Buku Teks Fizik Tingkatan 4/physics-form-4-chapter-2.pdf

Newtonian mechanics is a branch of
physics that studies the motion of an object.
The topics in this theme investigate the
concepts and factors that cause changes in
the motion of an object.

Newtonian
Mechanics

2323
CHA
PTE

R FORCE ANDFORCE AND
MOTION MOTION II2

Let’s StudyLet’s Study

2.1 Linear Motion2.1 Linear Motion
2.2 Linear Motion Graphs2.2 Linear Motion Graphs
2.3 Free Fall Motion2.3 Free Fall Motion
2.4 Inertia2.4 Inertia
2.5 Momentum2.5 Momentum
2.6 Force2.6 Force
2.7 Impulse and Impulsive Force2.7 Impulse and Impulsive Force
2.8 Weight2.8 Weight

How is linear motion of an How is linear motion of an
object investigated?object investigated?

What causes the changes in the What causes the changes in the
state of motion of an object? state of motion of an object?

Why are Newton’s Laws of Motion Why are Newton’s Laws of Motion
important in the study of motion important in the study of motion
of an object?of an object?

2424
Learning Standards and Learning Standards and
List of FormulaeList of Formulae

Have you seen a personal transporter (PT) before? PT
is a smart personal transporter as shown in the photograph.
The device is used by security personnel to patrol in airports or
shopping malls. This device is also used by tourists to make tours
in tourist areas such as Perdana Botanical Gardens and Putrajaya.

This device is environmentally friendly because it runs on
batteries. Fully charged batteries allow a rider to travel as far as
28 km on level roads.

Do you know that this transporting device does not have
an accelerator or brakes? So, how does this device accelerate
or stop? The rider needs to lean forward for the PT to start
moving. If turning is required, the rider needs to turn his body
to the left or to the right. To stop the device, the rider needs to
stand upright.

Page Information

Video on PT movement

http://bt.sasbadi.com/p4025a

PageInformation

2525
Yes, Miss.
Sir, can you drive
me to Sungkai?

Photograph 2.1 Various types of moving objects

Figure 2.1 Taxi at rest Figure 2.2 Journey from Bagan Datoh to Sungkai

Photograph 2.1 shows various objects in
motion. How do you describe motion in
daily life? Motion in a straight line is called
linear motion.

Linear motion can be described in terms of distance, displacement, speed, velocity and
acceleration. Figure 2.1 shows a taxi waiting for passengers in a parking lot. The position of the
taxi remains unchanged with time. Therefore, the taxi is said to be at rest.

Madam Chong wants to go to Sungkai from Bagan Datoh. Figure 2.2 shows the route
marked in blue to be taken by the taxi through a distance of 83.00 km. However, after reaching
Sungkai, the position of the taxi is 57.22 km due East from Bagan Datoh. In fact, 83.00 km and
57.22 km due East are the distance and the displacement respectively of the taxi movements.
Table 2.1 shows the comparison between distance and displacement.

Linear Motion2.1

Table 2.1 Comparison between distance and displacement

Distance Displacement

Length of route covered by an object Shortest distance between the initial position
and the final position in a specific direction

Its magnitude depends on the route covered by
the object

Its magnitude is the straight-line distance
between the initial position and the
final position

A scalar quantity A vector quantity

2626 2.1.12.1.1
Figure 2.3 Speedometer reading and motion of vehicle

Figure 2.4 illustrates the difference in motions between uniform and non-uniform velocity
of a car. Observe the displacement and the time intervals for both cars. Assume that the motion
to the right is positive and to the left is negative.

Assume the right direction as a positive direction

The car is moving at uniform
velocity because its rate
of change of displacement
remains the same.20 m

t1 = 0 s t2 = 1 s t3 = 2 s t4 = 3 s

20 m 20 m

The car is moving at
non-uniform velocity
because its rate of change of
displacement is increasing.20 m

t1 = 0 s t2 = 1 s t3 = 2 s

30 m

Figure 2.4 Motions at uniform and non-uniform velocity

In Figure 2.4, the displacement of the blue car increases at equal time intervals. Therefore,
the blue car is moving at increasing velocity. In this case, the blue car accelerates in the same
direction as the motion of the car.

Conversely, if the displacement decreases at equal time intervals, the velocity decreases as
shown in Figure 2.5. The car accelerates but in the opposite direction to the motion of the car.

20 m

t1 = 0 s t2 = 1 s t3 = 2 s

30 m

Figure 2.5 Motion with decreasing velocity

While riding in a taxi, Madam Chong observed that at times, the reading on the
speedometer remains constant and at other times, the reading on the speedometer varies even
along straight roads. Her observation can be summarised as shown in Figure 2.3.

Acceleration

Speedometer reading

Uniform velocity

Non-uniform velocity

Constant

Varies

Examples of motion at
non-uniform velocity

http://bt.sasbadi.com/p4027

2727

Chapter 2 Force and Motion I

2.1.12.1.1
Determining Distance, Displacement, Speed, Velocity
and Acceleration
The motion of an object is investigated by determining the values
of its distance travelled, displacement, speed, velocity and
acceleration. The formulae to determine speed, velocity and
acceleration are as follows:

Observe the following examples to understand how  to determine distance, displacement,
speed, velocity and acceleration in linear motion.

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The Smart Tunnel in
Kuala Lumpur has a dual
function. It functions as a
storage space and passage
of flood flow as well as an
alternative route with a
shorter distance.

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Speed =  rate of change of distance travelled

= distance travelled
time taken

v = d
t

Velocity =  rate of change of displacement

=
displacement

time taken
 v = s

t

Acceleration = rate of change of velocity

= final velocity – initial velocity
time taken for change of velocity

a = v – u
t

Example 1

Figure 2.6 shows Radzi’s run from A to B and then back to C. The total time taken is 20 s.
Determine the
(a) distance
(b) displacement
(c) speed
(d) velocity
of Radzi’s motion.

20 m
A C B

100 m

Figure 2.6

2828 2.1.22.1.2
Solution:

(a) Distance  = Length of route covered
= AB + BC
= 100 m + 20 m
= 120 m

(b) Displacement = Distance between initial position and final position in a specific direction
  = AB + BC
  = (100 m) + (–20 m)
  = 80 m (to the right)

(c) Speed = Distance travelled
Time taken

= 120 m
20 s

= 6 m s–1

(d) Velocity = Displacement
Time taken

= 80 m
20 s

= 4 m s–1 (to the right)

Muthu moves from O to B along the route OAB as shown in
Figure 2.7. The time taken is 15 s.
Determine the
(a) distance
(b) displacement
(c) speed
(d) velocity
of Muthu’s motion.

Solution:

(a) Distance  = OA + AB
= 5 m + 12 m
= 17 m

(b) Displacement = Shortest straight line distance from O to B
  = OB
  =   52 + 122

= 13 m (in the direction of OB)
�����

Example 2

A
O

B

12 m13 m

5 m
Muthu

Figure 2.7

2929

Chapter 2 Force and Motion I

2.1.22.1.2
(c) Speed = Distance travelled
Time taken

= 17 m
15 s

= 1.13 m s–1

(d) Velocity = Displacement
Time taken

= 13 m
15 s

= 0.87 m s–1 (in the direction of OB)

Contoh 3

A ticker timer works on alternating current of
frequency 50 Hz to make 50 ticks in 1 second
on a ticker tape.

The time interval between two adjacent dots is
called one tick.

Therefore, 1 tick: 1
50 s = 0.02 s

5 ticks: 5 × 0.02 s = 0.10 s
 10 ticks: 10 × 0.02 s = 0.2 s

After landing on the runway, a plane slows down so that
its velocity reduces from 75 m s–1 to 5 m s–1 in 20 s. What
is the acceleration of the plane?

Solution:

Initial velocity, u = 75 m s–1, final velocity, v = 5 m s–1,
time, t = 20 s

Acceleration, a = v – u
t

= 5 – 75
20

= –3.5 m s–2

A ticker timer fitted with ticker tape as shown in Figure 2.8 is used to investigate linear
motion of objects in the laboratory.

Figure 2.8 Ticker timer with ticker tape

Vibrating plate

Carbon
paper

Ticker tape

Pulled by moving object

Connected to an
alternating current
power supply

CCCoC ntoh 3CoCC ntoh 3CoCC ntoh 3Contoh 3Contoh 3CContoh 3Contoh 3Contoh 3Example 3

F ile
Deceleration is a condition in
which the velocity of motion of
an object is decreasing.

3030 2.1.22.1.2
Figure 2.9 shows part of a ticker tape that is pulled by
 an object moving linearly.

5th tick 4th tick 3rd tick 2nd tick 1st tick

Displacement over a 5 ticks period
Time = 5 ticks = 5 × 0.02 s = 0.10 s

Direction of
motion of
ticker tape

Figure 2.9 Part of a ticker tape pulled by an object

A ticker tape records the displacement of a moving object as well as the time taken.
Subsequently, the velocity and acceleration can be calculated. Figures 2.10 and 2.11 show  the
calculations of velocity and acceleration of the linear motion of an object.

