Focus TG4 KSSM (Physics) Terbitan Penerbitan Pelangi Sdn Bhd

CONTENTS

1Chapter 4.3 Specific Latent Heat 110
4.4 Gas Laws 119
Measurement 1
SPM Practice 4 134

1.1 Physical Quantities 2 5Chapter Waves 140
1.2 Scientific Investigation 6

SPM Practice 1 15

2Chapter Force and Motion I 19 5.1 Fundamentals of Waves 141

5.2 Damping and Resonance 148

2.1 Linear Motion 20 5.3 Reflection of Waves 151

2.2 Linear Motion Graphs 28 5.4 Refraction of Waves 155

2.3 Free Fall Motion 33 5.5 Diffraction of Waves 160

2.4 Inertia 36 5.6 Interference of Waves 165

2.5 Momentum 41 5.7 Electromagnetic Waves 174

2.6 Force 49 SPM Practice 5 179

2.7 Impulse and Impulsive Force 54

2.8 Weight 58 6Chapter

SPM Practice 2 60

Light and Optics 187

3Chapter Gravitation 67 6.1 Refraction of Light 188

6.2 Total Internal Reflection 200

3.1 Newton’s Law of Universal Gravitation 68 6.3 Image Formation by Lenses 207
3.2 Kepler’s Laws 80
3.3 Man-made Satellites 83 6.4 Thin Lens Formula 212

6.5 Optical Instruments 217

Praktis SPM 3 88 6.6 Image Formation by Spherical
Mirrors 222

4Chapter SPM Practice 6 230

Heat 93

PRE-SPM MODEL PAPER 235

4.1 Thermal Equilibrium 94 ANSWERS 252
4.2 Specific Heat Capacity 98

iv

2Chapter Physics  Form 4  Chapter 2 Force and Motion I  

Force and Motion I

CHAPTER FOCUS
2.1 Linear Motion
2.2 Linear Motion Graphs
2.3 Free Fall Motion
2.4 Inertia
2.5 Momentum
2.6 Force
2.7 Impulse and Impulsive Force
2.8 Weight

02 FOC PHYSICS F4 3P.indd 19 Have you heard of the story of the scientist Galileo
Galilea dropping two spheres of different masses from
the Leaning Tower of Pisa? He carried out this activity to
show that the time taken for two objects to fall from the
same height are the same even though they have different
masses. This is due to the property of gravity that you will
learn in this chapter.

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  Physics  Form 4  Chapter 2 Force and Motion I

2.1 Linear Motion 2. Non-linear motion is motion that is not in a
straight line.
My motion is a My motion is a
linear motion non-linear motion 3. When analysing linear and non-linear motion,
distance, displacement, speed, velocity,
Startt line acceleration and deceleration are commonly
encountered physical quantities.
Start End End Not a straigh
A straight line SPM Tips

Figure 2.1 Speed (scalar quantity)—magnitude only

1. Linear motion is a motion in a straight line. Velocity (vector quantity)—magnitude and direction

Chapter 4. Figure 2.2 shows a car moving on a straight road. Its speedometer shows the velocity of the car at

position A, B and C. Study the diagram to understand the meaning of stationary, uniform velocity and

non-uniform velocity. KPeodsuidtuioknanBB KePdousdiutikoann C

2 KePdousdiutikoann AA

Metre speed Metre speed Metre speed
The speed meter shows the car
The velocity of the car is zero. The car moving at a constant velocity of 60 The speed meter indicates a change
does not move (stationary). Object km per hour. of velocity from 60 to 90 km per
that is stationary has velocity equals hour. The car moves at a non-uniform
to zero. Figure 2.2 velocity. In this case, velocity is
increasing.

Distance, Displacement, Speed and Velocity

1. Figure 2.3 and Figure 2.4 show a car travelling from Delima to Mengkibol. We will use the figures to
study the physical quantities, distance, displacement, speed and velocity.

Distance and speed

1. (a) Distance is defined as the total length of the path travelled from one location

to another. N

(b) The distance travelled by the car is given by the following,

Distance travelled = (200 + 500 + 600 + 300) m WE Delima

= 1600m S 200 m
500 m
(c) Distance is a scalar quantity.

2. (a) Speed is the distance travelled per unit time. Speed can also be defined as 600 m
300 m
the rate of change of distance. Mengkibol

(b) Average speed of an object can be obtained by dividing the total distance Figure 2.3

travelled by the object by the total time taken for the travel.

(c) If the time taken by the car in Figure 2.2 to travel from Delima to Mengkibol

is 120 seconds, the average speed of the car is given by:

v = Total distance travelled, d
Time taken, t

1 600 m
(d) Speed is a scalar quantity. = 120 s = 13.3 m s–1

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Physics  Form 4  Chapter 2 Force and Motion I  

Displacement and velocity

3. (a) Displacement by an object is the distance travelled by the object in a N Delima
specific direction. WE 45°

(b) It is the distance travelled by the object from its initial position to its S 800 m
final position in a straight line.
Mengkibol
(c) For the car travelling from Delima to Mengkibol, the displacement is
determined by measuring the length of the straight line connecting 800 m
Delima to Mengkibol in the direction of Mengkibol from Delima.

(d) Hence from Figure 2.4, its displacement is given by the following,

Magnitude of the displacement Figure 2.4

=  8002 + 8002

= 1 131 m Chapter
Its direction is south-west of Delima (or S45°W of Delima)

(e) Hence, displacement is a vector quantity.

4. (a) Velocity is defined as the rate of change of displacement. 2
(b) In the case of the car travelling from Delima to Mengkibol, the average velocity, v, is given by:

v = Displacement,s
Time taken, t

1 131 m
= 120 s = 9.4 m s–1

(c) Velocity is a vector quantity.

EXAMPLE 2.1 EXAMPLE 2.2

A cow walked along a curved path from P to Q, An athlete runs from the P Q
which is 70 m away from P. Q lies to the south-west starting point O to P which is
of P. The distance travelled by the cow is 240 m 100 M
and the time taken is 160 s.
100 m north of O. After that he N
U runs to point Q which is 100 m 100 M

east of P. The total time taken

is 25 seconds. O

P What is the (a) distance, (b) Figure 2.6

average speed, (c) displacement, and (d) average

Q velocity traveled by the athlete?

Figure 2.5 Solution

Calculate the (a)  average speed, and (a) Total distance traveled
(b) average velocity = 100 + 100 = 200 m
of the cow moving from P to Q.
(b) Average speed
Solution
= ToTtoaltadlitsitmanece 200
Total distance travelled = 240 m = 25 = 8 m s–1

Displacement = 70 m (c) Displacement O
= 1002 + 1002
Time taken = 160 s

(a) Average speed = Total distance travelled = 100 2 m
Time taken
240 m
= 160 s = 1.5 m s–1 (d) Average velocity

(b) Average velocity = Displacement = 70 m = DTisoptlaalcteimmeent = 100 2 =4 2 m s–1
Time taken 160 s 25

= 0.44 m s–1 in the south- in the direction N45o E from O.

west direction of P

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  Physics  Form 4  Chapter 2 Force and Motion I

Acceleration and Deceleration Solution

25 30 35 25 30 35 25 30 35 25 30 35 Initial velocity, u = 10 m s–1
20 40 20 40 20 40 20 40
15 45 15 45 15 45 15 45 Final velocity, v = 20 m s–1

10 50 10 50 10 50 10 50

5 m s–1 55 5 m s–1 55 5 m s–1 55 5 m s–1 55 Time taken, t = 2.5 s
0 60 0 60 0 60 0 60

Figure 2.7 Acceleration, a = 20 – 10
2.5
1. Figure 2.7 shows a car moving along a straight = 4.0 m s–2
line. The speedometer of the car shows that it
is moving with increasing velocity. The car is EXAMPLE 2.4
accelerating.
Chapter A car travelling at

2. (a) Acceleration is defined as the rate of 24 m s–1 slowed down
change of velocity.
when the traffic light

turned red. After

2 Acceleration, a = Change of velocity undergoing uniform Figure 2.10
Time taken
deceleration for 4 s, it

stopped in front of the traffic light. Calculate the

(b) The formula for an object travelling with magnitude of the deceleration.
a uniform acceleration, a is given by:
Solution

Initial velocity, u = 24 m s–1

Final velocity, v = 0 m s–1

where u is the initial velocity of the object, Time taken = 4 s v – u 0 – 24
v is its final velocity and t is the time taken. t 4
Using the formula a = = = –6.0 m s–2
(c) The SI unit for acceleration is m s–2.
(d) Acceleration is a vector quantity. The negative value shows that the car is decelerating.
Hence, the magnitude of deceleration is 6.0 m s–2.

25 30 35 25 30 35 25 30 35 25 30 35 EXAMPLE 2.5
20 40 20 40 20 40 20 40
15 45 15 45 15 45 15 45

10 50 10 50 10 50 10 50

5 m s–1 55 5 m s–1 55 5 m s–1 55 5 m s–1 55 (a) A car moves from
0 60 0 60 0 60 0 60

BERHENTI

rest and accelerates

Figure 2.8 uniformly to a

3. (a) An object is said to be undergoing a velocity of 20 m s–1 Figure 2.11
deceleration or retardation when it is
slowing down. The rate of change of in 5 seconds. What
velocity of the object then has a negative
value. is the acceleration of the car?

(b) Figure 2.8 shows a car decelerating. (b) After that, the car slows down at a uniform
The speedometer of the car shows that
it is moving with decreasing velocity. rate until it stops in 5 seconds. What is the

acceleration of the car?

Solution

(a) Initial velocity, u = 0 m s–1

Final velocity, v = 20 m s–1

Time taken, t = 5 s

EXAMPLE 2.3 Therefore a = v – u = 20 – 0 = 4 m s–2
t 5
A van accelerates uniformly
from a velocity of 10 m s–1 (b) Initial velocity, u = 20 m s–1
to 20 m s–1 in 2.5 s. What is
the acceleration of the van? Final velocity, v = 0 m s–1

Time taken, t==v5–ts u = 0 – 20 = –4 m s–2
Therefore, a 5
Figure 2.9
The car decelerates with a magnitude of 4 m s–2.

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Physics  Form 4  Chapter 2 Force and Motion I  

ACTIVITY 2.1

Problem statement: What is the acceleration of the trolley when it moves down the inclined plane?

Aim: To determine the acceleration of a trolley

Apparatus and materials: Trolley, inclined plane, stopwatch, ticker timer, ticker tape, a.c. power supply,
cellophane tape, and metre rule.

Instructions: Ticker timer

Trolley

Ticker tape

Inclined plane Chapter

Power supply
a.c. 12 V

Figure 2.12

1. Figure 2.12 shows the arrangement of apparatus for this experiment. 2
2. An inclined plane is arranged at an angle such that a trolley can move down the plane without

being pushed.
3. The trolley attached to a ticker tape which passes through a ticker timer is released and moves

from the higher to the lower end of the inclined plane.
4. A few initial ticks of the tape are discarded and the rest cut into strips of 10 ticks each.
5. The tapes are arranged side-by-side from beginning to the end. Length/cm

6. (a) Total time of movement of the trolley, 7.0
(b) total displacement of the trolley, 6.0
5.0
(c) average velocity of the trolley, 4.0
(d) average initial velocity u, 3.0
(e) average final velocity v, 2.0
(f) acceleration of the trolley, a is determined. 1.0

Results: 10 20 30 40 50 60
Figure 2.13
(a) Number of ticks for each strip of tape = 10 Ticks

Time taken for each strip of tape = 10 × 0.02 = 0.2 s
Total time taken by the trolley = Number of strips × 0.2 s
= 6 × 0.2 = 1.2 s

(b) Total displacement of the trolley = Total length of all the strips

= (2.0 + 3.0 + 4.0 + 5.0 + 6.0 + 7.0) cm

= 27.0 cm

(c) Average velocity of the total motion of the trolley

= Total displacement = 27.0 cm = 22.5 cm s–1 2–1
Total time 1.2 s 1
Number of strips
(d) Average initial velocity, u = 2.0 = 10.0 cm s–1 of tapes
0.2

(e) Average final velocity, v = 7.0 = 35.0 cm s–1 1
0.2 1
1 1 1
(f) Time, t between u and v =  2 +1+1+1+1+ 2  × 0.2 second –21 v
u
= (6 – 1) × 0.2 second = 1.0 s
Figure 2.14
Hence, acceleration a = v – u = 35.0 – 10.0 = 25.0 cm s–2
t 1.0

Conclusion:
The trolley moves with an acceleration of 25.0 cm s–2.

