1. The document discusses concepts related to linear motion including distance, displacement, speed, velocity, acceleration, and their relationships. 2. Key concepts like uniform and non-uniform motion, positive and negative acceleration, and the use of graphs to represent motion are explained. 3. Inertia and its relationship to mass is defined, and examples are given to illustrate inertia. 4. Momentum is introduced, and the principles of conservation of momentum and its application to explosions and collisions are described through examples.

1. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.1 LINEAR MOTIONs
Physical Quantity Definition, Quantity, Symbol and unit
Distance is the ……
Distance, s Quantity: … SI unit : ..
(a) The distance in .. ….
(b) the distance between ….
Displacement, s ….direction.
(c) The distance of its final …..
specified ….
Quantity: SI unit:
Speed is the
Speed,v
Speed =
Quantity: SI unit:
Velocity is the
Velocity, v
Velocity =
Direction of velocity is
Quantity : SI unit:
Average speed v= Example: A car moves at an average
speed / velocity of 20 ms -1
On average, the car moves a distance/
displacement of
Average velocity Displacement
v
TotalTime
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2. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Uniform speed Speed that remains the same in
Uniform velocity Velocity that remains
An object has a non- (a) The direction of motion changes or the motion is not linear.
uniform velocity if
(b) The magnitude of its velocity changes.
Acceleration, a When the velocity of an object
v u Acceleration is defined as the
a
t Change in velocity
Acceleration=
Time taken
Final velocity,v - Initial velocity,u
Unit: ms-2 =
Time taken,t
The velocity of an object increases from an initial velocity, u, to a higher final
velocity, v
Acceleration is positive
Deceleration
acceleration is negative. The rate of decrease in speed in a specified direction.
Zero acceleration An object moving at a constants velocity, that is,
Constant acceleration Velocity increases at a uniform rate.
When a car moves at a constant or uniform acceleration of 5 ms -2, its velocity
1. Constant =
2. increasing velocity =
3. decreasing velocity =
4. zero velocity =
5. negative velocity = object moves at opposite direction
6. zero acceleration =
7. negative acceleration = deceleration
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3. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Comparisons between distance and displacement Comparisons between speed and velocity
Distance Displacement Speed Velocity
Total path length The distance between The rate of change of The rate of change of
travelled from two locations distance displacement
one location to measured along the Scalar quantity Vector quantity
another shortest path
connecting them in
specific direction
Scalar quantity
It has magnitude but no
direction
SI unit SI unit :
Fill in the blanks:
1. A steady speed of 10 ms -1 = A distance of __________________________________________
2. A steady velocity of -10 ms -1 = A displacement of _________________________________
3. A steady acceleration of 4 ms -2 = Speed _____________________________________________
4. A steady deceleration of 4 ms -2 = ____________________________________________________________
5. A steady velocity of 10 ms -1 = A displacement of 10 m is travelled every 1 second to the right.
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4. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Example 1 Example 2
Every day Rahim walks from his house to the junction Every morning Amirul walks to Ahmad’s house
which is 1.5km from his house. which is situated 80 m to the east of Amirul’s house.
Then he turns back and stops at warung Pak Din which is They then walk towards their school which is 60 m
0.5 km from his house. to the south of Ahmad’s house.
(a) What is the distance travelled by Amirul and his
displacement from his house?
(b) If the total time taken by Amirul to travel from
his house to Ahmad’s house and then to school
is 15 minutes, what is his speed and velocity?
Speed =
(a) What is Rahim’s displacement from his house,
• when he reaches the junction.
Velocity =
• when he is at warung Pak Din.
(b) After, Rahim walks back to his house. breakfast
When he reaches home,
(i) What is the total distance travelled by
Rahim?
(ii) What is Rahim’s total displacement from
his house?
Example 3 Example 4
Salim running in a race covers 60 m in 12 s.
An aeroplane flies towards the north with a
(a) What is his speed in ms-1 velocity 300 km hr -1 in one hour. Then, the plane
moves to the east with the velocity 400 km hr -1 in
one hour.
