NEWTON’S LAWS OF MOTION  
There are three Newton’s laws of motion. These include the following  
o
o
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Newton’s first law of motion  
Newton’s second law of motion  
Newton’s third law of motion  
1. Newton’s First Law of Motion (the law of inertia or the seat belt – law)  
This law is also referred to as the law of inertia or the seat belt – law  
➢
Inertia is the tendency of an object to resist changes in its state of motion  
OR  
Inertia is the ability of a body to resist changes in motion  
Types of Inertia  
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Inertia of rest  
Inertia of motion  
Inertia of direction  
Inertia of Rest  
•
Is the resistance of a body to change its state of rest  
Inertia of Motion  
•
I
s the resistance of a body to change its state of motion  
Inertia of Direction  
•
I
s the resistance of a body to change its direction of motion  
Therefore the Newton’s first law of motion state that  
“A body continues in its state of rest or uniform motion in a straight line unless  
external force act on it”  
OR  
“An object will remain at rest or in uniform motion in a straight line unless  
acted upon by an external force”  
Some examples of inertia in everyday life  
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When you shake a branch the leaves get detached  
When you beat a carpet the dust particles come out  
When you quickly pull a book from the bottom of a file of books ,the other  
books remain arranged  
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It is harder to stop a big vehicle like a bus than a smaller vehicle like a  
motorcycle .There is more inertia with the bigger object  
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If a car is moving forward ,it will continue to move forward unless friction or the  
brakes interfere with its movement  
If you jump from a car or bus that is moving, your body is still moving in the  
direction of the vehicle. When your feet hit the ground ,the grounds act on your  
feet and they stop moving .You will fall because the upper part of your body  
didn’t stop and you will fall in the direction you were moving  
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When a bus suddenly starts moving, the passengers sitting or standing in the  
bus tend to fall backwards  
When a bus suddenly stops, the passengers sitting or standing in the bus are  
thrown forward  
Momentum  
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A body is said to be in motion if it changes its position with time and when it  
has velocity  
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A body with zero velocity therefore it is not in motion and hence it is at rest  
The motion of a body can be measured by multiplying out its mass ‘m’ and its  
velocity ‘v’ the product M.V is known as the linear momentum of a body  
Momentum is the product of the mass and velocity of an object  
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Linear momentum = Mass x Velocity  
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... The S.I unit of momentum is kg m/s  
Example A man of mass 1000 kg is moving with a velocity of 60 km/h. find its  
momentum  
Solution:  
Given: Mass of a car, m = 1000 kg , Velocity of a car, v = 60km/h = 16.7 m/s  
Momentum of a car, p =?  
From: Momentum (p) = mass x velocity = mv = 1000 x 16.7 = 16700 kgm/s  
Newton’s Second Law of Motion  
It states that: “The rate of change momentum of a body is directly proportional  
to the applied force and takes place in the direction in which the force acts”  
•
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Suppose force F acts on a body of mass ‘m’ for time t. This force causes the  
velocity of the body to change from initial velocity ‘u’ to final velocity ‘v’ in that  
interval t  
The change in momentum will then be mv – mu  
풎풗−풎풖  
풕
•
The rate of change of momentum is  
, by Newton’s second law of motion  
풎풗−풎풖  
→ 푭 ∝ 풎 풗−풖  
풕
.
푭 ∝  
풕
풗−풖  
풕
But  
= a (acceleration of a body).  
∴ F ∝ ma  
•
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If a constant of proportionality  
k
is introduced in the above relation, then  
F = kma. This equation can be used to define unit of force. If m = 1kg and a =  
1m/s2, then the unit of force is chosen in such a way that when F = 1 the  
constant k = 1,  
hence F = ma  
If a mass of 1kg is accelerating with 1m/s2, then a force 1N is said to be acting  
on the body.  
∴
F = ma  
Example  
1. Suppose you exert an upward force of 10 N on a 3kg object. What will be the  
object acceleration?  
