A body of mass m has a kinetic energy equal to one-fourth kinetic energy of another body of mass m/4' If the speed of the heavier body is increased by 4 m/s, its new kinetic energy equals the original kinetic energy of the lighter body. The original speed of the heavier body in m/s is
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a
8
b
6
c
4
d
2
answer is C.
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Detailed Solution
According to the given problem12mv12=1412m4v22 ∴ v22=16v12 When v1, is increased by 4 m,/s, then 12mv1+42=12m4v22 v1+4=2v1v1=4m/s