Energy loss when a ball hits a wall

Thinking about this post : Conservation of energy in the breaking of glass

How is energy conserved when I throw a heavy ball at a piece of glass and cause it to shatter? The ball starts off with significant kinetic energy and ends up with very little. The glass sees what I think can be a small change in potential energy, depending on how you position things, but otherwise, I still see humongous energy losses. Is it heat energy during the process of shattering? Or am I neglecting something else...

I wanted to estimate the thermical loss when a ball was hitting the wall. I thought it came mainly from the air inside the ball being compressed.

I have little knowledge in continuum mechanics and was only interested in an estimate so I imagined a toy model:

A cylindrical piston of radius $r$, filled with air, is thrown toward a wall with a velocity $v$. It hits a wall so the air is compressed. The phenomenon was, in a first time, assumed so quick that thermic exchanges could be neglected. However, the equations we reached were symetrical for (t -> -t) so they described a reversible phenomenon which I don't want. So I added a convection term with a thermal flux $$\Phi = h(T-T_{0})\pi r l$$ where $h$ is the convection coefficient, and $\pi r l$ is the surface of the piston.

Let us isolate two systems. $S_1$ is the body of the piston, its mass will be denoted $m$. $S_2$ is the air inside the piston, the lid plus the stem.

The mass of $S_2$ will be neglected.

Applying Newton's law and the 1st principle of thermodynamics, I get

\begin{equation} \left\{ \begin{aligned} m \ddot{x_0} &=& (P_0-P)S \\ C \dot{T} &=& (P-P_0)S \dot{x_0} - h(T -T_{0}) \pi r l \end{aligned} \right. \end{equation} where $S$ is the section of the piston and $C$ the capacity of the air inside.

Then denoting $l = x_1 - x_0$, one reaches \begin{equation} \left\{ \begin{aligned} m \ddot{l} &=& (P-P_0)S \\ C \dot{T} &=& (P_0-P)S \dot{l} - h(T -T_{0}) \pi r l \end{aligned} \right. \end{equation}

Assuming the air is a perfect gas, one has: \begin{equation} \left\{ \begin{aligned} m \ddot{l} &=& (\frac{nRT}{V}-P_0)S \\ C \dot{T} &=& (P_0-\frac{nRT}{V})S \dot{l} - h(T -T_{0}) \pi r l \end{aligned} \right. \end{equation} with $V = Sl$ and $n= \frac{P_0 V_0}{RT_0}$ is the number of mole of air inside the piston

And finally: \begin{equation} \left\{ \begin{aligned} \ddot{l} &=& \frac{1}{m}(\frac{nRT}{l}-P_0S) \\ \dot{T} &=& \frac{\dot{l}}{C}(P_0S-\frac{nRT}{l}) - \frac{1}{C} h(T -T_{0}) \pi r l \end{aligned} \right. \end{equation}

I then did a numerical resolution in Python.

Remembering that at first this was inspired of a ball hitting a glass panel, I took

• $m$ the mass of a soccer ball : 500g
• $l(t=0)$ the radius of a soccer ball : $r = 10$cm
• $S$ the section of the ball : $\pi r^2$
• h = 10 $\mathrm{W m^{-2} K^{-1}}$ for air

I decided that the piston leaves the wall when $\ddot{l} = 0$ because the normal reaction of the wall cancels out.

I get the following plot : The problem is that when $\ddot{l}$ cancels out, the conditions are very similar to the ones at the beginning. - $U_0 = U_{end} = 2090$ J - $E_ 0= 1/2 m v^2 = E_{end} = 25$ J

What I don't get is where the thermical dissipation comes from. Did I miss something in my toy model ? Does it come from the stretching of the rubber ? Does it come from a phenomenon that occurs when the glass is broken ?

• In your example of compressing the gas rapidly by throwing the cylinder against the wall is not a symmetric situation. In a rapid compression (or expansion), the gas is not at thermodynamic equilibrium, and, during this time, its behavior is not described by the ideal gas law. In the rapid compression, viscous stresses within the gas must be considered in the force- and energy balances, and this results in an unsymmetrical situation. The viscous stresses result in a higher final temperature than if the gas were compressed slowly. – Chet Miller May 28 '18 at 12:55
• Ok, so in fact I should use Navier Stokes equation to describe the air inside the cylindre ? – Cabirto May 30 '18 at 10:33
• Sure. For a rapid deformation like this, you need to solve for the detailed pressure, temperature, stresses, and density variations within the cylinder. Equilibrium thermodynamics is not adequate to handle such a non-equilibrium situation. – Chet Miller May 30 '18 at 11:12