Standard content Learning standards

2.1 Linear motion • Describe the types of linear motion of objects:

• stationary
• uniform velocity
• non-uniform velocity
• Determine:
• distance and displacement
• speed and velocity
• acceleration / deceleration
• Solve the problem of linear motion using kinematic equations.

2.2 Linear Motion •

Interpreting the type of movement of the graph:
Graphs
• displacement-time
• velocity-time
• acceleration-time
• Analyzing the displacement-time graph to determine the distance,
displacement and velocity.
• Analyzing the velocity-time graph to determine the distance, displacement,
velocity and acceleration.
• Interpret and sketch:
• displacement-time graph to velocity-time graph and vice versa
• velocity-time graph to graph acceleration-time and vice versa
• Solve problems involving linear motion graphs.

2.3 Free Fall Motion • Explain with examples free fall motion and gravitational acceleration
• Experiment to determine the value of gravitational acceleration.

• Solve problems involving gravitational acceleration for any object in free fall.

2.4 Inertia • Explain the concept of inertia through example.

• Experiment to determine the relationship between inertia and mass.

• Justifying the effects of inertia in daily life.

2.5 Momentum • Explains momentum, p as a product of the mass, m, and velocity, v.

• Applying the Principle of Conservation of Momentum in collision and
explosion.
2.6 Force • Defines force as the rate of change of momentum.
• Solve problems involving the equation F = ma.

2.7 Impulse and • Communicate to explain impulse and impulsive force.

Impulsive Force • Solve problems involving impulse and impulsive force.
2.8 Weight • Describe weight as the gravitational force that acts on an object, W = mg.

1
2.1 Linear motion

This work is licensed under the Creative Commons Attribution NonCommercial-NoDerivatives 4.0
International License. Most of the image, vector or diagram in this module are either original
content or available from Freepik.com
• Linear motion: Motion in a straight line.
Distance and Displacement

Distance Displacement

Shortest distance between the initial and final

Length of the route traveled by an object position in a specific direction

Same value as shortest distance between initial

Depending on the route taken objects and final position

Scalar quantity Vector quantity

• lengths of Route 1 and 3 are distances

A B
• length of Route 2 is the displacement Route 2

Example
10 m 7m
X Y Z
a. distance b. displacement
Irfan walks from point X to Z through Y. After
10 + 7 + 7 = 24 m X to Y = 10 m
that, she walks back to Y. Determine distance and
displacement.
Speed and Velocity

Speed, v ( m s -1) V e l o c i t y , v ( m s -1)

Rate of change of distance Rate of change of displacement

2
 2.1 Linear motion

distance displacement
velocity =
speed = time time

Scalar quantity Vector quantity

Example A

Farah runs from A to C by passing through B. If the travelling

time is 60 s, calculate her speed and velocity.

130 + 100 130 m

a. speed = 150 m
60

= 3.83 𝑚𝑚𝑚𝑚 −1

150
b. velocity =
60
C B
= 2.5 𝑚𝑚𝑚𝑚 −1 (𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝐴𝐴𝐴𝐴 )
100 m
Acceleration

• Acceleration: Rate of change of velocity a : acceleration, ms-2

𝑣𝑣 − 𝑢𝑢 v : final velocity, ms-1
• Deceleration: object slows down 𝑎𝑎 = : initial velocity, ms -1
u
• At a constant velocity, acceleration is zero
𝑡𝑡 t : time, s

Example

A car accelerates from 20ms-1 to 36 ms-1 in 8 seconds. Calculate the acceleration .

36 − 20
𝑎𝑎 =
8
= 2 𝑚𝑚𝑚𝑚 −2

Exercise
15 m
1. Wulan walks to 15 m the right and then 10 m back towards her
X Y
original position . If she takes 15 s for the whole journey, calculate
her speed and velocity.
A

10 m
2. Nanda walks in a circle. If she starts at point
A and completes the path and back to A in 2
minutes, calculate her speed and velocity.

3
 2.1 Linear motion

Ticker tape

• Used to analyze linear motion of object

• 1 tick – time interval between two consecutive points
3. Yussup walks from A to C through B. A
Calculate his speed and velocity in SI unit if
the time taken is 20 minutes.

