A bullet of mass $0.01 k g$ and travelling at a speed of $500 m s^{- 1}$ strikes a block of $2 k g$ , which is suspended by a string of length $5 m$ . The centre of gravity of the block is found to rise through a vertical height of $0.1 m$ . The speed of the bullet after it emerges from the block is

200 m / s

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240 m / s

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220 m / s

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280 m / s

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Solution

The correct option is C $220 m / s$
Assuming that $v_{1}$ and $v_{2}$ is the velocity of the bullet before collision. From the law of conservation of energy,
Initial velocity $= g h_{2}^{2} = m_{2} v_{2}^{\frac{1}{2}} m = \sqrt{2} g h_{2} \partial v = \sqrt{2} \times 9.8 \times 0.1 \times \partial v = 1.4 m / s e c$
By law of conservation of momentum,
$m_{2} v_{2} + m_{1} v_{1} = m u_{1} \times m_{2} v_{2} - m_{1} u_{1} = \frac{(5 - 2.8)}{0.01} = \frac{2.2}{0.01} m / s e c = 220 m / s e c$

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