A system of three identical balls connected by identical springs is suspended on a thread. The thread is burned out. Find the acceleration of the balls immediately after burning the thread.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#2.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $2.1.14.$ A system of three identical balls connected by identical springs is suspended on a thread. The thread is burned out. Find the acceleration of the balls immediately after burning the thread.

    <h3>Solution</h3>
    <p>
        <center>

When the oscillations have already stopped, the balls move with the same acceleration $a$.

Let's write Newton's second law $$\left\{\begin{matrix} m_1a_1 = -k\Delta x \\ m_2a_2 = -F + k\Delta x \end{matrix}\right.$$ Socratic to $a$ $$\frac{m_1}{m_2} = \frac{-k\Delta x}{-F + k\Delta x}$$ We express $\Delta x$ $$\boxed{\Delta x = \frac{F}{k} \frac{m_1}{m_1+m_2}}$$ Immediately after the force ceases to act on the bodies, the elastic force $k \Delta x$ will act, which will cause acceleration $$\left\{\begin{matrix} m_1a_1 = -k\Delta x \\ m_2a_2 = k\Delta x \end{matrix}\right.$$ Where do we express $a_1$? $$\boxed{a_1 = -\frac{F}{m_{2}+m_{1}}}$$ And similarly $a_2$ $$\boxed{a_1 = \frac{Fm_{1}}{m_{2}(m_{1}+m_{2})}}$$

  </p>
    </p>

    <h4>Answer</h4>
    <p>
                <p>
        $$x=-\frac{Fm_{1}}{k(m_{1}+m_{2})}$$
        $$a_{1}=-\frac{F}{m_{2}+m_{1}}$$
        $$a_{2}=\frac{Fm_{1}}{m_{2}(m_{1}+m_{2})}$$
    </p>
  <p style="text-align: right; font-style: italic; font-size: 16;">Straton</p>
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>