Two particles of mass m and 2 m moving in opposite directions collide elastically with velocities v and 2v. Find their velocities after collision.

Explanation

Let the final velocities of m and 2m be `v_(1)` and `v_(2)` respectively as shown in the figure:

By conservation of momentum:
`m(2v)+2m(-v)=m(v_(1))+2m(v_(2))`
or `0+mv_(1)+2mv_(2)`
or `v_(1)+2v_(2)=0" "` .......(1)
and since the collision is elastic:
`v_(2)-v_(1)=2v-(-v)`
or `v_(2)-v_(1)=3v" "` ......(2)
Solving the above two equations, we get,
`v_(2) = v and v_(1) = –2v`
i.e., the mass 2m returns with velocity v while the mass m returns with velocity 2v in the direction shown in figure:

The collision was elastic therefore, no kinetic energy is lost, KE lost `=KE_(i)-KE_(f)`
or, `((1)/(2)m(2v)^(2)+(1)/(2)(2m)(-v)^(2))-((1)/(2)m(-2v)^(2)+(1)/(2)(2m)v^(2))`
`=0`