A pendulum consists of a wooden bob of mass m and length L. A bullet of mass m1 is fired towards the pendulum with speed v1. The bullet comes out of the bob with speed v1/3, the pendulum just completes motion along vertical circle. The velocity v1 is

Engineering

Physics

Vertical Circular Motion

Question

A pendulum consists of a wooden bob of mass m and length L. A bullet of mass m₁ is fired towards the pendulum with speed v₁. The bullet comes out of the bob with speed v₁/3, the pendulum just completes motion along vertical circle. The velocity v₁ is

$\frac{m}{m_{1}} \sqrt{5 gℓ}$

$\frac{3}{2} \frac{m}{m_{1}} \sqrt{5 gℓ}$

$\frac{3}{2} \frac{m}{m_{1}} \sqrt{gℓ}$

$\frac{3}{2} \frac{m_{1}}{m} \sqrt{5 gℓ}$

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To find the bullet's initial velocity, we analyze momentum conservation during collision and energy conservation for the pendulum's vertical circle motion.

Momentum conservation: $m_{1} v_{1} = m_{1} \frac{v_{1}}{3} + m v_{2}$ , where $v_{2}$ is the bob's velocity post-collision.

Solving gives: $v_{2} = \frac{2 m_{1} v_{1}}{3 m}$ .

For the bob to just complete the vertical circle, its minimum speed at the bottom is $\sqrt{5 g L}$ . Thus, $v_{2} = \sqrt{5 g L}$ .

Equating and solving for $v_{1}$ : $v_{1} = \frac{3}{2} \frac{m}{m_{1}} \sqrt{5 g L}$ .

Final answer: $\frac{3}{2} \frac{m}{m_{1}} \sqrt{5 g L}$