Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 35: Problem 90

A gold nucleus of rest mass \(183.473 \mathrm{GeV} / \mathrm{c}^{2}\) is accelerated from some initial speed to a final speed of \(0.8475 c .\) In this process, \(137.782 \mathrm{GeV}\) of work is done on the gold nucleus. What was the initial speed of the gold nucleus as a fraction of \(c ?\)

Short Answer

Expert verified
Answer: The initial speed of the gold nucleus is approximately 0.630c.

Step by step solution

01

Calculate the final kinetic energy of the gold nucleus

We are given the final speed of the gold nucleus, \(v_f = 0.8475 c\). To find the final kinetic energy, \(K_f\), we first need to find the final Lorentz factor, \(\gamma_f\) using the formula \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\). Then, we can calculate the final kinetic energy using the formula \(K = (\gamma - 1)mc^2\). \(\gamma_f = \frac{1}{\sqrt{1-\frac{(0.8475c)^2}{c^2}}}\) \(\gamma_f \approx 2.017\) \(K_f = (\gamma_f - 1)mc^2 \approx (2.017 - 1)(183.473 \,\mathrm{GeV})\) \(K_f \approx 185.058 \,\mathrm{GeV}\)
02

Calculate the initial kinetic energy of the gold nucleus

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. So, we can now find the initial kinetic energy, \(K_i\), using the formula \(K_i = K_f - W\), where \(W\) is the work done on the gold nucleus. \(K_i = 185.058 \,\mathrm{GeV} - 137.782 \,\mathrm{GeV}\) \(K_i \approx 47.276 \,\mathrm{GeV}\)
03

Calculate the initial Lorentz factor of the gold nucleus

Now that we have the initial kinetic energy, we can calculate the initial Lorentz factor, \(\gamma_i\), using the formula \(K_i = (\gamma_i - 1)mc^2\). \(\gamma_i = \frac{K_i}{mc^2} + 1 \approx \frac{47.276 \,\mathrm{GeV}}{183.473 \,\mathrm{GeV}} + 1\) \(\gamma_i \approx 1.258\)
04

Calculate the initial speed of the gold nucleus

Finally, we can calculate the initial speed, \(v_i\), as a fraction of \(c\) using the formula \(\gamma_i = \frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\). \(v_i^2 = c^2 \left(1 - \frac{1}{\gamma_i^2}\right) \approx c^2 \left(1 - \frac{1}{1.258^2}\right)\) \(v_i \approx c \sqrt{1 - \frac{1}{1.258^2}}\) \(v_i \approx 0.630 c\) The initial speed of the gold nucleus is approximately \(0.630 c\).

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

Consider two clocks carried by observers in a reference frame moving at speed \(v\) in the positive \(x\) -direction relative to Earth's rest frame. Assume that the two reference frames have parallel axes and that their origins coincide when clocks at that point in both frames read zero. Suppose the clocks are separated by a distance \(l\) in the \(x^{\prime}\) -direction in their own reference frame; for instance, \(x^{\prime}=0\) for one clock and \(x^{\prime}=I\) for the other, with \(y^{\prime}=z^{\prime}=0\) for both. Determine the readings \(t^{\prime}\) on both clocks as functions of the time coordinate \(t\) in Earth's reference frame.

A famous result in Newtonian dynamics is that if a particle in motion collides elastically with an identical particle at rest, the two particles emerge from the collision on perpendicular trajectories. Does the same hold in the special theory of relativity? Suppose a particle of rest mass \(m\) and total energy \(E\) collides with an identical particle at rest, and the two particles emerge from the collision with new velocities. Are those velocities necessarily perpendicular? Explain.

Calculate the Schwarzschild radius of a black hole with the mass of a) the Sun. b) a proton. How does this result compare with the size scale of \(10^{-15} \mathrm{~m}\) usually associated with a proton?

Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28 ), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, the observer sees that the particle is, in fact, moving toward the wire.)

In the twin paradox example, Alice boards a spaceship that flies to a space station 3.25 light-years away and then returns with a speed of \(0.650 c\). View the trip in terms of Alice's reference frame. a) Show that Alice must travel with a speed of \(0.914 c\) to establish a relative speed of \(0.650 c\) with respect to Earth when she is returning to Earth. b) Calculate the time duration for Alice's return flight to Earth at the speed of \(0.914 c\)
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Fundamentals of Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Wave Optics

Read Explanation

Famous Physicists

Read Explanation

Electricity

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