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Boxcar and cannonballs as point particles.

Not Important in this part.

.

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

\begin{large}\[ x_{Boxcar, final} = 

|!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:*  

{cloak:id=appa}

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

{cloak:id=diaga}


|!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

{cloak:diaga}

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

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}\[\ M_{Boxcar} x_{Boxcar, initial} + \frac{\sum M{m_{i} x_{i, initial}} = M- \sum{m_{Boxcari} x_{Boxcari, final} + \sum }}}{M_{i} x_{i, finalBoxcar}}  \]\end{large}

\begin{latex}

Here {*}x{~}i{~}{*}  is the position of the center of the *i*th cannonball and {*}x{~}Boxcar{~}{*} 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:

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

\begin{large}\[ < x >  =  \frac{M_{Boxcar} x_{Boxcar,initial} + \sum{m_{i}x_{i,initial}}}{M_{Boxcar} + M_{i}} \]\end{large}

\begin{large}\[ x_{i,initial} = -x_{Boxcar,initial} \frac{M_{Boxcar}}{M_{i}}\]\end{large}

 \sum{M_{i} x_{i, final}}}{M_{Boxcar}} \]\end{large}{latex}
\\
{latex}\begin{large}\[ x_{Boxcari, finalinitial} = x_{Boxcar, initial} - \frac{M_{iL}}{M2} + R \]\end{large}

\begin{large}\[ < x > = \frac{M_{Boxcar} + \sum{M_{i}(x_{Boxcar,initial} - \frac{L}{2} + R)}}_{Boxcar}} ( \sum{x_{i, final}} - \sum{x_{i, initial}}) \]\end{large}{latex}
\\
The location of the Center of Mass *<x>* is given by:

{latex}\begin{large}\[ < x > = \frac{M_{Boxcar} x+ M_{Boxcar,initiali}} + \]\sum{M_{i}x_{i,end{large}

\begin{large}\[ < x > = \frac{(initial}}}{M_{Boxcar} + \sum{M_{i}}} \]\end{large}{latex}


{cloak:matha}
{cloak:appa}

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

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

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