x=...............

Eureka!!! I got it :P

See if pivot impulse is zero then ang momentum about the point of application of impulse will be conserved.. and also about the pivot Jx=ml²/3ω.......... waise no need of this :P

also as pivot is axis of rotation therefore v_c=ωl/2

about point of application of J

mv_c(x-l/2)-ml²/12ω=0
where v_c=ωl/2

therefore mωl/2(x-l/2)=ml²/12ω
(x-l/2)=l/6

x=l/6+l/2...
x=2l/3...

Note it doesn't depend on J...

also if there have been any collision then it doesn't depends on coefficient of restitution and velocity or mass of colliding particle and not on mass of rod also..

It only depends on length of rod :P

Pink so fast :)

Good speed... :>

angular impulse abt axis of rotation can also be taken right ?

linear impulse

J = mv_{com}= \frac{m\omega L}{2} ..........(1)

angular impulse

Jx=\frac{mL^{2}\omega }{3}.............(2)
divide 2 by 1

x = \frac{2L}{3}