5 km

B C
3 km
4. A car brakes to a stop in 5 s. Calculate 5. An object moves with constant acceleration of
deceleration of the car if the initial speed is 4 ms-2. What is the speed after 15 s if its initial
from 30ms-1. speed is 3 ms-1?
1 tick = 0.02 s
Ticker tape with constant speed

• The further the distance between two consecutive points, the faster the speed

Ticker tape A
• Distance between two consecutive points in A is the same

• Speed is constant, zero acceleration

• Distance between two consecutive points in B is the same
• Speed is constant, zero acceleration
Ticker tape B
• Distance between two consecutive points in A is further than B
• Speed A > speed B

Ticker tape with varying speed

• Read arrow from tip to tail 2 tail tip 1

Ticker tape C Direction of motion

• Distance between two consecutive points in C

increase
• Speed increase, acceleration

4
2.1 Linear motion

Calculating acceleration of ticker tape

0.1
Direction of motion 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
1 tick 0.02
= 5 𝑐𝑐𝑐𝑐 𝑠𝑠 −1

0.5
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
0.02
0.5 cm 0.1 cm
= 25 𝑐𝑐𝑐𝑐 𝑠𝑠 −1
Total distance for initial speed = 0.1 cm
Ticker tape D Direction of motion
• Speed decrease, deceleration

Calculating speed of ticker tape

1 tick

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝑡𝑡𝑖𝑖𝑖𝑖𝑖𝑖
2 cm 2
=
Total distance = 2 cm 0.16
Total time = number of tick x 0.02 s
= 8 x 0.02 s = 12.5 cms-1
= 0.16 s
• Distance between two consecutive points in D decrease
Total distance for final speed = 0.5 cm
Total time = (tick – 1) x 0.02 s
= (6 – 1) x 0.02 s

= 0.1 s = 200 𝑐𝑐𝑐𝑐 𝑠𝑠−2

5
2.1 Linear motion

Displacement, cm Steps to find acceleration:

Tape length
1. Calculate initial speed, u =
Tick x 0.02

Tape length
2. Calculate final speed, v =
Tick x 0.02

3. Find time taken, t = (tape – 1) x time per tape

4. Calculate acceleration, a =
Time, s

1 2 3 4
4 24 240 − 40
𝑢𝑢 = 𝑣𝑣 = 𝑎𝑎 =
5 × 0.02 5 × 0.02 t = (6 – 1) x 0.1 0.5
= 0.5 s
= 40 𝑐𝑐𝑐𝑐 𝑠𝑠−1 = 240 𝑐𝑐𝑐𝑐 𝑠𝑠−1 = 400 𝑐𝑐𝑐𝑐 𝑠𝑠 −2

6
Exercise
1. Describe the motion represented by the ticker tape below.

Direction of motion Direction of motion

a. b.

2. Calculate the speed of the ticker tape below.

a. b.
2 cm 2 cm

3. Calculate the acceleration of the ticker tape below.

Direction of motion

0.4 cm 1.2 cm

4. Calculate the acceleration of the ticker tape below.

Displacement, cm

2.4

2.0

1.8

1.5
1.2

Time, s

7
2.1 Linear motion
Linear motion equation

s u v a t
Displacement Initial speed Final speed Acceleration Time

Equation1 Equation2 Equation3 Equation4

1 1
𝑣𝑣 = 𝑢𝑢 + 𝑎𝑎𝑎𝑎 𝑠𝑠 = 𝑢𝑢 + 𝑣𝑣 𝑡𝑡 𝑠𝑠 = 𝑢𝑢𝑢𝑢 + 𝑎𝑎𝑡𝑡 2 𝑣𝑣 2 = 𝑢𝑢2 + 2𝑎𝑎𝑎𝑎
2 2

use when there is no s use when there is no a use when there is no v use when there is no t

Example
u a v t
-1 -2
A car accelerates from 20 ms with acceleration 2 ms . What is the velocity after 8 seconds?

s u v a t use equation 1 𝑣𝑣 = 𝑢𝑢 + 𝑎𝑎𝑎𝑎

Exercise

1. A rocket accelerates from 1500ms-1 to 4000 ms-1 after moving 70 km. Calculate the time taken.

2. Wulan is driving a car. After hitting the brake, the speed decreases from 5 ms-1 to 4 ms-1 within a
distance of 30 m. Calculate deceleration experienced by the car.

3. Siti reaches maximum velocity from rest in 7 seconds. Distance travelled during the time interval is 80
m. Calculate her acceleration.

