Question

A ball with a mass of `2` `kg` and velocity of `5` `ms^-1` collides with a second ball with a mass of `7` `kg` and velocity of `- 4` `ms^-1`. If `40%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

Momentum is conserved, but in this instance kinetic energy is not... but we know how much is lost. The solution is `(-3.4, 2.9)`.

Explanation:

Let's call the `2` `kg` ball '1' and the `7` `kg` ball '2' for convenience.

Momentum before the collision:

`p=m_1v_1+m_2v_2=2*5+7*(-4)=10-28=-18` `kgms^-1`

Kinetic energy before the collision:

`E_k=1/2m_1v_1^2+1/2m_2v_2^2`
`=1/2*2*5^2+1/2*7*(-4)^2=25+56=81` `J`

The momentum after the collision will be the same as before, the kinetic energy after the collision will be 60% of the value before (due to the 40% loss): `48.6` `J`

Momentum after the collision:

`p=m_1v_1+m_2v_2=2v_1+7v_2=-18` : call this Equation 1

Kinetic energy after the collision:

`E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2*2*v_1^2+1/2*7*v_2^2=48.6` : call this Equation 2

We have two equations in two unknowns, which means we can solve them. Rearrange Equation 1 to find a value for `v_1`:

`v_1=-7/2v_2-9`

Substitute into Equation 2:

`1/2*2*(-7/2v_2-9)^2+1/2*7*v_2^2=48.6`

`(-7/2v_2-9)^2+7/2v_2^2=48.6`

`(49/4v_2^2+63v_2+81)+7/2v_2^2=48.6`

`63/4v_2^2+63v_2+81=48.6`

`63/4v_2^2+63v_2+32.4=0`

This is a quadratic equation, which can be solved using the quadratic formula or whichever method your prefer:

This yields the value `v_2=-0.61` or `-3.4` `ms^-1`

Substituting these values into Equation 1 yields:

`v_1=-6.9 or2.9` `ms^-1`

The possible solutions for `(v_1,v_2)` are `(-0.61, -6.9)` or `(-3.4, 2.9)`.

The first is physically impossible, since the ball on the right would be moving faster to the left than the one to the left of it, so the solution is `(-3.4, 2.9)`.