A bullet of mass $10 g$ moving with a horizontal velocity $100 m / s$ passes through a wooden block of mass $100 g$ . The block is resting on a smooth horizontal floor. After passing through the block the velocity of the bullet is $10 m / s$ . The velocity of the emerging bullet with respect to the block is:

10 m / s

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9 m / s

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1 m / s

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5 m / s

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Solution

The correct option is D $1 m / s$
Using momentum conservation,

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

$m_{1} = 10 g$ $, m_{2} = 100 g$ $, u_{1} = 100 m / s$ $, u_{2} = 0$ $, v_{1} = 10 m / s$

$\Rightarrow 10 \times 100 + 100 \times 0 = 10 \times 10 + 100 \times v_{2}$

$\Rightarrow 1000 = 100 + 100 v_{2} \Rightarrow v_{2} = 9 m / s$

Therefore $v_{1}$ with respect to $v_{2} = v_{1} - v_{2} = 1 m / s$

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A bullet of mass 20g travelling horizontally with a speed of 500 m/s passes through a wooden block of mass 10.0 kg initially at rest on a level surface. The bullet emerges with a speed of 100 m/s and the block slides 20 cm on the surface before coming to rest. The friction coefficient between the block and the surface is $g = 10 m s^{- 2}$