Posted by **Andy ** on Friday, April 16, 2010 at 11:40am.

Spaceship I, which contains students taking a physics exam, approaches Earth with a speed of 0.640 c, while spaceship II, which contains instructors proctoring the exam, moves away from Earth at 0.280 c, as in Figure P26.27. If the instructors in spaceship II stop the exam after 65 min have passed on their clock, how long does the exam last as measured by (a) the students (b) an observer on earth

I've been trying relative velocity addition followed but time dilation but have not gotten the problem right so far

- physics -
**drwls**, Friday, April 16, 2010 at 1:12pm
(a) The relative velocity of the students with respect to the proctors can be calculation from the relative velocity addition formula

Vsp = (Vs + Vp)/[1 + VsVp/c^2]

= 0.92c/1.1792 = 0.7802c

Rest-frame exam time as measured by the students will be longer than 65 minutes, as determined by the gamma factor for that relatve velocity.

1/sqrt[1 - (.7802)^2] = 1.599

Which makes the actual exam time 104 minutes.

(b) Apply the gamma factor with a (earth-proctor) relative velocity of

Vep = 0.28 c.

That would give you 65 minutes * 1/sqrt[1-(.28^2)] = 67.7 minutes

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