Consider a head-on elastic collision between two bodies possessing...

Let the masses of the two bodies be $m_1 = m_2 = m$.

Let their initial velocities be $u_1$ and $u_2$, and final velocities be $v_1$ and $v_2$.

From the conservation of linear momentum:

$m u_1 + m u_2 = m v_1 + m v_2 \Rightarrow u_1 + u_2 = v_1 + v_2$

For a perfectly elastic head-on collision, the coefficient of restitution is $e = 1$:

$v_2 - v_1 = u_1 - u_2$

Adding the two equations yields $2v_2 = 2u_1 \Rightarrow v_2 = u_1$.

Subtracting the second equation from the first yields $2v_1 = 2u_2 \Rightarrow v_1 = u_2$.

Thus, the velocities of the two bodies are interchanged.

Since the velocities are interchanged, their magnitudes (speeds) are also interchanged.

As the masses are identical, the momentum $\vec{p} = m\vec{v}$ of each body is also interchanged.

The faster body becomes the slower one and vice versa, meaning the faster body slows down and the slower body speeds up.