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

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.

**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.

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.

**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.

**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**