Solution  
:
Given: Force applied, f = 10 N, Mass of object, m = 3 kg = 30 N  
Acceleration of an object, a =?  
Net force (F) = 30 - 10 = 20 N  
From: F = ma --------------------- make  
a
the subject  
푭
ퟐퟎ  
ퟑ
∴ 풂 =  
=
= ퟔ. ퟔퟕ m/s2  
풎
2. A tennis ball whose mass is 150 g is moving at a speed of 20 m/s. it is then  
brought to rest by one player in 0.05 s. find average force applied  
Solution:  
Mass of tennis ball, m = 150 g = 0.15 kg,  
Final velocity, v = 0 m/s  
Initial velocity, u = 20 m/s  
Time taken, t = 0.05 s  
Force applied/average, f =?  
ퟎ−ퟐퟎ  
ퟐퟎퟎퟎ  
∴ 푭 = ퟎ. ퟏퟓ 풙  
= −ퟎ. ퟏퟓ 풙  
= −ퟔퟎ 푵  
ퟎ.ퟎퟓ  
ퟓ
NB:  
❖
The product of force and unit time is called impulse  
❖
Impulse is the change in momentum  
OR Impulse = Force x time  
→
Impulse = change in momentum  
∴ I = Ft  
•
Its SI unit is Newton second ( Ns)  
Example  
1. During a collision, a truck applies a force of 20000N on a 250 kg van for 0.5  
seconds. Determine the impulse experienced by the van  
Solution  
Given: F = 20000 N, t = 0.5 s , m = 250 kg  
From: Impulse (I) = Force x time = Ft = 20000 x 0.5 = 10000 Ns  
Application of Newton’s second law (Momentum & Impulse) in our daily life  
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The high jumpers usually bend their knees on landing. This increases the time  
of impact thus reducing the chance of injuries  
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This reduces the possibility of them cracking on sudden stop or start  
While goalkeeper catching a ball, he extends his hands forward so that he has  
enough room to let his hands move backward after impact to prevent bounce of  
ball  
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A person is better off falling on a wooden floor than a concrete floor. Because  
the wooden floor allows for a longer time of impact and, therefore a lesser force  
of impact than a concrete floor.  
Glass wares are wrapped in a paper before packing to avoid breakage.  
Because this increases the time of impact between various articles during  
jerks, thereby decreasing the force of impact on the articles  
When the car goes out of control while driving, you would prefer to hit  
something soft than something hard. It is because by hitting soft material you  
extend your time of impact ,thereby reducing the impact force  
Individual Task – 1  
1. A net force of 15 N is exerted on an encyclopedia to cause to accelerate at a  
rate of 3ms-2 .Determine the mass of the encyclopedia  
(
ANS: m = 5 kg  
)
2. A trolley of mass 400 g has a velocity of 600 cm/s. Calculate the momentum of  
the trolley (ANS: p = 2.4 kgm/s)  
3. Suppose that a sphere is accelerating at rate of 2 m/s2 .If the net force is tripled  
and the mass is halved, then what is the new acceleration of the sphere?  
(
ANS: acceleration will increase by six times)  
4. Determine the momentum of a 1000 kg truck moving Northwards at a velocity  
of 20 m/s (ANS: p = 20000 kgm/s --northward)  
5. An athlete has a westward momentum of 5000 kgm/s. If the athlete has a mass  
of 75 kg, at what velocity is he moving? (ANS: v = 66.7 m/s-1)  
6. A cricket ball of mass 180 g travelling at 25 m/s is hit towards the bowler at 15  
m/s .The impact lasts for 0.04 s. (ANS: I = 7.2 Ns, F = 180 N to the left)  
Find: (a) the impulse (b) the average force applied  
7. An unbalanced force of 12 N acts on a mass of 2 kg. Calculate  
a) The resulting acceleration (ANS: a = 6 m/s2)  
b) The force that would give a body of 10 kg the same acceleration  
(ANS: F = 60 N)  
Newton’s Third Law of Motion  
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Consider the block when kept on a top of a table it cannot move either  
downward or upward due to equal action (weight of block) and reaction (the  
force pulls the block upward).  
