Chapter 9, Problem 55A

a.

To determine

The molecule’s impulse on the wall.

Answer to Problem 55A

  $\left(5.17\times {10}^{-23}\right)\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

The mass of the nitrogen molecule, $m=4.7\times {10}^{-26}\text{kg}$

Initial velocity of the molecule, $v_i=550\text{}\text{m/s}$

Final velocity of the molecule, $v_f=550\text{}\text{m/s}$ in the opposite direction

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }F\Delta t=p_f-p_i$

Where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Assume the initial direction of motion of the molecule be positive, then the opposite direction will be negative. Then, the velocities will be,

  $\begin{array}{c}{v}_i=550\text{m/s}\\ v_f=-550\text{m/s}\end{array}$

Using the given values, the impulse on the molecule will be,

  $\begin{array}{c}\text{Impulse}=p_f-p_i\\ =\left(4.7\times {10}^{-26}\right)\left(-550-550\right)\\ =\left(-5.17\times {10}^{-23}\right)\text{kg}\cdot \text{m/s}\end{array}$

Then, the molecule’s impulse on the wall will be in opposite direction of above Impulse,

  $\begin{array}{c}{\left(\text{Impulse}\right)}_{\text{molecule on wall}}=-\left(-5.17\times {10}^{-23}\right)\text{kg}\cdot \text{m/s}\\ =\left(5.17\times {10}^{-23}\right)\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the molecule’s momentum on the wall is $\left(5.17\times {10}^{-23}\right)\text{kg}\cdot \text{m/s}$ .

b.

To determine

The average force on the wall if there are $1.5\times {10}^{23}$ of this collision each second.

Answer to Problem 55A

  $7.755\text{N}$

Explanation of Solution

Given:

The mass of the nitrogen molecule, $m=4.7\times {10}^{-26}\text{kg}$

Initial velocity of the molecule, $v_i=550\text{}\text{m/s}$

Final velocity of the molecule, $v_f=550\text{}\text{m/s}$ in the opposite direction

From part (a), the molecule’s impulse on the wall, $\left(5.17\times {10}^{-23}\right)\text{kg}\cdot \text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }F\Delta t=p_f-p_i$

Where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Then the force on the molecule due to the collisions will be,

  $\begin{array}{c}{F}_{avg}\Delta t=n\times \text{Impulse}\\ F_{avg}=\frac{n\times \text{Impulse}}{\Delta t}\end{array}$

Using all the given values in above,

  $\begin{array}{c}{F}_{avg}\Delta t=n\times \text{Impulse}\\ F_{avg}=\frac{1.5\times {10}^{23}\times \left(-5.17\times {10}^{-23}\right)}{1}\\ F_{avg}=-7.755\text{N}\end{array}$

Then, the force on the wall will be,

  $\begin{array}{c}{\left(F_{avg}\right)}_{wall}=-F_{avg}\\ =7.755\text{N}\end{array}$

Conclusion:

Thus, the average force on the wall is $7.755\text{N}$ .