Answer
(a) The ball will go up to a maximum height of $0.6~H_0$.
(b) 40% of the original potential energy has been converted into rotational kinetic energy.
Work Step by Step
(a) We can use conservation of energy to find the kinetic energy at the bottom of the rough part.
$mgH_0 = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$
$mgH_0 = \frac{1}{2}mv^2+\frac{1}{2}(\frac{2}{3}mr^2)(\frac{v}{r})^2$
$mgH_0 = \frac{1}{2}mv^2+\frac{1}{3}mv^2$
At the bottom, the translational kinetic energy is $\frac{1}{2}mv^2$, and the rotational kinetic energy is $\frac{1}{3}mv^2$. When the sphere rolls up the smooth part, the translational kinetic energy will be converted into potential energy, while the rotational kinetic energy remains the same because the torque is zero.
Since 40% of the energy is rotational kinetic energy, the potential energy on the smooth part will only reach 60% of its original value. Therefore, the ball will only go up to a maximum height of $0.6~H_0$.
(b) The ball does not return to the original height of $H_0$ because 40% of the original potential energy has been converted into rotational kinetic energy. The total energy in the system is conserved.
