine whether the equation is correct or
not 7 If an object moves with constant acceleration, this
C The unit of physical quantities can be means that ...
determined by using dimensional analysis A there is a change in velocity per unit time.
D The dimension of a dimensionless quantity is 0 B there is a change in distance per unit time.
C there is no change in velocity per unit time.
2 Which of the following statements is true about D there is no change in distance per unit time.
vector quantity?
A Time is an example of vector quantity 8 What is the physical quantity represented by the
B A quantity which has magnitude only gradient of the displacement-time graph?
C Velocity is an example of vector quantity A Velocity
D A quantity which has a value larger than 1 B Distance
C Acceleration
3 State the number of significant figures for 0.540. D Displacement
A 4 significant figures
B 3 significant figures 9 Which of the following statements is not true?
C 2 significant figures A Impulse occurs when there is a change in
D 1 significant figure velocity of an object
B Impulse is equal to the product of force and
4 What is an instantaneous velocity? displacement
A Rate of change of velocity C Impulse is equal to the product of force and
B Rate of change of distance contact time
C Rate of change of speed at a particular time D Impulse occurs when there is a change in
D Rate of change of displacement at a particular momentum
time
10 The principle of conservation of linear momentum
5 What is the average velocity of an object? states that ...
A Rate of change in displacement A total linear momentum is not the same if there
B Rate of change in acceleration is a change in velocity.
C Rate of change in distance B total linear momentum is constant in a closed
D Rate of change in speed system.
C total energy can be changed anytime and
anywhere.
D total energy is constant in a closed system.

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Rotation of Rigid Body 

Parameter Rotational motion Linear motion

Equations of α is constant a is constant
motion
ω = ω0 + αt v = u + at

ω2 = ω02 + 2αθ v 2 = u2 + 2as

PENERBIT ILMU θ = ω0t + 1 αt2 s = ut + 1 at2
BAKTI SDN. BHD. 2 2

θ = 1 (ω0 + ω)t s = 1 (u + v)t
2 2

Example 1

An ant is 0.28 m from the axle of a miniature rotating wheel. If the rotation Common Error Chapter 6
speed of the wheel decreases from 4.5 rad s‒1 to 1.9 rad s‒1 in 6 seconds,
calculate Students cannot
(a) the average rotational acceleration of the wheel, differentiate between
(b) the angle through which the ant rotates during the 6 seconds, initial angular velocity
(c) the total linear displacement of the ant during the period. and final angular
velocity.

Solution Alternative
Method
r = 0.28 m, ω0 = 4.5 rad s–1, ω = 1.9 rad s–1, t = 6 s Mid Semester Test 1
(a) ω = ω0 + αt Other rotational motion

1.9 = 4.5 + α(6)

α = – 0.43 rad s–2 equations can also be

(b) θ = 1(ω490..5t2)+6(6r12)ad+αt212 used to find the angle of
θ
= (– 0.43)(6)2 rotation.
=
θ = 1 (ωo + ω)t
2
1
(c) s = rθ = (0.28)(19.26) = 5.39 m θ = 2 (4.5 + 1.9)6

= 19.2 rad

Quick Check 6.1 Answers

1 A marble is attached to a 0.8-m string and makes 2 rotations in one second.
Find its period, tangential velocity and angular velocity.

2 A disc starts to spin from rest about the axis through its centre with a constant
angular acceleration of 1.5 rad s‒2. Find the linear speed of a point 4 cm from
its centre after 3 seconds.

3 A light string is wound around a disc with a diameter of 30 cm which can
rotate freely about the axis through its centre. The string is then pulled for
2.2 seconds with an acceleration of 1.1 m s‒2. Calculate the final angular speed
of the disc.

4 The angular velocity of a wagon wheel is 8 rad s‒1. After 4 seconds, its angular
velocity is 5 rad s‒1. Calculate the angular acceleration of the wheel and the
time taken for the wheel to stop.

5 A wheel starts to spin from rest about its axis of rotation to 35 rad s‒1 in
18 seconds. Determine, after 5 seconds
(a) its angular displacement
(b) its tangential acceleration at a distance of 15 cm from the axis of rotation.

