Perfectly Elastic Collisions and the Conservation of Kinetic Energy
Theorem
When the coefficient of restitution $e$ is $1$, it is said to be a perfectly elastic collision. There are two important characteristics of a perfectly elastic collision.
(a) The sum of the kinetic energy of each object before and after the collision is conserved.
(b) If the masses of the two objects are the same, their velocities are exchanged after the collision.
Proof
(a)
By the law of conservation of momentum,
$$m_{1}v_{1}+m_2v_2 = m_{1}{v_{1}}^{\prime}+m_2{v_2}^{\prime}$$
$$\implies m_{1}v_{1} - m_{1}{v_{1}}^{\prime} = m_2{v_2}^{\prime} - m_2v_2$$
$$\begin{equation}\implies m_{1}(v_{1} - {v_{1}}^{\prime}) = m_2 ( {v_2}^{\prime}-v_2)\end{equation}$$
In the case of a perfectly elastic collision, since the coefficient of restitution is $e=1$,
$$e= \dfrac{ {v_2}^{\prime} - {v_{1}}^{\prime} }{ v_{1} - v_2 }=1$$
$$ \implies {v_2}^{\prime} - {v_{1}}^{\prime} = v_{1} - v_2 $$
$$\begin{equation} \implies v_{1} + {v_{1}}^{\prime} = v_2 + {v_2}^{\prime}\end{equation}$$
If the same thing is multiplied to both sides, the equation still holds. Therefore, if we multiply the left and right sides of $(1)$ by the left and right sides of $(2)$ respectively,
$$m_{1}(v_{1} - {v_{1}}^{\prime}) (v_{1} + {v_{1}}^{\prime}) = m_2 ({v_2}^{\prime} - v_2) ( {v_2}^{\prime} - v_2)$$
$$\implies m_{1}({v_{1}}^2 - {{v_{1}}^{\prime}}^2) = m_2 ( {{v_2}^{\prime}}^2 - {v_2}^2)$$
$$\implies \dfrac{1}{2}m_{1}{v_{1}}^2 - \dfrac{1}{2}m_{1}{{v_{1}}^{\prime}}^2 = \dfrac{1}{2}m_2{{v_2}^{\prime}}^2 -\dfrac{1}{2}m_2 {v_2}^2$$
$$\implies \dfrac{1}{2}m_{1}{v_{1}}^2 + \dfrac{1}{2}m_2 {v_2}^2= \dfrac{1}{2}m_{1}{{v_{1}}^{\prime}}^2 + \dfrac{1}{2}m_2{{v_2}^{\prime}}^2$$
This means that the sum of the kinetic energy of each object is conserved before and after the collision.
(b)
When the masses are equal, per $m_{1}=m_2$, equation $(1)$ becomes
$$v_{1} - {v_{1}}^{\prime} = {v_2}^{\prime}-v_2$$
The equation derived from the coefficient of restitution is
$$v_{1} + {v_{1}}^{\prime} = v_2 + {v_2}^{\prime}$$
Adding both equations,
$$2v_{1} = 2{v_2}^{\prime}$$
$$\implies v_{1}={v_2}^{\prime}$$
Subtracting the two equations,
$$2{v_{1}}^{\prime}=2v_2$$
$$\implies {v_{1}}^{\prime}=v_2$$
Therefore, the velocities of the two objects are exchanged before and after the collision.