2.0 forces and motion

Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011

2.1 L I N E A R M O T I O N

Physical Quantity Definition, Quantity, Symbol and unit

Distance is the total path length travelled from one location to another.
Distance, s Quantity: scalar SI unit: meter (m)

(a) The distance in a specified direction.
(b) the distance between two locations measured along the shortest path
Displacement, s connecting them in a specific direction.
(c) The distance of its final position from its initial position in a
specified direction.
Quantity: vector SI unit: meter (m)

Speed is the rate of change of distance
Speed,v
Dis tan ce
Speed =
time

Quantity: scalar SI unit: m s -1
Velocity is the rate of change of displacement.
Velocity, v
Displacement
Velocity = time

Direction of velocity is the direction of displacement
Quantity : Vector SI unit: m s -1

Average speed TotalDis tan t Example: A car moves at an average
v= speed / velocity of 20 ms -1
TotalTime
On average, the car moves a distance/
displacement of 20 m in 1 second for the
whole journey.
Average velocity Displacement
v
TotalTime

Uniform speed Speed that remains the same in magnitude without considering its direction

Uniform velocity Velocity that remains the same in magnitude and direction

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 increases, the object is said to be accelerating.

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v u Acceleration is defined as the rate of change of velocity
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.

The velocity of an object decreases from an initial velocity, u, to a lower final
velocity, v.

Zero acceleration An object moving at a constants velocity, that is, the magnitude and direction of
its velocity remain unchanged – is not accelerating

Constant acceleration Velocity increases at a uniform rate.
When a car moves at a constant or uniform acceleration of 5 ms -2, its velocity
increases by 5 ms -1 for every second that the car is in motion.

1. Constant = uniform
2. increasing velocity = acceleration
3. decreasing velocity = deceleration
4. zero velocity = object at stationary / at rest
5. negative velocity = object moves in opposite direction
6. zero acceleration = constant velocity
7. negative acceleration = deceleration

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Comparisons between distance and displacement Comparisons between speed and velocity

Distance Displacement
Total path length The distance between Speed Velocity
travelled from two locations The rate of change of The rate of change of
one location to measured along the distance displacement
another shortest path Scalar quantity Vector quantity
connecting them in
specific direction It has magnitude but It has both magnitude
no direction and direction

-1 -1
Scalar quantity Vector quantity SI unit : m s SI unit : m s

It has magnitude but no It has both magnitude
direction and direction

SI unit meter SI unit : meter

Fill in the blanks:

1. A steady speed of 10 ms -1 = A distance of 10 m is travelled every second.
2. A steady velocity of -10 ms -1 = A displacement of 10 m is travelled every 1 second to the left.
-1
3. A steady acceleration of 4 ms -2 = Speed goes up by 4 ms every 1 second.
4. A steady deceleration of 4 ms -2 = speed goes down by 4 ms-1 every 1 second
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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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 They then walk towards their school which is 60 m
is 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?
Distance = (80 +60 ) m = 140 m
Displacement = 100 m
80
tan θ = =1.333 θ = 53.1º
60
(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?

140m
(a)What is Rahim’s displacement from his house Speed = =0.156 in ms-1
• when he reaches the junction. 1.5 km to the 15  60 s
right
• When he is at warung Pak Din. 0.5 km to the 100m
Velocity = = 0.111 ms-1
left. 15  60 s
(b)After breakfast, Rahim walks back to his house.
w hen he reaches home,
(i) what is the total distance travelled by
Rahim?
(1.5 + 1.5 + 0.5+0.5 ) km = 4.0 km
(ii) what is Rahim’s total displacement from
his house?
1.5 +( -1.5) +(- 0.5 )+0.5 km = 0 km

Example 3 Example 4
Salim running in a race covers 60 m in 12 s.
An aeroplane flies towards the north with
(a) What is his speed in ms-1
a velocity 300 km hr -1 in one hour.
60m Then, the plane moves to the east with
Speed = = 5 ms-1
12 s 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? Average speed = (300 km hr -1 +
4 00 km hr -1 ) / 2 = 350 km hr -1
distance covered = 40 s × 5 ms-1 = 200 m (b)What is the average velocity of the plane?

Average velocity = 250 km hr -1

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400
Tan θ = = 1.333 θ =
300

(c)What is the difference between average speed and
average velocity of the plane?

Average speed is a scalar quantity.
Average velocity is a vector quantity
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?

