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Elastic/inelastic collision - speed?

  1. Mar 12, 2007 #1
    1. The problem statement, all variables and given/known data

    The students set up a cart with a spring attached to the force probe.They weigh the cart assembly and find that its mass is 734.0 g. They launch the cart towards the target and collect the following force-vs.-time data.

    The students discover that Data Studio has an “area under the curve” feature. They use this and find the total area under the curve to be 0.4209 N·s.

    1) Assuming a perfectly inelastic collision, what is your best estimate of the speed of the cart before the collision?


    2) Assuming a perfectly inelastic collision, what is your best estimate of the speed of the cart after the collision?

    *what would you do for an elastic collision?



    2. Relevant equations

    elastic
    1/2m1v1^2 + 1/2m2v2^2= 1/2m1v1'^2 +1/2m2v2'^2


    3. The attempt at a solution

    very confused even on how to start this question...
    how would we calculate without the mass of the other object?
    cannot find the equations that go from a given impulse (force and time) to velocity. help??? :cry:

    :confused: :confused: :confused: :confused: :confused:
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 12, 2007 #2
    Besides [tex]F \Delta t[/tex]
    what else is impulse equal to?
    Hint: Impulse equals the change in...
     
    Last edited: Mar 12, 2007
  4. Mar 12, 2007 #3
    well the data is given in a graph of force vs. time so the area under the curve represents the impulse (0.4209 N·s).
    I just dont know how to get from that impulse value to velocity???
     
  5. Mar 12, 2007 #4
    impulse is also equal to the change in momentum:
    delta p = pf -pi
    and p = mv

    but how would you know pf or pi given just delta p?
     
  6. Mar 12, 2007 #5
    so for the velocity after an perfectly nonelastic collision,
    J = pf -pi
    pf = mv
    .4209 N·s.= .7340 kg (v)
    v = .5734 m/s

    correct?
     
  7. Nov 25, 2007 #6
    I don't believe so.

    Your implying that the impulse(.4209 Ns) = mv. False, it is the momentum (p) that equals mv, not the impulse.
    But if your assuming (for a perfectly nonelastic collision) p(final) = 0, thus J=pi=mv, than I believe your right. I guess you probably meant that.
     
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