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. $\frac{2}{3}\left(\frac{m_1}{m}\right)\sqrt{5gl}$
• D. $\left(\frac{m_1}{m}\right)\sqrt{5gl}$

Answer 1

The correct option is B $\frac{3}{2}\left(\frac{m}{m_1}\right)\sqrt{5gl}$

Let the velocity of the bob after the collision be $v$.

As the bob just completes the vertical circular motion $⟹v=\sqrt{5gl}$

Apply conservation of momentum just before and after the collision :

$m_1{v}_1+0=mv+m_1{v}_1$

$m_1{v}_1=m\times \sqrt{5gl}+m_1\times \frac{v_1}{3}$

$⟹v_1=\frac{3m}{2m_1}\sqrt{5gl}$