# After how many bounces will a ball's mechanical energy equal zero?

This was a question I asked myself for fun. It turned out to be more difficult than I initially imagined.

The Problem: Let's say a ball is dropped from h0. Air friction is negligible. The collisions between the ball and ground are inelastic, so mechanical energy is not conserved. After how many bounces will the ball come to a rest?

My thought process: I wasn't sure if the change in mechanical energy after one bounce is constant. My guess was that it wasn't, and an exponential function is more likely (although I have no idea how to prove this, other than conducting an experiment. Is there a theoretical proof? I tried thinking about it but couldn't come up with anything on paper.) If we model the change in mechanical energy as an exponential function, we can graph it with the mechanical energy of the ball-earth system on the $$y$$ axis, and the number of bounces $$n$$ on the $$y$$ axis. It then becomes obvious that the number of bounces approaches infinity as the energy approaches zero Joules. But the ball obviously cannot bounce an infinite amount of times!

So what gives?

• On the model of infinitely many bounces with exponentially reduced energy, their durations also exponentially decay, so take a finite time as in one of Zeno's paradoxes. They also reach exponentially small heights, so we might argue they stop eventually based on atoms' size.
– J.G.
Commented Jun 15 at 15:50
• The atoms in the floor are jiggling because of thermal energy. This adds some randomness to the bounces. Eventually the bounces become smaller than the size of the thermally induced motion. Commented Jun 15 at 16:13
• Depends on many factors. Sometimes in a single collision the ball can stick and therefore the entire mechanical energy would be transformed. This would be a completly inelastic collision. Commented Jun 15 at 16:57
• In an idealized system, there will be an infinite number of ever-smaller bounces. However, real systems eventually reach what is known as "inelastic collapse," in which surface adhesion replaces the bouncing behavior. See, e.g., journals.aps.org/pre/abstract/10.1103/PhysRevE.50.R28 or nature.com/articles/s42005-019-0184-y
– Buzz
Commented Jun 16 at 1:11
• This is somewhat similar to asking "when will a damped pendulum actually stop?" Commented Jun 16 at 18:56

Never, if you treat the problem with the laws of classical mechanics, treat both the ball and the ground as "macroscopically" rigid, and you don't set a threshold $$\overline{h}$$ under which you consider it at rest.

It's the same problem my professor gave us at High school, also asking us the space covered by the ball during these bounces. He put it in a test, and we knew nothing about series: as you'll see reading the whole answer, that could be an issue. Anyways, I'm so grateful to that professor who forced us to think, don't get used to standard tests (his tests were much more challenging than lots of exams in Engineering) and work properly in an orderly way, and I felt so proud to get (almost, I forgot the factor 2) the right answer in that test. I feel he's one of the most important people of my life.

Starting from height $$h_0$$ over the ground, the ball has total mechanical energy $$E_0 = m g h_0$$. If the inelastic collision dissipates $$\varepsilon E_{before}$$, the total mechanical energy after the first bounce (and before the second bounce) is $$E_1 = (1 - \varepsilon) E_0 = \eta E_0$$, and the maximum height is $$h_1 = \frac{E_1}{m g} = \frac{\eta E_0}{m g} = \frac{\eta \, m g h_0}{m g} = \eta h_0 \ .$$

The maximum height reached between the $$n^{th}$$ and $$(n+1)^{th}$$ bounce is

$$h_n = \eta^n h_0 \ .$$

If you set the threshold $$\overline{h}$$, you just need to solve the problem

$$\text{min} \, n \in \mathbb{N} \text{ s.t. } \overline{h} \ge h_n = \eta^n h_0 \ ,$$

i.e.

$$\text{min} \, n \in \mathbb{N} \text{ s.t. } \log_\eta \frac{\overline{h}}{h_0} \ge n \ .$$

or you can solve for $$n \in \mathbb{R}$$ and take the ceil, i.e. the smaller integer number greater than $$n$$.

Bonus. The answer to my professor question involves the geometric series $$\sum_{n=1}^{\infty} \eta^n = \frac{1}{1 - \eta}$$ and reads

$$y^{tot} = h_0 + 2 \sum_{n=1}^{\infty} h_n = h_0 + 2 h_0 \sum_{n=1}^{\infty} \eta^n = h_0 \left( 1 + 2 \frac{1}{1 - \eta} \right) \ .$$

• +1 Very good answer! Though, I feel like stating that "Never, if you treat the problem with the laws of classical mechanics and don't set threshold" is a bit too much. I believe you can get a finite number of bounces by using the proper laws of classical mechanics without the threshold height, if you consider the ball's elasticity and deformation when in contact with the ground. Commented Jun 16 at 19:16
• You're right. I was thinking about writing it, specifying that, at some point, it's likely that while the center of mass of the ball keeps oscillating, some part of the ball would be always in contact with the ground one takes into account deformations of the ball and the ground. But I thought it would have over-complicated the answer. Maybe I could update it Commented Jun 16 at 19:18
• Up to you! I just saying that, if you consider ball and ground deformation, it is still classical mechanics hehe. Commented Jun 16 at 19:20
• Let's do a minor modification now, and update the first sentence with "and if you consider both the ball and the ground (macroscopically) rigid" Commented Jun 16 at 19:22

First approximation: ball and surface are perfectly rigid. This implies the collision is instantaneous, and therefore elastic. The bounce height is always the same.

Second approximation: ball and surface are deformable, non-lossy elastic bodies. Now the collisions are not instantaneous, but still no energy is lost. The bounce height is always the same.

Third approximation: ball and surface are deformable, lossy elastic bodies (if you prefer: only the ball is deformable, the surface is rigid). Bouncing stops when the remaining energy is only sufficient to cause vibrations within the ball, but not to raise the ball above the surface. The total kinetic energy in the system is distributed between vibration and up and down motion; it decays roughly exponentially with time, but at some point the amplitude of the vibrations of the surface of the ball relative to its center becomes larger than the amplitude of the movement of the center up and down, and so the ball stays continuously in touch with the surface after that.

How many bounces is a very difficult question, because (1) the ball is vibrating while it is in the air as well as during a collision, and those vibrations are also damped because it is made of a lossy elastic medium, and (2) what happens during each collision with the ground depends on the relative phase of the ball's vibrations at that time. Sometimes it hits "in phase" (the collision is lower velocity because the surface of the ball was moving towards the center of the ball at the time it hits; after that the vibrations are stronger), and sometimes "out of phase" (the collision is higher velocity because the surface of the ball was moving away from the center of the ball at the time it hits; after that the vibrations are greatly reduced or reverse phase depending on how strong the collision is); or any relative phase in between. You can observe this effect when bouncing a basketball. I think "in phase" collisions lose the most energy during the collision itself, because the deformation is greatest. But the short version is that what happens during a collision cannot be predicted from the velocity of the center of mass of the ball alone.

Lastly, a solid elastic ball can have multiple vibrational modes, each collision would excite all of them (but not to the same degree). The real vibration is a superposition of these.

These last two phenomena are almost impossible to model without very careful finite element simulation of the ball; don't expect anything like a formula that tells you "how many bounces".