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Chapter 26: Problem 28

A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of \(0.60 c\) and \(0.80 c .\) What is the relative speed of the two ships as measured by a passenger on either one of the spaceships?

Short Answer

Expert verified
Answer: The relative speed of the two ships as measured by a passenger on either one of the spaceships is approximately \(0.9459c\).

Step by step solution

01

Identify the given velocities

The given velocities of the two spaceships are \(0.60c\) and \(0.80c.\) We will use these values in the formula for relative velocity in special relativity.
02

Use the relativistic velocity addition formula

To find the relative velocity of the spaceships, use the formula \(v_{rel} = \frac{v_1 + v_2}{1 + \frac{v_1 \cdot v_2}{c^2}},\) where \(v_1\) and \(v_2\) are the velocities of the spaceships. Substitute the given values: \(v_{rel} = \frac{0.60c + 0.80c}{1 + \frac{(0.60c) (0.80c)}{c^2}}.\)
03

Simplify the expression

Now, simplify the expression: \(v_{rel} = \frac{1.40c}{1 + \frac{0.48c^2}{c^2}}.\)
04

Cancel common terms

Cancel out the \(c^2\) terms in the denominator: \(v_{rel} = \frac{1.40c}{1 + 0.48}.\)
05

Calculate the relative velocity

Finally, calculate the relative velocity by dividing the numerator by the denominator: \(v_{rel} = \frac{1.40c}{1.48} = 0.9459c.\) The relative speed of the two ships as measured by a passenger on either one of the spaceships is approximately \(0.9459c.\)

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Most popular questions from this chapter

In a beam of electrons used in a diffraction experiment, each electron is accelerated to a kinetic energy of \(150 \mathrm{keV} .\) (a) Are the electrons relativistic? Explain. (b) How fast are the electrons moving?
A rectangular plate of glass, measured at rest, has sides \(30.0 \mathrm{cm}\) and \(60.0 \mathrm{cm} .\) (a) As measured in a reference frame moving parallel to the 60.0 -cm edge at speed \(0.25 c\) with respect to the glass, what are the lengths of the sides? (b) How fast would a reference frame have to move in the same direction so that the plate of glass viewed in that frame is square?
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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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