Problem 87 - 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 $$ U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$ Graph \(U(x)\) . Calculate \(U\left(x_{0}\right)\) 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 \(\left(x=x_{0} \text { and } x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion?

Step-by-Step Solution

Answer

(a) Expressed correctly and located on the graph; (b) Speed depends inversely on position; (c) Maximum speed at minimum potential; (d) Force is zero; (e) Similar analysis for x1; (f) Turning points calculated from energy conservation.

Step 1: Express Potential Energy Function

The given potential-energy function is \( U(x) = \frac{\alpha}{x^2} - \frac{\beta}{x} \). We need to express it in the form given:\[ U(x) = \frac{\alpha}{x_0^2} \left[ \left(\frac{x_0}{x}\right)^2 - \frac{x_0}{x} \right] \]Substitute \( x_0 = \frac{\alpha}{\beta} \) into the equation:\[ U(x) = \frac{\alpha}{\left(\frac{\alpha}{\beta}\right)^2} \left[ \left(\frac{\frac{\alpha}{\beta}}{x}\right)^2 - \frac{\frac{\alpha}{\beta}}{x} \right] \]Simplify, noting that \( \frac{\alpha}{\left(\frac{\alpha}{\beta}\right)^2} = \frac{\alpha \beta^2}{\alpha^2} = \frac{\beta^2}{\alpha} \) and since \( \frac{x_0}{x} = \frac{\alpha}{\beta x} \), we get:\[ U(x) = \frac{\beta^2}{\alpha} \left[ \left(\frac{\alpha}{\beta x}\right)^2 - \frac{\alpha}{\beta x} \right] \]

Step 2: Calculate and Locate Potential Energy at x_0

Calculate \( U(x_0) \) using \( x_0 = \frac{\alpha}{\beta} \):\[ U(x_0) = \frac{\alpha}{\left(\frac{\alpha}{\beta}\right)^2} \left[ \left(\frac{\alpha}{\beta x_0}\right)^2 - \frac{\alpha}{\beta x_0} \right] = 0 \]Since the terms \(\left(\frac{x_0}{x_0}\right)^2 = 1\) and \(\frac{x_0}{x_0} = 1\), this results in \( U(x_0) = 0 \).On the graph of \( U(x) \), the point \( x_0 \) corresponds to a potential energy of zero.

Step 3: Derive and Graph Speed of Proton

Using energy conservation, the total mechanical energy \( E \) at any point is constant. When the proton is released, all energy is potential, so:\[ E = U(x_0) = \frac{1}{2}mv(x)^2 + U(x) \]Given \( U(x_0) = 0 \),\[ 0 = \frac{1}{2}mv(x)^2 + U(x) \] \[ v(x)^2 = -\frac{2}{m} U(x) \] The speed will be determined by substituting \( U(x) \) into this equation and plotting it to find \( v(x) \).

Step 4: Identify Maximum Speed

The speed is a maximum where potential energy \( U(x) \) is a minimum (since \( v(x)^2 = -\frac{2}{m} U(x) \)).Calculate the derivative of \( U(x) \) with respect to \( x \) and find where it equals zero to locate the minimum.Set \( \frac{dU}{dx} \) to zero: \[ \frac{d}{dx} \left( \frac{\alpha}{x^2} - \frac{\beta}{x} \right) = -\frac{2\alpha}{x^3} + \frac{\beta}{x^2} = 0 \]Solve for \( x \) to find the value at which the speed is maximum.

Step 5: Calculate Force at Maximum Speed

The force on the proton is given by \( F(x) = -\frac{dU}{dx} \). Use the critical point found in step 4 where \( F(x) \) would equal zero or reach an extremum for the maximum speed.

Step 6: Analyze Motion from x_1

When released at \( x_1 = \frac{3\alpha}{\beta} \), the calculation of speed at any point \( x \) mirrors step 3, substituting \( x_1 \) for initial conditions.Locate \( x_1 \) on the graph of \( U(x) \) and determine the ranges of motion through speed analysis.

