The symbol for the coefficient of restitution is e. For two colliding bodies:

e = (speed of separation) / (speed of approach)

As a consequence of the principle of conservation of energy, e lies between 0 (for a completely inelastic collision, i.e. one where all kinetic energy is lost to the surroundings) and 1 (for a completely elastic collision).

To make the use of the coefficient of restitution clear, let us consider a typical mechanics problem.

Immediately before a collision, a particle A with mass 2m is travelling with speed 2 m/s to the right, and a particle B with mass 4m is travelling with speed 1 m/s also to the right. Immediately after the collision, A travels with speed u, and B travels with speed v, both to the right. Given that the coefficient of restitution between A and B is 1/2 and that the particles do not stick together, find the values of u and v.

In order to solve this problem, two laws need to be used: the law of conservation of linear momentum, and the law of restitution. Using the law of conservation of momentum first (taking left-to-right as the positive direction for velocity, and ignoring units for clarity):

(2m)(2) + (4m)(1) = (2m)(u) + (4m)(v)
8m = 2um + 4mv
Dividing by m:
8 = 2u + 4v
Rearranging for u:
2u = 8 - 4v
u = 4 - 2v

Now we need to use the law of restitution:

e = (speed of separation) / (speed of approach)
Before the collision, both particles were moving to the right, and collided. A was moving faster than B, so A must be on the left. Therefore the speed of approach is (2 - 1) = 1 m/s.
After the collision, both particles are still moving to the right, and they don't stick together. From this we can see that B must now be moving faster than A. Therefore the speed of separation is (v - u). Hence:
e = (v - u) / (1)
e = v - u
But e = 1/2, so:
1/2 = v - u
Substituting u = 4 - 2v:
1/2 = v - (4 - 2v)
1/2 = 3v - 4
3v = 9/2
v = 3/2 m/s

Now substitute the value of v into the equation for u:
u = 4 - 2(3/2)
u = 1 m/s

And we have our answers. Just for kicks, we can show that the total kinetic energy after the collision is less than the total kinetic energy before the collision, and that therefore e should indeed be less than 1. The equation for the kinetic energy of a body with mass m and speed v is: K.E. = 1/2m(v^2).

(Total K.E. before collision) = (1/2)(2m)(2)^2 + (1/2)(4m)(1)^2 = (m)(4) + 2m = 6m J

(Total K.E. after collision) = (1/2)(2m)(1)^2 + (1/2)(4m)(3/2)^2 = m + (9/2)(m) = 5.5m J

5.5m J is less than 6m J, therefore 0.5m J of energy were lost in the collision.

Iin golf it's known as COR.

It is the percentage that is derived from the speed at which the ball leaves the clubhead divided by the speed at which the clubhead strikes the ball.

This is to put a limit on the spring like effect that the newer thin metal faced clubheads are to have.

The highest COR allowed by the USGA is .830

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