- #1

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## Homework Statement

At exactly 00:00:00 hours, a group of convicts escape from a planet in a space-ship that travels at speed

##v=\frac{4}{5}c##.

After 11 min, a patrol spaceship goes after them with ##v_P=\frac{24}{25}c##.

Ignore all acceleration periods.

(i) The convicts immediately notice the patrol spaceship taking off, and release a stealth missile to destroy it. The missile has speed

##v_M=\frac{40}{41}c##

with respect to the convicts. Show that, when the missile hits the patrol ship, the time on a watch worn by the pilot of the patrol ship is 00 : 23 : 36.

## Homework Equations

##u_x' = \frac{u_x-v}{1-\frac{u_xv}{c^2}}## (1)

##t' = \gamma (t - \frac{vx}{c^2})## (2)

## The Attempt at a Solution

I should say this is from a past paper, so it won't get marked. There are way too many frames! I'm going to assume that all the speeds given are 'proper' speeds.

##u_x=\frac{40}{41}##, I think v is the speed of one frame relative to the other so that might be ##\frac{24}{25}-(-\frac{40}{41}) = \frac{1984}{1025}##. Then I realized that ##u_x'## might not be useful anyway, I was going to just use speed = distance/time, but can't do that because I don't know distances. And (2) probably isn't useful either. Do I try and work out at what time in the ship's frame the missile hits, and transform that? But then do I need to transform the 40/41 to see what the missile's speed is in the ship's frame? Don't I need distances for this approach as well?