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At what speed is the momentum of a particle three times as great as the result obtained from the nonrelativistic expression \(m v ?\)

Short Answer

Expert verified
The speed is \( \frac{2c\sqrt{2}}{3} \).

Step by step solution

01

Understand the Problem

We're being asked to determine the speed at which the momentum of a particle is three times its nonrelativistic momentum, given by the formula \( m v \). This involves considering both nonrelativistic and relativistic momentum expressions.
02

Write Down Nonrelativistic Momentum

The nonrelativistic momentum for a particle is given by \( p = m v \), where \( m \) is the mass of the particle, and \( v \) is its velocity.
03

Write Down Relativistic Momentum

The relativistic momentum is given by \( p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( c \) is the speed of light. We need this to find when it is three times the nonrelativistic momentum.
04

Set Up Equation for Momentum

According to the problem, the relativistic momentum is three times the nonrelativistic momentum: \( \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 m v \). We can simplify this equation by canceling \( m \) and \( v \) (assuming \( v eq 0 \)).
05

Simplify and Solve for Velocity

After canceling common terms, we get \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 \). Solving for \( v \), we first square both sides to get \( \frac{1}{1 - \frac{v^2}{c^2}} = 9 \). Transfer \( 1 \) to the other side: \( 1 - \frac{v^2}{c^2} = \frac{1}{9} \).
06

Final Calculation

Rearrange to find \( v^2 \): \( \frac{v^2}{c^2} = 1 - \frac{1}{9} = \frac{8}{9} \). Then \( v^2 = \frac{8c^2}{9} \), and thus \( v = c \sqrt{\frac{8}{9}} = \frac{2c\sqrt{2}}{3} \).

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Key Concepts

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

Nonrelativistic Momentum
In classical physics, momentum is a fundamental concept. Nonrelativistic momentum is the product of the mass of an object and its velocity. This simple formula is expressed as:
  • \( p = m v \)
Here, \( p \) represents momentum, \( m \) is the mass, and \( v \) is the velocity. This equation is straightforward. However, it's only accurate when the speed is much less than the speed of light.
At higher speeds, like those approaching the speed of light, we need to use relativistic physics instead. This ensures our calculations remain accurate.
Nonrelativistic momentum works well for most everyday applications. For example, calculating the momentum of a moving car or a baseball.
However, when dealing with particles moving at very high speeds, such as electrons in an accelerator, classical momentum doesn't suffice.
Relativity in Physics
Relativity in physics revolutionized our understanding of space and time. Introduced by Albert Einstein, it encompasses both the special and general theories of relativity.
Special relativity, which relates to our topic, describes how space and time are linked for objects moving at constant speeds close to the speed of light.
A key insight from special relativity is that the nonrelativistic expression for momentum must be adjusted to account for the effects of high velocity. This is where the concept of relativistic momentum comes in.
  • Relativistic momentum is expressed as: \( p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} \)
  • This equation considers the increase in an object's mass as it approaches the speed of light.
These advanced calculations demonstrate that at high speeds, time appears to slow, and objects contract in length, according to an outside observer.
The relativistic effects ensure that as speeds increase, the laws of physics remain consistent and the same for all observers, a fundamental principle of relativity.
Speed of Light Calculation
The speed of light, denoted as \( c \), is a constant value of approximately 299,792 kilometers per second (or about 186,282 miles per second).
It's not just any speed but a universal speed limit in the universe, as described by the theory of relativity. When we calculate speeds approaching the speed of light, we use relativistic equations.
In our exercise, we sought the speed at which a particle's momentum is three times its nonrelativistic value. By using the formula for relativistic momentum:
  • \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 3 \)
We discovered the sought velocity to be \( \frac{2c\sqrt{2}}{3} \). This value is a specific fraction of the speed of light, showcasing how relativistic effects play a significant role as velocities approach the speed of light.
As particles move faster, calculations incorporating the speed of light emphasize the nonlinear nature of velocity and momentum in these extreme conditions.

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Most popular questions from this chapter

An airplane has a length of 60 \(\mathrm{m}\) when measured at rest. When the airplane is moving at 180 \(\mathrm{m} / \mathrm{s}(400 \mathrm{mph})\) in the alternate uni- verse, how long would it appear to be to a stationary observer? A. 24 \(\mathrm{m}\) B. 36 \(\mathrm{m}\) C. 48 \(\mathrm{m}\) D. 60 \(\mathrm{m}\) E. 75 \(\mathrm{m}\)

\(\bullet\) You take a trip to Pluto and back (round trip 11.5 billion km), traveling at a constant speed (except for the turnaround at Pluto) of \(45,000 \mathrm{km} / \mathrm{h}\) . (a) How long does the trip take, in hours, from the point of view of a friend on earth? About how many years is this? (b) When you return, what will be the difference between the time on your atomic wristwatch and the time on your friend's? (Hint: Assume the distance and speed are highly precise, and carry a lot of significant digits in your calculation!)

\(\bullet\) The negative pion \(\left(\pi^{-}\right)\) is an unstable particle with an average lifetime of \(2.60 \times 10^{-8} \mathrm{s}\) (measured in the rest frame of the pion).(a) If the pion is made to travel at very high speed relative to a laboratory, its average lifetime is measured in the laboratory to be \(4.20 \times 10^{-7}\) s. Calculate the speed of the pion expressed as a fraction of \(c .\) (b) What distance, as measured in the laboratory, does the pion travel during its average lifetime?

\(\bullet\) Why are we bombarded by muons? Muons are unstable subatomic particles (more on them in Chapter 30 ) that decay to electrons with a mean lifetime of 2.2\(\mu \mathrm{s}\) . They are produced when cosmic rays bombard the upper atmosphere about 10 \(\mathrm{km}\) above the earth's surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth's surface. (a) What is the greatest distance a muon could travel during its 2.2\(\mu\) s lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the 2.2\(\mu\) lifetime is measured in the frame of the muon, and they are moving very fast. At a speed of \(0.999 c,\) what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far could the muon travel in this time? Does this result explain why we find muons in cos- mic rays? (c) From the point of view of the muon, it still lives for only \(2.2 \mu s,\) so how does it make it to the ground? What is the thickness of the 10 \(\mathrm{km}\) of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

A meterstick moves past you at great speed. Its motion relative to you is parallel to its long axis. If you measure the length of the moving meterstick to be \(1.00 \mathrm{ft}(1 \mathrm{ft}=0.3048 \mathrm{m})-\) for example, by comparing it with a l-foot ruler that is at rest relative to you, at what speed is the meterstick moving relative to you?

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