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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: |
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Answer» Solution :The bomb initially was at rest therefore Initial momentum of bomb = 0 Final momentum of SYSTEM =`m_1v_1 + m_2v_2` As there is no external FORCE `:.m_1v_1 + m_2v_2 = 0 implies 3 xx 1.6 + 6 xx v_2 = 0` Velocity of 6 kg mass, `v_2 = 0.8 m//s ("numerically")` Its kinetic energy `=1/2 m_2v_2^2 = 1/2 xx 6 xx (0.8)^2 = 1.92J` 
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