A uniform rod of mass M and length `a` 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 `a/4` from the centre and stops after the collision. Find (a). the velocity of the centre of the rod and (b). the angular velocity of the rod abut its centre just after the collision.

Explanation

The situation is shown in figure. Consider the rod and the particle together as the system. As there is no external resultant force, the linear momentum of the system will remain constant. Also there is no resultant external torque on the system and so the angular momentum of the system about any line will remain constant.
Suppose the velocity of the centre of the rod is V and the angular velocity about the centre is `omega` .
(a) The linear momentum before the collision is mv and that after the collision is MV. Thus,
mv = MV, or V ` = (m)/(M)v`
(b) Let A be the centre of the rod when it is at rest. Let AB be the line perpendicular to the plane of the figure. Consider the angular momentum of the rod plus the particle system about AB. Initially the rod is at rest. The angular momentum of the particle about AB is
L = mv (a/4)
After the collision, the particle mass to rest. The angular momentum of the rod about A is
`vec(L) = vec(L) _(cm) M (vec(r)_(0) xx vec(V))`
As ` vec(r)_(0) ||vec(V), vec(r)_(0) xx vec(L) = 0 `
Thus , `vec(L) = vec(L)_(cm)`
Hence the angular momentum of the rod about AB is
`L = l omega = (Ma^(2))/(12) omega `
Thus, ` (mva)/(4) = (Ma^(2))/(12) omega "or" omega = (3 mv)/(Ma) `
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