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What must be the momentum of a particle with mass \(m\) so that the total energy of the particle is \(3.00\) times its rest energy?

Short Answer

Expert verified
The momentum of the particle must be \(2\sqrt{2}mc\).

Step by step solution

01

Understand the rest energy

The rest energy of a particle with mass \(m\) is given by \(E_0 = mc^2\), where \(c\) is the speed of light.
02

Determine the total energy

Given that the total energy \(E\) is \(3.00\) times the rest energy, we have \(E = 3.00 \times E_0 = 3.00 \times mc^2\).
03

Use the relativistic energy-momentum relation

The relationship between total energy, momentum \(p\), and rest energy is given by \(E^2 = (pc)^2 + (mc^2)^2\). Now substitute \(E = 3.00mc^2\) into this equation.
04

Solve for momentum

Substitute \(E = 3.00mc^2\) and \(E_0 = mc^2\) into the energy-momentum relation: \[(3.00mc^2)^2 = (pc)^2 + (mc^2)^2\]This simplifies to:\[(9.00m^2c^4) = (pc)^2 + (m^2c^4)\]Subtract \(m^2c^4\) from both sides:\[8.00m^2c^4 = (pc)^2\]Now, solve for \(p\):\[p = \sqrt{8.00}mc\].
05

Simplify the expression

The expression \(\sqrt{8.00}mc\) simplifies to \(2\sqrt{2}mc\). Therefore, the momentum \(p\) is \(2\sqrt{2}mc\).

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Key Concepts

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

Momentum in Relativistic Physics
In relativistic physics, momentum takes on a special form compared to classical physics. The classical momentum is defined as the product of mass and velocity, given by:
  • Classical Momentum: \( p = mv \)
However, in the realm of relativistic speeds—those approaching the speed of light—this simple formula needs modification. This is because, as a particle's speed nears the speed of light, its effective mass increases. Hence, its momentum increases more than it would in classical terms.
The momentum in relativistic terms becomes:
  • Relativistic Momentum: \( p = \gamma mv \)
where \( \gamma \) (gamma) represents the Lorentz factor, defined as \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). This factor accounts for the effects of relativity when objects move at substantial fractions of the speed of light. These adjustments ensure the laws of physics remain consistent at all speeds.
Understanding Rest Energy
Rest energy is a concept introduced by Einstein that signifies the energy contained in a particle when it is not in motion—relative to the observer. This energy comes solely from the particle's mass. It is expressed by the famous equation:
  • Rest Energy: \( E_0 = mc^2 \)
\( m \) represents the rest mass, and \( c \) is the speed of light in a vacuum. This formula shows that mass itself is a form of energy, an insight that revolutionized our understanding of both energy and mass.
The rest energy also helps explain why it requires an enormous amount of energy to accelerate particles to high speeds. As particles reach higher velocities, their total energy surpasses their rest energy significantly, thus further impacting calculations involving momentum and energy.
Relativistic Energy-Momentum Relation
The relativistic energy-momentum relation is a fundamental formula that links a particle's total energy, momentum, and rest energy. This is crucial in calculations where high-speed particles are involved. The relation is given by:
  • \( E^2 = (pc)^2 + (mc^2)^2 \)
Here, \( E \) is the total energy, \( p \) is the momentum, and \( mc^2 \) is the rest energy. When analyzing the motion and energy of high-speed particles, this equation provides a complete picture by incorporating the effects of both energy and momentum.
In scenarios such as the exercise we examined, you substitute into this equation to find unknown variables like momentum when the total and rest energies are known. This highlights how all the components—energy, momentum, and mass—are interrelated at relativistic scales.

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Most popular questions from this chapter

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