The free-body diagrams in Fig. 9-27 give, from overhead views, the horizontal forces acting on three boxes of chocolates as the boxes move over a frictionless confectioner’s counter. For each box, is its linear momentum conserved along the x-axis and the y-axis?

Free body diagrams for three different chocolate boxes as the boxes move over a frictionless confectioner’s counter.

Linear momentum is conserved when there is no external force acting along that direction. In this case, we draw the free body diagram for each box, if the net force along that direction becomes zero, only then can we say that the linear momentum is conserved along that direction.

Formulae:

The net forces in the horizontal direction for conserved force,$\underset{}{\sum F_x=0}$ (1)

The net forces in the vertical direction for conserved force, $\underset{}{\sum F_y=0}$ (2)

Free body diagram for the box (a),

From the above diagram, we can say that the net horizontal forces can be given using equation (1):

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_{\mathrm{x}}=4\mathrm{N}+2\mathrm{N}-6\mathrm{N}} \\ =0\mathrm{N} \end{gathered}$$

From this, we can say that,

The linear momentum for Box (a) in the x direction is conserved.

From the above diagram, we can say that the net vertical forces can be given using equation (2):

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_{\mathrm{y}}=3\mathrm{N}-2\mathrm{N}} \\ =1\mathrm{N} \end{gathered}$$

From this, we can say that,

There is an external force acting on Box (a), so linear momentum along the y direction is not conserved.

Free Body Diagram for Box (b),

From the above diagram, we can say thatthe net horizontal forces can be given using equation (1):

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_x=(8\cos 60°)\mathrm{N}+5\mathrm{N}-2\mathrm{N}-3\mathrm{N}-(8\cos 60°)\mathrm{N}} \\ =0\mathrm{N} \end{gathered}$$

From this, we can say that,

The linear momentum for Box (b) in the x direction is conserved.

From the above diagram, we can say thatthe net vertical forces can be given using equation (2):

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_{\mathrm{y}}=(8\cos 60°)\mathrm{N}+(8\cos 60°)\mathrm{N}} \\ =13.85\mathrm{N} \end{gathered}$$

From this, we can say that,

There is an external force acting on Box (b), so linear momentum along the y direction is not conserved.

Free Body Diagram for Box (c),

From the above diagram, we can say thatthe net horizontal forces can be given using equation (1):

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_x=(6\cos 60°)\mathrm{N}+5\mathrm{N}-4\mathrm{N}-(6\cos 60°)\mathrm{N}} \\ =1\mathrm{N} \end{gathered}$$

From this, we can say that,

There is an external force acting on Box (c), so linear momentum in the x direction is not conserved.

From the above diagram, we can say thatthe net vertical forces can be given using equation (2):

$$\begin{gathered} \underset{}{\sum {\mathrm{F}}_{\mathrm{y}}=(6\cos 60°)\mathrm{N}-(6\cos 60°)\mathrm{N}} \\ =0\mathrm{N} \end{gathered}$$

From this, we can say that,

The linear momentum for Box (c) in the y direction is conserved.