A circular hoop of mass `m` and radius `R` rests flat on a horizontal smooth surface. A bullet of same mass `m` and moving with velocity `upsilon` strikes the hoop tangentially and gets embedded in it. Neglect the thickness of hoop in comparision to radius of hoop.

Mark the correct option

Options

A. energy of system (hoop+ bullet) is conserved
B. momentum of system is conserved
C. angular momentum of system is conserved about any point in the plane of hoop
D. angular momentum is not conserved due to torque of gravitational force.

Answer

C

Explanation

Let velocity of `CM` of hoop be `V_(1)` and angular velocity of hoop be `omega`, resultant velocity of bullet just after so collision is `V_(1)+Romega`
From conservation of momentum, we get
`Mupsilon_(0)=mV_(1)+m(V_(1)+omegaR)...(1)`

From conservation of angular momentum about `O`,
Which is a point on ground just below the point of impact
`L_(i)=0`
`L_(f)=-mV_(1)R+mR^(2)omega`
`L_(i)=L_(f)`
`V_(1)=R_(omega)`
from eqns (1) and (2), we get
`omega=(upsilon_(0))/(3R)` and `V_(1)=(upsilon_(0))/3`
Absolute velocity of bullet`=V_(1)+Romega=(2 upsilon_(0))/3`
Impulse on bullet
`=4vec(P)=(2 m upsilon_(0))/3 hati-m upsilon_(0)hati=-(m upsilon_(0))/3 i`