(b) If he takes 40 s to complete the race, what is his
(a) What is the average speed of the plane?
distance covered?
(b) What is the average velocity of the plane?
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5. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
(c) What is the difference between average speed
and average velocity of the plane?
Example 5
The speedometer reading for a car travelling due north
shows 80 km hr -1. Another car travelling at 80 km hr -1
towards south. Is the speed of both cars same? Is the
velocity of both cars same?
A ticker timer
 Use:
 1 tick = time interval
 The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2
consecutive dots is
 1 tick =
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6. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Relating displacement, velocity, acceleration and time using ticker tape.
VELOCITY FORMULA
Time, t = 10 dicks x 0.02 s
= 0.2 s
displacement, s = x cm
velocity =
ACCELERATION
Initial velocity, u =
final velocity, v =
acceleration, a =
Elapsed time, t = (5 – 1) x 0.2 s = 0.8 s or
t = (50 – 10) ticks x 0.02 s = 0.8 s
TICKER TAPE AND CHARTS TYPE OF MOTION
 Distance between the dots increases uniformly

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7. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
- Distance between the dots decrease uniformly
Example 6
The diagram above shows a ticker tape chart for a
moving trolley. The frequency of the ticker-timer
used is 50 Hz. Each section has 10 dots-spacing.
(a) What is the time between two dots?
(b) What is the time for one strips?
(c) What is the initial velocity?
(d) What is the final velocity?
(e) What is the time interval to change from initial
velocity to final velocity?
(f) What is the acceleration of the object?
vu 2
a=
t
THE EQUATIONS OF MOTION
v  u  at u=
1 v=
s  ut  at 2 t=
2 s=
v  u  2as
2 2
a=
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8. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.2 MOTION GRAPHS
DISPLACEMENT – TIME GRAPH Velocity is obtained from
A – B : gradient of the graph is
B – C : gradient of the graph =
object is
C – D : gradient of the graph
The object
VELOCITY-TIME GRAPH Area below graph
Positive gradient
Negative gradient
Zero gradient
GRAPH s versus t v versus t a versus t
Zero
velocity
Negative
constant
velocity
Positive
Constant
velocity
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9. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
GRAPH s versus t v versus t a versus t
Constant
acceleration
Constant
deceleration
Example 1: Example 2:
velocity/ m s-1
20
10
time/
0 10 20 30 40
s
Based on the s-t graph above: (a) Calculate the acceleration at:
(a) Calculate the velocity at (ii) JK (ii) KL (iii) LM
(i) AB (ii) BC (iii) CD
(b) Describe the motion of the object at: (b) Describe the motion of the object at:
(i) AB (ii) BC (iii) CD (ii) JK (ii) KL (iii) LM
(c) Find (c) Calculate
(i) total distance (iii) The total displacement
(ii) total displacement (iv) The average velocity
(d) Calculate
(i) The average speed
(ii) The average velocity of the
moving particle
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10. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.3 INERTIA
Inertia The inertia of an object is the tendency of the object
Newton’s first law Every object
Relation between inertia The larger the mass,
and mass
SITUATIONS INVOLVING INERTIA
SITUATION EXPLANATION
EEEEEEEEJNVJLKN drops straight into
When the cardboard is pulled away quickly, the coin
the glass. DNFLJKVNDFLKJNB
VJKL;DFN BLK;XC
NB[F
NDPnDSFJ[POJDE]O-
JBD]AOP[FKBOP[DF
LMB NOPGFMB
LKFGNKLB
FGNMNKL’ MCVL
Paste a picture Chilli sauce in the bottle can be BNM’CXLB out if the bottle is moved
easily poured
NFGNKEPLANATION
down fast with a sudden stop. The sauce inside the bottle moves
together with the bottle.
When the bottle stops suddenly,
Paste a picture Body moves forward when the car stops suddenly The passengers were in a
state of motion when the car was moving.
