A ball with a mass of `3 kg` is rolling at `4 ms^-1` and elastically collides with a resting ball with a mass of `4 kg`. What are the post-collision velocities of the balls?

Momentum is conserved in all collisions. Kinetic energy is conserved in elastic collisions but not in inelastic or partially elastic collisions.

The initial momentum of the whole system is `p = mv` for the `3kg` ball: the `4kg` ball has zero momentum because its has zero velocity.

`p=mv=3*4=12 kgms^-1`

The total momentum after the collision will be the same after the collision. If we call the `3kg` ball 1 and the `4kg` ball 2, the final momentum will be given by:

`p = m_1v_1 +m_2v_2 = 3v_1 +4v_2 = 12` (call this Equation 1)

The total kinetic energy before the collision will be `E_k=1/2mv^2` for the `3 kg` ball only - the `4 kg` ball has zero kinetic energy because it has zero velocity.

`E_k = 1/2mv^2 = 1/2*3*4^2 = 24 J`

Since kinetic energy is conserved, the final kinetic energy will be the same, and will be given by:

`E_k = 1/2m_1v_1^2 + 1/2m_2v_2^2 = 3/2v_1^2+4/2v_2^2 = 24`

Multiply both sides by 2 to make it tidier:

`3v_1^2 + 4v_2^2 = 48` (call this Equation 2)

We now have two equations and two unknowns, so we can solve them as simultaneous equations. Rearranging Equation 1 to express `v_2` in terms of `v_1`:

`v_2 = (12-3v_1)/4`

Substituting this into Equation 2:

`3v_1^2+4((12-3v_1)/4)^2 = 48`

I'll leave the algebra as an exercise for the reader, but this solves to give `v_1=0.9ms^-1 or 2.5 ms^-1`. Substituting this back into Equation 1 gives `v_2=2.7 ms^-1 or 1.1 ms^-1`, respectively.

Substituting these values should confirm that both momentum and kinetic energy were conserved.