A uniform rod of mass `M` and length `L` lies on a frictionless horizontal plane. A particle of same mass `M` moving with speed `v_(0)` perpendicular to the length of the rod strikes the end of the rod and sticks to it. Find the velocity of the center of mass and the angular velocity about the center of mass of (rod `+` particle) system.

Explanation

Since the mass of particle is comparable to the mass of the rod, distance of center of mass of system (rod `+` particle) from `A` is smaller than `L//2`.
Location of center of mass after collision
`x_(c.m)=(Mxx0+Mxx(L)/(2))/(M+M)=(L)/(4)`
Moment of inertia of the system about `C`
`I_(c)=Mx_(c.m.)^(2)+[(ML^(2))/(12)+M((L)/(2)-x_(cm))^(2)]`
`=M((L)/(4))^(2)+[(ML^(2))/(12)+M((L)/(4))^(2)]`
`=(5)/(24)ML^(2)`
After collision, the velocity of center of mass is `v_(c.m.)` and the angular velocity about center of mass is `omega`.
Conservation of linear momentum
`Mv_(0)x_(c.m.)=I_(c)omega`
`Mv_(0)(L)/(4)=(5)/(24)ML^(2)omega`
`omega=(6v_(0))/(5L)`