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A bullet travelling with a velocity of $\sqrt{2} m s^{- 1}$ pierces a plank of certain thickness and comes to a stop. What should be the velocity of the bullet to be stopped by double the thickness of the same plank?

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$1 m s^{- 1}$

$4 m s^{- 1}$

$2 m s^{- 1}$

$0.5 m s^{- 1}$

answer is A.

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Detailed Solution

$F s = \frac{1}{2} m v^{2}$ ( Acc to work energy theorem )

Since Fand m are constant

$S \propto v^{2}$

$\frac{s_{1}}{s_{2}} = \frac{v_{1}^{2}}{v_{2}^{2}}$

$\frac{S}{2 s} = \frac{{(\sqrt{2})}^{2}}{v_{2}^{2}}$

$v_{2}^{2} = 2 \times 2$

$v_{2} = 2 m s^{- 1}$

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