CONTENTS
Chapter 4.3 Specific Latent Heat 110
1 Measurement 1 4.4 Gas Laws 119
SPM Practice 4 134
1.1 Physical Quantities 2
1.2 Scientific Investigation 6
SPM Practice 1 15 Chapter
5 Waves 140
Chapter 5.1 Fundamentals of Waves 141
2 Force and Motion I 19
5.2 Damping and Resonance 148
5.3 Reflection of Waves 151
2.1 Linear Motion 20
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
SPM Practice 2 60 Chapter
6 Light and Optics 187
Chapter
3 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 6.4 Thin Lens Formula 212
3.3 Man-made Satellites 83 6.5 Optical Instruments 217
Praktis SPM 3 88 6.6 Image Formation by Spherical
Mirrors 222
Chapter SPM Practice 6 230
4 Heat 93
PRE-SPM MODEL PAPER 235
4.1 Thermal Equilibrium 94
4.2 Specific Heat Capacity 98 ANSWERS 252
iv
CONTENT FOC PHYSICF F4.indd 4 29/01/2020 12:26 PM
Chapter Physics Form 4 Chapter 2 Force and Motion I
2 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
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. Non-linear motion is motion that is not in a
2.1 Linear Motion straight line.
3. When analysing linear and non-linear motion,
My motion is a distance, displacement, speed, velocity,
linear motion My motion is a
non-linear motion Start acceleration and deceleration are commonly
Start End End Not a straight line encountered physical quantities.
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
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.
Chapter
2
Kedudukan A Kedudukan B Kedudukan C
Position B
Position A
Position C
Metre speed Metre speed Metre speed
The velocity of the car is zero. The car The speed meter shows the car The speed meter indicates a change
does not move (stationary). Object moving at a constant velocity of 60 of velocity from 60 to 90 km per
that is stationary has velocity equals km per hour. hour. The car moves at a non-uniform
to zero. velocity. In this case, velocity is
increasing.
Figure 2.2
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.
(b) The distance travelled by the car is given by the following, N
Distance travelled = (200 + 500 + 600 + 300) m W E Delima
= 1600m 200 m
(c) Distance is a scalar quantity. S 500 m
600 m
2. (a) Speed is the distance travelled per unit time. Speed can also be defined as 300 m
the rate of change of distance. Mengkibol
(b) Average speed of an object can be obtained by dividing the total distance
travelled by the object by the total time taken for the travel. Figure 2.3
(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:
Total distance travelled, d
v = Time taken, t
1 600 m
= 120 s = 13.3 m s –1
(d) Speed is a scalar quantity.
20
02 FOC PHYSICS F4 3P.indd 20 29/01/2020 1:39 PM
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
specific direction. N
(b) It is the distance travelled by the object from its initial position to its W E Delima
final position in a straight line.
(c) For the car travelling from Delima to Mengkibol, the displacement is S 45°
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, Mengkibol 800 m
Magnitude of the displacement Figure 2.4
2
= 800 + 800 2
= 1 131 m
Its direction is south-west of Delima (or S45°W of Delima)
(e) Hence, displacement is a vector quantity. Chapter
4. (a) Velocity is defined as the rate of change of displacement.
(b) In the case of the car travelling from Delima to Mengkibol, the average velocity, v, is given by:
Displacement,s 2
v = 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 100 M
of P. The distance travelled by the cow is 240 m 100 m north of O. After that he
and the time taken is 160 s. runs to point Q which is 100 m 100 M N
U east of P. The total time taken
is 25 seconds. O
What is the (a) distance, (b) Figure 2.6
P
average speed, (c) displacement, and (d) average
velocity traveled by the athlete?
Q
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 Total distance 200
–1
Total distance travelled = 240 m = Total time = 25 = 8 m s
Displacement = 70 m (c) Displacement O
Time taken = 160 s = 100 + 100 2
2
Total distance travelled
(a) Average speed = = 100 2 m
Time taken
240 m (d) Average velocity
= = 1.5 m s –1
160 s
Displacement 70 m = Displacement = 100 2 = 4 2 m s –1
(b) Average velocity = = Total time 25
Time taken 160 s
o
–1
= 0.44 m s in the south- in the direction N45 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 –1
20 40 20 40 20 40 20 40 Initial velocity, u = 10 m s
15 45 15 45 15 45 15 45 –1
10 50 10 50 10 50 10 50 Final velocity, v = 20 m s
5 –1 55 5 –1 55 5 –1 55 5 –1 55
m s m s m s m s
0 60 0 60 0 60 0 60 Time taken, t = 2.5 s
20 – 10
Acceleration, a =
Figure 2.7 2.5
= 4.0 m s –2
1. Figure 2.7 shows a car moving along a straight
line. The speedometer of the car shows that it
is moving with increasing velocity. The car is EXAMPLE 2.4
accelerating. A car travelling at
24 m s slowed down
–1
2. (a) Acceleration is defined as the rate of when the traffic light
change of velocity. turned red. After
Chapter
2 Acceleration, a = Change of velocity undergoing uniform Figure 2.10
deceleration for 4 s, it
Time taken
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
Time taken = 4 s
where u is the initial velocity of the object, Using the formula a = v – u = 0 – 24 = –6.0 m s –2
v is its final velocity and t is the time taken. t 4
–2
(c) The SI unit for acceleration is m s . The negative value shows that the car is decelerating.
–2
(d) Acceleration is a vector quantity. Hence, the magnitude of deceleration is 6.0 m s .
25 30 35 25 30 35 25 30 35 25 30 35
20 40 20 40 20 40 20 40 EXAMPLE 2.5
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
0 60 0 60 0 60 0 60 (a) A car moves from
BERHENTI
rest and accelerates
uniformly to a
Figure 2.8 velocity of 20 m s Figure 2.11
–1
3. (a) An object is said to be undergoing a in 5 seconds. What
deceleration or retardation when it is is the acceleration of the car?
slowing down. The rate of change of (b) After that, the car slows down at a uniform
velocity of the object then has a negative rate until it stops in 5 seconds. What is the
value. acceleration of the car?
(b) Figure 2.8 shows a car decelerating. Solution
The speedometer of the car shows that (a) Initial velocity, u = 0 m s –1
it is moving with decreasing velocity. Final velocity, v = 20 m s –1
Time taken, t = 5 s
v – u 20 – 0
EXAMPLE 2.3 Therefore a = t = 5 = 4 m s –2
A van accelerates uniformly (b) Initial velocity, u = 20 m s –1
from a velocity of 10 m s Final velocity, v = 0 m s –1
–1
to 20 m s in 2.5 s. What is Time taken, t = 5 s 0 – 20
–1
v – u
the acceleration of the van? Therefore, a = = = –4 m s –2
Figure 2.9 t 5
–2
The car decelerates with a magnitude of 4 m s .
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02 FOC PHYSICS F4 3P.indd 22 29/01/2020 1:39 PM
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
Power supply
a.c. 12 V Chapter
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
(c) average velocity of the trolley, 5.0
(d) average initial velocity u,
(e) average final velocity v, 4.0
(f) acceleration of the trolley, a is determined. 3.0
2.0
Results:
1.0
(a) Number of ticks for each strip of tape = 10 Ticks
Time taken for each strip of tape = 10 × 0.02 = 0.2 s 10 20 30 40 50 60
Total time taken by the trolley = Number of strips × 0.2 s Figure 2.13
= 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
Total time 1.2 s 1 –
Number of strips 2
(d) Average initial velocity, u = 2.0 = 10.0 cm s –1 of tapes
0.2 1
(e) Average final velocity, v = 7.0 = 35.0 cm s –1 1
0.2 1
1
1
(f) Time, t between u and v = +1+1+1+1+ × 0.2 second v
2 2 1
= (6 – 1) × 0.2 second = 1.0 s 1 –
2
Hence, acceleration a = v – u = 35.0 – 10.0 = 25.0 cm s –2
t 1.0 u
Conclusion:
–2
The trolley moves with an acceleration of 25.0 cm s . Figure 2.14
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Physics Form 4 Chapter 2 Force and Motion I
Analysis of Motion
EXAMPLE 2.6
1. A ticker timer is an apparatus used to study
the motion of an object within a short period A trolley pulled a ticker tape through a ticker timer
of time. while moving down an inclined plane. Figure 2.17
shows the ticker tape produced.
