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A pendulum of length $\mathrm{2m}$ consists of a wooden bob of mass $\mathrm{50g}$ . A bullet of mass $\mathrm{75 g}$ is fired towards the stationary Bob with a speed $\mathrm{v}$ . The bullet emerges out of the Bob with a speed $\mathrm{\frac{v}{3}}$ and the Bob just completes the vertical circle. The value of $\mathrm{v}$ is _______ $\mathrm{ms^{-1}}$ .

( if $\mathrm{g=10m/s^{2}}$ ).
Option: 1
10

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Linear momentum conservation

$\mathrm{75 \times V =50 \times V^{\prime}+75 \times \frac{V}{3}} \\$

$\mathrm{V^{\prime} \times 50 =2 \times \frac{75}{3} \times V} \\$

$\mathrm{V^{\prime} =\frac{2 \times 75}{50 \times 3} \times V} \\$

$\mathrm{=\frac{2 \times 3}{2 \times 3} V=V}$

The bob just completes the vertical circular motion

$\mathrm{\therefore V =\sqrt{5 \mathrm{gl}}} \\$

$\mathrm{V =\sqrt{5 \times 10 \times 2}} \\$

$\mathrm{V =10 \mathrm{~m} / \mathrm{s}}$

Hence answer is $\mathrm{10} \\$ .

Posted by

seema garhwal

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