A ball falls from a height of $h$ and, after elastic bounces on a horizontal surface, comes to a stop after a time $t$ from the moment of release. We assume that with each collision, a certain fraction of the ball's mechanical energy is converted into non-mechanical energy. What is this loss? The duration of the collisions can be neglected. (Let $h=20 \mathrm{~cm}$, $t=18 \mathrm{~s}$, and $g=10 \mathrm{~m} / \mathrm{s}^{2}$.)
From the moment of release, $\sqrt{2 h / g}$ time elapses until the first bounce. Upon collision, the ball loses $(1-y)$ times its mechanical energy. (We want to determine this $(1-y)$.) Only $y$ times the original energy remains, and therefore the ball will rise to a height of $y \cdot h$. The time until the next collision is thus $2 \sqrt{2 h y / g}$. The same will happen again, with the ball rising to a height of $y^{2} \cdot h$. The elapsed time is $2 \sqrt{2 h y^{2} / g}$, which is $\sqrt{y}$ times the previous time. The bounce repeats infinitely, with the time between bounces always being $\sqrt{y}$ times smaller (by definition $y<1$). The entire phenomenon lasts for $t$ time, so

$$
\sqrt{\frac{2 h}{g}}+2 \sqrt{\frac{2 h y}{g}}+2 \sqrt{\frac{2 h y^{2}}{g}}+2 \sqrt{\frac{2 h y^{3}}{g}}+\ldots=t
$$

(This equation expresses that the total time of infinitely many bounces is finite, as the height of ascent and thus the time between bounces decreases very quickly.)

Rearranging:

$$
2 \sqrt{\frac{2 h}{g}} \sqrt{y}\left(1+\sqrt{y}+\sqrt{y^{2}}+\ldots\right)=t-\sqrt{\frac{2 h}{g}}
$$

Inside the parentheses, a geometric series stands. Since $y<1$, the sum of the series is finite: $\frac{1}{1-\sqrt{y}}$. Solving for $\sqrt{y}$:

$$
\sqrt{y}=\frac{t-\sqrt{\frac{2 h}{g}}}{t+\sqrt{\frac{2 h}{g}}}
$$

Substituting the given data, $\sqrt{y}=4 / 5$, or $y=0.64$. At each bounce, the energy $1-y=0.36$ times, or $36\%$, is converted to non-mechanical energy.

András Szabó (Miskolc, 2nd Industrial Secondary School, 2nd grade)