NEWTONS1STLAW(law of inertia)

It states that “A body remains in its state of rest or uniform motion in

a straight line unless acted upon by an external force”
Inertia is the tendency for body not to change its state.
Newton’s first law is evident in;
(i) A coin is placed on top of a cardboard and then the cardboard is placed
on top of an open can
Coin
Cardboard
Can

When the cardboard is suddenly pulled, the coin just falls in to the can.
The coin maintains its state of rest and hence does not move with the
cardboard but falls inside
(ii) When a moving vehicle is suddenly stopped, the passengers are thrown
forward. The passengers in a vehicle move with the same velocity as the
vehicle, hence when the vehicle stops the passengers maintain the
uniform motion they were moving with.
Seat belts are used to hold passengers on their seats,

MOMENTUM
Momentum is the product of mass and velocity of a body. It is a vector quantity.
Momentum = mass x velocity
p= mv
The SI units are kilogram meter per second (kgms-1)
A body with big mass has a higher momentum is more difficult to stop and can
cause more damage than a lighter one.
EXAMPLES
1. Calculate the momentum of a car of mass 800kg is travelling at 50m/s.
2. A car of mass 600kg starts from the rest and accelerates at 2ms-2.
Determine its momentum after it has moved 400m from the starting
3. A lorry of mass 1200kg initially moving at14m/s accelerates uniformly
at4m/s2for9s.Calculate the final momentum it acquired.
4. A bus of mass900kg initially moving at72km/h accelerates uniformly for
5sfor 500m.Calculate the final momentum it acquired.

NEWTON’S 2nd LAW

It states that “The rate of change of momentum of a body is directly
proportional to the force applied and takes place in the direction of
the force”.
When a force is applied on a body of mass m its velocity increases from an initial
velocity U to final velocity V in time t.

FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 1

Initial velocity = mu
Final velocity = mv
Change in momentum = mv - mu
𝒎𝐯−𝒎𝒖
Hence rate of change of momentum =
𝒕
𝒎(𝐯−𝒖)
From Newton’s 2nd law…………………………𝑭 ∝ 𝒕
𝐯−𝒖
But 𝒂=
𝒕

Hence F ∝ ma
Thus F =kma
But for k = 1

Thus F =ma
EXAMPLES
1. A car of mass 800 kg is initially moving at 25 m/s. Calculate the force
needed to bring the car to rest over a distance of 20 m. (-12,500N)
2. A car of mass 1200 kg initially moving at 10 m/s accelerates uniformly to
a final velocity of 40m/s. Calculate the force acquired by the car over a
distance of 50 m. (18,000N)
3. A bullet traveling at 300ms-1 strikes a thick wall and escapes on the other
side of the wall in 0.005s at a velocity of 100 ms-1, if the mass of the
bullet is 20g, determine the average retarding force of the bullet.
(800N)
4. A bus of mass 2500kg initially moving at 20m/s is brought to rest over a
distance of100m. Determine the force required to achieve this.(-5,000N)
5. A lorry of mass 2400 kg initially moving at 20 m/s accelerates uniformly
to a final velocity of 50m/s. Calculate the force acquired by the car over a
distance of 35 m. (72,000N)

FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 2

 IMPULSE
Impulse is the change in momentum. Impulse occurs when two bodies
collides. When a force Facts on a body for a time t the momentum changes from
mu to mv. Impulse is given by.
Impulse =mv – mu
𝒎𝐯−𝒎𝒖
From Newton’s 2nd law……………………….𝑭 =
𝒕
Cross multiplying 𝒕
𝑭𝒕 = 𝒎𝐯 − 𝒎𝒖
Hence,
F x t=change in momentum = Impulse
The SI units are Newton second (Ns)
When a graph of force F against t is plotted, the area under the curve gives the
impulse of the body(Ft) or the change in momentum.
FORCE (N)

𝟏
𝑰𝒎𝒑𝒖𝒍𝒔𝒆 = 𝐅 𝐱 𝐭
𝟐

Times (s)
𝑰𝒎𝒑𝒖𝒍𝒔𝒆
Impulsive force depends with time taken i.e. 𝐅= 𝒕
Hence, this shows that the less the time of impulse the more the damage caused.
That is why
(i) A high jumper bends/flexes his knees when landing on the ground
so that he can increase the time of landing hence reduce the force
of impact which can cause accident (also A parachutist allows his
legs to bend and rolls over on the ground when he lands).
(ii) A high jumper usually lands on a thick soft mattress. The spongy
mattress increases the time of landing hence reduce the force of
impact which can cause accident.
(iii) Vehicles have plastic bumpers and not metallic. The plastic
bumper takes longer time to crush hence increases the time of
impact and hence a small force acts for the safety of passengers.
EXAMPLES
1. A force of 8kN acts on a body of mass 20kg at rest for 0.01s. Calculate
(a) Its momentum change. (80Ns)
(b) Its final velocity. (4m/s)
2. A force of 4000N acts on body for 0.3s. Find the impulse. (1200Ns)
3. The graph below shows the variation of velocity with time for a bullet of
mass 20g moving thru the trunk of a tree until it comes out on the other
side.
V (m/s)
300
FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N
200 Pg 3
Determine:
(i) The change of momentum of the bullet. ( 4Ns)
(ii) The diameter of the tree trunk. (0.4m)
(iii) The retarding force on the bullet. (2000N)

