Chapter 9, Problem 26A

The mass and speed of three vehicles are shown below.

a. Rank the vehicles by momentum from greatest to least.

b. Rank the vehicles by kinetic energy from greatest to least.

To determine

To rank: The vehicles on the basis of their momentum from greatest to least.

Answer to Problem 26A

The rank of the vehicles by momentum from greatest to least is $p_{\text{B}}>p_{\text{A}}>p_{\text{C}}$ .

Explanation of Solution

Given:

The given diagram representing the mass and speed of three vehicles is shown below.

  

Formula used:

The expression for momentum is,

  $p=mv$

Here, $m$ is the mass and $v$ is the speed.

Calculation:

The momentum for case A is,

  $\begin{array}{l}{p}_{\text{A}}=m_{\text{A}}{v}_{\text{A}}\\ p_{\text{A}}=\left(800\text{ kg}\right)\times \left(1.0\text{m}/\text{s}\right)\\ p_{\text{A}}=800\text{kg}\cdot \text{m}/\text{s}\end{array}$

The momentum for case B is,

  $\begin{array}{l}{p}_{\text{B}}=m_{\text{B}}{v}_{\text{B}}\\ p_{\text{B}}=\left(1000\text{ kg}\right)\times \left(2.0\text{m}/\text{s}\right)\\ p_{\text{B}}=2000\text{kg}\cdot \text{m}/\text{s}\end{array}$

Find the momentum for case C as follows:

  $\begin{array}{l}{p}_{\text{C}}=m_{\text{C}}{v}_{\text{C}}\\ p_{\text{C}}=\left(90\text{ kg}\right)\times \left(8.0\text{m}/\text{s}\right)\\ p_{\text{C}}=720\text{kg}\cdot \text{m}/\text{s}\end{array}$

By comparing the values of the vehicles by momentum is $p_{\text{B}}>p_{\text{A}}>p_{\text{C}}$ .

Conclusion:

Thus, the rank of the vehicles by momentum from greatest to least is $p_{\text{B}}>p_{\text{A}}>p_{\text{C}}$ .

To determine

To rank: The vehicles on the basis of their kinetic energy from greatest to least.

Answer to Problem 26A

The rank of the vehicles by kinetic energy from greatest to least is $K{E}_{\text{C}}>K{E}_{\text{B}}>K{E}_{\text{A}}$ .

Explanation of Solution

Given:

The given diagram representing the mass and speed of three vehicles is shown below.

  

Formula used.

The expression for kinetic energy is,

  $KE=\frac{1}{2}m{v}^2$

Here, $m$ is the mass and $v$ is the speed.

Calculation:

The kinetic energy for case A is,

  $\begin{array}{l}K{E}_{\text{A}}=\frac{1}{2}{m}_{\text{A}}{v}_{\text{A}}^2\\ K{E}_{\text{A}}=\frac{1}{2}\times \left(800\text{ kg}\right)\times {\left(1.0\text{m}/\text{s}\right)}^2\\ K{E}_{\text{A}}=400\text{ J}\end{array}$

The kinetic energy for case B is,

  $\begin{array}{l}K{E}_{\text{B}}=\frac{1}{2}{m}_{\text{B}}{v}_{\text{B}}^2\\ K{E}_{\text{B}}=\frac{1}{2}\times \left(1000\text{ kg}\right)\times {\left(2.0\text{m}/\text{s}\right)}^2\\ K{E}_{\text{B}}=2000\text{ J}\end{array}$

The kinetic energy for case C is,

  $\begin{array}{l}K{E}_{\text{C}}=\frac{1}{2}{m}_{\text{C}}{v}_{\text{C}}^2\\ K{E}_{\text{C}}=\frac{1}{2}\times \left(90\text{ kg}\right)\times {\left(8.0\text{m}/\text{s}\right)}^2\\ K{E}_{\text{C}}=2880\text{ J}\end{array}$

By comparing the values of the vehicles by kinetic energy is $K{E}_{\text{C}}>K{E}_{\text{B}}>K{E}_{\text{A}}$ .

Conclusion:

Thus, the rank of the vehicles by kinetic energy from greatest to least is $K{E}_{\text{C}}>K{E}_{\text{B}}>K{E}_{\text{A}}$ .