stefan10

## 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

$$u_x = \frac {u_x' + v}{1 + \frac{v}{c} \frac{u_x'}{c}}$$

$$u_y = \frac {u_y' + v}{\gamma (1 + \frac{v}{c} \frac{u_x'}{c})}$$

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 $$\sqrt {(\frac{4c}{5})^2 + (\frac{3c}{5})^2} = c$$

For the relativistic portion I use the Velocity Addition equations. I use $$v = \frac{4}{5}c = u_x'$$and $$u_y' = \frac{3}{5}c$$

This gives:

$$u_y = \frac{15}{41}c$$

and using the Pythagorean theorem with u_x = (4/5)c I get $$\frac{13}{15}c = .866 c$$

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!

Homework Helper
Gold Member
$$u_y = \frac {u_y' + v}{\gamma (1 + \frac{v}{c} \frac{u_x'}{c})}$$

This equation cannot be correct. For the case ##u_y' = 0## what would you expect ##u_y## to be?

Other than your result for ##u_y##, I think the rest of your work looks good (as far as method).

EDIT: I now see another error. You wrote
$$v = \frac{4}{5}c = u_x'$$

Shouldn't ##u_x' = 0##? Relative to the ship, the bottle moves only in the y' direction.
Maybe you meant to write $$v = \frac{4}{5}c = u_x$$. That would give you the value ##u_x## = ##\frac{4}{5}c## that you used later.

Last edited:
stefan10
Ah nevermind, I see what I did wrong the first time when I tried it with u_x' = 0. I must've been tired and reduced the fraction incorrectly, making it seem as if my final result was incorrect. Also the U_y equation I posted was not the one I used, it's a typo from copying and pasted the latex code for U_x and forgetting to delete (+v). Thank you very much, I have the correct answer now to be, (481)^(1/2)c/25 = .877 c