Q. A pendulum consists of a wooden bob of mass $m$ and length $l$. A bullet of mass $m_{1}$ is fired towards the pendulum with a speed $v_{1}$. The bullet emerges out with a velocity $\frac{v_{1}}{3}$ and the bob just completes the motion along a vertical circle. Then $v_{1}$ is

System of Particles and Rotational Motion Report Error

A

$\frac{3}{2}\left(\frac{m}{m_{1}}\right) \sqrt{5 g l}$

B

$\left(\frac{m}{m_{1}}\right) \sqrt{5 g l}$

C

$\frac{3}{2}\left(\frac{m}{m_{1}}\right) \sqrt{g l}$

D

$\frac{3}{2}\left(\frac{m_{1}}{m}\right) \sqrt{5 g l}$

Solution:

The speed acquired by bob after bullet comes out is
$v=\sqrt{5 g l}$
By conservation of linear momentum, we have
$m_{1} v_{1}=m \sqrt{5 g l}+m_{1} \frac{v_{1}}{3}$
$\Rightarrow \frac{2 m_{1} v_{1}}{3}=m \sqrt{5 g l}$
$\Rightarrow v_{1}=\frac{3}{2}\left(\frac{m}{m_{1}}\right) \sqrt{5 g l}$

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