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Newton’s third law is also known as the law of reciprocal actions or law of  
action and reaction  
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Since two forces are equal Isaac Newton establish a law which state that  
:
“To every action there is an equal and opposite reaction’’  
Question:  
1. Newton pair forces are equal in size and opposite in direction and yet they  
do not cancel each other out. Explain  
ANS: Forces can only cancel each other out if they act on the same  
object, Newton pair forces act on different objects  
Application of Newton’s Third Law  
The following are some practical examples involving Newton’s 3rd law of motion  
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Walking: When a person walks on the ground ,he or she exerts a force on the  
ground and in turn ground exerts an equal force on the person  
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If a car is accelerating forward , it is because its tyres are pushing backward on  
the road and the road is pushing forward on the tryres  
A block resting on a table exerts normal reaction which is equal to the weight of  
the block  
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The person firing the gun will feel the recoil (push back) when the bullet leaves  
the gun  
When a person throws a package out of a boat, the boat moves in opposite  
direction from the package. The package exerts an equal but opposite force on  
the person  
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A falling object exerts upward force on the earth as much as the earth is  
exerting a downward force on the object  
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A hammer driving a nail into block of wood  
When the air is released from balloon it rushed out (action) tend to give  
reaction balloon so it acquire the motion  
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An airplane pushes back on the air and the air pushes forward on the plane.  
Apparent weight of man in lift (elevator)  
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Suppose a man of mass  
elevator  
The man exerts on the floor a force vertically downward which is the weight of  
m
is standing on a weighing machine placed in lift or  
the man (W)  
On the other hand, the floor exerts an equal force  
R
on the man in the upward  
direction (Newton’s 3rd Law of motion)  
.
Therefore R = W  
Since the two forces are acting on the man ,Then the difference of the two  
forces will determine in which direction the net force acts  
(a) When the lift is at rest  
Acceleration is zero and net force is zero  
푹 − 풎품 = ퟎ  
∴ 푹 = 풎품  
(b) When the lift is moving with constant velocity  
Again in this case acceleration is zero ( a = 0 m/s2)  
∴ 푹 = 풎품 = 풂풄풕풖풂풍 풘풆풊품풉풕 풐풇 풎풂풏  
(c) When the lift is moving upward with acceleration, 풂  
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Here the net force is acting in the upward direction  
푵풆풕 풇풐풓풄풆 = 푹 − 풎품  
→ 풎풂 = 푹 − 풎품  
∴ 푹 = 풎풂 + 풎품 = 풎 (풂 + 품)  
∴ 푹 > 푊 ( = 푚푔)  
•
Therefore, apparent weight (R) is greater than the actual weight  
W (= mg) of the man.The man can feel it if he walks on the floor  
of lift (elevator),he will need more effort to walk naturally  
(d) When the lift is moving downward with acceleration a  
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Net force is acting downward  
푵풆풕 풇풐풓풄풆 = 풎품 − 푹 → 풎풂 = 풎품 − 푹  
∴ 푹 = 풎품 − 풎풂 = 풎 (품 − 풂)  
∴ 푹 < 푊 ( = 푚푔)  
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Therefore the apparent weight (R) of the man is less than the  
actual weight W(= mg) of the man .The man will feel lighter as  
he walks about on the floor of the lift (elevator)  
(e) For free fall.  
