Special Relativity Ship Problem

[harrietstowe]

Homework Statement

A ship (attached to reference frame S') passes us (we are standing in reference frame S) with velocity = 0.920c. A proton is fired at speed 0.975c relative to the ship from the front of the ship to the rear. The proper length of the ship is 775 m.
What is the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to us?

Homework Equations

u'=(u-v)/(1-((uv)/c^2))
Length contraction:
L=L_o/gamma
t=x/v

The Attempt at a Solution

I thought to let u be .975c
Let v be -.920c
solve for u' and get 2.99684e8 m/s
Observers on Earth will see a shorter length for the ship that is 775m/gamma
gamma = 1/sqrt(1-(.920^2)) and so the contracted length is 303.737m
This contracted length is x and divide it by u' to get 1.014 μs
This though was not the right answer.
Thanks

 

[vela]

You're not accounting for the fact that the rear end of the ship is moving forward toward the proton, so in S', the proton doesn't actually have to travel the entire distance equal to the contracted length of the ship.

Are you familiar with the Lorentz transformations? I usually find problems like these are easier to do using the transformations (not that there's anything wrong with your approach either).

 

[harrietstowe]

If I am interested in the velocity of the proton relative to observers on Earth I am struggling to see why we would have to take a length contraction into account for the proton. If I could get the velocity of the proton for the Earth observers I think i would have this solved and yes I am very familiar with Lorentz transformation equations.

 

[vela]

That's not what I'm saying. Suppose you and I were 10 meters apart and you throw a ball to me at with a horizontal velocity of 10 m/s. If we were both standing still, I'd catch the ball at t=1 s. If I'm running toward you, however, I will catch the ball before 1 second has elapsed, right?