/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 What is the speed of a particle ... [FREE SOLUTION] | 91影视

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Chapter 37: Problem 31

What is the speed of a particle whose kinetic energy is equal to (a) its rest energy and (b) five times its rest energy?

Short Answer

Expert verified
(a) \( v = \frac{\sqrt{3}}{2}c \); (b) \( v = \frac{\sqrt{35}}{6}c \).

Step by step solution

01

Understand the Rest Energy

The rest energy of a particle is given by the equation \( E_0 = mc^2 \), where \( m \) is the rest mass of the particle and \( c \) is the speed of light. This represents the energy the particle has when it is at rest.
02

Recall the Total Energy

The total energy \( E \) of a particle is given by \( E = \gamma mc^2 \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor, and \( v \) is the speed of the particle.
03

Relate Kinetic Energy with Rest Energy (Part a)

For part (a), the kinetic energy \( KE \) is equal to the rest energy \( E_0 \), so \( KE = mc^2 \). The kinetic energy can also be expressed as \( KE = E - E_0 = \gamma mc^2 - mc^2 \). Setting these equal gives \( mc^2 = \gamma mc^2 - mc^2 \), so \( \gamma mc^2 = 2mc^2 \). Thus, \( \gamma = 2 \).
04

Solve for Speed (Part a)

From \( \gamma = 2 \), we have \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2 \). Solving for \( v \) gives:1. Square both sides to get \( 1 - \frac{v^2}{c^2} = \frac{1}{4} \).2. Rearrange to find \( \frac{v^2}{c^2} = 1 - \frac{1}{4} = \frac{3}{4} \).3. Take the square root to find \( v = \frac{\sqrt{3}}{2}c \).
05

Relate Kinetic Energy with Rest Energy (Part b)

For part (b), the kinetic energy is five times the rest energy, so \( KE = 5mc^2 \). This gives \( \gamma mc^2 - mc^2 = 5mc^2 \), or \( \gamma mc^2 = 6mc^2 \). Therefore, \( \gamma = 6 \).
06

Solve for Speed (Part b)

From \( \gamma = 6 \), we have \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 6 \). Solving for \( v \) gives:1. Square both sides to get \( 1 - \frac{v^2}{c^2} = \frac{1}{36} \).2. Rearrange to find \( \frac{v^2}{c^2} = 1 - \frac{1}{36} = \frac{35}{36} \).3. Take the square root to find \( v = \frac{\sqrt{35}}{6}c \).

