91Ó°ÊÓ

The given data can be listed below as,

• ${\mathrm{m}}_1=2\mathrm{kg},{\stackrel{→}{\mathrm{v}}}_1=(8,-6,15)\mathrm{m}/\mathrm{s}$
• ${\mathrm{m}}_2=6\mathrm{kg},{\stackrel{→}{\mathrm{v}}}_2=(-12,9,-6)\mathrm{m}/\mathrm{s}$
• ${\mathrm{m}}_3=4\mathrm{kg},{\stackrel{→}{\mathrm{v}}}_3=(-24,34,23)\mathrm{m}/\mathrm{s}$

The expressions of relativistic energy include both rest mass energy and kinetic energy of motion.

The velocity of each particle is expressed as,

For particle one,

$$\begin{gathered} {\mathrm{m}}_1=2\mathrm{kg},{\stackrel{→}{\mathrm{v}}}_1=(8,-6,15)\mathrm{m}/\mathrm{s} \\ {\stackrel{→}{\mathrm{v}}}_1=(8,-6,15)\mathrm{m}/\mathrm{s} \\ {\stackrel{→}{\mathrm{v}}}_1=(8\stackrel{Áåœ}{\mathrm{i}},-6\stackrel{Áåœ}{\mathrm{j}},15\stackrel{Áåœ}{\mathrm{k}})\mathrm{m}/\mathrm{s} \\ {\stackrel{→}{\mathrm{v}}}_1=\sqrt{8^2+{\left(-6\right)}^2+{15}^2}\mathrm{m}/\mathrm{s} \\ =18.2\mathrm{m}/\mathrm{s} \end{gathered}$$

For particle two,

$$\begin{gathered} {\mathrm{m}}_2=6\mathrm{kg},{\stackrel{→}{\mathrm{v}}}_2=(-12,9,-6)\mathrm{m}/\mathrm{s} \\ {\stackrel{→}{\mathrm{v}}}_2=\left(-12,9,-6\right)\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_2=\left(-12\stackrel{Áåœ}{\mathrm{i}},9\stackrel{Áåœ}{\mathrm{j}},-6\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_2=\sqrt{\left({\left(-12\right)}^2+9^2+\left(-6\right)\left(2\right)\right)}\mathrm{m}/\mathrm{s} \\ =16.15\mathrm{m}/\mathrm{s} \end{gathered}$$

For particle third,

$$\begin{gathered} {\mathrm{m}}_3=4\mathrm{kg},{\stackrel{→}{\mathrm{v}}}_3=(-24,34,23)\mathrm{m}/\mathrm{s} \\ {\stackrel{→}{\mathrm{v}}}_3=(-24,34,23)\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_3=(-24\stackrel{Áåœ}{\mathrm{i}},34\stackrel{Áåœ}{\mathrm{j}},23\stackrel{Áåœ}{\mathrm{k}})\mathrm{m}/\mathrm{s} \\ =47.55\mathrm{m}/\mathrm{s} \end{gathered}$$

The total momentum of the system is expressed as,

role="math" localid="1657848421092" ${\stackrel{→}{P}}_{Total}=m_1{\stackrel{→}{v}}_1+m_2{\stackrel{→}{v}}_2+m_3{\stackrel{→}{v}}_3$

Here ${\stackrel{→}{P}}_{Total}$ is the total momentum of the system.

Substitute all the value in the above equation.

role="math" localid="1657848906971" $$\begin{gathered} {\stackrel{→}{\mathrm{P}}}_{\mathrm{Total}}=\left(2\mathrm{kg}\right)×\left(8\stackrel{Áåœ}{\mathrm{i}}-6\stackrel{Áåœ}{\mathrm{j}}+15\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}+\left(6\mathrm{kg}\right)×\left(-12\stackrel{Áåœ}{\mathrm{i}}+9\stackrel{Áåœ}{\mathrm{j}}-6\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}+\left(4\mathrm{kg}\right)×\left(-24\stackrel{Áåœ}{\mathrm{i}}+34\stackrel{Áåœ}{\mathrm{j}}-23\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ =\left(16\stackrel{Áåœ}{\mathrm{i}}-12\stackrel{Áåœ}{\mathrm{j}}+30\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}+\left(-72\stackrel{Áåœ}{\mathrm{i}}+54\stackrel{Áåœ}{\mathrm{j}}-36\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}+\left(-96\stackrel{Áåœ}{\mathrm{i}}+136\stackrel{Áåœ}{\mathrm{j}}+92\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =\left(-152\stackrel{Áåœ}{\mathrm{i}}+178\stackrel{Áåœ}{\mathrm{j}}+86\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The velocity of center of mass is expressed as,

$$\begin{gathered} {\stackrel{→}{V}}_{cm}=\frac{m_1{\stackrel{→}{v}}_1+m_2{\stackrel{→}{v}}_2+m_3{\stackrel{→}{v}}_3}{m_1+m_1+m_3} \\ {\stackrel{→}{V}}_{cm}=\frac{{\stackrel{→}{P}}_{cm}}{m_1+m_1+m_3} \end{gathered}$$

Here ${\stackrel{→}{V}}_{cm}$is the velocity of center of mass of the system.

