Question

A ball with a mass of `5 kg` and velocity of `6 m/s` collides with a second ball with a mass of `3 kg` and velocity of `- 2 m/s`. If `40%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

`((v_1^a = 2.4, v_2^a = 4), (v_1^a = 3.6, v_2^a = 2))`

Explanation:

Calling `v_1^b, v_2^b` and `v_1^a,v_2^a` the velocities `b`-before and `a`-after collision we can equate:
`m_1 v_1^b + m_2 v_2^b = m_1 v_1^a + m_2 v_2^a`
and
`(1/2m_1 (v_1^b)^2 + 1/2 m_2 (v_2^b)^2) mu = (1/2m_1( v_1^a)^2 + 1/2 m_2 (v_2^a)^2)` with `mu = 0.4`
Solving for `v_1^a, and v_2^a` we obtain two possible arrangements
`((v_1^a = 2.4, v_2^a = 4), (v_1^a = 3.6, v_2^a = 2))`