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I have learnt in collision mechanics about the term "coefficient of restitution": $$e=\frac{v_{2,\text{f}}-v_{1,\text{f}}}{v_{1,\text{i}}-v_{2,\text{i}}}.$$ But the sad part is that my book only contains the formula. How do I derive or how is this formula established? Will I have to take the principle of conservation of momentum or energy considerations, or both?

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    $\begingroup$ That's the way the COR is defined. The equation you give can't be derived because it's a definition not a derived quantity. $\endgroup$ Commented Sep 20, 2015 at 10:20
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    $\begingroup$ You can derive it because it is an empirical quantity. You can only define it. $\endgroup$ Commented Apr 11, 2016 at 19:13
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    $\begingroup$ The coefficient of restitution is an empirical quantity that idealizes a very complex problem. $\endgroup$ Commented May 15, 2019 at 22:46
  • $\begingroup$ @JohnAlexiou Does that mean coefficient of restitution is only defined when two bodies collide and not more than that? $\endgroup$
    – Satya
    Commented Sep 13, 2021 at 8:55
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    $\begingroup$ @satyamkumarjha - what I mean is that as a first approximation we have observed that the bounce speed is somewhat of a fraction of the impact speed for a broad range of situations, but not all. That is the empirical part. $\endgroup$ Commented Sep 13, 2021 at 12:03

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This cant be fully derived but a part of it can be.

In an elastic collision kinetic energy is conserved, so $$ \frac12 m_1 u_1^2 + \frac12 m_2u_2^2 = \frac12 m_1 v_1^2 + \frac12 m_2 v_2^2 $$ $$ m_1 u_1^2 + m_2 u_2^2 = m_1 v_1^2 + m_2 v_2^2 $$ $$ m_1u_1^2 - m_1v_1^2 = m_2v_2^2 - m_2u_2^2 $$ $$ m_1(u_1+v_1)(u_1-v_1)=m_2(v_2+u_2)(v_2-u_2)\tag{1} $$ Now, according to conservation of linear momentum, $$ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 $$ $$ m_1(u_1-v_1) = m_2(v_2-u_2)\tag{2} $$ Dividing equation 1 by 2 we get, $$u_1 + v_1 = u_2 + v_2$$

$$ u_1-u_2 = v_2-v_1 $$ $$ \frac{v_2-v_1}{u_1-u_2} = 1 $$ Thus, we get an equation where the coefficient of restitution equals 1. This proves that if the collision is elastic this will be equal to 1. Moreover, if it approaches 1 both the equations will be satisfied approximately, which will contribute to the fact that kinetic energy is conserved which in turn will increase the fact that collision happens in an elastic plane. Thus it’s a measure of elastic collision.

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    $\begingroup$ All those capitals hurt the eyes... $\endgroup$
    – Jon Custer
    Commented Apr 11, 2016 at 19:22

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