Spacecraft I, containing students taking a physics exam, approaches the Earth wi

Question

Spacecraft I, containing students taking a physics exam, approaches the Earth with a speed of 0.6c (relative to the Earth), while spacecraft II, containing professors proctoring the exam, moves at 0.3c (relative to the Earth) directly toward the students. The professors stop the exam after 50 min have passed on their clock.

(a) What is the speed of the professors relative to the students?

(b) How long does the exam last as measured by the students?

(c) How long does the exam last as measured by an observer on Earth?

I have some vague idea for b and c, but a is really what I need help with, since I don't really know if I'm using the right relative velocity equation for it.

Explanation / Answer

a)

the velocity of the professors as viewed by students is:

v’ = [v – u]/[1 – vu/c2] = [– 0.6 – 0.3]c/[1 + (0.6)(0.3)] = - 0.763 c

b) = 1/[1 – (v/c)2] = 1/[1 – (0.763)2] = 1.546

The 50 min in professors’ frame is the proper time, so in students’ frame, t = to = 1.546 (50) = 77 min

c) velocity of the earth as viewed by professors is – 0.3 c

= 1/[1 – (v/c)2] = 1/[1 – (0.3)2] = 1.048

in earth’s reference frame, t = to = 1.048 (50) = 52 min

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