Question

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?

Short answer

The relative kinetic energy is,3038.97 J .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

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

Significance of angular speed

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

Determination of the relative kinetic energy

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 .