Problem Set #8 Solutions

So algebraically this says xcmi xcmf xcm 0 now since

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Unformatted text preview: d duplication or distribution prohibited. Physics 1111 Spring 2012 PS #8 Solutions Page 7 of 9 Problem 8: We use the idea that, for the system consisting of the dog and boat, the center of mass doesn’t move because there are no external forces acting on the system. So, algebraically, this says xcm,i = xcm,f =⇒ ∆xcm = 0. Now, since the masses are constant, we can expand this “delta” equation as ∆xcm = 0 =⇒ md ∆xdog + mb ∆xboat = 0, where we’re measuring all positions relative to some fixed object like the shore. The problem is, we don’t know how far the boat moves; we just know how far the dog walks along the boat (call this L). Well, if the dog walks a distance L to the left relative to the boat, and the boat moves ∆xboat to the right (relative to the shore), then the dog’s actual motion relative to the shore can be written as ∆xdog = ∆xboat − L. We rearrange this to get ∆xboat = L + ∆xdog and then substitute into the previous centerof-mass equation: This document requires a JavaScript-enabled PDF reader md ∆xdog + mb ∆xboat = 0 md ∆xdog + mb (L + ∆xdog ) = 0 (md + mb )∆xdog + mb L = 0 mb L ∆xdog = − . md + mb such as Adobe Reader. Finally, we can put in numbers: ∆xdog = − 38.0(9.0 m = −7.43 m. 38.0 + 8.0 Given that the dog started at 25.9 m, his final position will be xf = xi + ∆x = 25.9 m − 7.43 m = 18.46 m ≃ 18.5 m. Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited. Physics 1111 Spring 2012 PS #8 Solutions Page 8 of 9 Problem 9: (a) The answer for this part is quite simply 0 m/s. There are no external forces acting on the boxcar, and it’s initially at rest. That means its center-of-mass velocity is zero both before and after the water leaks out of the tank. (b) For this part, we use the fact that the center of mass has to stay in the same spot. Let’s take the left side of the boxcar (before it moves) to be the origin of our coordinates. The center of the boxcar is at 4.65 m, and the center of the tank is at 0.5 m. So the initial center of mass for the system is mcar xcar + mwater xwater mtot (3600)(4.65 m) + (1700)(0.5 m) = 3600 + 1700 17590 m = = 3.32 m. 5300 xcm,i = Now, if the boxcar had been Super Glued to the tracks, then after the water leaked from the tank, the center of been at 4.65 m (the center of boxcar). document requires mass would havebut the center of mass stayed thethe same a JavaScript-enabled PDF reader Since the boxcar was actually free to move, in place, this...
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This note was uploaded on 08/15/2012 for the course PHYS 1111 taught by Professor Plascak during the Spring '08 term at University of Georgia Athens.

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