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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.
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.
The location of the Center of Mass <x> after Q cannonballs out of a total of N have been moved is given by:
\begin
[ < x > = \frac{M_
x_
+ \displaystyle\sum_
^
{m_
x_{i,j}} + \displaystyle\sum_
^
{m_
x_
}}{M_
+ \sum{m_
}} ]\end
\begin
[ < x > = \frac{M_
x_
+ \displaystyle\sum_
^
{m_
x_{i,j}} + \displaystyle\sum_
^
{m_
x_
}}{M_
+ M_{Cannonballs}} ]\end
The center of mass position, < x >, is the fixed point, since there are no external forces on the masses being considered. To simplify the expression, we relate all the cannonball positions to those of the boxcar. This enables us to put everything in terms of a single variable. Let us call the position of the center of the boxcar when Q cannonballs have been moved from one end to the other xBoxcar, Q. The Cannonballs that have not been moved are at one end of the boxcar, a distance L/2 - R to the left of the boxcar center, while the Q that have been moved are at xBoxcar,Q - (L/2) + R. The equation for the location of the Center of Mass this becomes:
\begin
[ < x > = \frac{M_
x_{Boxcar, Q}}{M_
+ M_{Cannonballs}} + Q \frac{m_{i}}{M_
+ M_{Cannonballs}} (x_
+ \frac
- R ) + ( N - Q ) \frac{m_{i}}{M_
+ M_{Cannonballs}} (x_
- \frac
+ R ) ]\end
This can be further re-arranged and condensed to give:
\begin
[ < x > = x_
+ \frac{m_{i}}{M_
+ M_{Cannonballs}} (2Q - N) (\frac
- R ) ]\end
Solving for xBoxcar,Q
\begin
[ x_
= < x > + \frac{m_{i}}{M_
+ M_{Cannonballs}} (N - 2Q) (\frac
- R ) ]\end
\begin
[ x_
= < x > + \frac{M_{Cannonballs}}{M_
+ M_{Cannonballs}} (\frac
- R ) - 2Q \frac{m_{i}}{M_
+ M_{Cannonballs}} (\frac
- R ) ]\end
For Q = 0 we have
\begin
[ x_
= < x > + \frac{M_{Cannonballs}}{M_
+ M_{Cannonballs}} (\frac
- R ) ]\end
In the limit that MCannonballs >> MBoxcar this becomes
\begin
[ x_
= < x > + (\frac
- R ) ]\end
Similarly, for Q = N we have, in the same limit,
\begin
[ x_
= < x > - (\frac
- R ) ]\end
These are the same results as in Part A, and once again the most that the Boxcar can move is L, and then only in the limit as R goes to zero. With the formulae developed in this part, however, we can see how much the boxcar moves as each cannonball is moved from one side to the other. Note that the center of the boxcar can only coincide with the overall center of mass if N is an even number, and half the cannonballs are moved from one side to the other.
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
Let's assume that we launch, throw, roll, or fire the cannonballs one at time from one end of the Boxcar to the other, and that the other Cannonballs are held rigidly in place in the Boxcar, so that they move with it. Let;s further assume that when the cannonball reaches the other side it sticks in place on the other side, coming instantly to rest (and doing no damage).
We write the equations of motion for the boxcar (with the rest of the cannonballs) and the one fired cannonball. Let us call the velocity of the velocity of the cannoball vi and the velocity of the Boxcar (with the remaining cannonballs) vBoxcar. Assuming Conservation of Momentum we have:
\begin
[ m_
v_
= - (M_
+ (N - 1) m_
) v_
]\end
\begin
[ v_
= - \frac{m_{i}}{M_
+ (N - 1) m_{i}} v_
]\end
So we can write the equations of motion for both the Boxcar (and Cannonballs) and the lone cannonball:
\begin
[ x_
(t) = x_
+ v_
t ]\end
\begin
[ x_
(t) = x_
+ v_
t ]\end
We can substitute for the Boxcar Velocity to get:
\begin
[ x_
(t) = x_
- v_
\frac{m_{i}}{M_
+ (N - 1)m_{i}}t ]\end
The Cannonball will stop when it strikes the far wall of the Boxcar at time tfinal. At this point the cannonball will have traveled, relative to the Boxcar, a distance L - 2R. But the Boxcar will have moved, as well. The relevant expression relating the distances is thus:
\begin
[ x_
(t_
) - x_
(t_
) = L - 2R ]\end
Substituting in the above equations and re-arranging, we get:
\begin
[ x_
- x_
+ v_
( 1 + \frac{m_{i}}{M_
+ (N - 1)m_{i}})t_
= L - 2R ]\end
But we can see that the difference between the initial position of the cannonball and the initial position of the boxcar is just half the car length, minus the cannonball radius:
\begin
[ x_
- x_
= \frac
- R ]\end
Solving for tfinal, we get:
\begin
[ t_
= \frac{(\frac
- R)}{v_{i}} \frac{M_
+ (N - 1) m_{i}}{M_
+ M_{Cannonballs}} ]\end
Inserting this into the above expressions for xBoxcar(tf) and xcannonball(tf) we get
\begin
[ x_
(t_
) = x_
- \frac{m_{i}}{M_
+ M_{Cannonballs}} (\frac
- R ) ]\end