4. From rest, Syahriana starts to run with increasing speed. If her acceleration is uniform at 0.02 ms-2,
calculate her speed after 1 minute.

8
2.2 Linear Motion Graphs
Displacement - time graph

• Gradient of graph: velocity 5−0

𝑚𝑚 =
2−0
• Example: Calculate velocity from graph.
v = 2.5 𝑚𝑚𝑚𝑚 −1

d Graph 1 d Graph 2 d Graph 3 d Graph 4

t t t t

gradient : ↑↓=0 gradient : ↑↓=0 gradient : ↑↓=0 gradient : ↑↓=0

velocity: ↑↓=0 velocity: ↑↓=0 velocity: ↑↓=0 velocity: ↑↓=0

Graph analysis

A to B (move 150 m to the right)

B to C
(at rest for 10 s)
E to F (move 150 m to the left) C to D (move 150 m to the left)
Initial
position
D to E (at rest for 15 s)

Displacement, m
Turning back
B C
150 (opposite direction)
100 Moving Into the opposite
50 A side, passes the initial point
D E
0 Time, s
10 20 30 40 50 60 70
- 50
- 100
- 150 F

From A to B From B to C From C to D From D to E From E to F

Constant velocity Object at rest Constant velocity Object at rest at Constant velocity
initial position
150 − 0 150 − 0 −150 − 0
𝑚𝑚 = 𝑚𝑚 = 𝑚𝑚 =
15 − 0 0−5 65 − 45
v = 10 𝑚𝑚𝑚𝑚 −1 v = −30 𝑚𝑚𝑚𝑚 −1 v = −7.5 𝑚𝑚𝑚𝑚 −1
10 ms-1 to the right 30 ms-1 to the left 7.5 ms-1 to the left

9
 of initial position
2.2 Linear Motion
acceleration: ↑↓=0 Graphs
acceleration: ↑↓=0 acceleration: ↑↓=0 acceleration: ↑↓=0

Velocity - time graph

Graph analysis
5−0
• Gradient of graph: acceleration Gradient = 2−0
Gradient = 2.5
• Velocity,
Area under -1
msthe graph: displacement Velocity decrease but 𝑎𝑎 = 2.5 𝑚𝑚/𝑠𝑠 2
B C moving in same direction
• 150
Example:
1
100
Find acceleration and distance from graph. Velocity increase but moving Area = 2 5 2
50 A in opposite direction Area = Distance:
5
D E d = 5 𝑚𝑚Area ABCD + EF
0 Time, s
10 20 30 40 50 60 70 Displacement:
- 50 Area ABCD - EF
v- 100 Graph 1 v Graph 2 v Graph 3 v Graph 4
- 150 F

t t t t
gradient
From :A↑↓=0
to B gradient
From B to C : ↑↓=0 From C togradient
D : ↑↓=0
From D to E gradient : ↑↓=0
From E to F

Velocity increase Velocity constant Velocity decrease Object at rest Velocity increase
Constant acceleration Zero acceleration Constant deceleration Constant acceleration
−150 − 0
150 − 0 𝑚𝑚 =
65 − 45
𝑚𝑚 = 150 − 0
15 − 0 𝑚𝑚 = v = −7.5 𝑚𝑚𝑚𝑚 −1
−2 0−5
v = 10 𝑚𝑚𝑚𝑚
v = −30 𝑚𝑚𝑚𝑚 −1 7.5 ms-2 to the left
10 ms to the right
-2
Negative acceleration Negative acceleration
due to decrease in due to motion in
velocity opposite direction
Exercise
1. Diagram below shows the velocity-time graph for an object.

Velocity, ms-1

150
A B
100
50
C
0 Time, s
10 20 30

a. Describe the motion of the object during;

i. AB ii. BC.
b. Calculate total distance travelled in 30 seconds by the object.

c. Calculate the average speed of the object.

10
 d. Sketch a displacement-time graph to represent the object’s motion.

2. Diagram below shows the displacement-time graph for objects A and B.

Displacement, m
A a. Compare displacement of object A and B.

b. Compare time taken by object A and B.

B
c. Compare velocity of object A and B.
Explain your answer.

