Chapter 7, Problem 73P

(a)

To determine

The direction and speed of the acrobats after they grab onto each other.

Answer to Problem 73P

The velocity of the acrobats after grabbing onto each other is $0.64\text{ m/s}$ and is at $73°$ above the x-axis.

Explanation of Solution

Write an expression using conservation of angular momentum in the x-direction

    $\begin{array}{c}{m}_1{v}_{\text{1}}\cos {\theta }_1-m_2{v}_{\text{2}}\cos {\theta }_2=\left(m_1+m_2\right)v_{\text{x}}{}^{\prime }\\ v_{\text{x}}{}^{\prime }=\frac{m_1{v}_{\text{1}}\cos {\theta }_1-m_2{v}_{\text{2}}\cos {\theta }_2}{\left(m_1+m_2\right)}\end{array}$        (I)

Here, $m_{\text{1}}$ is the mass of the first acrobat, $m_{\text{2}}$ is the mass of the second acrobat, $v_1$ is the velocity of the first acrobat and $v_{\text{2}}$ is the final velocity of the combined mass and ${\theta }_1$ is the angle of first acrobat with the horizontal and ${\theta }_{\text{2}}$ us the horizontal.

Write the expression for final velocity of the combined mass conservation of momentum in the y-direction

    $\begin{array}{c}{m}_1{v}_1\sin {\theta }_{\text{1}}+m_2{v}_{\text{2}}\sin {\theta }_2=\left(m_1+m_2\right)v_{\text{y}}{}^{\prime }\\ v_{\text{y}}{}^{\prime }=\frac{m_1{v}_{\text{1}}\sin {\theta }_{\text{1}}+m_2{v}_{\text{2}}\sin {\theta }_2}{\left(m_1+m_2\right)}\end{array}$        (II)

Here, $v_{\text{y}}{}^{\prime }$ is the y-component of the final velocity of the combined mass.

Write the expression for magnitude of the velocity of the combined mass

    $v^{\prime }=\sqrt{{\left(v_{\text{x}}{}^{\prime }\right)}^2+{\left(v_{\text{y}}{}^{\prime }\right)}^2}$        (III)

Here, $v^{\prime }$ is the final velocity of combined mass.

Write the expression for direction of the velocity of the combined mass

    ${\theta }^{\prime }={\tan }^{-1}\left(\frac{v_{\text{y}}{}^{\prime }}{v_{\text{x}}{}^{\prime }}\right)$        (IV)

Here, ${\theta }^{\prime }$ is the direction of the velocity of the combined mass.

Conclusion:

Substitute $60\text{ kg}$ for $m_1$, $80\text{ kg}$ for $m_2$, $3.0\text{m/s}$ for $v_1$, $2.0\text{m/s}$ for $v_2$, $10°$ for ${\theta }_1$ and $20°$ for ${\theta }_2$ in equation (I) to find $v_{\text{x}}{}^{\prime }$

    $\begin{array}{c}{v}_{\text{x}}{}^{\prime }=\frac{\left(60\text{ kg}\right)\left(3.0\text{m/s}\right)\cos \left(10°\right)-\left(80\text{ kg}\right)\left(2.0\text{m/s}\right)\cos \left(20°\right)}{\left(60\text{ kg}\right)+\left(80\text{ kg}\right)}\\ =0.19\text{ m/s}\end{array}$

Substitute $60\text{ kg}$ for $m_1$, $80\text{ kg}$ for $m_2$, $3.0\text{m/s}$ for $v_1$, $2.0\text{m/s}$ for $v_2$, $10°$ for ${\theta }_1$ and $20°$ for ${\theta }_2$ in equation (I) to find $v_{\text{y}}{}^{\prime }$

    $\begin{array}{c}{v}_{\text{y}}{}^{\prime }=\frac{\left(60\text{ kg}\right)\left(3.0\text{m/s}\right)\sin \left(10°\right)+\left(80\text{ kg}\right)\left(2.0\text{m/s}\right)\sin \left(20°\right)}{\left(60\text{ kg}\right)+\left(80\text{ kg}\right)}\\ =0.61\text{ m/s}\end{array}$

Substitute $0.19\text{ m/s}$ for $v_{\text{x}}{}^{\prime }$ and $0.61\text{ m/s}$ for $v_{\text{y}}{}^{\prime }$ in equation (III) to find $v_{\text{B}}{}^{\prime }$

    $\begin{array}{c}{v}^{\prime }=\sqrt{{\left(2.34\text{ m/s}\right)}^2+{\left(0.61\text{ m/s}\right)}^2}\\ =0.64\text{ m/s}\end{array}$

Substitute $0.19\text{ m/s}$ for $v_{\text{x}}{}^{\prime }$ and $0.61\text{ m/s}$ for $v_{\text{y}}{}^{\prime }$ in equation (IV) to find ${\theta }^{\prime }$

    $\begin{array}{c}{\theta }^{\prime }={\tan }^{-1}\left(\frac{0.61\text{ m/s}}{2.34\text{ m/s}}\right)\\ =73°\end{array}$

Therefore, the velocity of the acrobats after grabbing onto each other is $0.64\text{ m/s}$ and is at $73°$ above the x-axis.

(b)

To determine

The nature of the collision whether, elastic or inelastic.

Answer to Problem 73P

The collision is elastic.

Explanation of Solution

Write the expression for change kinetic energy

    $\Delta K=\frac{1}{2}m\left({\left(v_{\text{A}}{}^{\prime }\right)}^2+{\left(v_{\text{B}}{}^{\prime }\right)}^2-{\left(v_{\text{A}}\right)}^2\right)$

Here, $\Delta K$ is the change in kinetic energy.

Conclusion:

Substitute $1.0\text{ m/s}$ for $v_{\text{A}}{}^{\prime }$, $1.7\text{ m/s}$ for $v_{\text{B}}{}^{\prime }$ and $2.0\text{ m/s}$ for $v_{\text{A}}$ in the above equation to find $\Delta K$

    $\begin{array}{c}\Delta K=\frac{1}{2}m\left({\left(1.0\text{ m/s}\right)}^2+{\left(1.73\text{ m/s}\right)}^2-{\left(2.0\text{ m/s}\right)}^2\right)\\ \approx 0\end{array}$

Since the change in kinetic energy is zero, the collision is elastic.