When the car stopped suddenly,
Paste a picture A boy runs away from a cow in a zig zag motion. The cow has a large inertia
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11. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
 The head of hammer is secured tightly to its handle by
knocking one end of the handle, held vertically, on a hard
surface.
This causes the hammer head to continue on its
downward motion
when the handle has been stopped, so that the top
end of the handle is slotted deeper into the hammer
head.
• The drop of water on a wet umbrella will fall when the
boy rotates the umbrella.
• This is because the drop of water on the surface of the
umbrella moves simultaneously as the umbrella is rotated.
• When the umbrella stops rotating, the inertia of
the drop of water will continue to maintain its
motion.
Ways to reduce the negative 1. Safety in a car:
effects of inertia (a)Safety belt secure the driver to their seats.
When the car stops suddenly, the seat belt provides
the external force that prevents the driver from
being thrown forward.
(b)Headrest to prevent injuries to the neck during rear-
end collisions. The inertia of the head tends to
keep in its state of rest when
the body is moved suddenly.
(c)An air bag is fitted inside the steering wheel.
It provides a cushion to prevent the driver from
hitting the steering wheel or dashboard during a
collision.
2. Furniture carried by a lorry normally are tied up together by
string.
When the lorry starts to move suddenly, the furniture are
more difficult to fall off due to their inertia because
their combined mass has increased.
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12. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
• Two empty buckets which are hung with rope from the
Relationship between mass ceiling.
and inertia • One bucket is filled with sand while the other bucket is
empty.
• Then, both pails are pushed.
• It is found that
Push and compared to the bucket with sand.
• The bucket filled with sand offers more resistance to
movement.
• When both buckets are oscillating and an attempt is made
to stop them, the bucket filled with sand offers

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13. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.4 MOMENTUM
Definition Momentum =
SI unit:
Principle of In the absence of an external force,
Conservation of
Momentum
Elastic Collision Inelastic collision
ƒ Both objects move ƒ The two objects
ƒ Momentum ƒ Momentum
ƒ Kinetic energy ƒ Kinetic energy.
Total energy ƒ Total energy
Total Momentum Before = Total Momentum Before =
m1u1 + m2u2 = m1 v1 + m2 v2 m1 u1 + m2 u2 = ( m1 + m2 ) v
Explosion
Paste a picture Before explosion both object
Total Momentum Total Momentum after
before collision is collision :
zero m1v1 + m2v2
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14. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
From the law of conservation of momentum:
Total Momentum = Total Momentum
Before collision after collision
0 = m1v1 + m2v2
m1v1 = - m2v2
Negative sign means
EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum)
Paste a picture When a rifle is fired, the bullet of mass m,
moves with a high velocity, v. This creates a
momentum in the forward direction.
From the principle of conservation of
momentum,
Paste a picture Application in the jet engine:
The launching of rocket
Mixture of hydrogen and oxygen fuels
These high speed hot gases produce
By conservation of momentum,
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15. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Paste a picture In a swamp area, a fan boat is used.
The fan produces a high speed movement of air
backward. This produces a large momentum
backward.
By conservation of momentum, an equal but opposite
momentum is produced and acted on the boat. So the
boat will move forward.
Paste a picture A squid propels by expelling water at high velocity.
Water enters through a large opening and exits
through a small tube. The water is forced out at a
high speed backward.
Total Mom. before= Total Mom. after
0 =Mom water + Mom squid
0 = mwvw + msvs
-mwvw = msvs
The magnitude of the momentum of water and
squid are equal but opposite direction.
This causes the quid to jet forward.
Example
Example
Before collision After collision
Car A of mass 1000 kg moving at 20 ms -1
MA = 4 kg
collides with a car B of mass 1200 kg moving at MB = 2 kg
10 m s -1 in same direction. If the car B is UA = 10 ms -1 r i g h t
shunted forwards at 15 m s -1 by the impact, UB = 8 ms -1 l e f t VB 4 ms-1 right
what is the velocity, v, of the car A immediately
after the crash? Calculate the value of VA .
