A stick of length `l` lies on horizontal table. It has a mass `M` and is free to move in any way on the table. A bal of mass `m` moving perpendicularly to the stick at a distance d from its centre with speed `v` collides elastically with it as shown in figure. what quantities are conserved in the collision ? what must be the mass of the ball, so that it remains at rest immediately after collision?

Explanation

Linear momentum angular momentum and kinetic energy are conserved in the process. From conservation of linear momentum
`Mv'=mv` ltbr. Or `v'=(m)/(M)v` ..(i)
Conservation of angular momentum gives,
`mvd=sqrt((Ml^(2))/(12))omega`
or `omega=((12mvd)/(Ml^(2)))` ...(ii)
Collision is elastic, Hence
`e=1`
or relative speed of approach
`=` relative speed of separation
`thereforev=v'+domega`
Subsitituting the values, we have
`v=(m)/(M)v+(12mvd^(2))/(Ml^(2))`
Solving it we get
`m=(Ml^(2))/(12d^(2)+l^(2))`