Figure 2.10
Displacement, s = 18 cm
Time taken, t  = 6 ticks

= 6 × 0.02 s
= 0.12 s

Velocity, v = s
t

= 18 cm
0.12 s

= 150 cm s–1

Calculating velocity

123456

18 cm

Direction of motion

Calculating acceleration

Figure 2.11

Initial velocity, u = velocity at the first tick

= 1.2 cm
0.02 s

= 60 cm s–1

Final velocity, v = velocity at the seventh tick

= 3.6 cm
0.02 s

= 180 cm s–1

Time between the final and initial velocity,
 t = (7 – 1) ticks
  = 6 ticks
  = 6 × 0.02 s
  = 0.12 s

Acceleration, a = v – u
t

= (180 – 60) cm s–1

0.12 s

= 1 000 cm s–2

123457 6

3.6 cm 1.2 cm

Direction of motion

We already know how to calculate
velocity and acceleration of the linear
motion of an object.

Let us carry out an activity using a ticker
timer and ticker tape to determine the
velocity and acceleration of a trolley.

F ile
When an object moves  along
a straight line and does not
go backwards, its distance
and displacement are of  the
same value.

3131

Chapter 2 Force and Motion I

2.1.22.1.2
Activity 2.1

Aim:  To use a ticker tape to determine the displacement, velocity and acceleration
of a trolley

Apparatus:  Ticker timer, trolley, runway, alternating current power supply, retort stand and
wooden block

Materials: Ticker tape and connecting wires

Instructions:
1. Set up the apparatus as shown in Figure 2.12. Raise one end of the runway slightly so

that the trolley can move slowly down the runway.

Alternating
current
power supply

0

2
4 6

8

10

POWER
ON

DUAL MODE
POWER SUPPLY

Trolley

Ticker tape

20 cm

Ticker timer Retort stand

Runway
Wooden block

Figure 2.12

2. Attach a 100 cm ticker tape to the trolley, switch on the ticker timer and release the
trolley. Observe the ticker tape obtained.

3. From the ticker tape, determine the displacement and calculate the average velocity of
the trolley.

4. Raise the end of the runway further so that the trolley moves with a higher velocity
down the runway.

5. Repeat steps 2 and 3. Then, calculate the acceleration of the trolley.
6. Rearrange the apparatus as shown in Figure 2.13.

0

2
4 6

8

10

POWER
ON

DUAL MODE
POWER SUPPLY

Trolley

Alternating
current power
supply

Ticker tape

Ticker timer

Retort stand

Runway

Figure 2.13

7. Push the trolley from the bottom of the runway and allow it to move up.
8. Stop the trolley when it starts to descend the runway.
9. From the ticker tape obtained, determine the deceleration of the trolley.

Discussion:
Discuss the patterns of motion obtained from the ticker tape.

3232 2.1.22.1.2
Bab 2 Daya dan Gerakan I

If a long ticker tape is used, a large number of dots can be recorded on the ticker tape. As
such, the ticker tape can be divided into strips with an equal number of ticks. These strips are
cut and attached side by side on a piece of graph paper to form a ticker tape chart as shown in
Figure 2.14.

Uniform velocity Uniform acceleration Velocity decreases uniformly

Time / s Time / s Time / s

Length / cm Length / cm Length / cm

Figure 2.14 Ticker tape chart

Other than ticker timer, photogate system and electronic timer can be used to study
linear motion more accurately. Figure 2.15 shows a photogate system and an electronic timer
used with a non-motorised trolley moving on an inclined aluminium runway.

Runway

Weight

Shutter plate

Non-motorised
trolley

W

N
tr

Photogate

First photogate

Second photogate

Ph t t

Electronic timer

Figure 2.15 Photogate system and electronic timer

3333

Chapter 2 Force and Motion I

2.1.22.1.2
Activity 2.2

The use of a photogate system and electronic timer is more accurate because no ticker tape
is attached to the trolley. As such, there will be no friction between  the ticker tape and ticker
timer affecting the motion of the trolley. An electronic timer can measure the time interval to
an accuracy of 0.001 second compared to 0.02 second for a ticker timer. This short time interval
enables us to determine velocity and acceleration of the trolley more accurately.

Aim:  To use a photogate system and an electronic timer to determine the velocity and
acceleration of a moving trolley

Apparatus: Photogate system and electronic timer, trolley and runway

Instructions:

1. Set up the apparatus as shown in the manual of the
QR code.

2. Raise one end of the runway to a height of 15 cm.

3. Adjust the distance of separation between the two
photogates, s = 40.0 cm.

4. Turn the electronic timer switch to the symbol .
Release the trolley from the high end of the runway and
catch the trolley after it passes the second photogate.

5. Record the total time, t in Table 2.2.

6. Remove the fi rst photogate.

7. Turn the electronic timer switch to the symbol . Release the trolley again from its
original point.

8. Record the time interval, ∆t in Table 2.2.

9. Repeat steps 3 to 8 for s = 50.0 cm,  60.0 cm, 70.0 cm and 80.0 cm.

Results:
Table 2.2

Distance of separation between two photogates, s / cm 40.0 50.0 60.0 70.0 80.0

Total time, t / s

Time interval, Δt / s

Final velocity, v = s
Δt

/ cm s–1

Acceleration, a = v
t

/ cm s–2

Discussion:
1. Based on the results in the table, determine the average acceleration of the trolley.
2. When s increased from 40.0 cm to 80.0 cm, the total time, t increased but the time

interval Δt decreased. Why?

Manual on the use of  photogate
system and  electronic timer

http://bt.sasbadi.com/p4034

3434 2.1.22.1.2
Example 1

A school bus moves from rest with an acceleration of 2 m s–2 for 5 s. Calculate its velocity
after 5 s.

Solution:

Step �
List the given information in symbols.

Step �
Identify and write down the formula used.

Step �
Substitute numerical values in the formula and
perform the calculations.

Initial velocity, u = 0 m s–1
Time, t = 5 s
Acceleration, a = 2 m s–2
Final velocity, v = ?

v = u + at

v = 0 + (2)(5)
 = 10 m s–1

First linear motion equation

Acceleration = Final velocity – initial velocity
Time taken for change of velocity

a = v – u
t

at = v – u
 v = u + at  (1)

Third linear motion equation
Substitute equation (1) into equation (2)

s = 1
2 [u + (u + at)]t

s = 1
2 (2u + at)t

s = ut + 1
2

at2  (3)

Second linear motion equation
 Displacement = Average velocity × time

Displacement = �Initial velocity + final velocity
2 � × time

s = 1
2

(u + v)t  (2)

Fourth linear motion equation
Square equation (1)
v2  = (u + at)2

v2  = u2 + 2uat + a2t2

v2 = u2 + 2a �ut + 1
2 at2�

v2  = u2 + 2as  (4)

From equation (3)

s = ut + 1
2 at2

Solving Problems of Linear Motion Using Linear Motion Equations
Figure 2.16 shows a car moving with uniform acceleration.

Time = 0 s Displacement,
s / m

Acceleration, a / m s–2

Time = t s

Initial velocity, u / m s–1 Final velocity, v / m s–1

Time = t / sTime = 0 s Displacement,

Acceleration, a / m s–2

Figure 2.16 Car moving with uniform acceleration

Five physical quantities of linear motion with uniform acceleration can be expressed as
four linear motion equations.

3535

Chapter 2 Force and Motion I

2.1.32.1.3
Solution:

Example 2

As a sports car moves along
a straight track, its velocity
is 40 m s–1. After 3 seconds,
the sports car has reached
50 m s–1. Calculate
its displacement.

s = 1
2  (u + v)t

= 1
2  (40 + 50)(3)

= 135 m

Initial velocity, u = 40 m s–1

Final velocity, v = 50 m s–1

Time, t = 3 s
Displacement, s = ?

Solution:

Example 3

An athlete starts his run
from rest and achieves a
maximum velocity after
accelerating uniformly for
8.0 s. If the displacement
of the athlete is 40 m,
determine his acceleration.

Initial velocity, u = 0 m s–1

Time, t = 8.0 s
Displacement, s = 40 m
Acceleration, a = ?

Solution:

v2 = u2 + 2as
 02 = 82 + 2(a)(2)
 –4a = 64
 a = –16 m s–2

Example 4

Maria rides a bicycle at a
velocity of 8 m s–1. She
brakes suddenly and stops
after a distance of 2 m.
What is the acceleration of
Maria and her bicycle?

Initial velocity, u = 8 m s–1

Final velocity, v = 0 m s–1
Displacement, s = 2 m
Acceleration, a = ?
Negative value shows that Maria accelerates at 16 m s–2 in the
opposite direction to the motion of the bicycle.

s = ut + 1
2  at2

40 = (0)(8) + 1
2 (a)(82)

40 = 0 + 64a
2

a = 2 × 40
64

= 1.25 m s–2

Formative Practice 2.12.1

1. Explain the difference between
(a) distance and displacement (b) speed and velocity

2. A car moving along a straight road at a velocity of 30 m s–1 reduces its velocity at a
constant rate until it stops after 5 s. What is the acceleration of the car?

3. Aina rides a smart personal transporter at the Perdana Botanical Gardens. The transporter
accelerates uniformly from a velocity of 1 m s–1 to a velocity of 5 m s–1 in 0.5 minutes.
Calculate the displacement of the transporter.

3636 2.1.32.1.3
Figure 2.17  Interpretation of types of linear motion from graphs

Displacement
-time graph

Time, t / s

Acceleration, a / m s–2

0

0

0

Velocity, v / m s–1

Velocity-time
graph

Acceleration
-time graph

Time, t / s Time, t / s

Interpretation of graphs is important to understand the type of linear motion of an object.
Figure 2.17 shows the interpretation of the types of linear motion from graphs.

