Question

3. (6 marks) A rod of length L'-4m is moving in frame y s S' with respect to a stationary observer in frame S. The rod makes an angle e' 60 to the direction of travel in the S' frame. What is the length and orientation of the rod as measured by a stationary observer in the S frame if the relative speed of the frames is อ' 0.5c as shown in the diagram.

Answer 1

Given:-

Frame S', in which the rod ( Length L' = 4m ) is moving with an orientation of 60 degrees with respect to direction of travel

Frame S, in which there is a Stationary Observer.

According to Special theory of Relativity,

The length of the rod as observed by the Stationary observer in the S frame is as follows

$L=L'\left ( 1-\frac{v^{2}}{c^{2}} \cos^{2} \Theta '\right )^{\frac{1}{2}}$

Sustituting the values L'=4m,v=0.5c, $\Theta '=60^{o}$ ,

$L=4\left ( 1-\frac{(0.5c)^{2}}{c^{2}} \cos^260^{o}\right )^{\frac{1}{2}}$

$L=4\left ( 1-0.25 \times \frac{1}{4}\right )^{\frac{1}{2}}$

$L=3.873$ m

The Orientation of the rod is as follows:-

$\Theta =\arctan (\gamma \tan \Theta ')$ , here $\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

Substituting the value of v=0.5c $\gamma =1.154$

putting the value   $\gamma =1.154$ and $\Theta '=60^{o}$

$\Theta =\arctan (1.154\times \tan 60^{o})$

$\Theta =\arctan (1.154\times {\sqrt{3}})$

$\Theta =1.107 rad$