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Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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L I N E A R M O T I O N
Physical Quantity Definition, Quantity, Symbol and unit
Distance, s
Distance is the total path length travelled from one location to another.
Quantity: scalar SI unit: meter (m)
Displacement, s
(a) The distance in a specified direction.
(b) the distance between two locations measured along the shortest path
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,v
Speed is the rate of change of distance
Speed = time
ceDis tan
Quantity: scalar SI unit: m s -1
Velocity, v
Velocity is the rate of change of displacement.
Velocity = time
ntDisplaceme
Direction of velocity is the direction of displacement
Quantity : Vector SI unit: m s -1
Average speed
v =TotalTime
tTotalDis tan
Example: A car moves at an average speed / velocity of 20 ms
-1
On average, the car moves a distance/
displacement of 20 m in 1 second for the
whole journey. Average velocity
Displacement
TotalTimev
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-
uniform velocity if
(a) The direction of motion changes or the motion is not linear.
(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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Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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v ua
t
Unit: ms-2
Acceleration is positive
Acceleration is defined as the rate of change of velocity
Change in velocityAcceleration=
Time taken
Final velocity,v - Initial velocity,u =
Time taken,t
The velocity of an object increases from an initial velocity, u, to a higher final
velocity, v
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
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Speed Velocity
The rate of change of
distance
The rate of change of
displacement
Scalar quantity Vector quantity
It has magnitude but
no direction
It has both magnitude
and direction
SI unit : m s-1
SI unit : m s-1
Comparisons between distance and displacement Comparisons between speed and velocity
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.
3. A steady acceleration of 4 ms -2
= Speed goes up by 4 ms-1
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.
Distance Displacement
Total path length
travelled from
one location to
another
The distance between
two locations
measured along the
shortest path
connecting them in
specific direction
Scalar quantity Vector quantity
It has magnitude but no
direction
It has both magnitude
and direction
SI unit meter SI unit : meter
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Example 1
Every day Rahim walks from his house to the junction
which is 1.5km from his house.
Then he turns back and stops at warung Pak Din which
is 0.5 km from his house.
(a)What is Rahim’s displacement from his house
• when he reaches the junction. 1.5 km to the
right
• When he is at warung Pak Din. 0.5 km to the
left.
(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 2
Every morning Amirul walks to Ahmad’s house
which is situated 80 m to the east of Amirul’s house.
They then walk towards their school which is 60 m
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
tan θ = 60
80 =1.333 θ = 53.1º
(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 =140
15 60
m
s =0.156 in ms
-1
Velocity =100
15 60
m
s = 0.111 ms
-1
Example 3
Salim running in a race covers 60 m in 12 s.
(a) What is his speed in ms-1
Speed = s
m
12
60= 5 ms
-1
(b) If he takes 40 s to complete the race, what is his
distance covered?
distance covered = 40 s × 5 ms-1
= 200 m
Example 4
An aeroplane flies towards the north with
a 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.
(a)What is the average speed of the plane?
Average speed = (300 km hr -1
+
4 00 km hr -1
) / 2 = 350 km hr -1
(b)What is the average velocity of the plane?
Average velocity = 250 km hr -1
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Tan θ =
300
400 = 1.333 θ =
(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
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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
Elapsed time, t = (5 – 1) x 0.2 s = 0.8 s or
t = (50 – 10) ticks x 0.02 s = 0.8 s
Initial velocity, u =
final velocity, v =
acceleration, a =
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.
a = t
uv =
2
1060 ms
-2 = 25 ms
-2
THE EQUATIONS OF MOTION
2
2 2
1
2
2
v u at
s ut at
v u as
u = initial velocity
v = final velocity
t = time taken
s = displacement
a = constant acceleration
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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
2.2
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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GRAPH s versus t v versus t a versus t
Constant
acceleration
Constant
deceleration
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Example 1
Contoh 11
Based on the s – t graph above:
(a) Calculate the velocity at
(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:
(i) AB (ii) BC (iii) CD (i) constant velocity 5 ms
-1
(ii) at rest / 0 ms-1
(iii) constant velocity of 10 ms
-1in opposite
direction
(c)Find:
(i) total distance 50 m + 50 m = 100 m
(ii) total displacement 50 m + (- 50 m) = 0
(d) Calculate
(i) the average speed s
m
35
100= 2.86 ms
-1
(ii) the average velocity of the moving particle. 0
Example 2
(a) Calculate the acceleration at:
(i) JK (ii) KL (iii) LM
(i) 2 ms-2
(ii) -1 ms-2
(iii) 0 ms-1
(b) Describe the motion of the object at:
(i) JK (ii) KL (iii) LM
(i) constant acceleration of 2 ms
-2
(ii) constant deceleration of 1ms-2
(iii) (iii) zero acceleration or constant velocity
Calculate the total displacement.
