Chapter 7, Problem 7P

To determine

Arrange the situations according to the ascending order of change in momentum.

Answer to Problem 7P

$\left(\text{b}\right)\text{<}\left(\text{a}\right)\text{=}\left(\text{e}\right)\text{<}\left(\text{c}\right)\text{<}\left(\text{d}\right)\text{<}\left(\text{f}\right)$.

Explanation of Solution

Write the equation for change in momentum when initial and final velocities are in same direction.

    $\Delta p=m\left(v_{\text{f}}-v_{\text{i}}\right)$        (I)

Here, $\Delta p$ is the magnitude of change in momentum, $m$ is the mass, $v_{\text{f}}$ is the final velocity and $v_{\text{i}}$ is the initial velocity.

Write the equation for change in momentum when initial and final velocities are in opposite direction.

    $\Delta p=m\left(v_{\text{f}}+v_{\text{i}}\right)$        (II)

Write the equation for change in momentum when initial and final velocities are in perpendicular directions.

    $\Delta p=m\sqrt{v_{{}_{\text{f}}}^2+v_{{}_{\text{i}}}^2}$        (III)

Conclusion:

Consider case (a).

Substitute 200 g for $m$, $10\text{m}/\text{s}$ for $v_{\text{f}}$ and $10\text{m}/\text{s}$ for $v_{\text{i}}$ in equation (II).

    $\begin{array}{c}\Delta p=\left(200\text{ g}\right)\left(10\text{m}/\text{s}+10\text{m}/\text{s}\right)\\ =4000\text{g}\cdot \text{m}/\text{s}\end{array}$

Consider case (b).

Substitute 200 g for $m$, $10\text{m}/\text{s}$ for $v_1$ and $10\text{m}/\text{s}$ for $v_{\text{2}}$ in equation (III).

    $\begin{array}{c}\Delta p=\left(200\text{ g}\right)\left|\sqrt{{\left(10\text{m}/\text{s}\right)}^2+{\left(10\text{m}/\text{s}\right)}^2}\right|\\ =\left(200\text{ g}\right)\left(14.14\text{m}/\text{s}\right)\\ =2828\text{g}\cdot \text{m}/\text{s}\end{array}$

Consider case (c).

Substitute 200 g for $m$, $20\text{m}/\text{s}$ for $v_{\text{f}}$ and $10\text{m}/\text{s}$ for $v_{\text{i}}$ in equation (II).

    $\begin{array}{c}\Delta p=\left(200\text{ g}\right)\left(20\text{m}/\text{s}+10\text{m}/\text{s}\right)\\ =6000\text{g}\cdot \text{m}/\text{s}\end{array}$

Consider case (d).

Substitute 400 g for $m$, $20\text{m}/\text{s}$ for $v_{\text{f}}$ and $10\text{m}/\text{s}$ for $v_{\text{i}}$ in equation (II).

    $\begin{array}{c}\Delta p=\left(400\text{ g}\right)\left(20\text{m}/\text{s}+10\text{m}/\text{s}\right)\\ =12000\text{g}\cdot \text{m}/\text{s}\end{array}$

Consider case (e).

Substitute 400 g for $m$, $20\text{m}/\text{s}$ for $v_{\text{f}}$ and $10\text{m}/\text{s}$ for $v_{\text{i}}$ in equation (I).

    $\begin{array}{c}\Delta p=\left(400\text{ g}\right)\left(20\text{m}/\text{s}-10\text{m}/\text{s}\right)\\ =4000\text{g}\cdot \text{m}/\text{s}\end{array}$

Consider case (f).

Substitute 400 g for $m$, $30\text{m}/\text{s}$ for $v_1$ and $10\text{m}/\text{s}$ for $v_{\text{2}}$ in equation (III).

    $\begin{array}{c}\Delta p=\left(400\text{ g}\right)\left|\sqrt{{\left(30\text{m}/\text{s}\right)}^2+{\left(10\text{m}/\text{s}\right)}^2}\right|\\ =\left(400\text{ g}\right)\left(14.14\text{m}/\text{s}\right)\\ =12649\text{g}\cdot \text{m}/\text{s}\end{array}$

Thus, the ascending order of change in momentum is $\left(\text{b}\right)\text{<}\left(\text{a}\right)\text{=}\left(\text{e}\right)\text{<}\left(\text{c}\right)\text{<}\left(\text{d}\right)\text{<}\left(\text{f}\right)$.