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(B) Let the mass of the ring $A$ be $m$ and the mass of the body $B$ be $m$. Initially,the ring is rolling without sliding,so its angular velocity is

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$\frac{v^2}{4g}$

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(B) Let the mass of the ring $A$ be $m$ and the mass of the body $B$ be $m$. Initially,the ring is rolling without sliding,so its angular velocity is $\omega = v/R$. The total initial kinetic energy is $K_i = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(v/R)^2 = mv^2$.At the maximum height $h$,the ring and the body $B$ move together with a common horizontal velocity $v'$. By conservation of linear momentum in the horizontal direction: $mv = (m + m)v'$,which gives $v' = v/2$.Since all surfaces are smooth,there is no friction,so the angular velocity $\omega$ of the ring remains constant throughout the motion.Applying the conservation of mechanical energy:$K_i = K_f + U_f$$mv^2 = \frac{1}{2}(m+m)(v')^2 + \frac{1}{2}I\omega^2 + mgh$$mv^2 = \frac{1}{2}(2m)(v/2)^2 + \frac{1}{2}(mR^2)(v/R)^2 + mgh$$mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}mv^2 + mgh$$mv^2 = \frac{3}{4}mv^2 + mgh$$mgh = mv^2 - \frac{3}{4}mv^2 = \frac{1}{4}mv^2$$h = \frac{v^2}{4g}$

In the figure shown,a ring $A$ is initially rolling without sliding with a velocity $v$ on the horizontal surface of the body $B$ (of the same mass as $A$). All surfaces are smooth. $B$ has no initial velocity. What will be the maximum height reached by $A$ on $B$?

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