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  Physics  Form 4  Chapter 2 Force and Motion I EXAMPLE 2.6

Analysis of Motion A trolley pulled a ticker tape through a ticker timer
1. A ticker timer is an apparatus used to study while moving down an inclined plane. Figure 2.17
shows the ticker tape produced.
the motion of an object within a short period
of time.

2. Figure 2.15 shows a ticker timer connected to
an a.c. power supply.

Power supply POWER SUPPLY 12 cm
a.c.12 V Figure 2.17

FUSE 10 12 Volts
8 plus
O 6 1 Determine the average velocity of the trolley.
0
Ticker timer 42 0

I AC OVLD DC

Chapter Ticker Solution
tape
Displacement = 12 cm

There are 10 spaces between the dots,

2 Vibrating bar Therefore, time taken = 10 × 0.02 = 0.2 s
Displacement
Average velocity = Time taken

Figure 2.15  Ticker timer = 12 cm = 60 cm s–1
0.2 s
3. A carbonised ticker tape is passed through the
ticker timer and is pulled by a moving object. EXAMPLE 2.7

4. When the a.c. power supply is switched on, A trolley is pushed Length/cm t /s
the vibrating bar vibrates 50 times per second to move up the slope 4.5
(following the frequency of the power supply of an inclined plane. 4.0
which is 50 Hz). Each time the vibrating bar Figure 2.18 shows the 3.5
hits the tape, a dot is produced on the tape. arrangement of the 3.0
ticker tapes. Each strip 2.5
5. The period of time between 2 adjacent dots is consists of 5 ticks or 0.1 2.0
known as 1 tick. second. Determine the 1.5
acceleration of the trolley. 1.0
0.5

0.2 0.4 0.6 0.8
Figure 2.18

Solution

Average initial velocity of the trolley,

Dot 1 tick u = 4.5 cm = 45.0 cm s–1
Figure 2.16  1 tick for carbonised ticker tape 0.1 s

6. A ticker timer can be used to determine Average final velocity of the trolley,
(a) time,
(b) displacement, v = 1.5 cm = 15.0 cm s–1
(c) average velocity, 0.1 s
(d) acceleration, and
(e) type of motion Time taken, t = (4 – 1) × 0.1
of an object.
= 0.3 s v–u
t
Therefore, acceleration, a =

= 15.0 – 45.0
0.3

= –100 cm s–2

Because the acceleration has a negative value,
therefore the trolley experienced a deceleration
with the magnitude of 100.0 cm s–2.

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Physics  Form 4  Chapter 2 Force and Motion I  

EXAMPLE 2.8 Solution

Sometimes there are too few ticks to be cut into Average initial velocity, u = 0.6 cm
strips. In such cases, direct analysis is needed to 0.02 s
determine the acceleration. The following ticker
tape shows the motion of a trolley. The ticker = 30 cm s–1
timer vibrates at a frequency of 50 Hz.
Average final velocity, v = 1.6 cm
0.02 s
Initial Final
= 80 cm s–1

0.6 cm Uniform acceleration 1.6 cm Time taken, t = (11 – 1) × 0.02 s Chapter
Figure 2.19 = 0.2 s

Therefore, acceleration, a = v–u
t

= 80 – 30 2
0.2

= 250 cm s–2

Determining Acceleration with Photogate and Electronic Timer

A pair of photogates A and B is Photogate and electronic 0000 x cm
arranged at two different locations timer system A
along an inclined plane. This type
of photogate consists of optical B Cardboard
sensors that can measure the time, Trolley
t, taken by an object passing Inclined plane
through it. In this case, the object Figure 2.20
is a cardboard with a length of x cm
attached to the top portion of a
trolley.

1. Average initial velocity, u, when the cardboard passes through photogate A is x where tA is the time
taken by the cardboard to pass through photogate A. tA

2. Average final velocity, v, when the cardboard passes through photogate B is x where tB is the time
taken by the cardboard to pass through photogate B. tB

3. The time taken for the trolley to move between photogate A and photogate B, t, can be measured by
the time difference recorded when the trolley moved between photogate A and photogate B.

Results

u= x dan v = x
tA tB

Therefore, the acceleration, a, is given by the following:

v–u x – x
t tB tA
a= = cm s–2
t

This value can be divided by 100 to convert to the unit m s–2.

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  Physics  Form 4  Chapter 2 Force and Motion I EXAMPLE 2.9

Relationship between Displacement, Hisham drove his car from home with a uniform
Velocity, Acceleration and Time acceleration and achieved a velocity of 15.0 m s–1
in 5.0 seconds. What is the
Motion with Uniform Velocity (a) acceleration of Hisham’s car?
1. For an object moving with uniform velocity, v, (b) displacement of Hisham’s car 5.0 seconds

its displacement, s after time, t is given by: after starting the journey?
(c) velocity of Hisham’s car at t = 4.0 s?
where v is the uniform velocity of the (d) velocity of Hisham’s car when he has
object.
travelled 20.0 m from the starting point?
Chapter Motion with Uniform Acceleration

1. The equations of motion of an object with Solution
uniform acceleration, a, are as follows:
Summary of information:
2 Displacement, s is given by: Initial velocity, u = 0 m s–1
s = (Average velocity) × time Final velocity, v = 15.0 m s–1
Time taken, t = 5.0 s
=  Initial velocity + final velocity  × time
2
=  u + v t v – u
2 (a) Acceleration, a = t .............................. 

Hence, = 15.0 – 0
5.0
.................................. 
= 3.0 m s–2

2. Acceleration, a = v – u................................ (b) Displacement, s = 1 (u + v)t......................
t 2

Rearranging equation, = 1 (0 + 15.0)(5.0)
........................................  2

= 37.5 m

(Can also be solved by equation )

3. Substitue equation  into equation . (c) Hisham’s velocity at t = 4.0.
1 v = u + at.....................................................
s = 2 (u + u + at)t
= 0 + 3.0 (4.0)
Simplifying .................. = 12.0 m s–1

4. Rearrange equation  into (d) If the velocity of Hisham’s car after moving

v – u 20.0 m is v.
a
t = With the information u = 0 m s–1

and substitue into equation , a = 3.0 m s–2

1 s = 20.0 m
2
hence, s = (u+v) v – u and using the equation 
a
v2 – u2 v2 = u2 + 2as
2a
= Therefore v2 = 02 + 2 (3.0 × 20.0)

Simpliying, = 120.0

Hence, v = 10.95 m s–1

.................................... 

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SPM Highlights HOTS Physics  Form 4  Chapter 2 Force and Motion I  

A cat sees a rat running towards a hole with a uniform Checkpoint 2.1
velocity 0.4 m s–1. At that time the position of the cat, the
rat and the hole are shown in the diagram respectively. If Q1 Farid leaves his home at 10:00 am and walks to
the cat runs towards the rat with a uniform acceleration the post office to post a letter. After that, he walks
of 1.0 m s–2, can the cat catch the rat before it enters to a bookstore to buy a present. From there, he
the hole? walks to KEC Fried Chicken to pack some food
before going to Ken’s house to celebrate his
birthday. Farid reaches Ken’s house at 12:00
noon.

1.8 km 0.9 km Chapter

Farid’s Post office
house U

6.0 m 2.0 m 1.2 km BT
Figure 2.21 S
Ken’s house Bookstore 2
1.6 km 0.7 km

Answer: KEC Fried
Distance between the cat and the hole is 8.0 m and Chicken
distance between the rat and the hole is 2.0 m. If
time taken for the cat to reach the hole is longer than Figure 2.22
the time taken for the rat to reach the hole, the cat
will not able to catch the rat or vice versa. By the time Farid reaches Ken’s house,
(a) what is the total distance travelled?
For the rat, its uniform velocity, (b) what is his average speed?
v = 0.4 m s–1 (c) calculate his displacement.
s = 2.0 m (d) calculate his average velocity.

Since Q2 Figure 2.23 shows four strips of ticker tapes
st == vsv× t for uniform velocity, HOTS obtained from an experiment using a trolley.
= 02..04
= 5.0 s End Beginning
Tape P
For the cat, Tape Q
initial velocity, u = 0 m s–1
acceleration, a = 1.0 m s–2 Tape R
displacement, s = 8.0 m Tape S

Using the formula

s = ut + 1 at2, Figure 2.23
2
(a) Compare tape P with tape Q and describe
8.0 = 0 + 1 (1.0)t2 (i) one common characteristic,
2 (ii) one difference between them.
t 2 = 16.0
(b) Describe the motion of the trolley for
Therefore, t = 16.0 (i) tapes R,
= 4.0 s (ii) tapes S.

Since the rat needs 5.0 s to reach the hole while the
cat needs only 4.0 s, the cat will be able to catch the
rat before it enters the hole.

02 FOC PHYSICS F4 3P.indd 27 27

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  Physics  Form 4  Chapter 2 Force and Motion I Q5 A car which is moving with a constant acceleration
HOTS passes by lamp post A and lamp post B with a
Q3 A student carried out an experiment with
a trolley running down an inclined runway. velocity of 18 m s–1 and 20 m s–1 respectively.
Figure 2.24 shows a chart representing the
motion of the trolley. Lamp post B

x / cm Lamp post A

Chapter 11 Figure 2.26

10 If the distance between the two lamp posts is 10 m,
calculate
9
(a) the acceleration of the car,
8 (b) the time taken by the car to travel from lamp
7
post A to lamp post B.
6
2.2 Linear Motion Graphs
5
Displacement-Time Graphs
4
Displacement, s 0m 0.5 m 1.0 m 1.5 m 2.0 m
23 Time, t 0s 2s 4s 6s 8s

2 Z 2.0 m

1 Z
Z
0 t / 5-tick
Figure 2.24 Displacement, s 0 m 0.5 m 1.0 m 1.5 m
Time, t 0 to 8 s
If the ticker timer made 50 ticks per second,
Figure 2.27 The tortoise and hare race
determine the
(a) initial velocity, 1. Figure 2.27 shows a tortoise moving at a slow
(b) final velocity,
(c) acceleration
of the trolley.

Q4 A cyclist cycled in a straight line at a velocity
of 8 m s–1 on a hard surface. When he was
6 m from a soft surface, he braked to slow the
bicycle down so that it entered the soft surface
at a velocity of 4 m s–1.

and steady speed while a hare is sleeping

soundly from the time t = 0 to 8 s.

Displacement, s / m Displacement, s / m

Hard surface 6 m Soft surface 2.0 2.0

1.5 1.5

Figure 2.25 1.0 1.0

(a) What is the time taken by the bicycle to slow 0.5 0.5
down from 8 m s–1 to 4 m s–1?
0 2 468 0 2 468
(b) Calculate the deceleration of the bicycle. Time, t / s Time, t / s
(c) The bicycle decelerated to a stop, with a (b) Hare
(a) Tortoise
uniform deceleration, in 3 seconds after
entering the soft surface. Calculate the Figure 2.28 s-t graph
distance travelled by the bicycle on the soft
surface.

