# Coefficient of restitution

A bouncing basketball captured with a stroboscopic flash at 25 images per second: Ignoring air resistance, the square root of the ratio of the height of one bounce to that of the preceding bounce gives the coefficient of restitution for the ball/surface impact.

The coefficient of restitution (COR) is a measure of the "restitution" of a collision between two objects: how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects.

The coefficient, e is defined as the ratio of relative speeds after and before an impact, taken along the line of the impact:

${\displaystyle {\text{Coefficient of restitution }}(e)={\frac {\text{Relative velocity after collision}}{\text{ Relative velocity before collision}}}}$ [1]

Alternatively, this may be expressed as: ${\displaystyle {\text{Relative velocity of separation}}=e\times {\text{Relative velocity of approach}}}$

The mathematics were developed by Sir Isaac Newton[2] in 1687. It is also known as Newton's experimental law.

## Further details

Line of impact- It's 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 1-D parameter which is unit-less.

### 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.[3][4][5] 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 ${\displaystyle e=\infty }$ 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.[6]

### 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[7] 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 ${\displaystyle m_{1}}$, ${\displaystyle m_{2}}$ be the mass of object 1 and object 2 respectively. Let ${\displaystyle u_{1}}$, ${\displaystyle u_{2}}$ be the initial velocity of object 1 and object 2 respectively. Let ${\displaystyle v_{1}}$, ${\displaystyle v_{2}}$ be the final velocity of object 1 and object 2 respectively.

${\displaystyle {\begin{cases}{\frac {1}{2}}m_{1}u_{1}^{2}+{\frac {1}{2}}m_{2}u_{2}^{2}={\frac {1}{2}}m_{1}v_{1}^{2}+{\frac {1}{2}}m_{2}v_{2}^{2}\\m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\end{cases}}}$

From the first equation,

${\displaystyle m_{1}(u_{1}^{2}-v_{1}^{2})=m_{2}(v_{2}^{2}-u_{2}^{2})}$
${\displaystyle m_{1}(u_{1}+v_{1})(u_{1}-v_{1})=m_{2}(v_{2}+u_{2})(v_{2}-u_{2})}$

From the second equation,

${\displaystyle m_{1}(u_{1}-v_{1})=m_{2}(v_{2}-u_{2})}$

After division,

${\displaystyle u_{1}+v_{1}=v_{2}+u_{2}}$
${\displaystyle u_{1}-u_{2}=-(v_{1}-v_{2})}$
${\displaystyle -{\frac {v_{1}-v_{2}}{u_{1}-u_{2}}}=1}$

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 x 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."[8]

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.[9] For a hard linoleum floor with concrete underneath, a leather basketball has a COR around 0.81-0.85.[10]

## Equations

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

${\displaystyle C_{R}={\frac {v_{b}-v_{a}}{u_{a}-u_{b}}}}$, where:
${\displaystyle v_{a}}$ is the final velocity of object A after impact
${\displaystyle v_{b}}$ is the final velocity of object B after impact
${\displaystyle u_{a}}$ is the initial velocity of object A before impact
${\displaystyle u_{b}}$ is the initial velocity of object B before impact

Though ${\displaystyle C_{R}}$ 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, ${\displaystyle C_{R}}$ is defined as the ratio of the object's speed after the impact to that of prior to impact:

${\displaystyle C_{R}={\frac {v}{u}}}$, where
${\displaystyle v}$ is the speed of the object after impact
${\displaystyle u}$ 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:

${\displaystyle C_{R}={\sqrt {\frac {h}{H}}}}$, where
${\displaystyle h}$ is the bounce height
${\displaystyle H}$ 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, ${\displaystyle C_{R}}$ is a comparison between the kinetic energy, KE, of the object immediately before impact with that immediately after impact:

${\displaystyle C_{R}={\sqrt {\frac {KE_{\text{(after impact)}}}{KE_{\text{(before impact)}}}}}={\sqrt {\frac {{\frac {1}{2}}mv^{2}}{{\frac {1}{2}}mu^{2}}}}={\sqrt {\frac {v^{2}}{u^{2}}}}={\frac {v}{u}}}$

In a cases where frictional forces can be neglected (nearly every student laboratory on this subject[11]) 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:

${\displaystyle C_{R}={\sqrt {\frac {PE_{\text{(at bounce height)}}}{PE_{\text{(at drop height)}}}}}={\sqrt {\frac {mgh}{mgH}}}={\sqrt {\frac {h}{H}}}}$

