**Elastic Collision Formula **

An encounter between two bodies in which the total kinetic energy of both the bodies after the encounter is equal to their total kinetic energy before the encounter is called Elastic collision. Elastic collisions occur only when there is no net conversion of kinetic energy into different forms. A perfectly elastic collision is one wherein there no loss of kinetic energy during the collision. In any elastic collision, conservation of momentum and conservation of kinetic energy can be observed.

If two elastic bodies of masses m1, m2 with initial velocity u1 and u2 approaching towards each other undergo collision.

The Elastic Collision formula of momentum is given by

Where,

m1 = Mass of 1st body

m2 = Mass of 2nd body

u1 =Initial velocity of 1st body

u2 = Initial velocity of second body

v1 = Final velocity of first body

v2 = Final velocity of second body

It says “Momentum before the collision is equal to momentum after the collision.”

The Elastic Collision formula of kinetic energy is given by

Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

**Example 1**

If the ball has a mass 5 Kg and moving with the velocity of 12 m/s collides with a stationary ball of mass 7 kg and comes to rest. Calculate the velocity of the ball of mass 7 Kg ball after the collision.

**Solution:**

Given parameters are

Â Mass of 1st ball, m1 is 5 kg

Â Initial Velocity of the first ball, U1Â is 12 m/s

Â Mass of the second ball, m2 is 7 kg

Â Initial velocity of the second ball, u2 is 0

Â Final Velocity of first ball, v1 is 0

Final Velocity of the second ball, v2 =?

The elastic collision formula is given as

1 / 2 m1u1^{2 }+ 1 / 2 m2u2^{2}Â =Â 1 / 2 m1v1^{2} +1 / 2 m2 v2^{2}

1 / 2 Ã— 5 Ã— 122 + 1 / 2 x 7 Ã— 0Â =Â 1 / 2 Ã— 5Ã— 0^{2} + 1 / 2 Ã— 7 Ã— v2^{2}

360 = 3.5 v^{2}

288 = 3 v^{2}

v^{2} = 102.85

v = âˆš102.85

= 10.141 m/s

**Example 2**

A 10 Kg block is moving with an initial velocity of 12 m/s with 8 Kg wooden block moving towards the first block with velocity 4 m/s. The second body comes to rest after the collision. Determine the final velocity of first body

**Solution:**

Given parameters are

Mass of 1st ball, m1 is 10 kg,

Initial Velocity of the first ball, u1 is 12 m/s,

Mass of the 2nd ball, m2 is 8 kg,

Initial velocity of the second ball, u2 is 4 m/s,

Final Velocity of 2nd ball, v2 is 0

Final Velocity of the first ball, v1 =?

The Elastic collision formula is given as

m1u1 + m2u2 = m1v1 + m2v2

10 Ã— 12 + 8 Ã— 4 = 10 Ã— v1 + 8 Ã— 0

120 + 32 = 10 v1 + 0

152 = 10 v1

âˆ´Â v1 = 15.2 m/s