2. The motion of the tortoise and hare is
represented by the two displacement-time
graphs above.

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Physics  Form 4  Chapter 2 Force and Motion I  

(a) The graph in Figure 2.28(a) shows the Velocity-Time Graph
displacement of the tortoise from the
starting line from time t = 0 to 8 s, 1. The gradient of the velocity-time graph gives
the value of the rate of change of velocity, that
(b) The graph in Figure 2.28(b) shows that is, the acceleration of the object.
throughout the time from t = 0 to 8 s, the
hare had a displacement of 0.5 m from the 2. The area under the graph gives the value of
starting point. Throughout the 8 seconds, the displacement of the object.
it did not move at all as it was sleeping.
Table 2.2

3. The displacement-time graph of an object Graph of v against t Explanation
allows the displacement of the object at a (a)
specific time to be determined. v / m s–1 Chapter

4. The gradient of the s-t graph of an object gives The object is not moving. Velocity
the rate of change of displacement, which is of the object is zero.
the velocity of the object.
2t /s
Displacement, s / m Displacement, s / m 0
(b)
v / m s–1 The object is moving with uniform

Velocity = 0 velocity.
Gradient of (c)
the s -t graph v / m s–1 Acceleration = gradient of graph

0 = 0 m s–2.
(d) t /s
v / m s–1
0 Time, t / s 0 Time, t / s
0
Figure 2.29 The velocity increases with
a uniform rate. The gradient
5. Some common graphs are shown below. is positive and constant. The
object is moving with uniform
Table 2.1 t /s acceleration.

Graph of s against t Explanation

(a) s / m The displacement of the object The velocity decreases with
a uniform rate. The gradient
0 is always the same. Hence, is negative and constant. The
(b) s / m object is moving with uniform
the object is not moving or deceleration.

stationary. t /s

Velocity = gradient of graph

t /s = 0 m s-1

The displacement increases at 3. The area under the following graphs of v
a uniform rate. The gradient
of the graph is a constant. against t gives the displacement of the object
Hence, the object moves with
a uniform velocity. respectively.

0 t /s (a) v / m s–1 (b) v / m s–1
(c) s / m
The gradient of the graph is vv
0 increasing. The velocity of the
(d) s / m object is increasing. Hence, 0 t t /s 0 t t /s
the object moves with an
0 acceleration. (c) v / m s–1 t t /s
t /s
v
The gradient of the graph u
is decreasing. The velocity 0
of the object is decreasing.
Hence, the object moves with
t /s a deceleration.

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  Physics  Form 4  Chapter 2 Force and Motion I

Acceleration-Time Graphs EXAMPLE 2.11

Table 2.3

Graph of a against t Explanation Figure 2.31 shows the velocity-time graph of a
motorcycle along a straight road.
(a)
a / m s–2 v / m s–1

a Object moves with positive 25.0 AB
uniform acceleration. 20.0
15.0
Chapter 0 t /s 10.0
(b)
a / m s–2 Object moves with negative 5.0
uniform acceleration or
0 t /s deceleration. 0 5.0 10.0 15.0 C t /s
–5.0 20.0 25.0

–10.0 D

2 –a Figure 2.31

EXAMPLE 2.10 (a) Explain the state of motion of the motorcycle
represented by the portion of the graph
Figure 2.30 shows a displacement-time graph of a (i) OA (ii) AB
runner during a warming up session. (iii) BC (iv) CD

s /m C (b) Determine the displacement of the motorcycle
s1 (i) after 20.0 s.
(ii) after 25.0 s.
s2 A B (c) Sketch a acceleration-time graph for the

motorcycle during the first 25.0 seconds.

0 t1 D t /s Solution
t4
t2 t3 (a) (i) OA
Figure 2.30 25.0 – 0
Gradient of graph = 10.0 – 0

Explain the state of motion of the runner at the = 2.5 m s–2
portion of the graph represented by (a) OA (b) AB
(c) BC (d) CD. The motorcycle moves with a constant

acceleration of 2.5 m s–2.

Solution (ii) AB

(a) OA The motorcycle moves with a constant
The gradient of the graph is positive and
velocity of 25.0 m s–1.
constant. The runner runs with a uniform velocity.
(iii) BC:
(b) AB
The displacement does not change. The runner Gradient = (0 – 25.0)
(20.0 – 15.0)
is not moving.
= –5.0 m s–2
(c) BC
The gradient of the graph is increasing. The The motorcycle moves with a deceleration

runner runs with increasing velocity, or the with a magnitude of 5.0 m s–2 until it stops
runner accelerates.
at the time t = 20.0 s.
(d) CD
The gradient of the graph is negative and (iv) CD

constant. The runner is running with a uniform The motorcycle moves with a negative
velocity but in the opposite direction. The
runner goes back to his original position. uniform velocity. This means the

motorcycle is now moving in the opposite

direction, on its way back.

30

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Physics  Form 4  Chapter 2 Force and Motion I  

Gradient = (–10.0 – 0) SPM Highlights
(25.0 – 20.0)
The v-t graph in Figure 2.32 shows the situation of
= –20.0 m s–2 a skydiver when jumping from a plane until he lands
on the ground. Analyse the v-t graph and explain the
Note: In this case, even though the velocity motion of the skydiver when
of the motorcycle has an increasing (a) the parachute is not open,
magnitude, its acceleration is negative. (b) the parachute is open,
This is because the negative indicates (c) the skydiver lands,
the motorcycle moves in the opposite
direction. v / m s–1

50

(b) (i) The displacement of the motorcycle after 40 Chapter

20.0 s = area under the graph for t = 0 s to 30 Parachute Parachute
not opened opens
t = 20.0 s
1
s =  2 × 10.0 × 25.0 + (5.0 × 25.0) + 20

   1 × 5.0 × 25.0 10 Landing 2
2
0 t /s
= 312.5 m
5 10 15 20 25 30 35 40

Figure 2.32

t=0s t = 20.0 s Answer:
s = 312.5 m
t = 25.0 s (a) When the parachute is not open, the skydiver
falls with increasing velocity until it reaches
(ii) For t = 20.0 s to t = 25.0 s, the motorcycle a maximum velocity which is known as the
terminal velocity.
moves in the opposite direction as shown
(b) When the parachute is open, the skydiver fall
in the diagram. The distance travelled by with a uniform deceleration until it reaches
minimum velocity upon which he continues to
the motorcycle for t = 20.0 s to 25.0 s fall with this velocity.

s = 1 (5.0 × 10.0) (c) When the skydiver lands, stop falling and his
2 velocity descreases until 0 m s–1.

= 25.0 m

Therefore, the displacement of the

motorcycle after 25.0 s

s = 312.5 – 25.0 Checkpoint 2.2

= 287.5 m Q1 The s-t graph in Figure 2.33 shows the movement
of a crate that is being hoisted vertically by a
(c) a / m s–2 crane. s is the displacement of the crate from the

3.0 ground.
2.0
1.0 s/m

0 t /s 2.0
–1.0 5.0 10.0 15.0 20.0 25.0
–2.0
–3.0 1.5
–4.0
–5.0 1.0
s 0.5

0 1234 5 t/s

SPM Tips Figure 2.33

Ma a=kev s–tuuretoyocualccaunlauteseacthceelearcacteiolenrafrtoiomn formula, (a) Briefly describe the movement of the crate
• ticker tape, from t = 0 to 5 s.
• ticker chart, and
• velocity-time graph. (b) Calculate the velocity of the crate from t = 2.4
to 3.8 s.
(c) What is the average velocity of the crate

from t = 0 to 5 s?

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  Physics  Form 4  Chapter 2 Force and Motion I

Q2 A bus travelled along a straight road and stopped Q4 A lift travelled up from the ground floor to the fifth
at ‘Bus Stop 1’ to pick up some passengers. It floor. It stopped for a while before moving down

then continued its journey until it stopped at ‘Bus and then finally stopped at the first floor.
Stop 2’. The v-t graph in Figure 2.34 shows its
motion.

Bus stop 1 Bus stop 2

v / m s–1

20

Chapter 15 Velocity

10 4
2
25 t/s 0 20 Time
0 10 20 30 40 50 60 –2
–4 5 10 15

Figure 2.34 Figure 2.36

(a) How far did the bus travel before it stopped (a) (i) When did the lift stop at the fifth floor?
at ‘Bus Stop 1’? (ii) Explain your answer in (a)(i).
(b) (i) For which period of time did the lift travel
(b) For how long did the bus stop at ‘Bus
Stop 1’? down?
(ii) Explain your answer in (b)(i).
(c) Calculate the deceleration of the bus before (c) Calculate the total distance travelled by the
it stopped at ‘Bus Stop 2’.
lift when it finally stopped at the first floor.
(d) Find the distance between ‘Bus Stop 1’ and (d) Calculate the displacement of the lift at t = 20 s.
‘Bus stop 2’.

Q3 The s-t graph in Figure 2.35 shows the
displacement, s, of a boy from his classroom
along a corridor.

s/m

35 Physics of 100 m World Record
The v-t graph of Usain Bolt when breaking the
30
100 m record in Berlin in 2009
25 v / m s–1

20 12

15 10

10

5 8

0 t/s 6
5 10 15 20 25 30 35 40 45 50 55

Figure 2.35 4

2

(a) (i) What is the total distance travelled by the 0 t/s
boy? 2 4 6 8 10

(ii) Calculate the average speed of the boy. Q5 Without doing the actual calculation, what is the
(b) Halfway along the corridor, he stopped to area under the v–t graph shown above?

pick up some books that he had dropped
accidentally on the floor. When was the time,
t, he picked up the books?
(c) Sketch the v-t graph from t = 0 to 55 s.

32

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Physics  Form 4  Chapter 2 Force and Motion I  

2.3 Free Fall Motion

Motion of object falling with and without air resistance

Valve SPM Tips
Rubber stopper
To show the motion of free fall without vacuum, two objects
Air of the same size but different mass can be used.

Glass tube 2
Feather

Coin Feather Chapter

Rubber Coin
stopper

Figure 2.37

1. At the beginning, the glass tube consists of air inside it. When the tube is inverted, the coin
reaches the base faster than the feather even though both started to fall at the same time.

2. Next, the tube is connected to a vacuum pump to draw Connect to Coin
out the air from it. When all the air has been pumped vacuum pump Feather
out, a vacuum is resulted in the tube. Vacuum
Air being sucked
3. When the tube is inverted, both the coin and the feather out of the tube
fall to the base at the same time.

• Discussion: In this case, both the objects fall without
air resistance.

4. An object falling under the pull of the gravitational
force without any other forces acting on it is said to
be undergoing free fall.

5. An object undergoing free fall will accelerate. This
acceleration is known as gravitational acceleration.
The value of the gravitational acceleration does not
depend on the shape and mass of the object.

Figure 2.38

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  Physics  Form 4  Chapter 2 Force and Motion I

ACTIVITY 2.2

Aim: To determine the value of gravitational Diameter = d cm Steel ball
acceleration Photogate A
Time passing through = t1 s
Apparatus and Materials: photogate A Photogate B
Photogate kit and electronic timer, steel ball,
Chapter micrometer screw gauge Time passing through t2
photogate B
Instructions: = s

1. The arrangement of the apparatus is
shown in Figure 2.39.

2. A pair of photogates is arranged at
different heights.

3. A steel ball is dropped and falls through

2 both the photogates, each equipped with
optical sensors that can measure the
time, t, of an object falling through it.

4. The diameter of the steel ball, d, is
measured with a micrometer screw
gauge.

Results: Electronic timer

Diameter of the steel ball = d cm Figure 2.39

Time for the steel ball to pass through

photogate A = t1 s
Time for the steel ball to pass through photogate B = t2 s
Time for the steel ball to travel from photogate A to photogate B = t s

Calculations: d
t1
Average velocity of steel ball in passing photogate A, u= cm s–1.

