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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
So a particle whose kinetic energy equals its rest energy moves at a speed of approximately 0.866 times the speed of light, while when its kinetic energy is five times its rest energy, the speed is about 0.994 times the speed of light.

Step by step solution

01

Understand the concept of rest energy and kinetic energy

The rest energy of an object is given by \(E_0 = mc^2\), where \(m\) is the mass of the object and \(c\) is the speed of light in vacuum. According to the special theory of relativity, the kinetic energy \(K\) of a particle can be written as \(K = (\gamma - 1) mc^2\), where \(\gamma\) is the Lorentz factor, given by \(\gamma = 1 / \sqrt{1 - (v/c)^2}\), and \(v\) is the speed of the object.
02

Solve for (a) particle speed when kinetic energy equals its rest energy

When kinetic energy equals rest energy, it implies that \(K = E_0\). This means that \((\gamma - 1) mc^2 = mc^2\) or \(\gamma - 1 = 1\). Solving for \(\gamma\) gives \(\gamma = 2\). Substituting the value of \(\gamma\) in the Lorentz factor equation, solve for \(v\): \(\sqrt{1 - (v/c)^2} = 1/ \gamma \), which gives \(v = c \sqrt{1 - 1/\gamma^2} = c \sqrt{1 - 1/2^2} = 0.866c\). So when kinetic energy equals its rest energy, the particle speed is approximately 0.866 of the speed of light.
03

Solve for (b) particle speed when kinetic energy is five times its rest energy

When kinetic energy is five times the rest energy, it implies \(K = 5E_0\) or \((\gamma - 1) mc^2 = 5mc^2\), which gives \(\gamma - 1 = 5\). Solving for \(\gamma\), we get \(\gamma = 6\). Substituting the value of \(\gamma\) in the Lorentz factor equation, solve for \(v\): \(\sqrt{1 - (v/c)^2} = 1/ \gamma\), which gives \(v = c \sqrt{1 - 1/\gamma^2} = c \sqrt{1 - 1/6^2} = 0.994c\). So when kinetic energy is five times its rest energy, the particle speed is approximately 0.994 times the speed of light.

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Key Concepts

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

Rest Energy
In the world of physics, rest energy plays a fundamental role in understanding how mass and energy are related. According to Einstein's famous equation, \(E_0 = mc^2\), the rest energy (\(E_0\)) of an object is the energy inherent in its mass (\(m\)) when it is not moving. 'Rest' implies that the object is at relative stillness with respect to the observer.

The speed of light (\(c\)), approximately \(3 \times 10^8 \text{ m/s}\), is crucial in this equation as it relates mass to energy. This equation suggests that even when an object is not in motion, it possesses energy due to its mass. The concept of rest energy is vital in particle physics, as it helps to explain phenomena like particle creation and annihilation, where mass is converted to energy and vice versa.
Lorentz Factor
The Lorentz factor, often denoted by \(\text{\text{\text{gamma}}})\), is a quantity that emerges from the special theory of relativity and is crucial when describing the time, length, and relativistic mass of objects moving close to the speed of light.

It is defined by the equation \(\text{\text{\text{gamma}}}) = 1 / \text{\text{sqrt}}{1 - (v/c)^2}\), where \(v\) is the velocity of the object and \(c\) is the speed of light. As the speed of an object approaches the speed of light, the Lorentz factor increases significantly, leading to effects such as time dilation and length contraction. In terms of kinetic energy, the Lorentz factor is used to adjust the classical formula to fit the relativistic framework, revealing that kinetic energy increases more rapidly as we approach the speed of light.
Special Theory of Relativity
The special theory of relativity, formulated by Albert Einstein in 1905, fundamentally changed our understanding of space, time, and matter. One of its core principles is the constancy of the speed of light (\text{c}) in a vacuum, meaning that it is the same for all observers, regardless of their relative motion.

