Solving Problem 2-8: Relative Velocity and Time Elapses

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SUMMARY

The discussion revolves around Problem 2-8 from David Hogg's book, which involves calculating the relative velocity and time elapsed for two spaceships traveling in opposite directions. The calculated relative velocity of the two spaceships is 20,000,000 m/s, determined using the formula distance/time with a distance of 100 m and a time of 5x10^-6 s. For the time elapsed on spaceship B, the time dilation formula is applied, resulting in a time dilation factor of 1.00223, indicating that for every second on B, 1.00223 seconds pass on A. The clarity of the problem statement is questioned, particularly regarding the interpretation of "traversing the full length of A."

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JasonRox
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I think I'm thinking to hard for this one, so I'll wait till tomorrow before checking up with you guys. I'll let my brain freshen up first. Yes, I have a "thing" for figuring it out on my own.

Anyways, it's a question from David Hogg's book, so it is NOT my homework.

Problem 2-8

Two spaceships, each measuring 100 m in its own rest frame, pass by each other traveling in opposite directions. Instruments on board spaceship A determine that the front of spaceship B requires 5x10^-6 s to traverse the full length of A.

a) What is the relative velocity of the two spaceships?
b)How much time elapses on a clock from spaceship B as it traverses the full length of A?

Now, that I typed it out, I believe I got it now, so I'll check it out later, and see you guys in 12 hours.

Thanks.
 
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a) In the frame of A, synchronized clocks can be setup at the front and back ends of A's spaceship. Hence the speed that A measures for B is determined simply by distance / time.
b) The spaceships are symmetrical. What exactly does the phrase "traversing the full length of A" - the time it takes for the entire ship to pass by one stationary point (eg. the front) in the frame of B?
 
I thought the same for a).

Question b) What would the time be in B's Frame. B wouldn't measure 100m because of length contraction.

If we are right about a), than v is:

[tex]100m/5.0*10^-6 = 20,000,000m/s[/tex]

To find out how much time went by for spaceship B, we use the time dilation(pythagorean) formula.

[tex]\gamma \equiv \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]

If I did the calculations right, for every second that passes on B, 1.00223 seconds passes on A, relative to A.

This doesn't seem like it is the answer. The question isn't clear enough.

The above question are the EXACT words out of the book, so questions relating to what does "this" actually mean I do not know anything. This is where I think too deep, and I'm sure some of you get the same problem.

I'll move on, and if something comes up, I'll try it.
 

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