Average velocity of steel ball to passing photogate B, v = d cm s–1.
t2
Time taken for the steel ball to accelerate from u to v = t s
v – u,
Using the formula a = t the experimental value of gravitational acceleration is,

g = v – u = d – d cm s–2
t t2 t t1

This value can be divided by 100 to convert to m s–2 for easier comparison with standard value.

Discussion:
The scientific estimation for the value of g at the equator is around 9.78 m s–2 and at the poles is
about 9.83 m s–2.
Usually, the experimental value is lower than the scientific value because of the existence of air
resistance as the steel ball falls.

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EXAMPLE 2.12 Physics  Form 4  Chapter 2 Force and Motion I  

A student throws a stone Checkpoint 2.3 Chapter
vertically upwards.
The stone moves to a Q1 What is the time taken for a stone to undergo
height of 5 m before free fall of 100 m? (Assume the gravitational
stopping and then falls acceleration, g = 10 m s–2)
downwards. Figure 2.40
shows the path moved Q2 Figure 2.41(a) shows an astronaut standing on
by the stone from A to B the surface of the Moon. He releases a piece of
and C. paper and a rubber ball from the same height at
(a) What is the the same time. Figure 2.41(b) is a stroboscopic
photograph that shows how the piece of paper
magnitude of the and the rubber ball undergo free fall and reach
velocity of the stone the ground at the same time. HOTS
when it leaves the
hand of the student? Ball
(b) How long is the time Paper
taken for the stone to
travel from A to C? (a) (b) 2
(Take g = 10 m s–2) Figure 2.41

Figure 2.40 (a) What is meant by free fall ?
(b) What inference can you make based on the
Solution
Height, s = 5 m. observations?
(c) When the activity is repeated on the Earth,
(a) The stone moves in the opposite direction from
the gravitational force. Hence, the value of the which of the two objects will reach the
gravitational acceleration is g = –10 m s–2. At ground first? Explain your answer.
B, the stone come to a rest. Hence, v = 0 m s–1.
Q3 Figure 2.42 shows a ball is thrown vertically
upwards from the position A with a velocity of
24 m s–1. The ball moves to position B, and then
falls downwards.

If its speed when in A is vA , therefore by using

the formula, B

v2 = u2 + 2as

0 = v2A + 2 (–10)(5)

v2A = 100

Therefore, vA = 10 m s–1

(b) The displacement of the stone from A at position 24 m s–1

C is 0. By using the formula, A
s = ut + 12at2
Figure 2.42
(a) What is the distance between A and B?
0 = 10t + 21(–10)t2 (b) What is the time taken for the ball to move
t) = 0
5t(2 – from A to B?
(c) State the assumption made when doing the
t = 2 s
calculations.
Hence the total time taken by the stone to travel (Take g = 10 m s–2)
from A until C is 2 seconds.

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  Physics  Form 4  Chapter 2 Force and Motion I

2.4 Inertia Original State of Object that is in Motion
Situation 1
1. Inertia is a natural tendency of an object
to oppose any changes to its original state,
whether in the state of rest or state of motion.

Original State of Object Inertia
that is at Rest
VIDEO
Situation I

Chapter Figure 2.45

2 Cardboard Pull 1. When the car is moving, the passengers in the
Glass car move together.

Figure 2.43 2. When the car stops suddenly, the passengers
are thrown forward.
1. Figure 2.43 shows a 50 cent coin resting on a
cardboard that is placed on top of a glass. 3. The inertia of the passengers resists change
of state of motion. As a result, the passengers
2. When the cardboard is pulled quickly, the coin continue to move forward.
hovers over the top of the glass for an instant
before dropping into the glass. Situation II

3. The inertia of the coin resists change of its state 1. Figure 2.46 shows how tomato sauce can be
of rest and hence does not move together with forced out of its bottle. When the bottle is
the cardboard. Gravitational force pulls it down moved quickly towards the food, the sauce in
into the glass. the bottle moves together.

Situation II
1. A book placed in the middle of a stack of books

is pulled out horizontally with a quick jerk.



Figure 2.46

Pull 2. When the bottle is stopped suddenly, the sauce
will be forced out of the bottle into the food.
Figure 2.44
3. The inertia of the sauce resists change of its
2. Initially, the books above it tend to stay at rest. state of motion and continues to move out of
the bottle.
3. The inertia of these books resists change of
state of rest and do not move together with Relationship Between Mass and Inertia
the book that is being pulled out. The books
then fall vertically downwards due to the 1. The mass of an object is the amount or
gravitational force. quantity of matter in it.

2. The mass of an object depends on the number
and type of atoms of the object.

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Physics  Form 4  Chapter 2 Force and Motion I  

EEkxsppeerirmimeennt 2.1

Problem statement: What is the relationship between mass and inertia?
Aim: To investigate the relationship between inertia and mass.

Inference: The inertia of an object is influenced by its mass.

Hypothesis: The inertia of a body increases when its mass increases.

Variables:

(a) Manipulated variable : Mass

(b) Responding variable : Inertia

(c) Fixed variable : Type of hacksaw blade Chapter

Operational Definition: 2
Period of oscillation is an indicator for inertia, the responding variable. The bigger the period of
oscillation, the bigger the inertia.

Apparatus and material: Hacksaw blade, G-clamp, stopwatch, and plasticine

Procedure:

1. A hacksaw blade is clamped with a G-clamp to the leg of a table as shown in Figure 2.47.

2. A lump of plasticine with a mass of 30 g is attached to the G-clamp
free end of the hacksaw blade.

3. The hacksaw blade is displaced sideways slightly and
released so that it would oscillate horizontally.

4. The time taken for 10 complete oscillations, t1, is determined Plasticine
using a stopwatch and recorded. This step is repeated for
another reading t2. The readings of t1 and t2 are recorded in
Table 2.4.

5. Steps 3 and 4 are repeated with mass of plasticine, m = 40 Table leg Hacksaw
g, 50 g, 60 g and 70 g. blade

6. A graph of period against mass is drawn. Figure 2.47

Results:

1. Tabulation of results. Table 2.4

Mass of plasticine, Time taken for 10 oscillations / s Period, T = t /s
m/g t1 t2 Average, t 10

30 4.1 4.1 4.1 0.41

40 4.8 4.8 4.8 0.48

50 5.5 5.4 5.5 0.55

60 6.0 6.1 6.1 0.61

70 6.5 6.5 6.5 0.65

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  Physics  Form 4  Chapter 2 Force and Motion I

2. Graph of period, T against mass, m.

T/s

0.70

0.60

0.50

Chapter 0.40

2 0.30 0 10 20 30 40 50 60 70 m/g

Figure 2.48

Discussion:

1. From the graph of period against mass, it is observed that the bigger the mass of the load, the
bigger the period of oscillation of the hacksaw.

2. This means that the bigger the mass of an object, the bigger is its inertia. Hence, the hypothesis
is accepted.

Conclusion:
When the mass of an object increases, it is more difficult for the object to change its state of rest or
motion. This means the inertia of an object increases with the increase of its mass.

Precaution:
The period of a single full oscillation is a very small value and is difficult to measure accurately. Instead,
the time for 10 full oscillations is taken and upon dividing by 10, the average value for a single full
oscillation is obtained. This method of measurement will ensure greater accuracy. The measurement
of time is repeated and the average value is taken to reduce random errors.

Newton’s First Law of Motion 4. An object moving at uniform velocity will
continue to move with uniform velocity in a
1. The characteristic of inertia of an object can straight line. An external force will be required
be stated in the form of Newton’s First Law to change this state of motion.
of Motion.
5. Newton’s laws of motion were discovered by
2. Newton’s First Law of Motion states that Sir Isaac Newton (1642—1727). He wrote
an object will remain at its original state, a book ‘Philosophiae Naturalis Principia
whether it is at rest or in motion with uniform Matematica’ that became the foundation for
velocity, if there is no net force acting on it. the field of mechanics. This book also explains
many important laws, including Newton’s Law
3. An object at rest will remain at rest. An of Motion I, II and III.
external force is needed to move it.

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02 FOC PHYSICS F4 3P.indd 38

PPhhyyssicicss  FFoorrmm 4  Chapter 2 Force and Motion I  

Branch of tree Rear-end collision

A stick is used to jerk the branch When a rear-end Headrest
of a guava tree. At the end of collision occurs, the
the branch, there is a big guava. car and the body of the
The guava tends to remain in its driver move forward
original state of rest due to its suddenly. The headrest
inertia. This will cause its stalk to supports the head of
snap and the guava will fall to the the driver when it is
ground. thrown backwards.

Chapter

Guava 2

Tissue paper Effects of Steel structure of lorry
Inertia
When a sheet of tissue A steel structure is fitted in the
paper is pulled quickly space between the driver and the
from a tissue box, the load of a timber lorry. This steel
box will not move. The structure prevents the logs from
inertia of the box causes moving forward and knocking
it to resist motion and against the driver compartment
remain at rest. when the lorry stops suddenly.

Steel structure

Drying hair 39

A girl dries her hair by turning her head 29/01/2020 1:39 PM
fast. Due to inertia, the water from the
hair continues to move when the hair has
stopped moving and hence is removed from
the hair.

02 FOC PHYSICS F4 3P.indd 39

  Physics  Form 4  Chapter 2 Force and Motion I Checkpoint 2.4

SPM Tips Q1 Figure 2.53 shows two similar tins P and Q each
tied to a long rope.
To remember Newton’s First Law of Motion
when F = 0 (No resultant force) Rope

At rest – Always stays
at rest.

Sand

PQ

Chapter Figure 2.49 Push

In motion – Figure 2.53
Always moves
• at a constant speed, When both the tins are pushed with the same
force,
2 • in a straight line. (a) which tin is easier to be pushed?
(b) which tin is easier to be stopped?
(c) Explain your answer in (a) and (b).

Figure 2.50 Q2 Explain why
(a) a loaded lorry takes a longer time to stop.
SPM Highlights
(b) a train requires a long stopping distance.
Figure 2.51 shows an
(c) an oil drill platform is made of steel tubes
elephant and a tiger. that attach huge floats to the seabed.

Why is the inertia of the

elephant bigger than Q3 Figure 2.54 shows a moving toy car with a pencil
box placed on top of it. It is moving towards a
that of the tiger? Figure 2.51
A The size of the stationary obstacle.

elephant is bigger than the tiger.

B The mass of the elephant is bigger than the tiger. Toy car

C The elephant is taller than the tiger. Pencil box

D The elephant is longer than the tiger. Nail

Answer: B

Obstacle

SPM Highlights Figure 2.54

Figure 2.52 shows 2 (a) State what will happen when the car hits the
packets of the same size obstacle.
in space. One packet has
a small mass and another (b) Explain your answer in (a)
big mass. How can the
astronaut differentiate (c) Compare with the effect observed if the car
between the two packets is moving at a higher velocity.
when both are under the
condition without gravity? Figure 2.52

Answer:

Inertia depends on mass and not weight. Hence,
the packet with bigger mass is more difficult to be
moved from its initial state of rest.

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02 FOC PHYSICS F4 3P.indd 40

Physics  Form 4  Chapter 2 Force and Motion I  

2.5 Momentum Solution

Figure 2.55 (a) Momentum of the car = 800 × 20

1. Every moving object has momentum. However, = 16 000 kg m s–1
the momentum of a timber lorry is very much
bigger than that of a cyclist, despite both Momentum of the lorry = 5 000 × 20
vehicles moving at the same speed.
= 100 000 kg m s–1
2. The linear momentum of an object is defined
as the product of its mass and velocity (b) The lorry, due to its bigger momentum.

