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(A) The kinetic energy $K$ of a body of mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.Given that the kinetic energies of the

Vedclass · Accepted Answer

$1: 2$

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(A) The kinetic energy $K$ of a body of mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.Given that the kinetic energies of the two masses are equal,we have $K_1 = K_2$.Substituting the expression for kinetic energy,we get $\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}$.Rearranging the terms to find the ratio of momenta,we get $\frac{p_1^2}{p_2^2} = \frac{m_1}{m_2}$.Taking the square root on both sides,the ratio of momenta is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.Given $m_1 = 1 \ g$ and $m_2 = 4 \ g$,we substitute these values:$\frac{p_1}{p_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,the ratio of their momenta is $1: 2$.

Two masses of $1 \ g$ and $4 \ g$ are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

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