The speed of both cars are the same but the velocity of
both cars are different with opposite direction

A ticker timer

 Use: 12 V a.c. power supply
 1 tick = time interval between two dots.
 The time taken to make 50 ticks on the ticker tape is 1 second. Hence, the time interval between 2
consecutive dots is 1/50 = 0.02 s.
 1 tick = 0.02 s

2-5

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

Constant velocity
– slow moving

Constant velocity
– fast moving

 Distance between the dots increases uniformly
 the velocity is of the object is increasing uniformly
 The object is moving at a uniform / constant
acceleration.

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- Distance between the dots decrease uniformly

- The velocity of the object is decreasing Uniformly

- The object is experiencing uniform / constant

decceleration

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.

Time = 1/50 s = 0.02 s

(b) What is the time for one strips.

0.02 s × 10 = 0.2 s

(c) What is the initial velocity

2 cm / 0.2 s = 10 ms-1

(d) What is the final velocity.

12 cm / 0.2 s = 60 ms-1

(e) What is the time interval to change from initial
velocity to final velocity?

( 11 - 1) × 0.2 s = 2 s

(f) What is the acceleration of the object.

vu 60  10 -2
a= = ms = 25 ms-2
t 2

THE EQUATIONS OF MOTION

v  u  at u = initial velocity
1 v = final velocity
s  ut  at 2 t = time taken
2 s = displacement
v  u  2as
2 2
a = constant acceleration

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2.2 M O T I O N G R A P H S

DISPLACEMENT – TIME GRAPH Velocity is obtained from the gradient of the graph.

A – B : gradient of the graph is positive and constant
velocity is constant.
B – C : gradient of the graph = 0
the velocity = 0, object is at rest.
C – D : gradient of the graph negative and constant.
The velocity is negative and object moves
in the opposite direction.

VELOCITY-TIME GRAPH Area below graph Distance / displacement

Positive gradient Constant Acceleration
(A – B)
Negative gradient Constant Deceleration
(C – D)
Zero gradient Constant velocity /
zero acceleration
(B – C)

GRAPH s versus t v versus t a versus t
Zero
velocity

Negative
constant
velocity

Positive
Constant
velocity

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GRAPH s versus t v versus t a versus t
Constant
acceleration

Constant
deceleration

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Example 1
Example 2
Contoh 11

(a) Calculate the acceleration at:
Based on the s – t graph above: (i) JK (ii) KL (iii) LM
(a) Calculate the velocity at (i) 2 ms-2 (ii) -1 ms-2 (iii) 0 ms-1
(i) AB (ii) BC (iii) CD
(i) 5 ms-1 (ii) 0 ms-1 (iii) - 10 ms-1
(b) Describe the motion of the object at:
(b) Describe the motion of the object at: (i) JK (ii) KL (iii) LM
(i) AB (ii) BC (iii) CD
(i) constant velocity 5 ms-1 (i) constant acceleration of 2 ms-2
(ii) at rest / 0 ms-1 (ii) constant deceleration of 1ms-2
(iii) constant velocity of 10 ms-1in opposite (iii) (iii) zero acceleration or constant velocity
direction
Calculate the total displacement.
(c)Find: Displacement = area under the graph
(i) total distance 50 m + 50 m = 100 m = 100 m + 150 m + 100 m + 25 m
= 375 m
(ii) total displacement 50 m + (- 50 m) = 0
(c) Calculate the average velocity.
Average velocity = 375 m / 40 s
(d) Calculate
= 9.375 ms-1
100m
(i) the average speed = 2.86 ms-1
35s
(ii) the average velocity of the moving
particle.
0

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2.3 I N E R T I A
Inertia The inertia of an object is the tendency of the object to remain at rest or, if
moving, to continue its motion.
Newton’s first law Every object continues in its state of rest or of uniform motion unless
it is acted upon by an external force.

Relation between inertia The larger the mass, the larger the inertia
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
The inertia of the coin maintains its NB[Fat rest.
state
NDPnDSFJ[POJDE]O-
The coin falls into the glass due to gravity.
JBD]AOP[FKBOP[DF
LMB NOPGFMB
LKFGNKLB
FGNMNKL’ MCVL
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, the sauce continues in its state of
motion due to the effect of its inertia.

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, the inertia in the passengers made them
maintain their state of motion. Thus when the car stop, the passengers
moved forward.

A boy runs away from a cow in a zig- zag motion. The cow has a large inertia
making it difficult to change direction.

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 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.
1. Safety in a car:
Ways to reduce the negative (a)Safety belt secure the driver to their seats.
effects of inertia 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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2.4 M O M E N T U M

Definition Momentum = Mass x velocity = mv
SI unit: kg ms-1

Principle of Conservation of Momentum In the absence of an external force, the total
momentum of a system remains unchanged.