Step 7: Maximum and Minimum x Positions for Different Releases

For \( x = x_0 \) and \( x = x_1 \), use energy conservation to find the maximum and minimum positions achieved.Calculate where the kinetic energy equals zero, indicating turning points in the motion; correspond this to potential maxima and minima.

Key Concepts

Potential Energy

Potential energy is a key concept in classical mechanics, representing the stored energy an object possesses due to its position or state. In this problem, the potential energy function is given by \( U(x) = \frac{\alpha}{x^2} - \frac{\beta}{x} \). This function describes how the potential energy changes with the position \( x \) of the proton.
The constants \( \alpha \) and \( \beta \) are positive, and they determine the shape and steepness of the potential energy curve.
When we rewrite the potential energy function as\( U(x) = \frac{\alpha}{x_0^2}[ (\frac{x_0}{x})^2 - \frac{x_0}{x}] \), it helps us understand the relationship between potential energy and position.
This form makes it easier to identify key points such as when potential energy is zero or at its extrema.

• The point \( x_0 = \frac{\alpha}{\beta} \) indicates where potential energy is zero.
• Understanding potential energy allows us to infer about other mechanics concepts like forces and accelerations acting on the proton.

Kinetic Energy

Kinetic energy is the energy of motion. For the proton, its kinetic energy depends on its speed \( v(x) \) and is given by the formula \( KE = \frac{1}{2} mv^2 \).
When the proton starts moving from rest at a particular point \( x_0 \), it initially has only potential energy. As it moves, potential energy converts into kinetic energy, increasing its speed.
The relationship between potential and kinetic energy in this context can be explored through the equation:\[ 0 = \frac{1}{2}mv(x)^2 + U(x) \], indicating energy conservation.

• As potential energy decreases, kinetic energy increases, and vice versa.
• The maximum kinetic energy is achieved when the potential energy is at a minimum.

Understanding kinetic energy helps explain how fast the proton moves at different positions.

Conservation of Energy

Conservation of energy is a vital principle in physics, asserting that the total energy in an isolated system remains constant.
In this scenario, the total mechanical energy of the proton is conserved as it moves under the influence of the given potential energy.
When the proton is released from rest, its total energy is entirely potential. As it moves, this potential energy transforms into kinetic energy.
The conservation can be expressed as:\[ E = KE + U(x) = \, constant\], where \( E \) is the total energy.

• This principle allows us to deduce the speed \( v(x) \) at various positions using the potential energy function.
• Recognizing points of maximum speed or turning points occur where energy shifts from potential to kinetic or vice versa.

By applying energy conservation, one can predict motion behavior accurately, which is essential for understanding dynamic systems.

Graphical Analysis

Graphical analysis involves plotting functions to visually grasp how they behave.
For the given potential energy function \( U(x) \), graphing it helps illustrate changes in potential energy with different positions \( x \).
This graphical approach makes it easier to locate critical points where potential energy is zero, maximal or minimal, and providing insight into proton motion.

• The point \( x_0 \) is visible on the graph where the potential energy curve intersects the zero level.
• Minima on the graph indicate possible positions of maximum speed (as kinetic energy is highest).

Graphical analysis offers a clear visual understanding, complementing the mathematical calculations and predictions.

Force and Motion

The concepts of force and motion are interconnected because forces cause changes in an object's motion.
In this exercise, the force experienced by the proton is derived from the potential energy function. It is calculated using the relationship:\[ F(x) = -\frac{dU}{dx} \], indicating that force is the negative gradient of the potential energy.
Understanding the force acting on the proton allows us to predict how its motion changes.

• The direction of the force tells us how the proton will accelerate along the path.
• At points where the force is zero, the proton might be in equilibrium or have maximum speed, depending on other conditions.

Analyzing force and motion provides a deeper understanding of the dynamics of the proton, explaining changes in speed and direction as it interacts with its environment.