1. Given data

• Mass of each block: $m=250\text{ g}=0.25\text{ kg}$
• Spring constant: $k=2\,\text{N m}^{-1}$
• Both blocks are given equal speeds $v$ in opposite directions.

2. Physical idea

Since the two masses are equal and move with equal speeds in opposite directions, the center of mass remains at rest.

So the entire initial kinetic energy of the system gets converted into spring potential energy at the instant of maximum elongation.

3. Initial kinetic energy

Each block has kinetic energy $\frac12 mv^2$

For two blocks, $K_i=2\times \frac12 mv^2=mv^2$

Substitute $m=0.25$ kg: $K_i=0.25v^2$

4. Spring potential energy at maximum elongation

If maximum elongation is $x$, then $U=\frac12 kx^2$

Given $k=2$: $U=\frac12(2)x^2=x^2$

5. Apply conservation of energy

At maximum elongation, relative motion momentarily stops, so $K_i=U$

Thus, $0.25v^2=x^2$

Taking positive root, $x=\frac{v}{2}$

6. Check options

• A: $\dfrac{v}{2\sqrt{2}}$ ❌
• B: $\dfrac{v}{2}$ ✅
• C: $\dfrac{v}{4}$ ❌
• D: $\dfrac{v}{\sqrt{2}}$ ❌

Therefore, the maximum elongation of the spring is $\boxed{\frac{v}{2}}$