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Relative velocity problem

  1. Feb 25, 2007 #1
    Suppose two bodies A and B are moving perpendicular to each other with velocities 'a' and 'b'.

    A with 'a'
    ^
    |
    |
    |
    |
    |
    |
    |
    |
    |
    ---------------------------------------------->B with 'b'


    Now how would they measure their relative velocitites?As A's velocity in the direction of motion of B = 0 ,B would feel that A is moving away(linearly) from it with a velocity'b'
    And A would feel that B is moving away with velocity 'a'.

    Now how do we compare their relative velocity? This is in order to aply the Lorentz transfomation.
     
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  3. Feb 25, 2007 #2

    Meir Achuz

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    To find A's velocity with respect to B, make a LT with velocity b, using the transformation laws for velocity a (separate law for a perp and a parallel to b).
     
  4. Feb 25, 2007 #3
    Or rotate the system so that A and B are moving away from each other along an axis.
     
  5. Feb 25, 2007 #4
    But the velocities of A and B have components that hav the value of 0 in their respective axes.

    Like if A measures velocity of B in his direction of motion,it comes out to be zero :(
    Im confused
     
  6. Feb 25, 2007 #5

    Hootenanny

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    No it doesn't. Remember that A is moving upwards in your diagram.
     
  7. Feb 25, 2007 #6
    Yea i get that,but in his DIRECTION of motion wouldnt he be measuring the velocity of B as zero?
     
  8. Feb 25, 2007 #7

    Hootenanny

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    Okay, try it this way. How fast is B moving away from A from A's rest frame?
     
  9. Feb 25, 2007 #8
    with velocity 'a'? and with veolcity 'b' in the direction of the perpendicular axis
     
  10. Feb 25, 2007 #9

    Hootenanny

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    Yes, so what is the magnitude of this velocity?
     
  11. Feb 25, 2007 #10
    umm..
    a + bcos(theta)
     
  12. Feb 25, 2007 #11

    Hootenanny

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    Perhaps, the hypotenuse of your triangle...
     
  13. Feb 25, 2007 #12
    ok ok sry its the resultant of the two vectors sry :p.How foolish of me
     
  14. Feb 25, 2007 #13

    Hootenanny

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    No problem, and the direction of the resultant vector is...?
     
  15. Feb 25, 2007 #14
    the positive direction of the x axis.BBut one more question..wudnt we do resultantcos(theta) to get the magnitude of the vector in the reference frame's exact direction?
     
  16. Feb 25, 2007 #15

    Hootenanny

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    No it isn't.
     
  17. Feb 25, 2007 #16
    y isnt it ??????

    r(x)=a(x) +b(x)
    r(y)=a(y) +b(y)

    r=r(x)i + r(y)j

    r(x)=b
    r(y)=a

    so y isnt it?the answer is positive
     
  18. Feb 25, 2007 #17

    Hootenanny

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    Think about it for a minute. What is the direction of the resultant velocity (think about the 'poeple swimming across a river' questions that I'm sure you've done.)
     
  19. Feb 25, 2007 #18
    umm i havent done the above stated question.Im n grade 10.


    but wats wrong with the resultant as i said in my previous post?
     
  20. Feb 25, 2007 #19

    Hootenanny

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    I asked for the direction of the resultant velocity and you haven't given it, you've just given two components, which is what we were given at the start of the problem.
     
    Last edited: Feb 25, 2007
  21. Feb 25, 2007 #20
    r(x) is positive and so is r(y)

    so
    r=b i+a j

    So whats the problem if theyre in the positive direction of the x and y axes?
     
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