Three (A,B,C) particle each of mass m are connected by three mass-less rod of length l. All three particle lies on smooth horizontal plane. A particle of mass moving along one of the rod with velocity `v_(0)` strikes on a particle and stops as shown in diagram. Angular velocity of eachh particle after collision is.

Options

A. `(V_(0))/(sqrt(3)l)`
B. `(V_(0))/(l)`
C. `(V_(0))/(2sqrt(2)l)`
D. `(V_(0))/(3sqrt(3)l)`

Answer

C

Explanation

initial angular momentum=final angular momentum
`L_(1)=mv_(0)(l)/(2sqrt(3))=I_(cm)omega`
`mv_(0)=(l)/(2sqrt(2))=3m((l)/(sqrt(3)))^(2)omega,omega=(v_(0))/(2sqrt(3)l)`
`y=(l)/(2)tan30^(@)=(1)/(2sqrt(3))`