1. Jun 30, 2008

### ajinx999

Coefficient of Restitution (cor) of colliding bodies describes how elastic or inelastic a collsion is. But, most internet sources (the golf-related sites) and http://en.wikipedia.org/wiki/Coefficient_of_restitution" [Broken], describes cor as a property of a particular object. Isn't this wrong? The cor depends on the velocities of both the colliding bodies. So, the cor should be described as a property of a collision. For example, if I drop a golf ball on fairway and then, in sand; obviously, the ball will bounce more on the fairway. Here, it has 2 different cors (one close to elastic & other close to inelastic).
In one-dimensional collsion, the cor is given by
cor = ( v2f - v1f )/( v1i - v2i )
Here, we use positve and negative signs to deal with directions of velocity vectors.

Last edited by a moderator: May 3, 2017
2. Jun 30, 2008

### spideyunlimit

Yep, the cor is a property of a collision. You always say that a particular collision occurred and the coefficient of restitution was this.

3. Jun 30, 2008

### Staff: Mentor

The coefficient of restitution is a joint property of both colliding objects, not of a particular collision. For the same colliding objects, within a range of possible collision speeds, the COR should be the same.

4. Jul 13, 2008

### ajinx999

I, still, don't agree on cor as a joint property of colliding bodies. Its something like saying 'velocity' is a property of a body in motion. The cor changes accordingly with different characteristics of collision. A body can have different cors based on different conditions of collision.
It can be compromised with golfers, since they understand by what cor means as per their convention. But, I see no imperative reason to consider cor as a property of a particular body in the language of physics.
For example, in physics, the potential energy is always meaningful and ascribed to a system as a whole. But, many people do sometimes say "The potential energy of the apple might be 8 J." What actually meant is; 'potential energy of the apple-Earth system'.

5. Jul 13, 2008

### Staff: Mentor

Like what?
The COR depends on both objects.
Neither do I. It's a joint property of both objects. (Of course, if you standardize one of the objects, then you could attribute the COR to one object--but that's just a practical convention.)

6. Jul 13, 2008

### ajinx999

OK, I agree, coefficient of restitution is a joint property of both colliding bodies. It can't be attributed to one particular body.
Thanks!

How to measure cor when the collision is not one dimensional? For one dimensional collisions, we do take into consideration the directions of velocities by using positive-negative convention. What for collisions which aren't one dimensional?

7. Jul 13, 2008

### spideyunlimit

Consider the components of their velocities along the common normal of the two bodies ie. the line joining their centers of mass.

8. Jul 14, 2008

### clem

The COR has to be applied in the center of mass system of the two particles.
Then it is the ratio of the magnitude of their relative rebound velocity to
the magnitude of their relative approach velocity.

9. Jul 14, 2008

### nucleus

Last edited by a moderator: Apr 23, 2017
10. Jul 19, 2008

### ajinx999

Last edited by a moderator: Apr 23, 2017
11. Aug 1, 2008

### Carl H

I understand that the COR is a joint property of both colliding bodies but aside from
the materials involved, what is the role of shape? A one ounce steel sphere dropped
from one foot onto a steel plate would seem to have a consistent COR but what about
a one ounce cube? Would the COR be different with each drop according to its orientation at impact? As with an out of round ball? Is COR, strictly speaking, a
measurement of each and every individual collision and not a number assigned, as seen
so often, to baseballs, golf balls, billiard balls and such?

12. Aug 1, 2008

### ajinx999

Shape does affect cor. cor is just the ratio of the velocities of separation to velocities of approach. In determining cor, the colliding bodies can be any shape, one just need to know the velocities. Although, cor is a joint property of the colliding bodies, it does change when the shape of one or more of the colliding bodies is changed or when their initial velocities are altered.

In determining cor, masses aren't considered. So, it doesn't matter if its one ounce or zillion ounces. Also, the cor between steel sphere and steel plate will be closer to 1 than between steel cube and steel plate, provided their initial velocities are same.
In general, cor of colliding bodies remains same irrespective of the collision being head-on or glancing.

cors for baseballs, golf balls & billiard balls are attributed as per certain conventions. For example, the second colliding body (first is the ball) is referenced as a plain surface (ground or something...). The ball is dropped form a specific known height... etc
And since, the golfers & many other, understand the convention, you will often hear them saying 'the cor of the ball...' (rather than the cor between the ball and the ground).

13. Aug 1, 2008

### Carl H

Understood. Thanks.

14. Aug 8, 2008

### ajinx999

cor, actually, tells us how much of the kinetic energy of the system was converted into other form of energy after collision.
cor is not a (joint) property of colliding 'material', but is joint property of colliding 'bodies'. Shape affects cor. When we talk of a particular body, we consider its mass and shape. Taking the example of dropping steel balls; if both the steel balls are dropped from the same height (independently), they will attain the same velocity just before collision (with the ground). What would differ will be their kinetic energies just before collision. Similar scenario, probably, will be observed after collision. I haven't experimented this, but the cor for 1 lb steel ball and that of 1 oz steel ball would be same, if and only if, their velocities after collision are same. The only difference would be that in kinetic energies.

According to me the cor is independent of the dropped height, if you are measuring the velocities (The approach velocity will change with height, but the rebound velocity will also change proportionately (discarding air resistance)). In my last post, I mentioned "the ball is dropped from a specific known height", just to tackle the air resistance. I agree, the statement does seem a bit inept, here. But, I mentioned it, since one has to conform as per strict standards, in case of a convention.

Last edited: Aug 8, 2008
15. Dec 4, 2008

### fredthegolfer

I have been offer to buy a new golf driver. The seller pretend that the c.o.r. of this driver is 0.91. USGA rule limit c.o.r to 0.83, the seller pretend the ball will go longer with this club. reading on the subject, it make sense. how effective this difference in C.O.R. could be if my headspeed is 95 mph. I uunderstand that higher the c.o.r is the more trempoline effect you have at the colision with the ball. Does the same principle apply for the ball, the ball with a higher c.o.r should be the ball that go longer.

16. Dec 5, 2008

### clem

Yes...........