91Ó°ÊÓ

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\left(\alpha / x^{2}\right)-(\beta / x),\) 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 $$ 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 \(\left(x=x_{0}\right.\) and \(\left.x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion? U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$

Step by step solution

01

Rewrite potential energy function

First we need to rewrite the potential energy function \(U(x)\) as a fraction of \(\frac{\alpha}{x_{0}^{2}}\), as it is required in the exercise. The equation results in \(U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right]\).

02

Calculate the force on the proton

The next step is to calculate the force on the proton. We use the fact that the force \(F\) on an object due to potential energy is the negative derivative of the potential, that is, \(F = -\frac{dU(x)}{dx}\). Differentiating \(U(x)\) and simplifying, we find \(F = -\frac{1}{m} \frac{2\alpha}{x^3} + \frac{1}{m} \frac{\beta}{x^2}\).

03

Apply conservation of mechanical energy

Before releasing the proton, the total mechanical energy is just potential energy, \(E_{0} = U(x_{0}) = \frac{1}{2} \frac{\beta^2}{\alpha}\). Because the system does not lose energy, the total energy when the proton is at arbitrary position \(x\) will still be the same, so \(E = U(x) + \frac{1}{2} m V(x)^2\). Given that the total energy is constant, we can solve for \(V(x)\): \(V(x) = \sqrt{\frac{2}{m}[E_0 - U(x)]}\).

04

Determine maximum speed and related position

The speed \(V(x)\) is maximum when the potential energy \(U(x)\) is at its minimum. We can find the position of minimum \(U(x)\) by taking the derivative \(\frac{dU(x)}{dx}\) and solving for \(x\). This gives \(x = \frac{\beta}{\alpha}\). The maximum speed then is obtained by substituting the value of \(x\) into \(V(x)\).

05

Evaluate the force at maximum speed

The force on the proton at the point of maximum speed can be obtained by substituting the value of \(x\) into the force equation. That is, \(F = -\frac{1}{m} \frac{2\alpha}{x^3} + \frac{1}{m} \frac{\beta}{x^2}\).

06

Apply the principle of conservation of energy for second release point

If the proton is released at a point \(x_{1}\) instead of \(x_{0}\), follow the same procedure as outlined above. This would give the new \(U(x)\) as a function of \(x_{1}\) and speed function \(V(x)\) as a function of \(x_{1}\). Substituting \(x = x_{1}\) into these functions will give the maximum and minimum values of \(x\) during the motion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. For a proton moving at speed \(v(x)\), kinetic energy \(K\) is given by the formula:\[ K = \frac{1}{2} m v(x)^2 \]This equation tells us that kinetic energy is directly proportional to the mass \(m\) of the object and the square of its velocity. When the speed of the proton changes, its kinetic energy changes accordingly. This is important to understand in the context of the question, because the kinetic energy depends on the position \(x\), through its relation with \(v(x)\).

• The faster the proton moves, the higher its kinetic energy.
• The slower it moves, the lower its kinetic energy.

In our exercise, the kinetic energy of the proton depends on its speed which varies with position due to the forces acting on it. For this reason, it is crucial to calculate \(v(x)\) to understand the proton's dynamics as it moves through different positions.

Mechanical Energy Conservation

The principle of mechanical energy conservation states that the total mechanical energy of a system remains constant if only conservative forces (such as gravitational or elastic forces) are acting. Mechanical energy is the sum of kinetic energy and potential energy. For the proton in our example, the conservation can be expressed as:\[ E = U(x) + \frac{1}{2} m v(x)^2 \]where \(E\) is the total mechanical energy, \(U(x)\) is the potential energy at position \(x\), and \(v(x)\) is the velocity of the proton at that position.

• When the proton is released and moves, its potential energy \(U(x)\) will convert into kinetic energy \(\frac{1}{2} m v(x)^2 \), and vice versa, maintaining a constant \(E\).
• This interchange allows us to solve for the speed \(v(x)\) in terms of the potential energy difference: \[ v(x) = \sqrt{\frac{2}{m}[E_0 - U(x)]} \]

In this expression, \(E_0\) is the initial total energy which is equal to the potential energy at the starting point, \(x_0\). Understanding this concept helps to predict how the proton accelerates or decelerates.

Force Calculation

Force is a vector quantity that causes an object to accelerate. In the context of potential energy, the force acting on an object is obtained by taking the negative derivative of the potential energy function with respect to position. For our proton, this is expressed as:\[ F = -\frac{dU(x)}{dx} \]In the previous exercise, we calculated the force as:\[ F = -\frac{1}{m} \frac{2\alpha}{x^3} + \frac{1}{m} \frac{\beta}{x^2} \]This force describes how the proton is influenced as it moves along the potential energy landscape. Force can tell us important information about motion and how the proton interacts with the environment.

• A positive force can increase the speed of the proton, while a negative force can decrease it.
• The derivative of force with respect to potential energy shows us that it's related to changes in the potential landscape, influencing the proton's path.

By understanding and calculating these forces, one can predict how and where the proton will move, or when it achieves maximum velocity.