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Beginner Momentum

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A dynamics cart with a mass of 2.2kg is moving at 33 cm/s to the right when it has a head-on collision with a second cart with a mass of 1.2kg moving in the opposite direction also at 33 cm/s. After the nearly elastic collision the first cart continues forward at 13 cm/s. What is the new velocity of the second cart?


    2. Relevant equations

    conservation of momentum mv initial=mv final

    3. The attempt at a solution

    (mv of cart 1- mv of cart 2- mv final of cart 1)/(m of cart 2)= 3.6 cm/s to the right

    However that's not right, the answer should be 5.3 cm/s to the left, which honestly makes no sense. The teacher hinted that the solution would be counter-intuitive. However, when using the correct values I do not get the desired answer.

    It would really help if someone could either confirm my findings or prove me wrong. Also, I know that in elastic collisions kinetic energy is conserved, and the in inelastic collisions, KE is not conserved. I wanted to confirm that in elastic collisions the two masses do not become one, as opposed to inelastic collisions in which they do become one single mass.
     
  2. jcsd
  3. Mar 1, 2009 #2
    5.3 is the answer I get.

    .5*m1*v1^2 + .5*m2*v2^2 = .5*m1f*v1f^2 + .5*m2f*v2f^2

    All of the .5's cancel out.
    Be sure to convert cm to m.

    Note that m1f = m1 and m2f = m2 in this case, because the problem says nothing about mass being lost.
     
    Last edited: Mar 1, 2009
  4. Mar 1, 2009 #3

    Hmm, interesting. Why didn't you represent the value of m2v2 as negative? Shouldn't it have the opposite value of m1v1 since it is traveling in the opposite direction?
     
  5. Mar 1, 2009 #4
    You do when you plug it in, but the square should make it positive anyway.

    And I made an error in that equation, im sorry, the velocities are supposed to be squared.

    MY MISTAKE!
     
  6. Mar 1, 2009 #5
    alright thanks a ton! so you chose to use conservation of kinetic energy rather than conservation of momentum? if you wouldn't mind expanding on your answer, what made you choose conservation of energy rather than momentum?
     
  7. Mar 1, 2009 #6
    After an elastic collision, no kinetic energy is lost. So basically this is a system in which the energy will be constant (not realistic). Also I just looked at what they gave you. Knowing the given data kind of throws that switch about which equation to use, at least for me.
     
  8. Mar 1, 2009 #7
    wow, thanks! your help is much appreciated :)
     
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