NEWTON’S thirdLAW
It states that “Action and reaction force are equal and opposite”.
i.e. When a body exerts a force on another body, the other body creates
an equal and opposite force of reaction on the first body.

Action and reaction force is evident in;

(i) When walking, you push the floor backwards with your toes (action)
and the floor offers reaction force making us step forward.
(ii) When a shot is fired from a gun (action) an equal and opposite force
(reaction) makes the gun to recoil backwards.
(iii) When launching a rocket the fuel burns and the compressed hot gases are
thrown out downwards with a huge force (action). The reaction force acts
on the rocket in the upward direction and it rises.
(iv) When an inflated balloon is suddenly released it rises as the air comes out.
As the air moves out it produces an action force and hence the reaction
force makes the balloon to rise.
(v) A body of mass m placed on a table is attracted towards the earth with a
force F =mg (action). The earth also is attracted towards the body with
the same reaction force R =mg
R = mg
Block
Table

W = mg

Weight of a body on a lift

Lift at rest
For lift at rest, the spring balance would read the actual weight of the person
hence the person exerts on the floor of the lift ofW=mg
Lift moving upwards with acceleration “a”
For an upward motion, the acceleration aof the liftis +vewhile for a downward
motion a is-ve. Someone inside the lift experiences two forces; ieforce due to
acceleration a(F = ma)and the force due to gravityg(F= mg).The person
presses harder on the lifts floor. The resultant force exerted on the floor is:

FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 4

 𝑾 = 𝒎𝒈 + 𝒎𝒂
𝑾 = 𝒎(𝒈 + 𝒂)
Lift moving downwards with an acceleration“a’’less than“g”
The acceleration a is –ve hence the person feels lighter and presses lightly on
the lifts floor with a weight of
𝑾 = 𝒎(𝒈 − 𝒂)
Lift moving downwards with acceleration “a”greater than“g”
If the acceleration a is greater than g the person would hit the roof of the lift and
exert a force on the roof with his head. Since the force is acting in the opposite
direction, the resultant weight is negative.
Lift moving upwards or downwards with a uniform velocity
For a uniform velocity the acceleration a is zero and hence the person only
experiences a force due to gravity alone. i.e
F =mg
Lift falling freely
If the lift ropes breaks and it starts falling freely, then the lift and the person
experience a free fall. The person will only be accelerated downwards by gravity.
Hence g = a. the resultant weight exerted on the floor is zero.
𝑾 = 𝒎(𝒈 − 𝒂) But g = a
Hence
𝑾=𝒎𝒙𝟎=𝟎

EXAMPLES
1. A man of mass 80kg is standing in a lift. What force does he exert on the
floor when the lift is
(i)At rest.
(ii)Accelerating upwards at 4m/s2
(iii)Moving downwards at an acceleration 3m/s2
(iv)Moving with upwards with a constant velocity of 6m/s.
2. A girl of mass 60kg is standing in a lift. What force does she exert on the
floor when the lift is
(i) Accelerating upwards at 7m/s2
(ii) Moving downwards at an acceleration 1m/s2
3. A mass of 100kg is attached to a wire and raised vertically. What is the
tension in the wire when the mass is.
(i) Moving downwards at an acceleration 15m/s2
(ii) Accelerating upwards at 6m/s2
(iii) The lift breaks down and is under free fall.

Law of conservation of linear momentum

It states that “For a system of colliding bodies, the total linear
momentum remains constant provided no external force acts”
i.e. Momentum before collision = momentum after collision
m1u1 = m2v2
Conservation of linear momentum is observed in;
FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 5
 (a) When a bullet of mass m1is fired from a gun of mass m2the bullet
moves forward with a momentum m1u1 and the gun recoils backwards
with an equal momentum m2u2.
(b) When a trolley is moving at uniform speed along a track and a piece of
plasticine is dropped on the trolley and sticks on it the velocity of the
trolley decreases. The plasticine increases the mass of the trolley hence
the velocity decreases to maintain the initial momentum.
(c) A cyclist carrying a bag of maize tied a cross back seat is traveling at
uniform velocity. If the bag suddenly falls down, the velocity of the bike
increases in order to maintain its initial momentum.
EXAMPLES
1. Calculate the recoil velocity of a gun of mass 0.4kg which fires a bullet of
mass 0.0045kg at a velocity of 400ms-1 (4.5m/s)
2. A bullet of mass40g is fired at a velocity of 400ms from a gun of mass 8kg.
-1

Determine the corresponding velocity of the gun. (2m/s)

COLLISIONS
For any colliding bodies, after the collision bodies may;
(i) Fuse and move together in one direction.
(ii) Separate and move on different direction.
(iii) Separate and move in the same direction.

