A ball of mass m travelling at a speed v undergoes a head-on...

The loss in kinetic energy during a head-on collision is given by:

$\Delta K = \dfrac{1}{2} \dfrac{m_1 m_2}{m_1 + m_2} (u_1 - u_2)^2 (1 - e^2)$

Here, $m_1 = m_2 = m$, $u_1 = v$, and $u_2 = 0$.

$\Delta K = \dfrac{1}{2} \dfrac{m \cdot m}{m + m} v^2 (1 - e^2) = \dfrac{1}{4} m v^2 (1 - e^2)$

Given that the final kinetic energy is three-fourths of the initial kinetic energy, the loss in kinetic energy is one-fourth of the initial kinetic energy:

$\Delta K = \dfrac{1}{4} K_i = \dfrac{1}{4} \left( \dfrac{1}{2} m v^2 \right) = \dfrac{1}{8} m v^2$

Equating the two expressions for $\Delta K$:

$\dfrac{1}{4} m v^2 (1 - e^2) = \dfrac{1}{8} m v^2$

$1 - e^2 = \dfrac{1}{2}$

$e^2 = \dfrac{1}{2} \Rightarrow e = \dfrac{1}{\sqrt{2}}$