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Question

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 the heaviest fragment will be


A

11.5 m/s

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B

14.0 m/s

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C

7.0 m/s

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D

9.89 m/s

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Solution

The correct option is D

9.89 m/s


Px=m×vx=1×21=21kgm/s

Py=m×vy=1×21=21kgm/s

Resultant = P2x+P2y=212kg m/s

The momentum of heavier fragment should be numerically equal to resultant of Px and Py.

3×v=P2x+P2y=212v=72=9.89m/s


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