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Oscillation (conservation)

  1. Dec 6, 2006 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    E = 1/2mv^2 + 1/2kx^2 = 1/2kA^2
    omega = Sqrt(k/m)
    f = omega/(2pi) = 1/(2pi)*Sqrt(k/m)

    3. The attempt at a solution
    m = 150kg
    M = 150kg + 50kg = 200kg
    x = 15ft = 4.57m
    h1 = 3.04 m
    h_real = ?
    x = A?

    mgh = (1/2)Mv^2 + (1/2)kx^2
    Since there is no velocity given, I use:
    mgh = (1/2)kA^2

    I need to find k:
    3 cycles every 10 seconds
    f = 0.3 cycles/s

    f = (1/2pi)Sqrt(k/M)
    0.3 = (1/2pi)Sqrt(k/M)
    k = 710 N/m

    (150kg)(9.8m/s^2)(h) = (1/2)(710N/m)(4.57m)^2
    h = 5.04m

    h_real > h1

    Did I do that correctly?
     
  2. jcsd
  3. Dec 6, 2006 #2
    Almost: you have a rounding error in the final answer.
     
  4. Dec 6, 2006 #3

    Doc Al

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    Staff: Mentor

    You have assumed that mechanical energy is conserved during the collision. Rethink that assumption, given that the car and block stick together.

    Good.
     
  5. Dec 6, 2006 #4
    Hmm...
    Before the collision, the block would be:
    mgh = (1/2)mv^2
    Then
    mv = Mv_2
    Is there a way to use this relationship without velocity data (actual height)?
     
  6. Dec 6, 2006 #5

    Doc Al

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    Staff: Mentor

    Sure. Start with the SHM. That should enable you to find the post-collision speed, then use that relationship (conservation of momentum) to find the pre-collision speed. Then you can deduce the height.
     
  7. Dec 6, 2006 #6
    (1/2)Mv^2 = (1/2)kA^2 right?

    v = Sqrt(k/M)A
    v = Sqrt(710/200kg) * (4.57m)
    v = 8.6 m/s

    mv = Mv
    (150kg)v = (200kg)(8.6 m/s)
    v= 11.5 m/s

    mgh = (1/2)mv^2
    (150kg)(9.8)h = (1/2)(150kg)(11.5m/s)^2
    h = 6.75 m?
     
  8. Dec 6, 2006 #7

    Doc Al

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    Staff: Mentor

    Looks good.
     
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