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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?

Solution

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:

\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}

\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}

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:

\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}

This can be further re-arranged and condensed to give:

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

Solving for x_Boxcar,Q

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

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

For Q = 0 we have

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

In the limit that M_Cannonballs >> M_Boxcar this becomes

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

Similarly, for Q = N we have, in the same limit,

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

These are the same results as in Part A, and once again the most that the Boxcar can move is L, and then only in the limit as R goes to zero. With the formulae developed in this part, however, we can see how much the boxcar moves as each cannonball is moved from one side to the other. Note that the center of the boxcar can only coincide with the overall center of mass if N is an even number, and half the cannonballs are moved from one side to the other.

Part C

Image Added

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

Let's assume that we launch, throw, roll, or fire the cannonballs one at time from one end of the Boxcar to the other, and that the other Cannonballs are held rigidly in place in the Boxcar, so that they move with it. Let;s further assume that when the cannonball reaches the other side it sticks in place on the other side, coming instantly to rest (and doing no damage).

We write the equations of motion for the boxcar (with the rest of the cannonballs) and the one fired cannonball. Let us call the velocity of the velocity of the cannoball v_i and the velocity of the Boxcar (with the remaining cannonballs) v_Boxcar. Assuming Conservation of Momentum we have:

\begin{large}\[ m_{i} v_{i} = - (M_{Boxcar} + (N - 1) m_{i}) v_{Boxcar}\]\end{large}

\begin{large}\[ v_{Boxcar} = - \frac{m_{i}}{M_{Boxcar} + (N - 1) m_{i}}  v_{i} \]\end{large}

So we can write the equations of motion for both the Boxcar (and Cannonballs) and the lone cannonball:

\begin{large}\[ x_{cannonball}(t) = x_{cannonball, initial} + v_{i}t  \]\end{large}

\begin{large}\[ x_{Boxcar}(t) = x_{Boxcar, initial} + v_{Boxcar}t  \]\end{large}

We can substitute for the Boxcar Velocity to get:

\begin{large}\[ x_{Boxcar}(t) = x_{Boxcar, initial} - v_{i}\frac{m_{i}}{M_{Boxcar} + (N - 1)m_{i}}t  \]\end{large}

The Cannonball will stop when it strikes the far wall of the Boxcar at time t_final. At this point the cannonball will have traveled, relative to the Boxcar, a distance L - 2R. But the Boxcar will have moved, as well. The relevant expression relating the distances is thus:

\begin{large}\[ x_{cannonball}(t_{final}) - x_{Boxcar}(t_{final})  = L - 2R  \]\end{large}

Substituting in the above equations and re-arranging, we get:

\begin{large}\[ x_{cannonball,initial} - x_{Boxcar, initial} + v_{i}( 1 + \frac{m_{i}}{M_{Boxcar} + (N - 1)m_{i}})t_{final} = L - 2R  \]\end{large}

But we can see that the difference between the initial position of the cannonball and the initial position of the boxcar is just half the car length, minus the cannonball radius:

\begin{large}\[ x_{cannonball,initial} - x_{Boxcar,initial} = \frac{L}{2} - R \]\end{large}

Solving for t_final, we get:

\begin{large}\[ t_{final} = \frac{(\frac{L}{2} - R)}{v_{i}} \frac{M_{Boxcar} + (N - 1) m_{i}}{M_{Boxcar} + M_{Cannonballs}} \]\end{large}

Inserting this into the above expressions for x_Boxcar(t_f) and x_cannonball(t_f) we get

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

\begin{large}\[ x_{cannonball}(t_{f}) = x_{cannonball, initial} + \frac{M_{Boxcar} + (N - 1)m_{i}}{M_{Boxcar} + M_{Cannonballs}} (\frac{L}{2} - R )  \]\end{large}