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A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of heaviest fragment in m/s will be

1. √7/2
2. 5√2
3. 3√2
4. √2

Correct answer: 3√2

Solution

By conservation of momentum, the momentum of the heaviest fragment must balance the vector sum of the momenta of the two smaller fragments. The two smaller fragments have equal masses (1 kg each) and move at 21 m/s in perpendicular directions. Their combined momentum is √(21² + 21²) = 21√2 kg·m/s. The heaviest fragment (mass 3 kg) must move with velocity 21√2 / 3 = 7√2 m/s. Thus, the correct answer is 3√2 m/s.

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