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Conservation of Momentum problem

  1. Oct 20, 2004 #1
    I'm not sure what I've done wrong on this problem, so anyone who sees my mistake, please let me know!

    Two blocks are moving on a frictionless surface. The first block's mass is 1.17 kg and its initial velocity is 5.50 m/s. The second block's mass is 2.73 kg and its initial velocity is 2.15 m/s.

    What is the speed of the 1.17 kg block after the collision if the speed of the second block is 5.26 m/s after the collision?

    (For clarification, before the collision the blocks are moving to the right, which I stated as the positive direction, and after the collision they are both still moving to the right.)

    To solve, I used the equation Vfinal= (m1 x v1initial) + (m2 x v2initial) / (m1+m2)

    so my work shows Vfinal = (1.17x5.50) + (2.73x2.15) / (1.17+2.73)
    which worked out to approx. 3.16 m/s

    But this approach didn't get me the right answer, which I don't understand since it seemed really straightforward.

    The second part of the question asks,

    If the initial velocity of the 2.73 kg block is the same magnitude but in the opposite direction, what is the velocity of the 1.17 kg block after the collision?

    I used the exact same approach for this, but made the initial velocity of the second block negative. So,

    Vfinal= (1.17x5.50) + (2.73x - 2.15) / (1.17+2.73)

    which worked out to 0.145 m/s

    this wasn't the right answer either.

    Any help would be greatly appreciated!
  2. jcsd
  3. Oct 20, 2004 #2
    you must remember...that after the collision they do not stick together so their final velocity is not the same.....remember how you have the velocity of the second afterwards....

    it should be more like....
    vf = [ (m1 x v1 initial) + (m2 v v2 initial) - (m2 x v2 final) ] / m1
  4. Oct 20, 2004 #3


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    Dearly Missed

    You are confusing the general conservation law of momentum with the special case of INELASTIC COLLISION.
    Remember: At the end of an inelastic collision, the two objects stick together with the same velocity.
    That is what you've used, rather than the general law of conservation of momentum:
    (The notation ought to be self-explanatory)
    Use this approach and see if you get better results..
  5. Oct 20, 2004 #4
    actually yeah, WOW, I just looked at my work and was hoping to scramble here and edit this.... totally missed a step there. i'm sorry! I'm a bonehead.
    thanks though :)
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