2. Figure 2.15 shows a ticker timer connected to
an a.c. power supply.
12 cm
Power supply Figure 2.17
a.c.12 V POWER SUPPLY
FUSE 10 12 Volts 1
8 plus
O 6 Determine the average velocity of the trolley.
Ticker timer 4 AC 2 0 0
I OVLD DC
Solution
Ticker
tape Displacement = 12 cm
There are 10 spaces between the dots,
Chapter
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
A trolley is pushed Length/cm
4. When the a.c. power supply is switched on, to move up the slope 4.5
the vibrating bar vibrates 50 times per second of an inclined plane. 4.0
(following the frequency of the power supply Figure 2.18 shows the 3.5
3.0
which is 50 Hz). Each time the vibrating bar arrangement of the 2.5
hits the tape, a dot is produced on the tape. ticker tapes. Each strip 2.0
1.5
5. The period of time between 2 adjacent dots is consists of 5 ticks or 0.1 1.0
0.5
known as 1 tick. second. Determine the 0.2 0.4 0.6 0.8 t /s
acceleration of the trolley. Figure 2.18
Solution
Average initial velocity of the trolley,
4.5 cm
Dot 1 tick u = = 45.0 cm s –1
Figure 2.16 1 tick for carbonised ticker tape 0.1 s
Average final velocity of the trolley,
1.5 cm
6. A ticker timer can be used to determine v = 0.1 s = 15.0 cm s –1
(a) time, Time taken, t = (4 – 1) × 0.1
(b) displacement, = 0.3 s
(c) average velocity, Therefore, acceleration, a = v – u
(d) acceleration, and t
(e) type of motion = 15.0 – 45.0
of an object. 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
Sometimes there are too few ticks to be cut into Solution
strips. In such cases, direct analysis is needed to Average initial velocity, u = 0.6 cm
determine the acceleration. The following ticker 0.02 s
tape shows the motion of a trolley. The ticker = 30 cm s –1
timer vibrates at a frequency of 50 Hz. 1.6 cm
Average final velocity, v = 0.02 s
Initial Final –1
= 80 cm s
Time taken, t = (11 – 1) × 0.02 s
Uniform acceleration = 0.2 s
0.6 cm 1.6 cm v – u Chapter
Figure 2.19 Therefore, acceleration, a = t
80 – 30
= 0.2 2
= 250 cm s –2
Determining Acceleration with Photogate and Electronic Timer
A pair of photogates A and B is Photogate and electronic x cm
arranged at two different locations timer system 0000 A
along an inclined plane. This type B Cardboard
of photogate consists of optical Trolley
sensors that can measure the time,
t, taken by an object passing
through it. In this case, the object Inclined plane
is a cardboard with a length of x cm
attached to the top portion of a
trolley. Figure 2.20
x
1. Average initial velocity, u, when the cardboard passes through photogate A is where t A is the time
taken by the cardboard to pass through photogate A. t A
x
2. Average final velocity, v, when the cardboard passes through photogate B is where t B is the time
taken by the cardboard to pass through photogate B. t B
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
x x
u = dan v =
t A t B
Therefore, the acceleration, a, is given by the following:
x x
v – u t B – t A
a = = cm s –2
t 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
Relationship between Displacement,
EXAMPLE 2.9
Velocity, Acceleration and Time
Hisham drove his car from home with a uniform
Motion with Uniform Velocity acceleration and achieved a velocity of 15.0 m s –1
1. For an object moving with uniform velocity, v, in 5.0 seconds. What is the
its displacement, s after time, t is given by:
(a) acceleration of Hisham’s car?
(b) displacement of Hisham’s car 5.0 seconds
after starting the journey?
where v is the uniform velocity of the (c) velocity of Hisham’s car at t = 4.0 s?
object.
(d) velocity of Hisham’s car when he has
Motion with Uniform Acceleration travelled 20.0 m from the starting point?
1. The equations of motion of an object with Solution
uniform acceleration, a, are as follows: Summary of information:
Chapter
2 Displacement, s is given by: Initial velocity, u = 0 m s –1 –1
s = (Average velocity) × time
Final velocity, v = 15.0 m s
= Initial velocity + final velocity × time Time taken, t = 5.0 s
2
= u + v t (a) Acceleration, a = v – u .............................
2 t
Hence, = 15.0 – 0
5.0
................................. = 3.0 m s –2
1
2
2. Acceleration, a = v – u ............................... (b) Displacement, s = (u + v)t .....................
t
= 1 (0 + 15.0)(5.0)
Rearranging equation, 2
= 37.5 m
....................................... (Can also be solved by equation )
3. Substitue equation into equation . (c) Hisham’s velocity at t = 4.0.
1
s = (u + u + at)t v = u + at.....................................................
2 = 0 + 3.0 (4.0)
Simplifying .................. = 12.0 m s –1
(d) If the velocity of Hisham’s car after moving
4. Rearrange equation into 20.0 m is v.
v – u With the information u = 0 m s –1
t = a a = 3.0 m s –2
and substitue into equation , s = 20.0 m
1
hence, s = (u+v) v – u and using the equation
2
a
v = u + 2as
2
2
2
2
2
= v – u 2 Therefore v = 0 + 2 (3.0 × 20.0)
2a = 120.0
Simpliying, Hence, v = 10.95 m s –1
...................................
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Physics Form 4 Chapter 2 Force and Motion I
SPM Highlights HOTS Checkpoint 2.1
A cat sees a rat running towards a hole with a uniform Q1 Farid leaves his home at 10:00 am and walks to
velocity 0.4 m s . At that time the position of the cat, the the post office to post a letter. After that, he walks
–1
rat and the hole are shown in the diagram respectively. If to a bookstore to buy a present. From there, he
the cat runs towards the rat with a uniform acceleration walks to KEC Fried Chicken to pack some food
–2
of 1.0 m s , can the cat catch the rat before it enters before going to Ken’s house to celebrate his
the hole? birthday. Farid reaches Ken’s house at 12:00
noon.
1.8 km
0.9 km
Farid’s Post office
house U Chapter
6.0 m 2.0 m B T
1.2 km
Bookstore
Figure 2.21 S 2
0.7 km
1.6 km
Answer:
Distance between the cat and the hole is 8.0 m and Ken’s house KEC Fried
distance between the rat and the hole is 2.0 m. If Chicken
time taken for the cat to reach the hole is longer than
the time taken for the rat to reach the hole, the cat Figure 2.22
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
s = 2.0 m (c) calculate his displacement.
(d) calculate his average velocity.
Since
s = v × t for uniform velocity, Q2 Figure 2.23 shows four strips of ticker tapes
t = s
v HOTS obtained from an experiment using a trolley.
2.0
=
0.4
= 5.0 s End Beginning
Tape P
For the cat,
initial velocity, u = 0 m s –1 Tape Q
acceleration, a = 1.0 m s –2
displacement, s = 8.0 m Tape R
Using the formula Tape S
1
s = ut + at , Figure 2.23
2
2
1
8.0 = 0 + (1.0)t 2
2 (a) Compare tape P with tape Q and describe
t = 16.0 (i) one common characteristic,
2
Therefore, t = 16.0 (ii) one difference between them.
= 4.0 s (b) Describe the motion of the trolley for
(i) tapes R,
Since the rat needs 5.0 s to reach the hole while the (ii) tapes S.
cat needs only 4.0 s, the cat will be able to catch the
rat before it enters the hole.
27
02 FOC PHYSICS F4 3P.indd 27 29/01/2020 1:39 PM
Physics Form 4 Chapter 2 Force and Motion I
Q3 A student carried out an experiment with Q5 A car which is moving with a constant acceleration
a trolley running down an inclined runway. HOTS passes by lamp post A and lamp post B with a
–1
–1
Figure 2.24 shows a chart representing the velocity of 18 m s and 20 m s respectively.
motion of the trolley.
x / cm Lamp post B
Lamp post A
11
10
9
8 Figure 2.26
7
If the distance between the two lamp posts is 10 m,
6
calculate
5 (a) the acceleration of the car,
(b) the time taken by the car to travel from lamp
Chapter
4
2 3 post A to lamp post B.
