A bomb of mass 9 kg explodes into 2 pieces of mass 3 kg and 6 kg. The velocity of mass 3 kg is 1.6 m/s. The K.E. of mass 6 kg is

As the bomb initially was at rest thereforeInitial momentum of bomb = 0Final momentum of system = m1v1 + m2v2As there is no external force∴  m1v1+m2v2 = 0 ⇒ 3×1.6+6×v2 = 0velocity of 6 kg mass v2 = 0.8 m/s (numerically)Its kinetic energy = 12m2v22 = 12×6×(0.8)2  = 1.92 J