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+ <h2>Solutions of Savchenko Physics Textbook</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../#2.5">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.5.9^*.$ Beads of mass $m_1$, $m_2$, $m_3$ can slide along the horizontal spoke without friction, with $m_1 \gg m_2$ and $m_3 \gg m_2$. Determine the maximum velocities of the extreme beads, if at first they were at rest, and the middle bead had a velocity $v$. The blows are elastic.
+
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+ <img src="statement.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ For problem $2.5.9^*$
+ </figcaption>
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+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+ Let's consider the velocity of the center of mass $v_c$:
+$$
+v_c = \frac{m_2 v}{m_1 + m_2 + m_3}
+$$
+According to the problem we know that $m_2 \ll m_1$ and $m_2 \ll m_3$, hence:
+$$
+v_c \approx 0
+$$
+Since the velocity of the center of mass is approximately zero, it follows that:
+$$
+m_1 v_1 = m_3 v_3 \quad (1)
+$$
+Now let's write down the law of conservation of energy, since the collisions are elastic:
+$$
+\frac{m_2 v^2}{2} = \frac{m_1 v_1^2}{2} + \frac{m_3 v_3^2}{2} \quad (2)
+$$
+By expressing the velocity $v_3$ in equation (1) and substituting it into equation (2), we obtain the answer:
+$$
+\boxed{v_1 = v \sqrt{\frac{m_2 m_3}{m_1 (m_1 + m_3)}}}
+$$
+and
+$$
+\boxed{v_3 = v \sqrt{\frac{m_2 m_1}{m_3 (m_1 + m_3)}}}
+$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ $$v_1=v\sqrt{\frac{m_2m_3}{m_1(m_1+m_3)}}; \quad v_3=v\sqrt{\frac{m_2m_1}{m_3(m_1+m_3)}}$$
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