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Chapter 26: Problem 27

Particle \(A\) is moving with a constant velocity \(v_{\mathrm{AE}}=+0.90 c\) relative to an Earth observer. Particle \(B\) moves in the opposite direction with a constant velocity \(v_{\mathrm{BE}}=-0.90 c\) relative to the same Earth observer. What is the velocity of particle \(B\) as seen by particle \(A ?\)

Short Answer

Expert verified
Answer: The velocity of particle B as seen by particle A is approximately -0.9945c.

Step by step solution

01

Write down the given velocities relative to Earth

We are given that the velocities of particles A and B relative to the Earth observer are: \(v_{\mathrm{AE}} = +0.90c\) \(v_{\mathrm{BE}} = -0.90c\)
02

Use the velocity addition formula

To find the velocity of particle B as seen by particle A, we use the velocity addition formula: \(v_{\mathrm{BA}} = \frac{v_{\mathrm{BE}} - v_{\mathrm{AE}}}{1 - \frac{v_{\mathrm{AE}} v_{\mathrm{BE}}}{c^2}}\)
03

Substitute the given values

Substitute the given values for \(v_{\mathrm{AE}}\) and \(v_{\mathrm{BE}}\) into the equation: \(v_{\mathrm{BA}} = \frac{-0.90c - 0.90c}{1 - \frac{(-0.90c)(0.90c)}{c^2}}\)
04

Simplify the equation

Simplify the equation by performing the algebraic operations: \(v_{\mathrm{BA}} = \frac{-1.8c}{1 - 0.81}\)
05

Solve for \(v_{\mathrm{BA}}\)

Solve for the velocity of particle B as seen by particle A: \(v_{\mathrm{BA}} = \frac{-1.8c}{0.19} \approx -9.474c\) Since the calculated value of \(v_{\mathrm{BA}}\) is greater than the speed of light, it means that we made a mistake while solving the problem.
06

Step 4(Revised): Simplify the equation using the correct formula

The correct simplification of the equation is: \(v_{\mathrm{BA}} = \frac{-0.90c - 0.90c}{1 + \frac{(-0.90c)(0.90c)}{c^2}}\)
07

Step 5(Revised): Solve for \(v_{\mathrm{BA}}\)

Solve for the velocity of particle B as seen by particle A: \(v_{\mathrm{BA}} = \frac{-1.8c}{1 + 0.81} = \frac{-1.8c}{1.81} \approx -0.9945c\) Thus, the velocity of particle B as seen by particle A is approximately \(-0.9945c\).

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Most popular questions from this chapter

An unstable particle called the pion has a mean lifetime of 25 ns in its own rest frame. A beam of pions travels through the laboratory at a speed of $0.60 c .$ (a) What is the mean lifetime of the pions as measured in the laboratory frame? (b) How far does a pion travel (as measured by laboratory observers) during this time?
The rest energy of an electron is \(0.511 \mathrm{MeV} .\) What momentum (in MeV/c) must an electron have in order that its total energy be 3.00 times its rest energy?
A charged particle is observed to have a total energy of \(0.638 \mathrm{MeV}\) when it is moving at \(0.600 c .\) If this particle enters a linear accelerator and its speed is increased to \(0.980 c,\) what is the new value of the particle's total energy?
An astronaut has spent a long time in the Space Shuttle traveling at $7.860 \mathrm{km} / \mathrm{s} .$ When he returns to Earth, he is 1.0 s younger than his twin brother. How long was he on the shuttle? [Hint: Use an approximation from Appendix A.5 and beware of calculator round off errors.]
The Tevatron is a particle accelerator at Fermilab that accelerates protons and antiprotons to high energies in an underground ring. Scientists observe the results of collisions between the particles. The protons are accelerated until they have speeds only \(100 \mathrm{m} / \mathrm{s}\) slower than the speed of light. The circumference of the ring is \(6.3 \mathrm{km} .\) What is the circumference according to an observer moving with the protons? [Hint: Let \(v=c-u\) where $v \text { is the proton speed and } u=100 \mathrm{m} / \mathrm{s} .$]
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