# Inelastic ball bounce (sound vs. heat)

Today in my son's physics class, his teacher asked why a superball does not bounce as high as the point from which it is dropped. At dinner when my son repeated this question, I answered that a small portion of the energy is converted to heat in the ball (and floor). His answer, however, which was rewarded as correct by the teacher was acoustic energy. Of course, both answers are correct, but..... whose response represents the greatest energy conversion that robs the ball of its height, thermal or acoustic? I have not been able to find a quantitative answer to what I would think could be easily discovered through a quick search on line.

Greg

A bouncing superball is just not that loud and it takes a surprisingly small amount of energy to make a loud sound. On the other hand it takes a fair bit of energy to heat things up so I think that its safe to say that the vast majority of the lost kinetic energy is being converted to thermal energy. Let's use some high school physics to do a back of the envelope calculation.

Assumptions The superball has a mass of 50 grams and that it is dropped from a height of 1.0 m and bounces back up to a height of 0.9 m. The ball compresses 2 mm when it bounces.

First we calculate how much sound + thermal energy is generated by calculating how much gravitational potential energy is lost. If the ball bounces to 90% of its initial height it will lose 10% of its initial gravitational energy during the bounce. $$10\% \times mgh = 0.1 \times 0.05\,\mathrm {kg} \times 9.8 \,\mathrm {N/m}\times 1.0\,\mathrm m = 0.049\,\mathrm J$$

So, how long does it take to lose this much energy? Time for a quick bit of kinematics.
\begin{align}\Delta d &= -1.0\,\mathrm m \\ v_1 &= 0\\ v_2 &= \,?\\ a &= -9.8\,\mathrm {m/s^2}\\ \Delta t &= \,-\end{align}

\begin{align}v_2^2 &= v_1^2+2a\Delta d\\ v_2^2&=0+2(-9.8)(-1.0)\\ v_2&=\sqrt {19.6}\\ v_2&=4.43\,\mathrm {m/s^2} \end{align}

The ball hits the ground moving down at 4.43 m/s. A similar calculation reveals the speed leaving the ground to be 4.20 m/s. So how long does this collision last?

For the first part of the collision while the ball is compressing:

\begin{align}\Delta d &= -0.002\,\mathrm m \\ v_1 &= -4.43\\ v_2 &= 0\\ a &= -\\ \Delta t &= \,?\end{align}

\begin{align}\Delta d &=\frac{v_1+v_2}{2}\Delta t\\ -0.002&=\frac{-4.3+0}{2}\Delta t\\ \Delta t&=0.00093\,\mathrm s\\ \end{align}

A similar calculation for the second half of the collision while the ball is decompressing reveals that it takes 0.00095 s so a total collision time of 0.0019 s or just under 2 milliseconds.

Now that we have the power dissipated and the time taken we can calculate the power.

\begin{align}P &=\frac{\Delta E}{\Delta t}\\ P &=\frac{0.049\,\mathrm J}{0.0019\, \mathrm s}\\ P&=26 \, \mathrm {watts}\\ \end{align}

Intensity of sound is measured in watts per square meter so assuming 100% of the energy was sound energy and that the bouncing ball is heard by the person dropping it about 1.5 from the point of impact, and that the sound spreads out equally in all directions the intensity would be:

\begin{align}I&=\frac {P}{A}\\ I&=\frac{P}{4\pi r^2}\\ I&=\frac{26 \mathrm W}{4 \pi \times (1.5 \mathrm m)^2}\\ I&=0.92 \,\mathrm{W/m^2} \end{align}

Lastly, sound intensity is converted to loudness, or sound level, by the formula: $$\beta (\mathrm {dB})=10 \log_{10}\frac{I}{10^{-12}\mathrm {(W/m^2)}}$$ So, $$0.92 \,\mathrm{W/m^2}$$ converts to an astonishing 120 dB, a sound level often compared to pneumatic drills. What this tells us is that clearly much less than 100% of the gravitational energy ended up as sound energy. Maybe it was 50%? That would lead to a sound intensity of $$0.46 \,\mathrm{W/m^2}$$ but due to the logarithmic nature of the sound level formula that's still 117 dB. If 10% of the energy was sound the sound level would be 110 dB. If 1% of the energy was sound the sound level would be 100 dB and so on. Each ten fold reduction in the sound energy yields a 10 dB reduction in the sound level. This tells us that the fraction of the energy being converted to sound is very tiny compared to the amount of thermal energy produced.

• As long as I can remember, I've 'known' (taken on trust) that the sound energy emitted in collisions is negligible compared with the energy transferred to thermal, so it's really good to see such a clear and simple quantitative treatment. I wonder if a half period of shm might give a better value for the contact time, but I'm sure that your value is good enough for the purpose. Many thanks! Commented Feb 2, 2022 at 15:39
• @PhilipWood True, the acceleration during the collision wouldn't be constant. Commented Feb 2, 2022 at 15:42
• @PhilipWood I just did the math, the half period of shm works to to 4.5 ms which gives a average power 11 watts - same order of magnitude. Commented Feb 2, 2022 at 16:11
• Bravo! I thought it wouldn't make a whole lot of difference. Commented Feb 2, 2022 at 16:13
• Amazing! Thank you for your in-depth answer to my question. Commented Feb 2, 2022 at 22:44