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.

Answer

(a) `(m)/(M) = (1)/(4)` ;
(b) `x = (2L)/(3)` ;
( c) `(v_(0))/(2 sqrt(2))`

Explanation

Let just after collision, velocity of `C.M` of rod is `v` and angular velocity about `C.M` is `omega`.

from conservation of linear momentum in `x-direction`
`mv_(0) = Mv` ….(1)
from conservation of angular momentum about `C.M` of rod
`mv_(0) xx ((L)/(2)) = I omega. rArr mv_(0) xx (L)/(2) = (ML^(2))/(12) omega^(2)`
`mv_(0) = (ML omega^(2))/(6)` ....(2)
since, the collision is elastic kineticenergyis alos conserved.
`(1)/(2) mv_(0)^(2) = (1)/(2) Mv^(2) + (1)/(2) I omega^(2)`
`mv_(0) = Mv^(2) + (ML^(2))/(12) omega^(2)` ...(3)
From equation (1),(2) and (3)
`(m)/(M) =(1)/(4)`
`v = (mv_(0))/(M) = (v_(0)/(4)` and `omega = (6 mv_(0))/(ML)`
(b) Point `P` will be at rest if `x omega = v_(CM)`
`x = (v)/(omega) rArr x = (mv_(0)//M)/(6 mv_(0)//ML) rArr x = L//6`
`AP = (L)/(2) + (L)/(6) rArr AP = (2)/(3) L`
( c)
after time `t = (pi L)/(3 v_(0))`
angle rotated by rod `theta = omega t = (6 mv_(0))/(ML). (pi L)/(3 v_(0))`
`theta = 2 pi ((m)/(M))` put `(m)/(M) = (1)/(4) rArr theta = (pi)/(2)`
resultant velocity of point `P` will be
`|vec V_(p)|= sqrt(2) v rArr |vec V_(p)| = (v_(0))/(2 sqrt(2))rArr |vec V_(p)| = (v_(0))/(2 sqrt(2))`.