A.) Two spaceships are traveling, one in front of the other, in the same directi

Question

A.) Two spaceships are traveling, one in front of the other, in the same direction, at the same speed of 1.60 108 m/s relative to the earth. A laser pulse is emitted by one of the ships and is reflected back by a mirror on the other ship. The total round trip time for the laser pulse is 4.24 10-3 s. How long would it take for a small landing module to travel between the ships if its speed, relative to the ships, is 1.09 103 m/s

B.) A spaceship with a proper length of 104 m passes an observer at a speed of
0.99105c.
What is the length of the ship as seen by the observer?

c.) How fast must a meter stick move relative to an observer so that she would measure its length to be 1.5 cm?

d.) Spaceships A and B move in opposite directions at the same speed of
0.694c
relative to earth, with A moving away from earth and B moving toward earth. Find the velocity of B relative to A

Explanation / Answer

A. is a trick question, they are moving at relativistic speed with respect to the earth, but motionless with respect to each other. Since it takes 2.12 * 10-3 s for a laser pulse to go from one to the other, they are (3.00 * 108 m/s)(2.12 * 10-3 s) = 6.355 * 105 m apart. It will take (6.355 * 105 m)/(1090 m/s) = 583 s to make the trip. No relativity, normal kinematics.

B. Since v/c = 0.99105, = 7.491 so measured length is 104m / = 13.88 m

C. We want to be 100/1.5 = 66.67; = 1/(1 - v2/c2)1/2

1 - v2/c2 = (1/66.67)2 = 0.0152 = 0.000225

v2/c2 =  0.999775

v= 0.999887c

D. Don't really have time right now, but it's simple relativistic velocity addition.

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