Page History

|!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 0102.PNG!|
|!Boxcar and Cannonballs 0203.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}\[\ M_{Boxcar} x_{Boxcar, initial} + \sum M_{i} x_{i, initial} = M_{Boxcar} x_{Boxcar, final} + \sum M_{i} x_{i, final}  \]\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.  

{latex}\begin{large}\[ N = F_{A} = \mbox{300 N} \]\end{large}{latex}

{cloak:matha}
{cloak:appa}

h2. Part B

!thatnormal2.jpg|width=500!

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}Box as [point particle].{cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}External influences from the earth ([gravity|gravity (near-earth)]), the wall ([normal force]) and the person ([applied force]).{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}

 We begin with a free body diagram for the box:

!thatfbd2.jpg!

{cloak:diagb}

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

{cloak:id=mathb}

From the free body diagram, we can write the equations of [Newton's 2nd Law|Newton's Second Law]. 

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

Because Because the box is held against the wall, it has no movement (and no acceleration) in the _x_ direction (_a_~x~ = 0).  Setting _a_~x~ = 0 in the _x_ direction equation gives:

{latex}\begin{large}\[ N = F_{A}\cos\theta = \mbox{150 N} \]\end{large}{latex}

{cloak:mathb}

{cloak:appb}

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}Box as [point particle].{cloak}

{toggle-cloak:id=intc} *Interactions:* {cloak:id=intc}External influences from the earth ([gravity|gravity (near-earth)]), the ceiling ([normal force]) and the person ([applied force]).{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}

We begin with a free body diagram for the box:

!thatfbd3.jpg!

{note}The ceiling must push down to prevent objects from moving up through it.{note}

{cloak:diagc}

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

{cloak:id=mathc}

From the free body diagram, we can write the equations of [Newton's 2nd Law|Newton's Second Law]. 

{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}

Because Because the box is held against the ceiling, it has no movement (and no acceleration) in the _y_ direction (_a_~y~ = 0).  Setting _a_~y~ = 0 in the _y_ direction equation gives:

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

which we solve to find:

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

{tip}We can check that the _y_ direction is in balance.  We have N (52 N) and mg (98 N) on one side, and _F_~A,y~ on the other (150 N).{tip}

{cloak:mathc}
{cloak:appc}