Question

A ball with a mass of `6` `kg` and velocity of `7` `ms^-1` collides with a second ball with a mass of `4` `kg` and velocity of `- 8` `ms^-1`. If `15%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

Momentum is conserved, but because this is an inelastic collision kinetic energy is not. The final velocities of the masses are:

`6` `kg` mass: `6.5` or `-4.5` `ms^-1`

`4` `kg` mass: `-7.25` or `9.25` `ms^-1`

Explanation:

Before the collision:

`p=m_1v_1+m_2v_2=6xx7+4xx(-8)`
`=42-32=10` `kgms^-1`

`E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2xx6xx7^2+1/2xx4xx(-8)^2`
`=275` `J`

After the collision `15%` of the energy is lost, so `85%` remains:

`0.85xx275=233.75` `J`

Because momentum is always conserved, the momentum of the system after the collision will still be `10` `kgms^-1`.

We can now set up a system of simultaneous equations.

After the collision:

`p=10=6v_1+4v_2`

`E_k=233.75=1/2xx6xxv_1^2xx4xxv_2^2=3v_1^2+2v_2^2`

We can use the first equation to express `v_2` in terms of `v_1`:

`v_2=(10-6v_1)/4=2.5-1.5v_1`

We can then substitute that expression into the second equation, so that it will be in only one variable:

`233.75=3v_1^2+2(2.5-1.5v_1)^2`

The algebra is a bit messy to do and mark up here, so I might leave that as an exercise for you to do. It's just math, not physics.

Because it's a quadratic equation it yields two solutions, `v_1=6.5` and `v_1=-4.5` `ms^-1`. These correspond to the `6` `kg` ball either bouncing back and therefore having a negative velocity or continuing in the direction it was initially moving at a slightly slower velocity.

Let's see what they lead to in terms of `v_2`:

`v_2=2.5-1.5v_1`

If `v_1=6` then `v_2=-7.25` `ms^-1`

If `v_1=-4.5` then `v_2=9.25` `ms^-1`

Both sets of solutions actually make sense.