Chapter 7, Problem 71P

(a)

To determine

The velocity of puck B after collision.

Answer to Problem 71P

The speed of puck B is $1.7\text{m/s}$ and is at $30°$ below the x-axis.

Explanation of Solution

Write the expression for conservation of momentum in the x-direction

    $\begin{array}{c}m{v}_{\text{Ax}}+m{v}_{\text{Bx}}=m{v}_{\text{A}}{}^{\prime }\cos \theta +m{v}_{\text{Bx}}{}^{\prime }\\ v_{\text{Bx}}{}^{\prime }=v_{\text{Ax}}+v_{\text{Bx}}-v_{\text{A}}{}^{\prime }\cos \theta \end{array}$        (I)

Here, $m$ is the mass of puck A which is same as puck B, $v_{\text{Ax}}$ is the x-component of the initial velocity of puck A, $v_{\text{Bx}}$ is the x-component of the initial velocity of puck B, $v_{\text{A}}{}^{\prime }$ is the final velocity of puck A, $v_{\text{Bx}}{}^{\prime }$ is the x-component of the final velocity of puck B and $\theta$ is the angle at which puck A moves after collision.

Write the expression for conservation of momentum in the y-direction

    $\begin{array}{c}m{v}_{\text{Ay}}+m{v}_{\text{By}}=m{v}_{\text{A}}{}^{\prime }\sin \theta +m{v}_{\text{By}}{}^{\prime }\\ v_{\text{By}}{}^{\prime }=m{v}_{\text{Ay}}+m{v}_{\text{By}}-m{v}_{\text{A}}{}^{\prime }\sin \theta \end{array}$        (II)

Here, $v_{\text{Ay}}$ is the y-component of the initial velocity of puck A, $v_{\text{By}}$ is the y-component of the initial velocity of puck B, $v_{\text{A}}{}^{\prime }$ is the final velocity of puck A and $v_{\text{By}}{}^{\prime }$ is the y-component of the final velocity of puck B.

Write the expression for magnitude of the velocity of puck B

    $v_{\text{B}}{}^{\prime }=\sqrt{{\left(v_{\text{Bx}}{}^{\prime }\right)}^2+{\left(v_{\text{By}}{}^{\prime }\right)}^2}$        (III)

Here, $v_{\text{B}}{}^{\prime }$ is the final velocity of puck B

Write the expression for direction of the velocity of puck B

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

Conclusion:

Substitute $0$ for $v_{\text{Bx}}$, $2.0\text{m/s}$ for $v_{\text{Ax}}$, $1.0\text{m/s}$ for $v_{\text{A}}{}^{\prime }$ and $60°$ for $\theta$ in equation (I) to find $v_{\text{Bx}}{}^{\prime }$

    $\begin{array}{c}{v}_{\text{Bx}}{}^{\prime }=\left(2.0\text{m/s}\right)+\left(0\text{ m/s}\right)-\left(1.0\text{m/s}\right)\cos \left(60°\right)\\ =1.5\text{}\text{ m/s}\end{array}$

Substitute $0$ for $v_{\text{By}}$, $0$ for $v_{\text{Ay}}$, $1.0\text{m/s}$ for $v_{\text{A}}{}^{\prime }$ and $60°$ for $\theta$ in equation (II) to find $v_{\text{By}}{}^{\prime }$

    $\begin{array}{c}{v}_{\text{By}}{}^{\prime }=\left(0\right)+\left(0\right)-\left(1.0\text{m/s}\right)\sin \left(60°\right)\\ =-0.866\text{}\text{ m/s}\end{array}$

Substitute $1.5\text{}\text{ m/s}$ for $v_{\text{Bx}}{}^{\prime }$ and $-0.866\text{}\text{ m/s}$ for $v_{\text{By}}{}^{\prime }$ in equation (III) to find $v_{\text{B}}{}^{\prime }$

    $\begin{array}{c}{v}_{\text{B}}{}^{\prime }=\sqrt{{\left(1.5\text{}\text{ m/s}\right)}^2+{\left(-0.866\text{}\text{ m/s}\right)}^2}\\ =1.729\text{ m/s}\\ \approx \text{1}\text{.7 m/s}\end{array}$

Substitute $1.5\text{}\text{ m/s}$ for $v_{\text{Bx}}{}^{\prime }$ and $-0.866\text{}\text{ m/s}$ for $v_{\text{By}}{}^{\prime }$ in equation (IV) to find ${\theta }^{\prime }$

    $\begin{array}{c}{\theta }^{\prime }={\tan }^{-1}\left(\frac{-0.866\text{ m/s}}{1.5\text{}\text{ m/s}}\right)\\ =-30.0°\end{array}$

Therefore, the speed of puck B is $1.7\text{m/s}$ and is at $30°$ below the x-axis.

(b)

To determine

The nature of the collision whether, elastic or inelastic.

Answer to Problem 71P

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:

for $v_{\text{A}}$

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.