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There was problem in my high school test which says,

There is a ball of mass $m$ between two walls (seperated at a distance $d$) moving with speed $v$. The collision of ball with the walls is perfectly elastic. What is the force on the wall averaged over a long time?

According to the newton's $2^{nd}$ law, force exerted by the wall on the particle for a very short time interval $\Delta t$ (time gap between just before and just after collision) will be, $F=m\frac{2v}{\Delta t}$

I asked my teacher about this he said, for long time say $T$, the time after which the ball collide with the same wall again will be $\frac{2d}{v}$. This time will be time gap between just before and just after collision as it is small as compared to $T$. So, put $\Delta t = \frac{2d}{v}$

Which gives $F=\frac{mv^2}{d}$

But, I am unable to understand why is this so, say $d=1$ km, $v=100$ m/s, $m=100$ kg, and I think this can easily break a cemented wall, but the above formula gives just $F=1000$ N , which obviously isn't enough to break a wall.

So, either formula not correct or I am unable to understand what is averaged force. Can someone help me figuring it out?

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  • $\begingroup$ "and I think this can easily break a wall,". Tell us why you think that. Doesn't it depend on the wall? $\endgroup$
    – Bob D
    Dec 7, 2023 at 11:47
  • $\begingroup$ @BobD I mean normal cemented wall which is used in house. $\endgroup$
    – Chesx
    Dec 7, 2023 at 12:02

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The average force will be the force averaged over many collisions. So the average force is averaged over the whole time that the ball is moving, including the time taken to travel from one wall to the other between collisions. If the walls are far apart then the time taken by each collision (when the force is applied to the wall) is much smaller than the time between collisions, so the peak force will be much greater than the average force. Indeed, by making $d$ large, we can make the average force as small as we like.

This question is like asking for the average value of a sequence of $100$ followed by $99$ zeros. The peak value is $100$ but the average value is only $1$. And if we have $100$ followed by $999$ zeros then the average value is now only $0.1$.

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  • $\begingroup$ Do you mean $F_{avg}=\frac{\int F\cdot dt}{\int dt}$? But what is the significance of this force? Since the walls aren't experiencing this force? $\endgroup$
    – Chesx
    Dec 7, 2023 at 12:05
  • $\begingroup$ @chesx The time averaged force seems fairly meaningless if you only have one ball. But if you have a large number of balls then you could derive an expression for the average "pressure" on the walls. Maybe this problem is meant as an introduction to the methods of statistical mechanics - but that's just a guess. $\endgroup$
    – gandalf61
    Dec 7, 2023 at 13:42
  • $\begingroup$ Can you tell me where can I see the expression of average pressure, using the average force? I want to see how this force can be used in real life. $\endgroup$
    – Chesx
    Dec 8, 2023 at 9:56
  • $\begingroup$ @Chesx Try this page from an on-line undergraduate Chemistry course: chem.libretexts.org/Courses/New_York_University/… $\endgroup$
    – gandalf61
    Dec 8, 2023 at 10:07
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There is method in this madness as I hope to explain in simple terms.

This is the sort of calculation that you might encounter when considering the kinetic theory of gasses.
The gas molecules are constantly rebounding off the walls of the container and each time a molecule hits a wall it applies an impulsive force on a wall.
The wall in turn imparts an equal magnitude and opposite direction impulsive force on the molecule which changes the momentum of the molecule, the velocity of a molecule is reversed.

Assuming that the molecule hits a wall at right angles and the collision was elastic then the molecule travels all the way to another wall a distance $d$ away rebounds and then cames back to the wall it hit before.
The time of travel for the molecule to go from one one, to the other wall, and then return the first wall is $2d/v$ assuming that this time is much less than the time during which the molecule was actually colliding with the wall.

The change in momentum of the molecule is $2mv$ and so the magnitude of the average force exerted by the wall on the molecule, and so the average force exerted by the molecule on the wall is $\dfrac{2mv}{2d/v}= \dfrac{mv^2}{d}$ - Newton's second law.

Agreed that this will be a very small force but then when considering all the enormous number of molecules in the container also rebounding from the walls of the container, all the very small average forces exerted by the individual molecules exerted on the walls add up to a significant force which then cam be equated to the pressure acting on a wall by dividing the force by the area of the wall.

Please note for brevity and clarity I have missed out the for a gas it is a three dimensional system, the molecules collide with one another, etc and that that will no doubt be covered when you come onto considering a gas as being made up of molecules moving in random directions.

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Average forces don't break a wall, peak forces do. The ball only exerts force on the wall during the very brief period of contact. Unless the walls are extremely close together, most of the time the ball is traveling between the walls and exerting zero force on them. The graph of force per time is a flat line at 0 with large peaks occurring at each collision. The average of this curve with many low values and few high values is a fairly low value. Each time the ball hits the wall, there is a peak of force, and if this force exceeds the wall's breaking point, the wall breaks. But you can decrease the average force on the wall however low you want simply by spacing out the collisions.

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  • $\begingroup$ What is the physical significance of calculating this force? $\endgroup$
    – Chesx
    Dec 8, 2023 at 9:55
  • $\begingroup$ @Chesx Rather than the force-per-time graph mostly being a flat line at 0 with sharp peaks during collisions, you could replace it with a constant force with a magnitude equal to the average force, and get the same aggregate behavior. It would represent the ball starting at a wall and smoothly accelerating at a constant rate to the midway point, where it would then smoothly decelerate to a speed of 0 at the opposite wall. The average force is averaged over time, therefore not varying over time, unlike sudden spikes of force with the wall collisions. $\endgroup$ Dec 8, 2023 at 13:10

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