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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