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Conservation of Momentum law problem

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    m1= 3kg m2= 6kg
    v1 = 5m/s v2= 2m/s
    d = 3m
    DIAGRAM : http://gyazo.com/17b3126726b109256a35f65b86708bb0
    2. Relevant equations
    Conservation of momentum:
    m1+v1 = m1v1'+m2v2'

    Conservation of elastic:
    m1v1^2+m2v2^2 = m1v1'^2+m2v2'^2
    3. The attempt at a solution
    Now my attempt was use both equations & the answers I've gotten so far are 27=3v1'+6v2'(for conservation of momentum) & the 2nd one I've got for the conservation of elastic is 99= 3v1'^2 + 6v2'^2 since they will collide at some point. The only thing I'm stuck at is what to do next afterwards. My teacher has done a similar question but I can never understand how he did it.
     
  2. jcsd
  3. Nov 25, 2014 #2
    Is the conservattion of momentum law correct ??!
    I mean Is it llike what you wrote or like the following
    m1v1 + 0 = m1v1+m2v2

    Zero indecates that the momentum is zero because the vilocity of the second ball before collision is zero ..
    Am I right ..
     
  4. Nov 25, 2014 #3
    Sorry v1 has a value
    so the will be
    m1v1 +m2v2 = m1v1+m2v2
     
  5. Nov 25, 2014 #4
    Im just stuck because the example that my teacher gave was like "50=2(10-3v2'/2)^2+3v2'^2" that bracket part confuses me
     
  6. Nov 25, 2014 #5
    Try to equte the two equation or write v1 in term of v2..
     
  7. Nov 25, 2014 #6
    WOW WHAT THE ****, OK I DID THAT ONE BEFORE. Its this right? " V1+V1' = V2+V2'" but can u explain why that is that cus it made some sense to me. & I showed my friend this process but, she said it made no sense so.
     
  8. Nov 25, 2014 #7
    I think your teacher has written v1 in term of v2
    try to do so from the first equation your v1 will be (27 - 6 v2) /3 which is 9-2v2

    It seems reasonable
    ;)
     
  9. Nov 25, 2014 #8
    but why is that though? :( pls
     
  10. Nov 25, 2014 #9
    Because v1 and v2 are related by these two equations and their value must satisfy both of them
    so we make this substitution
    I hope that I understand your question and answer it ..
    o_O
     
  11. Nov 25, 2014 #10
    alright thaanks dawg really appreciate that, last question in the example he gave so the last line was 50-50 = -30v2'+7.5v2'^2
    the left side what if it wasnt 0
     
    Last edited: Nov 25, 2014
  12. Nov 25, 2014 #11
    If the lift side isnt zero move it to the right side and solve the quadratic equation..

    By the way what does dawg mean in your last comment
    o_O
     
  13. Nov 26, 2014 #12
    means like dude lol kids these days
     
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