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If the lift (elevator) cable breaks , then the lift(elevator) will fall  
freely and 풂 = 품  
(
)
From: 푹 = 풎품 − 풎풂 = 풎 품 − 풂 = ퟎ  
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Therefore the apparent of the man will become zero  
(f) When the lift moves downward with 풂 > 푔  
(
)
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In this case 푹 = 풎 품 − 풂 becomes negative  
Therefore the apparent weight of the man becomes negative and  
this indicates that the man will rise from the floor of the lift  
(elevator) and stick to the ceiling of the lift (elevator)  
Individual task – 2  
1. A 80 kg man stands in a lift .Calculate the force he exerts on the floor of the lift  
when the lift is (Assume g = 9.8 m/s2)  
a) Stationary  
(
(
(
ANS: F = 780)  
ANS: F = 940 N)  
ANS: F = 780 N)  
b) Ascending upward at 2 m/s2  
c) Moving with a constant velocity (upward) 4 m/s  
Conservation of Linear Momentum  
Collision  
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Is an occurrence where momentum or kinetic energy is transferred from one  
object to another  
Types of Collisions  
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Elastic collisions  
Inelastic collisions  
Elastic Collision  
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I
s a type of collision in which both kinetic energy and momentum are  
conserved after collision  
Inelastic Collision  
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I
s a type of collision in which the kinetic energy is not conserved after collision  
Difference between elastic and inelastic collision  
Elastic collision  
Inelastic collision  
Only momentum is conserved  
(kinetic energy is changed into  
other energies such as heat and  
sound energy)  
B
oth momentum and kinetic energy are  
conserved  
Forces during collision are conservative  
and mechanical energy is not  
transformed into some other form of  
energy such as sound and thermal  
energy  
Forces during collision are not  
conservative and mechanical  
energy is transformed into some  
other form of energy such as  
sound and thermal energy  
Bodies move apart after collision  
Bodies stick together after collision  
Bodies move with common velocity  
Each body moves with individual velocity  
An example of elastic collision is the  
An example of inelastic collision is  
movement of the swinging balls  
an automobile collision  
Principle of Conservation of linear Momentum  
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It states that: “when two or more bodies collide, their total momentum  
remains constant provided no external forces are acting”  
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Consider the case of firing a gun, as the bullet leaves the gun (reaction), the  
one holding it feels a backward force (reaction from the bullet of the gun)  
According to Newton’s 3rd law of motion, these two forces are equal and opposite.  
Since these two forces act at the same time, the impulses (i.e. change in  
momentum) produced must be equal in magnitude and opposite in direction.  
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Consider the collision of two balls moving in a straight line  
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The balls have the masses m1 and m2 and they are approaching each other  
with velocity u1 and u2 in fig (a)  
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Then the balls have the velocities V1 and V2 after collision in fig (b)  
Let F1 and F2 be the forces acting on M1 and M2 during collision.  
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By Newton’s third law of motion the forces are equal and opposite since the  
two forces act during the same time t, the impulses produced are therefore  
equal and opposite  
... 푭ퟏ풕 = − 푭ퟐ풕  
• But 푭ퟏ풕 = 풎ퟏ풗ퟏ − 풎ퟏ풖ퟏ and 푭ퟐ풕 = 풎ퟐ풗ퟐ − 풎ퟐ풖ퟐ  
From: 푭ퟏ풕 = − 푭ퟐ풕  
Thus: 풎ퟏ풗ퟏ − 풎ퟏ풖ퟏ = −(풎ퟐ풗ퟐ − 풎ퟐ풖ퟐ)  
풎ퟏ풗ퟏ − 풎ퟏ풖ퟏ = −풎ퟐ풗ퟐ + 풎ퟐ풖ퟐ  
풎ퟏ풗ퟏ + 풎ퟐ풗ퟐ = 풎ퟏ풖ퟏ + 풎ퟐ풖ퟐ  
∴ 풎ퟏ풖ퟏ + 풎ퟐ풖ퟐ = 풎ퟏ풗ퟏ + 풎ퟐ풗ퟐ  
•
This shows that the total momentum before collision is equals to the total  
momentum after collision  
.
NB:  
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If the two bodies are moving in opposite side (after or before collision),  
then minus sign (– ve) is introduced in the formula  
Example  
1. A bullet of mass 10 g leaves a gun of mass 500 g with a velocity of 100 m/s.  
Find the velocity of the gun coil.  
