Chapter 8, Problem 21A

Below are before-and-after pictures of a car’s speed. The mass of the car doesn’t change.

Rank the following from greatest to least.

a. the magnitude of momentum change

b. the magnitude of the impulse producing the momentum change

To determine

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

Answer to Problem 21A

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

Explanation of Solution

Formula used:

The expression for the change in velocity as follows:

  $\Delta v=v_{\text{f}}-v_{\text{i}}$

Here, $v_{\text{f}}$ is the final velocity and $v_{\text{i}}$ is the initial velocity.

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:

For case A:

The change in velocity as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{A}}=\left(20\text{m}/\text{s}\right)-\left(10\text{m}/\text{s}\right)\\ \Delta v_{\text{A}}=10\text{m}/\text{s}\end{array}$

The change in momentum as follows:

  $\begin{array}{l}\Delta p_{\text{A}}=m\Delta v_{\text{A}}\\ \Delta p_{\text{A}}=\left(10\text{m}/\text{s}\right)m\end{array}$

For case B:

The change in velocity as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{B}}=\left(15\text{m}/\text{s}\right)-\left(20\text{m}/\text{s}\right)\\ \Delta v_{\text{B}}=-5\text{m}/\text{s}\end{array}$

The change in momentum as follows:

  $\begin{array}{l}\Delta p_{\text{B}}=m\Delta v_{\text{B}}\\ \Delta p_{\text{B}}=\left(-5\text{m}/\text{s}\right)m\end{array}$

For case C:

The change in velocity as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{C}}=\left(10\text{m}/\text{s}\right)-\left(0\text{m}/\text{s}\right)\\ \Delta v_{\text{C}}=10\text{m}/\text{s}\end{array}$

The change in momentum as follows:

  $\begin{array}{l}\Delta p_{\text{C}}=m\Delta v_{\text{C}}\\ \Delta p_{\text{C}}=\left(10\text{m}/\text{s}\right)m\end{array}$

For case D:

The change in velocity as follows:

  $\begin{array}{l}\Delta v=v_{\text{f}}-v_{\text{i}}\\ \Delta v_{\text{D}}=\left(15\text{m}/\text{s}\right)-\left(15\text{m}/\text{s}\right)\\ \Delta v_{\text{D}}=0\text{m}/\text{s}\end{array}$

The change in momentum as follows:

  $\begin{array}{l}\Delta p_{\text{D}}=m\Delta v_{\text{D}}\\ \Delta p_{\text{D}}=\left(0\text{m}/\text{s}\right)m\\ \Delta p_{\text{D}}=0\end{array}$

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

Hence, the rank of the magnitude of momentum change from greatest to least is $\Delta p_{\text{A}}=\Delta p_{\text{C}}>\Delta p_{\text{D}}>\Delta p_{\text{B}}$ .

Conclusion:

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

To determine

The rank of the magnitude of the impulse producing the momentum change from greatest to least.

Answer to Problem 21A

The rank of the magnitude of the impulse producing the momentum change from greatest to least is ${\left(\text{impulse}\right)}_{\text{A}}={\left(\text{impulse}\right)}_{\text{C}}>{\left(\text{impulse}\right)}_{\text{D}}>{\left(\text{impulse}\right)}_{\text{B}}$ .

Explanation of Solution

Formula used:

The expression for the impulse as follows:

  $\text{impulse}=\Delta p$

Here, $\Delta p$ is the momentum.

Calculation:

Impulse exerted is the change in momentum. That is impulse producing is directly proportional to the change in momentum.

Refer part (a).

The change in momentum for case A is $\Delta p_{\text{A}}=\left(10\text{m}/\text{s}\right)m$ .

The change in momentum for case B is $\Delta p_{\text{B}}=\left(-5\text{m}/\text{s}\right)m$ .

The change in momentum for case C is $\Delta p_{\text{C}}=\left(10\text{m}/\text{s}\right)m$ .

The change in momentum for case D is $\Delta p_{\text{D}}=0$ .

By comparing the values, the change in velocity is $\Delta v_{\text{A}}=\Delta v_{\text{C}}>\Delta v_{\text{D}}>\Delta v_{\text{B}}$ .

Hence, the rank of the magnitude of the impulse producing the momentum change from greatest to least is ${\left(\text{impulse}\right)}_{\text{A}}={\left(\text{impulse}\right)}_{\text{C}}>{\left(\text{impulse}\right)}_{\text{D}}>{\left(\text{impulse}\right)}_{\text{B}}$ .

Conclusion:

Thus, the rank of the magnitude of the impulse producing the momentum change from greatest to least is ${\left(\text{impulse}\right)}_{\text{A}}={\left(\text{impulse}\right)}_{\text{C}}>{\left(\text{impulse}\right)}_{\text{D}}>{\left(\text{impulse}\right)}_{\text{B}}$ .