Coefficient of restitution

Further details

Line of impact – It is the line along which e is defined or in absence of tangential reaction force between colliding surfaces, force of impact is shared along this line between bodies. During physical contact between bodies during impact its line along common normal to pair of surfaces in contact of colliding bodies. Hence e is defined as a dimensionless one-dimensional parameter.

Range of values for e – treated as a constant

e is usually a positive, real number between 0 and 1.0:

e = 0: This is a perfectly inelastic collision. The objects do not move apart after the collision, but instead they coalesce. Kinetic energy is converted to heat or work done in deforming the objects.

0 < e < 1: This is a real-world inelastic collision, in which some kinetic energy is dissipated.

e = 1: This is a perfectly elastic collision, in which no kinetic energy is dissipated, and the objects rebound from one another with the same relative speed with which they approached.

e < 0: A COR less than zero would represent a collision in which the separation velocity of the objects has the same direction (sign) as the closing velocity, implying the objects passed through one another without fully engaging. This may also be thought of as an incomplete transfer of momentum. An example of this might be a small, dense object passing through a large, less dense one – e.g., a bullet passing through a target, or a motorcycle passing through a motor home or a wave tearing through a dam.

e > 1: This would represent a collision in which energy is released, for example, nitrocellulose billiard balls can literally explode at the point of impact. Also, some recent articles have described superelastic collisions in which it is argued that the COR can take a value greater than one in a special case of oblique collisions. These phenomena are due to the change of rebound trajectory caused by friction. In such collision kinetic energy is increased in a manner energy is released in some sort of explosion. It is possible that e = ∞ for a perfect explosion of a rigid system.

Maximum deformation phase – In any collision for 0 < e ≤ 1, there is a condition when for short moment along line of impact colliding bodies have same velocity when its condition of kinetic energy is lost in maximum fraction as heat, sound and light with deformation potential energy. for this short duration this collision e=0 and may be referred as inelastic phase.

Paired objects

The COR is a property of a pair of objects in a collision, not a single object. If a given object collides with two different objects, each collision would have its own COR. When an object (singular) is described as having a coefficient of restitution, as if it were an intrinsic property without reference to a second object, the definition is assumed to be with respect to collisions with a perfectly rigid and elastic object.

Generally, the COR is thought to be independent of collision speed. However, in a series of experiments performed at Florida State University in 1955, the COR was shown to vary as the collision speed approached zero, first rising significantly as the speed drops, then dropping significantly as the speed drops to about 1 cm/s and again as the collision speed approaches zero. This effect was observed in slow-speed collisions involving a number of different metals.

Relationship with conservation of energy and momentum

In a one-dimensional collision, the two key principles are: conservation of energy (conservation of kinetic energy if the collision is perfectly elastic) and conservation of (linear) momentum. A third equation can be derived from these two, which is the restitution equation as stated above. When solving problems, any two of the three equations can be used. The advantage of using the restitution equation is that it sometimes provides a more convenient way to approach the problem.

Let m 1 , m 2 be the mass of object 1 and object 2 respectively. Let u 1 , u 2 be the initial velocity of object 1 and object 2 respectively. Let v 1 , v 2 be the final velocity of object 1 and object 2 respectively.

{ 1 2 m 1 u 1 2 + 1 2 m 2 u 2 2 = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2

From the first equation,

m 1 ( u 1 2 − v 1 2 ) = m 2 ( v 2 2 − u 2 2 ) m 1 ( u 1 + v 1 ) ( u 1 − v 1 ) = m 2 ( v 2 + u 2 ) ( v 2 − u 2 )

From the second equation,

m 1 ( u 1 − v 1 ) = m 2 ( v 2 − u 2 )

After division,

u 1 + v 1 = v 2 + u 2 u 1 − u 2 = − ( v 1 − v 2 ) − v 1 − v 2 u 1 − u 2 = 1

The equation above is the restitution equation, and the coefficient of restitution is 1, which is a perfectly elastic collision.

Example

Q: A cricket ball is bowled at 50 km/h towards a batsman who swings the bat at 30 km/h. How fast, approximately, does the ball move after impact?

Step 1: Speed of separation = e × speed of approach. Speed of approach = relative closing speed of ball and bat = 50 km/h + 30 km/h = 80 km/h.

Step 2: Approximating that the collision is perfectly elastic (e = 1), therefore speed of separation is approximately 80 km/h.

Step 3: Approximating the ball as being of much smaller mass than the bat, the momentum of the bat is (almost) unchanged by the impact, therefore the bat continues to move at (nearly) the same speed (30 km/h) after impact.

