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Elastic Collision and Conservation of Energy

  1. Nov 30, 2009 #1
    SOLVED

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

    A 120g granite cube slides down a 40 degree frictionless ramp. At the bottom, just as it exits onto a horizontal table, it collides with a 225g steel cube at rest. How high above the table should the granite cube be released to give the steel cube a speed of 185cm/s?

    2. Relevant equations

    Ug=mgh
    K=0.5mv2
    P=mv

    3. The attempt at a solution

    I first calculated the momentum of the steel block (block 2):
    P=mv
    P2=(.225)(1.85)
    P2=0.41625kg*m/s

    Then, using conservation of momentum I deduced that the momentum of the granite block (block 1) before the collision must be equal to the momentum of the second block after the collision (P1=0.41625kg*m/s). Using this I can find the velocity of block 1 just before the collision:
    P=mv
    0.41625=(0.120)v
    v=3.46875m/s

    Now I can find the kinetic energy of block 1 right before it hit block 2:
    K=0.5mv2
    K=0.5(0.120)(3.46875)2
    K=0.722J

    Since energy is conserved, Ug at the top of the ramp is equal to K at the bottom of the ramp (Ug=0.722J). We can now find the height of the block:
    Ug=mgh
    0.722=(0.120)(9.8)h
    0.722=1.176h
    h=0.6139m=61.39cm

    Where did I go wrong?
     
    Last edited: Nov 30, 2009
  2. jcsd
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