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    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.

    Part A

    Solution One

    System:

    Interactions:

    Model:

    Approach:

    Diagrammatic Representation

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    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

    Mathematical Representation

    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. .

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    \begin

    Unknown macro: {large}

    [\ M_

    Unknown macro: {Boxcar}

    x_

    Unknown macro: {Boxcar, initial}

    + \sum M_

    Unknown macro: {i}

    x_

    Unknown macro: {i, initial}

    = M_

    x_

    Unknown macro: {Boxcar, final}

    + \sum M_

    Unknown macro: {i}

    x_

    Unknown macro: {i, final}

    ]\end

    Here xi is the position of the center of the *i*th cannonball and xBoxcar 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.

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    \begin

    Unknown macro: {large}

    [ N = F_

    Unknown macro: {A}

    = \mbox

    Unknown macro: {300 N}

    ]\end

    Part B

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    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?

    Solution

    System:

    Interactions:

    Model:

    Approach:

    Diagrammatic Representation

    We begin with a free body diagram for the box:

    Unable to render embedded object: File (thatfbd2.jpg) not found.

    Mathematical Representation

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

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    \begin

    Unknown macro: {large}

    [\sum F_

    Unknown macro: {x}

    = F_

    Unknown macro: {A}

    \cos\theta - N = ma_

    ]
    [ \sum F_

    Unknown macro: {y}

    = F_

    Unknown macro: {A}

    \sin\theta - mg = ma_

    ]\end

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

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    \begin

    Unknown macro: {large}

    [ N = F_

    Unknown macro: {A}

    \cos\theta = \mbox

    Unknown macro: {150 N}

    ]\end

    Part C

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    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?

    Solution

    System:

    Interactions:

    Model:

    Approach:

    Diagrammatic Representation

    We begin with a free body diagram for the box:

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

    Mathematical Representation

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

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    \begin

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    [\sum F_

    Unknown macro: {x}

    = F_

    Unknown macro: {A}

    \cos\theta = ma_

    ]
    [ \sum F_

    Unknown macro: {y}

    = F_

    Unknown macro: {A}

    \sin\theta - mg - N = ma_

    ]\end

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

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    \begin

    Unknown macro: {large}

    [ F_

    Unknown macro: {A}

    \sin\theta - mg - N = 0 ]\end

    which we solve to find:

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    \begin

    Unknown macro: {large}

    [ N = F_

    Unknown macro: {A}

    \sin\theta - mg = \mbox

    Unknown macro: {52 N}

    ]\end

    We can check that the y direction is in balance. We have N (52 N) and mg (98 N) on one side, and FA,y on the other (150 N).

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