Chapter 7, Problem 75P

To determine

The velocity of the birds after collision.

Answer to Problem 75P

The ratio of the speed of the balls is $0.781$.

Explanation of Solution

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

    $\left(m_1+M_1\right)u_1-\left(m_2+M_2\right)u_2=\left(m_1+m_2\right)v_{\text{x}}-M_1{v}_{\text{1}}\sin {\theta }_1+M_2{v}_{\text{2}}\sin {\theta }_2$

Here, $m_1$, $m_2$ are the masses of the swallows, $M_1$, $M_2$ are the masses of the coconut, $u_1$ is the mass of the first swallow and the first coconut, $u_2$ is the mass of the second swallow and the second coconut, $v_{\text{x}}$ is the horizontal velocity of the combined mass of the bird, ${\theta }_{\text{1}}$ is the angle subtended by the first coconut and ${\theta }_{\text{2}}$ is the angle subtended by the second coconut.

Rearrange the above equation

    $v_{\text{x}}=\frac{\left(m_1+M_1\right)u_1-\left(m_2+M_2\right)u_2+M_1{v}_1\sin {\theta }_1-M_2{v}_2\sin {\theta }_2}{\left(m_1+m_2\right)}$        (I)

Write an expression using conservation of momentum in the y-direction

    $0=\left(m_1+m_2\right)v_{\text{y}}-M_1{v}_1\cos {\theta }_1+M_2{v}_2\cos {\theta }_2$

Here, $v_{\text{y}}$ is the vertical velocity of the combined mass of the birds.

Rearrange the above equation

    $v_{\text{y}}=\frac{M_1{v}_1\cos {\theta }_1-M_2{v}_2\cos {\theta }_2}{\left(m_1+m_2\right)}$        (II)

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

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

Here, $v$ is the magnitude of the velocity of the combined mass of the birds

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

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

Here, $\theta$ is the direction of the velocity of the combined mass of the birds

Conclusion:

Substitute $20\text{ m/s}$ for $u_1$, $15\text{ m/s}$ for $u_2$, $0.270\text{ kg}$ for $m_1$, $0.220\text{ kg}$ for $m_2$, $0.80\text{ kg}$ for $M_1$, $0.70\text{ kg}$ for $M_2$, $10°$ for ${\theta }_1$, $30°$ for ${\theta }_2$, $13\text{m/s}$ for $v_{\text{1}}$ and $14\text{m/s}$ for $v_{\text{2}}$ in equation (I) to find $v_{\text{x}}$

    $\begin{array}{c}{v}_{\text{x}}=\frac{\left[\begin{array}{l}\left(\left(0.270\text{ kg}\right)+\left(0.80\text{ kg}\right)\right)\left(20\text{ m/s}\right)-\left(\left(0.220\text{ kg}\right)+\left(0.70\text{ kg}\right)\right)\left(15\text{ m/s}\right)+\\ \left(0.80\text{ kg}\right)\left(13\text{m/s}\right)\sin 10°-\left(0.70\text{ kg}\right)\left(14\text{m/s}\right)\sin 30°\end{array}\right]}{\left(0.270\text{ kg}+0.220\text{ kg}\right)}\\ =-1.27\text{ m/s}\end{array}$

Substitute $0.270\text{ kg}$ for $m_1$, $0.220\text{ kg}$ for $m_2$, $0.80\text{ kg}$ for $M_1$, $0.70\text{ kg}$ for $M_2$, $10°$ for ${\theta }_1$, $30°$ for ${\theta }_2$, $13\text{m/s}$ for $v_{\text{1}}$ and $14\text{m/s}$ for $v_{\text{2}}$ in equation (I) to find $v_{\text{x}}$

    $\begin{array}{c}{v}_{\text{y}}=\frac{\left(0.80\text{ kg}\right)\left(13\text{m/s}\right)\cos 30°-\left(0.80\text{ kg}\right)\left(14\text{m/s}\right)\cos 30°}{\left(0.270\text{ kg}+0.220\text{ kg}\right)}\\ =-2.39\text{ m/s}\end{array}$

Substitute $1.542\text{ m/s}$ for $v_{\text{x}}{}^{\prime }$ and $-1.828\text{ m/s}$ for $v_{\text{y}}{}^{\prime }$ in equation (III) to find $v_{\text{B}}{}^{\prime }$

    $\begin{array}{c}{v}_{\text{B}}{}^{\prime }=\sqrt{{\left(1.542\text{ m/s}\right)}^2+{\left(-1.828\text{ m/s}\right)}^2}\\ =2.39\text{ m/s}\end{array}$

Substitute $1.542\text{ m/s}$ for $v_{\text{x}}{}^{\prime }$ and $-1.828\text{ m/s}$ for $v_{\text{y}}{}^{\prime }$ in equation (IV) to find $\theta$

    $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-1.828\text{ m/s}}{1.542\text{ m/s}}\right)\\ =-49.8°\end{array}$

Therefore, the ratio of speed of the balls is $0.781$.