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Chapter 20: Problem 2

A \(100-\mathrm{MeV}\) electron, for which \(v=0.999987 c\), moves along the axis of an evacuated tube that has a length of \(2.86 \mathrm{~m}\) as measured by a laboratory observer \(S\) with respect to whom the tube is at rest. An observer \(S^{\prime}\) moving with the electron, however, would see this tube moving past with speed \(v\). What length would this observer measure for the tube?

Short Answer

Expert verified
The observer moving with the electron would measure the length of the tube to be approximately \(0.0404266 m\).

Step by step solution

01

Identify the given information

The problem provides that \(L_0 = 2.86 m\) (the length of the tube measured by observer \(S\)), \(v = 0.999987 c\) (the speed of the electron which is also the speed of observer \(S'\)), and \(c\) (the speed of light, a universal constant).
02

Use the formula of length contraction

We should use the {\it length contraction formula} which is \(L=L_0\sqrt{1-v^2/c^2}\) .
03

Substitute the given values into the formula

Plug the given values into the length contraction formula: \(L = 2.86 \sqrt{1 - (0.999987)^2}\).
04

Calculate the length

Perform the calculation inside the square root first, then multiply by the length of the tube as measured by observer \(S\): \(L = 2.86 \sqrt{1 - (0.999987)^2} = 2.86 \times 0.01415.\)
05

Compute the final value

Carrying out the multiplication yields the length of the tube as measured by observer \(S'\): \( L = 0.0404266 m\).

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