Hi Olin, On Mon, Oct 12, 2009 at 6:05 PM, Olin Lathrop w= rote: > Sean Breheny wrote: >> If you have a perfectly elastic material, the maximum force happens at >> the end of the compression, when the spring is completely compressed. >> If the material is perfectly inelastic, but still compresses at a >> constant rate for a given force, then the force will be constant >> during the deceleration. > > I agree, but don't see what this has to do with bouncing. =A0Your stateme= nt > above is a reply to mine where I debunked your assertion that bouncing > increases the G force beyond what it takes to get to a stop. Those statements were just to set up the argument. Also, I think it is a little early to assume that you have debunked my argument. > > Maybe you would have seen this yourself if you had properly snipped and > replied to specific points in my post. =A0As it is, your comments don't m= ake > sense within the context. > >> If the spring and the inelastic material >> compress at the same rate initially, then the (constant) force of the >> inelastic compression will be equal to the force of the spring at some >> point early in its compression, which will be less than the force >> later in the compression. > > I don't agree with this, but since it has nothing to do with the point at > hand I won't go into it. It has everything to do with the point at hand. Consider that the total impulse imparted by the floor upon the impacting object is mv for something which does not bounce and 2mv for something which does bounce. If we consider both events to take about the same total time, then the average force is higher for the bounce. My attempt at a more detailed explanation would be this: If the entire bounce (deceleration and then acceleration) takes twice the time that the inelastic collision takes, then the average force will be the same for both. However, in many (most?) cases, the inelastic collision will take longer to slow the object than a pure spring would (when the constant is the distance required to stop the object). So, the total time which the spring takes to impart 2mv of impulse is less than twice the time which the inelastic collision takes to impart mv, so the average force is higher. If the average force is higher for the spring AND the spring has a progressively increasing force over time (until the object is stopped), while the inelastic collision's force vs time is either decreasing over time or at least not ever increasing faster than the spring's, then the peak force must be higher for the spring. Sean -- = http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist