Yet another question on relativity

[stunner5000pt]

i posted this question a while ago but it used classical mechanics to solve it and your help in that one is greatl appreciated, now i need to solve using special relativity!

If two trains leave a station on the same track. An observer missed both these trains and is standing close to the track sees the westbound train recede at 0.6c and sees the eastbound train recede at 0.8c. There is a ticket collector on the westbound train going from the back of the train to the front at 0.4c, with respect to a passenger on the westbound train.

What would the speed of the eastbound train wit hrespect to the westbound train (call it Ur) according to:

a) Observer on the station: it's just as if the trains were approaching each other... right?

in taht case using Ux' = UW - UE / (1 - UW UE / C^2) yields 0.38c

b)Passenger seated on the westbound train?

The same framework as the previous question (which leads me to doubt part A) and i get 0.38c

c) Ticket collector on the westbound train?
The ticket collector would see hte same as the passenger, no?

Similarly what is the speed of the ticket collector :

d) According to the observer on the station?
I would think calculating the speed of the ticket collector with respect tothe train first, and then the observer to the train and then adding the velocities up

e) According to the passenger on the east train?
First find the relative velocities of the two trains and then add the velocity of the ticket collector with respect to the west train as done in the previous one

F) Relative to a passenger seated on the east train according tothe observer on the station?

STumped even more...

Please please help I'm desperate for help!

i would appreciate your help

 

[wais]

in solving this question using special relativity.Hello,

Thank you for reaching out for help with your question on relativity. It is great to see that you are exploring different approaches to solving this problem and incorporating special relativity into your analysis.

To address your first question, you are correct in thinking that the speed of the eastbound train with respect to the westbound train (Ur) would be calculated by taking the relative velocity (UW - UE) and then using the special relativity equation Ux' = (UW - UE) / (1 - UW UE / C^2). This gives a result of 0.38c, which is in agreement with your calculations.

For part B, you are correct in using the same framework as part A. The passenger on the westbound train would also see the eastbound train approaching at a relative speed of 0.38c. This is because, in special relativity, the relative velocity between two objects is the same for all observers.

Similarly, the ticket collector on the westbound train would also see the eastbound train approaching at a relative speed of 0.38c. This is because, from their perspective, they are stationary on the westbound train and the eastbound train is approaching them at 0.8c.

For part D, you are correct in thinking that the speed of the ticket collector with respect to the observer on the station would be calculated by first finding the speed of the ticket collector with respect to the train (0.4c) and then adding it to the speed of the train with respect to the observer (0.6c). This gives a result of 1c, which is the speed of light. This is an important result in special relativity, as it shows that the speed of light is the same for all observers regardless of their relative motion.

For part E, you are correct in first finding the relative velocity of the two trains (0.38c) and then adding the speed of the ticket collector with respect to the westbound train (0.4c). This gives a result of 0.78c for the speed of the ticket collector with respect to the eastbound train.

Finally, for part F, you would use the same approach as part D, but with the observer on the station instead of the ticket collector. This would give a final result of 0.78c for the speed of the ticket collector with respect