/*! 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 9 A Lincoln Continental is twice a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 9

A Lincoln Continental is twice as long as a VW Beetle, when they are at rest. As the Continental overtakes the \(\mathrm{VW}\), going through a speed trap, a (stationary) policeman observes that they both have the same length. The \(V W\) is going at half the speed of light. How fast is the Lincoln going? (Leave your answer as a multiple of \(c .)\)

Short Answer

Expert verified
The Lincoln's speed is \(\frac{1}{2}c\).

Step by step solution

01

Understanding the Problem

The problem involves special relativity, where the length contraction effect occurs when objects move at high speeds close to the speed of light. At rest, the Lincoln Continental is twice as long as a VW Beetle. When moving, both cars appear to have the same length due to length contraction.
02

Relating Lengths with Velocity

For the VW Beetle, moving at a speed of \(v = \frac{1}{2}c\), the length contraction formula is \(L' = L_0 \sqrt{1 - \frac{v^2}{c^2}}\), where \(L_0\) is the rest length of the Beetle. For the Lincoln, \(L'_L\) is the same observed length due to its velocity \(v_L\). Since the lengths appear equal to the observer, the formula \(L'_L = 2L'_B\) must hold true.
03

Expressing Length Contraction for Beetle

The length contraction for the VW is given as \(L'_B = L_0 \sqrt{1 - \left(\frac{1}{2}\right)^2}\), which simplifies to \(L'_B = L_0 \sqrt{\frac{3}{4}} = L_0 \frac{\sqrt{3}}{2}\).
04

Setting Up the Equation for the Lincoln

Since the observed lengths are equal, we have: \(L'_L = 2L_0\frac{\sqrt{3}}{2}\). The length contraction formula for the Lincoln is given as \(L'_L = 2L_0 \sqrt{1 - \frac{v_L^2}{c^2}}\). Set these equal: \(2L_0\sqrt{1 - \frac{v_L^2}{c^2}} = 2L_0\frac{\sqrt{3}}{2}\).
05

Solving for Lincoln's Speed

Cancel \(2L_0\) from both sides, simplifying to \(\sqrt{1 - \frac{v_L^2}{c^2}} = \frac{\sqrt{3}}{2}\). Now square both sides to get \(1 - \frac{v_L^2}{c^2} = \frac{3}{4}\). Rearranging gives \(\frac{v_L^2}{c^2} = \frac{1}{4}\), and taking the square root yields \(v_L = \frac{1}{2}c\).

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 91影视!

Key Concepts

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

Understanding Length Contraction
Length contraction is a fascinating concept from special relativity. It revolves around how the length of an object changes depending on its speed relative to an observer. 
At high speeds, close to the speed of light (\(c\)), an object's length appears shorter than when it is at rest. This is known as length contraction. 
Mathematically, it is expressed as:
  • \(L' = L_0 \sqrt{1 - \frac{v^2}{c^2}}\)
In this equation:
  • \(L'\) is the contracted length observed,
  • \(L_0\) is the actual rest length, and
  • \(v\) is the velocity of the object relative to the observer.
The closer an object's speed is to the speed of light, the more its length contracts. This means a car traveling very fast will appear shorter than it would if it were just parked on the street. It's important to note this effect only becomes noticeable at relativistic speeds, which are speeds close to the speed of light.
The Unchanging Speed of Light
One of the pillars of special relativity is the idea that the speed of light (\(c\)) is absolute. This means it remains constant at approximately \(299,792,458\) meters per second. 
Regardless of the observer's motion or the source of light's motion, the speed of light remains unchanged. This has profound implications on our understanding of time and space.
  • The speed of light is the ultimate speed limit in the universe.
  • No object with mass can accelerate to the speed of light or exceed it.
This constancy leads to the interesting phenomena observed in relativity, such as time dilation and length contraction. It also means all observers, regardless of their own speeds, will always measure light to be traveling at \(c\).
Exploring Relativistic Velocities
Relativistic velocities occur when objects move at speeds close to light, altering our intuitive understanding of time and distance. Handling these velocities requires thinking in terms of Einstein's theory, where classical physics' rules no longer apply without modifications.
  • As speeds approach the speed of light, relativistic effects like time dilation and length contraction become significant.
  • The formula used for addition of velocities is different in relativity:
The relativistic velocity addition formula is: 

\( w = \frac{u + v}{1 + \frac{uv}{c^2}} \)
  • Here, \(w\) is the resulting velocity observed,
  • \(u\) and \(v\) are two velocities being combined,
  • \(c\) is the speed of light.
This formula highlights that as velocities increase, they do not simply add up as they would in everyday experience. Instead, as they near the speed of light, the increase in resultant velocity diminishes, keeping it below the speed of light.

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

In a certain inertial frame \(\mathcal{S}\), the electric field \(\mathbf{E}\) and the magnetic field \(\mathbf{B}\) are neither parallel nor perpendicular, at a particular space-time point. Show that in a different inertial system \(\overline{\mathcal{S}}\), moving relative to \(\mathcal{S}\) with velocity \(\mathbf{v}\) given by $$ \frac{\mathbf{v}}{1+v^{2} / c^{2}}=\frac{\mathbf{E} \times \mathbf{B}}{B^{2}+E^{2} / c^{2}}, $$ the ficlds \(\overline{\mathrm{E}}\) and \(\check{\mathrm{B}}\) are parallel at that point. Is there a frame in which the two are perpendicular?

Find the invariant product of the 4 -velocity with itself, \(\eta^{\mu} \eta_{\mu}\).

Inertial system \(\bar{S}\) moves in the \(x\) dircction at speed \(\frac{3}{5} c\) relative to system \(\mathcal{S}\). (The \(\bar{x}\) axis slides long the \(x\) axis, and the origins coincide at \(t=\bar{t}=0\), as usual.) (a) On graph paper set up a Cartesian coordinate system with axes \(c t\) and \(x\). Carefully draw in lines representing \(\bar{x}=-3,-2,-1,0,1,2\), and \(3 .\) Also draw in the lines corresponding to \(c t=-3,-2,-1,0,1,2\), and \(3 .\) Label your lines clearly. (b) In \(\overline{\mathcal{S}}\), a free particle is observed to travel from the point \(\bar{x}=-2\) at timc \(c \bar{t}=-2\) to the point \(\bar{x}=2\) at \(c \bar{t}=+3 .\) Indicate this displacement on your graph. From the slope of this line. determine the particle's speed in \(\mathcal{S}\). (c) Use the velocity addition rule to determine the velocity in \(\mathcal{S}\) algebraically, and check that your answer is consistent with the graphical solution in (b).

A particle of mass \(m\) whose total encrgy is twice its rest energy collides with an identical particle at rest. If they stick togcther, what is the mass of the resulting composite particle? What is its velocity?

(a) Draw a space-time diagram representing a game of catch (or a conversation) between two people at rest, \(10 \mathrm{ft}\) apart. How is it possible for them to communicate, given that their separation is spacelike? (b) There's an old limerick that runs as follows: There once was a girl named Ms. Bright. Who could travel much faster than light. She departed one day, The Einsteinian way, And rcturned on the previous night. What do you think? Even if she could travel faster than the speed of light, could she return before she set out? Could she arrive at some intermediate destination before she set out? Draw a space-time diagram representing this trip.
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Translational Dynamics

Read Explanation

Physics of Motion

Read Explanation

Fields in Physics

Read Explanation

Famous Physicists

Read Explanation
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.