The total kinetic energy of a rolling body is the sum of its translational and rotational kinetic energies. $K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ For a hollow sphere, the moment of inertia about an axis passing through its centre is $I = \frac{2}{3}mR^2$. For pure rolling, $\omega = \frac{v}{R}$. Therefore, $K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{3}mR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{3}mv^2 = \frac{5}{6}mv^2$ Given $m = 2 \text{ kg}$ and $v = 0.5 \text{ m/s}$. $K_{total} = \frac{5}{6} \times 2 \times (0.5)^2 = \frac{5}{6} \times 2 \times 0.25 = \frac{2.5}{6} \approx 0.4167 \text{ J} \approx 0.42 \text{ J}$.