Momentum = Mass × Velocity Principle of Conservation of Chapter
Momentum
3. The SI unit for momentum is kg m s–1 or N s. 2
Momentum is a vector quantity. The direction 1. According to the principle of conservation
of momentum is the same as the direction of of momentum, when two or more bodies act
velocity. on each other, their total momentum remains
constant, provided that there is no external
EXAMPLE 2.13 force acting on them.

The total mass of a lorry is 20 000 kg and the total SPM Tips
mass of a car is 2000 kg.
If both the lorry and the car are travelling at a Elastic Collision vs Inelastic Collision
velocity of 25 m s–1, calculate the momentum of Q: Elastic and inelastic collisions are two very common
the lorry and the car respectively.
Solution types of collision. How is momentum related to
Momentum of the lorry = 20 000 × 25 these two types of collision and what are the
= 5 × 105 kg m s–1 differences between both types of collision?
Momentum of the car = 2 000 × 25
= 5 × 104 kg m s–1 (a) Collision between electric bumper cars in an elastic collision

EXAMPLE 2.14 (b) Collision between two cars is an inelastic collision
Figure 2.56
A car has a mass of 800 kg and a lorry has a mass
of 5000 kg. A: • In an elastic collision, the total kinetic energy of
(a) Compare the momentum of both vehicles when the system is conserved. The objects elastically
spring back to their original shapes and the
moving at a velocity of 20 m s–1. system retains all its original kinetic energy.
(b) If both vehicles move at the same velocity,
• In an inelastic collision, the total kinetic energy
which vehicle will cause greater destruction is not conserved . One or more of the colliding
when hitting a lamp post? objects may not spring back to its original shape
or heat or sound may be generated. The total

kinetic energy of the system after the collision
is less than that before the collision.

41

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  Physics  Form 4  Chapter 2 Force and Motion I 2. Types of collision
There are two types of collision, that is, elastic
Collision
1. The principle of conservation of momentum collision and inelastic collision.
(a) In an elastic collision
can be studied using the collision of objects.
(i) momentum is conserved,
2.0 kg 3.0 m s–1 1.0 kg At rest v m s–1 (ii) kinetic energy is conserved,
A B B (iii) the objects move separately after
A
collision ,
(a) (iv) total energy is conserved.
(b) In an inelastic collision.
2.0 kg 3.0 m s–1 1.0 kg 2.0 m s–1 v m s–1 (i) momentum is conserved,
A BA B (ii) kinetic energy is not conserved,
(iii) the objects may combine and move
Chapter (b)
Figure 2.57 together after collision,
(iv) total energy is conserved.
For case (a)
SPM Tips
2 bTeoftoalremcoomlliesniotunm = Total momentum
after collision In the explanation of a law or principle, the conditions
and limitations must be mentioned together. In the case
(2.0 × 3.0) + (1.0 × 0) = (2.0 + 1.0)v of stating the principle of conservation of momentum,
the condition of no external force acting on the system
6.0 kg m s–1 = 3.0 v needs to be mentioned.

Therefore, v = 2.0 m s–1

For case (b)

(2.0 × 3.0) + (1.0 × (–2.0)) = (2.0 + 1.0)v
4.0 kg m s–1 = 3.0 v

Therefore, v = 4.0
3.0

= 1.3 m s–1

EEkxsppeerirmimeennt 2.21

Problem statement: Is the total momentum constant in a closed system?

Aim: To show the total momentum in a closed system is constant for the case of collision.

Hypothesis: The total momentum before collision is the same as that of after collision if there is no
external force acting on the system.

Variables:

• Manipulated variable : Total momentum before collision

• Responding variable : Total momentum after collision

• Fixed variable : Friction compensated inclined plane

Apparatus and materials: Ticker timer, 12 V a.c. power supply, triple beam balance, 4 trolleys, inclined
plane, ticker tape, cellophane tape, and plasticine.

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02 FOC PHYSICS F4 3P.indd 42

I Non-elastic collision Ticker Trolley B Physics  Form 4  Chapter 2 Force and Motion I  
Procedure: tape Trolley A Friction compensated
inclined plane
Ticker Timer Plasticine

12 V a.c. power Support
supply

Figure 2.58

1. The apparatus is arranged as shown in Figure 2.58. Chapter

2. The inclined plane is adjusted to be friction compensated. When this is achieved, the trolley given 2
a slight push will move down the inclined plane with a uniform velocity which can be confirmed
by analysing the ticker tape attached to the trolley.

3. Trolley A and B are weighed to determine their mass. The values are recorded.

4. The ticker timer is switched on and trolley A is given a slight push to move towards the stationary
trolley B.

5. Trolley A then collides with trolley B and both stick together and move down the inclined plane.

6. The ticker tape is analysed to determine the velocity before and after collision, u and v respectively.
These values are recorded.

7. 2 other trolleys are weighed and the experiment is repeated with
(a) 1 trolley colliding with 2 stationary trolleys,
(b) 2 trolleys colliding with 1 stationary trolley, and
(c) 3 trolleys colliding with 1 stationary trolley,

Results: Table 2.5

Before collision After collision

Mass of Velocity, Total momentum, Jisim troli Velocity, Mass of trolley
trolley, u / m s–1 m1u / kg m s–1 (m1 + m2) u / m s–1
m1 / kg momentum,
/ kg (m1 + m2) v / kg m s–1
1
1+1=2
1
1+2=3
2
2+1=3
2
3+1=4

Conclusion:
The total momentum before collision is the same as that of after collision. The hypothesis is accepted.

II Elastic collision

Procedure:

Ticker timer Ticker Springed

tape Trolley A dowel rod Trolley B

Friction compensated
inclined plane

12 V a.c. power Support
supply

Figure 2.59

43

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  Physics  Form 4  Chapter 2 Force and Motion I

1. The procedure is modified with both the trolleys A and B each pulling a ticker tape separately.
This is required as the trolleys are moving separately after collision.

2. The resultant tape is as shown below.

Direction of motion

After collision During Before collision
Velocity = v collision Velocity = u

Chapter Figure 2.60

3. The experiment is repeated with
(a) 1 trolley colliding with 2 stationary trolleys,
(b) 2 trolleys colliding with 1 stationary trolley, and
(c) 3 trolley colliding with 1 stationary trolley.

2 Results: Table 2.6

Before collision After collision

m1 m2 u Total momentum, v1 v2 Total momentum
m1u m1v1 + m2v2
11
12
21
31

Conclusion:
The total momentum before collision is the same as that of after collision. The hypothesis is accepted.

EXAMPLE 2.15

Figure 2.62 shows two rocks moving towards each Let v be the velocity of the rocks after collision.
other along a straight line. After colliding with each Total momentum before collision
other, the two rocks lump together. = (2 000 × 750) + [500 × (–2 000)]
= 500 000 kg m s–1
750 m s–1 2 000 m s–1
Total momentum after collision
A = (2 000 + 500) × v
B = 2 500v

2 000 kg 500 kg According to the principle of conservation of
momentum, the total momentum before and after
Figure 2.61 the collision are the same.

Calculate the speed of the rocks after the collision. Therefore,
2 500v = 500 000
Solution vB = –2 000 m s–1 v = 200 m s–1
mB = 500 kg
vA = 750 m s–1
mA = 2 000 kg

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Physics  Form 4  Chapter 2 Force and Motion I  

EXAMPLE 2.16 Solution

A lorry with a mass of 8 000 kg moves with a Total momentum before coupling
velocity of 30 m s–1 . The lorry then collides with = (2.5 × 104 × 10) + (2.5 × 104 × 0.8)
a car with a mass of 1500 kg and travelling with = 4.5 × 104 kg m s–1
a velocity of 20 m s–1. After the collision, the two
vehicles stick and move together with a velocity v. Total momentum after coupling
Determine the value of v. = 5.0 × 104 × v

According to the principle of conservation of

Solution momentum,

30 m s–1 20 m s–1 Tbeoftoalremcoomupelnitnugm = Total momentum
after coupling

5.0 × 104 × v = 4.5 × 104 kg m s–1 Chapter

v = 0.9 m s–1

Momentum of lorry before collision Assumption: The sum of the external forces acting 2
= 8 000 × 30 on the system consisting of the two boxcars is zero.

= 240 000 kg m s–1

Momentum of car before collision Note:
= 1 500 × 20
1. For experiments in this chapter, a friction
= 30 000 kg m s–1 compensated inclined plane is used.

Tobteaflomreocmoellnistuiomn = Total momentum Friction
after collision
Component of
240 000 + 30 000 = (8 000+1 500)v weight of the
trolley
270 000 kg m s–1 = 9 500v

Therefore, v = 270 000 Weight component cancels the effect of friction
9 500
Figure 2.63
= 28 m s–1
2. A friction compensated inclined plane has an
EXAMPLE 2.17 inclination in such a way that is the friction on
the surface of the inclined plane is balanced
Figure 2.64 shows two identical boxcars of a train out by the component of the weight along the
trolley.
being coupled at a railway station. Each boxcar has
a mass of 2.5 × 104 kg. The velocity of the boxcar 3. The two forces cancel out each other and hence
1 and boxcar 2 before the coupling are 1.0 m s–1 it is as if the inclined plane is frictionless.
and 0.8 m s–1 respectively.
4. When this condition is achieved, the trolley
1.0 m s–1 0.8 m s–1 will move with a uniform velocity when given
a small push.
12
Before Explosion
v
1. When a rocket is launched, the exhaust gas is
12 forced out from the back of the rocket and the
rocket moves upwards. Initially the rocket is
After stationary at the launching pad. The launching
Figure 2.62 of a rocket is an example of explosion.

Find the velocity, v, of the boxcars after they are 2. In explosion, the total momentum is conserved,
coupled. State one assumption in your calculation. that is the total momentum before and after the
explosion is zero.

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Chapter   Physics  Form 4  Chapter 2 Force and Motion I

EEkxsppeerirmimeennt 2.31

Problem statement: In the case of explosion like balloon that is held and then released, the total
initial momentum is zero. Is the total momentum after the balloon is released
also zero?

Aim: To show that the total momentum in a closed system is constant in the case of explosion

Hypothesis: The total momentum of a system before and after explosion are the same.

Variables:
• manipulated variable : Momentum before explosion
• responding variable : Momentum after explosion

Apparatus and materials: Triple beam balance, 4 trolleys, 2 wooden blocks, a mallet, and a metre

2 rule.

Procedure:

Wooden block Mallet Wooden block
d1 Plunger
AB d2

Trolley
Figure 2.64

1. The arrangement of the apparatus is shown in Figure 2.64.

2. Trolleys A and B are arranged in contact with each other with two wooden blocks at two ends.

3. When the plunger is knocked by the mallet, the dowel rod is released causing the trolleys to
move in opposite directions.

4. The position of the wooden blocks are adjusted so that the trolleys hit the respective blocks at
the same time.

5. The distance travelled by the trolleys, d1 and d2 are measured and recorded.

6. 2 other trolleys are weighed and the experiment is repeated for
(a) 1 trolley in contact with 2 trolleys, and
(b) 1 trolley in contact with 3 trolleys.

Results:

1. Since the trolleys hit the wooden block at the same time, the magnitude of the velocity is directly
proportional to the distance travelled.

2. Since the motion of the trolleys are in the opposite direction, the direction towards the right can
be assumed to be positive. With that, the following table will show the results of the experiment.

