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At what speed is the magnitude of the relativistic momentum of a particle three times the magnitude of the nonrelativistic momentum?

Short Answer

Expert verified
The speed is approximately 0.94c.

Step by step solution

01

Understand the Problem

We need to find the speed at which the relativistic momentum of a particle is three times its nonrelativistic momentum. The relativistic momentum is given by the formula \[ p_r = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \]where \( m \) is the mass of the particle, \( v \) is its velocity, and \( c \) is the speed of light. The nonrelativistic momentum is simply \( p_{nr} = mv \).
02

Set Up the Equation

We set up the equation based on the problem statement, which says the relativistic momentum is three times the nonrelativistic momentum:\[ \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} = 3mv \]
03

Simplify the Equation

Cancel \( m \) and \( v \) from both sides of the equation (assuming \( v eq 0 \)): \[ \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 \]
04

Solve for \( v \)

Square both sides of the equation to eliminate the square root:\[ 1 = 9(1 - \frac{v^2}{c^2}) \]\[ 1 = 9 - 9\frac{v^2}{c^2} \]
05

Rearrange and Solve for \( v^2 \)

Rearrange the equation to solve for \( \frac{v^2}{c^2} \):\[ 9\frac{v^2}{c^2} = 8 \]\[ \frac{v^2}{c^2} = \frac{8}{9} \]
06

Solve for \( v \)

Take the square root of both sides to find \( v \):\[ v = c \sqrt{\frac{8}{9}} \]\[ v = c \times \frac{2\sqrt{2}}{3} \]
07

Simplify the Result

Finally, express the speed as a fraction of the speed of light:\[ v \approx 0.94c \]

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Key Concepts

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

Nonrelativistic Momentum
Momentum is a measure of the motion of an object and is dependent on its mass and velocity. In nonrelativistic or classical physics, momentum is calculated simply as the product of mass and velocity. This is expressed in the formula:
  • \( p_{nr} = mv \)
where:
  • \( p_{nr} \) is the nonrelativistic momentum,
  • \( m \) represents the mass of the particle,
  • \( v \) is the velocity of the particle.

This formula works well for speeds that are much smaller than the speed of light. However, when the velocity of an object approaches the speed of light, relativistic effects become significant, and we need to use a different formula to calculate momentum. In such cases, the above formula no longer provides accurate results as it doesn't account for the relativistic effects that come into play.
Speed of Light
The speed of light, denoted as \( c \), is one of the fundamental constants in physics. It represents the maximum speed at which all matter and information in the universe can travel through a vacuum.
  • \( c \approx 3.00 \times 10^8 \) meters per second.

In calculations involving high-speed particles, such as those in particle physics, the speed of light is crucial. As objects approach this vast speed, their momentum and energy significantly deviate from classical predictions due to the relativistic effects.
When calculating relativistic momentum:
  • The formula incorporates \( c \) to accommodate these deviations from the nonrelativistic or classical case.
Understanding the significance of the speed of light allows us to appreciate why relativistic equations must be used instead of classical ones when dealing with objects moving at substantial fractions of \( c \).
Velocity Equation
When transitioning from nonrelativistic to relativistic physics, the equations need to incorporate special relativity concepts. The velocity equation is a part of finding how speed changes affect a particle's momentum.
For relativistic momentum, we use:
  • \( p_r = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \)
This formula considers:
  • The Lorentz factor \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), which accounts for time dilation and length contraction effects.

In the example problem, we equate the relativistic momentum to three times the nonrelativistic counterpart and solve for \( v \).
By re-arranging our equation, we understand:
  • The expression \( v = c \frac{2\sqrt{2}}{3} \) gives a fractional multiple of \( c \).
This value informs us how close to the speed of light the particle must travel, emphasizing the need to shift from classic to relativistic physics for accurate results.

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Most popular questions from this chapter

Two spaceships A and B are exploring a new planet. Relative to this planet, spaceship A has a speed of \(0.60 c,\) and spaceship \(B\) has a speed of \(0.80 c .\) What is the ratio \(D_{\Lambda} / D_{B}\) of the values for the planet's diameter that each spaceship measures in a direction that is parallel to its motion?

A woman is \(1.6 \mathrm{m}\) tall and has a mass of \(55 \mathrm{kg} .\) She moves past an observer with the direction of the motion parallel to her height. The observer measures her relativistic momentum to have a magnitude of \(2.0 \times 10^{10} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) What does the observer measure for her height?

The crew of a rocket that is moving away from the earth launches an escape pod, which they measure to be \(45 \mathrm{m}\) long. The pod is launched toward the earth with a speed of \(0.55 c\) relative to the rocket. After the launch, the rocket's speed relative to the earth is \(0.75 c .\) What is the length of the escape pod as determined by an observer on earth?

ssm Starting from rest, two skaters push off against each other on smooth level ice, where friction is negligible. One is a woman and one is a man. The woman moves away with a velocity of \(+2.5 \mathrm{m} / \mathrm{s}\) relative to the ice. The mass of the woman is \(54 \mathrm{kg}\), and the mass of the man is \(88 \mathrm{kg}\). Assuming that the speed of light is \(3.0 \mathrm{m} / \mathrm{s},\) so that the relativistic momentum must be used, find the recoil velocity of the man relative to the ice. (Hint: This problem is similar to Example 6 in Chapter \(7 .\) )

A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft arrives, 12 years have elapsed on earth, and 9.2 years have elapsed on board the ship. How far away (in meters) is the planet, according to observers on earth?

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