Question

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 ball 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?

Answer 1

By conservation of linear momentum and angular momentum, we have
`mv+0=m.0+MVimplies=V=(mv)/M`
and `mvd=Iomegaimplies omega=(mvd)/I=(mvd)/((ML^(2))/12)`
since collision is elastic, therefore
`1/2mv^(2)=1/2 M((mv)/M)^(2)+1/2((ML^(2))/12)((mvd)/((mL^(2))/12))^(2)`
`implies m=(ML^(2))/(L^(2)+12d^(2))`