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Satyajeet Vispute
Mar 17, 2019

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

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.

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.

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 small amount of kinetic energy was lost during the collision. If the COR is low, i.e., close to 0, it means that a significant amount of kinetic energy was converted into heat or into the force required for the deformation of the bodies.

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.

The opposite of an elastic collision, an inelastic collision is one wherein most of the kinetic energy of the colliding bodies gets converted into other forms of energy during the collision. A perfectly inelastic collision occurs when all the kinetic energy in the system of the colliding bodies is converted into other forms of energy, and is lost.

The COR is 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.

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.

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

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 equation to calculate the COR is:

**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.

*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.*

Here, V

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?

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**

(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.

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.

The initial speed of the ball: U

The initial speed of club (before impact): U

M

M

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}

M

COR (e) = ( V

e(U

V

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

M

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

(210)(50) = V

V