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Momentum of carts of different masses

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data

    The drawing below shows two laboratory carts (each has a mass of 1.0 kg) X and Y in contact with a compressed exploder spring between them. The mass on cart Y is 2.84 kg, distance A is 12 cm, distance B is 18 cm. What mass must be placed on cart X, such that after the explosion both carts will hit the ends of the track at the same time?

    After the explosion has taken place, which of the following statements regarding the kinetic energies, EK, and the magnitudes of the momenta, pi, are correct or incorrect? (Note: The subscripts x and y refer to the carts, not the direction along x or y.)
    Ex = 2/3 Ey
    px > py
    px = py
    Ex < Ey
    px = py/2


    2. Relevant equations

    Final momentum = inital momentum = 0

    MxVx + MyVy = 0

    T = d/v

    3. The attempt at a solution

    z = unknown mass on cart x
    (1 + z)Vx + (1.00 + 2.84)Vy = 0

    T= 12/Vy = -18/Vy so Vx = -18/12 Vy

    (1+z) *-18/12 Vy + 3.84 Vy = 0

    Vy[(1+z)*-18/12 + 3.84] = 0

    -18/12 + -18/12 z + 3.84 = 0

    -1.5Z = -2.34

    Z = 1.54 kg


    After the explosion has taken place, which of the following statements regarding the kinetic energies, EK, and the magnitudes of the momenta, pi, are correct or incorrect? (Note: The subscripts x and y refer to the carts, not the direction along x or y.)
    Correct Incorrect Ex = 2/3 Ey
    incorrect energy should be the same
    Correct Incorrect px > py
    not sure: depends on answer in part 1 because p = mv
    Correct Incorrect px = py
    Incorrect unless Mx = 2.84
    Correct Incorrect Ex < Ey
    Incorrect energy should be the same
    Correct Incorrect px = py/2
     
  2. jcsd
  3. Oct 10, 2007 #2

    learningphysics

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    confused by this part:

    "T= 12/Vy = -18/Vy so Vx = -18/12 Vy"

    Do you mean 12/Vy = -18/Vx?
     
  4. Oct 10, 2007 #3
    yea.. and from that you can derive that Vx = (-18/12)Vy
     
  5. Oct 10, 2007 #4

    learningphysics

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    ok... So Ex = (1/2)(1+1.54)vx^2 = 1.27(1.5vy)^2 = 2.8575vy^2

    Ey = (1/2)(1+2.84)vy^2 = 1.92vy^2

    so it seems like out of all the choices the only right one is px = py.

    BTW Ex is not the same energy as Ey.
     
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