A thin ring of mass 2 ~kg and radius 0.5 ~m is rolling without...

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 $$\begin{aligned}& (-2 \times1)+(0.1 \times20) \\ & =(0.1 \times0)+(2 \timesv) \stackrel{\text{-ve }}{ ← } \stackrel{+v e}{ → } \end{aligned}$$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$$\begin{aligned}& 1 \times \left(\frac{\sqrt{3}}{2} \times \frac{1}{2}\right)-2 \times0.5 \times \frac{1}{2}\\ & =2 \times(0.5)^2 \left[\omega-\frac{1}{0.5}\right] \end{aligned}$$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.