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16. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Example
Example
A truck of mass 1200 kg moving at
30 ms-1 collides with a car of mass
1000 kg which is travelling in the opposite A man fires a pistol which has a mass of 1.5 kg.
direction at 20 ms-1. After the collision, the two If the mass of the bullet is 10 g and it reaches a
vehicles move together. What is the velocity of velocity of 300 ms -1 after shooting, what is the
both vehicles immediately after collision? recoil velocity of the pistol?
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17. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.5 FORCE
Balanced Force Example:
When the forces acting on an object are
balanced,
Effect : the object
[velocity ]
or
moves
[a= ]
Unbalanced Force/ Resultant Force When the forces acting on an object are not balanced,
there must be
The net force is known as
Weight, W = Lift, U Thrust, F = drag, G
Effect : Can cause a body to
-
Newton’s Second Law of Motion The acceleration produced by a force on an object is
Force = Mass x Acceleration
F = ma
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18. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Experiment to Find The Relationship between Force, Mass & Acceleration
Relationship a& F a&
between m
Situation
Both men are pushing the same mass Both men exerted the same strength.
but man A puts greater effort. So he But man B moves faster than man A.
moves faster.
Inference The acceleration produced by an
object depends on the net force
applied to it.
Hypothesis The acceleration of the object
increases when the force applied
increases
Variables:
Manipulated : Force
Responding :
Acceleration
Constant : Mass
Apparatus and Ticker tape and elastic cords, ticker timer, trolleys, power supply and friction
Material compensated runway and meter ruler.
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19. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Procedure : An elastic cord is hooked over the An elastic cord is hooked over a
- Controlling trolley. The elastic cord is stretched trolley.
manipulated until the end of the trolley. The
variables. trolley is pulled down the runway
with the elastic cord being kept
stretched by the same amount of
force
-Controlling Determine the acceleration by Determine the acceleration by analyzing
responding analyzing the ticker tape. the ticker tape.
variables. Acceleration
v u
v u Acceleration a 
Acceleration a t
t
-
Repeat the experiment by using two Repeat the experiment by
Repeating
experiment. , three, four and five elastic cords
Tabulation of Force, F/No of Acceleration, a/ ms-2 Mass, m/ Mass, 1/m, Acceleration/
elastic cord no of m/g g-1 ms-2
data 1 trolleys
2 1
3 2
4 3
5 4
5
Analysing
Result
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20. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
1. What force is required to move a 2 kg object
2 2. Ali applies a force of 50 N to move a 10 kg
with an acceleration of 3 m s- , if table at a constant velocity. What is the
(a) the object is on a smooth surface? frictional force acting on the table?
(b) The object is on a surface where the
average force of friction acting on the
object is 2 N?
3. A car of mass 1200 kg travelling at 20 ms -1 4. Which of the following systems will
is brought to rest over a distance of 30 m. produce maximum acceleration?
Find
(a) the average deceleration,
(b) the average braking force.
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21. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.6 IMPULSE AND IMPULSIVE FORCE
Impulse The change of m=
Unit : u =
v =
Impulsive Force The rate of change t=
change of momentum mv  mu

time t
Unit =
Effect of time Impulsive force Longer period of time →Impulsive force
is
Shorter period of time →
Situations for Reducing Impulsive Force in Sports
Situations Explanation
Thick mattress with soft surfaces are used in events such as high jump
so that
Goal keepers will wear gloves to
A high jumper will bend his legs upon landing. This is to
so as to
A baseball player must catch the ball in the direction of the motion of
the ball. Moving his hand backwards when catching the ball prolongs
the time for the momentum to change so as to reduce the impulsive
force.
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22. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Situation of Increasing Impulsive Force
Situations Explanation
A karate expert can break a thick wooden slab with his bare hand
that moves at a very fast speed. The short impact time results in
A massive hammer head moving at a fast speed is brought to rest
upon hitting the nail within a short time interval.
A football must have enough air pressure in it so
Pestle and mortar are made of stone. When a pestle is used to pound
chillies the hard surfaces of both the pestle and mortar cause the pestle
to be stopped in a very short time. A large impulsive force is resulted
and thus causes these spices to be crushed easily.
