Two bodies having kinetic energy in the ratio | vedclass

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(C) The kinetic energy $(K.E.)$ of a body is given by the formula $K.E. = \frac{1}{2}mv^2$.Since the bodies are moving with the same linear velocity $

Vedclass · Accepted Answer

$4: 1$

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(C) The kinetic energy $(K.E.)$ of a body is given by the formula $K.E. = \frac{1}{2}mv^2$.Since the bodies are moving with the same linear velocity $(v_1 = v_2 = v)$,the kinetic energy is directly proportional to the mass $(K.E. \propto m)$.Therefore,the ratio of their kinetic energies is equal to the ratio of their masses:$\frac{K.E._1}{K.E._2} = \frac{m_1}{m_2}$.Given that $\frac{K.E._1}{K.E._2} = \frac{4}{1}$,it follows that $\frac{m_1}{m_2} = \frac{4}{1}$.Thus,the ratio of their masses is $4: 1$.

Two bodies having kinetic energy in the ratio $4: 1$ are moving with the same linear velocity. The ratio of their masses is

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