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Physics  Form 4  Chapter 2 Force and Motion I  

Table 2.7

Before After explosion
explosion
v1 =  –d1  v2 = d2 m1 m2 Total left Total right Total
Total t t momentum
momentum momentum, momentum,
m1v1 m2v2

Note: The value of t need not be measured. Chapter
Conclusion:
The total momentum of a system before and after explosion are the same. The hypothesis is accepted. 2

EXAMPLE 2.18 SPM Tips

An astronaut with a mass of 80 kg throws a box of To answer questions related to the principle of conservation
instrument with a mass of 40 kg in order to return to of momentum:
the space capsule. If the box moves with a velocity • Sketch a diagram to show the motion of the objects
of 6 m s–1 , what is the velocity of the astronaut after
throwing the box? involved
• Label the mass and velocity (magnitude and direction)
Solution
of each object in the diagram.
Let v be the velocity of the astronaut. • Use the formula (Total initial momentum = Total final

The total momentum before the box is thrown is momentum).
zero because both the astronaut and the box are
stationary. EXAMPLE 2.19

Total momentum after the box is thrown Figure 2.65 shows 2 trolleys A and B with a mass
= momentum of the box + momentum of the of 1.5 kg and 1.0 kg respectively. A compression
astronaut spring is placed between the 2 trolleys and is
= (40 × 6) + (80 × v) compressed by forcing the 2 trolleys close to
= 240 + 80v each other. When the 2 trolleys are released, each
moves away from each other. What is the distance
According to the principle of conservation of travelled by trolley B when trolley A has moved
momentum, 40 cm?
total momentum before the box is thrown
= total momentum after the box is thrown AB

Therefore, Figure 2.65
240 + 80v = 0
v = –3.0 m s–1
(The negative sign shows that the astronaut is
moving in the opposite direction to the box.)

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  Physics  Form 4  Chapter 2 Force and Motion I big momentum. This causes a big momentum
with the same magnitude that moves the plane
Solution forward.
Let the time taken by trolley A to move 40 cm
be t seconds, and at the same time trolley B has 4. Satellites and space capsules use air jets to
moved d cm. By taking the direction travelled by control movement and position. This is based
A as positive, on the principle of conservation of momentum
1.5 × 4t0 + (1.0 × –td = 0 for the case of explosion.
(1.5 × 40) – d = 0
d = 60 cm

Chapter Applications of the Principle of
Conservation of Momentum in Daily
Life Figure 2.67 Water sprinkler system

2 1. In the field of sports and games like billiard, 5. A water sprinkler system uses this principle
to irrigate plants in a farm. The whole area can
bowling, and carrom, the principle fo be irrigated by the rotating sprinkler.
conservation of momentum is involved in
collision form whereas in shooting, the 6. Some animals like the octopus uses this
principle is involved in explosion. principle to move by squirting out a liquid
from its body.
2. In the launching of rocket, the combustion
of fuel and oxygen produces a big downward 7. This principle is also used to explain the
momentum and the rocket moves upwards following situation:
with the same magnitude of momentum. The (a) When a bullet comes out of a gun with
momentum is conserved and is equal to zero. a high velocity, the gun recoils in the
opposite direction.
Jet engine
(b) A few firemen are required to hold the
Figure 2.66 Jet engine in a fighter jet hose in place when it is jetting a huge
volume of water with a high velocity. This
3. The jet engine of an airplane uses this principle is because a big backward momentum is
to produce a forward thrust. Air is sucked into produced.
the engine, compressed and heated. Fuel is
then injected into the mixture and combustion (c) In swampy areas, huge fans are used to
occurs. The burnt gas rushing out of the propel the boats forward.
exhaust has a very high velocity and hence a

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Physics  Form 4  Chapter 2 Force and Motion I  

Checkpoint 2.5

Q1 Figure 2.68 shows two bumper cars moving in a straight line. Bumper car A 200 kg
with a total mass of 100 kg is moving at a velocity of 2 m s–1. Bumper car B 5 m s–2
with a total mass of 200 kg is moving at a velocity of 5 m s–1. After a while,
the two cars collide. After the collision, bumper car A moves with a velocity 2 m s–2 100 kg
of 6 m s–1.
(a) What is the total momentum of the bumper cars before the collision? B
(b) What is the total momentum of the bumper cars after the collision?
(c) State the principle used for your answer in (b). A
(d) Calculate the velocity of bumper car B after the collision.

Q2 A 1500 kg van travelling at a speed of 15 m s–1 collides with a 1000 kg car Figure 2.68
20 m s–1
travelling in the opposite direction with a speed of 20 m s–1. After the 15 m s–1 1 000 kg Chapter
collision, both vehicles get stuck and move together.
Figure 2.69
(a) What is the total momentum of the van and the car before the 1 500 kg

collision?

(b) What is the total momentum of the van and the car after the

collision? 2

(c) State the principle used in your answer in (b).

(d) Calculate the velocity of the van and the car after the collision.

Q3 A bullet of mass 12 g is fired at a speed of 360 m s–1 from a rifle of mass 6 kg.

Figure 2.70

(a) What is the recoil velocity of the rifle?

(b) The bullet travels towards a block of wood resting on a frictionless surface. The wooden block moves with

a velocity of 12 m s–1 after the bullet is embedded in it. Calculate the mass of the wooden block.

Q4 Figure 2.71 shows two situations. Explain the observations based on the figure.

(a) (b)
Figure 2.71

(a) When water rushes out of the hose with a very high speed and volume, the fireman holding the hose needs
the support of another fireman so that he does not fall backwards.

(b) A pair of identical twins are doing ice skating. Initially, they hold each other’s hands. When they push each
other and release their hands, both of them move in opposite directions with the same speed.

2.6 Force

1. Generally, a force is the push or pull on an object as result of interaction with other objects.
2. When two objects interact, a force will act on each other.

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  Physics  Form 4  Chapter 2 Force and Motion I

A man uses a Effects A cyclist uses a
force to move of a bigger force to pedal
a stationary force his bicycle and
lawnmower. increase its speed.
Chapter
A footballer A baker uses
uses a force a force on his
to change the dough to shape it
direction of into a curry puff.
motion of a ball.

2 Figure 2.72

Relationship between Force, Mass and Acceleration

EEkxsppeerirmimeennt 2.41

Problem statement: An object accelerates when acted upon by a force. What will happen to its
acceleration if its mass increases but the force is constant?

Aim: To find the relationship between acceleration and mass when the force is constant.

Hypothesis: If a force acting on an object is constant, the bigger the mass of the object, the smaller
is its acceleration.

Variables:

(a) manipulated variable : Mass

(b) responding variable : Acceleration

(c) fixed variable : Force

Apparatus and materials: Ticker timer, 12 V a.c. power supply, 3 trolleys, elastic string, inclined plane,
ticker tape.

Procedure:

Ticker timer Ticker tape

Elastic string Friction compensated
inclined plane

12V a.c. power Support Trolley
supply

Figure 2.73

1. The apparatus is arranged as shown in Figure 2.73

2. The inclined plane is adjusted for friction compensation.

3. A ticker tape is attached to the trolley and passes through the ticker timer.

4. The ticker timer is switched on and the trolley is pulled down the slope using the elastic string
that is stretched at a constant length in order to provide a constant force.

5. The tape is then analysed for the acceleration of the trolley. The value is recorded.

6. The experiment is repeated by pulling 2, then 3 trolleys stacked together.

50

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Physics  Form 4  Chapter 2 Force and Motion I  

Results: Table 2.8 2 3
1. Mass of trolley, m / unit
1

Initial velocity, u / cm s–1

Final velocity, v / cm s–1

Acceleration, a / cm s–2

2. A graph of aceleration a against 1 is plotted and the graph is a straight line
mass of trolley, m

passing through the origin. Therefore a ∝ 1 if F is constant. Acceleration, a / cm s–2 Chapter
m

Conclusion: 2
When the pulling force is constant, the bigger the mass of the trolley, the
smaller is its acceleration. The hypothesis is accepted.

0 3–1 –12 M–ass of 1trolley, m

Figure 2.74

EEkxsppeerirmimeennt 2.51

Problem statement: An object accelerates when acted upon by a force. What will happen to its
acceleration if its force increases but the mass is constant?

Aim: To find the relationship between acceleration and force when the mass is constant.

Hypothesis: If a mass of an object is constant, the bigger the force, the bigger is its acceleration.

Variables:

(a) manipulated variable : Force

(b) responding variable : Acceleration

(c) fixed variable : Mass

Apparatus and materials: Ticker timer, 12 V a.c. power supply, 2 trolleys, elastic string, inclined plane,
support, ticker tape, cellophane tape.

Procedure:

Ticker timer Elastic string

Ticker tape Friction
compensated
inclined plane

12V a.c. power Support Trolley
supply

Figure 2.75

1. The apparatus is arranged as shown in Figure 2.75.

2. The inclined plane is adjusted for friction compensation.

3. A system consisting of two trolleys stacked together is used. A ticker tape is attached to it and
passes through the ticker timer.

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  Physics  Form 4  Chapter 2 Force and Motion I

4. The ticker timer is switched on and the trolleys are pulled down the slope using the elastic string
that is stretched at a constant length in order to provide a constant force.

5. The tape is then analysed for the acceleration of the trolley. The value is recorded.

6. The experiment is repeated by using 2 then 3 elastic strings pulled to the same length.

Results: Table 2.9

1. Number of elastic string, 1 2 3
F / unit

Initial velocity, u / cm s–1

Chapter Final velocity, v / cm s–1

Acceleration, a / cm s–2

2 2. The graph plotted is a straight line passing through the Acceleration, a / cm s–2
origin, therefore a α F.

Conclusion: 3
If a mass of an object is constant, the bigger the force, the 2
bigger is its acceleration. The hypothesis is accepted.

1

0 Force, F / N
Figure 2.6

Newton’s Second Law of Motion SPM Tips

1. From the experiments, If a is directly proportional If a is inversely
to F, proportional to m,
(a) a∝ 1 when F is constant.
m a a

(b) a ∝ F, when m is constant

Combining both,

F = kma, where k = is a constant.

2. By defining 1 newton (N) as the unit force F m

that causes an object with a mass of 1 kg to a∝F a∝ 1
m
accelerate 1 m s-–1 and substituting into the

equation,

F = kma

where 1 = k (1)(1) 3. Newton’s Second Law of Motion describes the
relationship between acceleration of an object
Therefore, k = 1 and the force applied to it. Newton’s Second
Law of Motion states that when a net external
with this, force acts on an object, the acceleration of
the object is directly proportional to the net
force and has a magnitude that is inversely
proportional to its mass.

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Physics  Form 4  Chapter 2 Force and Motion I  

SPM Tips EXAMPLE 2.21

The definition of Newton’s second law of motion consists A box of mass of 5 kg moves with a uniform
of a few important facts: velocity when acted upon by a force of 10 N.
• The object will only accelerate if there is a net external Calculate,
(a) frictional force acting on the box, and
force acting on it. (b) acceleration of the box if the force is 20 N.
• The acceleration is directly proportional to the net
Solution
external force. (a)
• The direction of the acceleration is the same as the
Friction F = 10 N Chapter
direction of the external force.
• The magnitude of the acceleration of the object is Figure 2.78 2

inversely proportional to its mass. The force acting on the box in the horizontal
direction is the force F = 10 N and frictional
SPM Tips force. These forces are in equilibrium when the
box moves with uniform velocity. Hence, the
The meaning of F = ma is: magnitude of the frictional force = 10 N, in the
• The net force F, in the direction of the acceleration. opposite direction to F.
• The direction of the force and the direction of the
(b)
change of momentum are the same.
• The direction of the force F is the same as the direction a

of the acceleration, a.

EXAMPLE 2.20 Friction = 10 N F1 = 20 N

A 100 kg box is placed on a smooth floor and is
pulled by a force F N.

