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Very simple momentum question

  1. Jan 2, 2009 #1
    1. The problem statement, all variables and given/known data

    When an pea bounces off the surface, it exeriences a change of momentum of 0.0051.

    Show that the rebound velocity of this pea is about 5m/s

    Mass of pea = 5x10^-4

    2. Relevant equations

    m1v1 = m2v2

    3. The attempt at a solution

    So a change in momentum of 0.0051 means change in m1v1 = 0.0051

    as we know the mass of the pea we can do

    v2 = (m1v1)/m2
    = 0.0051 / 5x10^-4

    but my answer is a factor of 2 too big? What have I done wrong?

    Thanks :)
     
  2. jcsd
  3. Jan 2, 2009 #2

    Doc Al

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    Staff: Mentor

    No, the change in momentum = 0.0051, not the momentum.

    Assume that the pea bounces off with the same speed as it landed. Careful: While the speed hasn't changed, the velocity sure has. What's the change in momentum?

    Hint: mv1 ≠ mv2
     
  4. Jan 3, 2009 #3
    mv1 = -mv2
     
  5. Jan 3, 2009 #4

    Doc Al

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    Staff: Mentor

    Exactly. So what's the change in momentum?
     
  6. Jan 3, 2009 #5
    mv1 -- mv2 = mv1 + mv2

    mv1 + mv2 = 0.0051

    m(v1 + v2) = 0.0051
    v1 + v2 = 10.2
    if say the speed before = speed after then v1 = v2:

    2v1 = 10.2
    v1 = 5.1m/s

    TRADA!

    also is it true that in the change of momentum only speed will change as mass is constant?


    THANKS :)
     
  7. Jan 3, 2009 #6

    Doc Al

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    Staff: Mentor

    Good. Here's how I'd do it.

    If we call the initial velocity V, then the final velocity will be -V (since the direction reverses). Thus the initial momentum is mV and the final momentum is -mV. The change in momentum is -mV -mV = -2mV. (The magnitude of the change is just 2mV.)

    The mass is constant but the velocity changes (not the speed).
     
  8. Jan 3, 2009 #7
    sorted! Thanks alot :)
     
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