Substitute all the value in the above equation.

role="math" localid="1657848943249" $$\begin{gathered} {\stackrel{→}{\mathrm{V}}}_{\mathrm{cm}}=\frac{\left(-152\stackrel{Áåœ}{\mathrm{i}}+178\stackrel{Áåœ}{\mathrm{j}}+86\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}}{2\mathrm{kg}+6\mathrm{kg}+4\mathrm{kg}} \\ =\frac{\left(-152\stackrel{Áåœ}{\mathrm{i}}+178\stackrel{Áåœ}{\mathrm{j}}+86\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}}{12\mathrm{kg}} \\ =\left(12.67\stackrel{Áåœ}{\mathrm{i}}+14.83\stackrel{Áåœ}{\mathrm{j}}+7.167\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

The relative velocity of each particle is expressed as,

For particle one,

$\left({\stackrel{→}{v}}_{rel}\right)={\stackrel{→}{v}}_1-{\stackrel{→}{v}}_{cm}$

Substitute all the value in the above equation.

role="math" localid="1657849216872" $$\begin{gathered} {\left({\stackrel{→}{v}}_{rel}\right)}_1=\left(8\stackrel{Áåœ}{\mathrm{i}}-6\stackrel{Áåœ}{\mathrm{j}}+15\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}-\left(-12.67\stackrel{Áåœ}{\mathrm{i}}+14.83\stackrel{Áåœ}{\mathrm{j}}+7.167\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}_1=\left(20.67\stackrel{Áåœ}{\mathrm{i}}-20.83\stackrel{Áåœ}{\mathrm{j}}+7.833\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {{\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}^2}_1=922.49{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}_1=30.37\mathrm{m}/\mathrm{s} \end{gathered}$$

For particle two,

role="math" localid="1657849339830" ${\left({\stackrel{→}{v}}_{rel}\right)}_2={\stackrel{→}{v}}_2-{\stackrel{→}{v}}_{cm}$

Substitute all the value in the above equation.

role="math" localid="1657849381663" $$\begin{gathered} {\left({\stackrel{→}{v}}_{rel}\right)}_2=\left(-12\stackrel{Áåœ}{\mathrm{i}}+9\stackrel{Áåœ}{\mathrm{j}}-6\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}-\left(-12.67\stackrel{Áåœ}{\mathrm{i}}+14.83\stackrel{Áåœ}{\mathrm{j}}+7.167\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}_2=\left(0.67\stackrel{Áåœ}{\mathrm{i}}-5.83\stackrel{Áåœ}{\mathrm{j}}+13.167\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {{\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}^2}_2=207.8{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}_2=14.42\mathrm{m}/\mathrm{s} \end{gathered}$$

For particle third,

${\left({\stackrel{→}{v}}_{rel}\right)}_3={\stackrel{→}{v}}_3-{\stackrel{→}{v}}_{cm}$

Substitute all the value in the above equation.

$$\begin{gathered} {\left({\stackrel{→}{v}}_{rel}\right)}_3=\left(-244\stackrel{Áåœ}{\mathrm{i}}+34\stackrel{Áåœ}{\mathrm{j}}+23\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}-\left(-12.67\stackrel{Áåœ}{\mathrm{i}}+14.83\stackrel{Áåœ}{\mathrm{j}}+7.167\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}_3=\left(-11.33\stackrel{Áåœ}{\mathrm{i}}+19.17\stackrel{Áåœ}{\mathrm{j}}+15.833\stackrel{Áåœ}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {{\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}^2}_3=746.54{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\left({\stackrel{→}{\mathrm{v}}}_{\mathrm{rel}}\right)}_3=27.32\mathrm{m}/\mathrm{s} \end{gathered}$$

The relative kinetic energy to centre of mass is expressed as,

$K_{rel}=\frac{1}{2}{m}_1{\left(v_{rel}\right)}_1^2+\frac{1}{2}{m}_2{\left(v_{rel}\right)}_2^2+\frac{1}{2}{m}_3{\left(v_{rel}\right)}_3^2$ ..(i)

Substitute all the value in the equation (1).

$$\begin{gathered} {\mathrm{K}}_{\mathrm{rel}}=\frac{1}{2}×2\mathrm{kg}×{\left(30.37\mathrm{m}/\mathrm{s}\right)}^2+\frac{1}{2}×6\mathrm{kg}×{\left(14.42\mathrm{m}/\mathrm{s}\right)}^2+\frac{1}{2}×4\mathrm{kg}×{\left(27.32\mathrm{m}/\mathrm{s}\right)}^2 \\ =3038.97\mathrm{J} \end{gathered}$$

Hence the relative kinetic energy is,3038.97 J .