Two rockets A and B (they have an identical length) moving uniformly in the same
ID: 2029979 • Letter: T
Question
Two rockets A and B (they have an identical length) moving uniformly in the same direction in interstellar space. An observer outside the rockets measures the speed of rocket A to be VA 0.9c and the speed of rocket B to be VB 0.7c. At t 0 according to the clocks on board A and B as well as the observer, the front of both rockets line up together. They move towards a finish line a distance R 2000 km away, as measured by the observer. A. The observer uses a movie recorded in his frame to determine by how much time rocket A has won the race. How much time elapses on the movie from the time rocket A crosses the line, and the time rocket B crosses the line? B. What is the relative speed of rocket B, as measured by rocket A? Provide the answers in terms of c C. How much time elapses between the time rocket A and rocket B cross the line, according to the observers on rocket A? D. How much time elapses between the time rocket A and rocket B cross the line according to observers on rocket B? E. Do observers A and B agree with the observer in their measurements? If so, why? If not why not?Explanation / Answer
(a) The movie is in the frame of the observer.
Therefore,
tA = 2*106 / 0.9c
tB = 2*106 / 0.7c
Thus time difference,
Delta t = tB - tA = [2*106 / 0.7c] - [2*106 / 0.9c]
by using value of c = 3 * 108 m/s
(b) To measure velocity of B in frame of rocket A.
relative velocity
u = [vA - vB] / [ 1 - vAvB] = [0.9c - 0.7c] / [1 - (0.9c*0.7c)]
relative speed of rocket B as measured by rocket A = 0.54 c
(c) in the A frame, the length is contracted.
gamma = 1/sqrt[1- 0.542] = 1.1886
Therefore new contracted lenght for B
LBA = 2*106 /1.1886 = 1.68*106 m
implies the contracted length and thus the time taken is the same as calculated in part (c)
(e) observers agree with their measurement. The relative speed between the two non-inertial observers is the same.
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