For a solid sphere in rolling motion: Translational kinetic energy, $K_t = \frac{1}{2}mv^2$ Rotational kinetic energy, $K_r = \frac{1}{2}I\omega^2$ The moment of inertia of a solid sphere about its diameter is $I = \frac{2}{5}mr^2$. In pure rolling, $v = r\omega$. Substituting these values: $K_r = \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2$ Total kinetic energy, $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$ The required ratio is $\frac{K_t}{K_t + K_r} = \frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7}$. Therefore, the ratio is $5:7$.