To solve this problem, we need to analyze the elastic collision between two bodies.$\\$Given:$\\$• Mass of body $A = 4m \\$• Initial velocity of body $A = u \\$• Mass of body $B = 2m \\$• Initial velocity of body $B = 0 \\$• The collision is elastic.$\\$In an elastic collision, both momentum and kinetic energy are conserved.$\\$1. Conservation of momentum:$\\ 4m \cdot u + 2m \cdot 0 = 4m \cdot v_A + 2m \cdot v_B \\$Simplifying, we get:$\\ 4u = 4v_A + 2v_B \quad$(Equation 1)$\\$2. Conservation of kinetic energy:$\\ \frac{1}{2} \cdot 4m \cdot u^2 = \frac{1}{2} \cdot 4m \cdot v_A^2 + \frac{1}{2} \cdot 2m \cdot v_B^2 \\$Simplifying, we get:$\\ 4u^2 = 4v_A^2 + 2v_B^2 \quad$(Equation 2)$\\$From Equation 1, solve for $v_B: \\ v_B = 2u - 2v_A \\$Substitute $v_B$ in Equation 2:$\\ 4u^2 = 4v_A^2 + 2(2u - 2v_A)^2 \\$Simplify:$\\ 4u^2 = 4v_A^2 + 2(4u^2 - 8uv_A + 4v_A^2) \\ 4u^2 = 4v_A^2 + 8u^2 - 16uv_A + 8v_A^2 \\ 4u^2 = 12v_A^2 - 16uv_A + 8u^2 \\$Rearrange terms:$\\ 0 = 12v_A^2 - 16uv_A + 4u^2 \\$Divide the entire equation by 4:$\\ 0 = 3v_A^2 - 4uv_A + u^2 \\$This is a quadratic equation in $v_A.$ Solve using the quadratic formula:$\\ v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\$Here, $a = 3, \, b = -4u, \, c = u^2. \\ v_A = \frac{4u \pm \sqrt{(-4u)^2 - 4 \cdot 3 \cdot u^2}}{6} \\ v_A = \frac{4u \pm \sqrt{16u^2 - 12u^2}}{6} \\ v_A = \frac{4u \pm \sqrt{4u^2}}{6} \\ v_A = \frac{4u \pm 2u}{6} \\$This gives two solutions:$\\ v_A = \frac{6u}{6} = u \quad$(not possible as $A$ must lose some speed)$\\ v_A = \frac{2u}{6} = \frac{u}{3} \\$Thus, $v_A = \frac{u}{3}. \\$Now, calculate the initial and final kinetic energy of body $A: \\$Initial kinetic energy of $A = \frac{1}{2} \cdot 4m \cdot u^2 = 2mu^2 \\$Final kinetic energy of $A = \frac{1}{2} \cdot 4m \cdot \left(\frac{u}{3}\right)^2 = \frac{2mu^2}{9} \\$Energy lost by $A = 2mu^2 - \frac{2mu^2}{9} = \frac{18mu^2}{9} - \frac{2mu^2}{9} = \frac{16mu^2}{9} \\$Fraction of energy lost by $A = \frac{Energy \ lost}{Initial \ energy} = \frac{\frac{16mu^2}{9}}{2mu^2} = \frac{8}{9} \\$Therefore, the fraction of energy lost by body $A$ is $\frac{8}{9}. \\$This corresponds to Option 2.