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A bullet when fired at a target with a velocity of 100 ${ms}^{- 1}$ , penetrates one metre into it. If the bullet is fired with the same velocity at a similar target with a thickness 0.5 metre, then it will emerge from it with a velocity of

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50 $\sqrt{2}$ ${ms}^{- 1}$

50 ${ms}^{- 1}$

$\frac{50}{\sqrt{2}}$ ${ms}^{- 1}$

10 ${ms}^{- 1}$

answer is A.

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Detailed Solution

$\frac{1}{2} m v^{2} = F S$

$\frac{1}{2} m (100^{2})$ = F . 1

Again $\frac{1}{2} {mv}^{2} - \frac{1}{2} {mv}^{' 2} = f \times 0.5$

or m[ $v^{2} - v'^{2}$ ] = f $\times 1$

or $v^{' 2} = v^{2} - 5000$

or $v^{' 2} = 10000 - 5000$

or $v^{' 2} = 5000$

or $v^{'} = 50 \sqrt{2}$ ${ms}^{- 1}$

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