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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a
6E0
b
4E03
c
4E0
d
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