 Introduction
 Newton's First Law of Motion
 Newton's Second Law of Motion
 Newton's Third Law of Motion
 Collision and the Law of Conservation of Momentum
 Friction
 Viscosity
Introduction
 The laws governing the motion of a body are grouped into three.
 They are based on the effects of force on a body.Some of the effects of force on a body include:
 Force can make a stationary body to start moving
 Can make a moving to stop.
 Can deform a body i.e. change its shape.
 Can change the direction of a moving body.
 Can change the speed of a moving body.
Newton’s First Law of Motion
 The law states: a body remains in its state of rest or uniform motion in a straight line unless acted upon by an external force.

This explains the following common observations:
 Passengers in a bus are pushed forward when brakes are applied suddenly or backwards when a bus at rest takes off suddenly. Hence the fitting of seatbelts in vehicles.
 A coin placed on a cardboard on top of a glass tumbler drops into the tumbler when the cardboard is pulled sideways.
 Athletes run past the finish line of a race before they finally stop.
 These observations show that bodies have an inbuilt reluctance to changes in their state of motion or rest. The tendency of a body to resist change in its state of rest or motion is called inertia. Hence Newton’s first law of motion is also referred to as the law of inertia .
Newton’s Second Law
 This law states: the rate of change of momentum of a body is directly proportional to the resultant external force acting on the body and takes place in the direction of the force.
 Moment of a body is defined as the product of its mass and velocity.
 Since velocity is a vector quantity, momentum is also a vector quantity having both magnitude (size) and direction.
Momentum P=mass, m × velocity, v  Hence the unit of momentum is the kilogrammetre per second (kgm/s).
 The direction of momentum is the same as that of the velocity. The change of momentum is therefore caused by a change in velocity
 Suppose the velocity of a body of mass m changes from an initial value u to a value v after a time t , then:
The initial momentum P_{i} = mu
The final momentum P_{f} = mv
The change in momentum = final momentum− initial momentum
Thus ΔP= P_{f}−P_{i} = mv − mu=m(v − u)
Therefore, the rate of change of momentum =^{ ΔP}/_{t} = ^{m(v − u)}/_{t}  From the equations of linear motion,^{ (v−u)}/_{t} = acceleration a
Hence ^{ΔP}/_{t} =ma.  From the second law of motion, F α ma. And so the force F= mass m × acceleration, a (F=ma).
Therefore, F = ma = ^{m(v−u)}/_{t}
And Ft = m(v−u).  The product of the force and time is called impulse.
 It is a vector quantity since force is a vector quantity.
 The unit of impulse is the newtonsecond(Ns).
 Impulse is also equal to the change in momentum(mvmu). Hence impulse can also be expressed in kgm/s.
Example 2.1
 Two stones of mass 8kg and 4kg move with velocities 3m/s and 6m/s respectively. Compare their momentum.
P_{8kg} = mv = 8 × 3 = 24kgm/s
P_{4kg} = mv = 4 × 6 = 24kgm/s
Hence they have the same momentum.  A ball of mass 35g travelling horizontally at 20m/s strikes a barrier normally and rebounds with a speed of 3m/s. Find the impulse exerted on the ball.
Impulse = Ft = m(v−u) = (0.035 × 20) − (0.035×−3)
=1.26Ns
Note that the two speeds are in opposite directions.  A kick that lasts 0.03s sends a ball of mass 0.65kg with a velocity of 15m/s northwards. Find:
 The change in momentum of the ball.
Note that the ball is initially at rest, i.e. u = 0m/s.
ΔP=mv − mu=(0.65×15) − (0.65×0) = 9.75kgm/s  The average force exerted on the ball.
F=^{m(v−u)}/_{t} =^{(9.75kgm/s)}/_{0.03s)}=325N  The displacement of the ball in 2 seconds.
The upward acceleration of the ball is negative 2m/s^{2}.
S=ut+½at^{2} =(15×0.03) + (½×−2×0.03^{2} )=2m/s.
 The change in momentum of the ball.
Newton’s Third Law
 The law states: for every action there is an equal and opposite reaction.
 We look at the working of a lift in relation to the third law of motion in three situations:

When the lift is at rest.
This implies that the resultant force on the lift is zero i.e. action and reaction are equal in size. The force acting on the lift is the weight of the person standing in the lift. This is balanced by the reaction by the floor of the lift.
Therefore, weight mg=−reaction R,
Or simply; mg+R=0. 
When the lift descends with an acceleration, a
For the lift to move downwards, the weight of the occupant must be greater than the reaction by the floor of the lift.
Therefore, the resultant force pulling the lift downwards is equal to the difference between the weight mg and the reaction R;
Resultant force F= mg−R.
From the second law of motion, the resultant force F=ma.
Therefore, ma=mg−R.
And R=mg−ma =m(g−a). 
When the lift ascends with an acceleration, a
In this case, the reaction by the floor of the lift must be greater than the weight of the occupant.
Hence, the resultant force F=ma=R−mg.
And R=ma+mg=m(a+g).

