c
(c)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\( = \frac{1}{2}{m_2}v_2^2\)\( = \frac{1}{2} \times 6 \times {(0.8)^2} = 1.92\,J\)
