A ball with a mass of `3` `kg` and velocity of `1` `ms^-1` collides with a second ball with a mass of `5` `kg` and velocity of `- 8` `ms^-1`. If `25%` of the kinetic energy is lost, what are the final velocities of the balls?

Momentum is conserved in all collisions. In this instance, kinetic energy is not conserved: 25% is lost and 75% remains.

The initial momentum is given by:

`p=m_1v_1+m_2v_2=3*1+5*(-8)`
`=3-40=-37` `kgms^-1`

The final momentum will be the same as the initial momentum.

(if we define 'to the right' as the positive direction, the net momentum of the system is to the left)

The initial kinetic energy is given by:

`E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2*3*1^2+1/2*5*(-8)^2`
`=3/2+160=161.5` `J`

The final kinetic energy will be 75% of the initial kinetic energy:

`75/100*161.5=121.13` `J`

The expression for the final momentum is:

`p=m_1v_1+m_2v_2=3*v_1+5*v_2=-37` : call this Equation 1

The expression for the final kinetic energy is:

`E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2*3*v_1^2+1/2*5*v_2^2`
`3/2v_1^2+5/2v_2^2=121.13` : call this Equation 2

We now have two equations in two unknowns, so we can solve the system.

Divide Equation 1 by 3 and rearrange to give:

`v_1=-37/3-5/3v_2`

Substitute this value for `v_1` into Equation 2:

`3/2*(-37/3-5/3v_2)^2+5/2*v_2^2=121.13`

`3/2*(152.1+41.1v_2+2.78v_2^2)+5/2v_2^2=121.13`

`228.15+61.65v_2+4.17v_2^2+2.5v_2^2=121.13`

Rearranging:

`6.67v_2^2+61.65v_2+107.2=0`

We can solve this quadratic equation using the quadratic formula (or other methods):

`v_2=(-61.65+-sqrt(61.65^2-4*6.67*107.2))/(2*6.67)`
`v_2= -6.92` or `-2.32` `ms^-1`

That is, the two possible solutions are that Ball 2, the `5` `kg` ball, travels to the left at either `6.93` `ms^-1` or at `2.32` `ms^-1`. Remember that it was initially traveling to the left at `8` `ms^-1`.

Substituting these values back into Equation 1, we find that `3*v_1+5*v_2=-37` which means `v_1=-0.8` `ms^-1` (when `v_2=-6.92` `ms^-1`) or `v_1=-8.46` `ms^-1` (when `v_2=-2.32` `ms^-1`).