Ball, Fixed Curved Track, and Non-Fixed Curved Track

I wonder whether the way I approach and solve this question is correct or not.

Case 1

As illustrated, a track of mass $M$ is fixed on a horizontal table. $AB$ is horizontal while $CD$ is vertical. A ball of mass $m$ situated at point A is given an initial velocity $v$ in the direction $AB$. The ball then travels up the slope and passes through CD, reaching a height of $H$ above the horizontal surface of the track. (Assume that there is no frictional force between the track and the ball.)

Case 2

This time, the same track is not fixed to the table. As in Case 1, the same ball is given an initial velocity $v$ in the direction AB. The ball leaves the track and reaches a height of $h$ above the horizontal surface of the track. (Assume that there is no frictional force between the track and the table, and also between the ball and the track.)

Find an expression for $\frac{h}{H}$.

My solution:

1. When the track is fixed, by Conservation of Energy, Kinetic Energy of the Ball = Potential Energy of the Ball $$\frac{1}{2}mv^{2} = mgH\implies H = \frac{v^{2}}{2g}$$

2. When the track is not fixed, the ball and track will move as a single body with speed, $v_{1}$ to the right. By Conservation of Linear Momentum $$mv = (m + M)v_{1}\implies v_{1} = \frac{m}{(m + M)}v$$

By Conservation of Energy, Kinetic Energy of the Ball = Kinetic Energy of the Ball and the Track + Potential Energy of the Ball

$$\frac{1}{2}mv^{2} = \frac{1}{2}(m + M)(\frac{m}{(m + M)}v)^{2} + mgh$$

$$v^{2} = \frac{m}{m + M}v^{2} + 2gh$$

$$h = \frac{Mv^{2}}{2g(m + M)}$$

So, $$\frac{h}{H} = \frac{M}{m + M}$$

My questions:

1. Is it okay to assume that the ball and track will experience an inelastic collision in Case 2?
2. Are there any other ways to solve this question, with explanation?
3. Will the radius of curvature/type of curvature of the curved part of the track affect the outcome?
• Please don't use MathJax for simple text Jun 10 '17 at 16:06