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CALC Proton Bombardment. A proton with mass \(1.67 \times 10^{-27} \mathrm{kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) directly toward a uranium nucleus 5.00 \(\mathrm{m}\) 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 \(8.00 \times 10^{-10} \mathrm{m}\) 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 \(\mathrm{m}\) away from the uranium nucleus?

Step by step solution

01

Calculate Initial Kinetic Energy

The initial kinetic energy (KE) of the proton is given by the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the proton and \( v \) is its initial velocity. \[ KE = \frac{1}{2} \times 1.67 \times 10^{-27} \mathrm{kg} \times (3.00 \times 10^{5} \mathrm{m/s})^2 = 7.52 \times 10^{-17} \mathrm{J} \]

02

Calculate Potential Energy at 8.00×10^{-10} m

The potential energy (PE) between the proton and the uranium nucleus at a distance \( x \) is given by \( PE = \frac{\alpha}{x} \). At \( x = 8.00 \times 10^{-10} \mathrm{m} \), \[ PE = \frac{2.12 \times 10^{-26} \mathrm{N \cdot m}^2}{8.00 \times 10^{-10} \mathrm{m}} = 2.65 \times 10^{-17} \mathrm{J} \]

03

Apply Conservation of Energy to Find Speed at 8.00×10^{-10} m

By the conservation of energy, the total energy is constant: initial KE = KE at \( 8.00 \times 10^{-10} \mathrm{m} \) + PE at \( 8.00 \times 10^{-10} \mathrm{m} \). Thus, \( KE_\text{new} = KE_\text{initial} - PE_\text{at 8.00\times10^{-10}} \). \[ KE_\text{new} = 7.52 \times 10^{-17} \mathrm{J} - 2.65 \times 10^{-17} \mathrm{J} = 4.87 \times 10^{-17} \mathrm{J} \] Use \( KE = \frac{1}{2} m v^2 \) to find \( v \): \[ \frac{1}{2} \times 1.67 \times 10^{-27} \mathrm{kg} \times v^2 = 4.87 \times 10^{-17} \mathrm{J} \] Solving for \( v \): \[ v = \sqrt{\frac{2 \times 4.87 \times 10^{-17}}{1.67 \times 10^{-27}}} = 2.42 \times 10^{5} \mathrm{m/s} \]

04

Determine Closest Approach Using Conservation of Energy

The closest approach occurs when all initial kinetic energy has been converted to potential energy. Set initial KE equal to maximum PE at closest distance \( d \): \[ \frac{\alpha}{d} = 7.52 \times 10^{-17} \mathrm{J} \] Solving for \( d \): \[ d = \frac{2.12 \times 10^{-26}}{7.52 \times 10^{-17}} = 2.82 \times 10^{-10} \mathrm{m} \]

05

Calculate Speed when Proton is Again 5.00 m Away

The speed of the proton when it is again 5.00 m away can be calculated using conservation of energy. The initial energy equals the final kinetic energy since the potential energy at 5.00 m can be considered negligible compared to the distances when it was closer. Thus, the speed at 5.00 m away is the same as the initial speed: \[ v = 3.00 \times 10^{5} \mathrm{m/s} \]

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Key Concepts

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

Understanding Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. When you think about a moving car or a child on a swing, you're observing the concepts of kinetic energy in action. In physics, kinetic energy can be calculated using the formula:

• \( KE = \frac{1}{2}mv^2 \)

where \( m \) is the mass of the object, and \( v \) is its velocity. Imagine that you're rolling a ball. The faster and heavier the ball, the more kinetic energy it has.
For a proton in motion, like in our problem, its initial kinetic energy depends on how fast it's traveling and its mass. Initially, the proton has a velocity of \( 3.00 \times 10^{5} \) meters per second, which means it starts with a significant amount of kinetic energy. As the proton moves closer to the uranium nucleus, its speed decreases due to repulsive forces, reducing its kinetic energy.

Understanding Potential Energy

Potential energy is the energy stored in an object because of its position or configuration. Think of a book placed on a shelf or a compressed spring. In our proton problem, potential energy arises from the electrostatic force between the proton and the uranium nucleus.
The potential energy, in this case, is caused by the repulsive force as described by:

• \( PE = \frac{\alpha}{x} \)

Here, \( \alpha \) is a constant, and \( x \) is the distance between the proton and the nucleus. As the proton approaches the nucleus, the distance \( x \) decreases, and potential energy increases.
Potential energy is directly related to how close the proton gets to the nucleus; the closer it is, the more energy gets stored. This stored energy can later be converted back to kinetic energy when the proton is pushed away from the nucleus.

The Conservation of Energy Principle

The conservation of energy principle is fundamental in physics and tells us that energy cannot be created or destroyed, only transformed from one form to another. In our proton bombardment scenario, this principle helps us understand how the proton's kinetic energy gets converted into potential energy and vice versa.
When the proton moves toward the uranium nucleus, its kinetic energy decreases while its potential energy increases. The total energy of the system, kinetic plus potential, remains constant throughout:

• \( KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}} \)

This equation illustrates that when the proton slows down due to the repulsive force, its kinetic energy reduces, but the potential energy increases, keeping total energy balanced.
At its closest point to the nucleus, all kinetic energy is converted into potential energy, momentarily bringing the proton to rest before it reverses direction.