2
2.2 Linear Motion Graphs
1
0 t / 5-tick
Figure 2.24
Displacement-Time Graphs
If the ticker timer made 50 ticks per second,
determine the
(a) initial velocity, Displacement, s 0 m 0.5 m 1.0 m 1.5 m 2.0 m
(b) final velocity, Time, t 0 s 2 s 4 s 6 s 8 s
(c) acceleration
of the trolley. Z Z Z
Q4 A cyclist cycled in a straight line at a velocity Displacement, s 0 m 0.5 m 1.0 m 1.5 m 2.0 m
of 8 m s on a hard surface. When he was Time, t 0 to 8 s
–1
6 m from a soft surface, he braked to slow the
bicycle down so that it entered the soft surface Figure 2.27 The tortoise and hare race
at a velocity of 4 m s .
–1
1. Figure 2.27 shows a tortoise moving at a slow
and steady speed while a hare is sleeping
soundly from the time t = 0 to 8 s.
Displacement, s / m Displacement, s / m
2.0
2.0
Hard surface 6 m Soft surface
1.5 1.5
Figure 2.25
1.0
1.0
(a) What is the time taken by the bicycle to slow
down from 8 m s to 4 m s ? 0.5 0.5
–1
–1
(b) Calculate the deceleration of the bicycle.
(c) The bicycle decelerated to a stop, with a 0 2 4 6 8 0 2 4 6 8
uniform deceleration, in 3 seconds after Time, t / s Time, t / s
entering the soft surface. Calculate the (a) Tortoise (b) Hare
distance travelled by the bicycle on the soft Figure 2.28 s-t graph
surface.
2. The motion of the tortoise and hare is
represented by the two displacement-time
graphs above.
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02 FOC PHYSICS F4 3P.indd 28 29/01/2020 1:39 PM
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 1. The gradient of the velocity-time graph gives
starting line from time t = 0 to 8 s, 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
4. The gradient of the s-t graph of an object gives The object is not moving. Velocity Chapter
the rate of change of displacement, which is of the object is zero.
the velocity of the object.
Displacement, s / m Displacement, s / m 0 t /s 2
(b)
v / m s –1 The object is moving with uniform
Velocity = velocity.
Gradient of Acceleration = gradient of graph
–2
the s -t graph = 0 m s .
t /s
0
0 0 (c)
Time, t / s Time, t / s
v / m s –1 The velocity increases with
Figure 2.29 a uniform rate. The gradient
is positive and constant. The
5. Some common graphs are shown below. object is moving with uniform
Table 2.1 0 t /s acceleration.
Graph of s against t Explanation (d)
(a) v / m s –1 The velocity decreases with
s / m The displacement of the object
is always the same. Hence, a uniform rate. The gradient
the object is not moving or is negative and constant. The
stationary. object is moving with uniform
Velocity = gradient of graph deceleration.
= 0 m s -1
t /s t /s
0 0
(b)
s / m The displacement increases at
a uniform rate. The gradient 3. The area under the following graphs of v
of the graph is a constant. against t gives the displacement of the object
Hence, the object moves with respectively.
a uniform velocity.
(a) v / m s –1 (b) v / m s –1
t /s
0
(c)
s / m v v
The gradient of the graph is
increasing. The velocity of the
object is increasing. Hence,
the object moves with an 0 t /s 0 t /s
acceleration. t t
t /s (c)
0 v / m s –1
(d) s / m The gradient of the graph v
is decreasing. The velocity
of the object is decreasing. u
Hence, the object moves with 0 t /s
t /s a deceleration. t
0
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02 FOC PHYSICS F4 3P.indd 29 29/01/2020 1:39 PM
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
Object moves with positive 25.0 A B
a uniform acceleration.
20.0
0 t /s 15.0
(b) 10.0
a / m s –2 5.0
Object moves with negative C t /s
uniform acceleration or 0 5.0 10.0 15.0 20.0 25.0
0 t /s deceleration. –5.0
–10.0 D
Chapter
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
(b) Determine the displacement of the motorcycle
s /m
C (i) after 20.0 s.
s 1
(ii) after 25.0 s.
s A B
2 (c) Sketch a acceleration-time graph for the
motorcycle during the first 25.0 seconds.
Solution
D
0 t /s
t t t t
1 2 3 4 (a) (i) OA
Figure 2.30 Gradient of graph = 25.0 – 0
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 The motorcycle moves with a constant
(c) BC (d) CD.
acceleration of 2.5 m s .
–2
Solution (ii) AB
(a) OA The motorcycle moves with a constant
–1
The gradient of the graph is positive and velocity of 25.0 m s .
constant. The runner runs with a uniform velocity.
(b) AB (iii) BC: (0 – 25.0)
The displacement does not change. The runner Gradient =
is not moving. (20.0 – 15.0)
(c) BC = –5.0 m s –2
The gradient of the graph is increasing. The The motorcycle moves with a deceleration
–2
runner runs with increasing velocity, or the with a magnitude of 5.0 m s until it stops
runner accelerates. at the time t = 20.0 s.
(d) CD (iv) CD
The gradient of the graph is negative and The motorcycle moves with a negative
constant. The runner is running with a uniform uniform velocity. This means the
velocity but in the opposite direction. The motorcycle is now moving in the opposite
runner goes back to his original position. direction, on its way back.
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02 FOC PHYSICS F4 3P.indd 30 29/01/2020 1:39 PM
Physics Form 4 Chapter 2 Force and Motion I
(–10.0 – 0) SPM Highlights
Gradient =
(25.0 – 20.0)
= –20.0 m s –2 The v-t graph in Figure 2.32 shows the situation of
a skydiver when jumping from a plane until he lands
Note: In this case, even though the velocity on the ground. Analyse the v-t graph and explain the
of the motorcycle has an increasing motion of the skydiver when
magnitude, its acceleration is negative. (a) the parachute is not open,
(b) the parachute is open,
This is because the negative indicates (c) the skydiver lands,
the motorcycle moves in the opposite v / m s –1
direction.
50
(b) (i) The displacement of the motorcycle after 40
20.0 s = area under the graph for t = 0 s to Parachute Parachute
t = 20.0 s 30 not opened opens Chapter
1
s = × 10.0 × 25.0 + (5.0 × 25.0) + 20 Landing
2
1
× 5.0 × 25.0 10 0 t /s 2
2
= 312.5 m 5 10 15 20 25 30 35 40
Figure 2.32
t = 0 s
t = 20.0 s Answer:
s = 312.5 m
t = 25.0 s (a) When the parachute is not open, the skydiver
(ii) For t = 20.0 s to t = 25.0 s, the motorcycle falls with increasing velocity until it reaches
moves in the opposite direction as shown a maximum velocity which is known as the
terminal velocity.
in the diagram. The distance travelled by (b) When the parachute is open, the skydiver fall
the motorcycle for t = 20.0 s to 25.0 s with a uniform deceleration until it reaches
1 minimum velocity upon which he continues to
s = (5.0 × 10.0) fall with this velocity.
2
= 25.0 m (c) When the skydiver lands, stop falling and his
velocity descreases until 0 m s .
–1
Therefore, the displacement of the
motorcycle after 25.0 s
s = 312.5 – 25.0
= 287.5 m Checkpoint 2.2
Q1 The s-t graph in Figure 2.33 shows the movement
(c)
a / m s –2 of a crate that is being hoisted vertically by a
crane. s is the displacement of the crate from the
3.0 ground.
2.0
1.0 s / m
t /s
0 2.0
–1.0 5.0 10.0 15.0 20.0 25.0
–2.0 1.5
–3.0
1.0
–4.0
s
–5.0 0.5
0 1 2 3 4 5 t / s
SPM Tips Figure 2.33
Make sure you can use the acceleration formula, (a) Briefly describe the movement of the crate
v – u from t = 0 to 5 s.
a = to calculate acceleration from
t (b) Calculate the velocity of the crate from t = 2.4
• ticker tape, to 3.8 s.
