A uniform rod of mass `M` and length `L` lies on a smooth horizontal plane. A particle of mass `m` moving at a speed `v` perpendicular to the length of the rod strikes it at a distance `L//6` from the center and stops after the collision. Find
(a) the velocity of center of mass of rod and
(b) the angular velocity of the rod about its center just after collision.

Explanation

Since the floor is smooth, no external force is used.
By the conservation of linear momentum
`mv=Mv_(c.m.)implies v_(c.m.)=(m)/(M)v`
By the conservation of angular momentum about the center of mass (no external torque)
`mv(L)/(6)=I_(c)omega=(ML^(2))/(12)omegaimplies omega(6mv)/(ML)`