Time, s d. Sketch a velocity-time graph for A and B on the same

plane.
2.3 Free Fall Motion
• For an object near to the surface of the Earth, g is approximately constant at 9.81 ms-2
• Free fall: object falling only due to gravity, without any external force (friction or air resistance) • Air
resistance can be ignored when heavy object falls in the gravitational field (free fall motion)

Equation1 Equation2 Equation3 Upwards: motion is

in positive direction

1
𝑣𝑣 = 𝑢𝑢 + 𝑔𝑔𝑡𝑡 𝑠𝑠 = 𝑢𝑢𝑢𝑢 + 𝑔𝑔𝑡𝑡 2 𝑣𝑣 2 = 𝑢𝑢2 + 2𝑔𝑔𝑠𝑠
2

Downwards:
motion is in
use when there is no s use when there is no v use when there is no t negative direction

Example

Downwards motion Upwards motion

a. A ball is dropped from a height of 5 m b. A ball thrown upwards took 15 s to reach
from ground. Calculate its speed right maximum height. Calculate initial speed.
before it reaches the ground.

11
 u = 0 (not moving before dropped) v = 0 (speed at max height is 0)
g = -9.81 g = -9.81
s = -5 t = 15
v= u=

𝑣𝑣 2 = 𝑢𝑢2 + 2𝑔𝑔𝑔𝑔 𝑣𝑣 = 𝑢𝑢 + 𝑔𝑔𝑡𝑡

2 2
𝑣𝑣 = 0 + 2(−9.81)(−5) 0 = 𝑢𝑢 + (−9.81)(15)
−1
v = 9.9 𝑚𝑚𝑚𝑚 𝑢𝑢 = 147 𝑚𝑚𝑚𝑚 −1

2.4 Inertia

• Inertia is the tendency of an object to remain at rest or, if moving, to continue its motion in a straight
line at uniform velocity.
• The concept of inertia is explained in Newton’s First Law of Motion.
• Inertia is affected by mass. The bigger the mass, the bigger the inertia.
(refer to textbook page 53-54 for the experiment)
Inertiaineverydaysituation
(refer to textbook page 56-57 for details)

• Motion of passenger in a car

• Sauce move out of bottle
• Drying umbrella by spinning
• Oil tanker have smaller separated tanks
• Seat belt reduce effect of inertia
Exercise
1. Oli dropped a piece of paper and a ball from a height in a container filled with air. Then air is sucked out
of container and the experiment is repeated. Sketch the position of paper and ball after 3 seconds.
Container with air Container without air

2. Sketch graphs below to represent the motion of an object falling in a vacuum chamber.

a. Acceleration-time graph b. Velocity-time graph c. Displacement-time graph

12
3. When an object is thrown upwards with an 4. Alex dropped an object from a height of 20 m
initial speed of 40 ms , calculate the
-1
from ground. Calculate the time it takes to
maximum height reached by the object. reach the ground. Ignore air resistance.
5. Diagram shows the images of a falling object captured using a stroboscope.
0.01 m
Given that the time interval between each image is 0.1 s. a. Calculate;

i. initial speed

ii. final speed

iii. acceleration

0.5
m
b. Is the object free falling? Explain your answer.

13
2.5
Momentum

• Momentum is the product of mass and velocity.p : momentum, kgms-1

𝑝𝑝 = 𝑚𝑚𝑚𝑚
m : mass, kg
• Momentum is a vector quantity. v : velocity, ms-1
• Direction of momentum and velocity of an object are the same.

Principleofconservationofmomentum

• Total momentum in a collision stays the same if there is no external forces are acting on the objects.

Total momentum before collision =Total momentum after collision

𝑚𝑚 1 𝑢𝑢 1 + 𝑚𝑚 2 𝑢𝑢 2 = 𝑚𝑚 1 𝑣𝑣 1 + 𝑚𝑚 2 𝑣𝑣 2

• If objects are moving in opposite direction, assume that the motion to the right is positive and the
motion to the left is negative.
• Explosion refers to a situation where an object at rest breaks up into two or more parts.
Example

1. Diagram shows a 4 kg ball collided with a 2 kg ball that is at rest. After the collision, the balls stick
together and moved to the right. Calculate their velocity after collision.

Before collision After collision

4 kg 2 kg 4 kg 2 kg

12 ms-1 At rest v?

2. Diagram shows a 4 kg ball moving to the right collided with a 2 kg ball that is moving to the left. After
the collision, the balls rebounded and move in opposite direction from each other. Calculate velocity
v after collision.

