Figure 9-28 shows four groups of three or four identical particles that move parallel to either the x-axis or the y-axis, at identical speeds. Rank the groups according to center-of-mass speed, greatest first.

The figure for groups of three or four identical particles which move parallel to either x or y axis at identical speeds.

Each group consists of particles of equal mass and velocity but in different directions of velocity. We calculate the net velocity in the x and y-direction. Using this, we calculate the resultant velocity. From this, we rank them from greatest first and smallest to last.

Formulae:

The resultant of a vector,$V_{net}=\sqrt{v_x^2+v_y^2}$ (1)

The net velocity in the horizontal direction,$v_{net}^x=\underset{}{\sum v_x}$ (2)

The net velocity in the vertical direction, $v_{net}^y=\underset{}{\sum v_y}$ (3)

$$\begin{gathered} \underset{}{\sum v_x=v-v} \\ =0 \end{gathered}$$For diagram (a)

The net horizontal velocity is given using equation (2):

$$\begin{gathered} \underset{}{\sum v_x=v-v} \\ =0 \end{gathered}$$

The net vertical velocity is given using equation (3):

$\underset{}{\sum v_y=-v}$

We have, the resultant velocity using equation (1) as follows:

$$\begin{gathered} v_{net}=\sqrt{0^2+{(-v)}^2} \\ =v \end{gathered}$$

For diagram (b)

The net horizontal velocity is given using equation (2):

$$\begin{gathered} \underset{}{\sum v_x=v-v} \\ =0 \end{gathered}$$

The net vertical velocity is given using equation (3):

$$\begin{gathered} \underset{}{\sum v_y=v-v} \\ =0 \end{gathered}$$

We have, the resultant velocity using equation (1) as follows:

$$\begin{gathered} v_{net}=\sqrt{0^2+0^2} \\ =0 \end{gathered}$$

For diagram (c)

The net horizontal velocity is given using equation (2):

$$\begin{gathered} \underset{}{\sum v_x=v-v} \\ =0 \end{gathered}$$

The net vertical velocity is given using equation (3):

$$\begin{gathered} \underset{}{\sum {\mathrm{v}}_{\mathrm{y}}=-\mathrm{v}-\mathrm{v}} \\ =-2\mathrm{v} \end{gathered}$$

We have, the resultant velocity using equation (1) as follows:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{net}}=\sqrt{0^2+{(-2\mathrm{v})}^2} \\ =2\mathrm{v} \end{gathered}$$

For diagram (d)

The net horizontal velocity is given using equation (2):

$$\begin{gathered} \underset{}{\sum {\mathrm{v}}_{\mathrm{x}}=\mathrm{v}+\mathrm{v}} \\ =2\mathrm{v} \end{gathered}$$

The net vertical velocity is given using equation (3):

$$\begin{gathered} \underset{}{\sum {\mathrm{v}}_{\mathrm{y}}=-\mathrm{v}-\mathrm{v}} \\ =-2\mathrm{v} \end{gathered}$$

We have, the resultant velocity using equation (1) as follows:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{net}}=\sqrt{{\left(2\mathrm{v}\right)}^2+{(-2\mathrm{v})}^2} \\ =\left(2\sqrt{2}\right)\mathrm{v} \end{gathered}$$

From the above calculations, we rank the groups as d > c > a > b