Chapter 7, Problem 74P

To determine

The velocity of birds just after collision.

Answer to Problem 74P

The velocity is $20\text{m}/\text{s}\text{at}20°\text{north-west}$ direction.

Explanation of Solution

Speed of first swallow is $20\text{m}/\text{s}$, speed of second swallow is $15\text{m}/\text{s}$, mass of first swallow is $0.270\text{kg}$, mass of coconut of first swallow is $0.80\text{kg}$, mass of second swallow is $0.220\text{kg}$, mass of coconut of second swallow is $0.70\text{kg}$, speed of first coconut after collision is $13\text{m}/\text{s}$ at $10°$ south-west direction, speed of second coconut after collision is $14\text{m}/\text{s}$ at $30°$ north-east direction.

The free body diagram as per the data is shown below in figure 1.

Write the equation for momentum conservation in x-direction for the collision of birds.

$m_1{v}_{1x}+m_2{v}_{2x}=m_3{v}_{3x}+m_4{v}_{4x}+m_5{v}_{5x}$

Here,$m_1$ is the mass of first swallow, $m_2$ is the mass of second swallow, $m_3$ is the mass of coconut of first swallow, $m_4$ is the mass of coconut of second swallow, $m_5$ is the mass of tangled-up swallows, $v_{1x}$ is the x-component of velocity of first swallow, $v_{2x}$ is the x-component of velocity of second swallow, $v_{3x}$ is the x-component of velocity of coconut of first swallow, $v_{4x}$ is the x-component of velocity of coconut of second swallow, and $v_{5x}$ is the x-component of velocity of birds just after the collision.

Rewrite the above equation in terms of $v_{5x}$ by substituting $0\text{m}/\text{s}$ for $v_{1x}$ and $v_{2x}$.

$\begin{array}{c}{m}_1\left(0\text{m}/\text{s}\right)+m_2\left(0\text{m}/\text{s}\right)=m_3{v}_{3x}+m_4{v}_{4x}+m_5{v}_{5x}\\ v_{5x}=-\frac{m_3{v}_{3x}+m_4{v}_{4x}}{m_5}\end{array}$ (I)

Write the equation for momentum conservation in y-direction for the collision of birds.

$m_1{v}_1+m_2{v}_2=m_3{v}_{3y}+m_4{v}_{4y}+m_5{v}_{5y}$

Here, $v_1$ is the velocity of first swallow, $v_2$ is the velocity of second swallow, $v_{3y}$ is the y-component of velocity of coconut of first swallow, $v_{4y}$ is the y-component of velocity of coconut of second swallow, and $v_{5y}$ is the y-component of velocity of birds just after the collision.

$\begin{array}{c}{m}_1{v}_1+m_2{v}_2=m_3{v}_{3y}+m_4{v}_{4y}+m_5{v}_{5y}\\ v_{5y}=\frac{m_1{v}_1+m_2{v}_2-m_3{v}_{3y}-m_4{v}_{4y}}{m_5}\end{array}$ (II)

Write the equation to find the magnitude of velocity of birds just after the collision.

$v=\sqrt{v_{5x}^2+v_{5y}^2}$

Here,$v$ is the magnitude of velocity of birds just after the collision.

Rewrite the above equation by substituting equation (I) and (II).

$v=\sqrt{{\left(-\frac{m_3{v}_{3x}+m_4{v}_{4x}}{m_5}\right)}^2+{\left(\frac{m_1{v}_1+m_2{v}_2-m_3{v}_{3y}-m_4{v}_{4y}}{m_5}\right)}^2}$ (III)

Write the equation to find the direction of $v$.

$\theta ={\tan }^{-1}\left(\frac{v_{5y}}{v_{5x}}\right)$

Here,$\theta$ is the direction of $v$.

Rewrite the above equation by substituting equations (I) and (II).

