At a certain instant, object 1 is at location$\mathbf{(}\mathbf{10}\mathbf{,}\mathbf{-}\mathbf{8}\mathbf{,}\mathbf{6}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}$moving with velocity$\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$. At the same instant object 2 is at location$\mathbf{(}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}$, moving with velocity$\mathbf{(}\mathbf{-}\mathbf{8}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{7}\mathbf{)}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$. a) what is the location of the center of mass of the two equal mass objects? b) What is the velocity of the center of mass?

The given data can be listed below as,

• The location of object 1 is$(10,-8,6)\hspace{0.33em}m$ .
• The velocity of object 1 is$(4,6,-2)\hspace{0.33em}m/s$ .
• The location of object 2 is$(3,0,-2)\hspace{0.33em}m$ .
• The velocity of object 2 is$(-8,2,7)\hspace{0.33em}m/s$ .

The law states that if a rigid object is pushed at their center of mass, then a particular object will continue to move as it is a point mass.

The equation of the position of the center of mass gives the location and the equation of the velocity give the velocity of the center of mass.

a) From the law of the conservation of momentum, the equation of the position of the center of mass for the objects is expressed as:

${\overrightarrow{r}}_{CM}=\frac{m_1{\overrightarrow{r}}_1+m_2{\overrightarrow{r}}_2}{m_1+m_2}$

Here,${\overrightarrow{r}}_{CM}$is the location of the center of mass, $m_1+m_2$are the mass of the objects 1 and 2 respectively that is m and ${\overrightarrow{r}}_{1}and{\overrightarrow{r}}_2$are the location of the objects respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} {\overrightarrow{r}}_{CM}=\frac{m\left(\right(10,-8,6)m+(3,0,-2\left)m\right)}{2m} \\ {\overrightarrow{r}}_{CM}=(6.5,-4,2)\hspace{0.33em}m \end{gathered}$$

Thus, the location of the center of mass of the two equal mass objects are $(6.5,-4,2)\hspace{0.33em}m$

b) a) From the law of the conservation of momentum, the equation of the velocity of the center of mass for the objects is expressed as:

${\overrightarrow{v}}_{CM}=\frac{m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2}{m_1+m_2}$

Here, ${\overrightarrow{v}}_{CM}$is the velocity of the center of mass and ${\overrightarrow{v}}_{1}and{\overrightarrow{v}}_2$are the velocity of the first and the second object respectively.

Substituting the values in the above equation, we get-

$$\begin{gathered} {\overrightarrow{v}}_{CM}=\frac{m\left(\right(4,6,-2)\mathrm{m}/\mathrm{s}+(-8,2,7)\mathrm{m}/\mathrm{s})}{2m} \\ {\overrightarrow{\mathrm{v}}}_{\mathrm{CM}}=(-2,4,2.5)\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of the center of mass is$(-2,4,2.5)\mathrm{m}/\mathrm{s}$