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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. Assume that each Cannonball weighs mi and that there are N of them, totalling Mi.
\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_
+ M_{i}} ]\end
If we define the center of mass as the zero position, then <x> = 0 and we have
\begin
[ x_
= -x_
\frac{M_{Boxcar}}{M_{i}}]\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
+ R ]\end
Inserting the previous expression into this gives
\begin
[ < x > = \frac{M_
+ \sum{M_
(x_
- \frac
+ R)}}{M_
+ M_{i}} ]\end
Re-arranging this yields:
\begin
[ < x > = \frac{(M_
+ M_
)x_
- M_
(\frac
- R)}{M_
+ M_{i}} ]\end
Solving for the initial boxcar position yields
\begin
[ x_
= < x > + \frac{M_{i}}{M_
+ M_{i}}\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{M_{i}}{M_
+ M_{i}}\frac
]\end
Subtracting the Final position of the Boxcar from its Initial Position yields the total movement of the car:
\begin
[ x_
- x_
= \frac{M_{i}}{M_
+ M_{i}} ( L - 2 R ) ]\end
In the limit that the sum of the cannonballs weights are much less than that of the boxcar, the car is moved only negligibly by moving the cannonballs from one end to the other. In the limit that the cannonballs weigh much more than the Boxcar, the boxcar shifts by its own length minus the diameter of the cannonballs. For any other relative mass of the cannonballs to that of the boxcar, the result is between these two results, but it's seen that the maximum distance the car can be moved is its own length minus the size of the cannonballs.
