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A bullet of mass 0.005 kg moving with a speed of $200 m s^{- 1}$ enters a heavy wooden block and is stopped after a distance of 50 cm. What is the average force exerted by the block on the bullet ?

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–200 N

+ 200 N

400 N

–400 N

answer is A.

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Detailed Solution

$F = - \frac{m u^{2}}{2 s} = \frac{- 0.005 \times 200 \times 200}{2 \times 50}$

$= - 200 N$

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