Question

A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/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 data is incomplete. Let us assume that friction from ground on ring is not impulsive during impact. From linear momentum conservation in horizontal direction, we have
$(-2\times 1)+(0.1\times 20)$
$(0.1\times 0)+(2\times v)$
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
$J_1=\Delta p=0.1\times 20=2N-s$
(ii) In vertical direction $J_2=\Delta p=0.1\times 10=1N-s$
Writing the equation (about CM)
Angular impulse = Change in angular momentum
We have,
$1\times (\frac{\sqrt{3}}{2}\times \frac{1}{2})-2\times 0.5\times \frac{1}{2}=2\times (0.5{)}^2[\omega -\frac{1}{0.5}]$
Solving this equation co comes out to be positive or $\omega$ anti-clockwise. So just after collision rightwards slipping is taking place.
Hence, friction is leftwards.
Therefore, option (c) is also correct.