If a ball hits the floor after an acceleration then why does it bounces lower? I mean the Energy is passed to the floor then why does the floor give back less Energy?

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    $\begingroup$ Reductio ad absurdem proof: If it returned to the same height, you could make a mini windmill parallel to the ground where the sound from the ball emitted to, to spin the wheel. Even if just a little bit, this would be free, and created out of nothing energy. $\endgroup$ – Cruncher Jun 6 '14 at 20:45
  • $\begingroup$ @Cruncher: I don't think the OP is asking whether it's true, I think the OP is asking why it's true. $\endgroup$ – user541686 Jun 7 '14 at 5:27
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    $\begingroup$ @Mehrdad A proof is a fair idea of why. In this case, the why is really "Because of conservation of energy" $\endgroup$ – Cruncher Jun 9 '14 at 13:02

Assuming for a moment an infinitely hard and smooth surface, let's look at the energy of the ball.

When the ball is dropped from a height $h$, initial potential energy is $mgh$. You would expect it to accelerate to a velocity $v=\sqrt{2gh}$. However, during the fall, it will experience drag from the air. This will cause the dissipation of some of the energy of the ball into energy of the air (turbulence, heating, flow). How large this effect is will depend on the ball, the height, ... For example a ping pong ball (light for its size) will experience a much greater effect than a golf ball (same size, but heavier).

Then we get to the impact. The ball will deform during the impact - the center of mass tries to keep going, but the surface it hits try to stop it. This leads to elastic deformation like this:

enter image description here

(Image credit: http://ej.iop.org/images/0143-0807/34/2/345/Full/ejp450030f2_online.jpg )

The potential energy of the ball was converted to elastic energy. You can think of it as the mass of the ball being mounted to a spring that compresses when you hit the floor - but there will be some friction (both inside the ball, and particularly between the ball and the floor) which will dissipate energy:

enter image description here.

The picture on the left is the "in flight" state - the spring is uncoiled. The picture in the middle is the "fully compressed" state in which all energy of motion would have been converted to elastic energy. The picture on the right is the "actual" state: the spring did not compress all the way down because energy was lost due to friction (and thus was not available for compressing the spring).

Why do I say that friction between ball and floor is important for vertical impact? Look at this picture:

enter image description here

Image credit: http://deansomerset.com/wp-content/uploads/2011/11/tennis-ball-impact.jpg

This is a tennis ball bouncing on the ground. See how distorted it is? Imagine taking that spherical surface, and pushing it into this new shape. The distortion requires you to change the contact area. As you do so, ball rubs against floor. The lateral force that this generates dissipates heat - so energy is lost instead of being stored in the elasticity of the ball.

Of course for an air filled ball, there are losses associated with the compression of the air: while the air is (adiabatically) compressed, it heats up; while it is hot, it dissipates heat to the environment; and when it expands, it cools down again. This ought to mean that when you hit a ball it gets cold: but we know that a squash ball, for example, gets HOT, not cold, when you hit it (this is why squash players "warm the ball up" before playing: as it becomes hotter, the pressure in the ball rises and it becomes bouncier). This heating is due to the extreme distortion (and thus again friction) of the ball during impact:

enter image description hereenter image description here

Source: screen shots from https://www.youtube.com/watch?v=5IOvqCHTS7o

There are other loss mechanisms - internal friction of the rubber in the ball, internal friction in the surface you are hitting (sand vs concrete), ...

All this combines to give a particular ball and surface combination something called the coefficient of restitution - a number that expresses how much of the energy of the ball before the impact is "given back" (restitution (noun): the restoration of something lost or stolen to its proper owner.) after the impact. This coefficient is always less than 1 (unless you have flubber). Since the height to which the ball will bounce is directly proportional to its energy (barring effects of air friction), with a coefficient of restitution of less than one the ball will bounce less and less high.

  • $\begingroup$ How did you make/find the first image (the one where the ball changes shape in black and white)? $\endgroup$ – Peterix Jun 6 '14 at 21:25
  • $\begingroup$ @Peterix - Google is my friend. Search term was elastic deformation ball impact and this picture was the 146th (several pages down) under the "images" tab. $\endgroup$ – Floris Jun 6 '14 at 22:01

When the ball hits the floor, its center of mass needs to be decelerated. It is done by deforming the ball (and possibly the floor if it is soft enough compared to the ball). In vacuum, if the ball and floor were perfectly elastic, they would then recover their previous shape following exactly the reverse dynamics as the deformation ones: in doing so, the center of mass of the ball would be accelerated of the exact same amount it has been decelerated, thus it would bounce back to the same height.