Elastic Collision Inelastic collision

ƒ Both objects move independently at their ƒ The two objects combine and move together
respective velocities after the collision. with a common velocity after the collision.
ƒ Momentum is conserved. ƒ Momentum is conserved.
ƒ Kinetic energy is conserved. ƒ Kinetic energy is not conserved.
Total energy is conserved. ƒ Total energy is conserved.

Total Momentum Before = total momentum after Total Momentum Before = Total Momentum After
m1u1 + m2u2 = m1 v1 + m2 v2 m1 u1 + m2 u2 = ( m1 + m2 ) v

Explosion
Before explosion both object stick together and at
rest. After collision, both object move at opposite
direction.

Total Momentum Total Momentum after
before collision is collision :
zero m1v1 + m2v2

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From the law of conservation of momentum:
Total Momentum = Total Momentum
Before collision after collision
0 = m1v1 + m2v2
m1v1 = - m2v2

Negative sign means opposite direction

EXAMPLES OF EXPLOSION (Principle Of Conservation Of Momentum)
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, an equal but opposite
momentum is produced to recoil the riffle
backward.

Application in the jet engine:
A high-speed hot gases are ejected from the back
with high momentum.
This produces an equal and opposite
momentum to propel the jet plane forward.

The launching of rocket
Mixture of hydrogen and oxygen fuels burn
explosively in the combustion chamber.
Jets of hot gases are expelled at very high
speed through the exhaust.
These high speed hot gases produce a large
amount of momentum downward.
By conservation of momentum, an equal but
opposite momentum is produced and acted on
the rocket, propelling the rocket upwards.

2- 14


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.

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
Car A of mass 1000 kg moving at 20 ms -1
collides with a car B of mass 1200 kg moving at Before collision After collision
10 m s -1 in same direction. If the car B is MA = 4 kg
shunted forwards at 15 m s -1 by the impact, MB = 2 kg
what is the velocity, v, of the car A immediately UA = 10 ms -1 r i g h t
UB = 8 ms -1 l e f t VB 4 ms-1 right
after the crash?
1000 kg x 20 ms -1 + 1200 kg x 10 ms -1 =
Calculate the value of VA .
1000 kg x v + 1200 kg x 15 ms -1
[4 x 10 + 2 x (-8)]kgms -1 =[ 4 x v + 2 x 4 ] kgms -1
v= 14 ms -1
VA = 4 ms -1 right

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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 A man fires a pistol which has a mass of 1.5 kg.
the opposite direction at 20 ms-1. After the If the mass of the bullet is 10 g and it reaches a
collision, the two vehicles move together. What is velocity of 300 ms -1 after shooting, what is the
the velocity of both vehicles immediately after recoil velocity of the pistol?
collision?
0 = 1.5 kg x v + 0.01 kg x 300 ms -1
1200 kg x 30 ms
-1
+ 1000 kg x (-20 ms -1) v = -2 ms -1
= ( 1200 kg + 1000kg) v
v = 7.27 ms -1 to the right Or
it recoiled with 2 ms -1 to the left

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2.5 F O R C E

Balanced Force Example:
When the forces acting on an object are
balanced, they cancel each other out. The
net force is zero.

Effect : the object is at rest
[velocity = 0]
or
moves at constant velocity
[ a = 0]

Unbalanced Force/ Resultant Force When the forces acting on an object are not balanced,
there must be a net force acting on it.
The net force is known as the unbalanced force or
the resultant force.
Weight, W = Lift, U Thrust, F = drag, G

Effect : Can cause a body to
- change it state at rest (an object will
accelerate
- change it state of motion (a moving object
will decelerate or change its direction)

Newton’s Second Law of Motion The acceleration produced by a force on an object is
directly proportional to the magnitude of the net force
applied and is inversely proportional to the mass of the
object. The direction of the acceleration is the same as
that of the net force.

Force = Mass x Acceleration

F = ma

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Experiment to Find The Relationship between Force, Mass & Acceleration

Relationship between a& F a& 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 The acceleration produced by an
object depends on the net force object depends on the mass
applied to it.

Hypothesis The acceleration of the object The acceleration of the object
increases when the force applied decreases when the mass of the
increases object increases

Variables:
Manipulated : Force Mass
Responding :
Acceleration Acceleration

Constant : Mass Force
Apparatus and Ticker tape, elastic cords, ticker timer, trolleys, power supply, friction
Material compensated runway and meter ruler.