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 Rotation of Rigid Body

6.4 Conservation of Angular Momentum

Learning Outcomes

● Explain and use angular momentum, L = Iω
● State and use principle of conservation of angular momentum.
BAKTI SDN. BHD.
Chapter 6 Angular Momentum

Mid Semester Test 1 Smart Tips! 1 The angular momentum, L of a particle is defined as

L = r(mv) |L| = r × p sin θ
We know v = rω,
L = r m (rω) where
L = mr 2ω L = angular momentum
We know I = mr 2, r = vector from axis of rotation
L = Iω p = linear momentum of particle
θ = angle between r and p
2 The S.I. unit for angular momentum is kg m2 s–1.
3 Angular momentum, L is analogous to linear momentum.
Linear momentum, p = mv, thus

L = r × mv

4 Angular momentum, L is defined as the product of angular velocity, ω of a
body and its moment of inertia, I about the axis of rotation.

L = Iω

Principle of Conservation of Angular Momentum

1 The principle of conservation of angular momentum states that the total
momentum of a closed system about an axis of rotation is conserved if no
external torque acts on the system.

ΣLi = ΣLf , if Στ = 0
ILMU

AnswersPENERBIT Example 1

A 25-N m torque is used to rotate a cylindrical disc which has a moment of
inertia of 210 kg m2. If the disc starts from rest, what is its rotational speed

after 5 seconds?

Solution

τ = Iα,
τ
Common Error α = I

Students are unable α = 25
to relate equation 210
of torque, τ = Iα to = 0.12 rad s–2
equation of rotational
motion. ω = ω0 + αt
ω = 0 + (0.12)(5)
88
= 0.60 rad s–1

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Summative Assessment Test (UPS) 2 

Summative Assessment Test (UPS) 2

Time: 30 minutes
[13 marks]

Answers all questions.
PENERBIT ILMU
BAKTI SDN. BHD.1 Which of the following statements is true about5 Based on the principle of conservation of energy,
work done?
A Work is done when an object moves after it is total energy is always in a closed
pushed by a force
B Work is done by a lift force when the object system.
moves horizontally
C Work is done when the angle between force A decreasing
and displacement is 90°
D Work is done when an object remains at rest B increasing
after force is applied to it
C the same
2 Which of the following statements is true about
the principle of conservation of energy? D zero
A If there is no external force exerted on the
system, the initial total energy is equal to the 6 Which of the following statements is true about an
final total energy object in uniform circular motion?
B The total energy is the same at any point when A Speed is constant
it involves the same type of energy B Velocity is constant
C The kinetic energy is always equal to the C Momentum is constant
potential energy D Direction of acceleration is constant
D The total energy is not constant
7 Which of the following physical quantities UPS 2
3 What is the meaning of instantaneous power? changes when a body is in a uniform circular
A The rate of change of force over a time interval motion?
B The rate of change of force at a particular time A Mass
C The rate of change of work done over a time B Radius
interval C Kinetic energy
D The rate of change of work done at a particular D Direction of velocity
time only
8 What is the direction of acceleration in a circular Answers
motion?
A Along the circumference
B Perpendicular to the radius
C In the opposite direction to velocity
D Towards the centre of the circular path

4 Which of the following situations has zero work 9 Which of the following statements does not
done? describe about instantaneous angular acceleration?
A Force and displacement are perpendicular to
each other A Change of angular velocity per unit time at a
B Force and displacement are parallel to each particular time
other
C Force is in the opposite direction to B Instantaneous angular acceleration is a scalar
displacement quantity
D Force is in the same direction as displacement
C Differentiation of angular velocity with respect
to time

D The unit of angular acceleration is rad s–2

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PENERBIT ILMU7CHAPTER
BAKTI SDN. BHD.
Oscillations and Waves

7.1 Kinematics of Simple Harmonic Motion

Learning Outcomes

● Explain simple harmonic motion (S.H.M.).
● Apply S.H.M. displacement equation, y = A sin ωt

● Derive and use equations:
– velocity, v = ωA cos ωt = ± ω A2 – y 2

– acceleration, a = –ω2A sin ωt = –ω2y

– kinetic energy, K = 1 mω2(A2 – y 2) and
2
1
– potential energy, U = 2 mω2y 2

● Emphasise the relationship between total S.H.M. energy and amplitude.

