A block of mass 2.0 kg travelling at 2.0 m/s undergoes a head-on...

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

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

Given $m_1 = m_2 = m$ and the second block is at rest ($u_2 = 0$), the loss in kinetic energy becomes:

$\Delta K = \dfrac{1}{2} \dfrac{m \cdot m}{m + m} u_1^2 (1 - e^2) = \dfrac{1}{4} m u_1^2 (1 - e^2)$

The maximum possible loss in kinetic energy occurs when the collision is perfectly inelastic, i.e., $e = 0$:

\Delta K_{\text\max} = \dfrac{1}{4} m u_1^2

It is given that the actual loss in kinetic energy is half of the maximum possible loss:

\Delta K = \dfrac{1}{2} \Delta K_{\text\max}

Substituting the expressions, we get:

\dfrac{1}{4} m u_1^2 (1 - e^2) = \dfrac{1}{2} \left(\dfrac{1}{4}m u_1^2 \right)

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

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

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