2 - 18


Procedure : An elastic cord is hooked over the An elastic cord is hooked over a
- Controlling trolley. The elastic cord is stretched trolley. The elastic cord is stretched
manipulated until the end of the trolley. The until the end of the trolley. The trolley
variables. trolley is pulled down the runway is pulled down the runway with the
with the elastic cord being kept elastic cord being kept stretched by
stretched by the same amount of the same amount of force
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
-Repeating
Repeat the experiment by using two Repeat the experiment by using two,
experiment.
, three, four and five elastic cords three, four and five trolleys.

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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1. What force is required to move a 2 kg 2. Ali applies a force of 50 N to move a 10 kg
object with an acceleration of 3 m s-2, table at a constant velocity. What is the
if frictional force acting on the table?
(a) the object is on a smooth surface?
(b) The object is on a surface where the Answer: 50 N
average force of friction acting on the
object is 2 N?

(a) force = 6 N
(b) net force = (6 – 2) N
= 4N

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? D
Find
(a) the average deceleration,
(b) the average braking force.

(a) u = 20 ms -1 v = 0 s = 30 m a=?
a = - 6.67 ms-2
(b) force = 1200 x 6.67 N
= 8000 N

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2.6 IMPULSE AND IMPULSIVE FORCE

Impulse The change of momentum mv - mu m = mass
Unit : kgms-1 or Ns u = initial velocity
v = final velocity
Impulsive Force The rate of change of momentum in a t = time
collision or explosion
Impulsive force =
change of momentum mv  mu

time t

Effect of time Impulsive force Longer period of time →Impulsive force decrease
Unit = N
is inversely
proportional to Shorter period of time →Impulsive force increase
time of contact
Situations for Reducing Impulsive Force in Sports

Situations Explanation

Thick mattress with soft surfaces are used in events such as high jump
so that the time interval of impact on landing is extended, thus
reducing the impulsive force. This can prevent injuries to the
participants.

Goal keepers will wear gloves to increase the collision time. This
will reduce the impulsive force.

A high jumper will bend his legs upon landing. This is to increase the
time of impact in order to reduce the impulsive force acting on his legs.
This will reduce the chance of getting serious injury.

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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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
large impulsive force on the wooden slab.

A massive hammer head moving at a fast speed is brought to rest
upon hitting the nail within a short time interval.
The large change in momentum within a short time interval
produces a large impulsive force which drives the nail into the
wood.
A football must have enough air pressure in it so the contact time is
short. The impulsive force acted on the ball will be bigger and the
ball will move faster and further.

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 Answer:
(a) Impulse = 60 kg x ( 6 ms-1 - 0 )
A 60 kg resident jumps from the first floor of a burning house.
= 360 Ns
His velocity just before landing on the ground is 6 ms-1. 360 Ns
(a) Calculate the impulse when his legs hit the ground.
(b) Impulsive force = 0.5s
(b) What is the impulsive force on the resident’s legs if he =7200 N
bends upon landing and takes 0.5 s to stop? (c) He experienced a greater
(c) What is the impulsive force on the resident’s legs if Impulsive force of 7200 N and he
might injured his legs
he does not bend and stops in 0.05 s?
(d) Increase the reaction time so as to
(d) What is the advantage of bending his legs upon landing? reduce impulsive force

Example 2
Rooney kicks a ball with a force of 1500 N. The time of (a) Impulse = 1500N x 0.01 s
contact of his boot with the ball is 0.01 s. What is the impulse = 15 Ns
delivered to the ball? If the mass of the ball is 0.5 kg, what is
the velocity of the ball? 15 Ns
(b) velocity = = 30 ms-1
0.5kg

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2.7 S A F E T Y V E H I C L E

Safety features in vehicles
Head rest

Crash resistant door Windscreen
pillars

Crumple zones
Anti-lock brake system
(ABS)

Traction control Front bumper
Air bags

Component Function
Headrest To reduce the inertia effect of the driver’s head.
Air bag Absorbing impact by increasing the amount of time the driver’s head to come to the
steering. So that the impulsive force can be reduce

Windscreen To protect the driver (shattered proof)

Crumple zone Can be compressed during accident. So it can increase the amount of time the car
takes to come to a complete stop. So it can reduce the impulsive force.

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 Prevents the collapse of the front and back of the car into the passenger
compartment. Also gives good protection from a side impact

Seat belt To reduce the effect of inertia by avoiding the driver from thrown forward.

