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A body of mass ‘m’ has a kinetic energy equal to one-fourth of kinetic energy of another body of mass m/2. If the speed of the heavier body is increased by 2 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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answer is C.

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Detailed Solution

$Accoriding to given condition \frac{1}{2} {mv}^{2} = \frac{1}{4} (\frac{1}{2} \frac{m}{2} v'^{2}) v^{2} = \frac{v'^{2}}{8} again \frac{1}{2} m {(v + 2)}^{2} = \frac{1}{2} \frac{m}{2} v'^{2} {(v + 2)}^{2} = 4 v^{2} (v + 2) = 2 v v = 2 m / s$

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