After collision the 2.0 kg block will perform simple harmonic oscillation with time period

$$T = 2\pi \sqrt {{m \over k}} $$

Given m = 2.0 kg and k = 2.0 N m^$$-$$1, we have

T = 2$$\pi$$

Thus, the block returns to its original position in time

$$t = {T \over 2}$$ = $$\pi$$ s

That is, t = 3.14 s

Now, if v₁ and v₂ are velocities of 1.0 kg block and 2.0 kg block, respectively, before collision; v'₁ and v'₂ are velocities of 1.0 kg block and 2.0 block, respectively, after collision. So, by conservation of momentum

m₁v₁ + m₂v₂ = m₁v'₁ + m₂v'₂

Here, m₁ = 1.0 kg, v₁ is initial speed of 1.0 kg block = 2.0 m s^$$-$$1, m₂ = 2.0 kg, v₂ is initial speed of 2.0 kg block = 0.0 m s^$$-$$1, v'₁ is final speed of 1.0 kg block after collision and v'₂ is final speed of 2.0 kg block after collision. Then, 1.0 kg $$\times$$ 2.0 m/s + 2.0 kg $$\times$$ 0 m/s = 1.0 v'₁ + 2.0 v'₂

v'₁ + 2v'₂ = 2 ..... (1)

Also, using definition of coefficient of restitution

v'₂ $$-$$ v'₁ = $$\varepsilon $$(v₁ $$-$$ v₂)

Since collision is elastic, So $$\varepsilon $$ = 1

$$\Rightarrow$$ v'₂ $$-$$ v'₁ = v₁ $$-$$ v₂

$$\Rightarrow$$ v'₂ $$-$$ v'₁ = 2 $$-$$ 0

$$\Rightarrow$$ v'₂ $$-$$ v'₁ = 2 ...... (2)

From Eqs. (1) and (2), we get

$$v{'_2} = {4 \over 3}$$ m s^$$-$$1 and $$v{'_1} = {-2 \over 3}$$ m s^$$-$$1

Therefore, distance between the blocks is given as

$$s = v{'_1} \times t = {{ - 2} \over 3} \times 3.14 = 2.09$$ m