A slender rod of mass `M` and length `L` rests on a horizontal frictionless surface. The rod is pivoted about one of ends. The impulse of the force exerted on the rod by the pivot when the rod is struck by a blow of impulse `J` perpendicular to the rod at other end is

Options

A. `J`
B. `J/2`
C. `J/3`
D. Information is insufficient

Answer

B

Explanation

Let J' be the implse exerted by the pivot on the rod. Then from impulse mometum theorem `J+J'=mv_(0)`, where `v_(0)` is the velocity acquired by the centrte of mass.

Now apply, angular impulse `=` change in angular momentum about the pivot
`JxxL=(ML^(2))/3omega`
`Jxx(ML)/3omega`
where `omega` is the angular velocity of the rod
`v_(0)=(omegaL)/2impliesMv_(0)=(MomegaL)/2`
`J+J'=(3J)/2impliesJ'=J/2`