As shown in the figure, the coefficient of friction between the...

Let the initial velocity of the $2 \text{ kg}$ block be $u = 1.0 \text{ m/s}$. The retardation due to friction is $a = \mu g = 0.20 \times 10 = 2 \text{ m/s}^2$.

The velocity $v$ of the $2 \text{ kg}$ block just before the collision can be found using the kinematic equation:
$v^2 = u^2 - 2as$
$v^2 = (1.0)^2 - 2(2)(0.16) = 1 - 0.64 = 0.36$
$v = 0.6 \text{ m/s}$

Let $v_1'$ and $v_2'$ be the velocities of the $2 \text{ kg}$ and $4 \text{ kg}$ blocks respectively, just after the perfectly elastic collision. Using the formulas for one-dimensional elastic collision with the second body at rest:
$v_1' = \dfrac{m_1 - m_2}{m_1 + m_2}v = \dfrac{2 - 4}{2 + 4}(0.6) = -\dfrac{2}{6}(0.6) = -0.2 \text{ m/s}$
$v_2' = \dfrac{2m_1}{m_1 + m_2}v = \dfrac{2(2)}{2 + 4}(0.6) = \dfrac{4}{6}(0.6) = 0.4 \text{ m/s}$

The negative sign indicates that the $2 \text{ kg}$ block rebounds and moves to the left, while the $4 \text{ kg}$ block moves to the right.

Now, both blocks will slide and come to rest due to friction. The retardation for both is $a = \mu g = 2 \text{ m/s}^2$.

Distance traveled by the $2 \text{ kg}$ block before coming to rest:
$s_1 = \dfrac{|v_1'|^2}{2a} = \dfrac{(0.2)^2}{2(2)} = \dfrac{0.04}{4} = 0.01 \text{ m} = 1 \text{ cm}$

Distance traveled by the $4 \text{ kg}$ block before coming to rest:
$s_2 = \dfrac{|v_2'|^2}{2a} = \dfrac{(0.4)^2}{2(2)} = \dfrac{0.16}{4} = 0.04 \text{ m} = 4 \text{ cm}$

Since the blocks move in opposite directions from the point of collision, their final separation is:
$s = s_1 + s_2 = 1 \text{ cm} + 4 \text{ cm} = 5 \text{ cm}$

Answer: $5 \text{ cm}$