91Ó°ÊÓ

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

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

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,

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

Substituting the values in the above expression,

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

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

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

Substituting the values in the above expression,

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

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

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),

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

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),

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

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

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

Substituting from equation (a),

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

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

Substitute the values in the above expression,

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

Substituting from equation (a),

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

Substitute the values in the above expression,

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

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