A bullet of mass `m` moving with a velocity of `u` just grazes the top of a solid cylinder of mass `M` and radius `R` resting on a rough horizontal surface as shown and is embedded in the cylinder after impact. Assuming that the cylinder rolls without slipping, find the angular velocity of the cylinder and the final velocity of the bullet.

Explanation

Let `v` and `omega` be the velocity of the bullet and angular velocity of the cylinder, respectively. Applying onservation of angular momentum about the point of contact of the cylinder with the floor we get
`mu2R=mv2R+(I_(CM)+MR^(2))omega`

here, `v=2Romega` and `I_(CM)=1//2MR^(2)`
Substituting and solving, we get
`v=(8mu)/(8m+3M)`
and `omega=(4mu)/((8m+3M)R)`