Displacement = area under the graph = 100 m + 150 m + 100 m + 25 m = 375 m
(c) Calculate the average velocity.
Average velocity = 375 m / 40 s
= 9.375 ms-1
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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
and mass
The larger the mass, the larger the inertia
SITUATIONS INVOLVING INERTIA
SITUATION EXPLANATION
EEEEEEEEJNVJLKN
DNFLJKVNDFLKJNB
VJKL;DFN BLK;XC
NB[F
NDPnDSFJ[POJDE]O-
JBD]AOP[FKBOP[DF
LMB NOPGFMB
LKFGNKLB
FGNMNKL’ MCVL
BNM’CXLB
NFGNKEPLANATION
When the cardboard is pulled away quickly, the coin drops straight into
the glass.
The inertia of the coin maintains its state at rest.
The coin falls into the glass due to gravity.
Chilli sauce in the bottle can be easily poured out if the bottle is moved
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.
2.3
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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.
Ways to reduce the negative
effects of inertia
1. Safety in a car:
(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.
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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
respective velocities after the collision.
ƒ Momentum is conserved.
ƒ Kinetic energy is conserved.
Total energy is conserved.
ƒ The two objects combine and move together
with a common velocity after the collision.
ƒ Momentum is conserved.
ƒ Kinetic energy is not conserved.
ƒ Total energy is conserved.
Total Momentum Before = total momentum after
m1u
1 + m2u
2 = m1 v
1 + m2 v
2
Total Momentum Before = Total Momentum After
m1 u
1 + m
2 u
2 = ( m1 + m
2 ) v
Explosion
Before explosion both object stick together and at
rest. After collision, both object move at opposite
direction.
Total Momentum
before collision is
zero
Total Momentum after
collision :
m1v
1 + m2v
2
2.4
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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From the law of conservation of momentum:
Total Momentum = Total Momentum
Before collision after collision
0 = m1v
1 + m2v
2
m1v
1 = - m
2v
2
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.
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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 = mwv
w + msvs
- mwv
w = msvs
The magnitude of the momentum of water and
squid are equal but opposite direction.
This causes the quid to jet forward.
Example
Car A of mass 1000 kg moving at 20 ms -1
collides with a car B of mass 1200 kg moving at
10 m s -1
in same direction. If the car B is
shunted forwards at 15 m s -1
by the impact,
what is the velocity, v, of the car A immediately
after the crash?
1000 kg x 20 ms -1
+ 1200 kg x 10 ms -1
=
1000 kg x v + 1200 kg x 15 ms
-1
v= 14 ms -1
Example Before collision After collision
MA = 4 kg
MB
= 2 kg
UA = 10 ms
-1 r i g h t
UB = 8 ms
-1 l e f t V
B 4 ms-1
right
Calculate the value of VA .
[4 x 10 + 2 x (-8)]kgms
-1 =[ 4 x v
+ 2 x 4 ] kgms
-1
VA = 4 ms -1
right
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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 direction at 20 ms-1
. After the
collision, the two vehicles move together. What is
the velocity of both vehicles immediately after
collision?
1200 kg x 30 ms -1
+ 1000 kg x (-20 ms -1
)
= ( 1200 kg + 1000kg) v
v = 7.27 ms -1
to the right
Example A man fires a pistol which has a mass of 1.5 kg.
If the mass of the bullet is 10 g and it reaches a
velocity of 300 ms -1
after shooting, what is the
recoil velocity of the pistol?
0 = 1.5 kg x v + 0.01 kg x 300 ms
-1
v = -2 ms -1
Or
it recoiled with 2 ms -1
to the left
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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F O R C E
Balanced Force
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]
Example:
Weight, W = Lift, U Thrust, F = drag, G
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.
