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Partial Inelastic Collision

  1. Feb 23, 2008 #1
    [SOLVED] Partial Inelastic Collision

    Is it ok if I keep this here for my own reference (and others if it helps them)? I don't know why it helps to type everything out before I realize what I did wrong! But looking back on my work seems to reinforce what I learned.

    1. The problem statement, all variables and given/known data
    As shown in the figure (attached), a superball with mass m equal to 50 grams is dropped from a height of [tex] h_i [/tex] = 1.5 m. It collides with a table, then bounces up to a height of [tex] h_f [/tex] = 1.0 m. The duration of the collision (the time during which the superball is in contact with the table) is [tex] t_c [/tex] = 15 ms. In this problem, take the positive y direction to be upward, and use g = 9.8[tex] m/s^2 [/tex] for the magnitude of the acceleration due to gravity. Neglect air resistance.

    The problem has 5 parts; with some logic and work I solved the fist 4, but the last one is really getting to me. It's weird, because it should be the easiest:
    Find [tex] K_a - K_b[/tex], the change in the kinetic energy of the ball during the collision, in joules.

    2. Relevant equations
    1 kg/1000 g

    .5m[tex]v^2_a[/tex] - .5m[tex]v^2_b[/tex]

    .5m[tex]v^2[/tex] = mgh (This was used for both v before and after)

    3. The attempt at a solution
    Within the part of the problem I correctly solved were questions asking for the momentum immediately before and after the collision. Since p is solved for by multiplying mass and velocity, I used conservation of energy to relate gravitational potential energy (at [tex] h_i [/tex]) to kinetic energy after the collision. I got a velocity [tex]v_b [/tex] and multiplied it by the mass (.05 kg) to get a correct momentum for that part of the problem. The same logic was followed for the "after" momentum. The velocities I got from these two correct answers are what I used in this last question. ([tex]v_b [/tex]=5.4 m/s; [tex]v_a [/tex]=4.43 m/s)

    I solved for the kinetic energies by multiplying .5 * .05kg * 19.6[tex]m^2/s^2[/tex] and .5 *.05kg * 29.4[tex]m^2/s^2[/tex]. I then subtracted the "before" K (.729 J) by the "after" K (.491 J) to get .238 J.

    But.... it said "The ball bounces up to a lower height than that from which it was dropped. This implies that the kinetic energy after the collision was less than the kinetic energy before the collision." So I thought I had mixed up the two terms, and threw a negative sign in front of my answer, but that warranted a "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

    *edit* - It really was the wrong number of significant figure-rounding when calculating... that one is going to eat me forever if I don't stop it:shy:

    Attached Files:

    Last edited: Feb 23, 2008
  2. jcsd
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