● Apply equations of velocity, acceleration, kinetic energy and potential

energy for S.H.M.

1 Simple harmonic motion (S.H.M.) can be described as repetitive movement of
an object back and forth (oscillating) through an equilibrium position without
energy loss. The maximum displacement on one side is equal to the maximum
displacement on the other side.

2 The acceleration of the object is proportional to its displacement from the
equilibrium position but in the opposite direction. Mathematically,
ay

where a = – ω2y

a = acceleration of object

ω = angular velocity/angular frequency

y = displacement of the object from the equilibrium position

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 Oscillations and Waves Simple Harmonic Motion Displacement Equation

96 1 The equation for displacement as a function of time in simple harmonic motion
can be written as

where y = A sin (ωt +– ϕ)

A = amplitude BAKTI SDN. BHD.

(ωt +– ϕ) = phase or phase angle
ϕ = phase constant

2 If the oscillation starts at equilibrium, then ϕ = 0, and

y = A sin (ωt)

Chapter 7 Important terms in simple harmonic motion

1 The following are the important terms in simple harmonic motion:
(a) Amplitude, A
Amplitude is the maximum magnitude of the displacement from the
equilibrium position. Its S.I. unit is metre (m).
(b) Period, T
Period is defined as the time taken for one cycle. Its S.I. unit is second (s).

Equation: T = 1
f
Mid Semester Test 1
(c) Frequency, f

Frequency is defined as the number of cycles in one second. Its S.I. unit

is hertz (Hz).

1 Hz = 1 cycle s–1 = 1 s–1

Equation: From ω = 2πf,

ILMU f = ω
2π
(d) Phase, (ωt ± ϕ)

Phase is a time-varying quantity. Its unit is radian (rad).

(e) Initial phase angle or phase constant, ϕ

The phase constant indicates the starting point in simple harmonic

motion when the time, t = 0 s. When ϕ = 0, the starting point is at

AnswersPENERBIT equilibrium.

Mass at various positions

1 Figure 7.1 shows a mass, m undergoing simple harmonic motion in various
situations. In all cases, the mass is at the equilibrium position when it is at O
(points 2, 4). In Figures 7.1 (b) and (c), the mass is at an equilibrium position
when the spring is at its natural length and does not exert any force on the
mass.

3 42 1

+A 3

3 42 1 O 4
–A O +A 2
–A 1
(a) Simple pendulum –A O +A
(b) Vertical mass-spring
(c) Horizontal mass-spring

Figure 7.1 Mass, m undergoing simple harmonic motion in various situations

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 Semester Examination

Semester Examination

Time: 2 hours
[100 marks]
Answers all questions.
BAKTI SDN. BHD.
1 What is the unit for acceleration?

[2 marks]

Semester Examination 2 (a) A stationary car starts to move with a constant acceleration of 2 m s–2 through a distance of
1 500 m. Calculate
(i) the final velocity of the car,
(ii) the time taken by the car to travel the distance.

(b) A boy drops a stone from a height of 10 m above the ground.
Calculate
(i) the time taken for the stone to hit the ground,
(ii) the velocity just before it hits the ground.

(c) An arrow is launched from a height of 70 m above the ground. It flies with a velocity of
60 m s–1 at an angle of 40° above the positive x-axis.
(i) How long does it take to reach the ground?
(ii) Calculate its velocity just before it hits the ground.
[14 marks]

3 (a) A 0.40-kg ball moves to the right and hits a wall with a speed of 150 m s–1.
It bounces off the wall with a speed of 100 m s–1 in the opposite direction.
(i) Calculate the impulse delivered to the ball by the wall.
(ii) If the force exerted on the wall is 3 000 N, calculate the contact time between the ball
and wall.

(b) The graph of force against time for a ball of mass 50 g when struck by a racket is shown in
Figure 1.

