\(M_{bullet}\) = 0.05kg, \(V_{bullet}\) = 200m/s, \(M_{block}\) = 0.95kg and \(V_{block}\) = 0 for an inelastic collision \(M_{bullet} \times V_{bullet} + M_{block} \times V_{block} = (M_{bullet} + M_{block})\)v 0.05 x 200 + 0.95 x 0 = ( 0.05 + 0.95)v 10 + 0 = v v = 10m/s K.E of the block = \(\frac{1}{2}mv^2\) K . E = \(\frac{1}{2} \times(0.05 + 0.95) \times10^2\) ( bullet now part of the block) K . E = \(\frac{1}{2} \times1 \times100\) = 50J