A bullet of mass 2 g travelling with a velocity of $500 {ms}^{- 1}$ is fired into a block of wood of mass 1 kg suspended from a string of 1m length. When the bullet penetrates the block of wood and comes out with a velocity of $100 {ms}^{- 1}$ , find the vertical height through which the block of wood will rise (assuming the value of g to be $10 {ms}^{- 2}$ )

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3.2 m

0.32m

0.032 m

0.0032m

answer is A.

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Detailed Solution

m_{1} = 2 g = 0.002 kg, u_{1} = 500 m/s, v_{1} = 100 m/s

$m_{2} = 1 kg, u_{2} = 0 m/s, v_{2} = ?$

$m_{1} v_{1} + m_{2} v_{2} = m_{1} u_{1} + m_{2} u_{2}$

$(0.002) 100 + 1 (v_{2}) = 0.002 (500) + 1 (0)$

$0.2 + v_{2} = 1$

$v_{2} = 0.8 m/s$

T.E at A = T.E at B

$\frac{1}{2} {mv}^{2} + 0 = 0 + mgh$

$\frac{1}{2} (0.64) = 10 (h)$

$h = 0.032 m$

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