# CONSERVATION OF LINEAR MOMENTUM

CONSERVATION OF LINEAR MOMENTUM

This can be stated in 3 different ways;

1. In any system of colliding bodies, the total momentum is always conserved provided that there is no external force acting on the system.
2. It states that the total momentum of an isolated or closed system remains constant.
3. If 2 or more bodies collide in a closed or isolated system, the total momentum before collision is equal to the total momentum after collision.

Fa= – Fb

Mₐ x aₐ= -Mb x ab

Mₐ (vₐ – uₐ)/t =- Mb (vb – ub)/t

Mₐuₐ + Mbub = Mₐvₐ + Mbvb

Total momentum of bodies A and B before collision is equal to the total momentum of the two bodies after collision.

Example:- A body of mass 5kg moving with a velocity of 30m/s collides with another body moving in opposite direction with a velocity of 20m/s. If both bodies now move in the direction of the first body at a velocity of 10m/s. Calculate the mass of the second body.

Solution

From,

Total momentum before collision = Total momentum after collision

Mₐuₐ + Mbub = Mₐvₐ + Mbvb

5 (30) + Mb (-20) = 5(10) + Mb (10)

150 – 20Mb = 50 + 10Mb

100 = 30 Mb

Mb = 100/30

Mb = 3.33kg

COLLISION

Collision occurs when 2 or more bodies make an impact with each other. There are possibly 2 types of collision. They are;

1. Elastic Collision
2. Inelastic Collison
1. Elastic Collision:- A collision is said to be elastic, if the 2 bodies colliding together separate after impact and in this case, both momentum and kinetic energy are conserved.

If 2 bodies of masses m1 and m2 moving with initial velocities u1 and u2 before collision and final velocities v1 and v2 after collision which are in the same direction, from the law of conservation of linear momentum and that of conservation of kinetic energy, we can deduce that;

m1u1 + M2u2 = m1V1 + M2V2 – Conservation of Linear Momentum

½ m1u12 + ½ M2u22= ½ m1V12 +½ M2V22 – Conservation of Kinetic Energy

In this case, the collision is said to be perfectly elastic and its coefficient of restitution is 1.

Where coefficient of restitution denoted by “e” is

e = (V2 – V1)/ (u1 – u2)

Examples of perfectly elastic collision bodies include; collision of billiard balls, collision of molecules and atoms, collision of table tennis ball with the wall, etc.

Note:-The relative velocities of the 2 bodies is unchanged in magnitude but opposite in direction.

Figure: Elastic Collision

• Inelastic Collision:-  this is a form of collision in which the colliding bodies stick together and moves as a unit after impact. It is noted that the velocities of the 2 bodies after impact are the same i.e., V2 = V1 = V.

Kinetic energy after collision decreases but momentum is conserved. From conservation of linear momentum,

m1u1 + M2u2 = m1V1 + M2V2

But;

V2 = V1 = V

m1u1 + M2u2 = m1v + M2V= (m1 + M2)v

For;Kinetic Energy,

Kinetic energy before collision ˃ Kinetic energy after collision

½ m1u12 + ½ M2u22˃½ m1V2 +½ M2V2

½ m1u12 + ½ M2u22˃½ (m1 + M2) V2

Figure: Inelastic Collision

Example:-Two bodies of masses 4kg and 2kg move towards each other with velocities 2m/s and 3m/s and collide. If the collision is perfectly inelastic, find the velocity of the 2 bodies after collision. Find the kinetic energy of the system before and after collision, calculate the loss in kinetic energy.

Solution

1. From,

m1u1 + M2u2 = m1v + M2V= (m1 + M2)v

4 (3) + 2 (-2) = (4 +2) v

12 – 4 = 6v

8 = 6v

v = 8/6

v = 1.33m/s

1. Total Kinetic Energy before collision = ½ m1u12 + ½ M2u22

= ½ (4 x 32) + ½ (2 x(-2)2)

= 18 + 4

= 22J

1. Total Kinetic Energy after collision = ½ (m1 + M2) V2

= ½ (4 +2) 1.332

= 5.31 J

1. Loss in K.E. = 22 – 5.31

= 16.19 J

APPLICATION PRINCIPLES OF CONSERVATION OF LINEAR MOMENTUM

1. Launching of a rocket and a jet propulsion: The downward expulsion of the gas (action) from propulsion tanks at a very high speed induces an equal reaction in the opposite direction, causing the jet to move forward with a very high speed.
• Recoil of a gun: Recoiling is the backward momentum of a gun when it is discharged or it is the pushback motion the gun experiences when you fire it. The gun recoil is an example of the conservation of momentum. Before the shot, both the bullet and the gun are at rest so the total momentum is zero. After the shot, the bullet is travelling in one direction and the gun in the opposite direction such that the total momentum remains zero.

Let M = mass of the gun

m = mass of the bullet

v = velocity of the bullet

V = velocity of the gun.

From the conservation of momentum we know that MV + mv =0.

Such that MV= -mv

Make V the subject of the formula

V = -mv/M

The negative sign means the gun moves in opposite direction to the bullet.

Example: A machine gun with mass of 5kg fires a 50g bullet with a speed of 100ms-1. Calculate the recoil speed of the gun.

Solution

Given: M = 5kg; m = 50g = 0.05kg; v = 100m/s

Required: V =?

Analysis:

V = – mv/ M

= – 0.05 x 100/ 5

= -1m/s.

• Why walking is possible

In the process of walking, one has to use his foot to push backwards in other to move forward. The backward push of the legs provides an equal but opposite reaction which is the forward motion experienced when walking.