A lead bullet of mass $2 k g$ , travelling with a velocity of $20 m s^{- 1}$ , comes to rest after penetrating $20 m$ in a still target. Find the acceleration $(a)$ of the bullet.

a = 20 m s^{- 2}

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a = 40 m s^{- 2}

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a = - 10 m s^{- 2}

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a = 0 m s^{- 2}

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Solution

The correct option is C $a = - 10 m s^{- 2}$
Given:

Initial velocity, $u = 20 m s^{- 1}$
Final velocity, $v = 0$
Distance travelled, $s = 20 m$
Mass, $m = 2 k g$

From third equation of motion, we have
$v^{2} = u^{2} + 2 a s$
$0^{2} = 20^{2} + (2 \times a \times 20)$
$\Rightarrow a = - 10 m s^{- 2}$

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