/*! 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 33 What is the speed of a particle ... [FREE SOLUTION] | 91Ó°ÊÓ

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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

Identify Relevant Formulas

The rest energy of a particle is given by the formula \(E_0 = mc^2\), where \(m\) is the mass of the particle and \(c\) is the speed of light. The total energy \(E\) is the sum of the rest energy and kinetic energy \(K\), and it 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. The kinetic energy is \(K = (\gamma - 1)mc^2\).
02

Setting Kinetic Energy Equal to Rest Energy (Case a)

For part (a), we set the kinetic energy \(K\) equal to the rest energy \(E_0\): \(K = mc^2 = (\gamma - 1)mc^2\). Dividing both sides by \(mc^2\), we have \(1 = \gamma - 1\) or \(\gamma = 2\).
03

Solving for Speed (Case a)

With \(\gamma = 2\), we use the relation \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\). Solving for \(v\), we have \(2 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), which simplifies to \(\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{2}\). Solving for \(v^2\), we get \(\frac{v^2}{c^2} = \frac{3}{4}\), thus \(v = c\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}c\).
04

Setting Kinetic Energy as Five Times Rest Energy (Case b)

For part (b), the kinetic energy is five times the rest energy, \(K = 5mc^2\). Using \(K = (\gamma - 1)mc^2\), we have \(5 = \gamma - 1\) or \(\gamma = 6\).
05

Solving for Speed (Case b)

With \(\gamma = 6\), we use \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\). Solving for \(v\), we have \(6 = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), which simplifies to \(\sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{6}\). Solving for \(v^2\), we get \(\frac{v^2}{c^2} = \frac{35}{36}\), thus \(v = c\sqrt{\frac{35}{36}} = c\frac{\sqrt{35}}{6}\).

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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 and is defined as the energy possessed by an object due to its mass when it is at rest. The famous equation given by Albert Einstein, \(E_0 = mc^2\), describes this quantity, where \(E_0\) represents the rest energy, \(m\) is the mass of the object, and \(c\) is the speed of light. This equation highlights the idea that mass and energy are interchangeable. Rest energy forms the baseline energy level for any object, which becomes crucial when analyzing the behavior of particles at high velocities.
Understanding rest energy helps us grasp how energy transforms when particles gain speed and how this transformation relates to the total energy of a system.
In the case of high-energy physics, rest energy serves as a critical reference point for energy calculations.
Lorentz factor
The Lorentz factor, denoted as \(\gamma\), is an essential element in the study of relativity. It measures the effect of relativistic speeds on time, length, and mass. Use 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.
  • This factor becomes significant when objects move at a substantial fraction of the speed of light.
  • It reflects how time dilates and lengths contract as speeds increase.
As speeds approach the speed of light, \(\gamma\) increases dramatically, implying dramatic changes in physics observations. The Lorentz factor plays a pivotal role in calculating relativistic kinetic energy and understanding particle behavior in high-energy environments.
speed of light
The speed of light in a vacuum is a constant represented by \(c\), approximately equal to \(3 \times 10^8\) meters per second. As one of the universe's fundamental constants, it acts as the upper speed limit for the transmission of information and travel within our universe.
  • It is a crucial factor in Einstein’s equations relating to energy, mass, and speed.
  • The speed of light ensures that as particles accelerate, the effects of relativity must be considered, particularly time dilation and length contraction.
When calculating the kinetic energy of particles at velocities comparable to the speed of light, using \(c\) becomes necessary to understand their increasing influence on relativistic equations.
total energy
Total energy in relativistic physics is the sum of rest energy and kinetic energy. It accounts for all forms of energy an object possesses when it moves at significant speeds, integrating both the contributions of its mass at rest and its motion-induced energy. Expressed as \(E = \gamma mc^2\), total energy extends the traditional energy concepts to high-speed conditions typically observed in particle physics.
  • This definition broadens when particles approach light speeds, as their kinetic energy has substantial relativistic effects.
  • Total energy depends heavily on the Lorentz factor, \(\gamma\), which alters how energy behaves under rapid conditions.
The transition from rest energy to total energy includes incorporating the Lorentz factor and illustrates how traditional kinetic equations are modified to capture the true energy dynamics at play.
relativistic energy formulas
Relativistic energy formulas offer a framework for understanding energy dynamics at speeds comparable to the speed of light. These formulas adjust classical equations to incorporate relativistic effects observed at high velocities. Key among these is the kinetic energy formula given by: \(K = (\gamma - 1)mc^2\).
  • Such formulas ensure accurate calculations are made when dealing with energetic particles, removing discrepancies observed with classical mechanics.
  • They account for the energy needed to increase a particle's speed as it approaches light velocity, requiring exponentially more energy.
In practical terms, these formulas let scientists and engineers design systems that consider the limitations and results of relativistic movement, essential in fields like accelerator physics and advanced aerospace projects.

