A stationary body explodes into four identical fragments such that three of them flyoff mutually perpendicular to each other, each with same K.E., E0 . The energy released in the explosion is

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6E0

4E03

4E0

8E0

answer is A.

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Detailed Solution

By law of conservation of momentum, v''=−v' i^+v' j^+v'k^v''2=3v'2E0=12mv'2Total E = 312mv'2+12mv''2=3E0+3E0=6E0

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