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Freefall motion: velocity, acceleration, momentum, kinetic energy

  1. Jan 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider three ball bearings, one with a mass of 8 g, one with a mass of 16 g and one with a mass of 32 g.
    a.) What heights should the three ball bearings be dropped from so that at impact all three ball bearings will have the same velocity?
    b.) What heights should the three ball bearings be dropped from so that at impact all three ball bearings will have the same acceleration?
    c.) What heights should the three ball bearings be dropped from so that at impact all three ball bearings will have the same momentum?
    d.) What heights should the three ball bearings be dropped from so that at impact all three ball bearings will have the same kinetic energy?

    2. Relevant equations
    F=ma
    v2f=v2i+2ad
    p=mv
    KE=(1/2)mv2


    3. The attempt at a solution
    a.) I know that in free fall motion, neglecting air resistance, objects have the same and constant acceleration, 9.81 m/s2 downward. If I release all three objects from the same height at the same time, they will all hit the ground at the same time. Doesn't this mean that in order for the velocity of the three objects to be the same at impact, the height all three are dropped from must be the same? If not, I'm confused... I don't know how to relate mass and velocity in this case.
    b.) Since it is free fall motion and we can neglect air resistence, does the height really matter? Aren't the balls all going to have an acceleration of 9.81 m/s2 at every point in their motion?
    c.) p=mv, so obviously, since the masses are different, the velocities must also be different in order for all balls to have the same momentum. Building on what I said in a.), if the velocities are different it means they were dropped from different heights.

    I said m1 = 8 g, m2 = 16 g, and m3 = 32 g

    Then p=m1v1=m2v2=m3v3

    We know "m" and that "p" is constant for all balls, so I tried to condense all the velocity variables into one "v" like so...

    v1=v

    p=8v=16(1/2)v=32(1/4)v

    But that doesn't make sense because, solving for v, I get 1 m/s... and also because, using this, I could also say that v3 = v, then p=8(4v)=16(2v)=32v and end up with a different momentum.
    d.) Since kinetic energy involves a similar relationship to momentum I think I can try this again with some tips on how to approach c.)

    Thank you!
     
  2. jcsd
  3. Jan 13, 2007 #2

    AlephZero

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    Science Advisor
    Homework Helper

    Correct.

    Correct.

    It doesn't make sense because the velocities are not the same!

    What you mean is p = 8v1 = 16v2

    Therefore v2 = 1/2 v1.

    NB you are not given the actual value of the momentum, so you can only find the relative heights (i.e find h2 and h3 in terms of h1)
     
  4. Jan 14, 2007 #3
    Thanks so much! Now I think I can use the kinematics equation to find the relations between the heights, and do likewise for the kinetic energy question. Cool.
     
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