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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
| </div> | | </div> |
| </a> | | </a> |
| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#2.5">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#2.5">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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. | | $2.5.9^*.$ 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. |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $2.5.9^*$ | | For problem $2.5.9^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Let's consider the velocity of the center of mass $v_c$: | | Let's consider the velocity of the center of mass $v_c$: |
| $$ | | $$ |
| v_c = \frac{m_2 v}{m_1 + m_2 + m_3} | | 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: | | According to the problem we know that $m_2 \ll m_1$ and $m_2 \ll m_3$, hence: |
| $$ | | $$ |
| v_c \approx 0 | | v_c \approx 0 |
| $$ | | $$ |
| Since the velocity of the center of mass is approximately zero, it follows that: | | Since the velocity of the center of mass is approximately zero, it follows that: |
| $$ | | $$ |
| m_1 v_1 = m_3 v_3 \quad (1) | | 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: | | 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) | | \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: | | 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)}}} | | \boxed{v_1 = v \sqrt{\frac{m_2 m_3}{m_1 (m_1 + m_3)}}} |
| $$ | | $$ |
| and | | and |
| $$ | | $$ |
| \boxed{v_3 = v \sqrt{\frac{m_2 m_1}{m_3 (m_1 + m_3)}}} | | \boxed{v_3 = v \sqrt{\frac{m_2 m_1}{m_3 (m_1 + m_3)}}} |
| $$ | | $$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