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Key Concepts

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

Rest Energy
Rest energy is a fundamental concept in physics introduced by Albert Einstein. It is given by the famous equation \( E_0 = mc^2 \). Here, \( E_0 \) is the rest energy, \( m \) is the mass of the particle at rest, and \( c \) is the speed of light in a vacuum. This equation tells us that mass is a form of energy.
  • Rest energy is the energy that a particle possesses due to its mass alone.
  • It does not depend on the movement or the kinetic state of the particle.
  • This principle implies that even a particle at rest contains a significant amount of energy.
Understanding rest energy is vital because it serves as a baseline from which we calculate total energy and kinetic energy in relativistic contexts.
Lorentz Factor
The Lorentz factor, denoted as \( \gamma \), is a crucial component in understanding how time, length, and relativistic mass change as an object moves. It is described by the formula \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( v \) is the velocity of the object and \( c \) is the speed of light.
  • The Lorentz factor determines how much time dilation, length contraction, and relativistic mass increase occur at high speeds.
  • As an object's speed approaches the speed of light \( c \), \( \gamma \) increases dramatically, indicating significant relativistic effects.
  • For speeds much lower than \( c \), \( \gamma \) is approximately 1, showing negligible relativistic effects.
The Lorentz factor plays an essential role in computing the total energy of particles and in relativistic physics overall.
Speed of Light
The speed of light, symbolized as \( c \), is a constant central to the theories of special and general relativity. It is approximately \( 3 \times 10^8 \) meters per second.
  • Light speed is the maximum speed at which all energy, matter, and information in the universe can travel.
  • It acts as a cosmic speed limit due to its constant value in all inertial frames of reference.
  • It鈥檚 crucial for calculating energy in relativistic physics, as in the equation \( E = mc^2 \).
Understanding the speed of light helps students grasp how energy and mass are interchangeable and why the universe behaves differently at relativistic speeds.
Total Energy
Total energy in the context of relativity is the sum of a particle's rest energy and its kinetic energy. It can be expressed by the equation \( E = \gamma mc^2 \). Here, \( \gamma \) is the Lorentz factor, \( m \) is the rest mass, and \( c \) is the speed of light.
  • The total energy accounts for the energy due to motion when a particle moves at relativistic speeds.
  • It highlights the relationship between mass and energy, where even moving at high speeds adds significantly to a particle's energy.
  • Total energy increases dramatically as velocities approach the speed of light due to increasing \( \gamma \).
Recognizing total energy allows for a deeper understanding of energy conservation and transformation in high-speed scenarios.
Relativity
Relativity, primarily introduced through Einstein's theories, describes the fundamental physical laws that govern the universe at high speeds and large gravitational fields.
  • The theory includes special relativity, which focuses on objects moving at constant speeds, especially those approaching the speed of light.
  • It also explains general relativity, which deals with gravity's effect on space-time.
  • Key concepts include the constancy of light speed, time dilation, length contraction, and mass-energy equivalence.
Understanding relativity provides insight into how time and space are interconnected and how energy and mass interact at cosmic scales. This knowledge is essential for physics students to comprehend the universe's fundamental workings.

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Most popular questions from this chapter

An alien spacecraft is flying overhead at a great distance as you stand in your backyard. You see its searchlight blink on for 0.150 s. The first officer on the spacecraft measures that the searchlight is on for 12.0 ms. (a) Which of these two measured times is the proper time? (b) What is the speed of the spacecraft relative to the earth, expressed as a fraction of the speed of light \(c\)?

Two protons (each with rest mass \(M\) = 1.67 \(\times\) 10\(^{-27}\) kg) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that also produces an \(\eta_0\) particle (see Chapter 44). The rest mass of the \(\eta_0\) is m = 9.75 \(\times\) 10\(^{-28}\) kg. (a) If the two protons and the \(\eta_0\) are all at rest after the collision, find the initial speed of the protons, expressed as a fraction of the speed of light. (b) What is the kinetic energy of each proton? Express your answer in MeV. (c) What is the rest energy of the \(\eta_0\), expressed in MeV? (d) Discuss the relationship between the answers to parts (b) and (c). 37.39 .

The negative pion (\(\pi^-\)) is an unstable particle with an average lifetime of 2.60 \(\times\) 10\(^{-8}\)s (measured in the rest frame of the pion). (a) If the pion is made to travel at very high speed relative to a laboratory, its average lifetime is measured in the laboratory to be 4.20 \(\times\) 10\(^{-7}\) s. Calculate the speed of the pion expressed as a fraction of c. (b) What distance, measured in the laboratory, does the pion travel during its average lifetime?

An observer in frame \(S'\) is moving to the right (+\(x\)-direction) at speed \(u\) = 0.600c away from a stationary observer in frame S. The observer in \(S'\) measures the speed \(v'\) of a particle moving to the right away from her. What speed \(v'\) does the observer in S measure for the particle if (a) \(v'\) = 0.400c; (b) \(v'\) = 0.900c; (c) \(v'\) = 0.990c?

A spaceship moving at constant speed u relative to us broadcasts a radio signal at constant frequency \(f_0\) . As the spaceship approaches us, we receive a higher frequency \(f\) ; after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_0\) , and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_0\)? (\(Hint\): In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals 1 /T. Use the time dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_0\) = 345 MHz as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758\(c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f - f_0\)? (c) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift (\(f - f_0\)) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?
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