Consider the situation as shown in diagram. The bullet penetrates the block and emerges with a speed $\frac{V_{0}}{3}$ . It after collision the string becomes horizontal, V₀ will be

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$\frac{3 M}{2 m} \sqrt{2 g L}$

$\frac{2 m}{3 M} \sqrt{2 g L}$

$\frac{M}{m} \sqrt{2 g L}$

$\frac{m}{M} \sqrt{2 g L}$

answer is C.

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Detailed Solution

The velocity of the block in order to reach the horizontal position is given by $\sqrt{2 g l}$

From law of conservation of liner momentum we get $m V_{0} = \frac{m V_{0}}{3} + m \sqrt{2 g l}$

$V_{0} = \frac{3 M}{2 m} \sqrt{2 g l}$

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