7. Two small balls with charges $Q=-20 \cdot 10^{-6}$ C and $q=50 \cdot 10^{-6}$ C are located at the vertices $A$ and $B$ of a mountain slope (see figure). It is known that $AB=2$ m, $AC=3$ m. The masses of the balls are the same and equal to $m=200$ g each. At the initial moment of time, the ball with charge $q$ is released with zero initial velocity. Determine its speed at the moment when it reaches point $C$. The surface of the mountain slope is considered smooth. The acceleration due to gravity $g=10$ m/s², and the proportionality constant in Coulomb's law $k=9 \cdot 10^{9}$ N·m²/C². (10 points)

![](https://cdn.mathpix.com/cropped/2024_05_06_fbb46158359318ba2f7fg-11.jpg?height=431&width=625&top_left_y=1258&top_left_x=727)
Answer: $5 \mathrm{M} / \mathrm{c}$

Solution. The law of conservation of energy for this situation:

$k \frac{Q q}{A B}+m g \cdot A B=\frac{m v^{2}}{2}+k \frac{Q q}{B C}(5$ points $)$.

As a result, we get: $v=\sqrt{\frac{2 k Q q}{m \cdot A B}+2 \cdot g \cdot A B-\frac{2 k Q q}{m \cdot B C}}=\sqrt{\frac{2 \cdot 9 \cdot 10^{9} \cdot\left(-20 \cdot 10^{-6}\right) \cdot 50 \cdot 10^{-6}}{0.2 \cdot 2}+2 \cdot 10 \cdot 2-\frac{2 \cdot 9 \cdot 10^{9} \cdot\left(-20 \cdot 10^{-6}\right) \cdot 50 \cdot 10^{-6}}{0.2 \cdot 3}}=5 \cdot \mathrm{m} / \mathrm{c}$ $(2$ points $)$.