100 kg
F

Figure 2.77 Figure 2.79

(a) If F = 200 N, calculate the acceleration of the When F1 = 20 N, the net force acting in the
box. direction of the acceleration, a

(b) If the velocity of the box changes from rest to = F1 – friction
15.0 m s–1 in 10 seconds when acted upon by = (20 – 10) N
a force, what is the value of the force? = 10 N

Solution Using the formula F = ma

(a) Given, m = 100 kg, F = 200 N 10 = 5 × a

F = ma a = 2 m s–2

Therefore a = F = 200
m 100

= 2 m s–2 Checkpoint 2.6

(b) Acceleration, a = v – u
t
15.0 – 0 Q1 A box with a mass of 5.0 kg is pulled along a
= 10 smooth floor. If the force is 12.0 N, what is the

= 1.5 m s–2 acceleration of the box?

Force, F = ma

= 100 × 1.5

= 150 N

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  Physics  Form 4  Chapter 2 Force and Motion I

2.7 Impulse and Impulsive Force Hence, the = mv – mu
impulsive force t

1. From the equation = (1 000 × 0) – (1 000 × 25)
0.05
F = ma ............................. 
= –500 000 N
a nd a = v – u ............................. 
t
Note: The negative sign shows the the force is
Substitute  into , acting on the car.

F = m  v – u 
t

Chapter = mv – mu
t
Newton’s Third Law of Motion
Therefore, Ft = mv – mu
1. Newton’s Third Law of Motion states that
2. The product of force F and action time t is for every action, there is an equal but opposite
reaction. This means when an object exerts a
2 known as the impulse of a force. force on a second object, the second object
exerts a force of equal magnitude but in the
Impulse = Ft opposite direction on the first object.

3. Impulse is defined as change of momentum, 2. Examples of Newton’s Third Law of Motion.
that is, (a) A book placed on a table exerts a force on
the table. The table exerts a reaction force
Impulse,  Ft = mv – mu of the same magnitude on the book in the
opposite direction, that is, upwards.
= Change of momentum (b) When a person is walking, his sole pushes
the floor backwards. The floor reacts by
The SI unit for impulse is kg m s–1 or N s. pushing the sole forward with the same
Impulse is a vector quantity. magnitude. Hence, the person can move
forward.
4. Impulsive force is defined as the rate of (c) A boy pulls a tight drawer and he is
change of momentum resultant from an action pulled by that drawer. This is because
that occurs in a short period of time. An the force exerted by the boy pulling the
example is during a collision. drawer causes the drawer to exert a force
of the same magnitude but in the opposite
Impulsive force, F == mCvha–tngmeuo Tfimmeomentum direction to pull the boy towards it.



EXAMPLE 2.22 Figure 2.80

A car with a mass of 1000 kg moves with a
velocity of 25 m s–1. The car then hits a tree and
is stopped in 0.05 second. Calculate the impulsive
force acted on the car.

Solution

Mass of car = 1 000 kg
Initial velocity of car, u = 25 m s–1
Final velocity of car, v = 0 m s–1
Time taken to stop, t = 0.05 s

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Effects of Impulsive Force in Daily Physics  Form 4  Chapter 2 Force and Motion I  
Life
short time when it hits the floor. The resultant
impulsive force is big enough to crack the tile.

1. During collision, a car undergoes a big change

of momentum in a very short time. From

the equation F = change of momentum , the
time

impulsive force acting on the car is very big Figure 2.82
and can cause serious damage to the car and
danger to its passengers.

2. A bowl crashes when dropped on the floor. This The Effect of Time on the Chapter
is because the bowl undergoes a big change of Magnitude of the Impulsive Force
momentum in a very short time when it hits
the floor and the resultant impulsive force is 1. Consider the formula for impulsive force
very big.
F= mv – mu = Change of momentum 2
t Time

If the change of momentum is constant, the

magnitude of the impulsive force is inversely

proportional to the time of impact. That is,
1
F ∝ t .

Figure 2.81 2. Since F ∝ 1 , the shorter the time of impact,
t
3. The tile on the floor cracks after a hammer is the bigger the impulsive force. The longer the
accidentally dropped on it. The falling hammer
undergoes a big change of momentum in a very time of impact, the smaller the impulsive force.

Hence, by controlling the value of t, we can

control the value of the impulsive force, F.

SPM Highlights

Packaging is an important aspect in industry. The design and structure of packaging requires specific scientific expertise
in order to protect the manufactured products from damage. Figures 2.83 and 2.84 show two examples of packaging for
a computer and eggs respectively. Suggest suitable characteristics of the materials used in the packaging so that the
computer and eggs can be transported and stored safely.

Box
Computer

Polystyrene

Cardboard Eggs
            

Figure 2.83                        Figure 2.84

Answer:
Things that are easily broken are packed using polystyrene or other soft but strong materials. If by accident, these
products are dropped or knocked, the packaging materials can absorb the impulsive force. Their soft surfaces can
reduce the impulsive force by lengthening the time of impact or change of momentum.

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  Physics  Form 4  Chapter 2 Force and Motion I

Implusive force, F = mv – mu
t

Increasing the time of impact, t to reduce the Decreasing the time of impact, t to increase the
impulsive force, F impulsive force, F

Chapter Pestle
Mortar

Figure 2.85 Figure 2.88

1. When a high jumper falls on a mattress, the 1. Pestle and mortar are made of stone. The pestle

2 thick and soft mattress lengthens the time of is used to pound spices like chilli, onion and
impact. The impulsive force acting on the high garlic. The impact time between the pestle and

jumper is reduced and he is less likely to suffer the mortar is very short because of their hard

any injury. and rigid surfaces. The large impulsive force

causes the spices to be crushed easily.

Figure 2.86

2. A long jumper lands on a pit filled with loose Figure 2.89
sand. The loose sand lengthens the time of
impact and reduces the impulsive force acting 2. A blacksmith uses a heavy hammer and anvil to
on him. forge metals. An anvil, which is made of steel,
has a very hard surface. When the hammer
3. Figure 2.87 shows a knocks on a piece of metal placed on the anvil,
the short impact time causes a large impulsive
baseball player stopping force to act on the metal.

a fast moving ball with Figure 2.90

his hand. 3. A golf club has a very hard hitting surfaces
to give a big force on the golf ball. The hard
(a) The glove which surface gives a short impact time and a large
impulsive force which can make the ball go far.
is made of soft Figure 2.87
material lengthens

the time of impact

and reduces the impulsive force acting on

his palm.

(b) When catching the ball, the player moves

his hand backwards. This action will

lengthen the time of impact between the

ball and the glove and further reduce the

impulsive force acting on his palm.

(c) The combined effect of the glove and the

action of the hand prevents injury to the

player.

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Physics  Form 4  Chapter 2 Force and Motion I  

CONTOH 2.23 Important Safety Features of a Car Chapter

A billiard ball with a mass of 0.25 kg is resting on 1. While striving to improve the performance of 2
a smooth table. The ball is then given a horizontal their cars, car manufacturers also place great
impulse of 3.0 N s. What is the velocity of the ball importance on the safety features of the cars. A
after the hit? good car safety system that can prevent injuries
to passengers in the event of an accident is of
Solution utmost importance.
Initial velocity of the ball, u = 0 m s–1 (a) Air bags
Velocity of the ball after being hit = v m s–1 Air bags inflate within one hundredth
Mass of the ball, m = 0.25 kg second upon impact to cushion passengers
Impulse = Ft = 3.0 N s from direct impact with the steering wheel,
dashboard, or windscreen of the car. For
Since Ft = mv – mu safety protection by the air bags, the
passengers must also be wearing safety
Therefore, 3.0 = 0.25v belts at the same time.
H ence , v = 03.2.05
= 12 m s–1 (b) Safety belts
The inertia-reel safety belts lock to keep
EXAMPLE 2.24
wearers tightly held to their seats when a
A 60 kg resident jumps from a burning building. car decelerates rapidly. The passengers are
Before hitting the ground his velocity is 5.0 m s–1 . prevented from being thrown forward due
(a) Calculate the impulse when his legs hit the to their inertia.

ground. (c) Side-impact bar
(b) What is the impulsive force if he bends his leg Strong metal steel bars inside the door

and his velocity changes from 5 m s–1 to zero frame provide passengers with extra
in 0.5 s? protection in the event of a side impact.
(c) What is the impulsive force if he does not bend
his leg and his velocity changes from 5 m s–1 (d) Anti-lock braking system (ABS)
to zero in 0.1 s? The anti-lock braking system consists
(d) What is the advantage of bending his leg?
of a computerised system that rapidly
Solution holds and releases the brake alternately.
It prevents the wheels from locking when
(a) Impulse = Change of momentum the brake is applied and held by the driver.
This prevents the car from skidding.
= mv – mu
(e) Passenger safety cell
= (60 × 0) – (60 × 5) The passenger safety cell is a strong rigid

= –300 N s steel cage that will prevent the roof from
collapsing on the passengers in the event
(b) Impulse = Ft = –300 N s that the car overturns. The passengers
will be protected from direct impact of
Impulsive force, F = –300 = –600 N external forces.
0.5
(f) Safety glass
(c) Impulsive fprce, F = –300 = –3 000 N The windscreen glass is specifically
0.1
designed to fracture into small rounded
(d) Reduces the impulsive force and hence, reduces pieces upon impact instead of shattering
so that the passengers will not be cut by
injury. the pieces that easily.

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Chapter   Physics  Form 4  Chapter 2 Force and Motion I Figure 2.93 The book dropped by the student is
pulled towards the Earth
Checkpoint 2.7

Q1 A racing car with a mass of 600 kg is moving at
50 m s–1. Compare the magnitude of the impulsive
force acting on the car if it collides with
(a) a concrete pillar and stops in 0.02 seconds,
and
(b) the concrete pillar that has old tyres covering
it so that the car is stopped in 0.2 seconds.
(c) Compare the answer in (a) and (b) and
explain the role of the old tyres.
(d) The mass of a racing car is usually smaller
than that of an ordinary car. Using Newton’s
second law of motion, explain the advantage
of the racing car having a smaller mass.

Q2 Figure 2.91 shows a worker receiving a

2 watermelon thrown at him. Explain why he
moves his arms backwards when receiving the
watermelon.

Figure 2.91 Figure 2.94 Direction of the gravitational field
around Earth


Q3 (a) Explain the role of the front and rear 2. For an object with mass, m, its weight, w is
crumple zones of a car. given by w = mg where g is the gravitational
field strength.
(b) Explain why the passenger safety cell is
strong and rigid.

Q4 Figure 2.92 shows the action of an air bag in 3. The SI unit for weight is newton (N). Weight
cushioning a driver during a crash. is a vector quantity.

SPM Tips

Gravitational field strength,
1 kg g at this point is equal to

the gravitational force

Figure 2.92 acting on a mass of 1 kg

placed at this point.

(a) Why does the driver more forward when the g= F
car stops suddenly? m

(b) How can the air bag save the driver from SI unit = N kg–1
serious injury?