• ticker chart, and (c) What is the average velocity of the crate
• velocity-time graph. 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
15 Velocity
10 4
Chapter
2 5 2
0 10 20 30 40 50 60 t / s 0 5 10 15 20 Time
–2
Figure 2.34
(a) How far did the bus travel before it stopped – 4
at ‘Bus Stop 1’?
(b) For how long did the bus stop at ‘Bus
Stop 1’? Figure 2.36
(c) Calculate the deceleration of the bus before (a) (i) When did the lift stop at the fifth floor?
it stopped at ‘Bus Stop 2’. (ii) Explain your answer in (a)(i).
(d) Find the distance between ‘Bus Stop 1’ and (b) (i) For which period of time did the lift travel
‘Bus stop 2’.
down?
Q3 The s-t graph in Figure 2.35 shows the (ii) Explain your answer in (b)(i).
displacement, s, of a boy from his classroom (c) Calculate the total distance travelled by the
along a corridor. lift when it finally stopped at the first floor.
s / m (d) Calculate the displacement of the lift at t = 20 s.
35
Physics of 100 m World Record
30 The v-t graph of Usain Bolt when breaking the
100 m record in Berlin in 2009
25
v / m s –1
20
12
15
10
10
8
5
6
0 5 10 15 20 25 30 35 40 45 50 55 t / s
4
Figure 2.35
2
(a) (i) What is the total distance travelled by the
boy? 0 t / s
(ii) Calculate the average speed of the boy. 2 4 6 8 10
(b) Halfway along the corridor, he stopped to
pick up some books that he had dropped Q5 Without doing the actual calculation, what is the
accidentally on the floor. When was the time, area under the v–t graph shown above?
t, he picked up the books?
(c) Sketch the v-t graph from t = 0 to 55 s.
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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
Rubber stopper
Air
Glass tube
Feather SPM Tips
To show the motion of free fall without vacuum, two objects
of the same size but different mass can be used. Chapter
Coin
Feather
Rubber Coin 2
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
out the air from it. When all the air has been pumped
out, a vacuum is resulted in the tube.
3. When the tube is inverted, both the coin and the feather Connect to
vacuum pump
fall to the base at the same time.
• Discussion: In this case, both the objects fall without
air resistance. Coin
Feather
4. An object falling under the pull of the gravitational Air being sucked
force without any other forces acting on it is said to out of the tube
be undergoing free fall. Vacuum
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
acceleration Diameter = d cm Steel ball
Apparatus and Materials: Photogate A
Photogate kit and electronic timer, steel ball, Time passing through
micrometer screw gauge photogate A = t s
1
Instructions:
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 Time passing through = t s Photogate B
Chapter
2 both the photogates, each equipped with photogate B 2
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 = t 1 s
Time for the steel ball to pass through photogate B = t 2 s
Time for the steel ball to travel from photogate A to photogate B = t s
Calculations:
d
Average velocity of steel ball in passing photogate A, u = cm s .
–1
t 1
d
Average velocity of steel ball to passing photogate B, v = cm s .
–1
t 2
Time taken for the steel ball to accelerate from u to v = t s
v – u
Using the formula a = , the experimental value of gravitational acceleration is,
t
d – d
v – u
g = = t 2 t 1 cm s –2
t t
–2
This value can be divided by 100 to convert to m s for easier comparison with standard value.
Discussion:
The scientific estimation for the value of g at the equator is around 9.78 m s and at the poles is
–2
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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Physics Form 4 Chapter 2 Force and Motion I
EXAMPLE 2.12 Checkpoint 2.3
Q1 What is the time taken for a stone to undergo
A student throws a stone free fall of 100 m? (Assume the gravitational
–2
vertically upwards. acceleration, g = 10 m s )
The stone moves to a Q2 Figure 2.41(a) shows an astronaut standing on
height of 5 m before the surface of the Moon. He releases a piece of
stopping and then falls paper and a rubber ball from the same height at
downwards. Figure 2.40 the same time. Figure 2.41(b) is a stroboscopic
photograph that shows how the piece of paper
shows the path moved and the rubber ball undergo free fall and reach
by the stone from A to B the ground at the same time. HOTS
and C.
(a) What is the Ball
magnitude of the Paper Chapter
velocity of the stone
when it leaves the
hand of the student? 2
(b) How long is the time (a) (b)
taken for the stone to
travel from A to C? Figure 2.41
(Take g = 10 m s ) Figure 2.40 (a) What is meant by free fall?
–2
(b) What inference can you make based on the
Solution observations?
Height, s = 5 m. (c) When the activity is repeated on the Earth,
which of the two objects will reach the
(a) The stone moves in the opposite direction from ground first? Explain your answer.
the gravitational force. Hence, the value of the Q3 Figure 2.42 shows a ball is thrown vertically
gravitational acceleration is g = –10 m s . At upwards from the position A with a velocity of
–2
–1
B, the stone come to a rest. Hence, v = 0 m s . 24 m s . The ball moves to position B, and then
–1
falls downwards.
If its speed when in A is v A , therefore by using
the formula, B
v = u + 2as
2
2
0 = v A + 2 (–10)(5)
2
v A = 100
2
Therefore, v A = 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,
1
s = ut + at 2
2 A
1
0 = 10t + (–10)t 2
2
5t(2 – t) = 0
t = 2 s Figure 2.42
Hence the total time taken by the stone to travel (a) What is the distance between A and B?
(b) What is the time taken for the ball to move
from A until C is 2 seconds. from A to B?
(c) State the assumption made when doing the
calculations.
(Take g = 10 m s )
–2
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Physics Form 4 Chapter 2 Force and Motion I
Original State of Object that is in Motion
2.4 Inertia
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
Situation I VIDEO
Figure 2.45
1. When the car is moving, the passengers in the
Chapter
Pull
2 Cardboard car move together.
Glass 2. When the car stops suddenly, the passengers
are thrown forward.
Figure 2.43 3. The inertia of the passengers resists change
of state of motion. As a result, the passengers
1. Figure 2.43 shows a 50 cent coin resting on a continue to move forward.
cardboard that is placed on top of a glass.
2. When the cardboard is pulled quickly, the coin Situation II
hovers over the top of the glass for an instant 1. Figure 2.46 shows how tomato sauce can be
before dropping into the glass. forced out of its bottle. When the bottle is
3. The inertia of the coin resists change of its state moved quickly towards the food, the sauce in
of rest and hence does not move together with the bottle moves together.
the cardboard. Gravitational force pulls it down
into the glass.
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
2. When the bottle is stopped suddenly, the sauce
will be forced out of the bottle into the food.
Pull 3. The inertia of the sauce resists change of its
state of motion and continues to move out of
Figure 2.44
the bottle.
2. Initially, the books above it tend to stay at rest.
3. The inertia of these books resists change of Relationship Between Mass and Inertia
state of rest and do not move together with 1. The mass of an object is the amount or
the book that is being pulled out. The books quantity of matter in it.
then fall vertically downwards due to the 2. The mass of an object depends on the number
gravitational force. and type of atoms of the object.
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Physics Form 4 Chapter 2 Force and Motion I
Eksperimen 2.1
Experiment 2.1
Eksperimen 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
Operational Definition: Chapter
Period of oscillation is an indicator for inertia, the responding variable. The bigger the period of
oscillation, the bigger the inertia. 2
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
free end of the hacksaw blade. G-clamp
3. The hacksaw blade is displaced sideways slightly and
released so that it would oscillate horizontally.
4. The time taken for 10 complete oscillations, t 1 , is determined
using a stopwatch and recorded. This step is repeated for Plasticine
another reading t 2 . The readings of t 1 and t 2 are recorded in
Table 2.4.
