Chapter 7, Problem 7P

To determine

Rank the birds according to the smallest change in momentum to the largest.

Answer to Problem 7P

Rank the birds is $\left(\text{b}\right),\left(\text{a}\right)=\left(\text{e}\right),\left(\text{c}\right),\left(\text{d}\right),\left(\text{f}\right)$.

Explanation of Solution

Write an expression to calculate the change in momentum of the bird.

$\begin{array}{c}\Delta \overrightarrow{p}=m\Delta \overrightarrow{v}\\ =m\left({\overrightarrow{v}}_f-{\overrightarrow{v}}_i\right)\end{array}$

Here, $\Delta \overrightarrow{p}$ is the change in momentum, $m$ is the mass of the bird, $\Delta \overrightarrow{v}$ is the change in velocity of the bird, ${\overrightarrow{v}}_f$ is the final velocity and ${\overrightarrow{v}}_i$ is the initial velocity of the bird.

Conclusion:

Consider east and north as positive perpendicular axes and south and west as negative perpendicular axes.

Substitute $200\text{g}$ for $m$, $10\text{m/s}$ for $v_f$, $-10\text{m/s}$ for $v_i$ in equation (I) to find $\Delta \overrightarrow{p}$ for bird (a).

$\begin{array}{c}\Delta \overrightarrow{p}=\left(200\text{g}\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(10\text{m/s}-\left(-10\text{m/s}\right)\right)\\ =\left(200\times {10}^{-3}\text{kg}\right)\left(20\text{m/s}\right)\\ =4.0\text{kg}\cdot \text{m/s}\text{north}\end{array}$

Substitute $200\text{g}$ for $m$, $10\text{m/s }\widehat{\text{i}}$ for $v_f$, $-10\text{m/s}\widehat{\text{j}}$ for $v_i$ in equation (I) to find $\Delta \overrightarrow{p}$ for bird (b).

$\begin{array}{c}\Delta \overrightarrow{p}=\left(200\text{g}\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(10\text{m/s }\widehat{\text{i}}-\left(-10\text{m/s}\widehat{\text{j}}\right)\right)\\ =\left(200\times {10}^{-3}\text{kg}\right)\left(10\text{m/s }\widehat{\text{i}}+10\text{m/s}\widehat{\text{j}}\right)\\ =2.0\text{kg}\cdot \text{m/s}\left(\widehat{\text{i}}+\widehat{\text{j}}\right)\\ =2.0\text{kg}\cdot \text{m/s}\left(\text{east-south}\right)\end{array}$

Use Pythagoras theorem to calculate the magnitude of the change in momentum.

$\begin{array}{c}\left|\Delta \overrightarrow{p}\right|=\sqrt{{\left(2.0\text{kg}\cdot \text{m/s}\right)}^2+{\left(2.0\text{kg}\cdot \text{m/s}\right)}^2}\\ =\sqrt{2}\left(2.0\text{kg}\cdot \text{m/s}\right)\\ =2.8\text{kg}\cdot \text{m/s}\end{array}$

Substitute $200\text{g}$ for $m$, $20\text{m/s}$ for $v_f$, $-10\text{m/s}$ for $v_i$ in equation (I) to find $\Delta \overrightarrow{p}$ for bird (c).

$\begin{array}{c}\Delta \overrightarrow{p}=\left(200\text{g}\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(20\text{m/s}-\left(-10\text{m/s}\right)\right)\\ =\left(200\times {10}^{-3}\text{kg}\right)\left(30\text{m/s}\right)\\ =6.0\text{kg}\cdot \text{m/s}\text{north}\end{array}$

Substitute $400\text{g}$ for $m$, $20\text{m/s}$ for $v_f$, $-10\text{m/s}$ for $v_i$ in equation (I) to find $\Delta \overrightarrow{p}$ for bird (d).

$\begin{array}{c}\Delta \overrightarrow{p}=\left(400\text{g}\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(20\text{m/s}-\left(-10\text{m/s}\right)\right)\\ =\left(400\times {10}^{-3}\text{kg}\right)\left(30\text{m/s}\right)\\ =12\text{kg}\cdot \text{m/s}\text{north}\end{array}$

Substitute $400\text{g}$ for $m$, $-20\text{m/s}$ for $v_f$, $-10\text{m/s}$ for $v_i$ in equation (I) to find $\Delta \overrightarrow{p}$ for bird (e).

$\begin{array}{c}\Delta \overrightarrow{p}=\left(400\text{g}\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(-20\text{m/s}-\left(-10\text{m/s}\right)\right)\\ =\left(400\times {10}^{-3}\text{kg}\right)\left(-10\text{m/s}\right)\\ =-4.0\text{kg}\cdot \text{m/s}\\ =4.0\text{kg}\cdot \text{m/s}\text{south}\end{array}$

Substitute $400\text{g}$ for $m$, $-30\text{m/s }\widehat{\text{i}}$ for $v_f$, $-10\text{m/s}\widehat{\text{j}}$ for $v_i$ in equation (I) to find $\Delta \overrightarrow{p}$ for bird (f).

$\begin{array}{c}\Delta \overrightarrow{p}=\left(400\text{g}\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(-30\text{m/s }\widehat{\text{i}}-\left(-10\text{m/s}\widehat{\text{j}}\right)\right)\\ =\left(400\times {10}^{-3}\text{kg}\right)\left(-30\text{m/s }\widehat{\text{i}}+10\text{m/s}\widehat{\text{j}}\right)\\ =4.0\text{kg}\cdot \text{m/s}\left(-\text{3}\widehat{\text{i}}+1\widehat{\text{j}}\right)\\ =4.0\text{kg}\cdot \text{m/s}\left(-\text{3west-south}\right)\end{array}$

Use Pythagoras theorem to calculate the magnitude of the change in momentum.

$\begin{array}{c}\left|\Delta \overrightarrow{p}\right|=\sqrt{{\left(-3\left(4.0\text{kg}\cdot \text{m/s}\right)\right)}^2+{\left(4.0\text{kg}\cdot \text{m/s}\right)}^2}\\ =\left(\sqrt{3^2+1^2}\right)\left(4.0\text{kg}\cdot \text{m/s}\right)\\ =\sqrt{10}\left(4.0\text{kg}\cdot \text{m/s}\right)\\ =13\text{kg}\cdot \text{m/s}\end{array}$

Thus, rank the birds is $\left(\text{b}\right),\left(\text{a}\right)=\left(\text{e}\right),\left(\text{c}\right),\left(\text{d}\right),\left(\text{f}\right)$.