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. If we take rod and particle as a system, the impulse between them will be internal. We can conserve the linear momentum of rod `+` particle's system. Let `V` and `omega` be the velocity and angular velocity of the `CM` just after collision.
Using conservation of linear momentum, we have
`mv_(0)=0+Mv_(c)`..........i
Here is this case we can conserve angular momentum about any point on the rod.
And by coservation of angular momentum about the entire of mass of the rod, we get
`mv_(0)L/2=lomega` or `mv_(0)L/2=((ML^(2))/12)omega`...........ii
Form eqn i and ii `v_(c)=(mv_(0))/M` and `omega= (6mv_(0))/(ML)`
We can apply conservation of anngular momentum about point of collision.
`vecL_("initial")=vecL_("final")`
`O=I_(0)omega(-hatk)+Mv_(0)l/2(hatk)`
`(ML^(2))12 omega= (Mv_(c)L)/2` .............iii
which gives `omegaL=6v_(c)`
Solving eqn i and iii we get the same results.
Since collisiion is completely elastic, therefore `KE` before collision is equa to `KE` after collision
or `1/2mv_(0)^(2)+0=1/2mv_(c)^(2)+1/2omega^(2)`
or `1/2mv_(0)^(2)=1/2M((mv_(0))/M)^(2)+1/2((6mv_(0))/(ML))^(2)`
Which gives `m/M=1/4` and `v_(c)=(v_(0))/4`
Alternative method: We can apply coefficient of restitution equation at the positioinof point of collision, we have
`e=((v_(2)-v_(1))/(u_(1)-u_(2)))`
Here `u_(1),u_(2),v_(1)` and `v_(2)` are the velocities of particles and point on the rod at collision point.
`u_(1)=v_(0), u_(2)=0`
`v_(1)=0` and `v_(2)=v_(c)+(omegal)/2`
As collision in elastic `e=1`
Substituting the values in restitution equation we get
`I=((v_(c)+omegaL)/2)-0)/(v_(0)-0)impliesv_(0)=((mv_(0))/M)+((6mv_(0))/(ML))L/2`
Which gives `m/M=1/4`
b. Velocity of point `P` immediately after collisionn is zero, let if be at a distance `y` and from `CM`
`:. v_(c)-omegay=0` or `(mv_(0))/M-((6mv_(0))/(ML))y=0`
which gives `y=L/6`
`:. AP=L/2+L/6=(2L)/3`
c. Angle rotated by the rod at time `(piL)/(3v_(0))`
`theta=omegat=((6mv_(0))/(ML))xx(piL)/(3v_(0))=pi/2`
The rod turns through `pi/2` in this interval of time. the velocity of point `P` in `y`-direction will be
`v_(y)=omega=((6mv_(0))/M)xxL/6=(v_(0))/4`
The resultant velocity of point `P`,
`v=sqrt(v_(c)^(2)+v_(y)^(2))=sqrt(((v_(0))/4)^(2)+((v_(0))/4)^(2))implies(v_(0))/(2sqrt(2))`