How Does Newton's Third Law Apply to a Rebounding Nitrogen Molecule?

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SUMMARY

The discussion focuses on calculating the average acceleration and force exerted by a nitrogen molecule rebounding off a wall, utilizing Newton's Third Law. The average speed of the nitrogen molecule is 6.70x102 m/s, and its mass is 4.68x10-26 kg. The average acceleration calculated is -3.19x1015 m/s2. To find the average force exerted on the wall, the formula F=ma is applied, resulting in a positive force due to the cancellation of negative signs in the calculations.

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jantyme
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Here is the question I'm having trouble with:

The average speed of a nitrogen molecule is about 6.70x10^2 m/s, and its mass is about 4.68x10^ -26 kg.

(a) If it takes 4.20x10^ -13 s for a nitrogen molecule to hit a wall and rebound with the same speed but moving in an opposite direction (assumed to be the negative direction), what is the average acceleration of the molecule during this time interval?

(b) What average force does the molecule exert on the wall?


I've alread gotten the answer to part (a), I took the change in velocity over that interval of time ( -6.70x10^2 - 6.70x10^2) and devided it by the time. Which makes the answer to part (a) -3.19x10^15 m/s^2. I'm just lost on the second part. I must be using the wrong acceleration in the force equation or something. I know that F=ma but which acceleration do I use and maybe I didn't add all of the forces up? I don't know any help would be wonderful. Thanks in advance for your time. :confused:
 
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jantyme said:
Here is the question I'm having trouble with:

The average speed of a nitrogen molecule is about 6.70x10^2 m/s, and its mass is about 4.68x10^ -26 kg.

(a) If it takes 4.20x10^ -13 s for a nitrogen molecule to hit a wall and rebound with the same speed but moving in an opposite direction (assumed to be the negative direction), what is the average acceleration of the molecule during this time interval?

(b) What average force does the molecule exert on the wall?


I've alread gotten the answer to part (a), I took the change in velocity over that interval of time ( -6.70x10^2 - 6.70x10^2) and devided it by the time. Which makes the answer to part (a) -3.19x10^15 m/s^2. I'm just lost on the second part. I must be using the wrong acceleration in the force equation or something. I know that F=ma but which acceleration do I use and maybe I didn't add all of the forces up? I don't know any help would be wonderful. Thanks in advance for your time. :confused:
You found the average acceleration during that time so what is the average force beyond m * average a? Edit: weird the impulse (force * dt) should equal your original momentum but it's off by a factor of two. Hah nevermind I'm silly.
 
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Newton's Laws

I agree with your value for the acceleration. It sounds like you've done everything correctly there. To get part (b) you have to remember Newton's Third Law. In this case, Newton's Third Law should read: Force of wall on molecule = - (Force of molecule on wall). Now since we know what the acceleration of the molecule is, we know what the force on the molecule is via F=ma. From Newton's Third Law we know that the force on the wall is equal and opposite to this force on the molecule. Thus, F_wall = - (4.68x10^ -26 kg)*(-3.19x10^15 m/s^2). Notice that the force on the wall will be positive (the two negative signs cancel) as it should be since you picked the positive horizontal axis to be towards the wall.

This is the first time I posted anything on this forum and I wanted to try out some LaTeX so ignore the stuff that's below this line.

[tex]F_{molecule,wall} = -F_{wall, molecule}[/tex]
 

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