A pendulum consists of a wooden bob of mass m and length l. A bullet of mass m1 is fired towards the pendulum with a speed v1 and it emerges from the bob with speed v13. The bob just completes motion along a vertical circle. Then, v1 is

From conservation of linear momentum, m1v1=m5gl+m1v13 or v1=32mm15gl

A pendulum consists of a wooden bob of mass m and length l. A bullet of mass m1 is fired towards the pendulum with a speed v1 and it emerges from the bob with speed v13. The bob just completes motion along a vertical circle. Then, v1 is

From conservation of linear momentum, m1v1=m5gl+m1v13 or v1=32mm15gl

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A pendulum consists of a wooden bob of mass m and length l. A bullet of mass m₁ is fired towards the pendulum with a speed v₁ and it emerges from the bob with speed $\frac{v_{1}}{3}$ . The bob just completes motion along a vertical circle. Then, v₁is

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

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

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

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

answer is B.

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From conservation of linear momentum,

$m_{1} v_{1} = m \sqrt{5 g l} + m_{1} \frac{v_{1}}{3}$ or $v_{1} = \frac{3}{2} \frac{m}{m_{1}} \sqrt{5 g l}$

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A pendulum consists of a wooden bob of mass m and length l. A bullet of mass m1 is fired towards the pendulum with a speed v1 and it emerges from the bob with speed v13. The bob just completes motion along a vertical circle. Then, v1 is

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A pendulum consists of a wooden bob of mass m and length l. A bullet of mass m₁ is fired towards the pendulum with a speed v₁ and it emerges from the bob with speed $\frac{v_{1}}{3}$ . The bob just completes motion along a vertical circle. Then, v₁is