A disc of radius R and mass M is rolling horizontally without slipping with speed v. It then moves up an incline, as shown in fig.
The maximum height upto which it can reach is ( assume pure rolling on inclined plane ) :

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

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

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

$\frac{3 v^{2}}{4 g}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} [1 + \frac{K^{2}}{R^{2}}] = mgh \\ v^{2} [1 + \frac{1}{2}] = 2 gh \\ h = \frac{3}{4} \frac{v^{2}}{g} \end{array}$

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