By calculating numerical quantities for a multiparticle system. One can get a concrete sense of the meaning of the relationships ${\overrightarrow{\mathbf{p}}}_{\mathbf{sys}}\mathbf{=}{\mathbf{M}}_{\mathbf{tot}}\overrightarrow{\mathbf{v}}\mathbf{CM}$and ${\mathbf{K}}_{\mathbf{tot}}\mathbf{=}{\mathbf{K}}_{\mathbf{trans}}\mathbf{+}{\mathbf{K}}_{\mathbf{rel}}$. Consider an object consisting of two balls connected by a spring, whose stiffness is 400 N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant, the spring is stretched 0.3m, and the two balls at the ends of the spring have the following masses and velocities: $$\begin{gathered} \mathbf{1}\mathbf{:}\mathbf{5}\mathbf{}\mathbf{kg}\mathbf{.}\left(8,14,0\right)\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \\ \mathbf{2}\mathbf{:}\mathbf{3}\mathbf{}\mathbf{kg}\mathbf{}\left(-5,9,0\right)\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s} \end{gathered}$$

(a)For this system, calculate ${\overrightarrow{\mathbf{p}}}_{\mathbf{sys}}$. (b) Calculate ${\overrightarrow{\mathbf{v}}}_{\mathbf{CM}}$(c) Calculate ${\mathbf{K}}_{\mathbf{tot}}$3. (d) Calculate ${\mathbf{K}}_{\mathbf{trans}}$. (e) Calculate ${\mathbf{K}}_{\mathbf{rel}}$. (f) Here is a way to check your result for ${\mathbf{K}}_{\mathbf{rel}}$. The velocity of a particle relative to the center of mass is calculated by subtracting ${\overrightarrow{\mathbf{v}}}_{\mathbf{CM}}$from the particle’s velocity. To take a simple example, if you’re riding in a car that’s moving with ${\overrightarrow{\mathbf{v}}}_{\mathbf{CM}\mathbf{,}\mathbf{x}}\mathbf{=}\mathbf{20}\mathbf{m}\mathbf{/}\mathbf{s}$and you throw a ball with ${\overrightarrow{\mathbf{v}}}_{\mathbf{CM}\mathbf{,}\mathbf{x}}\mathbf{=}\mathbf{35}\mathbf{m}\mathbf{/}\mathbf{s}$, relative to the car, a bystander on the ground sees the ball moving with ${\mathbf{v}}_{\mathbf{x}}\mathbf{=}\mathbf{55}\mathbf{m}\mathbf{/}\mathbf{s}$So $\overrightarrow{\mathbf{v}}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{CM}}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{rel}}$and therefore we have$\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{rel}}\overrightarrow{\mathbf{v}}\mathbf{=}{\overrightarrow{\mathbf{v}}}_{\mathbf{CM}}$for each mass and calculate the corresponding${\mathbf{K}}_{\mathbf{rel}}$. Compare with the result you obtained in part (e).

• ${\overrightarrow{\mathrm{p}}}_{\mathrm{sys}}={\mathrm{M}}_{\mathrm{tot}}{\overrightarrow{\mathrm{v}}}_{\mathrm{CM}}$
  
• ${\mathrm{K}}_{\mathrm{tot}}={\mathrm{K}}_{\mathrm{trans}}+{\mathrm{K}}_{\mathrm{rel}}$
  
• role="math" localid="1657798446250" $$\begin{gathered} \mathrm{The}\mathrm{stiffness}\mathrm{of}\mathrm{a}\mathrm{spiring}\mathrm{is}400\mathrm{N}/\mathrm{m} \\ \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{Ball}1\mathrm{is}1.5\mathrm{kg}\mathrm{and}\mathrm{the}\mathrm{velocity}\mathrm{is}\left(8,14,0\right)\mathrm{m}/\mathrm{s} \\ \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{Ball}2\mathrm{is}2.3\mathrm{kg}\mathrm{and}\mathrm{the}\mathrm{velocity}\mathrm{is}\left(-5,9,0\right)\mathrm{m}/\mathrm{s} \\ \mathrm{Extension}\mathrm{e}=0.3\mathrm{m} \\ {\mathrm{m}}_1=5\mathrm{kg} \\ {\mathrm{m}}_2=3\mathrm{kg} \end{gathered}$$
  

• $$\begin{gathered} {\overrightarrow{\mathrm{v}}}_1=\left(8\stackrel{⏜}{\mathrm{i}}+14\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ {\overrightarrow{\mathrm{v}}}_2=\left(-5\stackrel{⏜}{\mathrm{i}}+9\stackrel{⏜}{\mathrm{j}}0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

The total momentum of the system is calculated by adding all momenta acting on the system. The following is the formula used to calculate the total momentum of the system,

${\overrightarrow{p}}_{sys}=m_1{v}_1+m_2{v}_2$

Substituting the values in the above expression,

$$\begin{gathered} {\overrightarrow{\mathrm{p}}}_{\mathrm{sys}}=5\mathrm{kg}\times \left(8\stackrel{⏜}{\mathrm{i}}+14\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}+3\mathrm{kg}\times \left(-5\stackrel{⏜}{\mathrm{i}}+9\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s} \\ =\left(40\stackrel{⏜}{\mathrm{i}}+70\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}-15\stackrel{⏜}{\mathrm{i}}+27\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\left(\mathrm{kg}.\mathrm{m}/\mathrm{s}\right) \\ =\left(25\stackrel{⏜}{\mathrm{i}}+97\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence,${\overrightarrow{p}}_{sys}$is $\left(25\stackrel{⏜}{\mathrm{i}}+97\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}$

