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Energy Conservation or Momentum?

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data
    A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-h frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.

    a) suppose the packages stick together. what is their common speed after the collision?


    b) suppose the collision between the packages is perfectly elastic. to what height does the package of mass m rebound?

    p10-42alt.gif


    2. Relevant equations
    KE = PE
    KE = 1/2 mv2
    PE = mgh



    3. The attempt at a solution
    Vi = 0m/s

    Ek = Ep

    1/2mv2 = mgh
    The masses cancel each other out, leaving me with

    v = [tex]\sqrt{}2gh[/tex]

    v = 58.8m/s, this is the velocity at which the box 1 hits box 2

    From this point i decided to use momentum to find the final velocity
    (2m+m)vf = mvi

    This simplifies to 3Vf = V/3 = 19.6m/s
    Why would using momentum be flawed?

    When I use KE is conserved throughout this system I get another answer... that is why one of the methods I'm using is wrong


    1/2mv2 = 1/2mv2

    1/2(2m+m)v2 = 1/2mv2

    Doing some algebra I get that V2 = V2/3

    For this answer the Vf = 33.95m/s
     
  2. jcsd
  3. Feb 26, 2009 #2

    Doc Al

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    Staff: Mentor

    You forgot the square root.
     
  4. Feb 26, 2009 #3

    D H

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    Using conservation of kinetic energy for the first problem will give you a different answer than conservation of momentum. You can't use conservation of kinetic energy here because kinetic energy is not conserved. The collision is inelastic.
     
  5. Feb 26, 2009 #4
    thanks for the input D H
    so, for all collisions momentum is conserved.
    for elastic collisions momentum + KE is conserved
    for inelastic collisions momentum is conserved but KE isn't
    so the square root of 58.8 = 7.668m/s
    and the final velocity of boxes stuck together is 2.56m/s
     
    Last edited: Feb 26, 2009
  6. Feb 26, 2009 #5

    D H

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    Whoa! That doesn't conserve momentum.
     
  7. Feb 26, 2009 #6
    sorry? i didnt understand what you meant by that
     
  8. Feb 26, 2009 #7

    Doc Al

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    Looks good to me. Now solve part b).
     
  9. Feb 26, 2009 #8
    alright for the next part we have perfect elastic collision

    Ek = conserved
    P = conserved

    This is my first equation for Momentum:
    m1v1+m2v2 = m1vf1+m2vf2

    This is my 2nd equation for kinetic energy
    1/2m1v12 + 1/2m2v22 = 1/2m1vf12+1/2m2vf22

    This should take some algebra :s


    Looking into my textbook I found these simplified equations

    Vf1 = Vi*(m1-m2/m1+m2)
    For this part I found that the Vf of the first object = -2.56m/s

    Vf2 = Vi*(2m1/m1+m2)
    For this part I found that the Vf of the object = 5.11m/s
     
    Last edited: Feb 26, 2009
  10. Feb 26, 2009 #9

    Delphi51

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    Homework Helper

    It checks out - momentum conserved, energy conserved.
    Also did it without the simplified equations and got the same answers.
     
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