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 than R. The angular velocity with which the system rotates after the bullet strikes the hoop is:
v/3R
Let system moves with v1 after impact & rotates with ω anti-clockwise
Consider any point P on a straight horizontal line passing through the centre of the hoop 0 and parallel to →v
Angular momentum initial
→Li=mvR^j (angular momentum of bullet about point p)
Angular momentum final
→Lf={m(v1+ωR)R+ICMω}^j
→Lf=(mv1R+2mωR2)^j
As there is no net external torque about p,
→Li=→Lf
⇒v1+2ωR=v - - - - - - (1)
Now by conservation of linear momentum,
mv1+m(v1+ωR)=mv
⇒2v1+ωR=v ....(2)
Subtracting (2) From (1) we get
v1=ωR
substituting the above in equation (2)
2ωR+ωR=v
ω=v3R
hence B is the correct option