5. Steps 3 and 4 are repeated with mass of plasticine, m = 40
g, 50 g, 60 g and 70 g. Hacksaw
Table leg blade
6. A graph of period against mass is drawn.
Figure 2.47
Results:
1. Tabulation of results.
Table 2.4
Time taken for 10 oscillations / s
Mass of plasticine, t
m / g Period, T = 10 / s
t 1 t 2 Average, t
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
0.40
Chapter
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
1. The characteristic of inertia of an object can 4. An object moving at uniform velocity will
be stated in the form of Newton’s First Law continue to move with uniform velocity in a
of Motion. straight line. An external force will be required
to change this state of motion.
2. Newton’s First Law of Motion states that
an object will remain at its original state, 5. Newton’s laws of motion were discovered by
whether it is at rest or in motion with uniform Sir Isaac Newton (1642—1727). He wrote
velocity, if there is no net force acting on it. a book ‘Philosophiae Naturalis Principia
3. An object at rest will remain at rest. An Matematica’ that became the foundation for
external force is needed to move it. the field of mechanics. This book also explains
many important laws, including Newton’s Law
of Motion I, II and III.
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Physics Form 4 Chapter 2 Force and Motion I
Physics Form 4 Chapter 2 Force and Motion I
Branch of tree Rear-end collision
A stick is used to jerk the branch
of a guava tree. At the end of When a rear-end
the branch, there is a big guava. collision occurs, the Headrest
The guava tends to remain in its car and the body of the
original state of rest due to its driver move forward
inertia. This will cause its stalk to suddenly. The headrest
snap and the guava will fall to the supports the head of
ground. the driver when it is
thrown backwards. Chapter
2
Guava
Effects of
Tissue paper Inertia Steel structure of lorry
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
A girl dries her hair by turning her head
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.
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Physics Form 4 Chapter 2 Force and Motion I
SPM Tips Checkpoint 2.4
To remember Newton’s First Law of Motion Q1 Figure 2.53 shows two similar tins P and Q each
when F = 0 (No resultant force) tied to a long rope.
At rest – Always stays Rope
at rest.
Sand
P Q
Figure 2.49
Push
In motion – Figure 2.53
Always moves
Chapter
• at a constant speed,
2 • in a straight line. When both the tins are pushed with the same
force,
(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
SPM Highlights (a) a loaded lorry takes a longer time to stop.
(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
that of the tiger? box placed on top of it. It is moving towards a
A The size of the Figure 2.51 stationary obstacle.
elephant is bigger than the tiger.
B The mass of the elephant is bigger than the tiger.
C The elephant is taller than the tiger. Toy car
D The elephant is longer than the tiger. Pencil box 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 (b) Explain your answer in (a)
a small mass and another (c) Compare with the effect observed if the car
big mass. How can the is moving at a higher velocity.
astronaut differentiate
between the two packets
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.
40
40
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Physics Form 4 Chapter 2 Force and Motion I
2.5 Momentum Solution
(a) Momentum of the car = 800 × 20
= 16 000 kg m s –1
Momentum of the lorry = 5 000 × 20
= 100 000 kg m s –1
(b) The lorry, due to its bigger momentum.
Figure 2.55
1. Every moving object has momentum. However, Principle of Conservation of
the momentum of a timber lorry is very much Momentum
bigger than that of a cyclist, despite both Chapter
vehicles moving at the same speed. 1. According to the principle of conservation
of momentum, when two or more bodies act
2. The linear momentum of an object is defined on each other, their total momentum remains 2
as the product of its mass and velocity constant, provided that there is no external
force acting on them.
Momentum = Mass × Velocity
SPM Tips
Elastic Collision vs Inelastic Collision
3. The SI unit for momentum is kg m s or N s. Q: Elastic and inelastic collisions are two very common
–1
types of collision. How is momentum related to
Momentum is a vector quantity. The direction these two types of collision and what are the
of momentum is the same as the direction of differences between both types of collision?
velocity.
EXAMPLE 2.13
The total mass of a lorry is 20 000 kg and the total
mass of a car is 2000 kg.
If both the lorry and the car are travelling at a
velocity of 25 m s , calculate the momentum of
–1
the lorry and the car respectively. (a) Collision between electric bumper cars in an elastic collision
Solution
Momentum of the lorry = 20 000 × 25
= 5 × 10 kg m s –1
5
Momentum of the car = 2 000 × 25
= 5 × 10 kg m s –1 (b) Collision between two cars is an inelastic collision
4
Figure 2.56
EXAMPLE 2.14 A: • In an elastic collision, the total kinetic energy of
the system is conserved. The objects elastically
A car has a mass of 800 kg and a lorry has a mass spring back to their original shapes and the
system retains all its original kinetic energy.
of 5000 kg. • In an inelastic collision, the total kinetic energy
(a) Compare the momentum of both vehicles when is not conserved . One or more of the colliding
moving at a velocity of 20 m s . objects may not spring back to its original shape
–1
(b) If both vehicles move at the same velocity, or heat or sound may be generated. The total
which vehicle will cause greater destruction kinetic energy of the system after the collision
when hitting a lamp post? is less than that before the collision.
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02 FOC PHYSICS F4 3P.indd 41 29/01/2020 1:39 PM
Physics Form 4 Chapter 2 Force and Motion I
Collision 2. Types of collision
1. The principle of conservation of momentum There are two types of collision, that is, elastic
can be studied using the collision of objects. collision and inelastic collision.
(a) In an elastic collision
(i) momentum is conserved,
2.0 kg 1.0 kg At rest
3.0 m s –1 v m s –1
A B A B (ii) kinetic energy is conserved,
(iii) the objects move separately after
(a) collision ,
(iv) total energy is conserved.
2.0 kg 3.0 m s –1 1.0 kg 2.0 m s –1 v m s –1 (b) In an inelastic collision.
A B A B
(i) momentum is conserved,
(b) (ii) kinetic energy is not conserved,
Figure 2.57 (iii) the objects may combine and move
For case (a) together after collision,
Chapter
2 Total momentum Total momentum (iv) total energy is conserved.
=
before collision after collision
(2.0 × 3.0) + (1.0 × 0) = (2.0 + 1.0)v SPM Tips
6.0 kg m s = 3.0 v
–1
Therefore, v = 2.0 m s –1 In the explanation of a law or principle, the conditions
and limitations must be mentioned together. In the case
For case (b) of stating the principle of conservation of momentum,
(2.0 × 3.0) + (1.0 × (–2.0)) = (2.0 + 1.0)v the condition of no external force acting on the system
needs to be mentioned.
4.0 kg m s = 3.0 v
–1
4.0
Therefore, v = 3.0
= 1.3 m s –1
Experiment 2.2
Eksperimen 2.1
Eksperimen 2.1
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 29/01/2020 1:39 PM
Physics Form 4 Chapter 2 Force and Motion I
I Non-elastic collision
Procedure:
Ticker
Ticker Timer
tape Trolley A Trolley B
Plasticine
Friction compensated
inclined plane
12 V a.c. power Support
supply
Figure 2.58
1. The apparatus is arranged as shown in Figure 2.58.
2. The inclined plane is adjusted to be friction compensated. When this is achieved, the trolley given Chapter
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. 2
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 m 1 u / kg m s –1 (m 1 + m 2 ) u / m s –1 momentum,
m 1 / kg / kg (m 1 + m 2 ) 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 dowel rod
Trolley A
Trolley B Friction compensated
inclined plane
12 V a.c. power Support
supply
Figure 2.59
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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 Before collision
During
Velocity = v collision Velocity = u
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.
Chapter
2 Results: Table 2.6
Before collision After collision
u Total momentum, Total momentum
m 1 m 2 v 1 v 2
m 1 u m 1 v 1 + m 2 v 2
1 1
1 2
2 1
3 1
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)]
750 m s –1 2 000 m s –1 = 500 000 kg m s –1
Total momentum after collision
A
B = (2 000 + 500) × v
= 2 500v
500 kg According to the principle of conservation of
2 000 kg
Figure 2.61 momentum, the total momentum before and after
the collision are the same.
Calculate the speed of the rocks after the collision.
