Question

A rod `AB` of mass `M` and length `L` is lying on a horizontal frictionless surface. A particle of mass `m` traveling along the surface hits the end `A` of the rod with a velocity `v_(0)` in a direction perpendicular to `AB`. The collision is completely elastic. After the collision the particle comes to rest. Find the ration `m//M`.

Answer 1

After collision the rod translates as well as rotates.
Conservation of linear momentum
`mv_(0)=Mv_(c.m.)impliesv_(c.m.)=(m)/(M)v_(0)` (`i`)
Conservation of angular momentum about the cneter of mass of rod
`mv_(0)(L)/(2)=I_(c.m.)omega=(ML^(2))/(12)omegaimpliesomega=(6mv_(0))/(ML)` (`ii`)
Since collision is completely elastic
`K_(i)=K_(f)`
`(1)/(2)mv_(0)^(2)=(1)/(2)Mv_(c.m.)^(2)+(1)/(2)I_(c.m.)omega^(2)` (`iii`)
`mv_(0)^(2)=M((m)/(M)v_(0))^(2)+(ML^(2))/(12)((6mv_(0))/(ML))^(2)`
`1=(m)/(M)+3(m)/(M)implies(m)/(M)=(1)/(4)`

Instead of using (`iii`), we can also use for point `A`.
Relative velocity of separtion `=-e` (relatice velocity of approach)
`v_(1)-v_(2)= -e(u_(1)-u_(2))`
`1:` ball, `2:` rod
`u_(1)=v_(0),u_(2)=0,v_(1)=0,v_(2)=v_(c.m.)+(L)/(2)omega,e=1`
`0-(v_(c.m.)+(L)/(2)omega)= -(v_(0)-0)`
`(m)/(M)v_(0)+(L)/(2)xx(6mv_(0))/(ML)=v_(0)`
`(m)/(M)+3(m)/(M)=1`
`(m)/(M)=(1)/(4)`