1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two Particles Collide - Momentum and Energy

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Q: A 3.000u object moving to the right through a laboratory at 0.8c collides with a 4.000u object to the left of the laboratory at 0.6c. Afterward, there are two objects, one of which is a 6.000u mass at rest.
    A) what are the mass and speed of the other object?
    B) determine the change in kinetic energy in the collision

    2. Relevant equations
    γ1m1v1 + γ2m2v2 = γ3m3v3+γ4m4v4
    γ1m1c2+γ2m2c2 = γ3m3c2+m4c2

    3.= The attempt at a solution
    I've been struggling with this for a while, and haven't really made much headway. The part I'm finding difficulty with is using both conservation of momentum and conservation of energy formulas when I need to consider that the third particle technically isn't moving. I don't necessarily want the answers handed to me - just a few tips to get me headed in that direction would be much appreciated.
     
  2. jcsd
  3. Feb 3, 2016 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Try working with momenta and energies instead of velocities.
     
  4. Feb 3, 2016 #3
    do you mean eliminating the v as a variable? i used the conservation of energy and momentum formulas, but there's just something not quite working. i end up with more unknowns than i know what to do with
     
  5. Feb 3, 2016 #4

    ChrisVer

    User Avatar
    Gold Member

    Aren't the unknowns in your equations the gamma3 and the m3 (for 2 equations this is solvable)?
     
  6. Feb 3, 2016 #5
    and whatever V3 would be in the momentum equation (unless I'm not supposed to use the momentum equation)
     
  7. Feb 3, 2016 #6

    ChrisVer

    User Avatar
    Gold Member

    The v3 and gamma3 are the "same" unknown... aren't they?
     
  8. Feb 3, 2016 #7
    gahhh, you're right. i think my issue was dealing with the one particle which was at rest (and has no momentum). i'll play around with it a bit
     
  9. Feb 3, 2016 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Try using ##E^2 = p^2c^2 + m^2c^4##

    These problems never work out if you try to solve for ##v## and ##\gamma##!
     
  10. Feb 3, 2016 #9

    ChrisVer

    User Avatar
    Gold Member

    Well they do, it's just a very rough way to do things... Either way you have to apply the conservation of energy/momentum at some point (in one way or another).
     
  11. Feb 3, 2016 #10

    ChrisVer

    User Avatar
    Gold Member

    But the expression for the momentum conservation is not quiet right...(except for if vi are vectors)
     
  12. Feb 3, 2016 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If Particle 4 is the one at rest after the collision we have
    [tex] \begin{array}{rcrcrc}
    \gamma_1 m_1 v_1 &+& \gamma_2 m_2 v_2 &=& \gamma_3 m_3 v_3 &+ &0 \\
    \gamma_1 m_1 &+& \gamma_2 m_2 &=& \gamma_3 m_3 &+& m_4
    \end{array}
    [/tex]
    In the second equation we dropped the common factor ##c^2## from all terms on both sides.
    From the known quantities, the equations allow us to find ##\gamma_3 m_3## and ##\gamma_3 m_3 v_3##; so we can get ##v_3## by taking a ratio of known quantities. Once we know ##v_3## we are almost done.
     
  13. Feb 3, 2016 #12
    @Ray Vickson - ok, so for the momentum I got
    γ3m3v3= 1uC
    for energy I got: γ3m3= 4u
    so I divide the first one by the second to get v3= 0.25c (?)
     
  14. Feb 3, 2016 #13
    ok, then I used that, calculated gamma, and inserted back into the momentum equation to solve for m3 and got 3.873 u. should be good to go. thanks!
     
  15. Feb 3, 2016 #14

    ChrisVer

    User Avatar
    Gold Member

    seems legit...
     
  16. Feb 3, 2016 #15
    unless you got something entirely different and I am to to lunch, hehe
     
  17. Feb 3, 2016 #16

    ChrisVer

    User Avatar
    Gold Member

    Yes I got something totally different because I omitted the m4, as if it was not there...
    As for the questionmark for the v3, it's true that [itex]\frac{p}{E}= v[/itex] for massive particles, and it's also the same for massless particles too...
     
  18. Feb 3, 2016 #17
    there are no massless particles in this equation though ... sorry, I'm just not entirely sure if you're agreeing with my end result, or implying I should make changes
     
  19. Feb 3, 2016 #18

    ChrisVer

    User Avatar
    Gold Member

    nah, I just made a general statement for your questionmarked phrase "so I divide the first one by the second to get v3= 0.25c (?)"
    That p/E happens to be always equal to the velocity.

    it seems I agree...
     
  20. Feb 3, 2016 #19
    welp, cheers!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted