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Impulse and average force

  1. Mar 23, 2009 #1
    1. The problem statement, all variables and given/known data

    A 2.18 kg steel ball strikes a massive wall at 12.6 m/s at an angle of a = 55.1o with the plane of the wall. It bounces off with the same speed and angle (as seen in the figure below). If the ball is in contact with the wall for 0.152 s, what is the average force exerted on the ball by the wall?

    2. Relevant equations

    mv/change in t = average force
    v=sq. root 2gh

    3. The attempt at a solution
    I've tried finding the impulse and then dividing it by the .152 sec. but I keep getting the wrong answer. Could someone please walk me through the equations and steps? Thanks!
     
  2. jcsd
  3. Mar 23, 2009 #2

    LowlyPion

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    What is the Δ in momentum?
     
  4. Mar 23, 2009 #3
    momentum= mass(velocity)...2.18*12.6=27.5
     
  5. Mar 23, 2009 #4

    LowlyPion

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    But what was the Δ in momentum.

    Think x,y components.
     
  6. Mar 23, 2009 #5
    Resolve the initial and final velocity vectors to find out which component undergoes a change.
     
  7. Mar 23, 2009 #6
    v1 prime would be -v1
    v2 prime would be 0

    how does that fit in to the problem?
     
  8. Mar 23, 2009 #7

    LowlyPion

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    Impulse requires knowing the Δ in momentum.

    Now not all the momentum changes here. For instance the momentum || to the wall is unchanged after contacting the wall.

    But the ⊥ component has a change. Figure that change as I think you will be needing it.
     
  9. Mar 23, 2009 #8
    I don't understand... Is that not just the negative of the velocity?
     
  10. Mar 23, 2009 #9

    LowlyPion

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    Negative of the original velocity means that it returned in the direction it came.

    It did not.
     
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