Data given  
Mass of a bullet, m1 = 100 g,  
Mass of a gun, m2 = 500 g  
Initial velocity of a bullet, u1 = 0 m/s  
Final velocity of a bullet, v1 = 100 m/s  
Solution  
Initial velocity of a gun, u2 = 0 m/s  
Final velocity of a gun, v2=?  
From: The Principle of conservation of linear momentum  
풎ퟏ풖ퟏ + 풎ퟐ풖ퟐ = 풎ퟏ풗ퟏ + 풎ퟐ풗ퟐ  
500 푥 0 + 10 푥0 = 100 푥 10 + 500푥 푣2  
= − 1000 = −2푚/푠  
0 = 1000 + 500V2 → 풗ퟐ  
500  
∴ 풕풉풆 풓풆풄풐풊풍 풗풆풍풐풄풊풕풚 풐풇 풕풉풆 품풖풏 풊풔 ퟐ 풎/풔 to the opposite side  
Individual task – 3  
1. Trolley A of mass 6 kg is rolling across a smooth horizontal desk with a velocity  
of 0.8 m/s. The trolley collides with a stationery trolley B of mass 2 kg .After the  
collision the trolleys couple and move off together in the direction in which A  
was travelling .Calculate the velocity of the trolleys after the collision  
(ANS: VC = 0.6 m/s)  
2. A body of mass 8 kg moving with a velocity of 20 m/s collides with another  
body of mass 4 kg moving with a velocity of 10 m/s in the same direction .The  
velocity of the 8 kg is reduced to 15 m/s after collision .If the bodies do not stick  
together after the collision ,calculate the velocity of the 4 kg body  
3. A 1000 kg cannon launches a cannon ball of mass 10 kg at a velocity of 100  
m/s. At what speed does the cannon recoil?  
(
ANS: VR = 1 m/s)  
4. During a shunting operation, a truck of total mass 15 metric tones (t) moving at  
1 m/s, collides with a stationary truck of mass 10 t .If the two trucks are  
automatically connected so that they move off together , find their velocity .  
Also calculate the kinetic energy of the trucks (a) before (b) after collision.  
Explain why these are not equal (ANS: VC = 0.6 m/s, KEB = 7.5 KJ, KEA = 4.5 KJ)  
5. A 350 kg van moving at a velocity of 20 m/s crashes on a lorry of mass 600 kg  
that was at rest .Assuming the van and the lorry stick together upon impact,  
how fast will they come to a rest? (ANS: VC = 7.4 m/s)  
6. A 4 kg object is moving to the right at 2 m/s where it collides elastically head on  
with a stationary 6 kg object as shown in the figure below after the collision, the  
velocity of the 6 kg object is 1.6 m/s to the right. Find  
(b) Velocity of 4 kg after the collision  
(ANS: V = 0.4 m/s)  
(c) Total kinetic energy before and after collision  
(ANS: KEB = KEA = 8 J)  
(d) Change in kinetic energy before and after collision (ANS: ∆ 푲. 푬 = ퟎ 푱)  
7. Consider the diagram below and answer the questions that follows  
(a) What is their velocity after the collision  
(ANS: VC = 0.8 m/s)  
(b) Total kinetic energy conserved  
(ANS: Since KEB≠KEA, Not conserved)  
Examples of conservation of linear momentum  
1. Recoil of a gun: When a gun is fired ,the bullet moves with a large velocity  
and the gun moves with a small velocity in opposite direction to that of the  
bullet  
2. When the bullet is fired ,the gun is always held close to the shoulder  
otherwise the shoulder may get hurt due to the recoil velocity of the gun  
3. When a man jumps from a boat , the boat slightly moves away from the  
shore  
4. Rocket propulsion: When the rocket is fired,fuel is burnt and very hot gases  
are formed. As the hot gases gain linear momentum to the rear on leaving  
the rocket, the rocket acquires equal linear momentum in the forward (i.e  
opposite direction) because linear momentum is conserved