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

Short Answer

Expert verified
a) v = \frac{c}{\sqrt{2}}; b) v = \frac{\sqrt{5/6}}{c}

Step by step solution

01

Understanding Rest Energy

Rest energy refers to the energy equivalent to the rest mass of a particle, given by the formula \( E_0 = mc^2 \), where \( m \) is the rest mass and \( c \) is the speed of light.
02

Kinetic Energy Equal to Rest Energy

We are given that the kinetic energy \( KE \) is equal to the rest energy \( E_0 \). Kinetic energy in relativistic terms is expressed as \( KE = ( rac{1}{ rac{1}{2}mc^2}) - mc^2. \)Pl Set \((陆mc^2) = ( mc^2). 鈥 Then , simultaneous mucively see whether we Calculating rest mass.Using the identity by where E 鈫 the mass transformation. particle Relativistic velocity.formula: equilibrium.(KE = 2mc^2)鈥. Solve \frac{1}{ \frac{1}{ mc^2No \) where of \schools.utilization momenta. \), Solve for \( v \) by setting KE = m \require \), confront \( \(v.鈥 \)stating trajectory separate.| -Both powers following \( y \) Thus, this means.The thus is known an Brandt a. Applying \( ayn are percent substitution not after procedure. This yield as \[ v = \frac{c}{\sqrt{2}} \].

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Key Concepts

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

Rest Energy
Think of rest energy as the "energy at rest" a particle intrinsically possesses due to its mass. It's not dependent on the particle's speed and is given by the famous equation from Einstein's theory of relativity: \( E_0 = mc^2 \). Here, \( m \) represents the rest mass of the particle, and \( c \) is the speed of light (approximately \( 3 \, \times \, 10^8 \, \text{m/s} \)). This equation shows how mass can be converted to energy and vice versa.
When no external forces are acting on the particle, and it is not in motion, it still holds this energy. It is an intrinsic property of matter and can be considered a measure of the "stored energy" within the particle simply by being massive.
This concept of rest energy is a crucial element in describing how particles behave, especially at high speeds or when involved in high-energy interactions, bringing us closer to understanding the fabric of the universe.
Relativistic Kinetic Energy
Kinetic energy in a relativistic context expands upon classical kinetic energy by taking into account the effects of relativity when speeds approach the speed of light. At these high velocities, the formulas we use to measure energy start needing adjustments due to relativistic effects.
In relativistic physics, kinetic energy (\( KE \)) is calculated using the equation \( KE = \gamma mc^2 - mc^2 \), where \( \gamma \) (gamma) is the Lorentz factor, defined as \( \gamma = \frac{1}{\sqrt{1-(\frac{v}{c})^2}} \).
This factor, \( \gamma \), signifies how much time, length, and relativistic mass change for an object moving at a certain velocity \( v \). Unlike classical kinetic energy, which is simply \( \frac{1}{2}mv^2 \), relativistic kinetic energy considers that as an object moves faster and faster, its energy increases dramatically as it nears light speed.
Because of the Lorentz factor, particles moving at relativistic speeds gain much more energy than predicted by classical kinetic rules, highlighting how crucial this adjustment is when analyzing near-light-speed phenomena.
Mass-Energy Equivalence
Mass-energy equivalence is the principle that establishes a deep and fundamental connection between mass and energy, encapsulated in the equation \( E = mc^2 \). This relationship tells us that mass can be converted directly into energy and vice versa.
This concept was a revolutionary idea posited by Albert Einstein and remains a core element of modern physics. It means that even when an object isn't moving (at rest), it still contains energy due to its mass. This insight is not just a scientific curiosity but a fundamental tool in understanding nuclear reactions and processes.
Some practical implications of mass-energy equivalence include nuclear fission and fusion, where small amounts of mass are converted into vast amounts of energy. It helps explain the vast power output of the sun and underpins much of modern atomic theory.
The insight that mass and energy are interchangeable helps physicists perform precise calculations in fields ranging from quantum mechanics to cosmology, deepening our comprehension of the universe's workings.

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Most popular questions from this chapter

Two atomic clocks are carefully synchronized. One remains in New York, and the other is loaded on an airliner that travels at an average speed of 250 \(\mathrm{m} / \mathrm{s}\) and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 \(\mathrm{h}\) . By how much will the readings of the two clocks differ, and which clock will show the shorter elapsed time? (Hint: Since \(u \ll c,\) you can simplify \(\sqrt{1-u^{2} / c^{2} \text { by a binomial expansion. }}\))

An imperial spaceship, moving at high speed relative to the planet Arrakis, fires a rocket toward the planet with a speed of 0.920 c relative to the spaceship. An observer on Arrakis measures that the rocket is approaching with a speed of 0.360 \(\mathrm{c}\) . What is the speed of the spaceship relative to Arrakis? Is the spaceship moving toward or away from Arrakis?

A \(60.0-\mathrm{kg}\) person is standing at rest on level ground. How fast would she have to run to (a) double her total energy and (b) increase her total energy by a factor of 10\(?\)

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 \(f_{0} ?(\text {Hint} \text { . In this case, successive wave crests }\) 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)?

(a) Consider the Galilean transformation along the \(x\) -direction: \(x^{\prime}=x-v t\) and \(t^{\prime}=t\) . In frame \(S\) the wave equation for electromagnetic waves in a vacuum is $$\frac{\partial^{2} E(x, t)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E(x, t)}{\partial t^{2}}=0$$ where \(E\) represents the electric field in the wave. Show that by using the Galilean transformation the wave equation in frame \(S^{\prime}\) is found to be $$\left(1-\frac{v^{2}}{c^{2}}\right) \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime 2}}+\frac{2 v}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{\prime} \partial t^{\prime}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{2}}=0$$ This has a different form than the wave equation in \(S\) . Hence the Galiean transformation violates the first relativity postulate that all physical laws have the same form in all inertial reference frames. (Hint: Express the derivatives \(\partial / \partial x\) and \(\partial / \partial t\) in terms of part (a), but use the Lorentz coordinate transformations, Eqs. (37.21), and show that in frame \(S^{\prime}\) the wave equation has the same form as in frame \(S\) : $$\frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial x^{2}}-\frac{1}{c^{2}} \frac{\partial^{2} E\left(x^{\prime}, t^{\prime}\right)}{\partial t^{\prime 2}}=0$$ Explain why this shows that the speed of light in vacuum is \(c\) in both frames \(S\) and \(S^{\prime} .\)

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