Therefore,
Solution 2 500v = 500 000
–1
v A = 750 m s v B = –2 000 m s –1 v = 200 m s –1
m A = 2 000 kg m B = 500 kg
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Physics Form 4 Chapter 2 Force and Motion I
Solution
EXAMPLE 2.16
Total momentum before coupling
A lorry with a mass of 8 000 kg moves with a = (2.5 × 10 × 10) + (2.5 × 10 × 0.8)
4
4
velocity of 30 m s . The lorry then collides with = 4.5 × 10 kg m s –1
–1
4
a car with a mass of 1500 kg and travelling with Total momentum after coupling
a velocity of 20 m s . After the collision, the two = 5.0 × 10 × v
–1
4
vehicles stick and move together with a velocity v.
Determine the value of v. According to the principle of conservation of
momentum,
Solution Total momentum Total momentum
30 m s –1 20 m s –1 before coupling = after coupling
5.0 × 10 × v = 4.5 × 10 kg m s –1
4
4
v = 0.9 m s –1 Chapter
Momentum of lorry before collision Assumption: The sum of the external forces acting
= 8 000 × 30 on the system consisting of the two boxcars is zero.
= 240 000 kg m s –1 2
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.
Total momentum = Total momentum Friction
before collision 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
270 000 Weight component cancels the effect of friction
Therefore, v = 9 500
= 28 m s –1 Figure 2.63
2. A friction compensated inclined plane has an
inclination in such a way that is the friction on
the surface of the inclined plane is balanced
EXAMPLE 2.17 out by the component of the weight along the
trolley.
Figure 2.64 shows two identical boxcars of a train 3. The two forces cancel out each other and hence
being coupled at a railway station. Each boxcar has
a mass of 2.5 × 10 kg. The velocity of the boxcar it is as if the inclined plane is frictionless.
4
1 and boxcar 2 before the coupling are 1.0 m s 4. When this condition is achieved, the trolley
–1
and 0.8 m s respectively. will move with a uniform velocity when given
–1
a small push.
1.0 m s –1 0.8 m s –1
1 2 Explosion
Before 1. When a rocket is launched, the exhaust gas is
v forced out from the back of the rocket and the
1 2 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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Physics Form 4 Chapter 2 Force and Motion I
Experiment 2.3
Eksperimen 2.1
Eksperimen 2.1
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
Chapter
Apparatus and materials: Triple beam balance, 4 trolleys, 2 wooden blocks, a mallet, and a metre
2 rule.
Procedure:
Mallet
Wooden block Wooden block
Plunger
A B
d d
1 2
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, d 1 and d 2 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
explosion After explosion
Total left Total right
Total –d 1 d 2 Total
momentum v 1 = t v 2 = t m 1 m 2 momentum, momentum, momentum
m 1 v 1 m 2 v 2
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
SPM Tips
EXAMPLE 2.18
To answer questions related to the principle of conservation
An astronaut with a mass of 80 kg throws a box of of momentum:
instrument with a mass of 40 kg in order to return to • Sketch a diagram to show the motion of the objects
the space capsule. If the box moves with a velocity involved
of 6 m s , what is the velocity of the astronaut after • Label the mass and velocity (magnitude and direction)
–1
of each object in the diagram.
throwing the box? • Use the formula (Total initial momentum = Total final
momentum).
Solution
Let v be the velocity of the astronaut.
The total momentum before the box is thrown is
zero because both the astronaut and the box are EXAMPLE 2.19
stationary.
Figure 2.65 shows 2 trolleys A and B with a mass
Total momentum after the box is thrown of 1.5 kg and 1.0 kg respectively. A compression
= momentum of the box + momentum of the spring is placed between the 2 trolleys and is
astronaut compressed by forcing the 2 trolleys close to
= (40 × 6) + (80 × v) each other. When the 2 trolleys are released, each
= 240 + 80v 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
Therefore, A B
240 + 80v = 0
v = –3.0 m s Figure 2.65
–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
Solution
Let the time taken by trolley A to move 40 cm with the same magnitude that moves the plane
forward.
be t seconds, and at the same time trolley B has
moved d cm. By taking the direction travelled by 4. Satellites and space capsules use air jets to
A as positive,
40 –d control movement and position. This is based
1.5 × t + (1.0 × t = 0 on the principle of conservation of momentum
(1.5 × 40) – d = 0 for the case of explosion.
d = 60 cm
Applications of the Principle of
Conservation of Momentum in Daily
Life
Chapter
2
1. In the field of sports and games like billiard,
bowling, and carrom, the principle fo
conservation of momentum is involved in
collision form whereas in shooting, the Figure 2.67 Water sprinkler system
principle is involved in explosion.
5. A water sprinkler system uses this principle
2. In the launching of rocket, the combustion to irrigate plants in a farm. The whole area can
of fuel and oxygen produces a big downward be irrigated by the rotating sprinkler.
momentum and the rocket moves upwards
with the same magnitude of momentum. The 6. Some animals like the octopus uses this
momentum is conserved and is equal to zero. principle to move by squirting out a liquid
from its body.
7. This principle is also used to explain the
following situation:
(a) When a bullet comes out of a gun with
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 . Bumper car B 5 m s –2
–1
with a total mass of 200 kg is moving at a velocity of 5 m s . After a while,
–1
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.
–1
Q2 A 1500 kg van travelling at a speed of 15 m s collides with a 1000 kg car Figure 2.68
travelling in the opposite direction with a speed of 20 m s . After the
–1
collision, both vehicles get stuck and move together. 15 m s –1 20 m s –1
(a) What is the total momentum of the van and the car before the 1 500 kg Chapter
collision? 1 000 kg
(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). Figure 2.69
(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 from a rifle of mass 6 kg.
–1
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
–1
a velocity of 12 m s 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 A cyclist uses a
force to move bigger force to pedal
a stationary his bicycle and
lawnmower. increase its speed.
Effects
of a
A footballer force
uses a force A baker uses
to change the a force on his
direction of dough to shape it
motion of a ball. into a curry puff.
Chapter
2 Figure 2.72
Relationship between Force, Mass and Acceleration
Eksperimen 2.1
Experiment 2.4
Eksperimen 2.1
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.
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Physics Form 4 Chapter 2 Force and Motion I
Results:
Table 2.8
1. Mass of trolley, m / unit 1 2 3
Initial velocity, u / cm s –1
Final velocity, v / cm s –1
Acceleration, a / cm s –2
1
2. A graph of aceleration a against is plotted and the graph is a straight line
mass of trolley, m
1 Acceleration, a / cm s –2
passing through the origin. Therefore a ∝ if F is constant.
m Chapter
Conclusion:
When the pulling force is constant, the bigger the mass of the trolley, the
smaller is its acceleration. The hypothesis is accepted. 2
0 1 1
1
– – –
3 2
Mass of trolley, m
Figure 2.74
Experiment 2.5
Eksperimen 2.1
Eksperimen 2.1
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
Number of elastic string,
1. 1 2 3
F / unit
Initial velocity, u / cm s –1
Final velocity, v / cm s –1
Acceleration, a / cm s –2
Chapter
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
bigger is its acceleration. The hypothesis is accepted.
2
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
1
(a) a ∝ m when F is constant. to F, proportional to m,
a
(b) a ∝ F, when m is constant a
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
1
accelerate 1 m s -–1 and substituting into the a ∝ m
equation,
F = kma
where 1 = k (1)(1) 3. Newton’s Second Law of Motion describes the
Therefore, k = 1 relationship between acceleration of an object
with this, and the force applied to it. Newton’s Second
Law of Motion states that when a net external
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
of a few important facts: A box of mass of 5 kg moves with a uniform
• The object will only accelerate if there is a net external velocity when acted upon by a force of 10 N.
force acting on it.
• The acceleration is directly proportional to the net Calculate,
external force. (a) frictional force acting on the box, and
• The direction of the acceleration is the same as the (b) acceleration of the box if the force is 20 N.
direction of the external force.
• The magnitude of the acceleration of the object is Solution
inversely proportional to its mass.
