Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 26: Problem 32

Electron A is moving west with speed \(\frac{3}{5} c\) relative to the lab. Electron \(\mathrm{B}\) is also moving west with speed \(\frac{4}{5} c\) relative to the lab. What is the speed of electron \(\mathrm{B}\) in a frame of reference in which electron \(\mathrm{A}\) is at rest?

Short Answer

Expert verified
Answer: The speed of electron B in a frame of reference where electron A is at rest is \(\frac{5}{13}c\).

Step by step solution

01

Identify given information

Electron A is moving west with a speed of \(\frac{3}{5}c\) relative to the lab and electron B is moving west with a speed of \(\frac{4}{5}c\) relative to the lab. Our goal is to find the speed of electron B in a frame of reference where electron A is at rest.
02

Apply the relativistic velocity addition formula

The relativistic velocity addition formula states: $$ V_{B/A} = \frac{V_B - V_A}{1 - \frac{V_A V_B}{c^2}} $$ Where \(V_A\) is the velocity of electron A in the lab frame, \(V_B\) is the velocity of electron B in the lab frame, \(V_{B/A}\) is the relative velocity of electron B with respect to electron A, and \(c\) is the speed of light.
03

Plug in the given values

Now, we will plug in the given values for electron A and electron B. Since they are both moving west, we can use their given speeds directly in the formula without worrying about their signs: $$ V_{B/A} = \frac{\frac{4}{5}c - \frac{3}{5}c}{1 - \frac{(\frac{3}{5}c)(\frac{4}{5}c)}{c^2}} $$
04

Solve for the relative velocity

Simplifying the equation, we get: $$ V_{B/A} = \frac{\frac{1}{5}c}{1 - \frac{12}{25}} $$ And finally, $$ V_{B/A} = \frac{\frac{1}{5}c}{\frac{13}{25}} = \frac{5}{13}c $$ Therefore, the speed of electron B in a frame of reference in which electron A is at rest is \(\frac{5}{13}c\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Derivation of the Doppler formula for light. A source and receiver of EM waves move relative to one another at velocity \(v .\) Let \(v\) be positive if the receiver and source are moving apart from one another. The source emits an EM wave at frequency \(f_{\mathrm{s}}\) (in the source frame). The time between wavefronts as measured by the source is \(T_{\mathrm{s}}=1 / f_{\mathrm{s}}\) (a) In the receiver's frame, how much time elapses between the emission of wavefronts by the source? Call this \(T_{\mathrm{T}}^{\prime} .\) (b) \(T_{\mathrm{T}}^{\prime}\) is not the time that the receiver measures between the arrival of successive wavefronts because the wavefronts travel different distances. Say that, according to the receiver, one wavefront is emitted at \(t=0\) and the next at \(t=T_{\mathrm{o}}^{\prime} .\) When the first wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}\) When the second wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}+v T_{\mathrm{r}}^{\prime} .\) Each wavefront travels at speed \(c .\) Calculate the time \(T_{\mathrm{r}}\) between the arrival of these two wavefronts as measured by the receiver. (c) The frequency detected by the receiver is \(f_{\mathrm{r}}=1 / T_{\mathrm{r}} .\) Show that \(f_{\mathrm{r}}\) is given by $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$
A particle decays in flight into two pions, each having a rest energy of 140.0 MeV. The pions travel at right angles to each other with equal speeds of \(0.900 c .\) Find (a) the momentum magnitude of the original particle, (b) its kinetic energy, and (c) its mass in units of \(\mathrm{MeV} / c^{2}\).
An extremely relativistic particle is one whose kinetic energy is much larger than its rest energy. Show that for an extremely relativistic particle $E \approx p c$.
A spaceship is traveling away from Earth at \(0.87 c .\) The astronauts report home by radio every \(12 \mathrm{h}\) (by their own clocks). At what interval are the reports sent to Earth, according to Earth clocks?
A lambda hyperon \(\Lambda^{0}\) (mass \(=1115.7 \mathrm{MeV} / \mathrm{c}^{2}\) ) at rest in the lab frame decays into a neutron \(n\) (mass \(=\) $939.6 \mathrm{MeV} / c^{2}\( ) and a pion (mass \)=135.0 \mathrm{MeV} / c^{2}$ ): $$\Lambda^{0} \rightarrow \mathrm{n}+\pi^{0}$$ What are the kinetic energies (in the lab frame) of the neutron and pion after the decay? [Hint: Use Eq. \((26-11)\) to find momentum.]
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Fundamentals of Physics

Read Explanation

Space Physics

Read Explanation

Solid State Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Work Energy and Power

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.

Sign-up for free