3
$\begingroup$

enter image description here

Since momentum is not changing in time t , but instantaneously in very small time at wall B.

Why do we write $F=\frac{2mv}{t}$? Where $t$ is (twice the distance AB)/ velocity.

Why don't we take the instantaneous time to write the expression of force?

$\endgroup$
  • $\begingroup$ do you know $F=\frac{dp}{dt} $? $\endgroup$ – jaromrax Mar 16 '17 at 15:01
  • $\begingroup$ Yes I know. dp change of momentum. $\endgroup$ – Ahmad Mar 16 '17 at 15:06
  • $\begingroup$ And what is the change of momentum after the time $t=2AB/v$ ? $\endgroup$ – jaromrax Mar 16 '17 at 15:14
  • $\begingroup$ 2mv but this doesn't happen all the way A to B to A but on wall B $\endgroup$ – Ahmad Mar 16 '17 at 15:40
  • $\begingroup$ i tried to answer - is that visible that i wanted to say, that you transfer $\Delta p$ every x microseconds or so per one particle? $\endgroup$ – jaromrax Mar 16 '17 at 15:59
0
$\begingroup$

enter image description here

I want to emphasize the fact that this is a very rough derivation and you can do more rigorous work using statistical mechanics. The pictorial representation i set up at the side is my interpretation based on how I learnt it and I am more than willing to correct it if it is wrong. Generally though, i think it's fine.

$\endgroup$
  • $\begingroup$ 2mv but this doesn't happen all the way A to B to A but on wall B $\endgroup$ – Ahmad Mar 16 '17 at 15:47
  • $\begingroup$ Of course. But if you think about it, the time period of contact between the particle and the wall is extremely short. So effectively, we say that the change in linear momentum occurs in the time period t=2L/u. Let me be clear that I don't like this derivation myself. I don't think it's very good. It has a lot of assumptions which I feel are unjustified. Statistical mechanics makes things a lot more clear and makes them more legitimate. I'm hoping someone gives a derivation based on that since I'm a little too bummed out at the moment to do it myself. $\endgroup$ – Abhijeet Vats Mar 16 '17 at 15:49
  • $\begingroup$ How the assumption that time t is taken, gives an approximation of the derivation? $\endgroup$ – Ahmad Mar 17 '17 at 14:53
0
$\begingroup$

I try to answer, not to solve the problem:

If you took an instantaneous time, you would get a quasi-infinite force and multiply it with very large number of particles, which is not going to help. And you get stuck with the mechanism of the recoil.

You should not care much about a single particle, rather care about what do all the particles and extract some macroscopic feature ... like $F=\sum_i \Delta p_i/\Delta t_i$ - one particle adds to the sum with the momentum transfered every x miliseconds and the details are not important in a large statistics.

If you wanted to $\lim_{\Delta t \rightarrow 0}$, you would end up again with one recoil dynamics, but what you really care about is an average behavior.

$\endgroup$
  • $\begingroup$ How do you extract macroscopic features in this case? $\endgroup$ – Ahmad Mar 16 '17 at 18:09
  • $\begingroup$ We know time at wall B is going to be very small, so we cannot take that, but why can we take the time from wall A to B to A? $\endgroup$ – Ahmad Mar 16 '17 at 18:15
  • $\begingroup$ You have "zilions" of impacts per second. This way you just realistically estimate, how many impacts per second you have. Estimating a velocity and a typical path.... $\endgroup$ – jaromrax Mar 22 '17 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.