Three space ships are seen flying past your classroom on a space station. When m

Question

Three space ships are seen flying past your classroom on a space station. When moving, Ship 1 is measured to be only 75% of its length compared to when it is at rest at the space station. Ship 2 is moving away from the space station at 0.4c and Ship 3 is moving toward the space station (in the opposite direction as Ship 2) at 0.5c. How fast is Ship 1 moving relative to the space station? After 5 hours pass on your watch, how much time do you measure to have passed on Ship 2's clocks? How fast does Ship 2 measure Ship 3 to be moving?

Explanation / Answer

1) c) 0.66c

Lo ---> length of ship when it is at rest.

L ---> length of ship when it is moving.

Apply Length contraction equation

L = Lo*sqrt(1 - (v/c)^2)

0.75*Lo = Lo*sqrt(1 - (v/c)^2)

0.75 = sqrt(1 - (v/c)^2)

0.75^2 = 1 - (v/c)^2

(v/c)^2 = 1 - 0.75^2

v/c = sqrt(1 - 0.75^2)

v = c*sqrt(1 - 0.75^2)

= 0.66*c

2) c) 4.58 hours

delta_t' = delta_t/sqrt(1 - (v/c)^2)

here, delta_t ---> proper time in the frame of reference moving with the closk

delta_t'----> relative time measured in laboratory.

given, delta_t' = 5 hours

delta_t = ?

delta_t = delta_t'*sqrt(1 - (v/c)^2)

= 5*sqrt(1 - (0.4)^2)

= 4.58 hours

3) c)0.75c

use relativistic addiation

relative speed between the two ships = (u + v)/(1 + uv/c^2)

= (0.4*c + 0.5*c)/(1 + 0.4*c*0.5*c/c^2)

= 0.75*c

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