Length Contraction - GRE Subject Test: Physics

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Question

A rocket of length 5 meters passes an observer on earth. The observer measures the passing rocket to be 3 meters long. What is the velocity of the rocket in the reference frame of the Earth-based observer?

Answer

Length contraction is given by

$L_{rest}=\gamma L_{move}$

Where in this case,

$L_{rest}=5\textup{m}, ,,,L_{move}=3\textup{m}$

The Lorentz factor is given by:

$\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Combining these two equations, we get:

$L_{rest}=\frac{L_{move}}{\sqrt{1-\frac{v^2}{c^2}}}$

Solving for v:

$v=c\sqrt{1-\left ( \frac{L_{move}}{L_{rest}}\right )^2}= c\sqrt{1-\left ( \frac{3}{5}\right )^2}= c\sqrt{1-\left ( \frac{9}{25}\right )}$

$= c\sqrt{\left( \frac{16}{25}\right )}=\frac{4}{5}c=0.8c$

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Question

A meter stick, at an angle of 30 degrees with the x-axis, is traveling at 0.6_c_ in the direction of the positive y-axis. To a stationary observer, how long does the meter stick appear to be?

Answer

Length contraction only occurs in the direction of motion. This means that the x component of the length, which is cos(30), does not change; length contraction only occurs the the y component, which is sin(30).

First, we find the Lorentz factor:

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-\frac{9}{25}}}=\frac{1}{\sqrt{\frac{16}{25}}}=\frac{5}{4}$

Next, we apply the time dilation equation to the length in the y direction:

$L'_y=L_{y}/ \gamma=\frac{4}{5}\sin(30)=\frac{2}{5}$

Finally, we find the total length by combining the length-contracted y component and the unchanged x component:

$L=\sqrt{L_x^2+L'_y^2}=$ $\sqrt{\frac{3}{4}+\frac{4}{25}}=\frac{\sqrt{91}}{10}$

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