Consider a system consisting of three particles:

$$\begin{gathered} {\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{kg}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{8}\mathbf{,}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{15}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s} \\ {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{6}\mathbf{kg}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{12}\mathbf{,}\mathbf{9}\mathbf{,}\mathbf{-}\mathbf{6}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s} \\ {\mathbf{m}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{kg}\mathbf{,}{\overrightarrow{\mathbf{v}}}_{\mathbf{3}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{24}\mathbf{,}\mathbf{34}\mathbf{,}\mathbf{23}\mathbf{)}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$

What is${\mathbf{K}}_{\mathbf{rel}}$, the kinetic energy of this system relative to the centre of mass?

The given data can be listed below as,

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

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,

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

For particle two,

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

For particle third,

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

The total momentum of the system is expressed as,

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

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

Substitute all the value in the above equation.

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

The velocity of center of mass is expressed as,

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

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

Substitute all the value in the above equation.

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

The relative velocity of each particle is expressed as,

For particle one,

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

Substitute all the value in the above equation.

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

For particle two,

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

Substitute all the value in the above equation.

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

For particle third,

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

Substitute all the value in the above equation.

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

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).

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

Hence the relative kinetic energy is,3038.97 J .