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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Infinity Learn · Accepted Answer

According  to the given $$problem12mv12=1412m4v22$$ ∴ $$v22=16v12$$ When $$v1$$, is increased by $$4$$ m,$$/s$$, then $$12mv1+42=12m4v22$$ $$v1+4=2v1v1=4m/s$$

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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