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stefan10
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Homework Statement
(d) A spaceship moves with velocity (4/5)c relative to Earth. The astronaut throws his empty beer bottle out the window with velocity (3/5)c relative to the ship in the sideways direction. What is the velocity of the bottle (magnitude and direction) relative to the Earth (i) classically, (ii) relativistically?
Homework Equations
Velocity Addition (relativistic)
[tex]u_x = \frac {u_x' + v}{1 + \frac{v}{c} \frac{u_x'}{c}} [/tex]
[tex]u_y = \frac {u_y' + v}{\gamma (1 + \frac{v}{c} \frac{u_x'}{c})} [/tex]
Pythagorean Theorem
The Attempt at a Solution
I just want to know if I've done this correctly. This is a problem on a practice exam.
For the classical portion I just used the Pythagorean theorem [tex] \sqrt {(\frac{4c}{5})^2 + (\frac{3c}{5})^2} = c [/tex]
For the relativistic portion I use the Velocity Addition equations. I use [tex]v = \frac{4}{5}c = u_x' [/tex]and [tex]u_y' = \frac{3}{5}c [/tex]
This gives:
[tex]u_y = \frac{15}{41}c [/tex]
and using the Pythagorean theorem with u_x = (4/5)c I get [tex] \frac{13}{15}c = .866 c [/tex]I am embarrassed to say that I only think it's the right answer because it looks like it would be. Conceptually I have no idea why I would set both u_x' and v equal to 4/5 c. Can anybody explain if I did get it correct, and if I didn't how I might get the correct answer? Thank you!