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Imagine that you have an indestructible boxcar sitting on frictionless railroad track. The boxcar has length L, height H, and width W. It has N cannonballs of radius R and mass M stacked up against one end. If I move the cannonballs in any fashion – slowly carrying them, rolling them, firing them out of a cannon – what is the furthest I can move the boxcar along the rails? Which method should I use to move the boxcar the furthest? Assume that the inside walls are perfectly absorbing, so that collisions are perfectly inelastic.
Part A
Solution One
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
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The system consists of the Boxcar on rails and the Cannonballs, plus whatever devices we use for propulsion inside.There are thus no external influences
Mathematical Representation
Since there are no external influences, which includes forces, the center of mass of the system is not affected, and by the Law of Conservation of momentum must remain fixed. .
\begin
[\ M_
x_
+ \sum M_
x_
= M_
x_
+ \sum M_
x_
]\end
Here xi is the position of the center of the *i*th cannonball and xBoxcar is the position of the center of the boxcar. The subscripts initial and final indicate the positions at the start and the end of our operation.
Re-arranging, we get that the final boxcar position is:
\begin
[ x_
= x_
+ \frac{\sum{M_
x_{i, initial}} - \sum{M_
x_
}}{M_{Boxcar}} ]\end
\begin
[ x_
= x_
- \frac{M_{i}}{M_{Boxcar}} ( \sum{x_{i, final}} - \sum{x_{i, initial}}) ]\end
The location of the Center of Mass <x> is given by:
\begin
[ < x > = \frac{M_
x_
+ \sum{M_
x_
}}{M_
+ \sum{M_
}} ]\end
If we define the center of mass as the zero position, then <x> = 0 and we have
\begin
[ x_
= -x_
\frac{M_{Boxcar}}{\sum{M_
}}]\end
Let's assume the Boxcar location is at its center of mass, in the middle of the Boxcar. The location of the Cannonballs relative to <x> is given by
\begin
[ x_
= x_
- \frac
+ \frac
]\end
Inserting the previous expression into this gives
\begin
[ < x > = \frac{M_
+ \sum{M_
(x_
- \frac
+ \frac
)}}{M_
+ \sum{M_
}} ]\end
Re-arranging this yields:
\begin
[ < x > = \frac{(M_
+ \sum{M_{i}})x_
- M_
(\frac
- \frac
Unknown macro: {R}
)}{M_
+ \sum{M_
}} ]\end
Solving for the initial boxcar position yields
\begin
[ x_
= < x > + \frac{\sum{M_
}}{M_
+ \sum{M_
}}\frac
]\end
One can in a similar way solve for the position of the Boxcar in its final position, assuming that the cannonballs are all moved from as close to one side to as close to the other side as possible. The steps are the same, with the final result:
\begin
[ x_
= < x > - \frac{\sum{M_
}}{M_
+ \sum{M_
}}\frac
]\end
Subtracting the Final position of the Boxcar from its Initial Position yields the total movement of the car:
\begin
[ x_
- x_
= \frac{ \sum{M_
}}{M_
+ \sum{M_
}} ( L - R ) ]\end
Part B
Now consider the intermediate stages as the cannonballs are moved slowly, one by one, from one side to the other. How does the car move as each ball is shifted?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
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Mathematical Representation
We calculate the Center of Mass as each Cannonball is shifted from one side to the other. Assume that each ball moves from as close to one side of the boxcar to as far one the other side as it can go.
\begin
[\sum F_
= F_
\cos\theta - N = ma_
]
[ \sum F_
= F_
\sin\theta - mg = ma_
]\end
\begin
[ N = F_
\cos\theta = \mbox
]\end
Part C
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Now imagine that the cannonballs are fired from one end to the other, one by one. What are the equations of motion during and after, and how does the car move along?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
Mathematical Representation
We write the equations of motion for the boxcar (with the rest of the cannonballs) and the one fired cannonball.
\begin
[\sum F_
= F_
\cos\theta = ma_
]
[ \sum F_
= F_
\sin\theta - mg - N = ma_
]\end
\begin
[ F_
\sin\theta - mg - N = 0 ]\end
\begin
[ N = F_
\sin\theta - mg = \mbox
]\end