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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 centerofmass equation:
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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 centerofmass 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
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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.
 Spring '08
 plascak
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