Checkpoint 2.8

2.8 Weight Q1 (a) Define weight and mass.
(b) Give three differences between weight and
1. The weight of an object is the force of gravity mass.
acting on the object. (c) What is the weight of a student on the
surface of the Earth if his mass is 47 kg?
(Assume the gravitational acceleration
= 10 m s–2)

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02 FOC PHYSICS F4 3P.indd 59 CONCEPT MAP

LINEAR MOTION

Types of motion Physical quantities Equations of Graphs of linear Momentum
involved linear motion motion

Stationary Non-uniform Scalar Vector v = u + at Displacement- Principle of Force Impulse
velocity time Conservation
Uniform s = 1 (u + v)t
velocity Free fall 2 Velocity-time of Momentum
1
Gravitational Distance Displacement s = ut + 2 at2
acceleration
v2 = u2 + 2as Newton’s Impulsive
Second force
Time Velocity Acceleration- Law of
time Motion Weight

Explosion
Speed Physics  Form 4  Chapter 2 Force and Motion I  

2 Collision Newton’s
Third Law
Acceleration Deceleration Newton’s of Motion
First Law of

Motion

59 Chapter

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  Physics  Form 4  Chapter 2 Force and Motion I

SPM Practice 2

Objective Questions 4. Figure 4 shows a portion of 8. Figure 8 shows a
ticker tape pulled by a toy velocity-time graph of an
1. Select the correct pair. object in motion. What
car. is the magnitude of the
displacement?
Scalar Vector

A Distance Speed 14.0 cm Velocity (m s–1)
Figure 4
B Speed Acceleration 5
If the ticker timer used has a
Chapter
C Velocity Acceleration frequency of 50 Hz, what is 4

D Acceleration Deceleration the average velocity of the 3
car indicated by this strip of

2 2. Which of the following ticker tape? 2
A 70 cm s–1 1
statements is true? B 100 cm s–1 0 1 2 3 4 5 Time (s)
A A stationary object C 117 cm s–1
D 140 cm s–1 Figure 8
undergoes either
5. Which of the following ticker
acceleration or

deceleration. tape shows a motion with A 10 m
B An accelerating object acceleration and then uniform B 15 m
velocity? C 20 m
has a velocity with big A D 25 m
magnitude.
C An object moving with a 9. Figure 9 shows a

uniform velocity will not

have an acceleration. B acceleration-time graph of a

D An object moving with car.

a non-uniform velocity C a / m s–1
will have a non-uniform

acceleration.

3. D

HOTS 0 t /s
Figure 9

P 9m OQ 6. A car moves with uniform Which of the following
West 12 m East acceleration from 0 m s–1 to
20 m s–1 in 4 seconds. What velocity-time graphs correctly
Figure 3 is the acceleration of the car?
A 5 m s–2 C 20 m s–2 shows the motion of the car?
B 10 m s–2 D 80 m s–2 A v

Figure 3 shows Rama 7. A lorry accelerates from rest 0 t
standing at O. He then walks with a uniform acceleration
to P. After reaching P, he of 10 m s–2 for 5 seconds. B v
moves to Q and stops there. What is the distance travelled
by the lorry in this period of 0t
What is the displacement of time?
A 50 m C 125 m
Rama?
B 100 m D 250 m
A 3 m to the East
B 3 m to the West
C 21 m to the East
D 21 m to the West

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Physics  Form 4  Chapter 2 Force and Motion I  

C v 11. A marble is dropped from C v
HOTS a height h m. Which of the D v
0
following graphs represent
D v the final velocity, v against
height, h of the marble if

t the air resistance can be t

neglected?

A v

0t h t Chapter
B v h
10. Figure 10 shows a 13. Figure 13 shows a teapot 2
HOTS displacement-time graph of a C v being pulled by a boy using a
thread.
motorcycle.
Thread
s /m
Teapot
t /s
Figure 13
Figure 10 h
h If the boy pulls the teapot
Which of the following D v with an increasing force
velocity-time graph shows the slowly, the thread does not
motion of this motorcycle? snap. But if he pulls the
A v teapot with a sudden strong
force the thread snaps. What
concept is related to this
observation?
A Impulse C Friction
B Momentum D Inertia

14. Figure 14 shows two identical
threads X and Y.

0 t 12. A ball is thrown vertically
upward. By neglecting the
B v X
air resistance, which of the Stone

following velocity against Y

time graph best explains

the motion of the ball until it

reaches a maximum height?

0t A v

C v Weight
Figure 14
t
If the mass of the load is
0 t B v increased slowly, state what
will happen to X and Y.
D v A Both X and Y will snap at
the same time.
t B X will snap first followed
by Y.
0t C Y will snap first followed
by X.
D Only X will snap.

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  Physics  Form 4  Chapter 2 Force and Motion I

15. Which of the following The rocket moves upwards. 21. Figure 21 shows a graph of
physical quantities will What concept can be used to velocity against time for an
explain this situation? aeroplane.
change when an object is A Inertia
B Impulse V
brought from Earth to the C Newton’s second law
D Newton’s Third Law BC
Moon?
20. D
A Mass AE
B Weight
C Inertia
D Length

16. What will happen when t
forces acting on a plane are
in equilibrium? Tomato Figure 21
A The plane will accelerate.
Chapter B The plane will decelerate. Tomato breaks Concrete Which part of the graph
C The plane moves with floor shows that the forces
uniform velocity. acting on the plane are in
equilibrium?
2 D The plane will change its A AB
height. B BC
C CD
17. A boy with a mass of 30 kg D DE
runs at a velocity of 6 m s–1.
He jumps on a stationary Tomato 22. Which is the unit for impulse?
trolley with a mass of 15 kg. A kg m s–2
Both the boy and the trolley B kg m2 s–2
move together in the original C kg2 m s–1
direction of the boy’s motion. D kg m s–1
What is the velocity of the
boy and trolley? Tomato does not break Sponge 23. Figure 22 shows a ball rolling
A 3.0 m s–1 down a smooth slope and
B 4.0 m s–1 continues to move on a
C 6.0 m s–1 rough horizontal surface.
D 8.0 m s–1
Figure 20 Smooth slope
18. A cannon ball with a mass of
20 kg is fired from a cannon Figure 20 shows two identical Rough surface
with a mass of 400 kg. If the tomatoes dropped from the
cannon ball moves with a same height onto a concrete Figure 22
velocity of 120 m s–1, what is floor and a piece of sponge
the magnitude of the recoil respectively. Which inference Which of the following
velocity of the cannon? is correct?
A 0.3 m s–1 C 3.0 m s–1 A The impulsive force velocity-time graphs best
B 2.0 m s–1 D 6.0 m s–1 depends on the velocity
of the tomato. explains the motion of the
19. Figure 19 shows a water B The impulsive force
rocket with water gushing out depends on the impact ball?
from its lower end. time. A v
C The impulsive force
depends on the
gravitational potential
energy.
Water D The impulsive force t
rocket depends on the height at B v
the tomato falls.
Movement

Water

t

Figure 19

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C v t Physics  Form 4  Chapter 2 Force and Motion I  
D v
24. A student has a mass of 45 kg. What is the weight of the student if
the gravitational field strength is 9.81 N kg–1?
A 54.8 N
B 220.7 N
C 441.5 N
D 450 N

t Chapter

2

Subjective Questions

Section A

1. Diagram 1(a) shows the motion of a black ball towards a white ball. Diagram 1(b) shows the motion of both
SPM
2012 balls after collision.

Stationary

AB AB

Before collision        After collision

(a) (b)

Figure 1

Table 1 shows the momentum of the balls before and after the collision.

Table 1

Momentum before collision (kg m s–1) Momentum after collision (kg m s–1)

ABAB

2.0 0.0 0.5 1.5

(a) What is the meaning of momentum? [1 mark]

(b) Based on the table, determine the total momentum of the balls [1 mark]
(i) before the collision, [1 mark]
(ii) after the collision.

(c) Compare the answers in b(i) and b(ii). [1 mark]

(d) (i) Based on the answers in b(i) and b(ii), state a conclusion about the total momentum. [1 mark]
(ii) Name the physics principle involved in d(i). [1 mark]

(e) State one condition needed in order to apply the physics principle stated in 1d(ii). [1 mark]

(f) The total kinetic energy of the balls decreases after the collision. What type of collision is this? [1 mark]

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  Physics  Form 4  Chapter 2 Force and Motion I

2. Figure 2.1 shows the arrangement of apparatus in an experiment to determine the acceleration due to gravity.
Figure 2.2 shows the ticker tape that is produced from the experiment. The ticker timer has a frequency of
50 Hz.

G-clamp

Ticker timer

Ticker tape

A.C. power
supply

Stool

Chapter 22.5 cm

Weight

Table

2 Polystyrene sheet 4.5 cm

   

Figure 2.1 Figure 2.2

(a) What is meant by acceleration? [1 mark]

(b) From Figure 2.2, determine the value of g, the acceleration due to gravity. [3 marks]

(c) Give two reasons why the value of g determined from 2(b) is smaller than the actual value. [2 marks]

3. Figure 3.1 shows a cyclist cycling from a road to a beach. In the first 5 seconds, he cycles on the road and
for the next 15 seconds, he cycles on the beach. Figure 3.2 shows the motion graph of his journey.

v / m s–1

15

10

Beach

5

Road

0 5 10 15 20 t/ s

Figure 3.1 Figure 3.2

(a) Calculate the acceleration of the cyclist when he is on [3 marks]
(i) the road, and
(ii) the beach.

(b) He uses the same force to cycle on both the road and the beach. Explain why there is a difference in
[2 marks]
HOTS the acceleration in the two parts of his journey.

(c) Calculate the total distance travelled by the cyclist in the 20 seconds. [3 marks]

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Physics  Form 4  Chapter 2 Force and Motion I  

Section B
4. (a) A boy plays with two punching bags, A and B, of different masses. The mass of punching bag A is 5 kg

and of punching bag B is 30 kg. Figure 4.1 and 4.2 illustrate the situation of both punching bags, A and B.

A Chapter

B 2

        

Figure 4.1 Figure 4.2

(i) What is meant by mass? [1 mark]

(ii) The boy needs to use force either to move or to stop the punching bags. He finds that it is more
HOTS difficult to move and to stop punching bag B compared with punching bag A. Relating the mass of

the punching bags and the resistance to the state of motion of the bags, deduce a relevant physics
[5 marks]
concept. [1 mark]

(iii) Name the physics concept.

(b) Explain how a bit of tomato sauce left inside a bottle can be forced out of the bottle. [3 marks]

(c) Figure 4.3 shows a car speeding on a road and Figure 4.4 shows the car crashing on a boulder. Due to
HOTS the lack of safety features, the driver sustained serious injuries.

   

Figure 4.3 Figure 4.4

The following are relevant facts concerning the car.
(i) The car stopped too suddenly upon hitting the boulder.
(ii) The driver was thrown out of the car, breaking the windscreen.
(iii) Sharp pieces of glass from the shattered windscreen cut the head of the driver.
(iv) The accident occurred at night. The driver did not see the boulder until it was very near.
HOTS Based on the above facts, suggest various features to be added to the car to improve its safety
standard and explain how each feature can reduce injury to the driver.
[10 marks]

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  Physics  Form 4  Chapter 2 Force and Motion I

Section C

5. When designing a remote control car, a few factors are taken into consideration. These include reaction time
and resistive force.

Chapter Figure 5.1

Definition:

2 Reaction time: The time between the pressing of the switch of the remote control and the starting of the
car from rest.
Resistive force: The total force against the motion of the car. This includes friction between the tyres
and the road and the air resistance.

(a) State one effect of a force. [1 mark]

(b) In a short distance competition, a car will normally accelerate from rest until it crosses the finishing line.

Sketch a graph of velocity against time for the situation described and explain how the acceleration and
[4 marks]
the distance travelled by the car can be determined from the graph.

(c) Table 5 shows the characteristics of four remote control racing cars, P, Q, R and S, that are taking part

in a 50 m dash competition. Table 5

Car Reaction time / s Mass / kg Engine thrust / N Resistive / N

P 0.25 1.6 10.5 3.8

Q 0.45 2.0 12.8 2.6

R 0.20 1.2 7.0 2.4

S 0.55 1.8 15.4 5.6

Based on Table 5;

HOTS (i) Explain the importance of each characteristic in the design of a remote control car.
(ii) By using the formula, t = Reaction time +
competition. 100m  , determine which car will win the 50 m dash
F [10 marks]

(d) Car Q is placed on a slope inclined at 30° with the horizontal.

Super [5 marks]

30°
Figure 5.2

By using the information concerning car Q from Table 5,
(i) deduce whether it can move up the slope,
(ii) determine its acceleration up the slope.

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