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This is the illustration for Case 1

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

Question:

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?
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  • $\begingroup$ Please don't use MathJax for simple text $\endgroup$ – Kyle Kanos Jun 10 '17 at 16:06
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  1. No. It is safe to assume that the "collision" is elastic - ie that mechanical energy is conserved. (I think this is probably what you intended but you made a typo.) Unlike the question velocity of an object after leaving a ramp, here the force acting on the ball is always normal to the surface of the track, which is continuous, and perpendicular to the velocity of the ball. The ball and track start in contact and remain in contact, so there is really no "impact" between them which could dissipate kinetic energy.

  2. Pedantically : Yes, but this is too broad a question. You can, for example, analyze the forces between the ball and track, but this would be unnecessarily complicated. Perhaps you are asking if there are simpler or equally simple methods? No, I don't think so.

  3. Provided that the track is smooth and the ball does not leave the track at any point, the shape of the track will not affect the outcome. If there is a discontinuity in the track or in the slope, this will result in an inelastic collision and a loss of some KE. See #1.

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  • $\begingroup$ Thank you. I have a few questions about your explanation to make sure I fully understand them. You said, "The ball and track start in contact and remain in contact, so there is really no "impact" between them which could dissipate kinetic energy." and "It is safe to assume that the "collision" is elastic." $\endgroup$ – user33727 Jun 11 '17 at 0:13
  • $\begingroup$ 1. Does that mean, right after the ball is given the initial velocity, the track and the ball started moving together? (When I was doing this question, I was imagining that the ball and track only started to move together when the ball reaches the end of the horizontal track and kind of pushes the track to move to the right.) $\endgroup$ – user33727 Jun 11 '17 at 0:14
  • $\begingroup$ 2. After that, the "elastic collision" happens when the ball leaves the track? (for the first time after passing through point D) $\endgroup$ – user33727 Jun 11 '17 at 0:14
  • $\begingroup$ 1. The ball remains in contact with the track because it rolls or slides along it. The track only moves when the ball starts to move upwards at B. 2. There is no "collision" - no "impact" - in the usual sense. Just as there is a force on a ball rolling inside a hemispherical cup and a continuous change of direction but no collision. $\endgroup$ – sammy gerbil Jun 11 '17 at 2:10
  • $\begingroup$ "Perhaps you are asking if there are simpler or equally simple methods? No, I don't think so." Of course there are. Assuming no friction, you can use conservation of momentum. $\endgroup$ – Peter Shor Nov 15 '17 at 15:43

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