Before collision After

4 kg 2 kg 4 kg 2 kg
collision
2 ms-1 5 ms-1 1 ms-1 v?

2. Diagram shows 4 kg and 2 kg ball at rest. Due to explosion, the two balls gain speed and move in opposite
direction Calculate velocity v after collision.

14
2.5
Momentum

Before collision After

4 kg 2 kg 4 kg 2 kg
collision
1 ms v? -1

Situationswheremomentumisconserved

Situation1:Launchofrocket

− Launching rocket involves explosion

Upward
− Before launch, the rocket and the fuel has zero momentum momentum as both are at rest

− Combustion produces hot gas downwards, creating momentum downwards

− Due to principle of conservation of momentum, the rocket

will propel upwards, generating momentum with equalDownward

momentum
magnitude in opposite direction

Situation2:Squidandink

− When encountering enemy, squid squirt ink to escape

− Ink produce momentum to the direction of its velocity
− Due to principle of conservation of momentum, the squid
will propel to the other direction
− Momentum with equal magnitude is generated in
opposite direction
1. Two objects with mass 500 g and 2 kg move with the
same speed of 3 ms-1. Calculate the momentum of each
object.

15
2.5
Momentum

2. Diagram below shows two objects before and after collision. After collision, they both stick together
and continue to move in the same direction. Calculate speed of 8 kg ball after the collision.

Before collision After collision

8 kg 6 kg 8 kg 6 kg

10 ms-1 At rest v ?

3. Diagram below shows two objects before and after collision. After collision, they separated but
continue to move in the same direction. Calculate speed of 5 kg ball after the collision.

Before collision After collision

4 kg 5 kg 4 kg 5 kg

8 ms-1 2 ms-1 5 ms-1 v?

4. Diagram below shows two objects before and after collision. After collision, they both stick together
and continue to move in the same direction. Calculate speed and state their direction of motion after
the collision.

Before collision After collision

5 kg 6 kg 5 kg 6 kg

10 ms-1 8 ms-1 v?

5. Diagram shows a gun and a bullet right after explosion.

a. State and explain the direction of motion of the gun right after the explosion.

Freepik.com

b. The speed of the bullet is higher than the gun.

Explain why.

16
2.6 Force
• Force is a push or pull
• Forces may produce changes in the size and shape of an object
velocity of an object by changing its
• direction
• speed

• Can be obtained using formula : a directly proportional to F

𝐹𝐹 = 𝑚𝑚 𝑎𝑎
F ↑a ↑ F ↓a ↓

F : force, N a inversely proportional to m

m : mass, kg
m ↑a ↓ m ↓a ↑
a : acceleration, ms -2

Example

1. An object at rest accelerates after 10 N force acted on it for 2 seconds. If the mass of the object is 0.5 kg,
calculate the final velocity of the object.
𝑣𝑣 − 𝑢𝑢
𝑎𝑎 = 𝑣𝑣 = 40 𝑚𝑚𝑚𝑚 −1
𝐹𝐹 = 𝑚𝑚 𝑎𝑎 𝑡𝑡
10 = 0.5 𝑎𝑎 𝑣𝑣 − 0
𝑎𝑎 = 20 20 =
2

Exercise
1. Ain kicked a ball with force 150 N. If the 2. Suriya pulled a box at rest with a force of 50 N ball accelerates
at 300 ms-2, calculate to reach a speed of 20 ms-1 within 4 seconds. mass of the ball. What is the mass
of the box?

3. Braking force of 200 N is exerted on a 400 4. From rest, an object travelled a distance of kg car. If the
speed at the time is 55 ms-1, 250 m while accelerating in 6 s. Calculate calculate the distance needed by
the car to force exerted on it if mass of object is 200 g. stop.
2.7 Impulse and Impulsive Force
Impulse

• Impulse, J: the change in momentumF : force, N

𝐽𝐽 = 𝐹𝐹 ×𝑡𝑡
• If objects rebounds or move in oppositet : time, s
m : mass, kg

17
 direction, assume that motion to the right isu : initial velocity, ms-1
𝐽𝐽 =mv−mu
positive and the motion to the left is negative.v : final velocity, ms-1

Example

1. Diagram shows a 0.5 kg ball initially at rest move to2. Diagram shows a 0.5 kg ball initially at rest
the right with speed 2ms-1. move to the right after 10 N force acting on it
Calculate impulse. for 3s. Calculate impulse.

