Problem 5. Three beads with masses $m_{1}=150$ g, $m_{3}=30$ g, $m_{2}=1$ g (see figure) can slide along a horizontal rod without friction.

Determine the maximum speeds of the large beads if at the initial moment of time they were at rest, while the small bead was moving with a speed of $V=10$ m/s.

Consider the collisions to be perfectly elastic.

![](https://cdn.mathpix.com/cropped/2024_05_06_810520cdb7af4c6bca44g-3.jpg?height=166&width=857&top_left_y=748&top_left_x=628)
Solution. After each collision, the magnitude of the velocity of the bead with mass $m_{2}$ decreases. After a certain number of collisions, its velocity will be insufficient to catch up with the next bead $m_{1}$ or $m_{3}$. After such a final collision, the velocities of the beads will no longer change. Let $V_{1}$ and $V_{3}$ be the magnitudes of the velocities of the beads $m_{1}$ and $m_{3}$, respectively, and $V_{2}$ be the projection of the velocity of the bead $m_{2}$ onto the direction of motion of the bead $m_{3}$ (after the final collision).

According to the laws of conservation of momentum and energy, we get:

$$
\begin{aligned}
& -m_{1} V_{1}+m_{2} V_{2}+m_{3} V_{3}=m_{2} V \\
& \frac{m_{1} V_{1}^{2}}{2}+\frac{m_{2} V_{2}^{2}}{2}+\frac{m_{3} V_{3}^{2}}{2}=\frac{m_{2} V^{2}}{2}
\end{aligned}
$$

Since, according to the problem statement, $m_{2} \ll m_{1}$ and $m_{2} \ll m_{3}$, the terms containing $m_{2}$ can be neglected in the left-hand sides of equations (1) and (2).

Indeed, if, for example, after the final collision, the bead $m_{2}$ moves in the direction of the bead $m_{1}$, then $\left|V_{2}\right| \leq V_{1}$ and therefore $\frac{m_{2}\left|V_{2}\right|}{m_{1} V_{1}} \ll 1, \frac{m_{2} V_{2}^{2}}{m_{1} V_{1}^{2}} \ll 1$. Similarly, the situation is considered when the bead $m_{2}$ moves after the final collision in the direction of the bead $m_{3}$.

Thus, we obtain the system of equations:

$$
\left\{\begin{array}{l}
-m_{1} V_{1}+m_{3} V_{3}=m_{2} V \\
m_{1} V_{1}^{2}+m_{3} V_{3}^{2}=m_{2} V^{2}
\end{array}\right.
$$

Denoting $m_{2}=m$, then $m_{3}=n m, m_{1}=5 n m$, where $n=30$, we write the system as:

$$
\left\{\begin{array}{l}
-5 n V_{1}+n V_{3}=V, \\
5 n V_{1}^{2}+n V_{3}^{2}=V^{2}
\end{array}\right.
$$

Solving this system, we find

$$
\begin{gathered}
\frac{V_{1}}{V}=-\frac{1}{6 n}+\sqrt{\frac{1}{30 n}-\frac{5}{30^{2} n^{2}}} \approx-\frac{1}{6 n}+\sqrt{\frac{1}{30 n}}=-\frac{1}{180}+\frac{1}{30}=\frac{1}{36} \\
\frac{V_{3}}{V}=\frac{1}{6 n}+\sqrt{\frac{30}{36 n}-\frac{5}{36 n^{2}}} \approx \frac{1}{6 n}+\sqrt{\frac{30}{36 n}}=\frac{1}{180}+\frac{1}{6}=\frac{31}{180}
\end{gathered}
$$

Answer: $V_{1} \approx 0.28 \mathrm{~m} / \mathrm{c} ; V_{3} \approx 1.72 \mathrm{~m} / \mathrm{c}$.