Displacement, s / m

Time, t / s

Acceleration, a / m s–2Velocity, v / m s–1

Time, t / s Time, t / s

Displacement, s / m

Time, t / s

Acceleration, a / m s–2Velocity, v / m s–1

Time, t / s Time, t / s

Displacement, s / m

(a) Object at rest (a) Object moving with
uniform velocity

(a) Object moving with zero
acceleration

(b) Object moving with
uniform velocity

(b) Object moving with
uniform acceleration

(b) Object moving with
uniform acceleration

(c) Object moving with
acceleration

(c) Object moving with
uniformly decreasing velocity

(c) Object moving with
increasing acceleration

For velocity-time graph:
Gradient of the graph = acceleration
Area under the graph = displacement

For displacement-time graph:
Gradient of the graph = velocity

Linear Motion Graphs2.2

3737

Chapter 2 Force and Motion I

2.2.12.2.1
Displacement, s / m

Time, t / s5 8 10 12 14 15 20 250

–50

100

A

B C

D E

F

Distance travelled = 100 + 100 + 50
 = 250 m
Displacement = 100 + (–100) + (–50)
 = –50 m

Figure 2.19  Analysis of displacement-time graph

Figure 2.18  Motion of a motorcycle being ridden

= 250
20

=12.5 m s–1

= –50
20

= –2.5 m s–1

50 m

Final point

Negative
direction

Initial point

100 m

= –50 – 0
20 – 14

= –8.33 m s–1

Positive
direction

Section AB:
Motorcycle is ridden for a distance of 100 m to
the right for 5 seconds.
Gradient of graph

Therefore, velocity of motorcycle is 20 m s–1

to the right.

Section BC:
Motorcycle stops for
3 seconds.
Gradient of graph = 0 m s–1

Therefore, velocity of
motorcycle is 0 m s–1.

Section CD:
Motorcycle is ridden back to initial
point A in 4 seconds.

Gradient of graph

Therefore, velocity of motorcycle is
25 m s–1 to the left.

= 0 – 100
12 – 8

= –25 m s–1

= 100 – 0
5 – 0

= 20 m s–1

Section DE:
Motorcycle is stationary at initial point
for 2 seconds.
Gradient of graph = 0 m s–1

Therefore, velocity of motorcycle is
0 m s–1.

Section EF:
Motorcycle is ridden to the left for a
distance of 50 m for 6 seconds.

Gradient of graph

Therefore, velocity of motorcycle is
8.33 m s–1  to the left.

How are average speed
and average velocity
determined from the
displacement-time graph?

Distance travelled is 250 m
in a total time of 20 s.
Therefore, its average speed is

Overall displacement is –50 m in
a total time of 20 s. Therefore,
average velocity is

FF
ment-time graph?

Analysis of Displacement-Time Graph to Determine Distance, Displacement and Velocity

Figure 2.18 shows the initial point and the final point of a motorcycle ridden to the right
(positive direction) and then to the left (negative direction). The displacement-time graph in
Figure 2.19 shows the linear motion of the motorcycle.

3838 2.2.22.2.2
Area, L1 = 500 m, L2 = 250 m, L3 = 225 m, L4 = 135 m
Total overall distance = L1 + L2 + L3 + L4
                                   = 500 + 250 + 225 + 135
                                   = 1 110 m
Displacement to the right = L1 + L2 + L3
                                         = 500 + 250 + 225
                                         = 975 m
Displacement to the left = L4
                                       = 135 m
Total displacement = L1 + L2 + L3 + L4

= 500 + 250 + 225 + (–135)
                               = 840 m to the right

Velocity, v / m s–1

Time, t / s 50 70

L1 L2 L3

L4
100

120
1500

–10
–9

15

10

Time interval: 0 – 50 seconds
Gradient of the graph = 0 m s–2

Therefore, the bicycle is moving to the
right with a uniform velocity of 10 m s–1.

Time interval: 50 – 70 seconds

Gradient of the graph = 15 – 10
70 – 50

= 0.25 m s–2

Therefore, the bicycle is moving
to the right with a uniform
acceleration of 0.25 m s–2.

Time interval: 70 – 100 seconds

Gradient of the graph = 0 – 15
100 – 70

= –0.5 m s–2
Acceleration of bicycle = –0.5 m s–2

The velocity of the bicycle gradually
decreases. The bicycle accelerates
at –0.5 m s–2 (in the opposite
direction to the motion of the bicycle).

Time interval: 100 – 120 seconds
Gradient of the graph = 0 m s–2

Its velocity is 0 m s–1. Therefore, the
bicycle has stopped and is at rest for
20 seconds.

Time interval: 120 – 150 seconds

Gradient of the graph = –9 – 0
150 – 120

= –0.3 m s–2
Acceleration of bicycle = –0.3 m s–2

The velocity of the bicycle gradually
increases. The bicycle accelerates at
–0.3 m s–2 (in the same direction as
the motion of the bicycle).

Analysis of Velocity-Time Graph to Determine Distance, Displacement, Velocity
and Acceleration

Figure 2.20 shows the linear motion of a bicycle. Figure 2.21 shows the velocity-time graph
showing the motion of the bicycle.

Final point

Negative
direction

Positive
direction

Initial point
Figure 2.20 Linear motion of a bicycle

Figure 2.21 Linear motion of a bicycle

Average speed and average velocity can be determined from the rate of change of the total
distance and also from the rate of change of the total displacement. Try to determine the average
speed and average velocity for the velocity-time graph above.

3939

Chapter 2 Force and Motion I

2.2.32.2.3
ICS

Aim: To use Tracker software to map the motion of a table tennis ball in the form of graphs

Apparatus: Retort stand, wooden block and metre rule

Materials: Table tennis ball and cellophane tape

Instructions:

1. Work in groups.

2. Scan the QR code to download Tracker software.

3. Set up the apparatus as shown in Figure 2.22 using two
metre rules arranged in the shape of a “V” to form an
inclined runway.

4. Record a video of the motion of the table tennis ball along a straight line.

5. Use Tracker software to analyse the motion of the table tennis ball in the video using
displacement-time, velocity-time and acceleration-time graphs.

6. Figure 2.23 shows examples of graphs that you can obtain using the software.

7. Discuss and interpret the motion of the table tennis ball based on the graphs obtained.

8. Analyse the motion of the table tennis ball from the graphs.

9. Present your interpretation and analysis of the graphs.

Figure 2.22

Graph of horizontal displacement
against time

Graph of vertical displacement
against time

Graph of speed against time

Figure 2.23 Examples of graphs obtained

2

0

0
1

cm

2
3

4
5

6
7

8
9

10
11

12
13

14
15

16
17

18
19

20
21

22
23

24
25

26
27

28
29

30

0
1

cm
2

3
4

5
6

7
8

9
10

11
12

13
14

15
16

17
18

19
20

21
22

23
24

25
26

27
28

29
30

Retort stand

Table tennis ball

Metre rules Wooden block

http://bt.sasbadi.com/p4040a

Download Tracker software

http://bt.sasbadi.com/p4040b

Instruction video on using
Tracker software

4040 2.2.32.2.3
0

1

2
3

4

5

6

7

8

9

10

2 4 6 8 10 12 14 16 18 20

0

0.5

1.0

2 4 6 8 10 12 14 16 18 20

Displacement, s / m

Time, t / s

Figure 2.24

Solution:

Table 2.3

Velocity, v / m s–1

v1 = 6 – 0
6 – 0

= 1 m s–1

v2 = 6 – 6
12 – 6

= 0 m s–1

v3 = 10 – 6
20 – 12

= 0.5 m s–1

0 s to 6 s

6 s to 12 s

12 s to 20 s

Figure 2.25

Time, t / s

Figure 2.24 shows the displacement-time graph of an object in linear motion.

(a) Determine the velocity of the object at each stage of its motion.
(b) Interpret the displacement-time graph in Figure 2.24 to sketch a velocity-time graph.

(a) Velocity = Gradient of
displacement-time graph

(b) Velocity-time graph

Example 1

Solution:

Interpreting and Sketching Graphs
Interpreting displacement-time graphs can help to sketch velocity-time graphs and vice versa.
Velocity-time graphs can also be interpreted into acceleration-time graphs and vice versa.
The skill in interpreting and sketching graphs is important to solve linear motion problems.
Study the examples given.

4141

Chapter 2 Force and Motion I

2.2.42.2.4
Figure 2.26 shows the velocity-time graph
that is plotted based on the linear motion of a
car driven by Encik Kassim. He drives his car
at a speed of 30 m s–1 and applies the brakes
when he sees an obstruction on the road.
Interpret the velocity-time graph of the
motion of his car and sketch
(a) displacement-time graph
(b) acceleration-time graph

Solution:

Displacement after 10 s,
s = 180 + 60

= 240 m

(a) Displacement-time graph
Displacement, s / m

Time, t / s0
1 2 3 4 5 6 7 8 9 10

20
40
60
80

100
120
140
160
180
200
220
240

Figure 2.28

(b) Acceleration-time graph

Figure 2.29

Acceleration, a / m s–2

Time, t / s
1 2 3 4 5 6 7 8 9 10

–8
–9

–10

–7
–7.5

–6
–5
–4
–3
–2
–1

0
1

Time, t / s0

10

20

BA

30

1 2 3 4 5 6 7 8 9 10

Velocity, v / m s–1

Figure 2.27

Example 2

0

10

20

30

1 2 3 4 5 6 7 8 9 10

Velocity, v / m s–1

Time, t / s

Figure 2.26
Table 2.4

Time Displacement Acceleration

0 s – 6 s Displacement
= area A
= 30 × 6
= 180 m

Acceleration
=  gradient of

the graph
= 0 m s–2

6 s – 10 s Displacement
= area B
= 1—2  × 30 × 4

= 60 m

Acceleration

= 0 – 30
10 – 6

= –7.5 m s–2

INFOSMARTSMART

To determine displacement, the area under the graph needs to
be calculated. To simplify the calculation of area, the graph
can be divided into several areas as shown in Figure 2.27.