10 N
0.5 kg 0.5 kg 0.5 kg 0.5 kg

𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = mv−mu 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝐹𝐹 × 𝑡𝑡

= 0.5(2)−0.5(0) = 10(3)
= 1𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚−1 @ 1 𝑁𝑁𝑁𝑁 = 30𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚−1@ 30 𝑁𝑁𝑁𝑁
Impulsive force

F : force, N
• Impulsive force : rate of change of : mass, kg
𝑚𝑚𝑚𝑚 − 𝑚𝑚𝑚𝑚 m
momentum in a collision or impact in a 𝐹𝐹 = v : final velocity, ms-1
𝑡𝑡 u : initial velocity, ms-1
short period of time t : time, s

Example

1. Diagram shows a 0.5 kg ball initially at rest move to the right with speed 2ms-1. Calculate impulsive
force if time of impact is 0.2 s.

0.5 kg 0.5 kg
= 5 𝑁𝑁

2. Diagram shows a 0.5 kg ball initially moving to the right with speed 4 ms-1 and
rebounds in opposite direction
with speed 3ms-1. Calculate
4 m/s
impulsive force if time of impact is 0.2 s.
0.5 kg
3 m/s 0.5(−3)−0.5(4)
== −17.5 𝑁𝑁
0.2
(in the opposite direction
to the velocity of the ball)
2.7 Impulse and Impulsive Force
(refer to textbook page 66-69 for details)

𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 −𝑚𝑚𝑚𝑚

F inversely proportional to t Impulse directly proportional to t

18
 t↑F↓ t↓ t↑J↑ t↓J↓

Short time of impact

• When collision happen on a hard surface, time of impact is short

• When time of impact is short, impulsive force increase
• Example: pile driver, pestle and mortar

Long time of impact

• When collision happen on a soft surface, time of impact is long

• When time of impact is long, impulsive force decrease
• Object sustain less damage
• Example: crumple zone of car, bending leg when landing

Follow through action

• Follow through increase time of impact

• Time increase, impulse is larger
• Object moves at a high velocity
• Example: sports such as tennis, football
Exercise
1. Diagram below shows two objects before and after collision. After collision, they both stick together and
continue to move in the same direction.
a. Calculate the momentum of the 4 kg object before the collision
b. Calculate the impulse experienced by the 4 kg object

Before collision After collision

4 kg 4 kg

10 ms-1 At rest 4 ms-1

2. Diagram shows a ball initially moving to the right and rebounds in opposite direction with speed.
Calculate impulsive force if time of impact is 50 ms.

6 m/s
0.2 kg
8 m/s

19
3. Diagram shows a ball initially at rest. Then, 25 N force acted on the ball for 0.8 s until its speed increases
to 5 ms-1.
a. Calculate change in momentum of the object
b. Calculate the mass of the object
25 N

At rest 5 ms-1
5. Your school is organizing a sports event. You are asked to help pick out
suitable mattress for the high jump event. Describe the mattress based
on the characteristics below. Justify your answer. a. Size
b. Thickness
c. Others

2.8 Weight

MASS WEIGHT
𝑊𝑊 = 𝑚𝑚 𝑔𝑔
measure of the quantity of matter gravitational force acting on an
in an object object due to its mass
W : weight, N
m : mass, kg
changes according to gravitational
constant at all time and location g : gravitational field
field strength
strength, Nkg-1

Object On Earth, g = 9.8 N/kg On Mars, g = 3.7 N/kg On Moon, g = 1.6 N/kg

Mass = 10 kg Mass = 10 kg Mass = 10 kg

10 kg
Weight = 10 x 9.8 = 98 N Weight = 10 x 3.7 = 37 N Weight = 10 x 1.6 = 16 N

Exercise
1. Mass of an astronaut on is 60 kg on Earth. Gravitational acceleration on
Earth is 9.81 ms-2. Gravitational acceleration on Moon is 1.64ms-2.
a. What is her weight on Earth?
b. What is her mass on the Moon?
c. What her weight on the Moon?

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2. A object weigh 47 N on Earth. When it is brought to the surface of Planet X, the
weight is 75 N.
a. Compare the gravitational field strength of Earth and Planet X
b. Calculate, the gravitational field strength on the surface of Planet X.

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