Question

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?

Answer 1

In an elastic collision, both momentum and kinetic energy are conserved. The velocity of the `3kg` ball is `0.9 or 2.7 ms^-1` and the velocity of the `4kg` ball is `2.7 or 1.1 ms^-1`.

Explanation:

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.