A mass $m$ is moving with velocity $v$ collides in elastically with a bob of simple pendulum of mass $m$ and gets embedded into it. The total height to which the masses will rise after collision is :

\frac{v^{2}}{8 g}

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

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

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

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Solution

The correct option is A $\frac{v^{2}}{8 g}$
For inelastic collision, linear momentum is conserved
$\Rightarrow m v_{1} = 2 m v_{2} \Rightarrow v_{2} = \frac{v_{1}}{2}$
$L o s s i n K . E . = G a i n i n P . E .$
$= \frac{1}{2} m v_{1}^{2} - \frac{1}{2} (2 m) v_{2}^{2} = 2 m g h$
$\Rightarrow 4 m g h = m v_{1}^{2} - \frac{m v_{1}^{2}}{2} = \frac{m v_{1}^{2}}{2} = \frac{m v^{2}}{2}$
$\Rightarrow h = \frac{v^{2}}{8 g}$

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A mass m is moving with velocity v collides in elastically with a bob of simple pendulum of mass m and gets embedded into it. The total height to which the masses will rise after collision is :

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