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 12-02-07, 09:38 PM #1 MattP. Obeying Gravity Thread Starter   Join Date: Oct 2004 Location: Bellingham, WA Bikes: Posts: 2,962 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) 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.
12-02-07, 09:39 PM   #2
Siu Blue Wind
I don't think so

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 12-02-07, 09:48 PM #3 p4nh4ndle coffeeeeee     Join Date: Aug 2007 Location: somewhere in Pennsyl-tucky Bikes: all that I ride Posts: 238 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) through change in momentum (impulse is defined as instantaneous change in momentum)
12-02-07, 09:52 PM   #4
MattP.
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 Originally Posted by p4nh4ndle 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/?

 12-02-07, 09:53 PM #5 BenLi Hardrocker   Join Date: Jul 2007 Bikes: Posts: 1,569 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) impulse is also delta F multiplied by t
 12-02-07, 10:01 PM #6 p4nh4ndle coffeeeeee     Join Date: Aug 2007 Location: somewhere in Pennsyl-tucky Bikes: all that I ride Posts: 238 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s)
12-02-07, 10:04 PM   #7
p4nh4ndle
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Quote:
 Originally Posted by MattP. 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)

12-02-07, 10:10 PM   #8
MattP.
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 Originally Posted by p4nh4ndle 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!