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Homework Help: This question is only meant for the professionals

  1. Mar 25, 2008 #1
    Modern physics special relativity

    1. The problem statement, all variables and given/known data
    Two spaceships,A and B move toward each other on a head-on collision course,According to an observer at rest,both spaceships have a velocity of 0.6c along the x-axis.At the time of observation(t=0),spaceship A has the same x value (x=0) as the observer,and spaceship B is at a distance of 100 km away from him.

    At what time will the collision occur for the observer at rest and for an observer on spaceship A?

    2. Relevant equations
    my thought ok dont laugh.... T=(2L/c)/((1-v/c)/(1+v/c))^(1/2) for the spaceship A
    and T=(2L/c)/(1-v^2/c^2)^(!/2) for the observer at rest and L= 50 km when the collision happen thanks in advance
     
    Last edited: Mar 25, 2008
  2. jcsd
  3. Mar 25, 2008 #2
    well thanks no one can solve it what can i say this question is meant for the professionals lol
     
  4. Mar 25, 2008 #3
    Use the formula for relative velocities (within Special Relativity)
     
  5. Mar 25, 2008 #4

    Doc Al

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    I don't quite understand how you arrived at those answers. Rather than just give your final conclusion, explain your thinking step by step.
     
  6. Mar 25, 2008 #5

    Doc Al

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    You're the one who has to solve it, not me! (I can solve it just fine. :wink:)
     
  7. Mar 25, 2008 #6
    relative velocities what do you mean by that theres F ,T ,Energies ,... but V i cant find it i have this book concepts of modern physics 6 th edition i cant find it thanks in advance please help!!!!
     
  8. Mar 25, 2008 #7
    ok never mind my answer is wrong or right if yes i will explain it dont worry my doctor will ask me the same thing uve asked and if yes i will reply how i thought about it if no let me think again about the question...
     
  9. Mar 25, 2008 #8

    Doc Al

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    Here's a hint: Distance = speed * time. From the observer at rest's point of view, what the distance that each ship must travel and what's the speed of each ship? From ship A's point of view, how far away is ship B and how fast is it coming at him? (That's the relative speed.)
     
  10. Mar 25, 2008 #9
    From the observer at rest's point of view, what the distance that each ship must travel and what's the speed of each ship?
    50 km,0.6c
    From ship A's point of view, how far away is ship B and how fast is it coming at him?
    100km,0.6c
    ok but whats the formula i have to use it d=vt no thers another one what it is thanks in advance
     
  11. Mar 25, 2008 #10

    Doc Al

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    That's correct. That's all you need to solve for the time.
    Incorrect: For the distance, consider length contraction; for the speed, consider the relativistic addition of velocity to find the relative speed.
     
  12. Mar 25, 2008 #11
    Thank youuuuuu ok L=L0(1-v^2/c^2)^(1/2) for lenght contration ok for the velocity i will try my best to find out how it will change any other hint it will be great (for the speed) thank youuuuuuuuu
     
  13. Mar 25, 2008 #12
    i have to use relativistic momentum ?
     
  14. Mar 25, 2008 #13

    Doc Al

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    Last edited by a moderator: Apr 23, 2017
  15. Mar 25, 2008 #14
    0.6c +0.6c but >c what can i do ?
     
  16. Mar 25, 2008 #15
    i have to use this formula vx=(0.6+0.6)/(1+0.6^2) ?
     
  17. Mar 25, 2008 #16
    ok vx=0 !!!!{*_*}!!!!!!
     
  18. Mar 25, 2008 #17

    Doc Al

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    That's the one.
     
  19. Mar 25, 2008 #18
    yoooooooooopiiiiiiii thank you then the answer for observer at rest T=50km/0.6c
    and for the observer at A is T= (l=50km((1-0.6^2)^1/2))/0=infinty if yes im going to give a big thanksssssss
     
  20. Mar 25, 2008 #19
    loooooooollllllllll i thought vx=0 think again
     
  21. Mar 25, 2008 #20
    my mistake the answer is ? for observer at rest T=50km/0.6c
    and for the observer at A is T= (l=50km((1-0.6^2)^1/2))/(vx=(0.6+0.6)/(1+0.6^2)) if yes im going to give a big thanksssssss
     
  22. Mar 25, 2008 #21
    just tell me if yes or no because its 10:00 pm over here and tomorow i have to give it to my doctor no offense but thank you just tell me yes and i will study more.....
     
  23. Mar 25, 2008 #22
    lol i will take that for a silent yes thank youuuuuuuuuuu
     
  24. Mar 25, 2008 #23

    Doc Al

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    correction

    Right.
    I was going to say almost right, but I'm afraid we neglected a key factor of this problem. (Almost right because from A's viewpoint, B must cover the entire distance not half of it.) While the relative velocity is correct, and the idea of length contraction is also correct, what is not correct is the idea that A agrees with the rest observer that B is at position x=100 km when A passes x=0. According to A, by the time A passes x=0, B has already passed the point x=100km and thus is even closer.

    The efficient way to solve for the time that the collision occurs according to A is to make use of the Lorentz transformation:
    [tex]\Delta t' = \gamma(\Delta t - v\Delta x/c^2)[/tex]

    (I realize that this is a bit subtle if you are just learning special relativity. But that's the way it is.)
     
  25. Mar 25, 2008 #24
    ok deltaT=T of observer at rest
    v=Vx(0.6+0.6)/(1+0.6^2))=0.88c
    and for the gamma the v inside it =0.88c ?
    deltaX= (l=50km((1-0.6^2)^1/2))/
    thanks in advance
     
  26. Mar 25, 2008 #25
    man what to do hm.m...
     
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