Example 1
A 60 kg resident jumps from the first floor of a burning house.
His velocity just before landing on the ground is 6 ms-1.
(a) Calculate the impulse when his legs hit the ground.
(b) What is the impulsive force on the resident’s legs if he
bends upon landing and takes 0.5s to stop?
(c) What is the impulsive force on the resident’s legs if
he does not bend and stops in 0.05 s?
(d) What is the advantage of bending his legs upon landing?
Example 2
Rooney kicks a ball with a force of 1500 N. The time of
contact of his boot with the ball is 0.01 s. What is the impulse
delivered to the ball? If the mass of the ball is 0.5 kg, what is
the velocity of the ball?
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23. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.7 SAFETY VEHICLE
Safety features in vehicles
Component Function
Headrest
Air bag
Windscreen
Crumple zone
Front Absorb the shock from the accident. Made from steel, aluminium, plastic or
bumper rubber.
ABS Enables drivers to quickly stop the car without causing the brakes to lock.
Side impact bar
Seat belt
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24. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.8 GRAVITY
Gravitational
Force Objects fall because they are
This force is known as the
The earth’s gravitational force
Free fall  An object is falling freely when it is falling under the force of gravity
only.

 An object falls freely only
 In vacuum,
 They fall with
Acceleration due to  Objects dropped
gravity, g

Gravitational field The gravitational field is the region around the earth in which an object
experiences a force towards the centre of the earth. This force is the
gravitational attraction between the object and the earth.
The gravitational field strength is defined as the gravitational force which acts
on a mass of 1 kilogram.
F -1
g= Its unit is N kg .
m
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25. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
-1
Gravitational field strength, g = 10 N kg
-2
Acceleration due to gravity, g = 10 m s
-2
The approximate value of g can therefore be written either as 10 m s
-1
or as 10 N kg .
Weight The gravitational force acting on the object.
Weight = mass x gravitational acceleration
W = mg SI unit : Newton, N and it is a vector quantity
Comparison Mass Weight
between weight The mass of an object is the The weight of an object is the force of
& amount of matter in the object gravity acting on the object.
mass
Constant everywhere Varies with the magnitude of gravitational
field strength, g of the location
A scalar quantity A vector quantity
A base quantity A derived quantity
SI unit: kg SI unit : Newton, N
The difference
between a
fall in air and
a free fall in a vacuum
of a coin and a
feather.
Both the coin and the
feather are released
simultaneously from
the same height.
At vacuum state: There is no air At normal state: Both coin and feather
resistance. will fall because of gravitational force.
The coin and the feather will fall Air resistance effected by the surface area of
freely. a fallen object.
Only gravitational force The feather that has large area will have
acted on the objects. Both will fall more air resistance.
at the same time. The coin will fall at first.
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26. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
(a) The two spheres are falling The two spheres are falling down with
with an acceleration. the same acceleration
The distance between two The two spheres are at the same level
successive images of the sphere at all times. Thus, a heavy object and
increases showing that the two a light object fall with the same
spheres are falling with increasing gravitational acceleration
velocity; falling with an Gravitational acceleration is
acceleration. independent of mass
Two steel spheres
are falling under
gravity. The two
spheres are dropped
at the same time
from the same
height.
Motion graph for free fall object
Free fall object Object thrown upward Object thrown upward and fall
Example 1
A coconut takes 2.0 s to fall to the ground. What
is
(a) its speed when it strikes the ground
(b) ) the height of the coconut tree
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27. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.9 FORCES IN EQUILIBRIUM
Forces in When an object is in equilibrium,
Equilibrium
rd
Newton’s 3 Law
Examples( Label the forces acted on the objects)
Paste more picture
Paste more picture
Resultant A single force that
Force
Addition of Forces
Resultant force, F = +
Resultant force, F = +
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28. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Two forces acting at a point at an angle [Parallelogram method]
STEP 1 : Using ruler and protractor, draw STEP 3
the two forces F1 and F2 from a point. Draw the diagonal of the parallelogram. The
diagonal represent the resultant force, F in
magnitude and direction.
