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 mass same 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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$\frac{3 V^{2}}{4 g}$

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

$\frac{V^{2}}{3 g}$

$\frac{V^{2}}{2 g}$

answer is B.

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Detailed Solution

$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m V^{1^{2}} + \frac{1}{2} I ω^{2} + m v^{1^{2}} + m g h$

$v^{1} = v / 2 \Rightarrow h = \frac{v^{2}}{4 g}$

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