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
2.5
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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
but man A puts greater effort. So he
moves faster.
Both men exerted the same strength.
But man B moves faster than man A.
Inference The acceleration produced by an
object depends on the net force
applied to it.
The acceleration produced by an
object depends on the mass
Hypothesis The acceleration of the object
increases when the force applied
increases
The acceleration of the object
decreases when the mass of the
object increases
Variables:
Manipulated :
Responding :
Constant :
Force
Acceleration
Mass
Mass
Acceleration
Force
Apparatus and
Material
Ticker tape, elastic cords, ticker timer, trolleys, power supply, friction
compensated runway and meter ruler.
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Procedure :
- Controlling
manipulated
variables. -Controlling
responding
variables. -Repeating
experiment.
An elastic cord is hooked over the
trolley. The elastic cord is stretched
until the end of the trolley. The
trolley is pulled down the runway
with the elastic cord being kept
stretched by the same amount of
force
An elastic cord is hooked over a
trolley. The elastic cord is stretched
until the end of the trolley. The trolley
is pulled down the runway with the
elastic cord being kept stretched by
the same amount of force
Determine the acceleration by
analyzing the ticker tape.
Acceleration
Acceleration v u
at
Determine the acceleration by analyzing
the ticker tape.
Acceleration v u
at
Repeat the experiment by using two
, three, four and five elastic cords
Repeat the experiment by using two,
three, four and five trolleys.
Tabulation of
data
Force, F/No of
elastic cord
Acceleration, a/ ms-2
1
2
3
4
5
Mass, m/
no of
trolleys
Mass,
m/g
1/m,
g-1
Acceleration/
ms-2
1
2
3
4
5
Analysing
Result
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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1. What force is required to move a 2 kg
object with an acceleration of 3 m s-2
,
if
(a) the object is on a smooth surface?
(b) The object is on a surface where the
average force of friction acting on the
object is 2 N?
(a) force = 6 N
(b) net force = (6 – 2) N
= 4 N
2. Ali applies a force of 50 N to move a 10 kg
table at a constant velocity. What is the
frictional force acting on the table?
Answer: 50 N
3. A car of mass 1200 kg travelling at 20 ms -1
is brought to rest over a distance of 30 m.
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
4. Which of the following systems will
produce maximum acceleration? D
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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I M P U L S E A N D I M P U L S I V E F O R C E
Impulse The change of momentum mv - mu
Unit : kgms-1
or Ns
m = mass
u = initial velocity
v = final velocity
t = time Impulsive Force The rate of change of momentum in a
collision or explosion
Impulsive force =
change of momentum
time
mv mu
t
Unit = N Effect of time Impulsive force
is inversely
proportional to
time of contact
Longer period of time →Impulsive force decrease
Shorter period of time →Impulsive force increase
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.
2.6
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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
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 0.5 s 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?
Answer:
(a) Impulse = 60 kg x ( 6 ms-1
- 0 )
= 360 Ns
(b) Impulsive force = s
Ns
5.0
360
=7200 N
(c) He experienced a greater
Impulsive force of 7200 N and he
might injured his legs
(d) Increase the reaction time so as to
reduce impulsive force
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?
(a) Impulse = 1500N x 0.01 s
= 15 Ns
(b) velocity = kg
Ns
5.0
15 = 30 ms
-1
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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S A F E T Y V E H I C L E
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
bumper
Absorb the shock from the accident. Made from steel, aluminium, plastic or
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.
Crash resistant door
pillars
Anti-lock brake system
(ABS)
Traction control Front bumper
Windscreen
Air bags
Head rest
Crumple zones
2.7
Safety features in vehicles
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2 - 24
G R A V I T Y
Gravitational
Force
Objects fall because they are pulled towards the Earth by the force of gravity.
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
gravity, g Objects dropped under the influence of the pull of gravity with
constant acceleration.
This acceleration is known as the gravitational acceleration, g.
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-2
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.
g = m
F Its unit is N kg
-1.
2.8
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Gravitational field strength, g = 10 N kg-1
Acceleration due to gravity, g = 10 m s-2
The approximate value of g can therefore be written either as 10 m s-2
or as 10 N kg-1
.
