A circular hoop of mass m and radius R rests flat on a horizontal frictionless surface A bullet also of mass m and moving with a velocity v strikes the hoop and gets embedded in it the thickness of the hoop is much smaller then R the angular velocity with which the system rotates after the bullet strikes the hoop is

Options

A. `V//(4R)`
B. `V//(3R)`
C. `2V//(3R)`
D. `3V//(4R)`

Answer

B

Explanation

By conservation of momentium `mV_(1)+m(V_(1)+omegaR)=mV_(0)`
`Rightarrow 2V_(1)+omegaR=V_(0)`
By applying conservation of angular momemtium about O(O is a point of ground frame near the point of collision)
`L_(i)=0`
`L_(f)=-mV_(1)R+mR^(2)omega` [only due to ring applying `vec(r)_(cm)xxvec(p)+I_(cm)vec(omega)` considering outward normals as +ve]
`L_(i)=0`
`L_(f)=-mV_(1)R+mR^(2)omega` [only due to ring applying `vec(r)_(cm)xxvec(p)+I_(cm)vec(omega) considering outward normal as + ve]
`L_(i)=L_(f) Rightarrow V_(1)=Romega`
Solving (1) and (2)
`omega=V_(0)/(3R)`