# What is the mechanism of heat exchange of a bouncing ball?

Imagine a falling ball on a perfectly hard ground.

The kinetic energy will be first converted into a deformation of the ball, then the ball will restore it into kinetic and heat energy and recover its previous shape.

Why would a part of the deformation energy convert into heat energy? Is the issue about how fast that energy is being restored?

Imagine two situations of restoration:

1. during the leaning to the ground, and

2. during and after the leaning to the ground

Maybe it is the case that, in the second situation, one can't use all the energy to lean to the ground in order to push the ball back. But how is the heat produced?

• Couldn't you compare the kinetic energies of the ball before and after the collision to calculate what percentage of original mechanical energy converted to heat energy? Assuming you had been provided the coefficient of restitution for the collision,of course. Commented Oct 28, 2015 at 6:00
• I'm not doing experiments. It's just a theoretical thinking. Also, I'd like to know the mechanism of heat creation. Please don't remove it. Commented Oct 28, 2015 at 6:13
• Then that's how you could do it. Compare the initial and final kinetic energies. Consider some data. Commented Oct 28, 2015 at 6:14
• And no, the issue is not about how fast the energy is restored, rather how much of it is restored. Collision durations are very difficult to analyse theoretically. Commented Oct 28, 2015 at 6:15
• :) That puzzles me Commented Oct 28, 2015 at 6:21

For the question of why a deformation of a bouncing ball would convert some of the energy of the system into heat energy, you need to think about what heat actually is in the ball. Heat will be transferred to the ball in the form of vibrations of the atoms that make up ball. When a collision occurs, the ball compresses, and upon restoration, there will be some leftover vibrations. Consider the following system:

Lets say you make an object out of a system of springs, and you hold it up and drop it onto the floor. When the system drops onto the floor, the springs will compress, then they will bounce back, putting the object back into the air, but it won't come back up as high as it originally started. So, the question is, what happened during this process?

Well, the impact is going to store the energy of the system in the springs as potential energy. As the springs restore, the potential energy will then be converted into new types of energy. One type of energy is the energy that it had before the impact, which is the kinetic energy of the falling system. However, spring restoration will also convert some of that energy into waves (and oscillations) throughout the spring system. Think about stretching out a slinky, and then compressing only one part of it. When you let go, it won't just go back to its original shape, but instead it will send a wave down the length of the slinky. These waves will bounce around the system until it reaches equilibrium, and this energy will not be converted back into the original kinetic energy, meaning that the system won't bounce back to its original height. The wave energy here in the spring system is the analog of the heat in your bouncing ball.

It is typical to using the analogy of modeling interactions between atoms in solids as many masses connected in a complicated system of springs. The analog of the spring constant will be related to how strong the bonds of the atoms are in the solid, and different values of this "spring constant" will determine how fast waves propagate through the system.

• If I understand correctly, the atomic matrix of the ball gains oscillation (energy), and since atoms are so small and many (and chaotic due to quantum effects?) that oscillation energy is unrecoverable as net kinetic energy (velocity of the ball) and dissipates randomly throughout (i.e. increased heat). Follow up question, is the % kinetic energy lost (to this heat) typically more or less for a higher velocity impact? Commented Jan 13, 2021 at 4:44

Provided that you have been provided or can calculate the coefficient of restitution $e$ for the collision, you could find $K_i$ the kinetic energy before the collision and $K_f$ the kinetic energy after the collision. Then the part which was transferred into heat is:

$$H=\Delta K=K_f-K_i$$

And as a percentage,

$$\% H=\frac {\Delta K}{K_i} * 100 \%$$