nbsp;9 kg દળનો એક બોમ્બના ફાટીને 3 kg અને

As the bomb initially was at rest therefore
Initial momentum of bomb $= 0$
Final momentum of system = ${m_1}{v_1} + {m_2}{v_2}$
As there is no external force
 ${m_1}{v_1} + {m_2}{v_2} = 0$ ==> $3 \times 1.6 + 6 \times {v_2} = 0$
velocity of 6 kg mass ${v_2} = 0.8\,m/s$ (numerically)
Its kinetic energy$ = \frac{1}{2}{m_2}v_2^2$$ = \frac{1}{2} \times 6 \times {(0.8)^2} = 1.92\,J$