0
$\begingroup$

The coefficient of restitution (COR) according to some sources (https://en.wikipedia.org/wiki/Coefficient_of_restitution) is equal to the square root of the ratio of final kinetic energy to the initial kinetic energy ?

1) Is this valid in each and every collision?

2) Do we consider the kinetic energy of the two bodies combined or only a single body involved in the collision?

3) In cases involving angular momentum also, should we only consider the translational kinetic energy or the total kinetic energy( translational + rotational) ?

I have tried to derive the COR from the square root of the ratio of final kinetic energy to the initial kinetic energy , but I couldn't solve it completely.

Let us consider the following example.

A particle of mass 1 kg moving with a speed of 20 m/s collides with another particle of same mass but at rest. Let us suppose the speed of the 1st particle becomes 8 m/s without any change in direction and the second particle attains a speed of 12 m/s in the same direction

From COR = speed of separation/speed of approach COR = (12-8)/20 = 1/5 = 0.2

But from COR = final Kinetic energy/initial kinetic energy COR = 0.72

They are not the same which implies one should be wrong or else the example which I have used might not be correct. Help me out with this.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

1) Yes, as long as the collision is a simple exchange of KE, and doesn't involve the release of PE stored in the colliding bodies.

2) You must consider the combined KE.

3) The COR is usually defined in terms of translational KE, so that rotational KE is considered to be one of the ways in which KE is lost during an inelastic collision.

The COR is defined as the ratio of the final relative velocity to the initial relative velocity; you cannot 'derive' the definition.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Can you show how it is valid in each and every case? (Through proof or any general example) $\endgroup$
    – sheshin
    Commented Oct 12, 2019 at 16:09
  • $\begingroup$ I have added an example in the question, kindly look into it and confirm. $\endgroup$
    – sheshin
    Commented Oct 12, 2019 at 16:30
  • $\begingroup$ @sheshin You need to calculate the relative KE, which is half the combined mass times the square of the relative speed. All the best. $\endgroup$
    – Professor Sushing
    Commented Oct 12, 2019 at 17:09
  • $\begingroup$ Why should we consider the relative kinetic energy and why is relative kinetic energy defined the way you mentioned? If there is something that I have to go through, mention the reference. $\endgroup$
    – sheshin
    Commented Oct 12, 2019 at 17:18
0
$\begingroup$

Actually $e$=velocity of sepration/velocity of approach

$K. E=1/2MV^2$ SO you can easily can convert $v$ in term of k. E

Now coming to your question yes it is valid for all equation. K. E energy refer here is the total k. E of the system before and after collision

For different bodies value are $e$ is different for perfectly elastic body $e=1$,and if the body stick then $e=0$.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I have added an example in the question, kindly look into it and confirm. $\endgroup$
    – sheshin
    Commented Oct 12, 2019 at 16:30
  • $\begingroup$ If you can add value of masses it will be better. $\endgroup$
    – आर्यभट्ट
    Commented Oct 12, 2019 at 17:48
  • $\begingroup$ The mass is 1kg for both the particles. $\endgroup$
    – sheshin
    Commented Oct 12, 2019 at 17:56
  • $\begingroup$ You will get the same value when you write v in term of k. E $\endgroup$
    – आर्यभट्ट
    Commented Oct 13, 2019 at 3:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.