This theory introduces several counterintuitive concepts. For example, it predicts that time can flow at different rates for observers who are in relative motion (time dilation), and that distances can contract (length contraction). While these effects are not noticeable at everyday speeds, they become significant as an object's speed approaches \(c\). Moreover, the theory lays the groundwork for the concept of mass-energy equivalence, the idea that mass can be converted into energy and vice versa, which is critical for understanding the full spectrum of kinetic and rest energy.
Speed of Light
The speed of light (\text{c}), approximately \(3 \times 10^8 \text{ m/s}\), is a fundamental constant in physics and serves as the cosmic speed limit. In the context of relativistic kinetic energy, the speed of light is especially important because no matter how much energy you put into accelerating an object, it cannot exceed this limit.

The faster an object moves, especially as it gets closer to \(c\), the more its mass appears to increase from the perspective of an outside observer. This increase in mass then makes it increasingly difficult to speed up the object further, creating a natural barrier to the speed of light. This core concept is essential for understanding why, as kinetic energy in a particle becomes comparable to or larger than its rest energy, its speed can be a significant fraction of \(c\) but never equal to or greater than \(c\).

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Most popular questions from this chapter

Two particles in a high-energy accelerator experiment approach each other head-on with a relative speed of \(0.890 c .\) Both particles travel at the same speed as measured in the laboratory. What is the speed of each particle, as measured in the laboratory?

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

(a) How much work must be done on a particle with mass \(m\) to accelerate it (a) from rest to a speed of \(0.090 c\) and (b) from a speed of \(0.900 c\) to a speed of \(0.990 c ?\) (Express the answers in terms of \(\left.m c^{2} .\right)\) (c) How do your answers in parts (a) and (b) compare?

The positive muon \(\left(\mu^{+}\right),\) an unstable particle, lives on average \(2.20 \times 10^{-6} \mathrm{~s}\) (measured in its own frame of reference) before decaying. (a) If such a particle is moving, with respect to the laboratory, with a speed of \(0.900 c\), what average lifetime is measured in the laboratory? (b) What average distance, measured in the laboratory, does the particle move before decaying?

Many of the stars in the sky are actually binary stars, in which two stars orbit about their common center of mass. If the orbital speeds of the stars are high enough, the motion of the stars can be detected by the Doppler shifts of the light they emit. Stars for which this is the case are called spectroscopic binary stars. Figure \(\mathbf{P 3 7 . 6 8}\) shows the simplest case of a spectroscopic binary star: two identical stars, each with mass \(m,\) orbiting their center of mass in a circle of radius \(R .\) The plane of the stars' orbits is edge-on to the line of sight of an observer on the earth. (a) The light produced by heated hydrogen gas in a laboratory on the earth has a frequency of \(4.568110 \times 10^{14} \mathrm{~Hz}\) In the light received from the stars by a telescope on the earth, hydrogen light is observed to vary in frequency between \(4.567710 \times 10^{14} \mathrm{~Hz}\) and \(4.568910 \times 10^{14} \mathrm{~Hz}\). Determine whether the binary star system as a whole is moving toward or away from the earth, the speed of this motion, and the orbital speeds of the stars. (Hint: The speeds involved are much less than \(c,\) so you may use the approximate result \(\Delta f / f=u / c\) given in Section \(37.6 .\) ) (b) The light from each star in the binary system varies from its maximum frequency to its minimum frequency and back again in 11.0 days. Determine the orbital radius \(R\) and the mass \(m\) of each star. Give your answer for \(m\) in kilograms and as a multiple of the mass of the sun, \(1.99 \times 10^{30} \mathrm{~kg} .\) Compare the value of \(R\) to the distance from the earth to the sun, \(1.50 \times 10^{11} \mathrm{~m}\). (This technique is actually used in astronomy to determine the masses of stars. In practice, the problem is more complicated because the two stars in a binary system are usually not identical, the orbits are usually not circular, and the plane of the orbits is usually tilted with respect to the line of sight from the earth.)

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