Worked exercise — Bouncing ball (semilog)

Problem

A ball bounces on the floor. The maximum height $h$ reached at each bounce is measured. We expect $h$ to follow an exponential law $h = a\cdot k^{n}$, where $n$ is the bounce number and $k < 1$ (a fixed percentage of energy is lost at each bounce). Measured data (simulated with $a = 3{,}00$ m, $k = 0{,}87$, plus experimental noise). Lab experience 3C and 3F, Fanti 2024–25.

$n$	$h$ (m)	$\log_{10} h$
0	2{,}917	0{,}465
1	2{,}485	0{,}395
2	2{,}220	0{,}346
3	1{,}566	0{,}195
4	1{,}686	0{,}227
5	1{,}870	0{,}272
6	1{,}228	0{,}089
7	1{,}191	0{,}076
8	0{,}468	−0{,}330
9	1{,}035	0{,}015
10	0{,}941	−0{,}026

Work out the parameters $a$ and $k$ of the law and give a physical interpretation of the result.

Solution

Set-up. The expected law $h = a\,k^{n}$ is exponential: on the linear graph $(n,h)$ the points trace a descending curve, but taking the logarithm turns it into a straight line. $\log_{10}(h) = \log_{10}(a) + n\cdot\log_{10}(k)$ Setting $Y = \log_{10} h$, $q = \log_{10} a$ and $m = \log_{10} k$ gives the line $Y = q + m\,n$: it is obtained by plotting the values $\log_{10} h$ (third column) against $n$, i.e. the semilogarithmic graph.

Linear regression. The best fit on the $\log_{10} h$ values gives $Y = 0{,}447 - 0{,}0577\,n$ hence slope $m = -0{,}0577$ and intercept $q = 0{,}447$.

Recovering the parameters. $k$ is obtained from the slope, $a$ from the intercept: $k = 10^{m} = 10^{-0{,}0577} \approx 0{,}875 \qquad a = 10^{q} = 10^{0{,}447} \approx 2{,}80\ \text{m}$ $\ev{a \approx 2{,}80\ \text{m}, \qquad k \approx 0{,}875}$

Physical interpretation. The slope is negative, so $k < 1$: this is a decay. The ball “starts” at $2{,}80$ m at the first bounce and at each bounce retains $87{,}5\%$ of the previous height, i.e. it loses about $12{,}5\%$ of its potential energy (height is proportional to energy). The single very low point at $n = 8$ is experimental noise: the best-fit line “averages” it with the others, and this is precisely the advantage of fitting a line rather than trusting a single measurement.

The two graphs. On the left, the data on a linear scale (curve); on the right, the same data on a semilogarithmic scale, where they become a straight line.

On the left: on a linear scale the points follow a descending exponential curve, with the fit $h = 2{,}80\cdot 0{,}875^{\,n}$.

On the right: on a semilogarithmic scale the same points lie along a line of slope $m = -0{,}0577$; $k$ and $a$ are read off from the slope and intercept.

Links

Topics: Metodo sperimentale Skills: Uso delle scale logaritmiche · Linearizzazione dei dati · Lettura dei grafici