4242 2.2.42.2.4
Solving Problems Involving Linear Motion Graphs

The velocity-time graph in Figure 2.30 shows
the motion of Hasri. Determine his
(a) acceleration
(b) displacement
(c) average velocity

Solution:

(a) Acceleration = gradient of the graph

From 0 s – 3 s:

Acceleration a1 = 6 – 0
3 – 0

= 2 m s–2
From 3 s – 6 s:
Acceleration a2 = 0 m s–2

From 6 s – 10 s:
Acceleration a3 = 0 – 6

10 – 6
 = –1.5 m s–2

(b) Displacement, s = area under the graph
  = area of trapezium

= 1
2 (3 + 10)(6)

= 39 m

(c) Average velocity, v = Total displacement
Time taken

= 39
10

= 3.9 m s–1

Figure 2.31 shows the velocity-time graph of the linear motion of a car. Interpret the
velocity-time graph and sketch a graph of:
(a) displacement against time
(b) acceleration against time

0
5 10 15 20 25 30 35 40 45 50

2
4
6
8

10
12
14
16
18
20

Velocity, v / m s–1

Time, t / s

Figure 2.31

0
1
2
3
4
5
6

1 2 3 4 5 6 7 8 9 10

Velocity, v / m s–1

Time, t / s

Figure 2.30

Example 1

Example 2

4343

Chapter 2 Force and Motion I

2.2.52.2.5
0
5 10 15 20 25 30 35 40 45 50

2
4
6
8

10
12
14
16
18
20

A B C

0
10 20 30 40 50

100

200

300

400

500

360

520

–0.8

–0.6

–0.4

–0.2

0
0.1

0.3
0.2

–0.9

–0.7

–0.5

–0.3

–0.1
20 40 503010

0 s – 10 s

10 s – 30 s

= 16 – 10
30 – 10

= 0.3 m s–2

30 s – 50 s

= 0 – 16
50 – 30

= –0.8 m s–2

Velocity, v / m s–1

Time,
t / s

Figure 2.32

Table 2.5

Time Displacement Acceleration

Acceleration

Acceleration

(a) Displacement-time graph (b) Acceleration-time graph

Displacement, s / m
Acceleration, a / m s–2

Time, t / s

Figure 2.34Figure 2.33

Time, t / s

Solution:

Displacement
= area A
= 10 × 10
= 100 m

Acceleration
= gradient of
 the graph
= 0 m s–2

Displacement
= area C

=       × 20 × 16

= 160 m

Displacement
= area B

=       (10 + 16)(20)

= 260 m

1
2

1
2Displacement after 30 s = 100 + 260

= 360 m
Displacement after 50 s = 100 + 260 + 160
 = 520 m

4444 2.2.52.2.5
Formative Practice 2.22.2

1. How do you determine
(a) velocity from a graph of displacement against time?
(b) acceleration from a graph of velocity against time?
(c) displacement from a graph of velocity against time?

2. Based on Figure 2.35, describe the motion of the object from O to D.

Velocity, v / m s–1

Time, t / st1 t2 t3 t4

u

O

A B

C

D

v

Figure 2.35

3. Figure 2.36 shows Rokiah taking 3 minutes to walk
to the sundry shop located 400 m to the right of
her house. After 1 minute, she buys an ice cream
and walks to the playground located 300 m from
the sundry shop in 2 minutes. She sits and rests on
a bench near the playground for 2 minutes. Then,
using a shortcut to return to her house, Rokiah
reaches her house in 2 minutes.
(a) What is the average velocity of the motion of

Rokiah from
(i) house to the sundry shop?
(ii) sundry shop to the playground?
(iii) playground to the house?

(b) Calculate Rokiah’s average speed.

4. A car is driven from rest and accelerated at 4 m s–2 for 8 s along a straight road. Th e car
is driven at the constant velocity for 20 s and then its brakes are applied. Th e car reduces
its velocity at the rate of 2 m s–2 till it stops. Sketch a graph of:
(a) acceleration against time
(b) velocity against time
(c) displacement against time

Figure 2.36

300 m

400 m

House Sundry shop

Bench

Shortcut

KEDAI

4545

Chapter 2 Force and Motion I
Free Fall Motion2.3

Photograph 2.2 Coconut
falling from a coconut tree

Free Fall Motion and Gravitational Acceleration
An object experiences free fall if the motion of the object
is affected only by gravitational force. This implies a free
falling object does not experience the action of other forces
such as air resistance or friction.

Photograph 2.2 shows a coconut falling from a coconut
tree. Is this a free fall motion?

Carry out Activity 2.4 and Activity 2.5 involving free
fall motion.

Activity 2.4

Aim: To watch a video on free fall motion

Instructions:
Scan the QR code or visit the website to watch a video on
free fall motion.

Discussion:
What are your observations on the free fall motion shown in
the video?

Video on free fall motion

http://bt.sasbadi.com/p4046

Activity 2.5

Aim: To investigate the motion of falling objects

Materials: Table tennis ball and a piece of A4 paper

Instructions:
1. Work in pairs.
2. Hold a piece of A4 paper in the right hand and a

table tennis ball in the left hand as shown in
Photograph 2.3.

3. Release the paper and table tennis ball simultaneously
from the same height.

4. Observe the motion of the paper and table tennis ball.
5. Repeat steps 2 to 4 with the A4 paper crumpled into a ball.

Discussion:
1. Why do the falling paper and table tennis ball reach the fl oor at different times in the

fi rst try?
2. The paper used in step 2 and step 5 is the same piece of paper. Why does the paper

before and after being crumpled fall at different rates?

Photograph 2.3

A4 paper Table
tennis ball

4646 2.3.12.3.1
HISTORY
INTEGRATION OF

From 384–322 B.C., Aristotle
concluded that the rate of an
object falls depends on the
weight, shape and orientation
of the object. However,
Galileo Galilei (1564–1642)
discovered that objects fall with
the same acceleration if air
resistance is ignored.

In daily life, we see that heavy objects fall and reach the
ground faster than light objects. This is due to other forces such
as air resistance. Observe Figure 2.37.

Feather Ball bearing

Figure 2.37 Motion of falling objects in air

Objects of different masses fall with the same acceleration if air resistance does not exist.
This can happen in a vacuum. Observe Figure 2.38. A feather and a ball bearing released in
a vacuum will reach the floor at the same time. The free fall that you watched in the video in
Activity 2.4 was actually performed in a vacuum.

Feather Ball bearing

Figure 2.38 Free fall in a vacuum

The acceleration of a free falling object caused by gravitational force is known as
gravitational acceleration, g. The average value of Earth’s gravitational acceleration is
9.81 m s–2. The velocity of a free falling object increases by 9.81 m s–1 per second in
uniform gravitational field close to the surface of the Earth. When a heavy object falls in
the gravitational field, air resistance can be ignored. The object is in free fall motion.

Video of object falling in
normal condition and
in a vacuum

http://bt.sasbadi.com/p4047

4747

Chapter 2 Force and Motion I

2.3.12.3.1
Determining the Value of Gravitational Acceleration
An object that falls freely in the gravitational field has
gravitational acceleration. As such, the value of gravitational
acceleration can be determined by measuring the acceleration
of a heavy object such as a steel ball in a physics laboratory. Let
us use a photogate system and an electronic timer to determine
gravitational acceleration, g. t

Experiment 2.1
Figure 2.39 Releasing a steel ball

Aim: To determine Earth’s gravitational acceleration

Apparatus:  Photogate system and electronic timer, electromagnetic release, steel ball and
container to catch steel ball released

Procedure:

1. Scan the QR Code or visit the website given to download the manual on the use of the
photogate and electronic timer.

2. Set up the apparatus as shown in Photograph 2.4.

Second photogate

First photogate

Electromagnetic release

Steel ball

Container to catch
steel ball

Tripod stand

Electronic
timer

Height
adjuster

Photograph  2.4

3. Place the second photogate so that the distance is 30.0 cm away from the fi rst photogate.
4. Ensure the steel ball can fall through both photogates into the container.
5. Release the steel ball from the electromagnetic release.
6. Record the time when the steel ball passes through the fi rst photogate as t1 and the

second photogate as t2 in Table 2.6.
7. Repeat steps 3 to 6 with distances 40.0 cm, 50.0 cm, 60.0 cm and 70.0 cm.

Manual for the use of
photogate and electronic
timer for the experiment on
free fall

http://bt.sasbadi.com/p4048a

Alternative method using
ticker tape

http://bt.sasbadi.com/p4048b

4848 2.3.22.3.2
Results:
Table 2.6

Distance between two
photogates, h / cm

Time when the steel ball passes
through photogates Gravitational

acceleration, g / m s–2

t1 / s t2 / s

30.0

40.0

50.0

60.0

70.0

Analysis of data:

1. Determine the value of g using the formula g = 2h
t2

2 – t1
2
.

