An impulse `J = mv` at one end of a stationary uniform frictionless rod of mass `m` and length `l` which is free to rotate in a gravity-free space. The impact is elastic. Instantaneous axis of rotation of the rod will pass through

Options

A. its centre of mass
B. the centre of mass of the rod plus ball
C. the point of impact of the ball on the rod
D. the point which is at a distance `2//3` from the striking end

Answer

D

Explanation

`J=mv=mv'`
`implies` Velocity of the `CM` of ther rod `=v`
applying impulse momentum equation about the `CM` of rod

`J=1/2=I_(CM)omegaimplies(mvl)/2=((ml^(2))/12)omegaimplies omega=(6V)/l`
About instantaneous axis of rotatioin the rod is considered to have pure rotation.
Let instantaneous axis of rotation be located at a distance `x` from the colliding end.
`((omegal)/2-v)/(l-x)= (v+(omegal)/2)/x`...i ltbr substituting the value of `omega=6V/l` in eqn i we get `x=2/3l`