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This might seem like a very absurd/ridiculous question, but this question suddenly popped on my head a few days ago. Is it that I'm missing something as I haven't reviewed this topic for quite a while?

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So my assumption is if an object exerts a force equal and opposite to the object which acts on it, shouldn't a ball dropped from a certain height exerts a force (m.g) on the ground and it reacts back by pushing the ball upwards and thus sending it back to its original position/height?

I then thought for a while and realized this would be a contradiction to the Energy Conservation Law and it wouldn't make any sense anymore. I've come to known that I've definitely forgotten something. Any help would be greatly appreciated.

  • Why should the reaction force be enough to send the ball back to its original height? what kind of logic did you follow to assume this? –  Apr 22 '19 at 11:44
  • @Wolphramjonny, So if this was a completely isolated system, no energy loss etc, the ball would never bounce to its original height and eventually get lower and lower? That doesn't really sound right to me –  Apr 22 '19 at 12:15
  • yes, but this is a particular case, and has nothing to do with the third law in way you think –  Apr 22 '19 at 12:32
  • I don't fully understand the question. Does a ball not bounce back to its original height? Ideally, it certainly would. And the explanation is the energy conservation law, as you mention. Realistically, some energy is always lost (either in surface, ball or air), so it does not necessarily reach the exact same height. But energy conservation still holds true. All this can be discussed without talking about Newton's 3rd law, so I'm not sure what the issue with that law is for this description? – Steeven Apr 22 '19 at 14:50
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    @Steeven The 3rd law is precisely the issue. The OP is interpreting "every action has an equal and opposite reaction" to mean that if the ball hits the ground at some speed then it must rebound with that same speed. The OP fails to realize that the 3rd law only tells us that forces arise due to interactions, but it doesn't tell us anything about the actual value of those forces. This is discussed in the potential duplicate question – BioPhysicist Apr 22 '19 at 15:19

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Yes, if the ball were to bounce elastically(no energy lost in the deformation of the ball) and there was no atmosphere(vacuum space around Earth) the ball would reach the initial height.

It is very useful in this situation to answer in terms of momentum and energy conservation. In real life the collision of the ball on the ground is not perfectly elastic, that means that some energy is lost in the form of heat and deformation of the ball.

Also, the initial energy of the ball is $mgh$, with m being the balls mass, g the gravitational field on earth and h being the heigh(assume h=0 the Earths surface). This energy is converted to kinetic $1/2 m u^2$. From that you can deduce the momentum of the ball and apply momentum conservation. Depending on the energy loss due to impact you can calculate the new height that the ball will reach when it bounces.

Does that help?

twisted manifold
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  • "that means that some energy is lost in the form of heat and deformation of the ball." And sound. Oh and deformation can be perfectly elastic. – Gert Apr 22 '19 at 15:31
  • Correct, if one is looking for energy loses, one can surely find a large number of them, from heat, sound, molecular friction down to quantum fluctuation effects. – twisted manifold Apr 22 '19 at 15:54
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The reaction force is not equal to mg because the ball is being accelerated, in addition once the ball leaves the floor there is no longer a force acting on it. When you drop the ball the force is mg all along the trajectory, even when going up. There is no reason why the reaction force should bounce the ball back to the original height. It will only happen when there is conservation of mechanical energy.

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    Shouldn't a perfectly elastic object (neglect drag, friction, energy loss) come back to its original position? –  Apr 22 '19 at 13:08
  • @KEVINIOP Yes, if everything was perfectly elastic with no energy loss then the ball would get back to its original height. This cannot be truly obtained though. Even if your ball was perfectly elastic, you would probably still lose some energy to the ground. – BioPhysicist Apr 22 '19 at 13:13
  • @AaronStevens, thanks for the clarification –  Apr 22 '19 at 13:14
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This happens because some energy that the ball had initially carried gets lost in the form of heat some energy goes into the ground and the atoms over there begin to vibrate faster and some energy goes into the atoms of the ball and the begin to vibrate faster. But if we can create a situation in which no energy goes in form of heat and that the ground in infinitely heavy then the ball will reach its original height

Vivek Singh
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The tax man takes his cut in energy in the form of heat. Accelerating electrons emit photons. When the ball hits the earth the sudden deceleration which is a reactionary acceleration causes the electrons to emits photons and tax the bounce back energy so the ball doesn't go back to the same height. If the ball is flexible like a basket ball then the flexing of the walls will add to the problem.

Mike Redman
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    Whilst this answer explains why the ball doesn't return to its original height, it fails to address OPs misconception that this has anything at all to do with Newton's third law (which can hold perfectly fine, and does here, despite energy not being conserved). – jacob1729 Apr 22 '19 at 12:50
  • This is misleading in several ways. Emission of photons does not have to happen (and usually doesn't) in order to explain the KE energy decrease due to conversion to thermal, sound and inelastic deformations. "Reactionary acceleration" is an unusual construction. Maybe you mean teh acceleration produced by the "reaction" force? – nasu Apr 22 '19 at 15:27
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    This is a poor answer indeed. – Gert Apr 22 '19 at 15:42