A super ball rolling on the floor enters a half circular track (radius $R$). The ball rolls without slipping around the track and leaves (velocity $v$) traveling horizontally in the opposite direction. Afterwards, it bounces on the floor. How far (horizontally) from the end of the track will the ball bounce for the second time? The ball’s surface has a theoretically infinite coefficient of static friction. It is a perfect sphere of uniform density. All collisions with the ground are perfectly elastic and theoretically instantaneous. Variations could involve the initial velocity being given before the ball enters the track or state that the normal force between the ball and the track right before leaving is zero (centripetal acceleration). 

[i]Problem proposed by Brian Yue[/i]
To solve this problem, we need to analyze the motion of the ball as it rolls around the half circular track, leaves the track, and bounces on the floor. We will use principles of mechanics, including conservation of energy, kinematics, and dynamics of rolling motion.

1. **Determine the velocity of the ball as it leaves the track:**
   - The ball rolls without slipping around the half circular track of radius \( R \). As it leaves the track, it has a horizontal velocity \( v \).
   - Since the ball rolls without slipping, the point of contact with the track has zero velocity relative to the track. The velocity of the center of mass \( v \) is related to the angular velocity \( \omega \) by \( v = R \omega \).

2. **Analyze the collision with the ground:**
   - The ball hits the ground with the point of contact moving at a velocity \( 2v \) with respect to the ground. This is because the ball's center of mass has velocity \( v \) and the point of contact has an additional velocity \( v \) due to rolling.
   - After hitting the ground, the ball exerts a force \( F \) on the ground, and the ground exerts an equal and opposite force on the ball. This force changes both the translational and rotational motion of the ball.

3. **Calculate the change in translational and rotational motion:**
   - The change in translational velocity \( \Delta v \) is given by \( \Delta v = -\frac{Ft}{M} \), where \( M \) is the mass of the ball and \( t \) is the duration of the collision.
   - The torque \( \tau \) exerted on the ball is \( \tau = FR \), which changes the angular velocity \( \Delta \omega \) by \( \Delta \omega = \frac{\tau t}{I} = \frac{FRt}{I} \), where \( I \) is the moment of inertia of the ball. For a sphere, \( I = \frac{2}{5}MR^2 \).

4. **Relate the change in angular velocity to the change in translational velocity:**
   - The change in angular velocity \( \Delta \omega \) corresponds to a change in the translational velocity of the surface of the ball relative to the center of the ball by \( \Delta v_{\text{surface}} = R \Delta \omega = \frac{5Ft}{2M} \).
   - The total change in the velocity of the surface of the ball relative to the ground is \( \Delta v_{\text{total}} = \Delta v + \Delta v_{\text{surface}} = -\frac{Ft}{M} + \frac{5Ft}{2M} = \frac{3Ft}{2M} \).

5. **Determine the force and time of collision:**
   - Since the velocity of the surface of the ball relative to the ground must be \( -2v \), we have \( \frac{3Ft}{2M} = -2v \). Solving for \( Ft \), we get \( Ft = \frac{4Mv}{3} \).

6. **Calculate the final horizontal velocity of the ball:**
   - The change in the translational velocity of the ball after the impact is \( \Delta v = -\frac{Ft}{M} = -\frac{4v}{3} \).
   - The final horizontal velocity of the ball is \( v_{\text{final}} = v + \Delta v = v - \frac{4v}{3} = -\frac{v}{3} \).

7. **Determine the horizontal distance for the second bounce:**
   - The ball bounces with a horizontal velocity of \( -\frac{v}{3} \). The time \( t \) it takes for the ball to hit the ground again can be found using the vertical motion. The ball falls a distance \( 2R \) (the diameter of the track) under gravity.
   - Using the kinematic equation \( y = \frac{1}{2}gt^2 \), we get \( 2R = \frac{1}{2}gt^2 \), so \( t = \sqrt{\frac{4R}{g}} \).
   - The horizontal distance \( d \) traveled in this time is \( d = v_{\text{final}} \cdot t = -\frac{v}{3} \cdot \sqrt{\frac{4R}{g}} = -\frac{2v}{3} \sqrt{\frac{R}{g}} \).

8. **Calculate the total horizontal distance for the second bounce:**
   - The ball travels a distance \( 2R \) horizontally while in the track, and an additional distance \( -\frac{2v}{3} \sqrt{\frac{R}{g}} \) after the first bounce.
   - The total horizontal distance is \( 2R + \left(-\frac{2v}{3} \sqrt{\frac{R}{g}}\right) \).

The final answer is \( \boxed{ 2R - \frac{2v}{3} \sqrt{\frac{R}{g}} } \).