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Marcos
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Does the coefficient of restitution between two particles stay the same if we place them both in some other medium (e.g. in the sea)? Also, can the law be mathematically derived from Newton's Laws of Motion (and/or other physical laws), or is it an 'experimental' law (as is, I think, )? Marcos |
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Arun Iyer
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coefficient of restitution is defined as ratio of separation velocity to approach velocity. hmm does this answer your two questions ?? love arun |
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James
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I think the coefficient will stay the same, but, for instance in the sea, the resistance and upthrust is far more significant than in air, and so will make a massive difference. It is an experimental law; Newtons experimental law for elastic collisions. v2 - v1 = e(u1 - u2 ) This is what Arun said but as an equation. |
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Nicola Coles
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Marcos
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Arun, I don't know if you misinterpreted what I meant by "does the coefficient remain constant?", but I was asking whether, when we calculate e for two particles, and then place them in water and make them collide, will the e be the same? If this is what you thought I meant then I'm not quite sure I get what you were referring to. In the definition of e there is nothing which implies it will remain the same however the two particles collide. Thanks James and Arun, Marcos P.S. Is there some sort of derivable formula for the resistance exerted by a stationary fluid on a moving object? I think air resistance is meant to be about kv2 (where I'm guessing k varies with surface area in contact with the air) |
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James
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The formula for resistance in air is much simpler than what you're asking for, because its for air only and a lot of factors make negligable difference and so are ignored, but a general formula for any fluid will have to consider many more factors which affect the resistance. |
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Arun Iyer
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hmm i was a leetl bit creeptic i think!! Quote : "In the definition of e there is nothing which implies it will remain the same however the two particles collide." this is exactly my point.(Sometimes having no information is a good information!!!).A look at the definition shows that it is more generalised and deals with only the separation and the approach velocities. Consider the equation james gives, v2 - v1 = e(u1 - u2) Now we bounce a particle in airy medium and calculate e for say a wooden plank. Now we will conduct the same expt under water.Let us put our imagination into play a little bit. What are the factors that might affect e? Thinks like adsorption,absorption etc will these play a role?? if yes, how? and if no why?? love arun P.S -> as far as the resistance is concerned, A general formula would be inconceivable because if it were , the guys with grey hairs and geeky glasses would have already found that out. unless you have some radically great idea that you would like to share with me. |
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Marcos
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Arun, Firstly, thanks for the response. I'm not sure if e would change. Intuitively, I'm inclined to say 'no' but there's too many factors to take into account (most of which I don't even know about). I think it may help me to look at it in terms of the ratio of kinetic energy before and after impact (ie. e2 ). The speed of separation and approach will obviously both be less (if we give it the same initial velocity) due to a number of resistive forces in the water. But I'm unsure as to whether the ratio will stay the same. (Or even, why it should/shouldn't) Perhaps I'm just being stupid and missing some important point. Marcos P.S. Maybe I've misunderstood Newton's law. Am I right that it states the coefficient of restitution is the same for any collision between two bodies (in the same medium)? |
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Arun Iyer
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i must say your intuition is pretty much good. e would not change because 'e' is inherent to the nature of the substance that undergo collision. so, e in question is only associated with the iron ball which bounces on the wooden plank and is not concerned whether this system is in air or in water. does this mean that the factors i mentioned don't affect e?? yeap they don't (under ideal conditions). What is an ideal condition for this problem?? The ideal condition would be "as long as the external factors don't change the nature of the materials that collide, the 'e' would remain same". What do i mean by the above? Consider a particle that bounces on some plank. Suppose the characteristic of the plank is such that it is hard when its dry and but becomes soft and spongy when its wet. Will the 'e' associated with this particle-plank stay the same inside air and water?? Obviously no. OTOH, if the plank were such that its nature does not change at all, then as per your intuition 'e' wouldn't change. love arun P.S -> you can omit the "in the same medium" in your P.S line.Hopefully my above post clarifies why this is so. |
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Marcos
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Ah right! (That wet/dry plank business didn't occur to me!) Thanks Arun (yet again!), Marcos P.S. Can you provide a link/information about some 'common' values of e? (I don't really have that much inclination to go out and bounce balls on different surfaces and measure their velocities) |
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James
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Just have a think about it, how high would whichever object bounce to if you dropped it from 1 meter? This is easier than measuring the actual velocities, but does the same job. |
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Arun Iyer
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hmmm, mathworld gave, COR of basketball as nearly 0.6 and that of baseball as nearly 0.55 At some other place i got, COR of tennis ball as nearly 0.85 however i don't seem to find any link with COR's of different materials given. love arun P.S-> James' idea could be easily followed.Then we can apply kinematic equations to find out our answer. |
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Matthew Smith
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I think part of the reason why these coefficients aren't generally available is that they can only be defined for two particular bodies, not two materials, because they depend on the shape of the bodies (as was implicit in Arun's posts discussing a specific wooden plank and iron ball, rather than wood and iron in general) So they couldn't be tabulated in Kaye and Laby as an intrinsic property of a material. Another reason might be that Newton's law of restitution isn't that accurate. It's a good first approximation to allow us to solve problems, but, like the coefficient of friction, if you tried to determine it experimentally you'd find that there was some dependence on velocity. I'm not sure how accurate it is, though I might try to find out if I've got some free time next week. As for the original question, I agree with Arun that the coefficient itself wouldn't change much in a fluid, but I think that in an experiment to measure it based on, for example, rebound heights, it might appear to. This would be because, even once you'd taken upthrust and viscous drag into account, you'd find that the formula for drag in the fluid would need to change near to the collision: Stokes' formula, for example, might stop being a good approximation, as you've no longer got an infinite fluid, but one near a boundary. These boundary effects on the drag formula could appear like a change to the coefficient of restitution. There would also be a slight change in the coefficient itself, as the rebounding doesn't happen instantaneously, but takes time and deforms both objects, and their deformation - an intrinsic property of the collision - will dissipate energy into the fluid as it pushes on it, so change e very slightly. |
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Arun Iyer
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yeap!! mostly that is the possible reason. ( if there is a demand ,one can expect them to be tabulated in Kaye and Laby soon enough ) love arun |
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Marcos
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Thanks guys, Marcos P.S. I'm guessing Kaye and Laby is some sort of physics statistical reference? P.S.2 Even though it's hard to find even a valid approximation, is there any approximation to the deceleration experienced in a fluid (no gravity) by a body which at least looks fine? ie. if I were to make a computer simulation of some body moving through water it would look realistic enough... |
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Matthew Smith
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Yes, Kaye and Laby is a standard reference book listing physical and chemical constants, especially properties of materials. As for a formula for drag, the best-known result is Stokes' formula. This states that, for a sphere of radius moving through a fluid of viscosity with velocity , the viscous drag force is given by . The viscosity (dynamic viscosity) is something that can be found in Kaye and Laby. The result, like most of the exact results of theoretical fluid dynamics, is valid only for laminar flow (low Reynolds number). This means that it works very well for the traditional A-level physics experiments involving dropping medium-sized ball bearings through glycerol (glycerine), but not, say, for calculating the effect of air resistance on something moving quickly. The effect of drag at high speeds can sometimes be modelled as being proportional to the square of the velocity, but it is very dependent on the shape of the object. |
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Marcos
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Thanks, Marcos |