A homogenous rod of length `l=etax` and mass `M` is lying on a smooth horizontal floor. A bullet of mass m hits the rod at a distance `x` from the middle of the rod at a velocity `v_(0)` perpendicular to the rod and comes to rest after collision. If the velocity of the farther end of the rod just after the impact is in the opposite direction of `v_(0)`, then:

Options

A. `etagt3`
B. `etalt3`
C. `etagt2`
D. `etalt6`

Answer

D

Explanation

Let after collision velocity of rod be `v` and angualr velocity be `omega` then
`mv_(0)=mv+MVimpliesmv_(0)x=mvx+(Ml^(2))/12 omega`
`mv_(0)=mv+(Momegaeta^(2)x)/12`
From above equations `(Momegaeta^(2)x)/12=MVimplies omegaeta^(2)x=12V`
Velocity of farther end of the rod `V=-(lomega)/2`
`=(omegaeta^(2)x)/12-(omegaetax)/2=(omegaetax)/2(eta/6-1)`
If this velocity is opposite to `v_(0)` then
`eta/6-1lt0implies etalt6`