Question

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

• A. $\left(\frac{m}{m_{1}}\right)\sqrt{5gl}$
• B. $\frac{3}{2}\left(\frac{m}{m_{1}}\right)\sqrt{5gl}$
• C. $\frac{2}{3}\left(\frac{m_{1}}{m_{2}}\right)\sqrt{5gl}$
• D. $\left(\frac{m_{1}}{m}\right)\sqrt{gl}$

Answer 1

Answer: (B)

Applying the law of conservation of momentum
$m_{1} v_{1}=m_{1} \frac{v_{1}}{3}+m v$
or $v=\frac{2 m_{1} \dot{v}_{1}}{3 m}$
To describe a vertical circle v should be $\sqrt{5gl}$
So $\frac{2 m_{1} v_{1}}{3 m}=\sqrt{5 g l}$
or $v_{1}=\left(\frac{m}{m_{1}}\right) \frac{3}{2} \sqrt{5 g l}$