PERFECTLY Elastic collisions

This is a collision in which both kinetic energy and momentum are
conserved. In this collision the bodies separate after collision.
When two bodies of mass m1 and m2 moving with velocities v1andv2respectively
collides, then

𝒎𝟏 𝒖𝟏 + 𝒎𝟐 𝒖𝟐 = 𝒎𝟏 𝐯𝟏 + 𝒎𝟐 𝐯𝟐
Since KE is conserved, then KE before collision is equal to KE after collision.

𝟏 𝟐 𝟏 𝟏 𝟏
𝒎1 𝑼 1 + 𝟐 𝒎 2 𝑼𝟐 2 𝒎1 𝐯 𝟐
1 + 𝟐𝒎2 𝐯 𝟐 2
𝟐 = 𝟐
⏟ 𝑲𝑬 𝒃𝒆𝒇𝒐𝒓𝒆 ⏟ 𝑲𝑬 𝒂𝒇𝒕𝒆𝒓

PERFECTLY In-Elastic collisions

This is a collision in which only momentum is conserved but kinetic
energy is not. In this collision the bodies fuse and move together after
collision.eg
~ Two pieces of clay
~ A bullet fired towards a block of wood and is embedded.
For this collision the bodies ends up with the same velocity (Vf)

𝒎𝟏 𝒖𝟏 + 𝒎𝟐 𝒖𝟐 = 𝐯𝒇 (𝒎𝟏 + 𝒎𝟐 )
The KE is not conserved because the bodies undergo some deformation and hence
loses some energy which is transformed into heat, sound and light.

FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 6

EXAMPLES
1. A bullet of mass 300g traveling with a velocity of 1000ms-1strikes a block
of mass 19.7kg which rests on a rough surface. If the bullet and the block
move together, calculate their combined velocity. (15m/s)
2. An arrow of mass 20g traveling horizontally strikes a block of wood of mass
1980g resting on a horizontal surface. After collision the two moves
together. Calculate their combined velocity. (ANS
500m/s)
3. Two trolleys of mass 2kg and 3kg are traveling towards each other at
2.5m/s and 4m/s respectively collides and combine on collision.
(i) Calculate the velocity of the combined trolleys. (-1.4m/s)
(ii) In what direction do the trolleys move after collision?
4. A car of mass 600kg traveling at 20m/s is involved in a head on collision
with a cyclist of mass 200kg traveling at 40m/s moving in the opposite
direction. The cyclist is thrown onto the bonnet of the car which continued to
move in its original direction. Calculate their combined velocity. (5m/s)
5. A minibus of mass 1200kg travelling at a velocity of 15m/s collides with a
stationary car of mass 600kg. The impact takes 1.5 seconds before the two
moves together at a constant velocity for 25s.Calculate.
i) The common velocity (10 m/s)
ii) Distance moved after impact. (125m)
iii) The impulsive force. (4000N)
6. A bus of mass 2400kg traveling at a velocity of 20m/s collides with a
stationary car of mass 600kg. The impact takes 2 seconds before two
moves together at a constant velocity for 30 seconds. Find
(i) The common velocity after impact . (16m/s)
(ii) The distance moved after impact. (240m)
(iv) K.E before collision. (480,000J)
(iv) K.E after collision. (384,000J)
(v) The impulsive force. (4800N)

Elastic collisions
7. A mini bus of mass 2500 kg traveling at a speed of 40m/s collides head on
with a lorry of mass 3000kg moving at a speed of 20m/s in the opposite
direction. The two vehicles separate on collision and the lorry moves with a
speed of 5m/s backwards along its initial path. Calculate the speed of the
minibus after collision. (10m/s)
8. A ball X of mass 0.1kg moving with a velocity of 6m/s collides directly with
a ball Y of mass 0.2kg at rest. X rebounds back with a velocity 2m/s in the
opposite direction after collision. Determine
(i) The velocity of Y after collision (4m/s)
(ii) Total KE after collision (1.8J)
9. A bullet of mass 20g traveling horizontally at a speed of 200m/s hits a
block of wood of mass 800g suspended by a light inextensible string so that
it swings freely. If the bullet emerges on the other side of the block at
80m/s, find:
(i) The initial velocity of the block after being hit by the bullet. (3m/s)
(ii) The maximum height through which the block may rise. (0.45m/s)

Application of the law of conservation of momentum

FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 7
 (a) Rocket jet propulsion
When launching a rocket, the fuel burns throwing out compressed hot gases
with a high velocity and hence gain momentum in the downward direction.
The rocket therefore gains an equal momentum in the opposite direction and
the rocket is thus accelerated upwards and rises.

