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Chapter 26: Problem 66

Two spaceships are observed from Earth to be approaching each other along a straight line. Ship A moves at \(0.40 c\) relative to the Earth observer, and ship \(\mathrm{B}\) moves at \(0.50 c\) relative to the same observer. What speed does the captain of ship A report for the speed of ship B?

Short Answer

Expert verified
Answer: The speed of ship B as observed by the captain of ship A is 0.125c.

Step by step solution

01

List the given information and the unknown

The given information is as follows: - Velocity of ship A relative to Earth, \(v_A = 0.40 c\) - Velocity of ship B relative to Earth, \(v_B = 0.50 c\) - We need to find the velocity of ship B relative to ship A, denoted as \(v_{BA}\)
02

Use the formula for relative velocity in special relativity

In special relativity, the formula for calculating the relative velocity is given by the following equation: $$v_{BA} = \frac{v_B - v_A}{1 - \frac{v_A v_B}{c^2}}$$
03

Solve for the unknown and give the final answer

Now, substituting the given values of \(v_A\) and \(v_B\) into the formula, we get: $$v_{BA}= \frac{0.50 c - 0.40 c}{1 - \frac{0.40 c \times 0.50 c}{c^2}}$$ Simplifying, we have: $$v_{BA}=\frac{0.10 c}{1 - 0.20} = \frac{0.10 c}{0.80}$$ Finally, we find the relative velocity of ship B as observed by the captain of ship A: $$v_{BA} = \frac{0.10}{0.80} c = 0.125 c$$ Therefore, the captain of ship A reports the speed of ship B to be \(0.125c\).

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Most popular questions from this chapter

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 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?
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?
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 $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$
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