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

Let us assume that speed of centre of mass of rod be `v_(0)` and angular speed about centre of mass be `omega` immediately after collision.

Taking the ball and rod to be part of a combined system, so for our system `sumF_("ext")=0`. So momentum can be conserved.
`(a) :. mv_(0)=MVimpliesv=(m)/(M)v_(0)`..........`(1)`
`sumtau=0` about centre of rod. So conserving angular momentum
`mv_(0)*(L)/(2)=I_(CM)*omegaimpliesmv_(0)(L)/(2)=(ML^(2))/(12)omega`
`:.omega=(6m)/(M)(v_(0))/(L)`..........`(2)`
Since the collision is perfectly elastic, K.E. of system will remain conserved.
`(1)/(2)mv_(0)^(2)=(1)/(2)MV^(2)+(1)/(2)I_(C.M.*omega^(2)` [Replacing the values of `v` and `omega` from equation `(1)` and `(2)` ]
`implies(1)/(2)mv_(0)^(2)=(1)/(2)M((m)/(M)v_(0))^(2)+(1)/(2)*(ML^(2))/(12)xx(36m^(2))/(M^(2))(v_(0)^(2))/(L^(2))`
or `1=(m)/(M)+(3m)/(M)=(4m)/(M)`
`(m)/(M)=(1)/(4)`
`(b)` For velocity of any point `P` to be zero immediately after the collision, the net velocity of point `P` is
`v+omegax=0`
`:.x=-(v)/(omega)`
`=-(mv_(0))/((M*6mv_(0))/(ML))`
`:.=-(L)/(6)`

(-ve sign shows that the point will be on opposite side of what we have taken and is quite-logical too)

So, `AP=(L)/(2)+(L)/(6)=(4L)/(6)=(2L)/(3)`
`(c )` Just after collision,
`v=(m)/(M)v_(0)=(v_(0))/(4)`
`omega=(6mv_(0))/(ML)=(3v_(0))/(2L)`
Since `v=(omegaL)/(6)`, both the translational speed as well as speed due to rotation will be equal to `v=(v_(0))/(4)` at any time after collision.
After time `t=(piL)/(3v_(0))`

`theta=omega t=(3v_(0))/(2L)xx(piL)/(3v_(0))=(pi)/(2)`
So rad has turned by an angle `(pi)/(2)`, till that time after collision.
The point `P` has two velocities one in the direction of motion of cenre of rod and another perpendicular to the rod.

Resultant speed `=sqrt(v^(2)+v^(2))=vsqrt(2)=(v_(0))/(4)sqrt(2)`