Step-by-step Solution:

1. Analyze the Elastic Collision

   Let's denote the mass of the incoming block as $m_1 = 1.0\,kg$ and the block attached to the spring as $m_2 = 2.0\,kg$. The initial velocity of $m_1$ is $u_1 = 2.0\,m/s$ and the initial velocity of $m_2$ is $u_2 = 0$. Let $v_1$ and $v_2$ be their velocities immediately after the elastic collision.

   We apply the principles of conservation of linear momentum and kinetic energy (or use the coefficient of restitution, $e=1$).

   • Conservation of Linear Momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ $(1.0)(2.0) + (2.0)(0) = (1.0)v_1 + (2.0)v_2$ $2 = v_1 + 2v_2 \quad ... (1)$

   • Coefficient of Restitution ($e=1$ for elastic collision): $e = \frac{v_2 - v_1}{u_1 - u_2} = 1$ $v_2 - v_1 = u_1 - u_2 = 2.0 - 0$ $v_2 - v_1 = 2 \quad ... (2)$

   Now, we solve the system of linear equations (1) and (2). From equation (2), we get $v_2 = v_1 + 2$. Substituting this into equation (1): $2 = v_1 + 2(v_1 + 2)$ $2 = v_1 + 2v_1 + 4$ $2 = 3v_1 + 4$ $3v_1 = -2 \implies v_1 = -\frac{2}{3}\,m/s$

   Substitute $v_1$ back into the expression for $v_2$: $v_2 = -\frac{2}{3} + 2 = \frac{-2+6}{3} = \frac{4}{3}\,m/s$

   So, after the collision, the $1.0\,kg$ block moves to the left with a speed of $2/3\,m/s$, and the $2.0\,kg$ block moves to the right with a speed of $4/3\,m/s$.

2. Analyze the Motion After Collision

   Let's define the position of the collision as the origin ($x=0$) and the time of collision as $t=0$.

   • The $1.0\,kg$ block ($m_1$) moves with a constant velocity $v_1 = -2/3\,m/s$. Its position at any time $t$ is given by $x_1(t) = v_1 t = -\frac{2}{3}t$.
   • The $2.0\,kg$ block ($m_2$) is attached to a spring and starts moving from the equilibrium position ($x=0$) with velocity $v_2 = 4/3\,m/s$. It will execute Simple Harmonic Motion (SHM).

3. Analyze the Simple Harmonic Motion of the $2.0\,kg$ Block

   The angular frequency ($\\omega$) of the SHM is given by: $\omega = \sqrt{\frac{k}{m_2}} = \sqrt{\frac{2.0\,N/m}{2.0\,kg}} = 1\,rad/s$

   The time period ($T$) of the SHM is: $T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi\,s$

   The question asks for the situation when the spring returns to its unstretched position for the first time after the collision. The block $m_2$ starts at the unstretched position ($x=0$) at $t=0$. It moves to the right, compresses the spring to its maximum, and then returns to the unstretched position. This process takes half of one time period.

   Time taken, $t_{SHM} = \frac{T}{2} = \frac{2\pi}{2} = \pi\,s$.

4. Calculate Positions at time $t = \pi\,s$

   We need to find the positions of both blocks at this specific time, $t = \pi\,s$.

   • Position of the $2.0\,kg$ block ($m_2$): As calculated, at $t=\pi$, it has returned to its equilibrium (unstretched) position. So, $x_2(\pi) = 0$.
   • Position of the $1.0\,kg$ block ($m_1$): It has been moving with a constant velocity $v_1 = -2/3\,m/s$. $x_1(\pi) = v_1 \times t = \left(-\frac{2}{3}\right) \times \pi = -\frac{2\pi}{3}\,m$

5. Calculate the Distance Between the Blocks

   The distance ($D$) between the two blocks at time $t=\pi\,s$ is the absolute difference in their positions: $D = |x_2(\pi) - x_1(\pi)| = |0 - \left(-\frac{2\pi}{3}\right)| = \frac{2\pi}{3}\,m$

   Now, we compute the numerical value: $D = \frac{2 \times 3.14159...}{3} \approx 2.09439\,m$

   Rounding to two decimal places, the distance is $2.09\,m$.