Finding relative velocity when there are multiple unknowns

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The problem involves calculating the relative velocity of two spaceships passing each other, with spaceship A taking 1 microsecond to traverse the length of spaceship B, which is 300 m. The contracted length of spaceship B is expressed using the Lorentz factor, leading to a complex equation with the unknown relative velocity (v) on both sides. A simplification is possible by recognizing that the distance divided by time (300 m/1 microsecond) equals the speed of light (c), allowing the equation to be rearranged. Ultimately, the calculated relative velocity is approximately 2.12 x 10^8 m/s, or about 70% the speed of light. This example highlights the utility of convenient numbers in relativity problems for easier calculations.
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Homework Statement



Two spaceships, each measuring 300 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 A requires 1 microsecond to traverse the full length of B. What is the relative velocity of the two spaceships (in units of the speed of light)?

Homework Equations



In this course, we are not using gamma in the Lorentz factor...just sqrt(1-v^2/c^2)



The Attempt at a Solution



I know that this is a relativity problem involving length contraction. There are two unknowns - relative velocity (v) and the contracted length of Spaceship B. Using the Lorentz factor, I know that the contracted length (L) of spaceship B is: L=300m(sqrt(1-v^2/c^2)). Because velocity is distance/time, I also know that v = L/1microsecond.
I substituted L=300m(sqrt(1-v^2/c^2)) for L in this second equation, to give:

v = ((300m(sqrt(1-v^2/c^2)/1 microsecond)

Unfortunately, here I end up with a messy equation with v on both sides, and I'm having trouble solving for v. Am I just not doing the algebra right? Is there a way to simplify this?
 
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You've done good so far. And yes, it is a bit messy, but if you rearrange, then you can solve for v.

Or there is a short-cut, which you wouldn't ordinarily be able to use, but it is possible in this case because of the numbers which they have given you. If you look at the numbers, you've got 300m / 1 microsecond so what constant could you divide this by to get a nice number?
 
Oh, ok, thanks! 300m/1 microsecond is the speed of light (c), so I can just separate that out...so v = c (sqrt(1-v^2/c^2). Solving for v, I get 2.12*10^8 m/s relative velocity, or about 70% the speed of light. ? Thanks so much!
 
yep, that's right :) Often in exams, they do use convenient numbers so that you can use short-cuts such as this, but you can also practise rearranging the equation, in case they give you a question with not-convenient numbers.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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