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Special relativity problem

  1. Dec 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Two students on the ground, separated 100m from each other, launch one rocket each vertically. The rocket explode at the same time for an observer, on the ground, that's midway between the rocket explosion. At the exact time the observer sees the rockets exploding a spaceship flies over his head, horizontally, at a speed of 0.7c. The ship length as seen by the observer is 15m.

    (c)How much time goes by, to a tripulant of the spaceship, for the ship to complete the 100m between the two students?
    (d)For a tripulant of the space ship is the explosion of the rockets simultaneous? What's the time interval? (time between the first rocket explosion and the second, as seen by the tripulant)


    2. Relevant equations

    After solving the first two questions I've reached this data:
    Actual distance between the two students = 71.4m
    Spaceship length as measured by the tripulant = 21m

    As to question (c) I used the equation t=d/v to solve it.

    3. The attempt at a solution

    (c) I think I've done her correctly. I've simply divided the length between the two students as seen by a tripulant in a ship by her speed:
    t=71.4/0.7c=3.4*10^-7

    (d) I know the explosions are not simultaneous, however I have no idea how I'm supposed to measure the time interval between the two explosions... If anyone could give me some hint, or some equation that could help me i'd be grateful. I've readen the whole relativity chapter of Serway's Physics for Scientists looking for something to help me without any luck.

    Thanks, and a happy new year to everyone.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 29, 2011 #2

    Nugatory

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    Staff: Mentor

    For part d:
    The time interval between any two events, such as the two explosions, is the difference between the times at which they happen... And those times may be different for different observers, which is why things that are simultaneous (happened at the same time, so difference is zero) for one observer may not be simultaneous for another observer.

    So you need to figure out the time of the two explosions, as observed from the spaceship. Then you can calculate the interval between those two times.
     
  4. Dec 30, 2011 #3
    I understand that, is the how that I'm having problems with :|
     
  5. Dec 30, 2011 #4
    Since light will travel at the same speed from both sources the ratio of the times is the same as the ratio of the distances, so can you find the distance the ship is from each explosion when they go off?
     
  6. Dec 30, 2011 #5
    When the explosion go off for the observer the ship is above his head, with the same x position
     
  7. Dec 31, 2011 #6
    If by actual you mean distance b/w boys in spaceship frame ....

    then why is d = 71.4 and not 71.4 + 21 :confused:
    and 0.7c in in ground frame :confused::confused:

    And wont this be a better way:
    find time in ground frame and them convert it for spaceship ...
     
  8. Dec 31, 2011 #7
    when you move, time slows down, but 0 time difference should remain 0

    even in time dilation formula, put any 1 time 0 and other becomes 0 ...
     
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