A thin ring of mass `2kg` and radius `0.5 m` is rolling without slipping on a horizontal plane with velocity `1m//s`. A small ball of mass `0.1 kg` moving with velocity `20 m//s` in the opposite direction, hits the ring at a height of `0.75 m` and goes vertically up with velocity `10 m//s`. Immediately after the collision

Options

A. the ring has pure rotation about its stationary CM
B. the ring comes to a complete stop
C. friction between the ring and the ground is to the left
D. there is no friction between the ring and the ground

Answer

A, C

Explanation

The data is incomplte. Let us asumethat friction from ground on ring is not impulsive during impact.
From linear momentum conservation in horizontal direction, we have
`(-2 xx 1) + (0.1 xx 20) = (0.1 xx 0) + (2 xx v)`

Here,v is the velocity of CM of ring after impact.
Solving, the above equation, we have v=0
Thus, CM becomes stationary, `therefore` Correct option is (A).
Linear impulse during impact
(i) In horizontal direction, `J_(1) = Deltap =0.1 xx 20 =2N-s`
(ii) In vertical direction `J_(2) =Delta_(p)=0.1 xx 10 =1N-s`

Angular impulse = Change in angular momentum
We have, `1 xx (sqrt(3)/2 xx 1/2) -2 xx 0.5 xx 1/2 =2 xx (0.5)^(2) [omega -1/(0.5)]`
Solving this equation `omega` comes out to the positive or `omega` anticlockwise. So just after collision rightwards slipping is taking place.
Hence, friction is leftwards. Therefore, option ( C) is also correct.