A stationary body explodes into four identical fragments such that three of them fly off mutually perpendicular to each other, each with same kinetic energy E 0 . The energy of the explosion will be K times of E 0 , then the value of K is

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Aicharya · Accepted Answer

P 1 → = m v 0 i ^ P 2 → = m v 0 j ^ P 3 → = m v 0 j ^ P 4 → = m v → 0 = P 1 → + P 2 → + P 3 → + P 4 → V → = - v 0 ( i ^ + j ^ + k ^ ) v = v 0 3 Total energy = 3 × 1 2 m v 0 2 + 1 2 m v 2 = 3 E 0 + 3 E 0 = 6 E 0

A stationary body explodes into four identical fragments such that three of them fly off mutually…

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