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(A) The total kinetic energy of a rolling sphere is given by $K = K_{tr} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since the moment of inert

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$2:1$

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(A) The total kinetic energy of a rolling sphere is given by $K = K_{tr} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since the moment of inertia of a solid sphere is $I = \frac{2}{5}mr^2$ and $\omega = \frac{v}{r}$,we have:$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.Given that both spheres have the same velocity $(v_1 = v_2)$,the ratio of their kinetic energies is:$\frac{K_1}{K_2} = \frac{\frac{7}{10}m_1v_1^2}{\frac{7}{10}m_2v_2^2} = \frac{m_1}{m_2}$.Given $\frac{K_1}{K_2} = 2:1$,it follows that $\frac{m_1}{m_2} = 2:1$.The radius ratio is irrelevant in this specific case because the velocity is the same for both spheres.

Two spheres are rolling with the same velocity (for their $C.M.$). Their ratio of kinetic energy is $2:1$ and the ratio of their radii is $2:1$. Their mass ratio will be:

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