(b) The garden sprinkler

The pressure of water causes water to be ejected out thru the nozzle at a
very high velocity. As it moves out it gains momentum and makes the
sprinkler to rotate in the opp direction.

Water moves out

Force of friction
This is the force that opposes or tends to oppose relative motion btn two
surfaces in contact. It occurs when one body moves or tends to move over
another.

TYPES OF FRICTION
(a) Static friction
Consider a block of wood resting on a table.
R = mg Spring balance

Block
F
Table

W = mg

The minimum force required for the block to just start moving is called
limiting static friction (FS)
This frictional force(FS) is directly proportional to the normal reaction force R of
the block. i.e
𝑭∝𝑹
𝑭𝑺 = 𝝁𝑺 𝑹
𝑭𝑺
= 𝝁𝑺 Where 𝝁𝑺 is the coefficient of static friction and R is mg
𝑹
(b) Kinetic friction
When the block above now starts moving at a steady velocity, the minimum force
required to pull it is called the limiting kinetic friction. (Fk)
This frictional force (Fk) is directly proportional to the normal reaction R of the
block. i.e
𝑭𝒌 = 𝝁𝒌 𝑹
FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 8
 𝑭𝒌
= 𝝁𝒌 Where 𝝁𝒌 is the coefficient of friction.
𝑹
Practically the force required to start a motion is larger than the force required to
maintains it i.e
𝝁 𝑺 > 𝜇𝒌
EXAMPLES
1. A wooden box of mass 40kg rests on a rough floor. If the coefficient of
friction btn the floor and the floor is 0.8.
(i) Calculate the force required to just move the box. ( 320N)
(ii) Calculate its acceleration when a force of 500N is applied to the box.
2. A block of metal of mass 50kg requires a force of 125N to pull it with a
uniform velocity along a horizontal surface. Calculate
(i) The coefficient of friction. (0.25)
(ii) The force that can pull with an acceleration of 3m/s 2 (275N)
3. A force of 9N acts on a 2.5kg trolley and accelerates at 2 m/s . Calculate the
2

retarding force acting on the trolley. (-4N)

4. A cart of mass 30kg is pushed along a horizontal path by a horizontal force
of 8N and moves with a constant velocity. The force is then increased to
14N. Determine
(a) The resistance force to the motion of the cart. (8N)
(b) The acceleration of the cart. (0.2m/s2)

Laws of friction
- Frictional force depends with the nature of the surface in contact.
- Frictional force is independent of the area of contact of the given surfaces
when the normal reaction is constant.
- Kinetic friction is independent of relative velocity.

Advantages of friction
(i) It makes walking easier ie prevents people from slipping.
(ii) Enables breaking system of a vehicle.
(iii) Help in ignition of a match stick.
(iv) Helps in chewing of food
(v) Helps in sharpening of pangas, knives etc

Disadvantages of friction
(i) Causes wear and tear btn moving parts.
(ii) Decreases the efficiency of a machine due to extra effort needed to
overcome friction.

Methods of minimizing friction

(i) Lubrication ~ this is applying oil.
(ii) Use of rollers ~ the two surfaces slides over the rollers.
(iii) Use of ball bearing.

Viscosity
Viscosity is the friction thru fluids. It causes resistance to any motion thru
fluids. The ball moves down with a weight W = mg. As it moves down it

FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 9

encounters upward resistance from the viscous drag force F and the upthrust
force U.
U F

Sphere
Viscous liquid

W = mg

The upthrust U is constant but the frictional force F of the fluid increases as
velocity of the ball increases. The ball first accelerates downwards until the weight
W = F +U, then its velocity reaches a constant value and moves with a uniform
velocity called terminal velocity up to the bottom.

Terminal velocity
Velocity (m/s)

Times (s)

Different fluids have different viscosity and hence have diff terminal velocities. Eg
glycerin is denser than water, hence a steel ball would take longer to reach the
bottom in glycerine than in water. The higher the viscosity the lower the terminal
velocity.
Terminal velocity
Water
Glycerin
Velocity (m/s)

Times (s)

FORM 3NOTES: N E W T O N’S L A WS O F M O T I O N Pg 10