A ball with a mass of `5 kg` is rolling at `1 m/s` and elastically collides with a resting ball with a mass of `4 kg`. What are the post-collision velocities of the balls?

Since the collision is elastic, we have conservation of momentum

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

And conservation of kinetic energy

`1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2`

The mass the first ball is `m_1=5kg`

The velocity of the first ball before the collision is `u_1=1ms^-1`

The mass of the second ball is `m_2=4kg`

The velocity of the second ball before the collision is `u_2=0ms^-1`

The velocity of the first ball after the collision is `=v_1ms^-1`

The velocity of the second ball after the collision is `=v_2ms^-1`

Therefore,

`5*1+4*0=5*v_1+4*v_2`

`4v_2+5v_1=5`....................................`(1)`

`1/2*5*1^2+1/2*4*0=1/2*5*v_1^2+1/2*4*v_2^2`

`4v_2^2+5v_1^2=5`..............................`(2)`

Solving for `v_1` and `v_2` in equations `(1)` and `(2)`

From equation `(1)`, `v_2=(5(1-v_1))/4`

Substituting in equation `(2)`

`5v_1^2+4*((5(1-v_1))/4)^2=5`

`4v_1^2+5+5v_1^2-10v_1=4`

`9v_1^2-10v_1+1=0`

`v_1=(10+-sqrt(((-10)^2)-4(9)(1)))/(18)=(10+-8)/18`

Therefore,

`v_1=1ms^-1` or `v_1=1/9=0.11ms^-1`

`v_2=0ms^-1` or `v_2=10/9=1.11ms^-1`

The first solutions are the initial conditions.

The post collisions velocities are `=011ms^-1` and `=1.411ms^-1`