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Impulse-steel ball dropped on steel slab

  1. Jul 13, 2012 #1
    1. The problem statement, all variables and given/known data

    mass of steel ball = 40g = .04 kg
    h1 = 2.00 m
    h2 = 1.60 m
    A steel ball is dropped from height of 2.00 m onto horizontal steel slab, rebounds to height of 1.60m. Calculate impulse of ball during impact.


    2. Relevant equations

    impulse J = F*dt
    J = P2 - P1
    K0 + U0 + Wother = Kf + Uf

    3. The attempt at a solution

    I tried to get velocities using conservation of energy in two steps; from drop to impact; from impact to rebound.

    .04kg(9.8m/s2)(2m) = 1/2(.04kg)v2
    v = 6.26m/s

    1/2(.04kg)v2 = .04kg(9.8m/s2)(1.6m)
    v = 5.6m/s

    then I used ΔP = .04kg(5.6m/s - 6.26m/s) = -.0264

    Book shows answer to be .47N*s

    Not sure where I'm botching this up. Any help is appreciated.
     
  2. jcsd
  3. Jul 13, 2012 #2

    TSny

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    Hi, Herbally. This is a common problem that is easy to slip up on. Remember that momentum and velocity are vector quantities (with direction). If you take upward as the positive direction, then what is the sign of the initial velocity (momentum) just before the collision?
     
  4. Jul 13, 2012 #3
    Naturally the initial velocity is negative. I'm failing to see the correlation. Then again, I've been doing physics for about 3 hours straight so all the numbers are beginning to run together.

    Perhaps another nudge? Thanks.
     
  5. Jul 13, 2012 #4
    WOW. NM. I'm rum-dumb at this point. I got it. Thanks a ton.
     
  6. Jul 13, 2012 #5

    TSny

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    OK. So, the initial velocity is -6.26 m/s. You will have to subtract this negative number from 5.6 m/s. What is 5.6-(-6.26)? [Edit: I see you got it! Great!]
     
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