Step 4: Therefore, the ball's final speed is (slightly less than) 30 km/h + 80 km/h = 110 km/h.

[This assumes that the ball is struck head-on by the bat, and that the collision is perfectly elastic. To obtain a more accurate answer, a measured value for the coefficient of restitution for cricket-ball-on-bat is needed, and use the equation for conservation of linear momentum simultaneously with the restitution formula.]

Sports equipment

The coefficient of restitution entered the common vocabulary, among golfers at least, when golf club manufacturers began making thin-faced drivers with a so-called "trampoline effect" that creates drives of a greater distance as a result of the flexing and subsequent release of stored energy, imparting greater impulse to the ball. The USGA (America's governing golfing body) has started testing drivers for COR and has placed the upper limit at 0.83. According to one article (addressing COR in tennis racquets), "[f]or the Benchmark Conditions, the coefficient of restitution used is 0.85 for all racquets, eliminating the variables of string tension and frame stiffness which could add or subtract from the coefficient of restitution."

The International Table Tennis Federation specifies that the ball shall bounce up 24–26 cm when dropped from a height of 30.5 cm on to a standard steel block thereby having a COR of 0.89 to 0.92. For a hard linoleum floor with concrete underneath, a leather basketball has a COR around 0.81–0.85.

Equations

In the case of a one-dimensional collision involving two objects, object A and object B, the coefficient of restitution is given by:

C R = v b − v a u a − u b , where: v a is the final velocity of object A after impact v b is the final velocity of object B after impact u a is the initial velocity of object A before impact u b is the initial velocity of object B before impact

Though C R does not explicitly depend on the masses of the objects, it is important to note that the final velocities are mass-dependent. For two- and three-dimensional collisions of rigid bodies, the velocities used are the components perpendicular to the tangent line/plane at the point of contact, i.e. along the line of impact.

For an object bouncing off a stationary target, C R is defined as the ratio of the object's speed after the impact to that of prior to impact:

C R = v u , where v is the speed of the object after impact u is the speed of the object before impact

In a case where frictional forces can be neglected and the object is dropped from rest onto a horizontal surface, this is equivalent to:

C R = h H , where h is the bounce height H is the drop height

The coefficient of restitution can be thought of as a measure of the extent to which mechanical energy is conserved when an object bounces off a surface. In the case of an object bouncing off a stationary target, the change in gravitational potential energy, PE, during the course of the impact is essentially zero; thus, C R is a comparison between the kinetic energy, KE, of the object immediately before impact with that immediately after impact:

C R = K E (after impact) K E (before impact) = 1 2 m v 2 1 2 m u 2 = v 2 u 2 = v u

In a cases where frictional forces can be neglected (nearly every student laboratory on this subject) and the object is dropped from rest onto a horizontal surface, the above is equivalent to a comparison between the PE of the object at the drop height with that at the bounce height. In this case, the change in KE is zero (the object is essentially at rest during the course of the impact and is also at rest at the apex of the bounce); thus:

C R = P E (at bounce height) P E (at drop height) = m g h m g H = h H

Speeds after impact

The equations for collisions between elastic particles can be modified to use the COR, thus becoming applicable to inelastic collisions, as well, and every possibility in between.

v a = m a u a + m b u b + m b C R ( u b − u a ) m a + m b and v b = m a u a + m b u b + m a C R ( u a − u b ) m a + m b

where

v a is the final velocity of the first object after impact v b is the final velocity of the second object after impact u a is the initial velocity of the first object before impact u b is the initial velocity of the second object before impact m a is the mass of the first object m b is the mass of the second object

Derivation

The above equations can be derived from the analytical solution to the system of equations formed by the definition of the COR and the law of the conservation of momentum (which holds for all collisions). Using the notation from above where u represents the velocity before the collision and v after, yields:

m a u a + m b u b = m a v a + m b v b C R = v b − v a u a − u b

Solving the momentum conservation equation for v a and the definition of the coefficient of restitution for v b yields:

m a u a + m b u b − m b v b m a = v a v b = C R ( u a − u b ) + v a

Next, substitution into the first equation for v b and then resolving for v a gives:

m a u a + m b u b − m b C R ( u a − u b ) − m b v a m a = v a m a u a + m b u b + m b C R ( u b − u a ) m a = v a [ 1 + m b m a ] m a u a + m b u b + m b C R ( u b − u a ) m a + m b = v a

A similar derivation yields the formula for v b .

Predicting the coefficient from material properties

When colliding objects do not have a center of gravity that is inline with their direction of motion and point of impact, energy that would have been available for the post-collision velocity difference will be lost to rotation and friction. This section will consider only spherical objects colliding directly either with other spherical objects or with a flat surface to avoid rotation and friction losses.