(a)
SPM Tips
Friction F = 10 N
The meaning of F = ma is: Chapter
• The net force F, in the direction of the acceleration. Figure 2.78
• The direction of the force and the direction of the
change of momentum are the same. The force acting on the box in the horizontal
• The direction of the force F is the same as the direction direction is the force F = 10 N and frictional 2
of the acceleration, a. force. These forces are in equilibrium when the
box moves with uniform velocity. Hence, the
magnitude of the frictional force = 10 N, in the
EXAMPLE 2.20 opposite direction to F.
A 100 kg box is placed on a smooth floor and is (b)
pulled by a force F N. a
100 kg Friction = 10 N F = 20 N
F 1
Figure 2.79
Figure 2.77
(a) If F = 200 N, calculate the acceleration of the When F 1 = 20 N, the net force acting in the
direction of the acceleration, a
box. = F 1 – friction
(b) If the velocity of the box changes from rest to = (20 – 10) N
15.0 m s in 10 seconds when acted upon by = 10 N
–1
a force, what is the value of the force?
Solution Using the formula F = ma
(a) Given, m = 100 kg, F = 200 N 10 = 5 × a
F = ma a = 2 m s –2
F 200
Therefore a = = 100
m
= 2 m s –2
v – u Checkpoint 2.6
(b) Acceleration, a = 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
acceleration of the box?
= 1.5 m s –2
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
t
impulsive force
(1 000 × 0) – (1 000 × 25)
1. From the equation = 0.05
F = ma ............................
v – u
and a = ............................ = –500 000 N
t
Note: The negative sign shows the the force is
Substitute into , acting on the car.
v – u
F = m
t
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
Chapter
2 known as the impulse of a force. reaction. This means when an object exerts a
Impulse = Ft force on a second object, the second object
exerts a force of equal magnitude but in the
3. Impulse is defined as change of momentum, opposite direction on the first object.
that is, 2. Examples of Newton’s Third Law of Motion.
Impulse, Ft = mv – mu (a) A book placed on a table exerts a force on
= Change of momentum the table. The table exerts a reaction force
of the same magnitude on the book in the
The SI unit for impulse is kg m s or N s. opposite direction, that is, upwards.
–1
Impulse is a vector quantity. (b) When a person is walking, his sole pushes
the floor backwards. The floor reacts by
4. Impulsive force is defined as the rate of pushing the sole forward with the same
change of momentum resultant from an action magnitude. Hence, the person can move
that occurs in a short period of time. An forward.
example is during a collision. (c) A boy pulls a tight drawer and he is
pulled by that drawer. This is because
Change of momentum
Impulsive force, F = Time the force exerted by the boy pulling the
mv – mu drawer causes the drawer to exert a force
= of the same magnitude but in the opposite
t direction to pull the boy towards it.
EXAMPLE 2.22
A car with a mass of 1000 kg moves with a
velocity of 25 m s . The car then hits a tree and
–1
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 Figure 2.80
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Physics Form 4 Chapter 2 Force and Motion I
Effects of Impulsive Force in Daily short time when it hits the floor. The resultant
Life 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
change of momentum
the equation F = , the
time
impulsive force acting on the car is very big
and can cause serious damage to the car and Figure 2.82
danger to its passengers.
2. A bowl crashes when dropped on the floor. This The Effect of Time on the
is because the bowl undergoes a big change of Magnitude of the Impulsive Force
momentum in a very short time when it hits 1. Consider the formula for impulsive force Chapter
the floor and the resultant impulsive force is
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
1
2. Since F ∝ , the shorter the time of impact,
t
Figure 2.81 the bigger the impulsive force. The longer the
3. The tile on the floor cracks after a hammer is time of impact, the smaller the impulsive force.
accidentally dropped on it. The falling hammer Hence, by controlling the value of t, we can
undergoes a big change of momentum in a very 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
mv – mu
Implusive force, F =
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
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
Chapter
2 thick and soft mattress lengthens the time of is used to pound spices like chilli, onion and
garlic. The impact time between the pestle and
impact. The impulsive force acting on the high
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
sand. The loose sand lengthens the time of
impact and reduces the impulsive force acting Figure 2.89
on him. 2. A blacksmith uses a heavy hammer and anvil to
3. Figure 2.87 shows a forge metals. An anvil, which is made of steel,
baseball player stopping has a very hard surface. When the hammer
a fast moving ball with knocks on a piece of metal placed on the anvil,
his hand. the short impact time causes a large impulsive
(a) The glove which force to act on the metal.
is made of soft
material lengthens Figure 2.87
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 Figure 2.90
lengthen the time of impact between the 3. A golf club has a very hard hitting surfaces
ball and the glove and further reduce the to give a big force on the golf ball. The hard
impulsive force acting on his palm. surface gives a short impact time and a large
(c) The combined effect of the glove and the impulsive force which can make the ball go far.
action of the hand prevents injury to the
player.
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Important Safety Features of a Car
CONTOH 2.23
1. While striving to improve the performance of
A billiard ball with a mass of 0.25 kg is resting on their cars, car manufacturers also place great
a smooth table. The ball is then given a horizontal importance on the safety features of the cars. A
impulse of 3.0 N s. What is the velocity of the ball good car safety system that can prevent injuries
after the hit? 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
Air bags inflate within one hundredth
Velocity of the ball after being hit = v m s –1 second upon impact to cushion passengers
Mass of the ball, m = 0.25 kg from direct impact with the steering wheel,
Impulse = Ft = 3.0 N s dashboard, or windscreen of the car. For
Since Ft = mv – mu safety protection by the air bags, the Chapter
Therefore, 3.0 = 0.25v passengers must also be wearing safety
Hence, v = 3.0 belts at the same time. 2
0.25 (b) Safety belts
= 12 m s –1 The inertia-reel safety belts lock to keep
wearers tightly held to their seats when a
car decelerates rapidly. The passengers are
prevented from being thrown forward due
EXAMPLE 2.24 to their inertia.
(c) Side-impact bar
A 60 kg resident jumps from a burning building.
Before hitting the ground his velocity is 5.0 m s . Strong metal steel bars inside the door
–1
(a) Calculate the impulse when his legs hit the frame provide passengers with extra
ground. protection in the event of a side impact.
(b) What is the impulsive force if he bends his leg (d) Anti-lock braking system (ABS)
and his velocity changes from 5 m s to zero The anti-lock braking system consists
–1
in 0.5 s? of a computerised system that rapidly
(c) What is the impulsive force if he does not bend holds and releases the brake alternately.
–1
his leg and his velocity changes from 5 m s It prevents the wheels from locking when
to zero in 0.1 s? the brake is applied and held by the driver.
(d) What is the advantage of bending his leg? This prevents the car from skidding.
Solution (e) Passenger safety cell
(a) Impulse = Change of momentum The passenger safety cell is a strong rigid
= mv – mu steel cage that will prevent the roof from
= (60 × 0) – (60 × 5) collapsing on the passengers in the event
= –300 N s that the car overturns. The passengers
will be protected from direct impact of
(b) Impulse = Ft = –300 N s
–300 external forces.
Impulsive force, F = 0.5 = –600 N (f) Safety glass
–300 The windscreen glass is specifically
(c) Impulsive fprce, F = 0.1 = –3 000 N designed to fracture into small rounded
(d) Reduces the impulsive force and hence, reduces pieces upon impact instead of shattering
injury. so that the passengers will not be cut by
the pieces that easily.
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Physics Form 4 Chapter 2 Force and Motion I
Checkpoint 2.7
Q1 A racing car with a mass of 600 kg is moving at
50 m s . Compare the magnitude of the impulsive
–1
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 Figure 2.93 The book dropped by the student is
of the racing car having a smaller mass. pulled towards the Earth
Q2 Figure 2.91 shows a worker receiving a
Chapter
2 watermelon thrown at him. Explain why he
moves his arms backwards when receiving the
watermelon.
Figure 2.94 Direction of the gravitational field
Figure 2.91 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.
(b) Explain why the passenger safety cell is given by w = mg where g is the gravitational
strong and rigid. field strength.
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
acting on a mass of 1 kg
Figure 2.92 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
Q1 (a) Define weight and mass.
2.8 Weight (b) Give three differences between weight and
mass.
(c) What is the weight of a student on the
surface of the Earth if his mass is 47 kg?
