# Perpetual bouncing question

1. Nov 9, 2004

### kilowatthour

I have a physics question that’s puzzling me. If say there was theoretical environments were there was absolutely no friction, but have a large solid celestial body that exerted significant gravity and there was an object that was perfectly elastic. If this object were dropped at say 10 meters from the surface of this celestial body, would not the object bounce exactly 10 meters, and continue to bounce forever? I asked somebody that seemingly should be very knowledgeable at this but they told me no because of gravity the object couldn’t go the same height. I don’t see how that makes any sense if it was in this theoretical situation because it is just like if a ball was rolled down a incline to a level surface then to a oppositely inclined plane, if no friction was present the ball would roll back and forth on this forever reaching the same height each time. Since energy can’t be created or destroyed why would the energy in this perfect bouncing object be lost? The only reason I see that the object would not bounce forever is because there is no way that a perfectly elastic object or friction free environment would exist. Please explain to me what is the truth and tell me detailed simple logical reasoning to it. Thanks!

2. Nov 9, 2004

### rcgldr

The celestial body also need to have a 100% (no energy loss) elastic collision with the dropped ball. If no energy is lost from the collision, then the sum of kinetic and potential energy remain constant, and the ball bounces back up to 10 meters again.

3. Nov 9, 2004

### kilowatthour

Thank you very much!! any more comments to add? I look forward to explaining this to my physics teacher.

Last edited: Nov 10, 2004
4. Nov 10, 2004

### kilowatthour

If anybody could prove this in other ways or point me to references it would be great. I am using this to prove to a somwhat stuburn teacher of mine that either doesnt understand what he is teaching or more likely is just not listening / thinking about what I said. Just a logical statement that I write and draw on paper probobly would not be effective enough.

Last edited: Nov 10, 2004
5. Nov 10, 2004

### krab

Your comparison to an inclined plane is a good one. Basically, energy is conserved. The only way that it would not bounce forever is if the energy leaks into other modes. Here, I can think of one complication. A realistic perfectly elastic ball will rebound, but some of the energy will be left in the ball in the form of other vibration modes; like the ball hitting the floor will squash, but when it rebounds it will continue to stretch and continue vibrate between a squashed form and a stretched form. Now you have two frequencies: the bouncing and the vibrating. Sometimes these will be in phase when the ball hits the floor, and sometimes not. So the rebound won't be exactly 10 m but will have some variation.

6. Nov 11, 2004

### Skomatth

I have an idea of how to show this mathematically.

An object if mass $$m$$ drops from a height $$h$$ at an initial speed of zero above a plant whose acceleration due to gravity is $$g$$. The object makes an instantaneous, elastic collision with the planet and then bounces to a height $$h^'$$. There is no transformation of energy to any other factors.

Since, as we defined in the parameters, the initial energy is transformed into only kinetic energy we can say that
$$PE_{initial} = KE_f + PE_f$$
$$mgh = \frac{1}{2}mv^2 + m g h_2$$
Where v and h are the object's speed and distance above the planet respectively.
Solving this equation we get a helpful equation, for the speed as a function of the object's intial height.
$$v = \sqrt{2gh}$$

Since the collision is perfectly elastic the velocity $$v = \sqrt{2gh}$$ ,positive or negative depending on your reference frame, is immediately reversed at the moment of the collision.

Let's also use the fact that the work done on an object is equal to its change in KE. The work done by gravity on the object is $$mgh^'$$ so
$$KE_{initial}-KE{final} = mgh^'$$ When the object is at the top of its bounce its speed is 0 so KE(final)=0.
Then we have
$$\frac{1}{2} m (\sqrt{2gh})^2 = -mgh^'$$
Solve and you get $$h= -h^'$$. This is basically the fact that the work done by gravity is the negative of the change in potential energy, which your teacher should really know. Since he didn't maybe I should have derived conservation of energy all the way from F=ma but maybe this could convince him.