Question

A ball with a mass of `4 kg` and velocity of `4 m/s` collides with a second ball with a mass of `2 kg` and velocity of `- 1 m/s`. If `15%` of the kinetic energy is lost, what are the final velocities of the balls?

Answer 1

Final velocities of balls are `0.94m/s` and `5.12m/s`.

Explanation:

As the velocity of first ball is`4m/s` and second ball is `-1m/s`, it is apparent that they are moving towards each other before collision.

Let the final velocity of firrst ball be `v_1m/s` and that of second ball be `v_2m/s`

Conservation of momentum - As momentum is always conserved, the initial momentum is equal too final momentum.

Hence `4xx4+2xx(-1)=4xxv_1+2xxv_2`

i.e. `4v_1+2v_2=14` i.e. `2v_1+v_2=7`

i.e. `v_2=7-2v_1` ..............(1)

Further initial kinetic energy is

`1/2xx4xx4^2+1/2xx2xx(-1)^2=32+1=33` joules

and final kinetic energy is

`1/2xx4xxv_1^2+1/2xx2xxv_2^2`

and as it is `15%` less we have

`2v_1^2+v_2^2=33xx0.85=28.05`

putting value of `v_2` from (1), we get

`2v_1^2+(7-2v_1)^2=28.05`

or `6v_1^2-28v_1+20.95=0`

and `v_1=(28+-sqrt(28^2-4xx6xx20.95))/12`

= `(28+-sqrt281.2)/12`

= `(28+-16.77)/12` i.e. `3.73` or `0.94`

and using (1), `v_2=-0.46` or `5.12`

However, second ball cannot move with a velocity of `-0.46` after collision

and hence final velocities of balls are `0.94m/s` and `5.12m/s`.