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Simple Relativity

  1. Dec 20, 2011 #1
    In frame S, two balls are travelling along the x axis directly to each other, both have the same speed c/2. They are 150x10^8m apart. being that c=3x10^8ms^-1
    a) in frameS, how long does it take to reach each other?
    b) given the answer to a), at what speed is the distance between the
    objects decreasing?
    c) one objects is at the rest frmae, how long does it take the objects to reach each other if thery are 150x10^8m apart?
    d)Now in frame S the two objects are again going towards each other, but now with different speed (v,u). They start "d" away from each other?
    In the same frame, how long do they take to reach each other?

    The way I solved this problem was
    a) v=d/t, therefore t=d/v, so i plugged in (150x10^8/2)/(c/2)
    b) I'm struggling. In this case i wrote down c, because is asking the whole distance so i just added both speeds (c/2+c/2). Am i right? can someone explain in details
    c)since it's a frame in rest and the other moving I used t'=γ(t-(v/c^2)x) and i plugged in the numbers since i know t,distace(150x10^8) and v(c/2).
    d) I'm lost. I know that if my b is correct the speed is going to decrease by that amount but then how should I approach this problem?
     
  2. jcsd
  3. Dec 20, 2011 #2

    BruceW

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    You've done part a) correctly. You are also correct with part b). It is simple addition (as you've done), because it is all within one frame of reference. You would only have to use a complicated formula if you were using velocities in different frames of reference.
     
  4. Dec 20, 2011 #3

    BruceW

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    This question is a bit more difficult. To start with, is it supposed to represent the same system as before, just according to a different frame of reference? Also, with respect to what frame of reference are they 150x10^8m apart? Is it this reference frame?
     
  5. Dec 20, 2011 #4
    Yes "is it supposed to represent the same system as before, now according to a different frame of reference". The question is the same as how long does it take the objects to reach each other if they start 150x10^8m apart in the frame rest frame of one of the objects. I just assumed they were different frames I used Lorentz equation for t'.
     
  6. Dec 21, 2011 #5

    BruceW

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    You shouldn't be using Lorentz transformation. The time difference for part c) is not just a Lorentz transform of the time difference for part a). This is because the events representing the 'initial' space-time coordinates of the balls are different for parts a) and c). In other words, its not just a change of frame of reference; we are also talking about different events.

    There is another equation you should use. In the rest frame, you know one ball is stationary, so what do you need to find out about the other ball to get the answer?
     
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