A uniform rod AB of length `L` and mass `m` is suspended freely at A and hangs vertically at rest when a particle of same mass `m` is fired horizontally with speed `v` to strike the rod at its mid point. If the particle is brought to rest after the impact. Then the impulsive reaction at A is horizontal direction is

Options

A. `mv//4`
B. `mv//2`
C. `mv`
D. `2mv`

Answer

A

Explanation

From conservation of angular momentum, just before and just after impact about point `O` we have
`L_(i)=L_(f)`
`thereforemv(L)/(2)=((mL^(2))/(3))omega`
`thereforeomega=(3v)/(2L)`
`v_(C)=(1)/(2)omega=(3)/(4)v`
Impulse J has changed the momentum of partic,le from mv to O. hence
for rod
`J_(H)=(3)/(4)mv-J=(3)/(4)mv-mv`
`=-(mv)/(4)` or `|J_(H)|=(mv)/(4)`