Step 1

We have given the following values:

The weight of each box A and B is $W = 160\;{\rm{lb}}$.

The weight of conveyor is ${W_C} = 500\;{\rm{lb}}$.

The initial velocity of conveyor is ${u_C} = 0\;{\rm{ft}}/{\rm{s}}$.

The initial velocity of box A is ${u_A} = 0\;{\rm{ft}}/{\rm{s}}$.

The initial velocity of box B is ${u_B} = 0\;{\rm{ft}}/{\rm{s}}$.

The relative velocity of conveyor with the box B or box A is ${v_{C/B}} = 3\;{\rm{ft/s}}$.

We are asked to calculate the final velocity of conveyor when (a) the boxes are not stacked and A falls off then B falls off, and (b) A is stacked on top of B and both fall off together.

 
Step 2

Part (a)

Apply conservation of linear momentum to the boxes and conveyor:

\[{m_A}{u_A} + {m_B}{u_B} + {m_C}{u_C} = {m_A}{v_A} + {m_B}{v_B} + {m_C}{v_C}\]

Substitute the value of ${u_A}$, ${u_B}$, and ${u_C}$ in the above equation:

\[\begin{array}{c} {m_A}\left( {0\;{\rm{ft}}/{\rm{s}}} \right) + {m_B}\left( {0\;{\rm{ft}}/{\rm{s}}} \right) + {m_C}\left( {0\;{\rm{ft}}/{\rm{s}}} \right) = {m_A}\left( { - {v_A}} \right) + {m_B}\left( { - {v_B}} \right) + {m_C}{v_C}\\ {m_A}{v_A} + {m_B}{v_B} = {m_C}{v_C}\\ \left( {\frac{W}{g}} \right){v_A} + \left( {\frac{W}{g}} \right){v_B} = \left( {\frac{{{W_C}}}{g}} \right){v_C}\\ W{v_A} + W{v_B} = {W_C}{v_C} \end{array}\]

Substitute the value $W$ and ${W_C}$ in the above equation:

\[\begin{array}{c} \left( {160\;{\rm{lb}}} \right){v_A} + \left( {160\;{\rm{lb}}} \right){v_B} = \left( {500\;{\rm{lb}}} \right){v_C}\\ 160{v_A} + 160{v_B} = 500{v_C} \end{array}\]......(1)

Here, the both box are places on the same conveyor. Then, the velocity of both boxes is equal.

Substitute ${v_A} = {v_B}$ in equation (1):

\[\begin{array}{c} 160\left( {{v_B}} \right) + 160{v_B} = 500{v_C}\\ 320{v_B} = 500{v_C}\\ {v_B} = 1.5625{v_C} \end{array}\]......(2)

Consider the velocities along the positive direction of x-axis as positive.

The velocity of box B can be written as:

\[{v_B} = \left( { - {v_C}} \right) + {v_{C/B}}\]

Substitute the value of ${v_{C/B}}$ in the above equation:

\[\begin{array}{l} {v_B} = \left( { - {v_C}} \right) + \left( {3\;{\rm{ft}}/{\rm{s}}} \right)\\ {v_B} = 3\;{\rm{ft}}/{\rm{s}} - {v_C} \end{array}\]......(3)

Substitute the value of ${v_B}$ from equation (2) into equation (3).

\[\begin{array}{c} 1.5625{v_C} = 3\;{\rm{ft}}/{\rm{s}} - {v_C}\\ 2.5625{v_C} = 3\;{\rm{ft}}/{\rm{s}}\\ {v_C} = 1.17\;{\rm{ft}}/{\rm{s}} \end{array}\]

Part (b)

If box A is stacked on the top of box B and both fall off together, this will not alter the momentum of the system and the momentum will be constant. Thus the final speed of the conveyor is:

\[{v_C} = 1.17\;{\rm{ft}}/{\rm{s}}\]