Path: ...!2.eu.feeder.erje.net!feeder.erje.net!fu-berlin.de!uni-berlin.de!news.dfncis.de!not-for-mail From: Mikko Newsgroups: sci.physics.research Subject: Re: Inertia and third principle Date: 8 Aug 2024 08:56:30 GMT Lines: 59 Approved: hees@itp.uni-frankfurt.de (sci.physics.research) Message-ID: References: X-Trace: news.dfncis.de 2nVbMzX4sG0ueZu6f2W4Ngu23Yd0nID1Wh0z2r9UMXlUll9tK8jF0oBe6kq8cdRv3b Cancel-Lock: sha1:GnJc78vJYPCz8AjU1GEAV0A2WPI= sha256:o0RQJIYvdV4vVhui/Hg8erCzwFpvuv+NrsstEQ1zEyk= Bytes: 3921 On 2024-08-08 07:03:32 +0000, Luigi Fortunati said: > Luigi Fortunati il 05/08/2024 20:41:20 ha scritto: >> [[Mod. note -- >> To analyze a collision at the level of granularity you're trying to >> achieve (i.e., forces acting on the individual mass particles which >> make up bodies A and B, you need to include the inter-particle forces >> between the particles which make up each body. That is, if we label the >> columns of particles in your animations from left to right as A5, A4, >> A3, A2, A1 (comprising body A) and B1, B2, B3 (comprising body B), >> then we need to take into account that the actual structure is >> effectively >> A5 <--> A4 <--> A3 <--> A2 <--> A1 B1 <--> B2 <--> B3 >> where each '<-->' denotes a spring... > > Exactly, between the particles there are elastic forces that act like springs. > > And as in all springs, if there is compression there are forces, if there is no compression there are no forces. > > Before the collision there are no forces because there is no compression, during the collision suddenly the forces are there because the compression is there: this is why I talk about forces that are activated. > >> To analyze the collision, we really need to write out Newton's 2nd >> law for all the particles, taking into account all the forces. It's >> a bit of a mess, but it will give a precise answer. >> -- jt]] > > Great, I use Newton's 2nd law F=ma. > > In my animation https://www.geogebra.org/m/qterew9m the force that brakes body A is F=ma=15*-3/4=-11.25. > > The one that stops body B and then accelerates it backwards is F=9*(1/4--1)=9*+5/4=+11.25. > > Therefore, the forces acting on bodies A and B during the collision are equal and opposite. > > This result was tormenting for me because I was certain (and I am certain!) that the action of body A on body B cannot be equal to the reaction of body B on body A. > > But my certainty was shattered on the rock of the mathematical result mentioned above. > > With these doubts, I spent days and days trying to clarify every little thing, I consumed an entire notepad filling it with notes, numbers, formulas and concepts. > > And finally, I understood where the solution was. > > The forces acting on bodies A and B are (as I have already said) certainly equal to +11.25 and -11.25. > > But how much of that +11.25 is action force exerted by A against B and how much of that -11.25 is reaction force exerted by B against A? The two forces are equal. If they are not equal to the number shown then there is no reason to show that numbmer. > Certainly, the force +11.25 on body B is exerted entirely by body A. > > However, not all of the -11.25 force on body A is exerted by body B. Yes, it is. You can prove that from the third law or from the conservation of momentum. Each momentum is changed by the product of the force and the duration of the interaction. -- Mikko