A uniform rod AB of mass m and length 5a is free to rotate on a smooth horizontal table about a pivot through P, a point on AB such that AP = a. A particle of mass 2m moving on the table strikes AB perpendicularly at the point 2a from P with speed v, the rod being at rest. If the
coefficient of restitution between them is` (1)/( 4)` , find
their speeds immediately after impact.

Explanation

Let the point of impact be Q so that
PQ = 2a
Let P be the point of pivot that AP = a
Let the velocities of point, Q and the particle after impact
be `v_(q) and v_(p)` respectively then from momentum conservation about point P.
`L_(1) = L_(f)`
`2a(2mv) = I_(p)omega + (2a) (2mv_(p))" ". . . (i)`
`l_(p) = (1)/(3) m ((5a)/(2))^(2)+ m ((3a)/(2))^(2) " " {{:("use parallel"),("axis theorem"):}}`
`= (13 ma^(2))/(3)" ". . . (ii)`
use equation (ii) in equation (i)
`4 ma (v - v_(p)) = (13 ma^(2))/(3) omega `
`12 (v - v_(p)) = 13 a omega " "(iii)`
coefficient of restitutione `e = ("velocity of seperation")/("velocity of approach")`
`(1)/(4) = (v_(q) - v_(p))/(v)`
`rArr v_(q) - v_(p) = (v)/(4) " ". . . (iv)`
`v_(q) = 2 a omega " ". . . (v)`
Put value of `omega ` from eq (iii) to equation(v)
`v_(q) = 2((12)/(13)) (v - v_(p))`
So now from equation (iv)
`(24)/(13) (v - v_(p)) - v_(p) = (v)/(4) rArr v_(p) = (83 v)/(148)`
So in this way get ` omega = (15 v)/(37 a)`