2. Calculate the average value of g.

Conclusion:
What conclusion can be made from this experiment?

Prepare a complete report on this experiment.

Discussion:

1. Compare the average value of g from this experiment with the standard value of g at the
equator, 9.78 m s–2. Why are these two values different?

2. Explain one precaution to improve the accuracy of the results in this experiment.

The value of gravitational acceleration, g changes from one place to another. For example,
the value of g at the equator is 9.78 m s–2 while the value of g at the Earth’s poles is 9.83 m s–2.
This is because Earth is actually not a perfect sphere.

Figure 2.40 shows that the distance from the equator to the centre of the Earth is further
than the distance from the poles to the centre of the Earth. As such, the value of g is smaller at
the equator than at the Earth’s poles. Generally, the value of gravitational acceleration on the
surface of the Earth, g used in calculations is 9.81 m s–2.

Figure 2.40 Different distances from the centre of the Earth

North pole
g = 9.83 m s–2

g = 9.78 m s–2

Distance from pole to the
centre of the Earth = 6 350 km

South pole

Earth
Distance from equator to the
centre of the Earth = 6 372 km

Equator

4949

Chapter 2 Force and Motion I

2.3.22.3.2
Solving Problems Related to Free Falling Objects

Objects thrown vertically upwards or released and falls
downwards experience gravitational acceleration, g. Therefore,
the equations of linear motion below can be applied to free
falling objects.

v = u + gt

s = ut + 1
2

gt 2

v 2 = u 2 + 2gs

In solving problems, our assumptions are upward motion
is motion in the positive direction and downward motion is
motion in the negative direction as shown in Figure 2.41.

Amirah threw a ball vertically upwards with an initial
velocity of 10 m s–1. Calculate the:
(a) time for the ball to achieve maximum height
(b) maximum height that can be reached by the ball
Ignore air resistance. [g = 9.81 m s–2]

Solution:

(a) Step �
List the given information in
symbols.

Initial velocity , u = 10 m s–1

Final velocity at maximum height, v = 0 m s–1

Acceleration, g = –9.81 m s–2

Step �
Identify and write down the
formula used.

v = u + gt

Step �
Substitute numerical values into
the formula and perform the
calculations.

0 = 10 – 9.81t
t = 10

9.81
 =  1.02 s

(b)  v 2 = u 2 + 2gs
  0 = 102 + 2(–9.81)s

s = 102

2 × 9.81

= 5.10 m

1
2

3
1

2
3

1
2

3

Figure 2.41 Assumption of direction
of an object’s vertical motion

Assumption:
Motion in the
positive direction

Object

Assumption:
Motion in the
negative direction

Object thrown
upwards

Object released
downwards

Example 1

INFOSMARTSMART

Ball moves upwards, u is
positive but g is negative.

INFOSMARTSMART

The equation s = ut + 12 gt 2
can also be used.
 s = ut + 12 gt 2

= 10(1.02) + 12 (–9.81)(1.02)2

= 5.10 m

5050 2.3.32.3.32.3.22.3.2
Formative Practice 2.32.3

Chan released a stone from a cliff of 10 m height. Determine
(a) the time taken for the stone to reach the bottom of the cliff
(b) the velocity of the stone just before it touches the ground
Ignore air resistance. [g = 9.81 m s–2]

Solution:

(a) s = ut + 1
2

gt 2

–10 = (0)t + 1
2

(–9.81)t 2

2 × (–10) = (–9.81)t 2

t = ±  –20
–9.81

= 1.43 s
 (t = –1.43 is ignored)

(b) v 2 = u 2 + 2gs
  = 2 × (–9.81) × (–10)
 v = ±  2 × (–9.81) × (–10)
  = ±14.0 m s–1

v = –14.0 m s–1

(v = 14.0 m s–1 is ignored because the stone moves downwards).

����

���������

INFOSMARTSMART

Formula  v = u + gt
can also be used
v = –9.81 × 1.43
 = –14.0 m s–1

1. What is meant by free fall?

2. A light hollow plastic ball and a solid steel ball of the same size are released from the cliff
of a hill. Will both balls reach the foothill at the same time? Explain your answer.

3. An object thrown vertically upwards reached a maximum height of 5.0 m. Calculate:
(a) the velocity of the object when thrown
(b) the time taken for the object to reach its maximum height
(c) the time required for the object to return to its original level
Ignore air resistance. [g = 9.81 m s–2]

4. A tennis ball that is released falls vertically from a building of height 50 m. Calculate:
(a) the time taken for the ball to reach the base of the building
(b) the velocity of the ball just before hitting the base of the building
(c) the vertical distance passed at the third second
Ignore air resistance. [g = 9.81 m s–2]

Example 2

5151

Chapter 2 Force and Motion I

2.3.32.3.3
Inertia2.4

Figure 2.42 An entertainer jerks a table cloth
without moving the objects on the table

Concept of Inertia
Figure 2.42 shows objects placed on a table
remain stationary even though the table cloth
under the objects is jerked by the entertainer.
This is due to inertia.

Inertia is the tendency of an object to
remain at rest or, if moving, to continue its
motion in a straight line at uniform velocity.
The concept of inertia is explained in Newton’s
First Law of Motion.

Newton’s First Law of Motion
Newton’s First Law states that an object will
remain at rest or move at uniform velocity
unless acted upon by an external force.

k bl l h

Activity 2.6

Aim: To demonstrate the concept of inertia

Materials: Glass filled with water, coin and thin cardboard

Instructions:
1. Arrange the materials as shown in Figure 2.43.

Coin

Thin
cardboard

Glass filled
with water

Figure 2.43

2. Jerk the cardboard under the coin horizontally.

Discussion:

1. Why is the coin not moving with the thin cardboard?

2. What happens if the cardboard is pulled slowly? Explain your answer by referring to
Newton’s First Law of Motion.

5252 2.4.12.4.1
Recognising the Relationship between Inertia and Mass
Photograph 2.5 shows a bowling ball and a football. Is it easier to move a bowling ball or a
football? Which ball is difficult to stop when in motion?

Photograph 2.5 Two balls of different masses

An object with a bigger mass is more difficult to set in motion or stop from moving as
compared to a lighter object. What is the relationship between mass and inertia? Let us carry out
an experiment using a simple inertial balance.

Experiment 2.2
Inference: Inertia of an object depends on its mass

Hypothesis: The larger the mass of an object, the larger the inertia of the object

Aim: To study the relationship between inertia and mass

Variables:
(a) Manipulated variable: Mass of plasticine, m
(b) Responding variable: Period of oscillation, T
(c) Constant variable: Distance between G-clamp and plasticine

Apparatus: Stopwatch, G-clamp, ruler and jigsaw blade

Materials: Plasticine of mass 20.0 g, 30.0 g, 40.0 g, 50.0 g and 60.0 g

Procedure:

1. Set up the apparatus as shown in Figure 2.44.

2. Fix a piece of 20.0 g plasticine to the
free end of the jigsaw blade.

3. Displace the free end of the jigsaw blade
horizontally and release it so that the
plasticine oscillates.

4. Record the time, t1 for 10 complete
oscillations of the plasticine in Table 2.7.

5. Repeat steps 3 and 4 and record the
time as t2.

6. Repeat steps 2 to 5 using plasticine of mass 30.0 g,
40.0 g, 50.0 g and  60.0 g.

G-clamp

Leg of table
20 cm20 cmPlasticine

Jigsaw blade

Figure 2.44

Bab 2 Daya dan Gerakan I

5353

Chapter 2 Force and Motion I

2.4.12.4.1 2.4.22.4.2
Results:

Table 2.7

Mass of plasticine
m / g

Time for 10 complete oscillations, t / s Period of oscillation
T / st1 t2 taverage

20.0

30.0

40.0

50.0

60.0

Analysis of data:

1. Determine the period of oscillation, T of the plasticine, with the following equation:

T =
taverage

10
, where taverage =

t1 + t2

2

2. Plot a graph of T 2 against m.

3. Based on your graph, state the relationship between period of oscillation, T and mass of
plasticine, m.

4. Explain how the relationship between mass and inertia is determined from this
experiment.

Conclusion:
What conclusion can be made from this experiment?

Prepare a complete report on this experiment.

Discussion:

1. How can this apparatus set up be used to determine the mass of an object?

2. Explain one precaution to improve the accuracy of the results.

Experiment 2.2 shows that the inertia of an object is related to its mass. The horizontal
oscillations in an inertial balance are not influenced by gravitational force. The period of horizontal
oscillation of the inertial balance depends only on the mass of the plasticine. The larger the mass
of an object, the larger the inertia of the object.

5454 2.4.22.4.2
Effects of Inertia in Daily Life
Astronauts in a spaceship such as the International Space Station
(ISS) are in a condition of zero gravity. Under the condition of
zero gravity, only the inertial balance can be used to measure
mass. Photograph 2.6 shows a special type of inertial balance used
by astronauts. The periods of oscillation of astronauts are used to
determine their body mass.

Photograph 2.6 An astronaut using an inertial balance

Inertia can have good and bad effects to our daily lives. Let us discuss situations that
involve inertia.

Activity 2.7

Aim: To discuss situations in daily life that involve inertia

Instructions:
1. Carry out a Hot Seat activity.

