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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
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			<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 $E$?
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			<h3>Solution</h3>
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		@@ -65,7 +65,7 @@

			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.
		−	$$E_{k1}\to~~\infty~~;\quad E_{k2}\to~~\infty~~$$
		+	$$E_{k1}\to 0;\quad E_{k2}\to 0$$
			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
			$$Er^2+2Vlr-Vl^2=0$$
		@@ -75,7 +75,7 @@
			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 $E>0$
		−	$$r\geq fl\frac VE\left(\sqrt{1+E/V}-1\right)$$
		+	$$r\geq l\frac VE\left(\sqrt{1+E/V}-1\right)$$
			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 $r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$
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			Almaskhan Arsen<br>
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	<h2>Solutions of Savchenko Problems in Physics</h2>		<h2>Solutions of Savchenko Problems in Physics</h2>
	<p class="author">		<p class="author">
	Aliaksandr Melnichenka <br/>		Aliaksandr Melnichenka <br/>
	October 2023		October 2023
	</p>		</p>
	</header>		</header>

	<h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#2.3">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<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 $E$?		$2.3.40.$ 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>		</p>

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

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="2.3.40.png"		<img src="2.3.40.png"
	loading="lazy" width="230" />		loading="lazy" width="230" />
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	</figcaption>		</figcaption>
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	</center>		</center>
	<p>		<p>
	@@ -65,7 +65,7 @@

	Let us consider the case where the system (the second particle) starts moving at a distance $r$ from the first particle.</p><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.		Since this is the beginning of the movement, the kinetic energy of the first and second particles tends to zero.
	$$E_{k1}\to~~\infty~~;\quad E_{k2}\to~~\infty~~$$		$$E_{k1}\to 0;\quad E_{k2}\to 0$$
	From where the total energy would be equal to the potential energy		From where the total energy would be equal to the potential energy
	$$E=E_p$$		$$E=E_p$$
	Substituting expression from the statement		Substituting expression from the statement
	$$\frac{E}{V}=\frac{l^2}{r^2}-\frac{2l}{r}$$		$$\frac{E}{V}=\frac{l^2}{r^2}-\frac{2l}{r}$$
	Transforming it into a quadratic equation		Transforming it into a quadratic equation
	$$Er^2+2Vlr-Vl^2=0$$		$$Er^2+2Vlr-Vl^2=0$$
	@@ -75,7 +75,7 @@
	Let's solve it		Let's solve it
	$$r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$$		$$r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$$
	Since the distance is not less than zero, we choose a positive sign for $E>0$		Since the distance is not less than zero, we choose a positive sign for $E>0$
	$$r\geq fl\frac VE\left(\sqrt{1+E/V}-1\right)$$		$$r\geq l\frac VE\left(\sqrt{1+E/V}-1\right)$$
	For $E<0$ it will be in the range between $r_1$ and $r_2$		For $E<0$ it will be in the range between $r_1$ and $r_2$
	$$l\frac VE\left(-1-\sqrt{1+E/V}\right)\leq r\leq l\frac VE\left(-1+\sqrt{1+E/V}\right)$$		$$l\frac VE\left(-1-\sqrt{1+E/V}\right)\leq r\leq l\frac VE\left(-1+\sqrt{1+E/V}\right)$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	With $E>0$ area of motion $r\geqslant l\frac VE\left(-1+\sqrt{1+E/V}\right)$</p><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 $r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$		if $E<0$ area of motion, $r$ between $r_{1,2}=l\frac VE\left(-1\pm\sqrt{1+E/V}\right)$
	</p>		</p>
	<p style="text-align: right; font-style: italic; font-size: 14;">		<p style="text-align: right; font-style: italic; font-size: 14;">
	Almaskhan Arsen<br>		Almaskhan Arsen<br>
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