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## Re: What happens at the interface of two colliding masses?

 Subject: Re: What happens at the interface of two colliding masses? "Edward Green" 23 Dec 2006 14:15:58 -0800 sci.physics
 ```[email protected]/* */ wrote: > In article <[email protected]/* */>, "Edward > Green" <[email protected]/* */> writes: <...> > It is really not easy to guess anything about this without solving > explicitly and even then it appears messy for a while till it > "magically" simplifies. As I recall, the peak strain obtained is > something like u/(2w) where u is the collision velocity and w the > speed of sound in the material. Mind you, for most materials once the > strain is of the order of 1% or so, you enter the inelastic region and > the solution no longer applies. I still think it is of interest, though you might lose interest far beyond the "physical" region. You could say it is of mathematical interest (to characterize the complete solution space of a model), and may have some parallel application in GR (where you might also say it is of mathematical interest), when we extend continuum solutions to extremes. I heard some interest in a MTW symposium, but that seems to have self-extinguished. Anyway, beyond the high-flown generalities... <...> > Now here comes an interesting tidbit. Lets assume (just for > simplicity) that the two blocks are of the same material, but of > different length. Say, block 1 is shorter than block 2. Then the > situation is as follows: > > 1) First, we've shock waves propagating in both blocks away from the > collision face, leaving a "permanent" compression in their wake. > > 2) The wave in block 1 reaches the back face and rebounds, relaxing > the material it is passing through. The wave in block 2 still > proceeds in the original direction. > > 3) The wave in block 1 arrives back at the collision face. This in > block 2 may be still going in the original direction or may be comiong > back (depending on the length of 2) but didn't make it all the way > back yet. > > 4) When 3 occurs the two block separate. Now, block 1 is back to its > relaxed state at this point. Block 2, on the other hand, is left with > the wave still going trhough it and this wave will keep bounding there > and back, indefinitely (in this approximation). Neat. So in this case one block flies off as it was, while the other is left ringing. Maybe you could draw some parallel in this to a clapper depositing energy in a larger bell, itself not significantly excited. Now, I haven't written down the framework for a quantitative solution yet (one of my favorite edicts is that you don't write down a formal solution until you know what the answer is going to be), but here are some thoughts. Again take our model of a semi-infinite slab of elastic solid moving with uniform speed u (to the left) in the lab frame, contacting an immovable wall at time t = 0. Suppose, based on your hints, we guess the following solution: At all positive times there is a static uniformly stressed thickness of material, adjoining, via a step boundary, the remaining unstressed and moving material. Momentum conservation: The stress step is moving to the right at some speed v in the lab frame. The moving material has momentum density r (rho) u and is arriving at the step at relative velocity u + v, whereupon it stops. Therefore the momentum flux across the step, which is the force/area necessary to stop the material (ultimately delivered by the static wall), is s = r u(u+v). So we have an expression for assumed uniform stress s. But what is v? It's tempting to say "Oh, v is w, the speed of sound in the material". But in which frame? Some of the material is stopped in the lab frame, the remainder is at rest in a frame moving to the left at u. At this point you might say "Well, we assume u << w, so this distinction is not important; otherwise the model is not applicable". I don't know if you would say that, but let's not, and see how far we can get. At least one consistency condition remains. Material conservation: At time t, the moving step boundary has swept out a length (u+v)t of uncompressed material. This material is now compressed to a length vt between the moving boundary and the wall. Without prejudicing ourselves to the form, stress is some function of the fractional deformation v/(u+v). Combining our two expressions for stress, we have s[v/(u+v)] = r u(u+v) . In principle then we have solved for the interface velocity v, and hence for stress s, in terms of density, impact velocity, and the relation of stress to strain, by assuming our simple form of the solution. Suppose now we assume our usual linear dependence of stress on strain, which is just one minus the fractional compression. Thus eu/(u+v) = ru(u+v), where e is an elastic constant (I notice that the assumption of small strains is already equivalent to the assumption u << v ; i.e., that the impact velocity is much less than the speed of propagation, whether or not "v" turns out to be "w"). Then v = sqrt(e/r) - u I take it sqrt(e/r) is the expected expression for the velocity of longitudinal sound waves, and the "-u" answers our question "speed of sound wrt what?". The wavefront propagates with the speed of sound wrt rest frame of the undeformed material, hence v, wavefront velocity in the lab frame, is corrected by u. Holy Moses! That was too good to be true! At this point I feel like doing one of those triumphant textbook "So we see ... as it must" things, but I am actually astounded what came out, given what went in. In recap, we made only two assumptions: (A) the solution had a certain simple form (B) linear relation of stress to (engineering) strain, and we used two conservation conditions: (1) conservation of momentum (2) conservation of material In return we showed that the velocity of the wavefront (step) was the expected velocity of sound, and that this velocity is to be measured realtive to the unstressed material. What we _didn't_ assume was that the strain was small (only that the given stress/strain relation held), or a priori what the velocity of propagation was relative to. I notice that this solution at least breaks down when u = w [or sqrt(e/r)], when v = 0, and the material just piles up at the impact point. But I'm not convinced the slightly more general solution -- where we assume a more realistic if still idealized stress/strain dependence -- does so; though you might not want to go there. The functional argument (engineering strain) which we employed in our Hooke's law suffers the material volume to go to zero at finite stress! We can do better than that, employing at least relations which prevent the material from disappearing. ```
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