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Impulse-Momentum Problem

  1. Nov 10, 2009 #1

    A 0.480-kg ball is dropped from rest at a point 1.10 m above the floor. The ball rebounds straight upward to a height of 0.880 m. Taking the negative direction to be downward, what is the impulse of the net force applied to the ball during the collision with the floor?

    m = 0.480 kg

    h0 = 1.10 m

    hf = 0.880 m

    vo = ?

    vf = ?

    a = 9.80 m/s2

    2. Relevant equations

    Impulse-Momentum Theorem

    J = mvf - mvo

    vf2 = vo2 + 2ay

    3. The attempt at a solution

    As we need the velocity of the ball before and after the collision, we can use vf2 = vo2 + 2ay. But I'm wondering, would we solve it once for the vf using ho = 1.10 m? As that would determine the final velocity of the ball just before it hits the floor. And then solve it again, but this time for vo using hf = 0.880 m? The would determine the initial velocity of the ball as soon as it rebounds up? Right....???

    Doing it this way, I came up with:

    vf = 4.64 m/s

    vo = 4.15 m/s

    Putting these into the Impulse-Momentum Theorem, vf from above would now be the vo as it is the velocity right before impact. This velocity would also be negative as it is pointing downwards. Then vo from above would be the new vf as it is the velocity right after impact. Did that make sense?

    J = mvf - mvo

    J = 0.480 kg (4.15 m/s) - [ - 0.480 kg (4.64 m/s)]

    J = 1.992 kg*m/s - (- 2.227 kg*m/s)

    J = 4.22 kg*m/s

    Ummm...Does that look right? I've tried this question a few different ways but haven't been successful yet.

  2. jcsd
  3. Nov 10, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me!

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