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Cart-Cart Collision, Conservation of Momentum

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A cart (m1 = 130 kg) is moving to the right along a track at v1i = 24 m/s when it hits a stationary cart (m2 = 350 kg) and rebounds with a speed of v1f = 4 m/s in the opposite direction.

    a) With what speed does the 350 kg cart move after the collision?
    A: 10.4 m/s

    An observer moves in the same direction as the incoming cart with a speed of 12 m/s.
    Using the convention that the positive direction is to the right, what are the following velocities with respect to this observer:

    b) v1i, ob... A: 12
    c) v2i, ob... A: -12
    d) v1f, ob... A: 16
    e) v2f, ob... A: -1.6

    This is the annoying part...

    f) What is the total momentum of the system before the collision as seen by this moving observer?

    g) What is the total momentum of the system after the collision as seen by this same observer?

    2. Relevant equations

    m1v1 = m2v2

    Any thoughts on how to solve this? I'm a little lost.
  2. jcsd
  3. Jun 12, 2010 #2
    Think about 1. the definition of momentum, and 2. conservation of momentum. :smile:
  4. Jun 12, 2010 #3
    yeah i tried using conservation of momentum but the answers im getting dont make sense to me. the equation doesnt even out.

    the way i think of it, from a stand-still perspective, the net momentum is 0, so if youre moving at 12 m/s, shouldnt the net momentum look like -12 m/s?

    i tried using the new velocities in the m1v1 = m2v2 equation, but i think i must be setting it up wrong somehow. any suggestions?
  5. Jun 12, 2010 #4
    Perhaps this diagram http://yfrog.com/45pf2cj will help somewhat (sorry it's somewhat crudely been done in MS Paint) :smile:

    Surely the momentum should always be the same, regardless of whether the observer is moving or not, right?! :wink:

    If you take the definition:

    [tex]m_{1}v_{1i}+m_{2}v_{2i} = m_{1}v_{2i}+m_{2}v_{2f}[/itex]

    Using all the values calculated in the first part, you should find that both sides of the equation are equal, hence momentum is conserved.

    Now if you also alternatively use the values calculated in respect to the observer, you should find that again both sides of the equation are equal and hence momentum conserved.

    Try doing those, and post you're calculations if you still have any problems. :smile:
    Last edited by a moderator: Apr 25, 2017
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