Solved: IP In Conceptual Checkpoint 29-2 we considered an

Chapter 29, Problem 93GP

(choose chapter or problem)

IP In Conceptual Checkpoint 29-2 we considered an astronaut at rest on an inclined bed inside a moving spaceship. From the point of view of observer 1, on board the ship, the astronaut has a length \(L_{0}\) and is inclined at an angle \(\theta_{0}\) above the floor. Ob-server 2 sees the spaceship moving to the right with a speed \(v\)

(a) Show that the length of the astronaut as measured by observer 2 is

                                           \(L=L_{0} \sqrt{1-\left(\frac{v^{2}}{c^{2}}\right) \cos ^{2} \theta_{0}}\)

(b) Show that the angle \theta\)

 the astronaut makes with the floor of the ship, as measured by observer 2, is given by

                                           \(\tan \theta=\frac{\tan \theta_{0}}{\sqrt{1-v^{2} / c^{2}}}\)

Equation Transcription:

Text Transcription:

L_0

phi_0

v

L=L_0 sqrt 1-(v^2/c^2))cos^2 theta_0

theta

tan =tan theta_0 sqrt 1-v^2/c^2