# Time dilation/Length contraction

## Homework Statement

Two powerless rockets are on a collision course. They are moving with speeds 0.800c (ship 1) and 0.600c (ship 2) and are initially 2.25*10^12 m apart. The ships are both 50m in length as observed by a stationary observer on earth.

a) what are their respective proper lengths?
b) what is the length of each rocket as measured by an observer on the other rocket?
c) How long before the rockets collide according the observer on earth, rocket 1 and rocket 2?

## Homework Equations

(1) L=Lp(1-v^2/c^2)^(1/2) (Lp = proper length)
(2) t'=(1/(1-v^2/c^2)) (t-v/c^2x) (Lorentz time transformation
(3) u'=(u-v)/(1-(uv/c^2)) (Relativistic velocity transformation)

(apologies for the clumsy notoation)

## The Attempt at a Solution

I've solved a and found the proper lengths to be 83.33m for ship 1 and 62.50 for ship 2.

I have also solved b, though i'm not 100% sure of my answer:

Using the Lorentz velocity transform (equation 3 above)
(-.6c-.8c)/[1-(-.6c*.8c)/c^2] = -2.838*10^8 m/s
I've taken ship 1 as reference frame S' and ship 2 is travelling in the -ve x direction according to 1, hence the -.6c and the -ve answer.
Substituting -2.838*10^8 into the Length contraction formula (eq 1) the length of ship 1 obswerved by ship 2 was 27.03m and ship2 observed by ship1 was 20.27m.
Is this correct?

As for part c, i am not sure how to apply the Lorentz transform in each case... In fact, i'm very confused! Is equation 2 (above) the correct one to apply?

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Also, is a question like this considered advanced undergrad? Or would i be better posting it in the other thread?

$$t' = \gamma\left(t - \frac{xv}{c^2}\right)$$