Question

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.

Short answer

The ranking of the groups according to the center of mass speed will be d > c > a > b, greatest first.

Step-by-step solution

The given data

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

Understanding the concept of the net speed and resultant speed

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)

Calculation of the ranking of the groups according to the center of massa

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