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Most popular questions from this chapter

In high-energy physics, new particles can be created by collisions of fast- moving projectile par- ticles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon \(\left(\mathrm{K}^{-}\right)\) and a positive kaon \(\left(\mathrm{K}^{+}\right)\) $$p+p \rightarrow p+p+\mathrm{K}^{-}+\mathrm{K}^{+}$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 \(\mathrm{MeV}\) , and the rest energy of each proton is 938.3 \(\mathrm{MeV}\) . (Hint: It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99540\(c\) relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 45.0 \(\mathrm{km} .\) (a) As measured by the scientist, how much time does it take the particle to travel the 45.0 \(\mathrm{km}\) to the surface of the earth? (b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle's frame. (c) In the particle's frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?

Two events observed in a frame of reference \(S\) have positions and times given by \(\left(x_{1}, t_{1}\right)\) and \(\left(x_{2}, t_{2}\right),\) respectively. (a) Frame \(S^{\prime}\) moves along the \(x\) -axis just fast enough that the two events occur at the same position in \(S^{\prime} .\) Show that in \(S^{\prime},\) the time interval \(\Delta t^{\prime}\) between the two events is given by $$\Delta t^{\prime}=\sqrt{(\Delta t)^{2}-\left(\frac{\Delta x}{c}\right)^{2}}$$ where \(\Delta x=x_{2}-x_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Hence show that if \(\Delta x>c \Delta t,\) there is \(n o\) frame \(S^{\prime}\) in which the two events occur at the same point. The interval \(\Delta t^{\prime}\) is sometimes called the proper time interval for the events. Is this term appropriate? (b) Show that if \(\Delta x>c \Delta t,\) there is a different frame of reference \(S^{\prime}\) in which the two events occur simultaneously. Find the distance between the two events in \(S^{\prime} ;\) express your answer in terms of \(\Delta x, \Delta t,\) and \(c\) This distance is sometimes called a proper length. Is this term appropriate? (c) Two events are observed in a frame of reference \(S^{\prime}\) to occur simultaneously at points separated by a distance of 2.50 \(\mathrm{m} .\) In a second frame \(S\) moving relative to \(S^{\prime}\) along the line joining the two points in \(S^{\prime},\) the two events appear to be separated by 5.00 \(\mathrm{m} .\) What is the time interval between the events as measured in \(S ?[\)Hint : Apply the result obtained in part (b).]

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 \(f_{0}=345 \mathrm{MHz}\) 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, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) 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)?

The French physicist Armand Fizeau was the first to measure the speed of light accurately. He also found experimentally that the speed, relative to the lab frame, of light traveling in a tank of water that is itself moving at a speed \(V\) relative to the lab frame is $$v=\frac{c}{n}+k V$$ where \(n=1.333\) is the index of refraction of water. Fizeau called \(k\) the draging coefficient and obtained an experimental value of \(k=0.44 .\) What value of \(k\) do you calculate from relativistic transformations?

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