A rod `AB` of mass `3m` and length `4a` is falling freely in a horizontal position and `c` is a point distance a from `A`. When the speed of the rod is `u`, the point `c` collides with a particle of mass `m` which is moving vertically upwards with speed `u`. if the impact between the particle and the rod is perfectely elastic find

The speed of `B` immediately after the impact is

Options

A. `19/27 u down`
B. `19/27 u up`
C. `27/19 u down`
D. `27/19 u up`

Answer

C

Explanation

In order to use the law of restitution. We need the speed of point `C`, which is `u_(1)-a omega` (down wards)

law of restitution now given relative velocity of separation at point of impact `=e`(relative velcity of approch) or `u_(2)-(u_(1)-aomega)=e(u+u)`
hence `2u=u_(2)-u_(1)+a omega`
For the rod the speed of the particle is `29/19 upsilon` (down wards)