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Old 12-02-07, 09:38 PM   #1
MattP.
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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.
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Old 12-02-07, 09:39 PM   #2
Siu Blue Wind
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ERrrrm. You need to ask Jschen about that stuff.

He's the brains here.
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Old 12-02-07, 09:48 PM   #3
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through change in momentum (impulse is defined as instantaneous change in momentum)
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Old 12-02-07, 09:52 PM   #4
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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/?
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Old 12-02-07, 09:53 PM   #5
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impulse is also delta F multiplied by t
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Old 12-02-07, 10:01 PM   #6
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see http://scienceworld.wolfram.com/physics/Impulse.html as a ref
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Old 12-02-07, 10:04 PM   #7
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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)
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Old 12-02-07, 10:10 PM   #8
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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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