Chapter 8, Problem 23A

Rick pushes crates starting at rest across a floor for 3 seconds with a net force as shown.

For each crate, rank the following from greatest to least.

a. change in momentum

b. final speed

c. momentum in 3 seconds

To determine

The rank of the change in momentum from greatest to least.

Answer to Problem 23A

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

Explanation of Solution

Given info:

The time $t=3\text{ s}$ .

Formula used:

The expression for change in momentum as follows:

  $\Delta p=Ft$

Here, $F$ is the force exerted and $t$ is the time interval.

Calculation:

For case A:

The change in momentum as follows:

  $\begin{array}{l}\Delta p_{\text{A}}=F_{\text{A}}t\\ \Delta p_{\text{A}}=\left(100\text{ N}\right)\left(3\text{ s}\right)\\ \Delta p_{\text{A}}=300\text{kg}\cdot \text{m}/\text{s}\end{array}$

For case B:

The change in momentum as follows:

  $\begin{array}{l}\Delta p_{\text{B}}=F_{\text{B}}t\\ \Delta p_{\text{B}}=\left(\text{75 N}\right)\left(3\text{ s}\right)\\ \Delta p_{\text{B}}=225\text{kg}\cdot \text{m}/\text{s}\end{array}$

For case C:

The change in momentum as follows:

  $\begin{array}{l}\Delta p_{\text{C}}=F_{\text{C}}t\\ \Delta p_{\text{C}}=\left(50\text{ N}\right)\left(3\text{ s}\right)\\ \Delta p_{\text{C}}=150\text{kg}\cdot \text{m}/\text{s}\end{array}$

By comparing the values, the change in momentum is $\Delta p_{\text{A}}>\Delta p_{\text{B}}>\Delta p_{\text{C}}$ .

Conclusion:

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

To determine

The rank of the final speed from greatest to least.

Answer to Problem 23A

The rank of the final speed from greatest to least is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

Explanation of Solution

Formula used:

The expression for change in momentum as follows:

  $\Delta p=m\Delta v$

Here, $m$ is the mass and $\Delta v$ is the change in velocity.

Calculation:

Refer part (a).

The change in momentum for case A is $\Delta p_{\text{A}}=300\text{kg}\cdot \text{m}/\text{s}$ .

The change in momentum for case B is $\Delta p_{\text{B}}=225\text{kg}\cdot \text{m}/\text{s}$ .

The change in momentum for case C is $\Delta p_{\text{C}}=150\text{kg}\cdot \text{m}/\text{s}$ .

For case A:

The final speed as follows:

  $\begin{array}{l}\Delta p_{\text{A}}=m{v}_{\text{A}}\\ v_{\text{A}}=\frac{300\text{kg}\cdot \text{m}/\text{s}}{30\text{ kg}}\\ v_{\text{A}}=10\text{m}/\text{s}\end{array}$

For case B:

The final speed as follows:

  $\begin{array}{l}\Delta p_{\text{B}}=m{v}_{\text{B}}\\ v_{\text{B}}=\frac{225\text{kg}\cdot \text{m}/\text{s}}{20\text{ kg}}\\ v_{\text{B}}=11.25\text{m}/\text{s}\end{array}$

For case C:

The final speed as follows:

  $\begin{array}{l}\Delta p_{\text{C}}=m{v}_{\text{C}}\\ v_{\text{C}}=\frac{150\text{kg}\cdot \text{m}/\text{s}}{10\text{ kg}}\\ v_{\text{C}}=15\text{m}/\text{s}\end{array}$

By comparing the values, the final speed is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

Hence, the rank of the final speed from greatest to least is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

Conclusion:

Thus, the rank of the final speed from greatest to least is $v_{\text{C}}>v_{\text{B}}>v_{\text{A}}$ .

To determine

The rank of the momentum in 3 seconds from greatest to least.

Answer to Problem 23A

The rank of the momentum in 3 seconds from greatest to least is $p_{\text{A}}>p_{\text{B}}>p_{\text{C}}$ .

Explanation of Solution

Formula used:

The expression for momentum as follows:

  $p=mv$

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

Calculation:

Refer part (a).

The speed for case A is $v_{\text{A}}=10\text{m}/\text{s}$ .

The speed for case B is $v_{\text{B}}=11.25\text{m}/\text{s}$ .

The speed for case C is $v_{\text{C}}=15\text{m}/\text{s}$ .

For case A:

The momentum as follows:

  $\begin{array}{l}{p}_{\text{A}}=m{v}_{\text{A}}\\ p_{\text{A}}=\left(30\text{ kg}\right)\left(10\text{m}/\text{s}\right)\\ p_{\text{A}}=300\text{kg}\cdot \text{m}/\text{s}\end{array}$

For case B:

The momentum as follows:

  $\begin{array}{l}{p}_{\text{B}}=m{v}_{\text{B}}\\ p_{\text{B}}=\left(20\text{ kg}\right)\left(11.25\text{m}/\text{s}\right)\\ p_{\text{B}}=\left(225\text{kg}\cdot \text{m}/\text{s}\right)\end{array}$

For case C:

The momentum as follows:

  $\begin{array}{l}{p}_{\text{C}}=m{v}_{\text{C}}\\ p_{\text{C}}=\left(10\text{ kg}\right)\left(15\text{m}/\text{s}\right)\\ p_{\text{C}}=150\text{kg}\cdot \text{m}/\text{s}\end{array}$

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

Conclusion:

Thus, the rank of the momentum in 3 seconds from greatest to least is $p_{\text{A}}>p_{\text{B}}>p_{\text{C}}$ .