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Impulse and height

  1. Oct 30, 2005 #1
    I found this to be the hardest question on my last test:

    A 5kg block is resting on two tables such that the middle isn't resting on a part of a table. A 10g bullet is fired at 1000m/s from directly below the block perpendicular. The bullet exits the block going 400m/s. What was the maximum height of the block?

    Here's what I did, the impulse was the change in momentum of the bullet, which would be .01kg(1000m/s)-.01(400m/s) = 6kgm/s. That impulse must have been imparted on the block, which would give it a final velocity of 1.2m/s (6kgm/s / 5kg). From there, y = [(v2^2 - v1^2)/-2a] letting a = -g = -9.8 and v2 = 0, v1 = 1.2, and I got 0.073m = 7.3cm.

    Did I go about this right?
  2. jcsd
  3. Oct 30, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good to me.
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