A uniform rod `AB` of mass `m` and length `l` is at rest on a smooth horizontal surface. An impulse `J` is applied to the end `B`, perpendicular to the rod in the horizontal direction. Speed of particlem `P` at a distance `(l)/(6)` from the centre towards `A` of the rod after time `t = (pi m l)/(12 J)` is.

Options

A. `2 J/m`
B. `J/(sqrt(2)m)`
C. `J/m`
D. `sqrt(2)J/m`

Answer

D

Explanation

Let v and `omega` be the linear and anglular speeds of the rod
after applying an impulse J at B. Then from
Impulse =change in momentum
we have
`mv=J`or `v=J/m`
1
`Iomega=J . l/2`
or `(ml^(2))/12omga=j/l/2 or omega=(6J)/(ml)`...........2
After the given time `t=(pi ml)/(12 J),` the rod will rotate an angle
`theta=omegat =((6J)/(ml))((pi ml)/(12J))=(pi)/2`
`l/6 . omega=(l/6)=((6J)/(ml))=J/m=v`
`:.|vecv_(P)|=sqrt(2)v=sqrt(2)J/m`