Chapter 8, Problem 22A

Jogging Jake runs along a train flatcar that moves at the velocities shown. In each case, Jake’s velocity is given relative to the car.

Rank the following from greatest to least.

a. the magnitude of Jake’s momentum relative to the car

b. Jake’s momentum to the right relative to an observer at rest on the ground

To determine

The rank of the magnitude of Jake’s momentum relative to the car from greatest to least.

Answer to Problem 22A

The rank of the magnitude of Jake’s momentum relative to the car from greatest to least is $p_{\text{B}}=p_{\text{D}}>p_{\text{A}}=p_{\text{C}}$ .

Explanation of Solution

Formula used:

The expression for momentum as follows:

  $p=mv$

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

Calculation:

For case A:

The momentum as follows:

  $\begin{array}{l}{p}_{\text{A}}=m{v}_{\text{A}}\\ p_{\text{A}}=m\left|-4\text{m}/\text{s}\right|\\ p_{\text{A}}=\left(4\text{m}/\text{s}\right)m\end{array}$

For case B:

The momentum as follows:

  $\begin{array}{l}{p}_{\text{B}}=m{v}_{\text{B}}\\ p_{\text{B}}=m\left|+6\text{m}/\text{s}\right|\\ p_{\text{B}}=\left(6\text{m}/\text{s}\right)m\end{array}$

For case C:

The momentum as follows:

  $\begin{array}{l}{p}_{\text{C}}=m{v}_{\text{C}}\\ p_{\text{C}}=m\left|+4\text{m}/\text{s}\right|\\ p_{\text{C}}=\left(4\text{m}/\text{s}\right)m\end{array}$

For case D:

The momentum as follows:

  $\begin{array}{l}{p}_{\text{D}}=m{v}_{\text{D}}\\ p_{\text{D}}=m\left|-6\text{m}/\text{s}\right|\\ p_{\text{D}}=\left(6\text{m}/\text{s}\right)m\end{array}$

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

Conclusion:

Thus, the rank of the magnitude of Jake’s momentum relative to the car from greatest to least is $p_{\text{B}}=p_{\text{D}}>p_{\text{A}}=p_{\text{C}}$ .

To determine

The rank of the Jake’s momentum to the right relative to an observer at rest on the ground from greatest to least.

Answer to Problem 22A

The rank of the Jake’s momentum to the right relative to an observer at rest on the ground from greatest to least is $p_{\text{D}}>p_{\text{C}}>p_{\text{A}}=p_{\text{B}}$ .

Explanation of Solution

Formula used:

The expression for momentum as follows:

  $p=m\left(v_{\text{p}}+v_{\text{t}}\right)$

Here, $m$ is the mass, $v_{\text{p}}$ is the velocity of the person, and $v_{\text{t}}$ is the velocity of the train.

Calculation:

For case A:

The momentum at right relative to an observer at rest on the ground as follows:

  $\begin{array}{l}{p}_{\text{A}}=m{\left(v_{\text{p}}+v_{\text{t}}\right)}_{\text{A}}\\ p_{\text{A}}=m\left|\left(-4\text{m}/\text{s}\right)+\left(+8\text{m}/\text{s}\right)\right|\\ p_{\text{A}}=\left(4\text{m}/\text{s}\right)m\end{array}$

For case B:

The momentum at right relative to an observer at rest on the ground as follows:

  $\begin{array}{l}{p}_{\text{B}}=m{\left(v_{\text{p}}+v_{\text{t}}\right)}_{\text{B}}\\ p_{\text{B}}=m\left|\left(+6\text{m}/\text{s}\right)+\left(-10\text{m}/\text{s}\right)\right|\\ p_{\text{B}}=\left(4\text{m}/\text{s}\right)m\end{array}$

For case C:

The momentum at right relative to an observer at rest on the ground as follows:

  $\begin{array}{l}{p}_{\text{C}}=m{\left(v_{\text{p}}+v_{\text{t}}\right)}_{\text{C}}\\ p_{\text{C}}=m\left|\left(+4\text{m}/\text{s}\right)+\left(+6\text{m}/\text{s}\right)\right|\\ p_{\text{C}}=\left(10\text{m}/\text{s}\right)m\end{array}$

For case D:

The momentum at right relative to an observer at rest on the ground as follows:

  $\begin{array}{l}{p}_{\text{D}}=m{\left(v_{\text{p}}+v_{\text{t}}\right)}_{\text{D}}\\ p_{\text{D}}=m\left|\left(-6\text{m}/\text{s}\right)+\left(+18\text{m}/\text{s}\right)\right|\\ p_{\text{D}}=\left(12\text{m}/\text{s}\right)m\end{array}$

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

Conclusion:

Thus, the rank of the Jake’s momentum to the right relative to an observer at rest on the ground from greatest to least is $p_{\text{D}}>p_{\text{C}}>p_{\text{A}}=p_{\text{B}}$ .