How do I derive the formula of coefficient of restitution? I have learnt in collision mechanics about the term Coefficient of restitution, $\mathrm{e=\frac{v_{2f}-v_{1f}}{v_{1i}-v_{2i}}}$. But the sad part is that, in my book only the formula is there. My question is this, that 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?
 A: 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.
