Two equally massive and charged balls are $l$ apart when we release them with zero initial velocity. After $t$ time, the distance between them doubles. How much time will it take for the distance between the balls to double if we release them from an initial distance of $3 l$? We can neglect air resistance.
I. Solution. Let's denote the distance between the balls at any time $t$ by $x$! According to the conservation of energy:

$$
2 \cdot(1 / 2) m v^{2}+k q^{2} / x=k q^{2} / x_{0}
$$

where $x_{0}$ is the distance between the balls when they come to a stop. ( $q$ is the charge of the balls, $m$ is their mass.) The magnitude of the balls' velocity is

$$
v=(l / 2) \mathrm{d} x / \mathrm{d} t
$$

Substituting this and solving for $\mathrm{d} x / \mathrm{d} t$,

$$
\frac{\mathrm{d} x}{\mathrm{~d} t}=\sqrt{\frac{4 k q^{2}}{m} \frac{x-x_{0}}{x x_{0}}}
$$

is obtained. The sought distance $x$ - as a function of $t$ - satisfies the above relationship at every time $t$. (We say that the function $x(t)$ satisfies a certain differential equation.) In addition, we know that $x(0)=x_{0}$. It can be proven that exactly one such function exists, and it can be heuristically obtained as follows.

From the equation, for a small $\Delta t$ we get:

$$
\Delta t=\Delta x \sqrt{\frac{m x_{0}}{4 k q^{2}} \cdot \frac{x}{x-x_{0}}}
$$

Divide the interval $\left(x_{0}, 2 x_{0}\right)$, and the corresponding time interval, into small parts. For the $j$-th part, the following holds:

$$
\Delta t_{j}=\Delta x_{j} \sqrt{\frac{m x_{0}}{4 k q^{2}} \cdot \frac{x_{j}}{x_{j}-x_{0}}}
$$

Summing these,

$$
t_{x_{0}}=\sum_{j-1}^{n} \Delta t_{j}=\sum_{j-1}^{n} \Delta x_{j} \sqrt{\frac{m x_{0}}{4 k q^{2}} \cdot \frac{x_{j}}{x_{j}-x_{0}}}
$$

The right side is an approximate sum of the integral

$$
\int_{x_{0}}^{2 x_{0}} \sqrt{\frac{m x_{0}}{4 k q^{2}} \cdot \frac{x}{x-x_{0}}} \mathrm{~d} x
$$

so by refining the partition, we get:

$$
t_{x_{0}}=\int_{x_{0}}^{2 x_{0}} \sqrt{\frac{m x_{0}}{4 k q^{2}} \cdot \frac{x}{x-x_{0}}} \mathrm{~d} x
$$

Introducing the variable $z = x / x_{0}$ instead of $x$:

$$
t_{x_{0}}=\sqrt{\frac{m x_{0}^{3}}{4 k q^{2}}} \cdot \int_{1}^{2} \sqrt{\frac{z}{z-1}} \mathrm{dz}
$$

In the first case, $x_{0}=l$, so

$$
t_{i}=\sqrt{\frac{m l^{3}}{4 k q^{2}}} \int_{1}^{2} \sqrt{\frac{z}{z-1}} \mathrm{~d} z
$$

In the second case, $x_{0}=3 l$, thus

$$
t_{3 l}=\left(\frac{m 3^{3} l^{3}}{4 k q^{2}}\right)^{1 / 2} \int_{1}^{2} \sqrt{\frac{z}{z-1}} \mathrm{~d} z=3 \cdot \sqrt{3} t_{l}
$$

Béla Faragó (Csongrád, Batsányi J. Gym., IV. grade)

II. Solution. In the problem, let's take the charge $(q)$, mass $(m)$, and initial distance $\left(x_{0}\right)$ of the balls as given, and find the time $t_{x_{0}}$ it takes for the initial distance to double! Let's examine in what form $t_{x_{0}}$ can depend on the given quantities $g, m$, and $x_{0}$, as well as the constant $k$ in Coulomb's law. Among the given quantities, there is no quantity with the dimension of time, so the expression for $t_{x_{0}}$ can only be of the form

$$
t_{x_{0}}=q^{\alpha_{1}} \cdot k^{\beta_{1}} \cdot x_{0}^{\gamma_{1}} \cdot m^{\delta_{1}} \cdot f\left(q^{\alpha_{2}} \cdot k^{\beta_{2}} \cdot x_{0}^{\gamma_{2}} \cdot m^{\delta_{2}}\right)
$$

where $\alpha_{1}, \beta_{1}, \gamma_{1}, \delta_{1}$ and $\alpha_{2}, \beta_{2}, \gamma_{2}, \delta_{2}$ are such that $q^{\alpha_{1}} \cdot k^{\beta_{1}} \cdot x_{0}^{\gamma_{1}} \cdot m^{\delta_{1}}$ has the dimension of time, and the combination $q^{\alpha_{2}} \cdot k^{\beta_{2}} \cdot x_{0}^{\gamma_{2}} \cdot$ $m^{\delta_{2}}$ is dimensionless. Examining the dimensions of the given quantities (e.g., in the MKS system $[q]=$
$\left.\mathrm{C},[k]=\frac{\mathrm{kg} \cdot \mathrm{m}^{3}}{\mathrm{C}^{2} \cdot \mathrm{sec}^{2}} ;\left[x_{0}\right]=m ;[m]=\mathrm{kg}\right)$, we find that only $\alpha_{1}=-1, \beta_{1}=-1 / 2, \gamma_{1}=3 / 2, \delta_{1}=$ $1 / 2$ and $\alpha_{2}=0, \beta_{2}=0, \gamma_{2}=0, \delta_{2}=0$ satisfy our conditions. Therefore,

$$
t_{x_{0}}=\left(\frac{m x_{0}^{3}}{k q^{2}}\right)^{1 / 2} \cdot f
$$

where the dimensionless number $f$ is determined by the problem but does not depend on the given quantities. If $x_{0}=l$,

$$
t=t_{i}=\left(\frac{m l^{3}}{k \cdot q^{2}}\right)^{1 / 2} f
$$

and if $x_{0}=3 l$,

$$
t_{3 l}=\left(\frac{m 27 l^{3}}{k \cdot q^{2}}\right)^{1 / 2} f=3 \sqrt{3} \cdot t_{l}
$$

Based on the work of András Tóth (Pécs, Nagy Lajos Gym., III. grade)

Note: In the second solution, we wrote the dimensions in the MKS system. It can be shown that the result of dimensional analysis does not depend on the chosen unit system; dimensional analysis is based on the independence of natural laws from the unit system.

The constant $f$ in the second solution can be obtained from the first solution:

$$
f=\frac{1}{2} \cdot \int_{1}^{2} \sqrt{\frac{z}{z-1}} \mathrm{~d} z
$$