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+ <meta name="description" content="Potential energy of interaction of a particle with a fixed point source U = V (\frac{l^2}{r^2} - \frac{2l}{r}), where r is the distance between the particle and the source, V and l are positive constants having the dimension of energy and distance, respectively. In what region does the rectilinear motion of a particle occur, if the total energy of the system is equal to E?">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <title>Potential energy of interaction of a particle with a fixed point source U = V (\frac{l^2}{r^2} - \frac{2l}{r}), where r is the distance between the particle and the source, V and l are positive constants having the dimension of energy and distance, respectively. In what region does the rectilinear motion of a particle occur, if the total energy of the system is equal to E?</title>
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+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <p>
+ $2.3.40.$ Potential energy of interaction of a particle with a fixed point source $U = V (\frac{l^2}{r^2} - \frac{2l}{r})$, where $r$ is the distance between the particle and the source, $V$ and $l$ are positive constants having the dimension of energy and distance, respectively. In what region does the rectilinear motion of a particle occur, if the total energy of the system is equal to $E$?
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
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+ <figure>
+ <img src="2.3.40.png"
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+<p>
+
+Let us consider the case where the system (the second particle) starts moving at a distance $r$ from the first particle.</p><p>
+Since this is the beginning of the movement, the kinetic energy of the first and second particles tends to zero.
+$$E_{k1}\to\infty;\quad E_{k2}\to\infty$$
+From where the total energy would be equal to the potential energy
+$$E=E_p$$
+Substituting expression from the statement
+$$\frac{E}{V}=\frac{l^2}{r^2}-\frac{2l}{r}$$
+Transforming it into a quadratic equation
+$$Er^2+2Vlr-Vl^2=0$$
+Let's solve it
+$$r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$$
+Since the distance is not less than zero, we choose a positive sign for $E>0$
+$$r\geq fl\frac VE\left(\sqrt{1+E/V}-1\right)$$
+For $E<0$ it will be in the range between $r_1$ and $r_2$
+$$l\frac VE\left(-1-\sqrt{1+E/V}\right)\leq r\leq l\frac VE\left(-1+\sqrt{1+E/V}\right)$$
+ </p>
+
+ <h4>Answer</h4>
+ <p>
+ With $E>0$ area of motion $r\geqslant l\frac VE\left(-1+\sqrt{1+E/V}\right)$</p><p>
+if $E<0$ area of motion, $r$ between $r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$
+ </p>
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ Almaskhan Arsen<br>
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