A particle of mass `m` moves with velocity `u_(1)=20m//s` towards a wall that is moving with velocity `u_(2)=5m//s`. If the particle collides with the wall without losing its energy, find the speed of particle just after the collision.

Explanation

Velocity of approach
`u_("approch")=(u_(1)+u_(2))`
Let the velocity of the particle just after the collision be `vecv_(1)`.
The velocity of separation,
`v_("seperation")=(v_(1)-v_(2))`
Since, according to Newton's experimental law
`e=(v_(1)-v_(2))/(u_(1)+u_(2))`
As there is no loss of energy, collision is elastic
Putting the value of `e=1`, we obtain
`(v_(1)-v_(2))/((u_(1)+u_(2)))=1` or `u_(1)+u_(2)=v_(1)-v_(2)`
or `v_(1)=u_(1)+u_(2)+v_(2)`
Putting `u_(1)=20m//s,u_(2)=5m//s` and `v_(2)=u_(2')` since the wall being very heavy (infinite mass).
moves with constant velocity, we obtain
`v_(1)=v_(0)+2v=20+2(5)=30m//sec`.
Solving (i)&(ii), we obtain `v_(1)=(12)/(7)m//s,v_(2)=(26)/(7)m//s`