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Q.

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

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a

11.5 m/s

b

14.0 m/s

c

7.0 m/s

d

9.89 m/s

answer is D.

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Detailed Solution

Px=m×vx=1×21=21 kg m​/​sPy=m×vy=1×21=21 kg m​/​s∴ Resultant = Px2+Py2=212 kg m/sThe momentum of heavier fragment should be numerically equal to resultant of P→x and P→y. 3×v=Px2+Py2=212   ∴v=72 = 9.89 m/s
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