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Boxcar and cannonballs as point particles.

Not Important in this part.

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|!Boxcar and Cannonballs 01.PNG!|


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.

h2. Part A


h4. Solution One

{toggle-cloak:id=sysa} *System:*  {cloak:id=sysa}Boxcar and cannonballs as [point particles|point particle].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta} Not Important in this part.{cloak}

{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*  

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{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation* {color}

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|!Boxcar and Cannonballs 02.PNG!|
|!Boxcar and Cannonballs 03.PNG!|

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

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

{latex}\begin{large}\[\ M_{Boxcar} x_{Boxcar, initial} + \sum Mm_{i} x_{i, initial} = M_{Boxcar} x_{Boxcar, final} + \sum Mm_{i} x_{i, final}  \]\end{large}{latex}

Here {*}x{~}i{~}{*}  is the position of the center of the 

{latex}
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The location of the Center of Mass *<x>* is given by:

{latex}\begin{large}\[ < x >  =  \frac{M_{Boxcar} x_{Boxcar,initial} + \sum{Mm_{i}x_{i,initial}}}{M_{Boxcar} + \sum{M_{i}}} \]\end{large}{latex}
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If we define the center of mass as the *zero* position, then {*}<x> = 0{*} and we have
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{latex}

\begin{large}\[ x_{i,initial} = -x_{Boxcar,initial} \frac{M_{Boxcar}}{\sum{M_{i}}}\]\end{large}{latex}
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{latex}\begin{large}\[ x_{i,initial} = x_{Boxcar,initial} - \frac{L}{2} + R \]\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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{latex}
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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:
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{latex}\begin{large} \[ x_{Boxcar,finalal} =  < x >  - \frac{\sum{M_{i}}}{M_{Boxcar,initial} + \sum{M_{i}}}\frac{(L - 2 R)}{2} \]\end{large}{latex}
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Subtracting the Final position of the Boxcar from its Initial Position yields the total movement of the car:

{latex}

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