Problem 72 - University Physics with Modern Physics

Question

A proton with mass 1.67 \(\times\) 10\(^{-27}\) kg is propelled at an initial speed of 3.00 \(\times\) 10\(^5\) m/s directly toward a uranium nucleus 5.00 m away. The proton is repelled by the uranium nucleus with a force of magnitude \(F = \alpha/x^2\), where \(x\) is the separation between the two objects and \(\alpha = 2.12 \times 10^{-26} \, \mathrm{N} \cdot \mathrm{m}^2\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10}\) m from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again 5.00 m away from the uranium nucleus?

Step-by-Step Solution

Answer

(a) Calculate \(v_f\) at \(8.00 \times 10^{-10}\) m, (b) closest distance \(x_{min}\), (c) speed is initial speed \(3.00 \times 10^5\) m/s at 5.00 m.

Step 1: Understanding the Problem

We are given that a proton is moving towards a uranium nucleus with an initial speed of \(3.00 \times 10^5\) m/s. A force repels the proton as it gets closer to the nucleus, and we need to solve for the proton's speed at a specific distance, determine the closest distance the proton gets to the nucleus, and its speed when it returns to the original starting distance.

Step 2: Conservation of Energy Principle

The principle of conservation of energy tells us that the initial kinetic energy plus the initial potential energy equals the final kinetic energy plus the final potential energy. This is expressed as: \[ \frac{1}{2} m v_i^2 + U_i = \frac{1}{2} m v_f^2 + U_f \]where \(v_i\) is the initial speed, \(v_f\) is the final speed, and \(U_i\) and \(U_f\) are the initial and final potential energies respectively, given by \(U = - \frac{\alpha}{x}\).

Step 3: Applying to Part (a)

For part (a), the proton's initial kinetic energy is \( \frac{1}{2} (1.67 \times 10^{-27}) (3.00 \times 10^5)^2 \) J. The initial potential energy at 5.00 m is \( - \frac{2.12 \times 10^{-26}}{5.00} \). The final potential energy at \(8.00 \times 10^{-10}\) m is \( - \frac{2.12 \times 10^{-26}}{8.00 \times 10^{-10}} \). Solve for \(v_f\):\[ \frac{1}{2} m v_f^2 = \frac{1}{2} m v_i^2 + \frac{\alpha}{8.00 \times 10^{-10}} - \frac{\alpha}{5.00} \]

Step 4: Calculating Part (a)

Substitute values into the conservation of energy equation:- \(\frac{1}{2} (1.67 \times 10^{-27}) (v_f^2) = \frac{1}{2} (1.67 \times 10^{-27}) (3.00 \times 10^5)^2 + \frac{2.12 \times 10^{-26}}{8.00 \times 10^{-10}} - \frac{2.12 \times 10^{-26}}{5.00} \)Solve for \(v_f\).

Step 5: Determine Closest Approach (Part b)

The closest distance \(x_{min}\) occurs when the kinetic energy is zero. Thus, solve for zero kinetic energy where:\[ \frac{1}{2} m v_i^2 = \frac{\alpha}{x_{min}} - \frac{\alpha}{5.00} \]Solve for \(x_{min}\):\[ x_{min} = \frac{\alpha}{\frac{1}{2} m v_i^2 + \frac{\alpha}{5.00}} \]

Step 6: Calculating final speed (Part c)

When the proton is 5.00 m away again, the initial and final potential energies are the same, thus:\[ \frac{1}{2} m v_i^2 = \frac{1}{2} m v_{final}^2 \]Hence, \(v_{final}\) must equal \(v_i\).

Step 7: Solve and Conclude

By solving each equation:- (a) Compute \(v_f\) using the formula from Step 4.- (b) Calculate \(x_{min}\) using the formula from Step 5.- (c) Recognize \(v_{final} = v_i = 3.00 \times 10^5\) m/s from Step 6.

Key Concepts

Coulomb's Law

Coulomb's law is fundamental in understanding the forces between charged particles. It describes the interaction force between two charges, stating that the force is directly proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them. The equation is given as:\[ F = k \frac{q_1 q_2}{r^2} \]where:

• \(F\) is the force between the charges,
• \(k\) is Coulomb’s constant,
• \(q_1\) and \(q_2\) are the amounts of the two charges,
• \(r\) is the distance between the centers of the two charges.

In the context of a proton being repelled by a uranium nucleus, the equation of force changes slightly, as given in the problem statement \( F = \alpha/x^2 \). Here, \(\alpha\) is a proportionality constant that reflects the charge interaction unique to the proton-uranium system. Understanding this force is crucial, as it determines how the proton slows down and changes direction when it approaches the uranium nucleus. This interaction shapes the potential energy landscape as well, altering the proton's journey.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. The formula for kinetic energy (\( KE \)), particularly relevant for particles such as protons, is:\[ KE = \frac{1}{2} mv^2 \]where:

• \(m\) is the mass of the object,
• \(v\) is the velocity of the object.

In our scenario, the proton starts off with an initial speed of \(3.00 \times 10^5\) m/s, which gives it a certain amount of kinetic energy. As it approaches the uranium nucleus, the kinetic energy changes due to the repelling electrostatic force described by Coulomb's law.
This repulsive force acts opposite to the motion of the proton, decreasing its kinetic energy as potential energy increases. At the closest point to the nucleus, all kinetic energy is temporarily converted to potential energy, meaning the proton's speed becomes zero before it reaccelerates in the opposite direction.

Potential Energy

Potential energy in physics refers to the stored energy of position possessed by an object. For charged particles interacting through an electrostatic force, potential energy (\( U \)) can be expressed as:\[ U = -\frac{\alpha}{x} \]In the context of this problem:

• \( \alpha \) is a constant characteristic of the charge interaction,
• \( x \) is the distance between the proton and the uranium nucleus.

As the proton moves closer to the uranium nucleus, it experiences changes in potential energy. Initially, at a distance of 5.00 m, the potential energy is relatively low. However, as it approaches within \(8.00 \times 10^{-10}\) m, the repulsion increases significantly, and so does the potential energy.
The principle of conservation of energy is crucial here, as it states that the total energy (kinetic plus potential) remains constant if only conservative forces are acting. Thus, any loss of kinetic energy as the proton slows due to the force is compensated by an equivalent gain in potential energy.

Proton-Uranium Interaction

The interaction between a proton and a uranium nucleus is guided by electrostatic forces, primarily the repulsive force due to their like charges. The characteristics of this interaction can be examined through energy principles and force laws discussed earlier.
The path a proton takes upon approaching a fixed uranium nucleus highlights key physical concepts:

• Initially, the proton has a high speed and high kinetic energy.
• As it moves closer, it experiences a growing repulsive force, as modeled by the specific force equation \( F = \alpha/x^2 \).
• This causes the proton to decelerate, and its kinetic energy is transformed into potential energy.
• Upon reaching a minimum distance, it comes momentarily to rest before being repelled.

The measurement of this closest approach, or how near the proton gets to the nucleus before turning around, is determined when its kinetic energy reaches zero, signifying maximum conversion to potential energy. Eventually, the proton travels back and reaches the original distance with its speed restored, an illustration of energy conservation in its full circle.