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Chapter 26: Problem 30

Relative to the laboratory, a proton moves to the right with a speed of \(\frac{4}{5} c,\) while relative to the proton, an electron moves to the left with a speed of \(\frac{5}{7} c .\) What is the speed of the electron relative to the lab?

Short Answer

Expert verified
Answer: The speed of the electron relative to the lab frame is \(\frac{c}{5}\).

Step by step solution

01

Identify the given velocities

The given velocities in the problem are as follows: - Velocity of the proton relative to the lab frame: \(v_1 = \frac{4}{5}c\) - Velocity of the electron relative to the proton frame: \(v_2 = -\frac{5}{7}c\) (negative sign because it's moving to the left)
02

Use the relativistic velocity addition formula

Now, we will use the relativistic velocity addition formula mentioned above: \(v_{total} = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}\) Substitute \(v_1 = \frac{4}{5}c\) and \(v_2 = -\frac{5}{7}c\) into the formula: \(v_{total} = \frac{\frac{4}{5}c - \frac{5}{7}c}{1 + \frac{(\frac{4}{5}c)(-\frac{5}{7}c)}{c^2}}\)
03

Simplify the expression

Now, we can simplify the expression to calculate the velocity of the electron relative to the lab: \(v_{total} = \frac{\frac{4}{5}c - \frac{5}{7}c}{1 - \frac{20}{35}}\) Calculate the numerator: \(Numerator = \frac{4}{5}c - \frac{5}{7}c = \frac{28c - 25c}{35} = \frac{3c}{35}\) Calculate the denominator: \(Denominator = 1 - \frac{20}{35} = \frac{15}{35}\) Finally, divide the numerator by the denominator: \(v_{total} = \frac{\frac{3c}{35}}{\frac{15}{35}} = \frac{3c}{15} = \frac{c}{5}\)
04

State the final result

The velocity of the electron relative to the lab frame is \(\frac{c}{5}\).

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Most popular questions from this chapter

A spaceship moves at a constant velocity of \(0.40 c\) relative to an Earth observer. The pilot of the spaceship is holding a rod, which he measures to be \(1.0 \mathrm{m}\) long. (a) The rod is held perpendicular to the direction of motion of the spaceship. How long is the rod according to the Earth observer? (b) After the pilot rotates the rod and holds it parallel to the direction of motion of the spaceship, how long is it according to the Earth observer?
Muons are created by cosmic-ray collisions at an elevation \(h\) (as measured in Earth's frame of reference) above Earth's surface and travel downward with a constant speed of \(0.990 c .\) During a time interval of \(1.5 \mu \mathrm{s}\) in the rest frame of the muons, half of the muons present at the beginning of the interval decay. If one fourth of the original muons reach Earth before decaying, about how big is the height \(h ?\)
A clock moves at a constant velocity of \(8.0 \mathrm{km} / \mathrm{s}\) with respect to Earth. If the clock ticks at intervals of one second in its rest frame, how much more than a second elapses between ticks of the clock as measured by an observer at rest on Earth? [Hint: Use the binomial approximation \(.(1-v^{2} / c^{2})^{-1 / 2} \approx 1+v^{2} /(2 c^{2}) .\)]
A lambda hyperon \(\Lambda^{0}\) (mass \(=1115 \mathrm{MeV} / c^{2}\) ) at rest decays into a neutron \(\mathrm{n}\) (mass \(=940 \mathrm{MeV} / \mathrm{c}^{2}\) ) and a pion (mass \(=135 \mathrm{MeV} / c^{2}\)): $$\Lambda^{0} \rightarrow \mathrm{n}+\pi^{0}$$ What is the total kinetic energy of the neutron and pion?
Two atomic clocks are synchronized. One is put aboard a spaceship that leaves Earth at \(t=0\) at a speed of \(0.750 c .\) (a) When the spaceship has been traveling for \(48.0 \mathrm{h}\) (according to the atomic clock on board), it sends a radio signal back to Earth. When would the signal be received on Earth, according to the atomic clock on Earth? (b) When the Earth clock says that the spaceship has been gone for \(48.0 \mathrm{h},\) it sends a radio signal to the spaceship. At what time (according to the spaceship's clock) does the spaceship receive the signal?
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