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2.8 G R A V I T Y

Gravitational Objects fall because they are pulled towards the Earth by the force of gravity.
Force

This force is known as the pull of gravity or the earth’s gravitational force.

The earth’s gravitational force tends to pull everything towards its centre.

Free fall  An object is falling freely when it is falling under the force of gravity
only.
 A piece of paper does not fall freely because its fall is affected by air
resistance.
 An object falls freely only in vacuum. The absence of air means
there is no air resistance to oppose the motion of the object.
 In vacuum, both light and heavy objects fall freely.
 They fall with the same acceleration i.e. The acceleration
due to gravity, g.

Acceleration due to  Objects dropped under the influence of the pull of gravity with
gravity, g constant acceleration.
 This acceleration is known as the gravitational acceleration, g.
-2
 The standard value of the gravitational acceleration, g is 9.81 m s .
-2
The value of g is often taken to be 10 m s for simplicity.
 The magnitude of the acceleration due to gravity depends on the
strength of the gravitational field.

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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-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.

2 - 25


(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) t = 2 s u = 0 g = 10 v=?
A coconut takes 2.0 s to fall to the ground. What v = u + gt = 0 + 10 x 2 = 20 ms-1
is
(b) s = ut + ½ at2 = 0 + ½ (10) 22 = 20 m
(a) its speed when it strikes the ground
(b) ) the height of the coconut tree

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2.9 F O R C E S I N E Q U I L I B R I U M

Forces in Equilibrium When an object is in equilibrium, the resultant force acting on it is zero.
The object will either be
1. at rest
2. move with constant velocity.

rd Action is equal to reaction
Newton’s 3 Law

Examples( Label the forces acted on the objects)

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Resultant Force A single force that represents the combined effect of two of more forces
in magnitude and direction.

Addition of Forces

Resultant force, F = F1 + F2

Resultant force, F = F1 + - F2

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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 = ……

A force F can be resolved into components
Resolution of Forces 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
Fy = F sin θ Component of weight normal to the plane = mg cos θ

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Find the resultant force

17 N

5N

(d) (e)

7N

FR

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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 40 -30 = 10 N 30-2 = 28 N

2. Find the moving mass, m 4 + 3 = 7 kg 3+ 4 = 4 kg

3. Find the acceleration, a 40 -30 = (3+4)a 30 -2 = (4+3 )a
10 =7a 28 = 7a
a =10/ 7 ms-2 a = 4 ms-2

4. Find string tension, T T- 3 (10) = 3 a 30 – T = 3 (a)
T = 30 + 3 (10/7) T =30- 12
=240 /7 N = 18 N

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2.10 W O R K , E N E R G Y , P O W E R & E F FI C I E N CY

Work Work done is the product of an applied force and the
displacement of an object in the direction of the applied
force

W = Fs W = work, F = force s = displacement

The SI unit of work is the joule, J
1 joule of work is done when a force of 1 N moves an
object 1 m in the direction of the force

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. 75 Nm What is the work done by the in pulling the crate?
girl? 12 Nm 40 N cos 50º x 6 Nm

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Concept D Formula & Unit
Power ef
The rate at which work is done, W
in P=
or the amount of work done per
t
second. iti p = power, W = work / energy
on t = time
Energy  Energy is the capacity to do work.
 An object that can do work has energy
 Work is done because a force is applied and the objects move.
This is accompanied by the transfer of energy from one object
to another object.
 Therefore, when work is done, energy is transferred from one
object to another.
 The work done is equal to the amount of energy
transferred.

Potential Energy Gravitational potential energy is m = mass
the energy of an object due to
h = height
its higher position in the
gravitational field. g = gravitational acceleration

E = mgh

Kinetic Energy Kinetic energy is the energy of an m = mass
object due to its motion. v = velocity

2
E = ½ mv

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
No work is done when: The direction of motion of the
object is perpendicular to that of movement of the satellite.
The object is stationary.
the applied force.
A student carrying his bag while
waiting at the bus stop

A waiter is 32
2 - carrying a tray of
food and walking


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 one form to another, total
energy in a system is constant.
Total energy before = total energy after

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.
[Px = Wh]

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?
mgh
[
t

Example 6 Example 7

A stone is thrown upward with
-1
initial velocity of 20 ms .
What is the maximum height
which can be reached by the
stone?
[ 10m ]

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?
[4 ms-1]

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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?
[ v2 = 30( ms-1)2]
[ v = 5.48 ms-1]

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?

[v2 = 30( ms-1)2
v = 5.48 ms-1]

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?

[6 ms-1]

2 - 34

2.0 forces and motion

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