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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

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’s 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

Everyday Time Dilation. 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 m/s and then returns to New York. When the plane returns, the elapsed time on the clock that stayed behind is 4.00 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 \(\sqrt{1 - u^2/c^2}\) by a binomial expansion.)

Two events are observed in a frame of reference S to occur at the same space point, the second occurring 1.80 s after the first. In a frame \(S'\) moving relative to \(S\), the second event is observed to occur 2.15 s after the first. What is the difference between the positions of the two events as measured in \(S'\)?

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

A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.600c. A scientist on Coruscant measures the length of the moving spacecraft to be 74.0 m. The spacecraft later lands on Coruscant, and the same scientist measures the length of the now stationary spacecraft. What value does she get?

Space pilot Mavis zips past Stanley at a constant speed relative to him of 0.800c. Mavis and Stanley start timers at zero when the front of Mavis's ship is directly above Stanley. When Mavis reads 5.00 s on her timer, she turns on a bright light under the front of her spaceship. (a) Use the Lorentz coordinate transformation derived in Example 37.6 to calculate x and t as measured by Stanley for the event of turning on the light. (b) Use the time dilation formula, Eq. (37.6), to calculate the time interval between the two events (the front of the spaceship passing overhead and turning on the light) as measured by Stanley. Compare to the value of \(t\) you calculated in part (a). (c) Multiply the time interval by Mavis's speed, both as measured by Stanley, to calculate the distance she has traveled as measured by him when the light turns on. Compare to the value of \(x\) you calculated in part (a).

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