|!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}\[\ M_{Boxcar} x_{Boxcar, initial} + \sum M_{i} x_{i, initial} = M_{Boxcar} x_{Boxcar, final} + \sum M_{i} x_{i, final} \]\end{large}{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}\[ x_{Boxcar, final} = x_{Boxcar, initial} + \frac{\sum{M_{i} x_{i, initial}} - \sum{M_{i} x_{i, final}}}{M_{Boxcar}} \]\end{large}{latex}
\\
{latex}\begin{large}\[ x_{Boxcar, final} = x_{Boxcar, initial} - \frac{M_{i}}{M_{Boxcar}} ( \sum{x_{i, final}} - \sum{x_{i, initial}}) \]\end{large}{latex}
\\
The location of the Center of Mass *<x>* is given by:
{latex}\begin{large}\[ < x > = \frac{M_{Boxcar} x_{Boxcar,initial} + \sum{M_{i}x_{i,initial}}}{M_{Boxcar} + \sum{M_{i}}} \]\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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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
{latex}\begin{large}\[ x_{i,initial} = x_{Boxcar,initial} - \frac{L}{2} + \frac{R}{2} \]\end{large}{latex}
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Inserting the previous expression into this gives
{latex}\begin{large}\[ < x > = \frac{M_{Boxcar} + \sum{M_{i}(x_{Boxcar,initial} - \frac{L}{2} + \frac{R}{2})}}{M_{Boxcar} + \sum{M_{i}}} \]\end{large}{latex}
\\
Re-arranging this yields:
{latex}\begin{large}\[ < x > = \frac{(M_{Boxcar} + \sum{M_{i}})x_{Boxcar,initial} - M_{i} (\frac{L}{2} - \frac{R}{2})}{M_{Boxcar} + \sum{M_{i}}} \]\end{large}{latex}
Solving for the initial boxcar position yields
{latex}\begin{large} \[ x_{Boxcar,initial} = , x . + \frac{\sum{M_{i}}}{M_{Boxcar,initial} + \sum{M_{i}}}(\frac{(L - R)}{2} \]\end{large}{latex}
\\
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