A small ball starts moving from $A$ over a fixed track as shown in the figure. Surface $AB$ has friction. From $A$ to $B$ the ball rolls without slipping. Surface $BC$ is frictionless. $K_A$, $K_B$ and $K_C$ are kinetic energies of the ball at $A$, $B$ and $C$, respectively. Then

Step 1: Analyze the motion from $A$ to $B$.

The ball starts from rest at $A$ (at height $h_A$), so $K_A = 0$. Since the ball rolls without slipping on the rough surface $AB$, the friction is static and does no work. Therefore, mechanical energy is conserved.

$$K_B = K_A + mgh_A = mgh_A$$

Step 2: Express the kinetic energy at $B$.

At the lowest point $B$, the total kinetic energy $K_B$ is the sum of translational and rotational kinetic energies.

$$K_B = K_{\text{trans}} + K_{\text{rot}}$$

For any rolling body with moment of inertia $I = k m r^2$ (where $k > 0$), both components are strictly positive.

$$K_{\text{rot}} > 0 \quad \text{and} \quad K_{\text{trans}} < K_B$$

Step 3: Analyze the motion from $B$ to $C$.

The surface $BC$ is frictionless, so there is no torque acting on the ball. Consequently, the angular velocity $\omega$ and the rotational kinetic energy $K_{\text{rot}}$ remain constant. Only the translational kinetic energy is converted into gravitational potential energy as the ball climbs to $C$. At the highest point $C$, the translational velocity is zero, meaning the kinetic energy at $C$ is purely rotational.

$$K_C = K_{\text{rot}}$$

Step 4: Compare the kinetic energies $K_B$ and $K_C$.

Using conservation of energy between $B$ and $C$:

$$K_B = mgh_C + K_C$$ $$K_B - K_C = mgh_C$$

Since $h_C > 0$, it follows that:

$$K_B > K_C$$

Step 5: Compare the heights $h_A$ and $h_C$.

The potential energy gained at $C$ comes entirely from the translational kinetic energy at $B$.

$$mgh_C = K_{\text{trans}}$$

Since $K_{\text{trans}} < K_B$ and $K_B = mgh_A$:

$$mgh_C < mgh_A$$ $$h_A > h_C$$

Answer: The correct relationships are $h_A > h_C$ and $K_B > K_C$.