# Here's a Brief Explanation of the Coefficient of Restitution

The coefficient of restitution is an important parameter pertaining to the collision of objects. Click here for a brief explanation regarding the same.

Did You Know?

Golf club manufacturers used to make clubs that had a thin face, which provided a trampoline-like effect, allowing the ball to travel further. However, noting that this would be an unfair advantage, the USGA has since begun testing clubs for COR, and placed the upper limit at 0.83 for them.

What is the Coefficient of Restitution?

The term restitution, in English, describes the act of restoring something that was lost or stolen, while the term coefficient means a single number that describes how much.

By definition, the coefficient of restitution is a number that indicates how much kinetic energy is restored after two or more objects collide with each other. This is dependent on several factors, including the property of the materials of the objects, their body geometry, and their impact velocities.

Mathematically, the coefficient of restitution (COR) is a number between 0 and 1. If the COR is high, i.e., close to 1, it means that a very small amount of kinetic energy was lost during the collision. On the other hand, if the COR is low, i.e., close to 0, it means that a significant amount of kinetic energy was converted into heat energy or into the force required for the deformation of the colliding bodies.

By definition, the coefficient of restitution is a number that indicates how much kinetic energy is restored after two or more objects collide with each other. This is dependent on several factors, including the property of the materials of the objects, their body geometry, and their impact velocities.

Mathematically, the coefficient of restitution (COR) is a number between 0 and 1. If the COR is high, i.e., close to 1, it means that a very small amount of kinetic energy was lost during the collision. On the other hand, if the COR is low, i.e., close to 0, it means that a significant amount of kinetic energy was converted into heat energy or into the force required for the deformation of the colliding bodies.

Types of Collisions

To understand the concept of COR better, let us examine the two specific types of collisions:

If all the kinetic energy is conserved, then it is called a perfectly elastic collision. A perfectly elastic collision can take place only if the kinetic energy of the colliding bodies doesn't get converted to any other form of energy during the collision.

**Elastic Collision:**When two bodies collide with one another in an elastic type of collision, most (if not all) of the kinetic energy of the bodies, before the collision, and after the collision, remains the same.If all the kinetic energy is conserved, then it is called a perfectly elastic collision. A perfectly elastic collision can take place only if the kinetic energy of the colliding bodies doesn't get converted to any other form of energy during the collision.

**Example:**A good example of an almost perfectly elastic collision is when you throw a rubber ball against a wall. The ball is able to bounce back from the wall at almost the same speed that you threw it with. Therefore, as most of its kinetic energy remains the same, this collision is an elastic collision.**Inelastic Collision:**The opposite of an elastic collision, an inelastic collision is one wherein most (if not all) of the kinetic energy of the colliding bodies gets converted into other forms of energy during the collision. A perfectly inelastic collision takes place when all the kinetic energy in the system of the colliding bodies is converted into other forms of energy, and is therefore lost.**Example:**A head-on collision between two cars is a good example of an inelastic collision. Almost all the kinetic energy of the colliding cars is used up for the generation of heat due to friction, and for deforming the bodies of the vehicles. This makes it an almost perfectly inelastic collision.Understanding the COR

From the discussion above, it can be stated that, the COR is in fact the measurement of the bounciness of an object.

A rubber ball bouncing on a wall rebounds with a large fraction of its initial kinetic energy before the impact. As such, the COR in this case is high. Thus, for elastic collisions, the COR is high - typically nearing the value '1'; with it being highest, i.e., '1' for a perfectly elastic collision.

On the other hand, when two cars collide, nearly all of their kinetic energy before the impact gets converted into other forms of energy, and is therefore lost. Hence, the COR in this case is low. Therefore, the COR for inelastic collision is low - nearing the value '0'; with it being lowest '0' for a perfectly inelastic type of collision.

A rubber ball bouncing on a wall rebounds with a large fraction of its initial kinetic energy before the impact. As such, the COR in this case is high. Thus, for elastic collisions, the COR is high - typically nearing the value '1'; with it being highest, i.e., '1' for a perfectly elastic collision.

On the other hand, when two cars collide, nearly all of their kinetic energy before the impact gets converted into other forms of energy, and is therefore lost. Hence, the COR in this case is low. Therefore, the COR for inelastic collision is low - nearing the value '0'; with it being lowest '0' for a perfectly inelastic type of collision.

How to Calculate the Coefficient of Restitution

Mathematically, the COR is obtained by calculating the ratio of the relative speed of separation of the colliding bodies to the relative speed at which they approach each other.

Speed of approach is obtained by taking the difference of the initial velocities of the colliding bodies before, while relative speed of separation is obtained by taking the difference of the speeds of the bodies after the collision. The following equation is used for calculating the COR.

Here, V

More accurately, the difference in the velocities of the colliding bodies after the collision is divided by the difference in their velocities before collision, to obtain the COR.

COR (e) | = | (speed of separation) | |

(speed of approach) |

COR (e) | = | ( | V_{B} | - | V_{A} | ) | |

( | U_{B} | - | U_{A} | ) |

Here, V

_{1}and V_{2}are the initial and final velocities of object 1, while U_{1}and U_{2}are the initial and final velocities of object 2.More accurately, the difference in the velocities of the colliding bodies after the collision is divided by the difference in their velocities before collision, to obtain the COR.

Practice Problems

**Ex. 1:**Particle A is traveling at 4 m/s. Particle B is traveling in the opposite direction to it at a speed of 3m/s. After the particles collide, A comes to rest. If the coefficient of restitution between the particles is 0.5, what will the speed of B be after the collision?

**Solution:**

A and B are approaching each other at a relative speed of (4 + 3) = 7 m/s

COR (e) | = | (speed of separation) |

(speed of approach) |

0.5 | = | (speed of separation) |

7 |

(speed of separation) = 7 × 0.5

speed of separation = 3.5 m/s

Since it is given that A is brought to rest after the collision, the speed of separation is equal to the speed of particle B.

**Speed of particle B = 3.5 m/s**

**Ex 2:**A 46-gm golf ball is struck with a golf club weighing 210 gm. The velocity of the golf club prior to impact with the ball is 50 m/s. If the coefficient of restitution between the club head and the ball is 0.8, find the speed of the ball immediately after impact.

**Solution:**

The initial speed of the ball: U

_{ball}= 0 m/s

The initial speed of club (before impact): U

_{club}= 50 m/s

M

_{ball}= 46 gm

M

_{club}= 210 gm

According to the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.

M

_{ball}U

_{ball}+ M

_{club}U

_{club}= M

_{ball}V

_{ball}+ M

_{club}V

_{club}--------(1)

COR (e) | = | ( | V_{club} | - | V_{ball} | ) |

( | U_{ball} | - | U_{club} | ) |

e(U

_{ball}- U

_{club})= V

_{club}- V

_{ball}

V

_{club}= (U

_{ball}- U

_{club}) + V

_{ball}

Substituting this in equation (1) we get:

M

_{ball}U

_{ball}+ M

_{club}U

_{club}= M

_{ball}V

_{ball}+ M

_{club}× [(U

_{ball}- U

_{club}) + V

_{ball}]

(46)(0)+(210)(50)=(46)V

_{ball}+ (210) x [0.8(0-50) + V

_{ball}]

(210)(50) = V

_{ball}(46 + 210) - (210)(0.8)(50)

V

_{ball}= (210)(90)/256

**V**

_{ball}= 74 m/s