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\begin{large}\[ < x >  =  \frac{M_{Boxcar} x_{Boxcar,Q} + \displaystyle\sum_{j = 0}^{Q}{m_{i}x_{i,j}} + \displaystyle\sum_{j = Q + 1}^{N}{m_{i}x_{i,j}}}{M_{Boxcar} + \sum{m_{i}}} \]\end{large}

\begin{large}\[ < x >  =  \frac{M_{Boxcar} x_{Boxcar,Q} + \displaystyle\sum_{j = 0}^{Q}{m_{i}x_{i,j}} + \displaystyle\sum_{j = Q + 1}^{N}{m_{i}x_{i,j}}}{M_{Boxcar} + M_{Cannonballs}} \]\end{large}

\begin{large}\[ < x >  =   \frac{M_{Boxcar} x_{Boxcar, Q}}{M_{Boxcar} + M_{Cannonballs}}   +   Q \frac{m_{i}}{M_{Boxcar} + M_{Cannonballs}} (x_{Boxcar,Q}  +  \frac{L}{2} -  R )  +  ( N - Q ) \frac{m_{i}}{M_{Boxcar} + M_{Cannonballs}} (x_{Boxcar, Q}  -  \frac{L}{2}  +  R ) \]\end{large}

\begin{large}\[ < x > = x_{Boxcar,Q} + \frac{m_{i=diagb} {color:red} *Diagrammatic Representation* {color}

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|!Boxcar and Cannonballs 02.PNG!|
|!Boxcar and Cannonballs 04.PNG!|
|!Boxcar and Cannonballs 05.PNG!|
|!Boxcar and Cannonballs 06.PNG!|
|!Boxcar and Cannonballs 07.PNG!|
|!Boxcar and Cannonballs 08.PNG!|
|!Boxcar and Cannonballs 03.PNG!|


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{toggle-cloak:id=mathb} {color:red} *Mathematical Representation* {color}

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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:

{latex}\begin{large}\[ < x >  =  \frac{M_{Boxcar} x_{Boxcar,Q} + \displaystyle\sum_{j = 0}^{Q}{m_{i}x_{i,j}} + \displaystyle\sum_{j = Q + 1}^{N}{m_{i}x_{i,j}}}{M_{Boxcar} + \sum{m_{i}}} \]\end{large}{latex}
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{latex}\begin{large}\[ < x >  =  \frac{M_{Boxcar} x_{Boxcar,Q} + \displaystyle\sum_{j = 0}^{Q}{m_{i}x_{i,j}} + \displaystyle\sum_{j = Q + 1}^{N}{m_{i}x_{i,j}}}{M_{Boxcar} + M_{Cannonballs}} (2Q - N) (\]\endfrac{largeL}{latex}

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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 {*}x{~}Boxcar, 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 {*}x{~}Boxcar,Q{~} - (L/2) + R{*}. The equation for the location of the Center of Mass this becomes:
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{latex}\begin{large}\[ < x >  =  - \frac{M_{Boxcar}}{M_{Boxcar} + M_{Cannonballs}}   +   Q \frac{m_{i}}{M_{Boxcar} + M_{Cannonballs}} (x_{Boxcar,Q}  +  \frac{L}{2} -  R )  +  ( N - Q ) \frac{m_{i}}{M_{Boxcar} + M_{Cannonballs}} (x_{Boxcar, Q}  -  \frac{L}{2}  +  R ) \]\end{large}{latex}
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Re-arranging terms yields: 

{latex}\begin{large}\[ < x > = x_{Boxcar,Q} \frac{(M_{Boxcar} +  Q m_{i} + (N - Q ) m_{i})}{M_{Boxcar} + M_{Cannonballs}} + \frac{m_{i}}{M_{Boxcar} + M_{Cannonballs}}(Q \frac{L}{2} - QR - N \frac{L}{2} + Q \frac{L}{2} + NR - QR )   \]\end{large}{latex} 

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Inserting the previous expression into this gives

{latex}\begin{large}\[ < x > = \frac{M_{Boxcar} + \sum{m_{i}(x_{Boxcar,initial} - \frac{L}{2} + R)}}{M_{Boxcar} + \sum{m_{i}}} \]\end{large}{latex}
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Re-arranging this yields:

{latex}\begin{large}\[ < x > = \frac{(M_{Boxcar} + \sum{m_{i}})x_{Boxcar,initial} - m_{i} (\frac{L}{2} - R)}{M_{Boxcar} + \sum{m_{i}}} \]\end{large}{latex}

Solving for the initial boxcar position yields

{latex}\begin{large} \[ x_{Boxcar,initial} =  < x >  + \frac{\sum{M_{i}}}{M_{Boxcar,initial} + \sum{M_{i}}}\frac{(L - 2 R)}{2} \]\end{large}{latex}



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2} - R )   \]\end{large}

\begin{large}\[  x_{Boxcar,Q} = < x > +  \frac{m_{i}}{M_{Boxcar} + M_{Cannonballs}} (N - 2Q) (\frac{L}{2} - R ) \]\end{large}

\begin{large}\[  x_{Boxcar,Q} = < x > +  \frac{M_{Cannonballs}}{M_{Boxcar} + M_{Cannonballs}} (\frac{L}{2} - R ) - 2Q \frac{m_{i}}{M_{Boxcar} + M_{Cannonballs}} (\frac{L}{2} - R ) \]\end{large}

\begin{large}\[  x_{Boxcar,Q} = < x > +  \frac{M_{Cannonballs}}{M_{Boxcar} + M_{Cannonballs}} (\frac{L}{2} - R ) \]\end{large}

\begin{large}\[  x_{Boxcar,Q} = < x > +  (\frac{L}{2} - R ) \]\end{large}

\begin{large}\[  x_{Boxcar,Q} = < x > -  (\frac{L}{2} - R ) \]\end{large}