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Deck of Cards
idbigdeck
Card
labelPart A
Wiki Markup

|!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. Assume that each Cannonball weighs {*}m{~}i{~}{*} and that there are *N* of them, totalling {*}M{~}i{~}{*}. 

{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}\[ < x >  =  \frac{M_{Boxcar} x_{Boxcar,initial} + \sum{m_{i}x_{i,initial}}}{M_{Boxcar} + M_{i}} \]\end{large}{latex}
\\
If we define the center of mass as the *zero* position, then {*}<x> = 0{*} and we have
\\
{latex}\begin{large}\[ x_{i,initial} = -x_{Boxcar,initial} \frac{M_{Boxcar}}{M_{i}}\]\end{large}{latex}
\\
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}\[ x_{i,initial} = x_{Boxcar,initial} - \frac{L}{2} + R \]\end{large}{latex} 
\\
Inserting the previous expression into this gives

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

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

Solving for the initial boxcar position yields

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

Subtracting the Final position of the Boxcar from its Initial Position yields the total movement of the car:

{latex}\begin{large} \[ x_{Boxcar, initial} - x_{Boxcar, final} = \frac{M_{i}}{M_{Boxcar} + M_{i}} ( L - 2 R ) \]\end{large}{latex}
\\
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. 
{cloak:matha}
{cloak:appa}
Card
labelPart B
Wiki Markup

h2. 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?
h4. Solution

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

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}None.{cloak}

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

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

{cloak:id=appb}

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

{cloak:id=diagb}

 
|!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!|


{cloak:diagb}

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

{cloak:id=mathb}

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

\\
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:
\\
{latex}\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}{latex}
\\
This can be further re-arranged and condensed to give: 

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

\\
Solving for {*}x{~}Boxcar,Q{~}{*}

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

Re-arranging this yields:

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



{cloak:mathb}

{cloak:appb}

Card
labelPart C
Wiki Markup

h2. Part C

!thatnormal3.jpg|width=500!

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?

h4. Solution

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

{toggle-cloak:id=intc} *Interactions:* {cloak:id=intc}Conservation of Momentum and equal and opposite forces.{cloak}

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

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

{cloak:id=appc}

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

{cloak:id=diagc}



{cloak:diagc}

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

{cloak:id=mathc}

We write the equations of motion for the boxcar (with the rest of the cannonballs) and the one fired cannonball.

{latex}\begin{large}\[\sum F_{x} = F_{A}\cos\theta = ma_{x}\]
\[ \sum F_{y} = F_{A}\sin\theta - mg - N = ma_{y}\]\end{large}{latex}



{latex}\begin{large}\[ F_{A}\sin\theta  - mg - N = 0 \]\end{large}{latex}



{latex}\begin{large}\[ N = F_{A}\sin\theta - mg = \mbox{52 N}\]\end{large}{latex}



{cloak:mathc}
{cloak:appc}