A bag of sand of mass 9 . 8 kg is suspended by a rope. A bullet of 200 g travelling with speed 10 ms - 1 gets embedded in it, then loss of kinetic energy will be

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If the final velocity of the bullet and the bag is $v$ , applying conservation of momentum,
$m u = (m + M) v \Rightarrow 0.2 \times 10 = (0.2 + 9.8) v \Rightarrow v = 0.2 m s^{- 1}$
Loss in kinetic energy will be,
$K_{i} - K_{f} = (\frac{1}{2} \times 0.2 \times 10^{2}) - (\frac{1}{2} \times 10 \times 0 . 2^{2}) = 9.8 J$

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