/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 (a) At what speed is the momentu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) At what speed is the momentum of a particle twice as great as the result obtained from the nonrelativistic expression \(m v ?\) Express your answer in terms of the speed of light. (b) A force is apphed to a particle along its direction of motion. At what speed is the magnitude of force required to produce a given acceleration twice as great as the force required to produce the same acceleration when the particle is at rest? Express your answer in terms of the speed of light.

Short Answer

Expert verified
Both speeds are \( v = \frac{\sqrt{3}}{2}c \).

Step by step solution

01

Understanding Momentum in Classical and Relativistic Mechanics

In classical mechanics, the momentum of a particle is given by the expression \( p = mv \) where \( m \) is the mass and \( v \) is the velocity. In relativistic mechanics, the momentum is given by \( p = \gamma mv \), where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor. Our goal is to find the speed at which the relativistic momentum is twice the classical momentum.
02

Equating Relativistic and Classical Momentum Expressions

We want \( \gamma mv = 2mv \). Simplifying, we get \( \gamma = 2 \). This means \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2 \). Solving this equation for \( v \) will give us the speed where the relativistic momentum is twice the classical momentum.
03

Solve for Speed using Lorentz Factor

We start from \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2 \). Squaring both sides, \( \frac{1}{1 - \frac{v^2}{c^2}} = 4 \). This leads to \( 1 - \frac{v^2}{c^2} = \frac{1}{4} \). Thus, \( \frac{v^2}{c^2} = 1 - \frac{1}{4} = \frac{3}{4} \). Solving for \( v \), we find \( v = \frac{\sqrt{3}}{2}c \).
04

Understanding Force in Classical and Relativistic Contexts

In classical mechanics, force \( F \) is related to acceleration \( a \) by \( F = ma \). In relativistic dynamics, the force required to produce the same acceleration increases as the speed of the particle approaches the speed of light due to the increase in relativistic mass or effective inertia.
05

Equating Relativistic and Classical Force Expressions

When asked to find when the force needed to produce a given acceleration is twice as much relativistically, we start with \( \gamma F = 2F = 2ma \), implying \( \gamma = 2 \). Just as earlier, we solve for the speed where this occurs.
06

Calculate Speed using Lorentz Factor for Force

Following the process we derive for momentum, we again use \( \gamma = 2 \) and solve \( \frac{1}{1 - \frac{v^2}{c^2}} = 4 \) leading to \( v = \frac{\sqrt{3}}{2}c \). This is the same speed as when the momentum is doubled relativistically.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Lorentz factor
In the realm of relativistic mechanics, the Lorentz factor plays a vital role in explaining how time, length, and physical mass are perceived differently when an object approaches the speed of light. The Lorentz factor, denoted as \( \gamma \), is given by the equation \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( v \) is the velocity of the object, and \( c \) is the speed of light.

This factor becomes significant when speeds are comparable to the speed of light. At such speeds, time dilates, lengths contract, and mass effectively increases from the observer's frame of reference. The Lorentz factor is essentially a measure of these relativistic effects.

For example, in the problem regarding momentum, the task is to find when the relativistic momentum is twice the classical. Equating \( \gamma mv = 2mv \) confirms that \( \gamma \) must be 2. Solving for speed reveals that at \( v = \frac{\sqrt{3}}{2}c \), the Lorentz factor equals 2, highlighting significant relativistic effects.
momentum in physics
Momentum serves as a cornerstone concept in both classical and relativistic physics. Classically, momentum \( p \) is the product of an object's mass \( m \) and velocity \( v \), expressed as \( p = mv \). It reflects how much motion an object possesses and is conserved in isolated systems.

In relativistic physics, momentum is adjusted to account for how motion appears at speeds nearing the speed of light. Here, momentum adopts the form \( p = \gamma mv \), incorporating the Lorentz factor \( \gamma \). This adjustment ensures momentum conservation even when relativistic speeds are involved.

Understanding momentum's variation at different speeds is crucial. For instance, the exercise’s question asks for the speed causing relativistic momentum to be twice the classical momentum. By working through the Lorentz factor, it is discovered that this speed is \( v = \frac{\sqrt{3}}{2}c \), emphasizing that even though formulas may look similar, relativistic mechanics require deeper considerations.
classical vs relativistic dynamics
Classical and relativistic dynamics serve as two fundamental branches in physics, often intersecting when discussing motion, force, and energy.

**Classical Dynamics:**
We begin with classical dynamics, governed by Newton's laws, where force \( F \) is related to mass \( m \) and acceleration \( a \) by \( F = ma \). It effectively describes motion at everyday speeds and remains a practical model until relativistic velocities come into play.

**Relativistic Dynamics:**
At velocities approaching the speed of light, relativistic dynamics come into focus. They require adjusting classical equations to account for changing perceptions of space and time. For force, this means acknowledging that as speed increases towards light speed, more force is needed to achieve the same acceleration. This adjustment is due to the Lorentz factor, affecting how mass and force are realized in this framework.

In our specific example, the question about force shows that at \( v = \frac{\sqrt{3}}{2}c \), the needed relativistic force becomes twice the classical. Thus, while classical dynamics provide an excellent approximation for most situations, relativistic dynamics offer crucial insights when speeds tread near light speed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As measured by an observer on the earth, a spacecraft runway on earth has a length of 3600 \(\mathrm{m}\) (a) What is the length of the runway as measured by a pilot of a spacecraft flying past at a speed of \(4.00 \times 10^{7} \mathrm{m} / \mathrm{s}\) relative to the earth? (b) An observer on earth measures the time interval from when the spacecraft is directly over one end of the runway until it is directly over the other end. What result does she get? (c) The pilot of the spacecraft measures the time it takes him to travel from one end of the runway to the other end. What value does he get?

When a particle meets its antiparticle, they annihilate each other and their mass is converted to light energy. The United States uses approximately \(1.0 \times 10^{19} \mathrm{J}\) of energy per year (a) If all this energy came from a futuristic ant-matter reactor, how much mass of matter and antimatter fuel would be consumed yearly? (b) If this fuel had the density of iron \(\left(7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) and were stacked in bricks to form a cubical pile, how high would it be? (Before you get your hopes up, antimatter reactors are a long way in the future\(- \)if they ever will be feasible.)

Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2\(\mu s\) . 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 -\mus lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the \(2.2-\mu\) s lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999 \(\mathrm{c}\) . What is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic 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? It is now clear how the muon is able to reach the ground?

Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520\(c\) as measured in the laboratory. What is the magnitude of the velocity of one particle relative to the other?

A spaceraft ties away from the earth with a speed of \(4.80 \times 10^{6} \mathrm{m} / \mathrm{s}\) relative to the earth and then returns at the same speed. The spacecraft carries an atomic clock that has been carefully synchronized with an identical clock that remains at rest on earth. The spacecraft returns to its starting point 365 days ( 1 year) later, as measured by the clock that remained on earth. What is the difference in the elapsed times on the two clocks, measured in hours? Which clock, the one in the spacecraft or the one on earth, shows the shortest elapsed time?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.