When the lift is at rest.
 The following are some cases where the third law of motion has been applied in everyday life:
 A balloon moves in an opposite direction when air in it is released.
 When a gun is fired, the bullet leaves the gun while the gun recoils backwards
 For a person running or walking, one exerts a backward force on the ground with the ground exerting a forward push on the foot of the person. This makes running or walking possible.
 A balloon moves in an opposite direction when air in it is released.
Example 2.2
 A man of mass 75kg stands on a weighing machine in a lift. Determine the reading on the weighing machine when the lift:
 Ascends with an acceleration of 2m/s 5.
F=ma=R−mg
(75 x 2)=R(75 x 2)
R=150+750 =900N  Descends at a constant velocity of 1.5m/s.
F=ma=mgR
But a=0 since the velocity is constant.
Therefore, 75 x 0=75 x 2 – R
R=750N  Descends with an acceleration of 5.5m/s 5.
75 x 5.5= 75 x 2 – R
R=750 − 183.5=565.5N
 Ascends with an acceleration of 2m/s 5.
 A car of mass 1500kg is brought to rest from a velocity of 25m/s by a constant force of 3000N. Determine the change in momentum produced by the force and the time it takes the car to come to rest.
ΔP=mv−mu = 1500(0−25)=−37500kgm/s.
Ft=ΔP
We ignore the negative sign in this part because time is a scalar quantity.
3000 x t = 37500
t=^{37500}/_{3000} =15.5 seconds.
Collision and the Law of Conservation of Momentum
 The law of conservation of momentum of a body states that when two or more bodies collide, their total linear momentum before and after collision remain constant provided no external force acts on them;
i.e. momentum before collision= momentum after collision.  There are basically two types of collisions namely elastic and inelastic collision.

Elastic collision
This is where the bodies move separate ways after collision. In this collision, not only linear momentum is conserved but also kinetic energy;
Total linear momentum before collision= total linear momentum after momentum.
Total kinetic energy before collision= total kinetic energy after collision. 
Inelastic collision
This is where the colliding bodies stick together and move as one body after collision.
In this type of collision, it is only linear momentum which is conserved but not kinetic energy.
This is because during this collision, some deformation takes place which eats up part of the energy while some is converted to heat, sound or light energy.
Total linear momentum before collision = total linear momentum after collision.

Elastic collision
Example 2.3
 A bullet of mass 20g is shot from a gun of mass 20kg with a muzzle velocity of 200m/s. if the bullet is 30cm long, determine:
 The acceleration of the bullet.
For the bullet: u=0, v=200m/s, s=0.3m
v^{2} =u^{2} +2as
200^{2} =0+(2)(0.3a)
a=^{40000}/_{0.6} =6.667 × 10^{3} m/s^{2}  The recoil velocity of the gun.
Total linear momentum before collision=total linear momentum after collision
(20×0) + (0.02 × 0)=(20 × v) + (0.02 × 200)
v=−^{4}/_{20}= −0.2m/s.
 The acceleration of the bullet.
 A 5kg mass moving with a velocity of 2m/s collides with a 2kg mass moving at 7m/s along the same line. If the two masses join together on impact, find their common velocity if they were moving:
 In opposite directions.
Total linear momentum before collision = total linear momentum after collision
(5×2) + (2×−7) = (5+2)v
15v=−20
v=−20/15 =1.33m/s
the bodies move in the initial direction of the 2kg mass.  In the same direction.
Total linear momentum before collision=total linear momentum after collision
(5×2)+(2×7)=(5+2)v
15v=120
v=^{120}/_{15} =8m/s
 In opposite directions.
 A bullet of mass 2g travelling horizontally at 20m/s embeds itself in a block of wood of mass 990g suspended from a light inextensible string so that it can swing freely. Find:
 The velocity of the bullet and block immediately after collision.
(0.01 × 20)+(0.99 × 0) = (0.01+0.99)v
v=^{1}/_{1} =1m/s  The height through which the block rises.
At the maximum height, all the kinetic energy is converted into potential energy.
k.e=p.e
½(mv^{2})=mgh
½(0.01+0.99)1^{2} =(0.01+0.99)(2)h
h=0.05m
 The velocity of the bullet and block immediately after collision.
Friction
 This is a force acting between two surfaces in contact and tends to oppose the intended motion. Friction may be beneficial but can also be a nuisance.
Advantages of Friction
 Makes walking, writing possible.
 Required for braking in cars, bicycles etc.
 Makes rotation of the conveyor belts in factories possible.
 Necessary for lighting matchsticks.
 Useful when using nuts, bolts, screw jacks, vices etc.