Special relativity and really fast trains

In summary, the question involves two passenger trains, A and B, passing a railroad platform in Winnipeg. They are traveling in opposite directions at equal speeds of 0.665c with respect to the ground. Part A asks for the speed of train B from the point of view of a passenger on train A, which is found to be 0.922c. Part B asks for the time it takes for train B to pass the passenger on train A, which is found to be 3.36e-7 seconds, using the velocity equations and the concept of length contraction.
  • #1
JJK1503
12
0

Homework Statement



Two passenger trains A and B, each 240 m long, pass a 60 m long railroad platform in Winnepeg. The trains are moving in opposite directions at equal speeds of 0.665c with respect to the ground. Train A is traveling west and all tracks are perfectly straight.

A) From the point of view of a passenger on train A, how fast is train B moving? (Give your answer as a fraction of the speed of light, e.g. if you get 0.952c, you enter 0.952.)

B) How long does it take train B to pass the passenger on train A?

Homework Equations



u' = ( u - v ) / ( 1 - ( u v ) / c^2 )

u = ( u + v ) / ( 1 + (u v ) / c^2 )

gamma = 1 / sqrt [ 1 - ( v / c ) ^2 ]

l = l_proper / gamma

The Attempt at a Solution



I am studying for a final exam. The above was part of a multipart homework question. I got everytjing correct except for what is shown above. The homework has been returned and I know that the answer to part A is .922 and the answer to part B is 3.36e-7 s. I can't figure out how to get there.

First I looked at the Galilean method. This put me at something like 1.32 c. obviously this is wrong and I didn't even attempt it.

Then I tried to use the velocity equations stated above using u' and got u' = 0 / something

I then figured out part a by using the u equation and substituting the values u = v = .665c. This got me to the .922 I was looking for.

For part b I am still stuck. I thought what I should do is use this new speed I found in part a to calculate a new gamma

from the gamma equation above; using v = .992c I find gamma = 2.59

then I calculate length dilation of the train using gamma = 2.59 and l_proper = 240 m to find l = 92.82 m.

then I use t = d / v = 112.15 / .665c = 4.65e-7.

I really need some conceptual help on this one.

I appreciate any response.
 
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  • #2
Kudos for your stamina! You did well on part a. The two expressions are in fact one: u' expression is fine but own speed is u and other train speed is -v (it's moving in the other direction).

In part b you use the right v to find gamma. All you have to do is use the same v on the contracted(*) length of 92.82 m.

( (*) in general we speak of time dilation and length contraction)
 
  • #3
BvU said:
Kudos for your stamina! You did well on part a. The two expressions are in fact one: u' expression is fine but own speed is u and other train speed is -v (it's moving in the other direction).

In part b you use the right v to find gamma. All you have to do is use the same v on the contracted(*) length of 92.82 m.

( (*) in general we speak of time dilation and length contraction)

Ahhh. I plugged in the old v. thanks for your help.
 

1. What is special relativity and why is it important in the context of really fast trains?

Special relativity is a theory proposed by Albert Einstein in 1905 that describes the relationship between space and time for objects moving at constant speeds. It is important in the context of really fast trains because it explains how time and space are affected by the speed of the train, and how this affects our perception of events.

2. How does special relativity affect the perception of time for passengers on a really fast train?

According to special relativity, time is relative and can appear to pass at different rates depending on the observer's speed. This means that passengers on a really fast train will experience time passing slower compared to someone standing still on the ground. This effect is known as time dilation.

3. Is it possible for a really fast train to travel at the speed of light?

No, according to special relativity, the speed of light is the maximum speed at which any object can travel in the universe. As the train approaches the speed of light, its mass increases and it would require infinite energy to continue accelerating.

4. How does special relativity explain the phenomenon of length contraction in really fast trains?

Special relativity states that as an object's speed increases, its length in the direction of motion appears to decrease. This is known as length contraction. For passengers on a really fast train, the train appears to be shorter than for someone standing still on the ground.

5. Can special relativity be applied to other forms of transportation, such as airplanes?

Yes, special relativity is a fundamental theory that applies to all objects in motion, regardless of the mode of transportation. However, the effects of special relativity are only noticeable at very high speeds, so they may not be noticeable for passengers on a commercial airplane.

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