This question is only meant for the professionals

  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. well thanks no one can solve it what can i say this question is meant for the professionals lol
     
  4. Use the formula for relative velocities (within Special Relativity)
     
  5. Doc Al

    Staff: Mentor

    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. Doc Al

    Staff: Mentor

    You're the one who has to solve it, not me! (I can solve it just fine. :wink:)
     
  7. 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. 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. Doc Al

    Staff: Mentor

    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. 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. Doc Al

    Staff: Mentor

    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. 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. i have to use relativistic momentum ?
     
  14. Doc Al

    Staff: Mentor

  15. 0.6c +0.6c but >c what can i do ?
     
  16. i have to use this formula vx=(0.6+0.6)/(1+0.6^2) ?
     
  17. ok vx=0 !!!!{*_*}!!!!!!
     
  18. Doc Al

    Staff: Mentor

    That's the one.
     
  19. 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. loooooooollllllllll i thought vx=0 think again
     
  21. 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
     
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