Posted by Andy on .
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
(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