Question

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 $V_1$ and it emerges from the bob with speed $\frac{V_1}{3}.$ The bob just completes motion along a vertical circle. Then $v_1$ is

• A. $\frac{m_1}{m}\sqrt{5gl}$
  
• B. $\frac{2m}{3m_1}\sqrt{5gl}$
  
• C. $\frac{3m}{2m_1}\sqrt{5gl}$
  
• D. $\frac{m}{m_1}\sqrt{5gl}$
  

Answer 1

The correct option is C $\frac{3m}{2m_1}\sqrt{5gl}$


Given,
mass of bob$=m$,
initial velocity of bob$=0$
mass of bullet$=m_1$
velocity of bullet just before collision$=V_1$
velocity of bullet just after collision$=\frac{V_1}{3}$

Let, velocity of $m$ just after collision is $v$.
Then, from conservation of momentum we have,
$m_1{V}_1=mv+m_1\frac{V_1}{3}$

$\Rightarrow v=\frac{2}{3}\frac{m_1{V}_1}{m}$

Now, for just completing the circle in vertical circular motion, the velocity of $m$ $$v=\sqrt{5gl}$$ Substituting the value of $v$, we have
$\frac{2}{3}\frac{m_1{V}_1}{m}=\sqrt{5gl}$

$\therefore V_1=\frac{3m}{2m_1}\sqrt{5gl}$

Hence, option (b) is correct answer.

Why this question ? This question checks the application of vertical circular motion and conservation of linear momentum.