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
between weight
&
mass
Mass Weight
The mass of an object is the
amount of matter in the object
The weight of an object is the force of
gravity acting on the object.
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
resistance.
The coin and the feather will fall
freely.
Only gravitational force
acted on the objects. Both will fall
at the same time.
At normal state: Both coin and feather
will fall because of gravitational force.
Air resistance effected by the surface area of
a fallen object.
The feather that has large area will have
more air resistance.
The coin will fall at first.
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2 - 26
(a) The two spheres are falling
with an acceleration.
The distance between two
successive images of the sphere
increases showing that the two
spheres are falling with increasing
velocity; falling with an
acceleration.
The two spheres are falling down with
the same acceleration
The two spheres are at the same level
at all times. Thus, a heavy object and
a light object fall with the same
gravitational acceleration
Gravitational acceleration is
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
(a) t = 2 s u = 0 g = 10 v = ?
v = u + gt = 0 + 10 x 2 = 20 ms-1
(b) s = ut + ½ at2 = 0 + ½ (10) 2
2 = 20 m
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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.
Newton’s 3rd
Law
Action is equal to reaction
Examples( Label the forces acted on the objects)
Paste more picture
Paste more picture
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
2.9
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Two forces acting at a point at an angle [Parallelogram method]
STEP 1 : Using ruler and protractor, draw
the two forces F1 and F2 from a point.
STEP 3
Draw the diagonal of the parallelogram. The
diagonal represent the resultant force, F in
magnitude and direction.
scale: 1 cm = ……
STEP 2
Complete the parallelogram
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
Fx = F cos θ
Fy = F sin θ
Inclined Plane
Component of weight parallel to the plane = mg sin θ
Component of weight normal to the plane = mg cos θ
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2 - 29
Find the resultant force
(d) (e)
17 N
5 N
FR
7N
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
2 - 30
Lift
Stationary Lift
Lift accelerate upward
Lift accelerate downward
Resultant Force = Resultant Force = Resultant Force =
The reading of weighing
scale =
The reading of weighing
scale =
The reading of weighing
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
10 = 7 a
a =10/ 7 ms-2
30 -2 = (4+3 )a
28 = 7a
a = 4 ms-2
4. Find string tension, T T- 3 (10) = 3 a
T = 30 + 3 (10/7)
=240 /7 N
30 – T = 3 (a)
T =30- 12
= 18 N
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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WORK, ENERGY, POWER & EFFICIENCY
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 object is in the direction of the force, F
The displacement , s of the
object is not in the direction of
the force, F W = Fs
s F
W = F s
W = (F cos θ) s
Example 1
A boy pushing his bicycle with a
force of 25 N through a distance
of 3 m.
Calculate the work done by the
boy. 75 Nm
Example 2
A girl is lifting up a 3 kg
flower pot steadily to a height
of 0.4 m. What is the work done by the
girl? 12 Nm
Example 3
A man is pulling a crate of fish
along the floor with a force of
40 N through a distance of 6 m.
What is the work done
in pulling the crate?
40 N cos 50º x 6 Nm
2.10
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Concept D
ef
in
iti
on
Formula & Unit
Power The rate at which work is done,
or the amount of work done per
second.
P = W
t
p = power, W = work / energy
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
the energy of an object due to
its higher position in the
gravitational field.
m = mass
h = height
g = gravitational acceleration E = mgh
Kinetic Energy
Kinetic energy is the energy of an
object due to its motion.
m = mass
v = velocity
E = ½ mv2
No work is done when:
The object is stationary.
A student carrying his bag while
waiting at the bus stop
The direction of motion of the
object is perpendicular to that of
the applied force.
A waiter is carrying a tray of
food and walking
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
movement of the satellite.
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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Principle of Conservation of
Energy
Energy can be changed from one form to another, but it cannot
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?
[ t
mgh
Example 6
A stone is thrown upward with
initial velocity of 20 ms-1
.
What is the maximum height which can be reached by the stone?
[ 10m ]
Example 7
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
]
Physics Module Form 4 Chapter 2 – Forces & Motion GCKL 2011
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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
with a velocity of 6 ms-1
. 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
]