**Momentum and Its Conservation**

Momentum is the tendency of the object to be in motion and therefore is a vector quantity. It is determined by the product of the mass of the object and its velocity. If both the objects are not experiencing any external forces, then the total momentum of the objects before and after the interaction is the same. This relation under the fundamental law of physics is known as the conservation of momentum.

This fundamental law of physics can be applied to explain the phenomenon of collision and explosion.The momentum formula is given by,

**Momentum (P) = mass (m) x velocity (v)**

Conservation of momentum can be described by

**P1 (before) + P2 (before) = P1 (after) + P2 (after)**

This equation is valid for the object that undergoes collision.

**Example 1**

A 0.55 Kg cricket ball moving at a speed of 3.5 m / s collides with a cricket bat of 3.5 Kg swung at a speed of 1.5 m / s. after the collision the cricket bat swings at a speed of 0.5 m / s. Calculate the magnitude and the velocity of the cricket ball.

**Solution:**

**Given:**

** m1 = 0.55 kg**

** m2 = 3.5 kg**

** v1 = 3.5 m/s**

** v2 = 1.5 m/s**

The relation between the colliding bodies is given by,

P1 ( before ) + P2 (before ) = P1 ( after ) + P2 ( after )

The momentum relation stands as,

( m1 v1 ) before + ( m2 v2 ) before = ( m1 v1 ) after + ( m2 v2 ) after

Therefore, v1 (after) = [(m1v1)before+(m2v2)before−(m2v2)after] / (m1)after

= 0.5 x 3.5 + 3.5 x -1.5 – 3.5 x 0.5 / 0.55

= 1.875 – 5.25 – 1.75 / 0.55

= -9.318

Therefore, v1 (after) = – 9.31 m/s

The negative sign shows that it travels in the opposite direction.

**Example 1**

A ball of 1 kg is moving at a speed of 5 m / s collides with a cricket bat of 2.5 Kg swung at a speed of 1.5 m / s. right after the collision the cricket bat swings at a speed of 1.5 m / s. Calculate the magnitude and the velocity of the cricket ball.

**Solution:**

**Given:**

** m1 = 1 kg**

** m2 = 2.5 kg**

** v1 = 5 m/s**

** v2 = 1.5 m/s**

The relation between the colliding bodies is given by,

P1 (before) + P2 (before ) = P1 ( after ) + P2 ( after )

The momentum relation is expressed by,

(m1 v1) before + ( m2 v2 ) before = ( m1 v1 ) after + ( m2 v2 ) after

v1 (after) = [(m1v1)before+(m2v2)before−(m2v2)after] / (m1)after

= 1 x 5 + 2.5 x -1.5 – 2.5 x 1.5 / 1

= -2.5

Therefore, v1 (after) = – 2.5 m/s

The negative sign shows that it travels in the opposite direction.