Special Relativity clarification required

[PsychonautQQ]

Homework Statement

Two spaceships, each 100m long when measured at rest, travel toward each other, each with a speed of .85c relative to the earth. At time t=0 on earth, the front ends of the ships are next to each other as they just begin to pass each other. At what time on Earth are their backs next to each other?

Homework Equations

L = 1/γ * Lp
t = L/u

The Attempt at a Solution

So the length of each ship from Earth's frame is 1/γ * Lp which comes out to 53 meters, and then apparently the time it takes for their back ends to be together is the time it takes either spaceship to move the length of the spaceship in Earth's frame. So it would be t = L/u which comes out to 2.1e-7 seconds.
The book says this is the correct answer, but I don't understand. Why do you only need to calculate the time it takes for one spaceship to travel the length of a spaceship from Earth's frame? Shouldn't the fact that they are both moving towards each other from Earth's frame mean something?

wtfz
Special relativity is confusing me in general actually... can somebody explain the relativistic velocity transformation equations to me??

ux = (u'x + v) / (1+ (vu'x/c^2))

and yet...

uy = u'y / γ(1+ (vu'x/c^2))

how come the relative velocity of the particle in the x direction effects it's velocity transformation in the y direction? I know it has to do with the fact that the frame S and S' are moving apart from each other in the x direction but I can't quite connect the dots for making sense of this equation, anyone think they can enlighten me?

 

[TSny]

Why do you only need to calculate the time it takes for one spaceship to travel the length of a spaceship from Earth's frame? Shouldn't the fact that they are both moving towards each other from Earth's frame mean something?

A picture in the Earth frame showing the locations of the ships when their front ends are coincident and then when their back ends are coincident should make this clear.

how come the relative velocity of the particle in the x direction effects it's velocity transformation in the y direction? I know it has to do with the fact that the frame S and S' are moving apart from each other in the x direction but I can't quite connect the dots for making sense of this equation, anyone think they can enlighten me?

u_y = Δy/Δt and u'_y = Δy'/Δt'

Although Δy = Δy', Δt differs from Δt' due to the relative motion of the frames in the x direction. It might help to review the derivation of the velocity transformation equations.

 

[Chestermiller]

Homework Statement

Two spaceships, each 100m long when measured at rest, travel toward each other, each with a speed of .85c relative to the earth. At time t=0 on earth, the front ends of the ships are next to each other as they just begin to pass each other. At what time on Earth are their backs next to each other?

Homework Equations

L = 1/γ * Lp
t = L/u

The Attempt at a Solution

So the length of each ship from Earth's frame is 1/γ * Lp which comes out to 53 meters, and then apparently the time it takes for their back ends to be together is the time it takes either spaceship to move the length of the spaceship in Earth's frame. So it would be t = L/u which comes out to 2.1e-7 seconds.
The book says this is the correct answer, but I don't understand. Why do you only need to calculate the time it takes for one spaceship to travel the length of a spaceship from Earth's frame? Shouldn't the fact that they are both moving towards each other from Earth's frame mean something?

wtfz
Special relativity is confusing me in general actually... can somebody explain the relativistic velocity transformation equations to me??

ux = (u'x + v) / (1+ (vu'x/c^2))

and yet...

uy = u'y / γ(1+ (vu'x/c^2))

how come the relative velocity of the particle in the x direction effects it's velocity transformation in the y direction? I know it has to do with the fact that the frame S and S' are moving apart from each other in the x direction but I can't quite connect the dots for making sense of this equation, anyone think they can enlighten me?

The problem is asking how long it takes the ships to pass each other as reckoned from the Earth's frame of reference. That is why the focus is on the Earth's frame of reference, and not the frames of reference of the two ships. You analyzed the problem flawlessly.