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A rod AB of mass 3m and length 4a is fal...

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 velocity of the particle immediately after the impact

A

`29/19 u down`

B

`19/29 u down`

C

`29/19,u up`

D

`27/19 udown`

Text Solution

Verified by Experts

The correct Answer is:
A

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)
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