Cosmic rays from space that strike Earth contain some charged particles with energies billions of times higher than any that can be produced in the biggest accelerator. One model that was proposed to account for these particles is shown schematically in the figure. Two very strong sources of magnetic fields move toward each other and repeatedly reflect the charged particles trapped between them. (These magnetic field sources can be approximated as infinitely heavy walls from which charged particles get reflected elastically.) The high- energy particles that strike the Earth would have been reflected a large number of times to attain the observed energies. An analogous case with only a few reflections demonstrates this effect. Suppose a particle has an initial velocity of \(-2.21 \mathrm{~km} / \mathrm{s}\) (moving in the negative \(x\) -direction, to the left), the left wall moves with a velocity of \(1.01 \mathrm{~km} / \mathrm{s}\) to the right, and the right wall moves with a velocity of \(2.51 \mathrm{~km} / \mathrm{s}\) to the left. What is the velocity of the particle after six collisions with the left wall and five collisions with the right wall?

Step by step solution

01

Establish the initial conditions

Initially, the particle has a velocity of \(-2.21 \; km/s\), the left wall has a velocity of \(1.01 \; km/s\), and the right wall has a velocity of \(-2.51 \; km/s\).

02

Calculate the relative velocities

Before each collision, we need to calculate the relative velocity of the particle with respect to each wall. The relative velocity of the particle with respect to the left wall is \(v_{p,left} = -2.21\; km/s - 1.01\; km/s = -3.22 \; km/s\). Similarly, the relative velocity of the particle with respect to the right wall is \(v_{p,right} = -2.21\; km/s - (-2.51\; km/s) = 0.3\; km/s\).

03

Reflect the particle after a collision

After each collision, the relative velocity of the particle with respect to the wall is reversed. Hence, after a collision with the left wall, the relative velocity of the particle with respect to the left wall is \(-v_{p,left} = 3.22 \; km/s\). Similarly, after a collision with the right wall, the relative velocity of the particle with respect to the right wall is \(-v_{p,right} = -0.3\; km/s\).

04

Repeat the process for 6 collisions with the left wall and 5 collisions with the right wall

After each collision, we need to add up the relative velocity of the particle with respect to the wall and the velocity of the wall itself to get the velocity of the particle after the collision. We'll do this repeatedly for 6 collisions with the left wall and 5 collisions with the right wall: For 6 collisions with the left wall: \(v_{1} = 3.22 \; km/s + 1.01 \; km/s = 4.23 \; km/s\) \(v_{2} = -3.22 \; km/s + 1.01 \; km/s = -2.21 \; km/s\) \(v_{3} = 3.22 \; km/s + 1.01 \; km/s = 4.23 \; km/s\) \(v_{4} = -3.22 \; km/s + 1.01 \; km/s = -2.21 \; km/s\) \(v_{5} = 3.22 \; km/s + 1.01 \; km/s = 4.23 \; km/s\) \(v_{6} = -3.22 \; km/s + 1.01 \; km/s = -2.21 \; km/s\) For 5 collisions with the right wall: \(v_{1} = -0.3\; km/s + (-2.51 \; km/s) = -2.81 \; km/s\) \(v_{2} = 0.3\; km/s + (-2.51 \; km/s) = -2.21 \; km/s\) \(v_{3} = -0.3\; km/s + (-2.51 \; km/s) = -2.81 \; km/s\) \(v_{4} = 0.3\; km/s + (-2.51 \; km/s) = -2.21 \; km/s\) \(v_{5} = -0.3\; km/s + (-2.51 \; km/s) = -2.81 \; km/s\)

05

Determine the final velocity of the particle

After six collisions with the left wall, the particle has a velocity of \(-2.21 \; km/s\). After five collisions with the right wall, the particle has a velocity of \(-2.81 \; km/s\). Therefore, the final velocity of the particle is \(-2.81\; km/s\).

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!