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Chapter 35: Problem 7

A square of area \(100 \mathrm{~m}^{2}\) that is at rest in the reference frame is moving with a speed \((\sqrt{3} / 2) c\). Which of the following statements is incorrect? a) \(\beta=\sqrt{3} / 2\) b) \(\gamma=2\) c) To an observer at rest, it looks like another square with an area less than \(100 \mathrm{~m}^{2}\) d) The length along the moving direction is contracted by a factor of \(\frac{1}{2}\)

Short Answer

Expert verified
Based on the analysis and calculations performed, all four statements, a), b), c), and d) are correct. Therefore, there is no incorrect statement among the given options, indicating an error in the problem itself.

Step by step solution

01

Determine \(\beta\)

The first statement says that \(\beta = \sqrt{3} / 2\). We know that \(\beta = \frac{v}{c}\), so let's check if this statement is correct: \[\beta = \frac{v}{c} = \frac{\frac{\sqrt{3}}{{~}}c}{c} = \frac{\sqrt{3}}{2}\] So, statement a) is correct.
02

Determine \(\gamma\)

The second statement says that \(\gamma = 2\). The Lorentz factor is given by \(\gamma = \frac{1}{\sqrt{1 - \beta^2}}\). Using the value of \(\beta\) we have just calculated: \[\gamma = \frac{1}{\sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2}} = \frac{1}{\sqrt{1 - \frac{3}{4}}} = \frac{1}{\sqrt{\frac{1}{4}}} = 2\] So, statement b) is correct.
03

Analyze Area under different reference frame

For an observer at rest, we need to find whether the area of the moving square is less than 100 m². The length contraction takes place only along the direction of motion, so we consider one side of the square and find the contracted length: \[L'_{x} = \frac{L_{x}}{\gamma} = \frac{\sqrt{100}}{2} = 5\sqrt{2}\] Since only one side is contracted, the area of the square in the rest frame of the observer will be: \[A' = L'_{x} * L_{y} = 5\sqrt{2} * 10 = 50\sqrt{2}\] As \(50\sqrt{2} < 100\), the statement c) is correct.
04

Check Length Contraction Factor

The fourth statement says that the length along the moving direction is contracted by a factor of \(\frac{1}{2}\). We have already calculated the contracted length \(L'_{x}\) and original length \(L_{x}\): \[\frac{L'_{x}}{L_{x}} = \frac{5\sqrt{2}}{10} = \frac{1}{2}\] So, statement d) is also correct. Since the problem asks for the incorrect statement and all four statements are correct, it seems there is an error in the problem itself. Each statement provided is accurate according to our calculations.

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Most popular questions from this chapter

You shouldn't invoke time dilation due to your relative motion with respect to the rest of the world as an excuse for being late to class. While it is true that relative to those at rest in the classroom, your time runs more slowly, the difference is likely to be negligible. Suppose over the weekend you drove from your college in the Midwest to New York City and back, a round trip of \(2200 .\) miles, driving for 20.0 hours each direction. By what amount, at most, would your watch differ from your professor's watch?

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