2. Gather information on situations in daily life that involve inertia.

3. Discuss whether the situations have good or bad effects on human.

4. Suggest a method to reduce the bad effects of inertia.

5. Choose a group representative to answer questions from other groups.

Inertial balance video

http://bt.sasbadi.com/p4055

ISS ICS

Inertial
balance

OscillationOscillation

5555

Chapter 2 Force and Motion I

2.4.32.4.3
Several examples of daily life situations that involve inertia and its effects are as follows.

Raindrops spun off when a wet umbrella is
rotated and suddenly stopped.

Raindrops on an umbrella are in motion as the
umbrella rotates. When the umbrella stops rotating,
the inertia of the raindrops causes the raindrops
to continue in motion and leave the surface of the
umbrella.

The inertia of the passengers keeps them in their initial state of rest or motion.

Chilli sauce or tomato sauce in a bottle flow out
when the glass is moved quickly downwards and
then suddenly stopped. The inertia of the sauce
causes it to continue moving downwards and out
of the bottle.

Passengers are thrown forward when a moving
bus suddenly stops.

Passengers are thrown backwards when a bus at
rest suddenly starts moving forward.

Oil tanker trucks have large inertia. In fact, an
oil tanker truck has its storage tank divided
into several separate tanks. The separated tanks
reduce the impact of the inertia of the petrol.

Situation 1

Situation 2

Situation 3 Situation 4

5656 2.4.32.4.3
When the carriage of the roller coaster changes
its direction of motion suddenly, the inertia of
the riders keeps them in their original state of
motion. The safety system ensures the riders
remain in their seats and are not thrown out
during the motion.

Riders on a roller coaster in a recreation park are
assigned to seats through a specialised safety system.
The roller coaster carriage moves at sudden changes of
speed and direction.

Driver and passengers of a car are advised
to wear seat belts.

Formative Practice 2.42.4

1. Explain the concept of inertia.

2. Brian wishes to pull a table cloth without toppling the things on top of the table cloth.
How can Brian do it? Explain your answer.

3. Study the following statements:

Statement 1: Objects can continue moving only if a force acts on them.
Statement 2: Rockets in outer space can move without any engine thrust.
Statement 3: Force is required to change the state of motion of objects.

(a) Which of the statements can be explained correctly using Newton’s First Law
of Motion?

(b) Give reasons of your choice.

When a car stops abruptly, the driver and
passengers in the car are thrown forward because
of inertia. Seat belts prevent them from being
thrown forward and hurting themselves.

Bab 2 Daya dan Gerakan I

Situation 5 Situation 6

5757

Chapter 2 Force and Motion I

2.4.32.4.3
Momentum2.5

Photograph 2.7 shows a fast moving car and a lorry
carrying a heavy load on the road. Which vehicle is more
difficult to stop if both are driven at the same velocity?

Momentum is a vector quantity. All moving objects
have momentum. The direction of momentum is the same
as the direction of the velocity of an object. An object
moving with high velocity or large mass has a large momentum.
Momentum, p of a moving object can be calculated using the
following formula:

p = mv, where p = momentum
 m = mass
 v = velocity

S.I. unit for momentum is kg m s–1

Activity 2.8

Photograph 2.7 A fast moving
car and a lorry on the road

LANGUAGE
iNtegraTION OF

The word momentum originates
from Latin that means
movement. Isaac Newton
described it as “quantity
of motion”.

Figure 2.45

Aim:  To study how the mass and velocity of an object influence the effect of stopping
the object

Materials:  Two marbles of different masses, ruler with a groove in the centre, two books of
equal thickness, thin cardboard and sticky tape

Instructions:
1. Arrange the materials as shown in

Figure 2.45. Raise one end of the ruler
using one thick book.

2. Release the marble from the top end of
the ruler so that it hits the cardboard.

3. Measure the displacement, s1 of the
cardboard and record it in Table 2.8.

4. Repeat steps 2 to 3 and record the
displacement as s2.

5. Calculate saverage =
s1 + s2

2
 and record it.

6. Repeat steps 1 to 5 using two books of equal thickness.

7. Repeat steps 1 to 5 using a marble of larger mass.

Ruler with a groove
in the centre Marble

Folded thin
cardboard

Line showing initial
position of cardboard

Thick book

5858 2.5.12.5.1
The marble that is released from a high position moves downwards at a high velocity
and displaces the cardboard further. This also happens when the marble has larger mass. The
displacement of the cardboard represents the difficulty in stopping the marble. Objects with larger
momentum are more difficult to stop.

Example

A lorry of mass 20 000 kg moves with a velocity of 22 m s–1. A car of mass 2 000 kg moves
with a velocity of 30 m s–1.
(a) What is the momentum of the lorry and the car?
(b) If the lorry moves with the same velocity as the car, what is the momentum of the lorry?

Solution:

(a) Step �
List the given information in symbols.

Step �
Identify and write down the formula used.

Step �
Substitute numerical values in the formula and
perform the calculations.

Momentum of the car, p = 2 000 kg × 30 m s–1
  = 60 000 kg m s–1

= 60 000 N s
(b) Momentum of the lorry moving at the velocity of the car
 = 20 000 kg × 30 m s–1

= 600 000 kg m s–1

= 600 000 N s

Momentum of lorry,
p = mv

p = 20 000 kg × 22 m s–1
 = 440 000 kg m s–1
 = 440 000 N s

Mass of lorry, m = 20 000 kg
Velocity of lorry, v = 22 m s–1

1
2

3
1

2
3

1
2

3

Results:
Table 2.8

Mass Number of books
Displacement of cardboard, s / cm

s1 s2 saverage

Small 1

Small 2

Large 1

Discussion:
1. What is represented by the displacement of the cardboard?
2. How does the velocity of the marble influence the displacement of the cardboard?
3. How does the mass of the marble influence the displacement of the cardboard?

INFOSMARTSMART

From Chapter 1, the unit newton
(N) in base unit is kg m s–2.
Unit for momentum:
kg m s–1 = (kg m s–2) s
 = N s

5959

Chapter 2 Force and Motion I

2.5.12.5.1
Activity 2.10

Applications of the Concept of Momentum in Daily Life
You have learned the concept of momentum. Carry out Activity 2.9 to understand the
applications of momentum in daily life.

Activity 2.9

Aim: To discuss the applications of the concept of momentum in daily life

Instructions:

1. Work in groups.

2. Visit websites to gather information on the applications of momentum in daily life and
present the results of your findings.

Applying the Principle of Conservation of Momentum in Collisions and Explosions
Figures 2.46(a) and (b) show two firemen putting out a fire. In Figure 2.46(b) both firemen are
thrown backwards when water at high speed gushes out of the hose. Why does this happen?

(a) (b)

Figure 2.46 Firemen putting out a fire

Water gushing out of the hose at high speed has a high forward momentum. Due to this,
two or more firemen are needed to balance the momentum by holding the hose tightly.

Aim: To study a situation involving the Principle of Conservation of Momentum in daily life

Instructions:

1. Work in groups.
2. Gather information on a situation which involves the Principle of Conservation of

Momentum in daily life.
3. Present your findings using multimedia presentation.

ICS

ICS

6060 2.5.22.5.22.5.12.5.1
Activity 2.11

Aim: To investigate the Principle of Conservation of Momentum

Apparatus:  Ticker timer, alternating current power supply, runway, trolleys, wooden block
and retort stand

Materials: Ticker tape, cellophane tape, plasticine, pin and cork

Instructions:
1. Work in groups.
2. Set up the apparatus as shown in Figure 2.47.

Retort stand

Friction compensated runway
Cork and pin

Cork
Ticker timer

Clamp

Ticker tape

Alternating current
power supply

Trolley A
Trolley B Wooden block

Figure 2.47

3. Raise one end of the runway so that it is friction compensated.
4. Switch on the ticker timer and push trolley A hard towards trolley B.
5. Determine the following velocity in cm per 10 ticks:

(a) Velocity of trolley A before collision, u1.
(b) Common velocity of trolleys A and B after collision, v

6. Record the results in Table 2.9 on page 62.
7. Repeat steps 4 to 6 for 1 trolley colliding with 2 trolleys at rest.
8. Repeat steps 4 to 6 for 2 trolleys colliding with 1 trolley at rest as shown in Figure 2.48.

Retort stand

Friction compensated runway
Cork and pin

Cork
Ticker timer

Ticker tape

Alternating current
power supply Trolley A

Trolley B Wooden blockClamp

Figure 2.48

9. Repeat steps 4 to 6 for 3 trolleys colliding with 1 trolley.

6161

Chapter 2 Force and Motion I

2.5.22.5.2
Results:

Table 2.9

Before collision After collision

Trolley A Trolley B Total
momentum

Trolley A and
trolley B

Total
momentum

m1

u1
(cm per
10 ticks)

m2 u2 m1u1 + m2u2 m1 + m2

v
(cm per
10 ticks)

(m1+ m2)v

1 1 0 2

1 2 0 3

2 1 0 3

3 1 0 4

Discussion:
1. What is a friction compensated runway?
2. Compare the total momentum before and after collision.
3. Is the total momentum conserved? Explain your answer.

In Activity 2.11, you may find that the total momentum before and after collision differs
slightly. This difference is due to external forces such as friction that cannot be compensated
completely.

