# Addition of Velocities Special Relativity Problem

[Mentor's note: this post does not use the homework template because it was originally posted in a non-homework forum.]

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A ship (attached to reference frame S') is passing us (standing in reference frame S) with velocity v= 0.933c. A proton is fired at speed 0.959c relative to the ship from the front of the ship to the rear. The proper length of the ship is 756 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 (a) a passenger in the ship and (b) us? Suppose that, instead, the proton is fired from the rear to the front. What then is the temporal separation between the time it is fired and the time it hits the front wall according to (c) the passenger and (d) us?

Attempt at a solution:
a) v=0.933c
u'=-0.959c
=> gamma = 3.52851
=> L'=214.255m
By time = distance/speed, temporal separation is 0.745microseconds

b)
v=0.933c
u'=-0.959c
Using velocity addition formula => u=-0.247024c
=> gamma = 1.03198
=> L'=712.571m
By time = distance/speed, temporal separation is 9.89microseconds

c) Same as a?

d) v=0.933c
u'=+0.959c
Using velocity addition formula => u=0.9985c
=> gamma = 26.2631
=> L'=28.7856m
By time = distance/speed, temporal separation is 0.136microseconds

Would love some hints on where I am going wrong, these are all incorrect.

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Chestermiller
Mentor
With regard to part (a), suppose you decided to measure the speed of the proton. So you would fire the proton from the front of the ship x' = 0 at t' = 0, and you would have another guy at the back of the ship at x' = -756 m record the time t' = Δt' at which the proton arrives. You would then determine that the speed of the proton was 756/Δt' m/s. This is how you reckon velocity from the frame of reference of the ship. So, what is the value of Δt' that would be measured?

Chet

Would length contraction not occur? Or do I simply need to multiply my time by gamma to account for time dilation?