Question

$A$ thin ring of mass 2 kg 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:

• 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 1

Answer: (A), (C)

The correct options are
A friction between the ring and the ground is to the left.
C the ring has pure rotation about its stationary $CM$.

Lets assume that friction between the ground and the ring gives no impulse during the collision with the ball.

Using conservation of momentum along the x-axis we get that the CM of the ring will come to rest.

Thus option A is correct.

Secondly the question tells us that the ball gets a velocity in the vertical direction hence there must be an impulse in the vertical direction. Thus on the ring at the point of contact there is a horizontal and a vertical impulse. These will have components along the tangent of the ring, which will provide angular impulses.

Using angular impulse = change in angular momentum we get:

$=2cos{30}^o\frac{1}{2}-1sin{30}^o\frac{1}{2}=2\frac{1}{4}({\omega }_2-{\omega }_1)$

note that we have assumed that direction of angular velocities is same before and after and since LHS of the above equation is positive ${\omega }_2>{\omega }_1$

thus the ring must be slipping to right and hence the friction will be to the left as it will be opposite to the direction of motion. Thus option C is correct