F (kN)
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AnswersPENERBIT 20

0 2.0 4.0 t (ms)

Figure 1

Calculate the impulse delivered to the ball.
(c) A bullet of mass 10 g travelling with a speed of 300 m s–1 to the right hits a stationary block

of mass 3 kg. The bullet gets embedded in the block and they both move together with a
speed, v. Determine their final velocity.

[9 marks]

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Answers 

Answers

PENERBIT ILMU1Physical Quantities and Dimensional analysis of m and a: 3 θ P = 30 N
BAKTI SDN. BHD.Measurements [m] = M
[a] = L T –2
Quick Check 1.1 [m][a] = M L T –2 Q = 20 N F
Comparing the results of LHS to RHS.
1 (a) Acceleration, a = ∆v The dimensions of both sides are the Magnitude of resultant vector,
∆t same.
Hence, the equation is homogeneous.
Putting the symbols of the base |F| = P 2 + Q 2

quantities of a into [ ],
[ ][a] =
∆v |F | = 302 + (–20)2
∆t = 36.06 N

= L T –1 Quick Check 1.2 Direction of resultant vector,
T
1y y | | | |tan θ =Q = –20
= L T –2 Py P 30
P = 45 kg m s–1
⸫ Unit for a is m s–2. P

(b) Force, F = ma θ = 33.69° below positive x-axis

Putting the symbols of the base 30° 0 x 30° 0x
Px
quantities of F into [ ], 4 v 20 m s–1

[F ] = [ma] = [m][a] Resolving P into x-component and
y-component,
= M L T –2 Mid Semester Test 1
P x = – 45 cos 30°
⸫ Unit for F is kg m s–2. = –38.97 kg m s–1
F
(c) Pressure, P = A P y = 45 sin 30° 15 m sv–1 20 m s–1
= 22.5 kg m s–1
Putting the symbols of the base v= vx2 + v 2
y

quantities of P into [ ], v =B =(1–11.55)N2 + (20F)2
[F]
[ ][P] =F = [A] 2 = 25 m s–115 m s–1
A

[P] = M L T –2 5 A = 13.5 N
L2 F
F Fx Fy B = 11.5 N

= M L–1 T–2 Let F 1 F 1x = 40 cos 60° F 1y = 40 sin 60°
= 40 N = 20 N = 34.64 N
⸫ Unit for P is kg m–1 s–2.

(d) Volume, V = l 3 A = 13.5 N

Putting the symbols of the base Let F 2 F 2x = –30 cos 85° F 2y = –30 sin 85° F= F 2 + F 2
= 30 N x y
quantities of V into [ ], = –2.61 N = –29.89 N

[V ] = [l 3] = L3 Sum of ∑F x = 17.39 N ∑F y = 4.75 N F = (13.5)2 + (11.5)2
components
⸫ Unit for V is m3. = 17.73 N Answers
(e) DPuetntisnitgy,thρe=symVmbols of the base
Magnitude of resultant force, Quick Check 1.3

quantities of ρ into [ ], |∑ F | = ∑(F x)2 + ∑(F y)2 1 (a) 2 significant figures because
trailing zero without decimal are
[ ][ρ] =m = [m] |∑ F | = (17.39)2 + (4.75)2 not significant.
V [V]
(b) 3 significant figures because all
= M = 18.03 N non-zero digits are significant.
L3 Direction of resultant force,
(c) 4 significant figures because
= M L–3 | | | |tan θ =Fy the zeros after the last non-zero
Fx = 4.75 digit after the decimal point are
⸫ Unit for ρ is kg m–3. 17.39 significant.

2 LHS, [s] = L θ = 15.28° above positive x-axis (d) 2 significant figures because
RHS, [vt] = [v][t] = (L T–1)(T) = L leading zeros are not significant.
LHS = RHS Fy = 4.75 N F = 18.03 N
Hence, the equation is dimensionally
correct. θ = 15.28° 2 (a) 45.9 (b) 3.85
(c) 16.7 (d) 2.5
3 Dimensional analysis of F: Fx = 17.39 N
[F ] = M L T –2

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ILMU
BAKTI SDN. BHD.