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

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
\(
\begin{aligned}
& (-2 \times 1)+(0.1 \times 20) \\
& =(0.1 \times 0)+(2 \times v) \stackrel{\text {-ve }}{\leftarrow} \stackrel{+v e}{\rightarrow}
\end{aligned}
\)
Here, \(v\) is the velocity of \(\mathrm{CM}\) of ring after impact.
Solving the above equation, we have
\(
v=0
\)
Thus, CM becomes stationary.
\(\therefore\) Correct answer is (a).
Linear impulse during impact
(i) In horizontal direction \(J_1=\Delta P=0.1 \times 20=2 \mathrm{Ns}\)
(ii) In vertical direction
\(
J_2=\Delta P=0.1 \times 10=1 \mathrm{Ns}
\)


Writing the equation (about \(\mathrm{CM}\) )
Angular impulse
\(=\) Change in angular momentum
\(
\begin{aligned}
& 1 \times\left(\frac{\sqrt{3}}{2} \times \frac{1}{2}\right)-2 \times 0.5 \times \frac{1}{2} \\
& =2 \times(0.5)^2\left[\omega-\frac{1}{0.5}\right]
\end{aligned}
\)
Solving this equation \(\omega\) 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.
\(\therefore\) Correct options are (a) and (c).
Note In JEE 2011 official answer key, correct option were given \(\mathrm{a}, \mathrm{ac}\).
Analysis of Question
(i) Question is moderately difficult.
(ii) In such type of problems impulse due to friction during collision is ignored.