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Question

A particle of mass 4m which is at rest explodes into three fragments. Two of the fragments, each of mass m are found to move with speed v each, in mutually perpendicular directions. The total energy released in the process of explosion is-

A
(23)mv2
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B
(32)mv2
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C
(43)mv2
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D
(34)mv2
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Solution

The correct option is A (23)mv2
By Conservation of momentum, the net momentum must be 0
The total momentum of 2 m masses is M=2m2v2=2mv
2mv=2mvv=v/2
hence KE=mv2+mv2/2=3mv2/2

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