A uniform rod `AB` of length `L` and mass `M` is lying on a smooth table. A small particle if mass `m` strike the rod with velocity `v_(0)` at point `C` at a distance comes to rest after collision. Then find the value of `x`, so that point `A` of the rod remains stationary just after collision.
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Answer

`L//6`

Explanation

Rod will have translational as well as rotational motion about centre of mass
from conservation of linear momentum
`mv_(o) = MV_(CM)` …(1)
from conservation of angular momentum about `C.M`.
`Mv__(o) = (ML^(2))/(12) omega`
`omega = ((mv_(o) x)12)/(ML^(2))` ...(2)
for point `A` to be at rest
`V_(CM) = (L)/(2) omega` ...(3)
from (1),(2) & (3)
`(mv_(o))/(M) = (L)/(2) xx((mv_(o) x)12)/(ML^(2)) rArr x = (L)/(6)`.
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