STEP 2
Complete the parallelogram
scale: 1 cm = ……
Resolution of Forces A force F can be resolved into components
which are perpendicular to each other:
(a) horizontal component , FX
(b) vertical component, FY
Inclined Plane
Fx = F cos θ
Component of weight parallel to the plane = mg sin θ
Component of weight normal to the plane = mg cos θ
Fy = F sin θ
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29. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Find the resultant force
(d) (e)
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30. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Lift
Stationary Lift Lift accelerate upward Lift accelerate downward
Resultant Force = Resultant Force = Resultant Force =
The reading of weighing The reading of weighing The reading of weighing
scale = scale = scale =
Pulley
1. Find the resultant force, F
2. Find the moving mass, m
3. Find the acceleration, a
4. Find string tension, T
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31. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.10 WORK, ENERGY, POWER & EFFICIENCY
Work Work done is
W = Fs W= ,F= s=
The SI unit of work is the joule, J
1 joule of work is done when
The displacement , s of the
The displacement, s of the object is in the direction of the force, F object is not in the direction of
the force, F
W = Fs
s F
W= F s
Example 1 Example 2 Example 3
A boy pushing his bicycle with a A girl is lifting up a 3 kg A man is pulling a crate of fish
force of 25 N through a distance flower pot steadily to a height along theW = (Fwith θ)force of
floor cos a s
of 3 m. of 0.4 m. 40 N through a distance of 6 m.
Calculate the work done by the What is the work done
boy. What is the work done by the in pulling the crate?
girl?
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32. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Concept D Formula & Unit
Power ef
The rate at which work is W
in P=
done,or
t
iti p = power, W = work / energy
on t = time
Energy  Energy is the capacity to do work.

Potential Energy m=
Gravitational potential energy is h=
the energy of an object due to
g=
its higher position in the
gravitational field. E=
Kinetic Energy Kinetic energy is the energy of an m=
object due to its motion. v=
E=
No force is applied on the object
in the direction of displacement
(the object moves because of its
own inertia)
A satellite orbiting in space.
There is no friction in space. No
force is acting in the direction of
A waiter is carrying a tray of movement of the satellite.
No work is done when: food. The direction of motion of
the object is perpendicular to
that of the applied force.
A waiter is carrying a tray of
food and2-32
walking

33. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Principle of Conservation of Energy can be changed from one form to another, but it cannot
Energy
be created or destroyed.
The energy can be transformed from
Example 4
A worker is pulling a wooden block of weight, W, with a force
of P along a frictionless plank at height of h. The distance
travelled by the block is x. Calculate the work done by the
worker to pull the block.
Example 5
A student of mass m is climbing
up a flight of stairs which has
the height of h. He takes t
seconds..
What is the power of the student?
Example 6 Example 7
A stone is thrown upward with
initial velocity of
-1
20 ms . What is the maximum
height which can be reached by
the stone?
A ball is released from point A of height
0.8 m so that it can roll along a curve frictionless track. What is the
velocity of the ball when it reaches point B?
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34. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Example 8
A trolley is released from rest at
point X along a frictionless track.
What is the velocity of the trolley at
point Y?
Example 9
A ball moves upwards along a
frictionless track of height 1.5 m
-1
with a velocity of 6 ms . What is
its velocity at point B?
Example 10
A boy of mass 20 kg sits at the top of a concrete slide of height 2.5 m. When he slides down the
slope, he does work to overcome friction of 140 J. What is his velocity at the end of the slope?
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35. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2.11 ELASTICITY
Elasticity A property of matter that enables an object
No external force is applied.
Molecules are at their equilibrium separation.
Intermolecular force is equal zero.
