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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

Rocket ship Able travels at \(0.400 c\) relative to an Earth observer. According to the same observer, rocket ship Able overtakes a slower moving rocket ship Baker that moves in the same direction. The captain of Baker sees Able pass her ship at \(0.114 c .\) Determine the speed of Baker relative to the Earth observer.
A cosmic ray particle travels directly over a football field, from one goal line to the other, at a speed of \(0.50 c .\) (a) If the length of the field between goal lines in the Earth frame is \(91.5 \mathrm{m}(100 \mathrm{yd}),\) what length is measured in the rest frame of the particle? (b) How long does it take the particle to go from one goal line to the other according to Earth observers? (c) How long does it take in the rest frame of the particle?
Suppose your handheld calculator will show six places beyond the decimal point. How fast (in meters per second) would an object have to be traveling so that the decimal places in the value of \(\gamma\) can actually be seen on your calculator display? That is, how fast should an object travel so that \(\gamma=1.000001 ?\) [Hint: Use the binomial approximation.]
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 $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$
For a nonrelativistic particle of mass \(m,\) show that \(K=p^{2} /(2 m) .\) [Hint: Start with the nonrelativistic expressions for kinetic energy \(K\) and momentum \(p\).]
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