When soft objects strike hard objects, most of the energy available for the post-collision velocity different will be stored in the soft object and the value of the COR will depend on how efficient the soft object is at storing the compression energy without losing it to heat and plastic deformation. A rubber ball will bounce a lot better off concrete than glass, but the COR of glass-on-glass is a lot better than rubber-on-rubber. So when wanting to know the COR of an object, it's good to impact it with an object that is much harder. For this reason the Leeb rebound hardness test impacts test samples with a tip of tungsten carbide, one of the hardest substances available. There is no perfectly hard material, and the COR depends on both objects, so for ideal testing and theory, determining the COR of a material depends on both objects being that same material.

Many materials are assumed to be perfectly elastic when their yield strength is not approached during impact. The impact energy is theoretically stored only in the spring-effect of elastic compression and results in e = 1. In practice a various stainless steels have a large variation well below e = 1. Amorphous metals can achieve e = 0.95 or higher. The elastic range can be exceeded at low velocities because all the kinetic energy is concentrated at the point of impact. If the velocity is above 1 m/s, the yield strength of metals is usually exceeded in part of the contact area, losing energy to "plastic deformation" by not remaining in the elastic region. To account for this, the following method estimates the percent of the initial impact energy that did not get lost. Approximately, it divides how easy a volume of the material can store energy in compression ( 1 / elastic modulus ) by how well it can stay in the elastic range ( 1 / yield strength ):

% impact energy available for restitution ∝ yield strength elastic modulus

For a given material density and velocity this results in:

coefficient of restitution ∝ yield strength elastic modulus

To be more precise, these and two more quantities can be shown to be important when predicting the COR at moderate velocities. A high yield strength allows the material to stay in the elastic region at higher energies. A lower elastic modulus allows a larger surface area of contact during impact so the energy is distributed to a larger volume at the contact point which helps prevent the yield strength from being exceeded. A lower velocity increases the coefficient by needing less energy to be absorbed. A lower density also means less initial energy needs to be absorbed. The density instead of mass is used because the volume of the sphere cancels out with the volume of the affected volume at the contact area.

Combining these four variables, a theoretical estimation of the coefficient of restitution can be made when a ball is dropped onto a surface of the same material.

e = coefficient of restitution

S_y = dynamic yield strength (dynamic "elastic limit")

E′ = effective elastic modulus

ρ = density

v = velocity at impact

μ = Poisson's ratio

e = 3.1 ( S y 1 ) 5 8 ( 1 E ′ ) 1 2 ( 1 v ) 1 4 ( 1 ρ ) 1 8

E ′ = E 1 − μ 2

This applies for a direct impact and when:

0.001 < ρ v 2 S y < 0.1

Although the accuracy of this equation is not good, it is easy to calculate and accurately predicts the relative coefficient for many materials, even at velocities above and below its intended range.

Theoretical coefficient of restitution solid spheres dropped 1 meter (v = 4.5 m/s). Values greater than 1.0 indicate that the equation has errors. Yield strength instead of dynamic yield strength was used.

Plastics and rubbers will give higher values than their actual values because they are not as ideally elastic as metals, glasses, and ceramics because of heating during compression. So the following is only a guide to ranking of polymers.

Polymers (overestimated compared to metals and ceramics):

polybutadiene (golf balls shell) 11.8

butyl rubber 6.24

EVA 4.85

silicone elastomers 2.80

polycarbonate 1.46

nylon 1.28

polyethylene 1.24

Teflon 1.21

polypropylene 1.14

ABS 1.12

acrylic 1.06

PET 0.95

polystyrene 0.87

PVC 0.86

For metals the range of speeds to which this theory can apply is about 5 to 100 m/s which is a drop of 1 to 500 meters, provided the sphere is small enough for Hertzian contact theory to apply (see page 366) But the above rankings that it provides remain accurate.

Dropping hard spherical objects onto a softer surface (lower elastic modulus) which also has a lower coefficient of restitution will reduce the apparent coefficient of restitution of the dropped object. For example, most rubber and plastic balls have a lower coefficient than glass and some metal alloys, but when dropped on wood or cement, the softer material will bounce higher. This is because the harder objects distribute the impact energy over a much smaller contact area, losing energy to heat by exceeding the elastic range of the floor.

For metals, the theoretically perfect elastic range (the coefficient theoretically equals 1.0 and the above equation does not apply) is when the velocity is less than

v = ( 26 S y ρ ( S y E ′ ) 4 ) 0.5

which is less than 0.1 m/s.