Question

A ball with a mass of `5 kg` and velocity of `2 m/s` collides with a second ball with a mass of `6 kg` and velocity of `- 3 m/s`. If `75%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

• Final velocity of the `5color(white)(l)"kg ball:" -1.94color(white)(l)"m"*"s"^(-1)`

• Final velocity of the `6color(white)(l)"kg ball:"-1.70color(white)(l)"m"*"s"^(-1)`

Explanation:

Let the final velocity be `v_1` for the first mass and `v_2` for the second mass.

`p_"initial"=5*2+6*(-3)=-8 color(white)(l) "N"*"s"^(-1)`
`p_"final"=5*v_1+6*v_2`

`E_(k, "initial")=1/2*5*2^2+1/2*6*3^2=37color(white)(l)"J"`
`E_(k, "final")=1/2*5*v_1^2+1/2*6*v_2^2`

`75%` of the kinetic energy has been lost so
`E_(k, "final")=(1-75%)*E_(k, "initial")=25%E_(k, "initial")=37/4color(white)(l)"J"`

Momentum conserves, hence
`p_"final"=p_"initial"=-8 color(white)(l) "N"*"s"^(-1)`

Construct a system of equations
`5*v_1+6*v_2=-8` (1)
`1/2*5*v_1^2+1/2*6*v_2^2=37/4` (2)

Substitute `v_1` in (1) with an expression of `v_2` derived from the first equation, `v_1=-1/5*(8+6*v_2)`

`5/2*(-1/5*(8+6*v_2))^2+3*v_2^2=37/4`
`132*v_2+192*v_2-57=0`

`v_2=0.253color(white)(l)"m"*"s"^(-1)` or `v_2=-1.70color(white)(l)"m"*"s"^(-1)`

Correspondingly,
`v_1=-1.651color(white)(l)"m"*"s"^(-1)` or `v_1=-1.94color(white)(l)"m"*"s"^(-1)`

The first scenario is not possible since the one ball won't be able to overtake another in a collision in one-dimensional motions.

Therefore
`v_1=-1.94color(white)(l)"m"*"s"^(-1),color(white)(m)v_2=-1.70color(white)(l)"m"*"s"^(-1)`