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 cente of the rod and b. the angular velocity of the rod abut its centre just after the collision.

Explanation

The situation is shown in ure. Consider the road and the particle together as the system. AS tehre is no exterN/Al resltant force, the linear momentum of the system will remain constant. Also ther eis no resultant exterN/Al torque on teh system and so the angular momentum of the system about any lilne will remain constant.
Suppose teh velocity of the centre of the rod is V and the angular velocity about the centre is `omega`.
a. the linear momentum before the colision is mv and that after teh collision is MV. thus,
`mv=mv,m or V=m/Mv`
b. Let A be the centre of teh rod whenit is ast rest. let AB be the line perpendicular to the plane of the ure. Consider the angular momentumof the rod plus the particle system abut 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 comes to rest. The angular momentum of the rod about A is
`vecL=vecL_(cm)+Mvecr_0xxvecV`
As `vecr_0||vecV, vecr_0xxvecV=0`
Thus, `vecL=vecL_(cm)` Hence the angular momentum of the rod about AB is
`L=Iomega=(Ma^2)/12omega`
Thus `(mva)/4=(Ma^2)/12omega or omega=(3mv)/(Ma)`