A bag of sand of mass 9 . 8kg is suspended by a rope. A bullet of...

If the final velocity of the bullet and the bag is $v$, applying conservation of momentum,

$mu = \left(m + M\right)v ⇒ 0 . 2 \times 10 = \left(0 . 2 + 9 . 8\right)v ⇒ v = 0 . 2m{s}^{- 1}$

Loss in kinetic energy will be,

${K}_{i} - {K}_{f} = \left(\dfrac{1}{2} \times 0 . 2 \times {10}^{2}\right) - \left(\dfrac{1}{2} \times 10 \times 0 . {2}^{2}\right) = 9 . 8J$