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

Substituting the values in the above expression,

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_{\mathrm{CM}}=\frac{5\mathrm{kg}\times \left(8\stackrel{⏜}{\mathrm{i}}+14\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}+3\mathrm{kg}\times \left(-5\stackrel{⏜}{\mathrm{i}}+9\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{m}/\mathrm{s}}{5\mathrm{kg}+3\mathrm{kg}} \\ =\frac{\left(25\stackrel{⏜}{\mathrm{i}}+97\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}}{8\mathrm{kg}} \\ =3.12\stackrel{⏜}{\mathrm{i}}+12.12\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\mathrm{m}/\mathrm{s}.........\left(\mathrm{a}\right) \end{gathered}$$

Hence,${\overrightarrow{v}}_{CM}$ is$3.12\stackrel{⏜}{\mathrm{i}}+12.12\stackrel{⏜}{\mathrm{j}}+0\stackrel{⏜}{\mathrm{k}}\mathrm{m}/\mathrm{s}.........\left(\mathrm{a}\right)$

Total kinetic energy${\mathrm{K}}_{\mathrm{tot}}$ is calculated by adding the Translational kinetic energy${\mathrm{K}}_{\mathrm{trans}}$ and Resolution kinetic energy ${\mathrm{K}}_{\mathrm{rel}}$. Then the expression is,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{tot}}={\mathrm{K}}_{\mathrm{trans}}+{\mathrm{K}}_{\mathrm{rel}} \\ =650\mathrm{J}+159\mathrm{J} \\ =809\mathrm{J} \end{gathered}$$

Hence,${\mathrm{K}}_{\mathrm{tot}}$ is 809 J

Translational kinetic energy is determined by the following formula,

$K_{trans}=\frac{1}{2}{m}_1{v^2}_1......\left(1\right)$

Here,

$$\begin{gathered} {\mathrm{v}}_1^2={\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2+{\mathrm{v}}_{\mathrm{z}}^2 \\ {\mathrm{v}}_1^2=8^2+{14}^2+0^2 \\ {\mathrm{v}}_1^2=260 \\ {\mathrm{v}}_1=16.12\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute these values in Equation (1),

$$\begin{gathered} K_{trans}=\frac{1}{2}\times 5\times 260 \\ =650J \end{gathered}$$

Hence,${\mathrm{K}}_{\mathrm{trans}}$ is 650 J

Resolution kinetic energy is determined by the following formula,

$K_{rel}=\frac{1}{2}{m}_2{v}_2^2.......\left(2\right)$

Here,

$$\begin{gathered} {\mathrm{v}}_2^2={\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2+{\mathrm{v}}_{\mathrm{z}}^2 \\ {\mathrm{v}}_2^2={(-5)}^2+9^2+0^2 \\ {\mathrm{v}}_2^2=106 \\ {\mathrm{v}}_2=10.30\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute these values in Equation (2),

$$\begin{gathered} K_{rel}=\frac{1}{2}\times 3\times 106 \\ =159J \end{gathered}$$

Hence,${\mathrm{K}}_{\mathrm{rel}}$ is 159 J

$$\begin{gathered} {\overrightarrow{\mathrm{v}}}_{\mathrm{rel}}=\overrightarrow{\mathrm{v}}-{\overrightarrow{\mathrm{v}}}_{\mathrm{CM}} \\ {\overrightarrow{\mathrm{v}}}_{1\mathrm{rel}}={\overrightarrow{\mathrm{v}}}_1-{\overrightarrow{\mathrm{v}}}_{\mathrm{CM}} \end{gathered}$$

Substituting from equation (a),

$$\begin{gathered} {\overrightarrow{v}}_{1rel}=\left(8,14,0\right)-\left(3.12,12.12,0\right) \\ {\overrightarrow{v}}_{1rel}=\left(4.88,1.88,0\right) \\ {\left|v_{1rel}\right|}^2=\left[{\left(4.88\right)}^2+{\left(1.88\right)}^2\right] \\ =27.33 \end{gathered}$$

$K_{1rel}=\frac{1}{2}{m}_1{v^2}_{1rel}$

Substitute the values in the above expression,

role="math" localid="1657800922425" $$\begin{gathered} {\mathrm{K}}_{1\mathrm{rel}}=\frac{1}{2}\times 5\times \left(27.34\right) \\ =68.34\mathrm{J} \\ {\overrightarrow{\mathrm{v}}}_{2\mathrm{rel}}=\overrightarrow{{\mathrm{v}}_2}-{\overrightarrow{\mathrm{v}}}_{\mathrm{CM}} \end{gathered}$$

Substituting from equation (a),

$$\begin{gathered} {\overrightarrow{v}}_{2rel}=\left(-5,9,0\right)-\left(3.12,12.12,0\right) \\ {\overrightarrow{v}}_{2rel}=\left(-8.12,-3.12,0\right) \\ \left|v_{2rel}\right|=\left[{\left(-8.12\right)}^2+{\left(-3.12\right)}^2\right] \\ =75.66 \\ {K}_{2rel}=\frac{1}{2}{m}_2{v}_{2rel}^2 \end{gathered}$$

Substitute the values in the above expression,

$$\begin{gathered} {\mathrm{K}}_{2\mathrm{rel}}=\frac{1}{2}\times 3\times \left(75.66\right) \\ =113.49\mathrm{J} \\ {\mathrm{K}}_{\mathrm{rel}}={\mathrm{K}}_{1\mathrm{rel}}+{\mathrm{K}}_{2\mathrm{rel}} \\ {\mathrm{K}}_{\mathrm{rel}}=68.35+113.49 \\ =181.84\mathrm{J} \end{gathered}$$

Hence,${\mathrm{K}}_{\mathrm{rel}}$ is higher than the${\mathrm{K}}_{\mathrm{rel}}$ obtained in part (e).