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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 neutron (mass 1.00866 u) disintegrates into a proton (mass $1.00728 \mathrm{u}\( ), an electron (mass \)0.00055 \mathrm{u}$ ), and an antineutrino (mass 0 ). What is the sum of the kinetic energies of the particles produced, if the neutron is at rest? $\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2} .\right)$
Show that each of these statements implies that \(v<c\) which means that \(v\) can be considered a nonrelativistic speed: (a) $\gamma-1<<1 \text { [Eq. }(26-14)] ;\( (b) \)K<<m c^{2}\( [Eq. \)(26-15)] ;(\text { c }) p<m c[\text { Eq. }(26-16)] ;\( (d) \)K \approx p^{2} /(2 m)$.
A muon with rest energy 106 MeV is created at an altitude of \(4500 \mathrm{m}\) and travels downward toward Earth's surface. An observer on Earth measures its speed as \(0.980 c .\) (a) What is the muon's total energy in the Earth observer's frame of reference? (b) What is the muon's total energy in the muon's frame of reference?
Derive the energy-momentum relation $$E^{2}=E_{0}^{2}+(p c)^{2}$$ Start by squaring the definition of total energy \(\left(E=K+E_{0}\right)\) and then use the relativistic expressions for momentum and kinetic energy $[\text{Eqs}. (26-6) \text { and }(26-8)]$.
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