However, nothing is perfectly elastic: there is some friction during deformation (solids are somewhat viscoelastic), dissipated as heat, and there's also damage (plastic deformations: if the ball is hard, it will notch or chip bits off the floor e.g.) This energy is not recovered when the ball takes back its original shape.

As Jim notes, another loss is due to friction with air: although this is generally small for a ball heavy enough, this would still prevent bouncing back to the same height. It occurs both during fall and rise, and when hitting the ground: that's when the sound of the ball hitting is produced.

  • $\begingroup$ It would not bounce back to the same height, there is still air resistance $\endgroup$ – Jim Jun 6 '14 at 14:02
  • $\begingroup$ @Jim: True! I'll add this. $\endgroup$ – Joce Jun 6 '14 at 14:14
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    $\begingroup$ I'd also like to take issue with saying that air resistance is usually small. That is entirely dependent on the initial height it is dropped from. If you dropped a ball out of an airplane, air resistance would be the most influential factor in why it doesn't return to the same height $\endgroup$ – Jim Jun 6 '14 at 14:38
  • $\begingroup$ @Jim: Well yes, but this is indeed not the most common situation, is it? Here we are facing a daily-life physics question, and answer it in these conditions. $\endgroup$ – Joce Jun 6 '14 at 14:41
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    $\begingroup$ @Floris: Following your comment, I re-read my answer and agree that the first paragraph was lacking some mention of air friction. I've added the words "in vacuum", thanks! $\endgroup$ – Joce Jun 20 '14 at 6:48

Right off the bat, as the ball falls, there is air resistance that slows it. This means the downwards acceleration it experiences as it falls is less than g. It also means that on the bounce back up, drag will be working with gravity so that the net acceleration is greater than g. So even in the case of purely elastic collision, the ball cannot reach the same height because of drag.

When it hits the ground, energy is not transferred perfectly. If you hear the bounce, that's some energy lost as sound. When it hits the ground, the act of deforming the surfaces along with frictions and compressing air all contribute to creating heat. That means some more energy is lost to heat. Then there is also the cases of inelastic collisions. If the collision is highly inelastic, the ball simply deforms like dough on a pizza tosser's head. This means that there is no return of energy as the ball resumes its original shape and thus no upward acceleration.

In a vacuum, with perfectly elastic collision, the ball will return to the same height (barring the effects of general relativity. So don't go bouncing balls close to a black hole). But alas, reality is often not ideal.


At impact, most of the kinetic energy is transferred to elastic energy in the ball (by its deformation) and not to the floor. Some energy is also converted to other forms like heat and sound. These other forms of energy, are mostly losses and they are not recovered thus making the ball bounce back to a lower height.

Surely there is some energy passed to the floor, but that's not very relevant for the process. Only the energy stored as elastic potential in the ball produces the force against the floor to send back up.

  • $\begingroup$ Thanks and is there a way to calculate how much Energy is lost and how much Energy is returned to the ball? $\endgroup$ – Peterix Jun 6 '14 at 10:58
  • $\begingroup$ As I said, there is no energy returned to the ball. There is just energy returned from elastic to kinetic, but that's all within the ball. I am not aware of any way to calculate how much energy is lost from first principles. $\endgroup$ – Ignacio Vergara Kausel Jun 6 '14 at 11:10
  • $\begingroup$ That's misleading. We generally say "energy is returned to the ball" to mean "potential energy is converted back to kinetic energy." You can't predict this value unless you already know the coefficient of elasticity, but you can easily measure it by dropping the ball and comparing the original drop height (and thus the starting potential energy) with the peak of the first bounce (the final potential energy). $\endgroup$ – Carl Witthoft Jun 6 '14 at 11:38
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    $\begingroup$ The other extreme of "floor elasticity" vs. "ball elasticity" is a shot putter's shot (or a cannonball) on a trampoline. $\endgroup$ – Jyrki Lahtonen Jun 6 '14 at 20:52
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    $\begingroup$ @CarlWitthoft I'd say the elasticy of the floor would be important in general, but it happens that in all the examples used here, the floor is so much less elastic (and more heavy) that it's right to ignore that part when implying this context; (And, isn't it that it's basically even symmetric?) $\endgroup$ – Volker Siegel Jun 7 '14 at 4:11

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