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Momentum including height

  1. Apr 10, 2007 #1
    Consider a frictionless track as shown in the diagram below. A block of mass m1=8.00kg is released from point A at a height of 5.00m. It collides with a block of mass m2=10.0 kg, initially at rest, at point B. The two blocks stick together. Determine the final velocity of the blocks

    Relevant equations

    M1V1+M2V2 = (M1+M2)(Vi)
    [1/2(MVf^2)-1/2(MVi^2) + (MGHf - MGHi)

    I havent attempted this problem, I wanted to make sure that it was the second equation and If I am right I think I should just plug into the second equation and solve for Vf.
  2. jcsd
  3. Apr 10, 2007 #2
    I think it would be easier to use the first equation. Just use kinematics to determine the initial velocity of the first mass, then you have v1 for the first mass and m1 and m2, just solve for vi.

    P.S. the kinematics equation your looking for is vf=v0+ad, accerleration is just gravity, and the distance is 5m.

    Also, it's possible to solve this equation using kinetic energy, but the first equation is much easier in my opinion.
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