To solve this problem, we will use the principles of conservation of momentum and the coefficient of restitution.$\\$Given:$\\$• Mass of the moving block $= m \\$• Mass of the stationary block $= 4m \\$• Initial velocity of the moving block $= v \\$• Final velocity of the moving block after collision $= 0 \\$• Initial velocity of the stationary block $= 0 \\$Let's denote:$\\$• Final velocity of the stationary block after collision as $v_2. \\$Using the conservation of momentum:$\\$Initial momentum $=$Final momentum$\\ m \cdot v + 4m \cdot 0 = m \cdot 0 + 4m \cdot v_2 \\$Simplify:$\\ m \cdot v = 4m \cdot v_2 \\$Solve for $v_2: \\ v_2 = \frac{v}{4} \\$Now, using the definition of the coefficient of restitution, which is $(e): e =$ Relative velocity of separation$\div$Relative velocity of approach$\\$Relative velocity of approach $= v - 0 = v \\$Relative velocity of separation $= v_2 - 0 = \frac{v}{4} \\$Substitute these values into the equation for $e: \\ e = \frac{\frac{v}{4}}{v} \\$Simplify:$\\ e = \frac{1}{4} \\$Therefore, the value of the coefficient of restitution $(e)$ is $0.25. \\$This corresponds to Option 2.