A bomb of mass 3m kg explodes into two pieces of mass m kg and 2m kg . If the velocity of m kg mass is 16 m/s, the total energy released in the explosion is
A
192mJ
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B
96mJ
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C
384mJ
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D
768mJ
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Solution
The correct option is A192mJ Given that, Initial velocity of mass 3m kg,(vi)=0 m/s after explosion
By the conservation of momentum, initial momentum of the systme = final momentum of the system initial momentum of the system = 0 0=mAvA−mBvB ⇒mAvA=mBvB ⇒m×16=2m×vB ⇒vB=8m/s Energy release during explosion is kinetic energy of system =12mAv2A+12mBv2B=192mJ ∴(a).