A rod AB of mass `M` and length L is lying on a horizontal frictionless surface. A particle of mass `m` travelling along the surface hits the end A of the rod with a velocity `v_(0)` in a direction perpendicular to AB. The collision in elastic. After the collision the particle comes to rest
(a). Find the ratio `m//M`
(b). A point P on the rod is at rest immediately after collision find the distance AP.
(c). Fid the linear speed of the point P a time `piL//3v_(0)` after the collision.

Explanation

(a). Suppose velocity of COM of the rod just after collision is v and angular velocity about is `omega` is `omega`. Applying following three laws
(1). External force on the system (rod `+` mass) in horizontal plane along x-axis is zero
`therefore` applying conservation of linear momentum in x-direction
`mv_(0)=Mv` ..(i)
(2). Net torque on the system COM of rod is zero
`therefore` Applying conservation of angular momentum about COM of rod, we get `mv_(0)((L)/(2))=Iomega`
or `mv_(0)(L)/(2)=(ML^(2))/(12)omega`
or `mv_(0)=(MLomega)/(6)` ...(ii)
(3) since, the collision is elastic, kinetic energy is also conserved
`therefore(1)/(2)mv_(0)^(2)=(1)/(2)Mv^(2)+(1)/(2)Iomega^(2)`
or `mv_(0)^(2)=Mv^(2)+(ML^(2))/(12)omega^(2)` ..(iii)
From Eqs (i), (ii) and (iii) we get the following result
`(m)/(M)=(1)/(4)`
`v=(mv_(0))/(M)` and `omega=(6mv_(0))/(ML)`
(b). point P will be rest if `xomega=v`
or `x=(v)/(omega)=(mv_(0)//M)/(6mv_(0)//ML)` or `x=L//6`
`thereforeAP=(L)/(2)+(L)/(6)` or `AP=(2)/(3)L)`
(c). After time `t=(piL)/(3v_(0))`
Angle rotated by rod `theta=omegat=(6mv_(0))/(ML).(piL)/(3v_(0))`
`=2pi((m)/(M))=2pi((1)/(4))`
`thereforetheta=(pi)/(2)`
Therefore, situation is as shown in figure
`therefore `resultant velocity of point `P` will be
`|V_(P)|=sqrt(2)v=sqrt(2)((m)/(M))v_(0)`
`=(sqrt(2))/(4)v_(0)=(v_(0))/(2sqrt(2))`
or `|V_(P)|=(v_(0))/(2sqrt(2))`