Problem 80 - University Physics with Modern Physics

Question

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x) = (\alpha/x^2) - (\beta/x)\), where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_0\) = \(\alpha\)/\(\beta\). (a) Show that \(U(x)\) can be written as $$U(x) = \frac{\alpha}{x^2_0}\ \Big[ \Big( \frac{x_0}{x}\ \Big)^2 - \frac{x_0}{x}\ \Big] $$ Graph \(U(x)\). Calculate \(U(x_0)\) and thereby locate the point \(x_0\) on the graph. (b) Calculate \(v(x)\), the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_1 = 3\alpha / \beta\). Locate the point \(x_1\) on the graph of \(U(x)\). Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point (\(x = x_0\) and \(x = x_1\)), what are the maximum and minimum values of \(x\) reached during the motion?

Step-by-Step Solution

Answer

Graph U(x), calculate points, and use energy conservation to find velocity and constraints.

Step 1: Expressing U(x)

Given the potential energy function \( U(x) = \frac{\alpha}{x^2} - \frac{\beta}{x} \), we need to show that it can be rewritten as \( U(x) = \frac{\alpha}{x_0^2}\left(\left(\frac{x_0}{x}\right)^2 - \frac{x_0}{x}\right) \). Substituting \( x_0 = \frac{\alpha}{\beta} \) gives \( \frac{\alpha}{x_0^2} = \frac{\beta^2}{\alpha} \). Thus, \( U(x) = \frac{\beta^2}{\alpha}\left(\frac{\alpha^2}{\beta^2}\frac{1}{x^2} - \frac{\alpha}{\beta}\frac{1}{x}\right) = \frac{\alpha}{x_0^2}\left(\frac{x_0^2}{x^2} - \frac{x_0}{x}\right) \).

Step 2: Calculating U(x_0)

To find \( U(x_0) \), substitute \( x = x_0 \) into \( U(x) \):\[ U(x_0) = \frac{\alpha}{x_0^2}\left(\left(\frac{x_0}{x_0}\right)^2 - \frac{x_0}{x_0}\right) = \frac{\alpha}{x_0^2}(1 - 1) = 0. \] Therefore, the point \( x_0 \) on the graph is where \( U(x) = 0 \).

Step 3: Graphing U(x)

Plot \( U(x) \) versus \( x \). The function decreases as \( x \) increases and passes through zero at \( x = x_0 \).

Step 4: Calculating v(x)

Apply conservation of energy: total energy \( E = U(x) + \frac{1}{2}mv^2(x) = 0 \). Hence, \[ \frac{1}{2}mv^2(x) = -U(x). \] Thus, \[ v(x) = \sqrt{\frac{2(-U(x))}{m}} = \sqrt{\frac{2}{m}\left(\frac{\alpha}{x^2} - \frac{\beta}{x}\right)}. \]

Step 5: Graphing v(x)

Use the expression obtained to graph \( v(x) \). The speed of the proton decreases gradually as \( x \) increases.

Step 6: Determining Maximum Speed

The speed is maximized when \( U(x) \) is minimized because \( v(x) = \sqrt{\frac{2(-U(x))}{m}} \). Set the derivative \( \frac{d}{dx}\left( \frac{\alpha}{x^2} - \frac{\beta}{x}\right) = 0 \) to find the critical points. Solve it to find the value of \( x \) where \( v(x) \) is maximum.

Step 7: Calculating Force at Maximum Speed

The force \( F \) is the negative gradient of the potential energy: \[ F(x) = -\frac{dU}{dx}. \] Compute this at the critical point determined in the previous step.

Step 8: Calculating Motion from x1

Set \( x_1 = \frac{3\alpha}{\beta} \) and repeat steps for \( U(x_1) \) and \( v(x_1) \). Using the same energy conservation method will provide insights into motion from \( x_1 \).

Step 9: Locating Maximum and Minimum x

For each release point, determine the constraints set by the initial total energy, ensuring \( U(x) \subseteq E \). Solve \( U(x) = E \) to get the turning points.

Key Concepts

Potential Energy Function

The potential energy function is a mathematical representation of how potential energy varies with position. It reflects the stored energy in an object based on its configuration or position in a force field.
In this exercise, we have the proton's potential energy as a function of its position, given by:\[ U(x) = \frac{\alpha}{x^2} - \frac{\beta}{x} \]where \( \alpha \) and \( \beta \) are positive constants. This equation shows us how potential energy changes based on the proton's position along a one-dimensional path.
The term \( \frac{\alpha}{x^2} \) indicates an inverse square relationship, which can be typical in certain physical situations, like electrostatics.
Meanwhile, \( \frac{\beta}{x} \) represents an inverse linear relationship, suggesting a different kind of influence on the potential energy.
By substituting a specific point like \( x_0 = \frac{\alpha}{\beta} \), we can simplify this function for analysis. Through substitution, one can express the function in terms of simpler components, making it easier to graph and understand the behavior at specific points.
This function is crucial for predicting the motion of the proton by calculating its potential energy at various points.

Conservation of Energy

The conservation of energy is a fundamental principle in physics stating that energy cannot be created or destroyed; it can only be transferred or converted from one form to another.
In our scenario, the total mechanical energy of the proton is conserved. Initially, when the proton is at rest, all the energy is stored as potential energy. As the proton begins to move, this potential energy is converted into kinetic energy.
At any point \(x\), the sum of the kinetic energy and potential energy remains constant:\[ E = U(x) + \frac{1}{2}mv^2(x) \]Here, \(E\) is the total energy, \(U(x)\) is the potential energy, and \( \frac{1}{2}mv^2(x) \) represents the kinetic energy of the proton with mass \(m\) and speed \(v(x)\).
Since the proton starts from rest, we can assume that initially, all energy is potential (U(x_0)\ = E).As the proton rolls down the potential energy graph, potential energy decreases and kinetic energy increases, maintaining the conservation equation.
By understanding conservation, we can express the proton's speed at any point \(x\) based on potential energy:\[ v(x) = \sqrt{\frac{2(-U(x))}{m}} \]The conservation principle provides a powerful method to evaluate the particle’s motion without directly solving complex equations of motion.

Maximum and Minimum Values

In the context of energy and motion, finding maximum and minimum values helps to predict where certain physical conditions, like maximum speed or turning points, occur.
For potential energy functions, crucial points occur where the derivative \( \frac{dU}{dx} = 0 \). These points, where the slope of the potential energy curve is zero, indicate local maxima or minima.
The maximum speed of the proton corresponds to where the potential energy \( U(x) \) is minimized. This results from the conservation of energy: less potential energy means more energy available for kinetic energy, hence higher speed.
By solving the derivative equation \( \frac{d}{dx}\left(\frac{\alpha}{x^2} - \frac{\beta}{x}\right) = 0 \), we find the positions where maximum and minimum values occur. This gives us critical insights into the motion of the proton.
Additionally, at these extremal points, calculating the force is essential:\[ F(x) = -\frac{dU}{dx} \]This computation aids in understanding how the potential energy landscape influences the proton’s motion, determining whether it accelerates or decelerates as it moves through these critical points.