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Chapter 28: Problem 23

Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy \(\left(\frac{1}{2} m v^{2}\right)\) when a particle has a speed of \(\left(\right.\) a) \(1.00 \times 10^{-3} c\) and \((\) b) \(0.970 c.\)

Short Answer

Expert verified
(a) Ratio is approximately 0. (b) Ratio is approximately 6.65.

Step by step solution

01

Understanding Kinetic Energy

Kinetic energy has two forms depending on if the particle's velocity is much less than the speed of light (non-relativistic) or close to the speed of light (relativistic). The non-relativistic kinetic energy is given by \( KE_{nr} = \frac{1}{2} mv^2 \). The relativistic kinetic energy is given by \( KE_r = (\gamma - 1)mc^2 \), where \( \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \).
02

Calculate Relativistic Factor (\(\gamma\)) for (a)

For (a), with speed \( v = 1.00 \times 10^{-3}c \), calculate the relativistic factor \( \gamma \) using \( \gamma = \frac{1}{\sqrt{1-(1.00 \times 10^{-3})^2}} \). This approximation leads to \( \gamma \approx 1 \), as the speed is very small compared to \( c \).
03

Calculate Kinetic Energies for (a)

For speed \( v = 1.00 \times 10^{-3}c \), calculate the kinetic energies: - Non-relativistic: \( KE_{nr} = \frac{1}{2} mv^2 \)- Relativistic: \( KE_r \approx (1 - 1)mc^2 = 0 \). As the velocity is very small, the approximated relativistic kinetic energy simplifies further.
04

Determine the Ratio for (a)

For (a), the ratio of relativistic to non-relativistic kinetic energy is \( \frac{KE_r}{KE_{nr}} \approx \frac{0}{\frac{1}{2} mv^2} \approx 0 \).
05

Calculate Relativistic Factor (\(\gamma\)) for (b)

For (b), with speed \( v = 0.970 c \), calculate the relativistic factor \( \gamma \) using \( \gamma = \frac{1}{\sqrt{1-(0.970)^2}} \). This gives \( \gamma \approx 4.125 \).
06

Calculate Kinetic Energies for (b)

For speed \( v = 0.970 c \), calculate the kinetic energies: - Non-relativistic: \( KE_{nr} = \frac{1}{2} m(0.970c)^2 \)- Relativistic: \( KE_r = (4.125 - 1)mc^2 \).
07

Determine the Ratio for (b)

For (b), the ratio of the relativistic kinetic energy to the non-relativistic one is \( \frac{(4.125-1)mc^2}{0.47005 \times mc^2} \approx 6.65 \). The mass \( m \) and \( c \) cancel out.

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Key Concepts

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

Non-relativistic Kinetic Energy
Non-relativistic kinetic energy is what we use when objects move at speeds much slower than the speed of light. This energy is given by the formula \( KE_{nr} = \frac{1}{2} mv^2 \), where \( m \) represents mass, and \( v \) is velocity. It is a straightforward calculation for everyday speeds, such as vehicles moving on the highway or a basketball being thrown. In the non-relativistic world, this energy grows quadratically as speed increases. This means, if you double the speed, the kinetic energy quadruples. However, this only holds true when the velocities in question are much less than the speed of light. Once speeds approach light-speed, we have to deal with relativity, which significantly changes the picture.
Relativistic Factor Gamma
The relativistic factor, often symbolized as \( \gamma \), is a crucial concept in special relativity. It helps us understand how the physics of an object changes as its speed increases and approaches the speed of light, denoted by \( c \). This factor is calculated with the formula \( \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \), where \( v \) is the velocity of the object.
  • When \( v \) is very small compared to \( c \), \( \gamma \) is close to 1, meaning relativistic effects are negligible.
  • As \( v \) approaches \( c \), \( \gamma \) increases, impacting mass, time, and energy significantly.
The increase in \( \gamma \) means that even small additions to speed demand much more energy, a key aspect when considering relativistic kinetic energy. This factor also demonstrates why nothing with mass can reach the speed of light, as \( \gamma \) would become infinite, requiring infinite energy.
Speed of Light in Physics
The speed of light, represented as \( c \), is a fundamental constant in physics, approximately equal to \( 3 \times 10^8 \) meters per second. It serves as the cosmic speed limit, dictating that nothing can travel faster. In the context of relativity:
  • It provides a universal speed limit. No information or matter can exceed this speed.
  • The speed of light is essential in the formulation of relativity equations and affects how we perceive time and space.
Understanding the speed of light is crucial because it marks the boundary where classical physics gives way to relativistic physics. This transition changes how energy and time work at high velocities and is vital for accurate calculations in high-speed situations, such as those encountered in particle physics and astrophysics.

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Most popular questions from this chapter

ssm An unstable particle is at rest and suddenly breaks up into two fragments. No external forces act on the particle or its fragments. One of the fragments has a velocity of \(+0.800 c\) and a mass of \(1.67 \times 10^{-27} \mathrm{~kg},\) and the other has a mass of \(5.01 \times 10^{-27} \mathrm{~kg}\). What is the velocity of the more massive fragment? (Hint: This problem is similar to Example 6 in Chapter \(7 .\) )

An unstable particle is at rest and suddenly breaks up into two fragments. No external forces act on the particle or its fragments. One of the fragments has a velocity of \(+0.800 c\) and a mass of \(1.67 \times 10^{-27} \mathrm{~kg}\), and the other has a mass of \(5.01 \times 10^{-27} \mathrm{~kg}\). What is the velocity of the more massive fragment? (Hint: This problem is similar to Example 6 in Chapter 7.)

Multiple-Concept Example 6 explores the approach taken in problems such as this one. Quasars are believed to be the nuclei of galaxies in the early stages of their formation. Suppose a quasar radiates electromagnetic energy at the rate of \(1.0 \times 10^{41} \mathrm{~W}\). At what rate (in \(\mathrm{kg} / \mathrm{s}\) ) is the quasar losing mass as a result of this radiation?

Review Conceptual Example 11 as an aid in answering this question. A person is approaching you in a truck that is traveling very close to the speed of light. This person throws a baseball toward you. Relative to the truck, the ball is thrown with a speed nearly equal to the speed of light, so the person on the truck sees the baseball move away from the truck at a very high speed. Yet you see the baseball move away from the truck very slowly. Why? Use the velocity- addition formula to guide your thinking.

Two atomic particles approach each other in a head-on collision. Each particle has a mass of \(2.16 \times 10^{-25} \mathrm{~kg} .\) The speed of each particle is \(2.10 \times 10^{8} \mathrm{~m} / \mathrm{s}\) when measured by an observer standing in the laboratory. (a) What is the speed of one particle as seen by the other particle? (b) Determine the relativistic momentum of one particle, as it would be observed by the other.
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