A bullet having a mass of $10 g$ and moving with a speed of $1.5 m s^{- 1}$ penetrates a thick wooden plank of mass $90 g$ . The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.

$0.10 m s^{- 1}$

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0.15 m s^{- 1}

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0.25 m s^{- 1}

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0.35 m s^{- 1}

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Solution

The correct option is B $0.15 m s^{- 1}$
Given, the mass of bullet, $m_{1} = 10 g = 0.01 k g$ , initial velocity of bullet, $u_{1} = 1.5 m s^{- 1}$ , mass of wooden plank, $m_{2} = 90 g = 0.09 k g$ , final velocity of plank and bullet (since bullet gets embedded in the plank) = $v$ and initial velocity of wooden plank, $u_{2} = 0$

Applying, the law of conservation of momentum, i.e. Final momentum = Initial momentum.

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

$0.01 \times 1.5 + 0.09 \times 0 = 0.01 \times v + 0.09 \times v$

$0.015 = 0.1 v$

$\Rightarrow$ $v = 0.15 m s^{- 1}$

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