Chapter 7, Problem 10P

To determine

Rank the momentum from smallest to largest.

Answer to Problem 10P

The momentum from smallest to largest is $\underset{\_}{\left(\text{e}\right)\text{=}\left(\text{d}\right)\text{<}\left(\text{a}\right)\text{=}\left(\text{b}\right)\text{<}\left(\text{c}\right)}$

Explanation of Solution

Write the expression for kinetic energy.

  $E_{\text{kinetic}}=\frac{1}{2}m{v}^2$        (I)

Here, $m$ is the mass, $v$ is the velocity.

Write the expression for momentum.

  $p=mv$        (II)

Conclusion:

Substitute, $8\text{kg}$ for $m$, and $400\text{J}$ for $E_{\text{kinetic}}$ in equation (I) to find the velocity.

  $\begin{array}{c}400\text{J}=\frac{1}{2}\left(8\text{kg}\right)v^2\\ v_a=10\text{m}/\text{s}\end{array}$

Substitute, $8\text{kg}$ for $m$, and $10\text{m}/\text{s}$ for $v_a$ in equation (II) to find $p_a$.

  $\begin{array}{c}{p}_a=\left(8\text{kg}\right)\left(10\text{m}/\text{s}\right)\\ =80\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute, $2\text{kg}$ for $m$, and $1600\text{J}$ for $E_{\text{kinetic}}$ in equation (I) to find the velocity.

  $\begin{array}{c}1600\text{J}=\frac{1}{2}\left(2\text{kg}\right)v^2\\ v_b=40\text{m}/\text{s}\end{array}$

Substitute, $2\text{kg}$ for $m$, and $40\text{m}/\text{s}$ for $v_b$ in equation (II) to find $p_b$.

  $\begin{array}{c}{p}_b=\left(2\text{kg}\right)\left(40\text{m}/\text{s}\right)\\ =80\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute, $4\text{kg}$ for $m$, and $1600\text{J}$ for $E_{\text{kinetic}}$ in equation (I) to find the velocity.

  $\begin{array}{c}1600\text{J}=\frac{1}{2}\left(4\text{kg}\right)v^2\\ v_c=28.3\text{m}/\text{s}\end{array}$

Substitute, $4\text{kg}$ for $m$, and $28.3\text{m}/\text{s}$ for $v_c$ in equation (II) to find $p_c$.

  $\begin{array}{c}{p}_c=\left(4\text{kg}\right)\left(28.3\text{m}/\text{s}\right)\\ =113.2\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute, $16\text{kg}$ for $m$, and $100\text{J}$ for $E_{\text{kinetic}}$ in equation (I) to find the velocity.

  $\begin{array}{c}100\text{J}=\frac{1}{2}\left(16\text{kg}\right)v^2\\ v_d=3.54\text{m}/\text{s}\end{array}$

Substitute, $16\text{kg}$ for $m$, and $3.54\text{m}/\text{s}$ for $v_b$ in equation (II) to find $p_d$.

  $\begin{array}{c}{p}_d=\left(16\text{kg}\right)\left(3.54\text{m}/\text{s}\right)\\ =56.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

Substitute, $1\text{kg}$ for $m$, and $1600\text{J}$ for $E_{\text{kinetic}}$ in equation (I) to find the velocity.

  $\begin{array}{c}1600\text{J}=\frac{1}{2}\left(1\text{kg}\right)v^2\\ v_e=56.6\text{m}/\text{s}\end{array}$

Substitute, $1\text{kg}$ for $m$, and $56.57\text{m}/\text{s}$ for $v_c$ in equation (II) to find $p_c$.

  $\begin{array}{c}{p}_c=\left(1\text{kg}\right)\left(56.57\text{m}/\text{s}\right)\\ =56.6\text{kg}\cdot \text{m}/\text{s}\end{array}$

Hence the momentum from smallest to largest is $\underset{\_}{\left(\text{e}\right)\text{=}\left(\text{d}\right)\text{<}\left(\text{a}\right)\text{=}\left(\text{b}\right)\text{<}\left(\text{c}\right)}$.