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  1. #1
    Obeying Gravity
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    Physics Lab Question

    Hey all.

    In this lab, we select different types of balls (bouncy, tennis, ect.) and drop each from 1m, onto a hard surface, and record how high the first bounce is. Then we have to determine the velocity of the ball before it hit (neglecting air resistance), and after it hit, the total energy the ball had before and after it hit, the momentum it had before and after it hit, and the change in momentum.

    Then I am asked to "determine the impulse that acted on the ball" Uhh, how?

    Case in point: Bouncy Ball
    Mass: .00856 kg
    Momentum before collision: -.038 kg/m/s
    Momentum after collision: .020 kg/m/s
    Change in Momentum: .058 kg/m/s

    How do you find the impulse that acted on the ball? If velocities are needed let me know.

  2. #2
    Hey guyz? Guyz? Wait up!! Siu Blue Wind's Avatar
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    ERrrrm. You need to ask Jschen about that stuff.

    He's the brains here.
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  3. #3
    coffeeeeee p4nh4ndle's Avatar
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    through change in momentum (impulse is defined as instantaneous change in momentum)

  4. #4
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    Quote Originally Posted by p4nh4ndle View Post
    through change in momentum (impulse is defined as instantaneous change in momentum)
    So if the change in momentum is .058 kg/m/s the impulse that acted upon the ball would be .058 kg/m/s/?

  5. #5
    Hardrocker
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    impulse is also delta F multiplied by t

  6. #6
    coffeeeeee p4nh4ndle's Avatar
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  7. #7
    coffeeeeee p4nh4ndle's Avatar
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    Quote Originally Posted by MattP. View Post
    So if the change in momentum is .058 kg/m/s the impulse that acted upon the ball would be .058 kg/m/s/?
    yes. you can see that the units work (from BenLi's eq).
    *edit* except it's Kgm/s not Kg/ms (or kg/m/s)

  8. #8
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    Quote Originally Posted by p4nh4ndle View Post
    yes. you can see that the units work (from BenLi's eq).
    *edit* except it's Kgm/s not Kg/ms (or kg/m/s)
    Excellent, appreciate everyone's help!

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