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