Before collision

m
1

m
2

A B

u
1

u
2

After collision

m
1

m
2

A B

v
1

v
2

Figure 2.49 A system of two objects before and after collision

Figure 2.49 shows a system of two objects before and after collision. The Principle of
Conservation of Momentum states that the total momentum before collision is equal to the total
momentum after collision if no external force is acting on it.

m1u1 + m2u2 = m1v1 + m2v2

6262 2.5.22.5.2
Formative Practice 2.52.5

1. What is meant by momentum and conservation of momentum?

2. A lorry of mass 1 000 kg moves at a velocity of 5.0 m s–1. It collides with a car of mass
800 kg moving at a velocity of 2.0 m s–1 in the same direction. If the lorry moves at a
velocity of 3.4 m s–1 in the same direction aft er the collision, calculate the velocity of
the car.

Figure 2.50 shows the launching of a rocket. Launching of a
rocket is an example of explosion. Explosion refers to a situation
where an object at rest breaks up into two or more parts. Before
launch, the rocket is at rest at the base of the launch pad with zero
momentum. During launch, hot gases move downwards and the
rocket moves upwards. Explosion is also a closed system which
does not involve external forces. As such, total momentum is also
conserved in explosion.

Total momentum = total momentum
 before explosion    after explosion
 0 = m1v1 + m2v2
 m1v1 = –m2v2

Activity 2.12

Aim: To build and launch a water rocket

Instructions:

1. Work in groups.

2. Gather information on the following:
(a) materials needed to build and launch a water rocket
(b) ways to build a water rocket
(c) safety precautions to be taken

3. Discuss the required information and complete the K-W-L Chart as a guide in the search
for information.

4. Design, build and launch your group’s water rocket in your school field.

5. Prepare a report on how the Principle of Conservation of Momentum is applied in the
water rocket launching technology.

K-W-L Chart

http://bt.sasbadi.com/p4063

m
1
v

1

m
2
v

2

Rocket
moves
upwards

Hot gases
move
downwards

Figure 2.50 Launching of
a rocket

ICS STEM

6363

Chapter 2 Force and Motion I

2.5.22.5.2
Force2.6

Most of our daily activities involve forces. A force can change
the state of motion of an object. How does a force change the
momentum of an object in motion along a straight line?

Recall

Force and Motion

Activity 2.13

Figure 2.51

Aim:  To investigate the relationship between force and acceleration and the relationship
between mass and acceleration

Apparatus:  Ticker timer, alternating current power supply, runway, three trolleys and
retort stand

Materials: Ticker tape, cellophane tape, three elastic strings with a knotted loop at each end

A  To investigate the relationship between force and acceleration with a fixed mass

Instructions:
1. Set up the apparatus as shown in

Figure 2.51.
2. Switch on the ticker timer and pull

the trolley down the runway with
one elastic string (one unit of force).

3. Calculate the acceleration of
the trolley using the ticker tape
obtained and record it in Table 2.10.

4. Repeat steps 2 to 3 using two elastic strings and three
elastic strings with each of the strings stretched to the
same length as that of the first elastic string in step 2.

5. Plot the graph of acceleration, a against force, F
and state the relationship between acceleration, a
and force, F.

Results:

Table 2.10

Force, F u / cm s–1 v / cm s–1 t / s a / cm s–2

1 elastic string

2 elastic strings

3 elastic strings

B  To investigate the relationship between mass and acceleration at a constant force

Instructions:
1. Repeat steps 1 and 2 in (A) by pulling the trolley using two elastic strings stretched together.
2. Repeat step 1 in (B) using two trolleys as shown in Figure 2.52 and then three trolleys.

Trolley

Elastic string

Alternating current
power supply

Pulled Friction
compensated
runway

Ticker timer
Ticker tape

Retort stand

Guide to carrying out
Activity 2.13

http://bt.sasbadi.com/p4064b

6464 2.6.12.6.1
3. Calculate the acceleration of the trolley from the ticker
tape obtained and record it in Table 2.11.

4. Plot a graph of acceleration, a against the reciprocal of mass,
1
m

and state the relationship between acceleration, a and

mass, m.

Results:

Table 2.11

Mass u / cm s–1 v / cm s–1 t / s a / cm s–2

1 trolley, m

2 trolleys, 2m

3 trolleys, 3m

Two trolleys

Two elastic strings

The results of Activity 2.13 shows that the acceleration of
an object depends on the applied force and the mass of the object.

Figure 2.53 Acceleration-force graph

Figure 2.54 Acceleration-reciprocal of mass graph

Acceleration is directly
proportional to the applied
force when the mass of an
object is fixed

In S.I. units, 1 N is the force that
produces an acceleration of 1 m s–2
when applied on a mass of 1 kg.
As such,
1 N = k × 1 kg × 1 m s–2

k = 1
Therefore, F = ma

Newton’s Second Law of Motion states that
the rate of change of momentum is directly
proportional to the force and acts in the direction
of the applied force. From the relationship
F ∝ ma
F = kma, k is a constant

F

a

0

a

0 1–m

F – Force
m – Mass
a – Acceleration

Combining the two
relationships:

a ∝ F
m

Therefore, F ∝ ma

Figure 2.52

The relationship between force, F, mass, m and
acceleration, a for an object in motion is

F ∝ ma

F ∝ m (v – u)
t

F ∝ (mv – mu)
t

Expression of Newton’s
Second Law of Motion

a ∝ F
m constant

Acceleration is inversely
proportional to mass
of an object when a
constant force is applied
on the object

a ∝ 1
m

F constant

F ile
Change of momentum = mv – mu

Rate of change of momentum = (mv – mu)
t

6565

Chapter 2 Force and Motion I

2.6.12.6.1
Formative Practice 2.62.6

1. A force, F acts on a body of mass 5 kg.
(a) If the body accelerates uniformly from 2 m s–1 to 8 m s–1, determine the value of F.
(b) If F = 10 N, determine the displacement of the body 6 seconds after the body starts to

move from rest.

2. A force of 80 N acts on a stationary object for 7 seconds and causes the object to reach a
velocity of 35 m s–1. Calculate:
(a) the mass of the object.
(b) the displacement of the object.

Solving Problems Involving the Formula F = ma
Example 1

A worker pulls a load of mass 80 kg along a horizontal surface by a force of 160 N. If the
surface is smooth and without any resistance that opposes the motion of the object, what is
the acceleration of the load?

Solution:

Step �
List the given information in symbols.

1
4

2
4

3

Mass, m = 80 kg
Force, F = 160 N

Step �
Identify and write down the formula used.

Step �
Substitute numerical values into the formula and
perform the calculations.

1
4

2
4

3
 160 = 80 × a

a = 160
80

= 2 m s–2

A car of mass 1 200 kg moves with a velocity of 30 m s–1. When the brakes of the car are
applied, the car stops in  5 seconds. Calculate the average braking force of the car.

Solution:

Deceleration of car, a = v – u
t

= 0 – 30
5

= –6 m s–2

Average braking force, F = ma
 = 1 200 kg(–6 m s–2)
 = –7 200 N (Negative sign shows that the force acts in the opposite
  direction to the motion of the car)

F = ma

Example 2

6666 2.6.22.6.2
Activity 2.14

Impulse and Impulsive Force2.7

Photograph 2.8 shows the action of a long jump athlete
bending his legs while landing. What is the effect of
this action?
 The action of bending his legs is to reduce the
magnitude of the impulsive force acting on his body.
Impulse is a change of momentum.

Impulse, J = mv – mu
 = Ft
F = applied force
t = time of action of the force

Impulsive force is the rate of change of momentum in a collision or impact in a short
period of time. The formula for impulsive force is as follows.

Impulsive force, F = mv – mu
t

t = impact time
mv – mu = change of momentum

If change of momentum, mv – mu is constant, then F ∝ 1
t

. If t is small, then the
magnitude of F is big or vice versa.

Aim: To discuss impulse and impulsive force

Instructions:

1. Work in groups.

2. Gather the following information from appropriate
websites.
(a) effect of follow through action on magnitude of

impulse.
(b) situations and applications in daily life that involve

impulse.
(c) situations and applications in daily life that involve impulsive force, including safety

features in vehicles.

3. Prepare a brief multimedia presentation and present it.

Video on impulse, momentum
and impulsive force

http://bt.sasbadi.com/p4067

Photograph 2.8 Long jump athlete bending
his legs

ISS ICS

Bab 2 Daya dan Gerakan I

6767

Chapter 2 Force and Motion I

2.7.12.7.1
In fact, all the situations that you studied in Activity
2.14 involve a pair of forces, action and reaction. The
relationship between action and reaction is explained in
Newton’s Third Law of Motion. Newton’s Third Law of
Motion states that for every action there is a reaction of equal
magnitude, but in the opposite direction. Study the examples
of situations and the explanations given below.

Increasing the magnitude of impulse by follow through action

Figure 2.55 Follow through in football

Reducing impulsive force by extending impact time

Photograph 2.9 Car collision test

Increasing impulsive force by reducing impact time

Photograph 2.10 Using a mortar and pestle

Action and reaction force

http://bt.sasbadi.com/p4068

Collision causes a car to stop and experience a
change in momentum. The crumple zone of the
car extends the impact time during collision. As
a result, the magnitude of the impulsive force
on the car is reduced.

Reaction on carReaction on car

Action on objectAction on object

A follow through action
after a strong kick produces
a large impulse. As a result,
the ball experiences a large
change in momentum and
moves at a high velocity.

A pestle moving at high velocity is stopped by a hard
mortar in a short interval of time. A large impulsive
force is produced.

