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3 years ago

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Spaceship A is 10 meters long and approaching you from the south at a speed of.7c while spaceship B, which is also 10 meters lon

g, is approaching you from the north at a speed of.7c. a. What is the relative speed of A as measured by B? b. What is the length of A as measured by you? c. What is the length of A as measured by B? d. An event on A lasts 1 second as measured by A. How long does the event last as measured by you? e. An event on A lasts 1 second as measured by A. How long does the event last as measured by B?

1 answer:

Valentin [98]3 years ago

3 0

Answer:

a) 0.94 C

b) 7.14 m

c) 3.11 m

d) 1.40 s

e) 2.93 s

Explanation:

First we need to set up a coordinate system. This will have the positive X axis pointing north. So spaceship A has positive speed, and spaceship B has negative speed.

The Lorentz transformation for speed is:

$u' = \frac{u - v}{1 - \frac{u*v}{c^2}}$

u: speed of spaceship A as observed by you

v: speed of spaceship B as observed by you

In the case of the speed of spaceship A as observed by spaceship B:

$u' = \frac{0.7c - (-0.7c)}{1 - \frac{0.7c*(-0.7c)}{c^2}} = 0.94c$

The transform for lengths is:

$L = L0 * \sqrt{1 - \frac{v^2}{c^2}}$

For the case of spaceship A as observed by you:

$L = 10 m * \sqrt{1 - \frac{(0.7c)^2}{c^2}} = 7.14 m$

For the case of spaceship A as observed by spaceship B:

$L = 10 m * \sqrt{1 - \frac{(0.94c)^2}{c^2}} = 3.12 m$

The time dilation equation is:

$T = \frac{T0}{\sqrt{1-\frac{v^2}{c^2}}}$

For the case of the event as observed by you:

$\frac{1 s}{\sqrt{1-\frac{(0.7c)^2}{c^2}}} = 1.40 s$

For the case of the event as observed by spaceship B:

$\frac{1 s}{\sqrt{1-\frac{(0.94c)^2}{c^2}}} = 2.93 s$

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Read 2 more answers

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