Question

A ball with a mass of `2 kg` and velocity of `6 m/s` collides with a second ball with a mass of `1 kg` and velocity of `- 3 m/s`. If `20%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

See below

Explanation:

firstly, momentum (`mv`) is always conserved so we can say that

`2*6 + 1 *(-3) = 2 u + v`

Or `9 = 2 u + v qquad triangle`

In terms of KE (`1/2 m v^2`), we can also say that

`color(red)(0.8) ( 1/2*2*6^2 + 1/2 1 *(-3)^2 )= 1/2*2*u^2 + 1/2 1 v^2`

.....the `0.8` indicating that the final amount of energy is only `80`% of the original amount.

`implies 64.8= 2u^2 + v^2`

These solve to give:

In terms of the second possible solution, for the first ball to have positive (right to left) velocity `u approx 0.5` after the collision and the second ball to continue to have negative (left to right) velocity after the collision, the balls would need to pass through each other. We can reject that as a practical solution.

In the first solution, both balls are travelling right to left, and the lighter ball is travelling with greater speed which conforms with intuition.