Reaction
on foot Follow through action

Action on ball

Action on mortarAction on mortar

Reaction on pestleReaction on pestle

6868 2.7.12.7.1
Example 1

A plasticine ball of mass 0.058 kg is thrown at a velocity of 10 m s–1 and hits the wall. It sticks
to the wall. What is the impulse of the plasticine?

Solution:

Step �
List the given information in symbols.

1
2

3

Mass, m = 0.058 kg
Initial velocity, u = 10 m s–1

Final velocity, v = 0 m s–1

Step �
Identify and write down the formula used.

1
2

3 Impulse, J = mv – mu

Step �
Substitute numerical values into the
formula and perform the calculations.

1
4

4
2

4
3

J = 0.058(0) – 0.058(10)
 = 0 – 0.058(10)
 = 0 – 0.58
 = –0.58 N s (in the opposite direction to the

velocity of the plasticine)

A golfer hits a golf ball of mass 45.93 g at a velocity of 50 m s–1. If the impact time is 0.005 s,
what is the impulsive force acting on the golf ball?

Solution:

m = 0.04593 kg, u = 0 m s–1, v = 50 m s–1, t = 0.005 s

Impulsive force, F = mv – mu
t

= 0.04593(50) – 0.04593(0)
0.005

= 459.3 N (acting in the same direction as the velocity of the golf ball)

Formative Practice 2.72.7

1. In a car collision test, a car of mass 1 500 kg hits the wall with a speed of 15 m s–1. The car
bounces back with a speed of 2.6 m s–1. If the collision time is 0.15 s, calculate the:
(i) impulse in the collision
(ii) impulsive force acting on the car

2. A football player kicks a ball of mass 450 g with a force of 1 500 N. Th e contact time of his
shoe with the ball is 0.008 s. What is the impulse on the ball? If contact time is increased to
0.013 s, what is the velocity of the ball?

Solving Problems Involving Impulse and Impulsive Force

Example 2

6969

Chapter 2 Force and Motion I

2.7.22.7.2
Weight2.8

Photograph 2.11 shows a man lifting a barbell.
Earth’s gravitational pull acting on the barbell
contributes to the weight of the barbell. The weight
of the barbell makes it difficult for the man to lift it.

The barbell will drop to the ground with an
acceleration if the man releases it. According to
Newton’s Second Law of Motion,

F = ma

W = mg

• Gravitational force acting on the barbell is its weight, W.
• Acceleration of the barbell is gravitational acceleration, g.

Weight is a vector quantity
which has the same
direction with gravitational
force, that is towards the
centre of Earth.

Unit for weight: N

Unit for mass: kg

W = mg

g = W
m

Unit for g: N kg–1

Physical quantity, g with unit N kg–1 is the gravitational field strength. Gravitational field
strength, g is the force acting per unit mass caused by gravitational pull. For objects on the surface
of the Earth, gravitational field strength, g is 9.81 N kg–1. Each 1 kg mass experiences a force of
gravity of 9.81 N. Can you calculate your weight on the surface of the Earth?
 Photograph 2.12 shows an astronaut wearing a full spacesuit when exploring the Moon. It is
easier for the astronaut to walk on the Moon compared to walking on Earth. Why is this so?

Astronaut’s movements on
the Moon

http://bt.sasbadi.com/p4070

Photograph 2.11 Lifting a barbell

Photograph 2.12 Astronaut wearing a
full spacesuit on the Moon

MalaysiakuMalaysiaku
MY PRIDE

Dato. Dr. Sheikh Muszaphar
Shukor bin Sheikh Mustapha
is the first astronaut of Malaysia.

7070 2.8.12.8.1
Formative Practice 2.82.8

Activity 2.15

Aim: To create a vehicle model by applying Newton’s Laws of Motion

Instructions:

1. Work in groups. Gather information on the applications
of Newton’s Laws of Motion in the creation of cars. Give
emphasis to the following:
(a) shape of vehicle model
(b) type of engine, transmission system, suspension

system, steering system and braking system
(c) safety aspects of driver and passengers
(d) aspects of comfort of driver and passengers in the vehicle
(e) type of fuel

2. Discuss the required information and complete the K-W-L Chart as a guide in your
search for information.

3. Create the vehicle model.

4. Present the applications of Newton’s Laws of Motion of your group’s vehicle model.

Mass of astronaut’s suit on
Earth is 81.65 kg.
WEarth = 81.65 kg × 9.81 N kg-1

= 800.99 N

Gravitational field strength on the Moon is 1
6

of the
gravitational field strength on Earth.
WMoon = 1

6
 × 800.99 N

= 133.50 N

1. What is the meaning of gravitational field strength?

2. State the differences between mass and weight.

3. A 10 kg object has a weight of 150 N on a planet.
(a) What is the gravitational field strength on the planet?
(b) Is the planet bigger than Earth? Give reasons for your answer.

4. An astronaut of mass 60 kg is assigned to explore the Moon. What is the astronaut’s weight
on the Moon’s surface?

K-W-L Chart

http://bt.sasbadi.com/p4071

ISS STEM

7171

Chapter 2 Force and Motion I

2.8.12.8.1
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7272
SELF REFLECTIONSELF-REFLECTION

Performance EvaluationPerformance Evaluation

1. New things I learnt in this chapter on force and motion are ✎ .

2. The most interesting thing I learnt in this chapter on force and motion is ✎ .

3. Things I still do not fully understand or comprehend are ✎ .

4. My performance in this chapter,

Poor  1 2 3 4 5   Excellent

5. I need to ✎  to improve my performance in
this chapter.

Download and print
Self-reflection Chapter 2

http://bt.sasbadi.com/p4073

1. A car moves from rest with an acceleration of 2.0 m s–2. Calculate:
(a) velocity of car after 5.0 s.
(b) distance travelled in 5.0 s.
(c) distance travelled in the fifth second.

2. Encik Nizam drives a car at a speed of 108 km h–1. Suddenly he sees a car in front moving
very slowly. Th erefore, Encik Nizam slows down his car to a speed of 72 km h–1. The
displacement made by the car is 125 m. If the acceleration of the car is uniform, calculate
(a) acceleration of Encik Nizam’s car
(b) time taken for the speed of the car to reduce from 108 km h–1 to 72 km h–1.

3. Swee Lan rows a boat forward. She uses an oar to push the water backwards. Why is the
boat able to move forward?

4. A car of mass 1 200 kg at rest is moved by a force of 150 N. Determine the acceleration of
the car and time taken for the car to reach a velocity of 1.5 m s–1.

5. Gravitational fi eld strength on the Moon is 6 times lower than that on Earth. If a stone
weighing 2 N is carried back from the Moon to Earth, calculate the weight of the stone
on Earth.

6. A bullet of mass 10 g is fi red from a gun of mass 2.0 kg. If the recoil velocity of the gun
aft er fi ring the shot is 0.5 m s–1, calculate the velocity of the bullet.

7373

Chapter 2 Force and Motion I
7. Photograph 1 shows a car moving along a road. Initially, the car moves at a uniform
velocity of 18 m s–1 for 15 s. Th en, the car accelerates at 1.5 m s–2 for 5 s. Subsequently, the
car starts to decelerate to 15 m s–1 in 5 s. Th e car continues to move at this speed for 10 s
and fi nally decelerates again and stops at time, t = 50 s.

Photograph 1
  Based on the given information, sketch the velocity-time graph for the motion of the car.

Show the values of velocity and time in your sketch.

8. A rubber ball is released from a height, H. The ball drops vertically downwards and upon
hitting the floor, bounces back a height, h (h < H). If the velocity of downward motion is
negative, sketch the velocity-time graph for the motion of the rubber ball.

9. A car at rest starts to move when a bus moves at a uniform velocity of 15 m s–1 passes
by its side. The car reaches a velocity of 20 m s–1 in 10 seconds and continues to move at
a constant velocity in the same direction with the bus. The graph in Figure 1 shows the
motion of the car and the bus along a straight road.

v / m s–1

Car

Bus

t / s0
10 50

15

20

Figure 1

(a) Calculate the time taken for the car to reach the same speed as the bus.

(b) What is the displacement of the car when it reaches the speed of the bus?

(c) Calculate the distance travelled by the car and bus at time, t = 50 s.

(d) At time, t = 50 s, is the car in front of the bus?

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Enrichment CornerEnrichment Corner

10. Photograph 2 shows the launching of a spacecraft using a rocket at a launch pad.

Photograph 2
(a) Explain how the release of hot gases through the rocket’s exhaust enables the rocket to

accelerate upwards.
(b) How can the acceleration of this rocket be increased?

11. Photograph 3 shows a hovercraft  which can move
on land or on the surface of water swift ly because
of the support from a layer of air trapped under the
hovercraft . Th e hovercraft  of mass of 25 000 kg
starts from rest and its engine produces a thrust, F
of 22 000 N.
(a) Determine the initial acceleration of the hovercraft

if there is no friction at that instant.
(b) What is the function of the layer of air trapped

under the hovercraft?

12. Kok Chew and Zulkefli wish to determine Earth’s gravitational acceleration. They plan
to release a table tennis ball from the third level of their school building. Discuss the
suitability of the table tennis ball in this experiment.

13. Assume that you are an engineer who is assigned to invent a model of fast train in
Malaysia. Th is train is required to travel at high speeds using levitation above the railway.
Draw your model of the fast train and list its properties by taking into consideration its
shape, materials, mechanism and safety aspects.

Photograph 3

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Chapter 2 Force and Motion I
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