## 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.

${\displaystyle v_{a}={\frac {m_{a}u_{a}+m_{b}u_{b}+m_{b}C_{R}(u_{b}-u_{a})}{m_{a}+m_{b}}}}$
and
${\displaystyle v_{b}={\frac {m_{a}u_{a}+m_{b}u_{b}+m_{a}C_{R}(u_{a}-u_{b})}{m_{a}+m_{b}}}}$

where

${\displaystyle v_{a}}$ is the final velocity of the first object after impact
${\displaystyle v_{b}}$ is the final velocity of the second object after impact
${\displaystyle u_{a}}$ is the initial velocity of the first object before impact
${\displaystyle u_{b}}$ is the initial velocity of the second object before impact
${\displaystyle m_{a}}$ is the mass of the first object
${\displaystyle m_{b}}$ 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 ${\displaystyle u}$ represents the velocity before the collision and ${\displaystyle v}$ after, yields:

{\displaystyle {\begin{aligned}&m_{a}u_{a}+m_{b}u_{b}=m_{a}v_{a}+m_{b}v_{b}\\&C_{R}={\frac {v_{b}-v_{a}}{u_{a}-u_{b}}}\\\end{aligned}}}

Solving the momentum conservation equation for ${\displaystyle v_{a}}$ and the definition of the coefficient of restitution for ${\displaystyle v_{b}}$ yields:

{\displaystyle {\begin{aligned}&{\frac {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}\\\end{aligned}}}

Next, substitution into the first equation for ${\displaystyle v_{b}}$ and then resolving for ${\displaystyle v_{a}}$ gives:

{\displaystyle {\begin{aligned}&{\frac {m_{a}u_{a}+m_{b}u_{b}-m_{b}C_{R}(u_{a}-u_{b})-m_{b}v_{a}}{m_{a}}}=v_{a}\\&\\&{\frac {m_{a}u_{a}+m_{b}u_{b}+m_{b}C_{R}(u_{b}-u_{a})}{m_{a}}}=v_{a}\left[1+{\frac {m_{b}}{m_{a}}}\right]\\&\\&{\frac {m_{a}u_{a}+m_{b}u_{b}+m_{b}C_{R}(u_{b}-u_{a})}{m_{a}+m_{b}}}=v_{a}\\\end{aligned}}}

A similar derivation yields the formula for ${\displaystyle v_{b}}$.

## References

1. ^ McGinnis, Peter M. (2005). Biomechanics of sport and exercise (2nd ed.). Champaign, IL [u.a.]: Human Kinetics. p. 85. ISBN 9780736051019.
2. ^ "'A' level Revision:Newton's Law of Restitution". Retrieved 12 March 2013.
3. ^ Louge, Michel; Adams, Michael (2002). "Anomalous behavior of normal kinematic restitution in the oblique impacts of a hard sphere on an elastoplastic plate". Physical Review E. 65 (2). Bibcode:2002PhRvE..65b1303L. doi:10.1103/PhysRevE.65.021303.
4. ^ Kuninaka, Hiroto; Hayakawa, Hisao (2004). "Anomalous Behavior of the Coefficient of Normal Restitution in Oblique Impact". Physical Review Letters. 93 (15): 154301. arXiv:. Bibcode:2004PhRvL..93o4301K. doi:10.1103/PhysRevLett.93.154301. PMID 15524884.
5. ^ Calsamiglia, J.; Kennedy, S. W.; Chatterjee, A.; Ruina, A.; Jenkins, J. T. (1999). "Anomalous Frictional Behavior in Collisions of Thin Disks". Journal of Applied Mechanics. 66 (1): 146. Bibcode:1999JAM....66..146C. doi:10.1115/1.2789141.
6. ^ "IMPACT STUDIES ON PURE METALS" (PDF). Archived from the original (PDF) on March 19, 2015.
7. ^
8. ^
9. ^ "ITTF Technical Leaflet T3: The Ball" (PDF). ITTF. December 2009. p. 4. Retrieved 28 July 2010.
10. ^ "UT Arlington Physicists Question New Synthetic NBA Basketball". Retrieved May 8, 2011.
11. ^ Mohazzabi, Pirooz. "When Does Air Resistance Become Significant in Free Fall?".
• Cross, Rod (2006). "The bounce of a ball" (PDF). Physics Department, University of Sydney, Australia. Retrieved 2008-01-16. In this paper, the dynamics of a bouncing ball is described for several common ball types having different bounce characteristics. Results are presented for a tennis ball, a baseball, a golf ball, a superball, a steel ball bearing, a plasticene ball, and a silly putty ball.
• Walker, Jearl (2011). Fundamentals Of Physics (9th ed.). David Halliday, Robert Resnick, Jearl Walker. ISBN 978-0-470-56473-8.