1. The weight of an object is the force of gravity (Assume the gravitational acceleration
acting on the object. = 10 m s )
–2
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Physics Form 4 Chapter 2 Force and Motion I
Physics Form 4 Chapter 2 Force and Motion I
Impulse Impulsive force Weight
Momentum Force Newton’s Second Law of Newton’s Third Law of Motion
Principle of Conservation of Momentum Motion Collision Chapter
Graphs of linear motion Displacement- time Velocity-time Acceleration- time Explosion 2
Newton’s First Law of Motion
LINEAR MOTION Equations of linear motion v = u + at s = 1 (u + v)t 2 s = ut + 1 at 2 2 v 2 = u 2 + 2as Deceleration
Physical quantities involved Vector Scalar Displacement Distance Time Velocity Speed Acceleration
CONCEPT MAP CONCEPT MAP Types of motion Non-uniform Stationary velocity Uniform velocity Free fall Gravitational acceleration
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Physics Form 4 Chapter 2 Force and Motion I
SPM Practice 2
SPM Practice
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. car. object in motion. What
is the magnitude of the
Scalar Vector displacement?
A Distance Speed 14.0 cm Velocity (m s )
–1
Figure 4
B Speed Acceleration 5
If the ticker timer used has a
C Velocity Acceleration frequency of 50 Hz, what is 4
the average velocity of the
D Acceleration Deceleration car indicated by this strip of 3
ticker tape? 2
Chapter
2 2. Which of the following A 70 cm s –1 1
–1
statements is true? B 100 cm s
C 117 cm s
–1
A A stationary object D 140 cm s –1 0 1 2 3 4 5 Time (s)
undergoes either
acceleration or 5. Which of the following ticker Figure 8
deceleration. tape shows a motion with A 10 m
B An accelerating object acceleration and then uniform B 15 m
has a velocity with big velocity? C 20 m
magnitude. A D 25 m
C An object moving with a
uniform velocity will not 9. Figure 9 shows a
have an acceleration. B acceleration-time graph of a
D An object moving with car.
a non-uniform velocity
will have a non-uniform C a /m s –1
acceleration.
3.
HOTS D
0 t /s
Figure 9
P O Q 6. A car moves with uniform Which of the following
9 m acceleration from 0 m s to
–1
12 m velocity-time graphs correctly
–1
West East 20 m s in 4 seconds. What shows the motion of the car?
is the acceleration of the car?
Figure 3 A v
A 5 m s –2 C 20 m s –2
–2
Figure 3 shows Rama B 10 m s D 80 m s –2
standing at O. He then walks
to P. After reaching P, he 7. A lorry accelerates from rest
moves to Q and stops there. with a uniform acceleration t
of 10 m s for 5 seconds.
–2
What is the displacement of B 0 v
Rama? What is the distance travelled
A 3 m to the East by the lorry in this period of
B 3 m to the West time?
C 21 m to the East A 50 m C 125 m
D 21 m to the West B 100 m D 250 m
0 t
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Physics Form 4 Chapter 2 Force and Motion I
C v 11. A marble is dropped from C v
a height h m. Which of the
HOTS
following graphs represent
the final velocity, v against
height, h of the marble if
0 t the air resistance can be t
neglected?
D v v
A v D
0 t
h t
10. Figure 10 shows a B v 13. Figure 13 shows a teapot Chapter
HOTS displacement-time graph of a being pulled by a boy using a
motorcycle. thread.
s /m
Thread 2
h Teapot
C v Figure 13
If the boy pulls the teapot
with an increasing force
t /s slowly, the thread does not
Figure 10 snap. But if he pulls the
h teapot with a sudden strong
Which of the following D v force the thread snaps. What
velocity-time graph shows the concept is related to this
motion of this motorcycle? observation?
A v A Impulse C Friction
B Momentum D Inertia
h 14. Figure 14 shows two identical
threads X and Y.
12. A ball is thrown vertically
0 t upward. By neglecting the
B v air resistance, which of the X
following velocity against Stone
time graph best explains
the motion of the ball until it Y
reaches a maximum height?
A v
0 t
C v
Weight
Figure 14
If the mass of the load is
t
increased slowly, state what
B v will happen to X and Y.
0 t
D v A Both X and Y will snap at
the same time.
B X will snap first followed
by Y.
C Y will snap first followed
t
by X.
0 t 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
change when an object is explain this situation? aeroplane.
brought from Earth to the A Inertia V
Moon? B Impulse
A Mass C Newton’s second law B C
B Weight D Newton’s Third Law
C Inertia D
D Length 20. A E
t
16. What will happen when
forces acting on a plane are Tomato Figure 21
in equilibrium? Which part of the graph
A The plane will accelerate. Concrete shows that the forces
B The plane will decelerate. Tomato breaks floor acting on the plane are in
C The plane moves with equilibrium?
Chapter
uniform velocity. A AB
2 D The plane will change its B BC
height. C CD
D DE
17. A boy with a mass of 30 kg
runs at a velocity of 6 m s . 22. Which is the unit for impulse?
–1
He jumps on a stationary A kg m s
–2
trolley with a mass of 15 kg. B kg m s –2
2
Both the boy and the trolley C kg m s
2
–1
move together in the original Tomato D kg m s –1
direction of the boy’s motion.
What is the velocity of the 23. Figure 22 shows a ball rolling
boy and trolley? Tomato does not break Sponge down a smooth slope and
A 3.0 m s continues to move on a
–1
–1
B 4.0 m s rough horizontal surface.
C 6.0 m s –1
D 8.0 m s –1
18. A cannon ball with a mass of Figure 20 Smooth slope
20 kg is fired from a cannon
with a mass of 400 kg. If the Figure 20 shows two identical Rough surface
cannon ball moves with a tomatoes dropped from the Figure 22
velocity of 120 m s , what is same height onto a concrete
–1
the magnitude of the recoil floor and a piece of sponge Which of the following
velocity of the cannon? respectively. Which inference velocity-time graphs best
A 0.3 m s –1 C 3.0 m s –1 is correct? explains the motion of the
B 2.0 m s D 6.0 m s –1 A The impulsive force ball?
–1
depends on the velocity A v
19. Figure 19 shows a water
rocket with water gushing out of the tomato.
from its lower end. B The impulsive force
depends on the impact
time.
Water
rocket C The impulsive force t
Movement depends on the B v
gravitational potential
energy.
D The impulsive force
depends on the height at
Water the tomato falls.
t
Figure 19
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Physics Form 4 Chapter 2 Force and Motion I
24. A student has a mass of 45 kg. What is the weight of the student if
C v
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
D v
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 balls after collision.
2012
Stationary
A B A B
Before collision After collision
(a) (b)
Figure 1
Table 1 shows the momentum of the balls before and after the collision.
Table 1
–1
Momentum before collision (kg m s ) Momentum after collision (kg m s )
–1
A B A B
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
(i) before the collision, [1 mark]
(ii) after the collision. [1 mark]
(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
22.5 cm
Weight
Table
Chapter
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 t / s
5 10 15 20
Figure 3.1 Figure 3.2
(a) Calculate the acceleration of the cyclist when he is on
(i) the road, and
(ii) the beach. [3 marks]
(b) He uses the same force to cycle on both the road and the beach. Explain why there is a difference in
HOTS the acceleration in the two parts of his journey. [2 marks]
(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
difficult to move and to stop punching bag B compared with punching bag A. Relating the mass of
HOTS
the punching bags and the resistance to the state of motion of the bags, deduce a relevant physics
concept. [5 marks]
(iii) Name the physics concept. [1 mark]
(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.
Based on the above facts, suggest various features to be added to the car to improve its safety
HOTS
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.
Figure 5.1
Definition:
Chapter
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
the distance travelled by the car can be determined from the graph. [4 marks]
(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.
100m
(ii) By using the formula, t = Reaction time + , determine which car will win the 50 m dash
competition. F [10 marks]
(d) Car Q is placed on a slope inclined at 30° with the horizontal.
Super
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. [5 marks]
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02 FOC PHYSICS F4 3P.indd 66 29/01/2020 1:39 PM
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