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A proton has momentum with magnitude \(p_{0}\) when its speed is \(0.400 c .\) In terms of \(p_{0},\) what is the magnitude of the proton's momentum when its speed is doubled to \(0.800 c ?\)

Short Answer

Expert verified
The magnitude of the proton's momentum when its speed is doubled to \(0.800 c\) is a bit more than two times its initial momentum \(p_{0}\).

Step by step solution

01

Expression for Initial Momentum

Express the initial momentum of the proton using the formula for relativistic momentum. Let's denote the mass of the proton as \(m\). We have: \(p_{0} = \frac{mv_{0}}{\sqrt{1-v_{0}^{2}/c^{2}}}\), where \(v_{0} = 0.400c\).
02

Expression for Final Momentum

Next, express the momentum of the proton when its speed is doubled using the same formula. We have: \(p_{f} = \frac{m\cdot 2v_{0}}{\sqrt{1-(2v_{0})^{2}/c^{2}}}\), where \(2v_{0} = 0.800c\).
03

Solve for the Proton's Final Momentum in Terms of Initial Momentum

Now, the task is to express \(p_{f}\) in terms of \(p_{0}\). Take the ratio \(p_{f}/p_{0}\). With some algebra, this simplifies to: \(p_{f}/p_{0} = \frac{\frac{2v_{0}}{\sqrt{1-(2v_{0})^{2}/c^{2}}}}{\frac{v_{0}}{\sqrt{1-v_{0}^{2}/c^{2}}}}\). The denominator and numerator of the right side will cancel out some terms, and the expression simplifies to \(p_{f}/p_{0} = \frac{2}{\sqrt{(1+ v_{0}^{2}/c^{2})}}\). Given that \(v_{0} = 0.400c\), substituting and simplifying further leads to a final ratio of little over 2.

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Key Concepts

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

Proton Dynamics
Understanding the behavior of protons at high speeds is pivotal in advanced physics. Protons, being subatomic particles, have a large mass compared to their size. When a proton moves at a significant fraction of the speed of light, its dynamics need a more complex description than when it's moving slow. At these high speeds, classical mechanics isn't enough. Instead, we use concepts from relativity to accurately describe their motion.

Protons exhibit relativistic properties when they approach speeds like 0.400c or 0.800c, where "c" is the speed of light. These speeds are so high that the usual equations of momentum and dynamics in classical mechanics fail to apply accurately. For understanding proton dynamics, familiarity with the governing principles of special relativity becomes essential.

Real-world applications include particle accelerators, where protons are accelerated to near-light speeds, aiding in fields like particle physics and medical therapies by analyzing proton dynamics.
Special Relativity
Special relativity changes the way we think about speed, time, and momentum, especially as speeds approach the speed of light. Developed by Albert Einstein, it introduces corrections to the classical equations of motion when dealing with high velocities.

A key idea is the concept of time dilation and length contraction, which states that as an object moves faster, time slows down for it relative to a stationary observer, and lengths contract in the direction of motion. This is critical when calculating the momentum of fast-moving particles like protons.

The relativistic momentum is expressed as \(p = \frac{mv}{\sqrt{1-v^{2}/c^{2}}}\), accounting for the increase in mass and decrease in time intervals at high speeds. This ensures that the momentum of an object never surpasses infinite and stays consistent within the laws of physics. As speed increases, the momentum increases non-linearly due to the relativistic effects. This reflects the conservation of energy and momentum in a relativistic framework.
Momentum in Physics
Momentum is a foundational concept in physics, representing the quantity of motion an object possesses. It combines both mass and velocity to determine how hard an object is moving. In classical physics, it’s simply \(p = mv\).

In the realm of relativistic speeds, momentum requires modification to account for near-light velocities. The formula changes to \(p = \frac{mv}{\sqrt{1-v^{2}/c^{2}}}\) in special relativity, recognizing that as speed approaches the speed of light, momentum becomes much higher than predicted by classical physics.

This effect has profound implications for objects moving at relativistic speeds, such as protons in accelerators. The relativistic formula ensures that the momentum is always finite and provides an accurate description of high-speed dynamics.

Understanding momentum helps in developments like calculating forces in collisions, propelling spacecrafts, and enhancing our knowledge of cosmic particles. Its importance stretches across many scientific and engineering disciplines, highlighting its relevance in both theoretical and practical applications.

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Most popular questions from this chapter

The distance to a particular star, as measured in the earth's frame of reference, is 7.11 light-years ( 1 light-year is the distance that light travels in \(1 \mathrm{y}\) ). A spaceship leaves the earth and takes \(3.35 \mathrm{y}\) to arrive at the star, as measured by passengers on the ship. (a) How long does the trip take, according to observers on earth? (b) What distance for the trip do passengers on the spacecraft measure?

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A particle has rest mass \(6.64 \times 10^{-27} \mathrm{~kg}\) and momentum \(2.10 \times 10^{-18} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} .\) (a) What is the total energy (kinetic plus rest energy) of the particle? (b) What is the kinetic energy of the particle? (c) What is the ratio of the kinetic energy to the rest energy of the particle?

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