# Example: Momentum conservation

## Understanding the situation

The pressure in a container of gas arises from the molecules of the gas hitting the wall. Let's consider in some detail the collision of a single molecule with a wall. (For this problem, we'll ignore the effects of gravity.)

Suppose that a single molecule is moving with a speed and has a mass m. The molecule bounces off the wall and rebounds with a speed that is practically the same as it hit the wall with. As a result of being hit, we expect that the wall will recoil. Assume just for now that the wall is much larger than the molecule and free-floating — not connected to anything else.

## Presenting a sample problem

A. If after the collision, the wall moves to the right with a speed, $V$, write an equation that would allow you to calculate $V$if you were given values for $m$, $M$and $v$. Explain what principles you used to obtain your equation.

B. Suppose the collision occurs in a time interval $\Delta t$. That is, the wall and the molecule are in contact for that amount of time. Can you find the average force, $F_{wall \rightarrow molecule}$ that the wall exerts on the molecule? If you can, give an equation for it in terms of the given parameters of the problem, $m$, $M$, and $v$, and show how you got it. If you can't, explain why not.

C.The wall will also feel a force from the molecule, $F_{molecule \rightarrow} wall$. How does its magnitude compare to the magnitude of $F_{wall \rightarrow molecule}$? Explain why.

D. Suppose the molecule did not hit the wall head on but hit it at an angle, but going at the same speed. Would the force it exerts on the wall be bigger, the same, or less than when it hit the wall head on? Why?

## Solving this problem

We have a system consisting of only two interacting objects so there are no external forces in our system and we can use momentum conservation. Let's use the "initial total momentum = final total momentum" form, remembering that momentum is a vector.

A. For the first three parts for the problem we only have motion in the x direction so we won't bother to specify which component of the vector we are talking about. We'll take motion to the right as being positive.

Our momentum conservation equation for the x component of the two objects is (we label them "m" for "molecule" and "W" for "Wall")

$$p^i_m + p^i_W = p^f_m + p^f_W$$

Putting in the symbols for the initial and final values, our equation becomes

$$mv + 0 = -mv + MV$$

Solving for $V$ give

$$V = \frac{2mv}{M}$$

The principles we used were the conservation of momentum, which implicitly uses N3.

B. Since there is only one force on the molecule — the force from the wall — the Impulse-Momentum form of N2 for the molecule is

$$\Delta p_m = F_{wall \rightarrow molecule} \Delta t$$

We know that

$$\Delta p_m = p^f_m - p^i_m = (-mv) - (mv) = -2mv$$

so the force had to be

$$F_{wall \rightarrow molecule} = \frac{\Delta p_m}{\Delta t} = -\frac{2mv}{\Delta t}$$

The minus sign means that the force was to the left, as we know it had to be.

C. The magnitudes of  $F_{molecule \rightarrow} wall$ and of $F_{wall \rightarrow molecule}$ have to be the same, by N3.

D. When the molecule approaches the wall at an angle, if the wall is smooth, it will bounce off at the same angle that it entered at as shown by the black dotted line. The initial momentum of the molecule now has both x and y components. When it reflects, only the x-component will change. It will reverse. The y-component will stay the same. Therefore, our change in momentum will be (using an x component label instead of an object label)

$$\Delta p_x = p^f_x - p^i_x = (-mv_x) - (mv_x) = -2mv_x$$

Since the magnitude of the x-component is less than the full magnitude of the velocity, the change will be less than it was when it head head-on and so the force (equal to the change in momentum over the interaction time) will be less.

Joe Redish 2/7/19

Article 538
Last Modified: February 24, 2019