Physics Problem : Relativity (Length Contraction)
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Physics Problem : Relativity (Length Contraction)

[From: ] [author: ] [Date: 11-05-25] [Hit: ]
)-Since I cant really draw a diagram, this is going to be close to impossible to understand, but here goes.Lets take the meterstick that Bob has and draw a triangle around it so that the x side is next to the 60˚ angle and the y side is the other.Because of how relativity works, only the x side,......
I do not understand how to do relativity problems (for length contractions) with angles? Can you please help me?

Bob is flying at some speed v (relative to John) in a spaceship, holding a 12 meter (rest length) flagpole at an angle of 60 degrees with the horizontal. John, at rest in space, measures Bob's flagpole to be 11 meters. How quickly is Bob going relative to John? (Draw the picture first. Label both Bob and John.)

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Since I can't really draw a diagram, this is going to be close to impossible to understand, but here goes.
Let's take the meterstick that Bob has and draw a triangle around it so that the x side is next to the 60˚ angle and the y side is the other. Because of how relativity works, only the x side, moving along the same dimension as Bob will experience relativistic effects. Therefore, we need only calculate the relativistic effects on the x side. We know that y will always be 6√3 because it doesn't change and this is a 30-60-90 triangle. However, x will change from 6 to something so that the length of the meterstick changes. So, here comes the hard part.

a^2 + b^2 = c^2 --> (6*√1-(v^2/c^2)) + (6√3)^2 = 11^2 --> (36 * (1-(v^2/c^2))) = 13 --> 1-(v^2/c^2) = 13/36
Multiply everything by c^2: c^2 - v^2 = (13/36)*c^2 --> c^2 - (13/36)c^2 = v^2, solve for v
v = √(23/36)c^2 = 2.398e08

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The pole contracts in the x direction only:

Triangle altitude a = 12*√3/2; Hypotenuse h = 11
base b = √[11² - (12√3/2)²] = 3.605
Original base bo = 12*sin30 = 6.0

Contraction ratio γ = 3.605/6 = .600 = √[1 - v²/c²] → v = c*√[1 - .600²]

v = 2.4E8 m/s
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