Compressing a solid causes its molecules
Repulsive intermolecular force
Stretching a solid
Its molecules are
Stretching a wire by an external
force:
When the external force is removed:
 The attractive intermolecular forces
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36. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Hooke’s Law The extension of a spring
F= k x where
F=
x=
k=
Force extension graph Based on the graph:
Relationship between F & x :
The gradient of the graph represent =
Area under the graph
2
= elastic potential energy = ½ F x = ½ k x
The elastic limit of a spring The maximum force that
If a force stretches a spring beyond its elastic limit, the spring
Force constant of the spring, k The force required to produce one unit of extension of the
spring.
k is a measurement of the stiffness of the spring
 The spring with a larger force constant is
 A spring with a smaller force constant is
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37. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Factors that effect elasticity
Factor Change in factor How does it affects the elasticity
Length Shorter spring
Longer spring
Diameter of spring wire Smaller diameter
Larger diameter
Diameter spring Smaller diameter
Larger diameter
Type of material Springs made of different materials
Elasticity changes according to the type of material
Arrangement of the spring
In series In parallel
The same load is applied to each spring. The load is shared equally among the springs.
Tension in each spring = W Extension of W
Tension in each spring =
each spring = x 2
Total extension = 2x x
Extension of each spring =
If n springs are used: The total 2
extension = n x If n springs are used:
x
The total extension =
n
Example 1
The original length of each
spring is 10 cm.
With a load of 10 g, the extension
of each spring is 2 cm.
What is the length of the spring
system for (a),
(b) and (c)?
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38. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Example 2 Using the diagram shown and the
information given about the weight of
Diagram below represent a the two objects,
50 cent coin and a leaf falling
in a vacuum container. compare the mass of the coin and the
The coin is heavier than the leaf, the time taken to fall, the position
leaf. of the coin and the leaf and
finally deduce the physical quantity
which causes the objects to fall.
coin leaf
Mass
of the taken to fall in a vacuum
Time coin
Position of the coin and the leaf
Coin and leaf of different mass reach the bottom of the container at the same time.
Coin and leaf fall down due to gravitational force. The magnitude of gravitational pull is constant.
It does not depend on the mass
Example 3 Diagram 10 shows a student trying to launch a water rocket.
You are required to give suggestion on how to design a water rocket for National
Competition. Based on your knowledge on forces, motion and properties of materials,
explain your suggestion based on the following aspect:
(i) material used
(ii) shape of the rocket
(iii) suitable angle to launch the rocket
(iv) volume of water in the rocket
(v) added structure for the motion of the rocket
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39. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
Water Rocket
Aspect Structure Explanation
Acceration of the Rocket Make from light material
Structure of the Rocket Aerodynamic
Upthrust force Fill with erated water or gasy
drink with water
Stability of the rocket during flight Use plasticine to make the head
of the rocket
Add fins at the back portion of
the rocket
Example 4
Diagram 4.1 shows a cradle with a baby in it is oscillating vertically. Diagram 4.2
shows another identical cradle with a heavier baby in it is oscillating vertically. It
is observed that the cradle with a heavier mass baby oscillates at a higher
frequency.
Design an experiment to test the hypothesis using spring, slotted weight and other
suitable apparatus.
4 (a) Inference : The extension of the spring depends
1
(b) Hypothesis : As the 1
c (i) Aim : To investigate the relationship between 1
Variables :
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40. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
c(ii) Manipulated variable : 1
Responding variable :
Constant variable : 1
c(iii) Apparatus : Metre rule, retort stand with clamp ,steel spring, slotted weight and 1
pin.
Set-up the apparatus
1
c(iv)
c (v) Method of controlling the manipulated variables : 1
1. Arrangement the apparatus as shown in the diagram.
2.Mark the initial
Method of measuring the responding variables :
Record 1
Extension of spring :
Measure the
Repeat the 1
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41. Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
c ( vi) Tabulate Results 1
Initial length , l0 = cm
Mass of the Slotted weight , m / g 40 80 120 160 200
Weight of slotted mass / N 0.4 0.8 1.2 1.6 2.0
Length of the spring ,l / cm
Extension of the spring x = l- l0
(vii) 1
Total 12
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