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\left(\frac{m_1{v}_1+m_2{v}_2-m_3{v}_{3y}-m_4{v}_{4y}}{m_5}\right)}{\left(-\frac{m_3{v}_{3x}+m_4{v}_{4x}}{m_5}\right)}\right)\\ ={\tan }^{-1}\left(\frac{m_1{v}_1+m_2{v}_2-m_3{v}_{3y}-m_4{v}_{4y}}{m_3{v}_{3x}+m_4{v}_{4x}}\right)\end{array}$ (IV)

Conclusion:

Substitute $0.80\text{kg}$ for $m_3$, $\left(13\text{m}/\text{s}\right)\left(\cos 260°\right)$ for $v_{3x}$, $0.70\text{kg}$ for $m_4$, $\left(14\text{m}/\text{s}\right)\left(\cos 60°\right)$ for $v_{4x}$, $1.07\text{kg}$ for $m_1$, $0.92\text{kg}$ for $m_2$, $20\text{m}/\text{s}$ for $v_1$,$-15\text{m}/\text{s}$ for $v_2$, $\left(13\text{m}/\text{s}\right)\left(\sin 260°\right)$ for $v_{3y}$,$\left(1.07\text{kg+}0.92\text{kg}\right)$ for $m_5$, and $\left(14\text{m}/\text{s}\right)\left(\sin 60°\right)$ for $m_{4y}$ in equation (III) to find $v_5$.

$\begin{array}{c}v=\sqrt{\begin{array}{l}{\left(-\frac{\left(0.80\text{kg}\right)\left(\left(13\text{m}/\text{s}\right)\left(\cos 260°\right)\right)+\left(0.70\text{kg}\right)\left(\left(14\text{m}/\text{s}\right)\left(\cos 60°\right)\right)}{\left(1.07\text{kg}+0.92\text{kg}\right)}\right)}^2+\\ {\left(\frac{\left(1.07\text{kg}\right)\left(20\text{m}/\text{s}\right)+\left(0.92\text{kg}\right)\left(-15\text{m}/\text{s}\right)}{\left(1.07\text{kg}+0.92\text{kg}\right)}\right)}^2\\ -{\left(\frac{\left(0.80\text{kg}\right)\left(\left(13\text{m}/\text{s}\right)\left(\sin 260°\right)\right)+\left(0.70\text{kg}\right)\left(\left(14\text{m}/\text{s}\right)\left(\sin 60°\right)\right)}{\left(1.07\text{kg}+0.92\text{kg}\right)}\right)}^2\end{array}}\\ =\sqrt{400}\text{m}/\text{s}\\ =20\text{m}/\text{s}\end{array}$

Substitute $0.80\text{kg}$ for $m_3$, $\left(13\text{m}/\text{s}\right)\left(\cos 260°\right)$ for $v_{3x}$, $0.70\text{kg}$ for $m_4$, $\left(14\text{m}/\text{s}\right)\left(\cos 60°\right)$ for $v_{4x}$, $1.07\text{kg}$ for $m_1$, $0.92\text{kg}$ for $m_2$, $20\text{m}/\text{s}$ for $v_1$,$-15\text{m}/\text{s}$ for $v_2$, $\left(13\text{m}/\text{s}\right)\left(\sin 260°\right)$ for $v_{3y}$,$\left(1.07\text{kg+}0.92\text{kg}\right)$ for $m_5$, and $\left(14\text{m}/\text{s}\right)\left(\sin 60°\right)$ for $m_{4y}$ in equation (IV) to find $\theta$.

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{\begin{array}{l}\left(1.07\text{kg}\right)\left(20\text{m}/\text{s}\right)+\left(0.92\text{kg}\right)\left(-15\text{m}/\text{s}\right)-\left(0.80\text{kg}\right)\left(\left(13\text{m}/\text{s}\right)\left(\sin 260°\right)\right)\\ -\left(0.70\text{kg}\right)\left(\left(14\text{m}/\text{s}\right)\left(\sin 60°\right)\right)\end{array}}{\left(0.80\text{kg}\right)\left(\left(13\text{m}/\text{s}\right)\left(\cos 260°\right)\right)+\left(0.70\text{kg}\right)\left(\left(14\text{m}/\text{s}\right)\left(\cos 60°\right)\right)}\right)\\ ={\tan }^{-1}\left(-3.077\right)\\ =-72°\end{array}$

Negative value of $\tan$ function indicates that the angle lies in second quadrant. That is, the above angle is measured with respect to negative x axis in clockwise direction.

Write the equation to find the angle made by $v_5$ with respect to positive x-axis in counterclockwise direction.

$\varphi =90°-\left|\theta \right|$

Here, $\varphi$ is the angle made by $v_5$ with respect to positive y-axis in counterclockwise direction.

Substitute $-72°$ for $\theta$ in the above equation to find $\varphi$.

$\begin{array}{c}\varphi =90°+\left|-72°\right|\\ =18°\end{array}$

Therefore, the velocity is $20\text{m}/\text{s}\text{at}20°\text{north-west}$ direction.