1. Interpret the motion before collision

Both particles move in the same original direction.

• First particle: mass $m$, speed $2v$
• Second particle: mass $2m$, speed $v$

So total initial momentum along the original direction is

$p_i = m(2v) + 2m(v) = 2mv + 2mv = 4mv$

There is no momentum perpendicular to the original direction initially.

2. Interpret the motion after collision

• The first mass $m$ stops completely, so its final momentum is $0$.
• The second mass $2m$ splits into two particles of mass $m$ each.
• These two particles move symmetrically at angles $+45^\circ$ and $-45^\circ$ with respect to the original direction.
• Let the speed of each fragment be $u$.

Because the two fragments move symmetrically, their perpendicular momentum components cancel.

So we only need to conserve momentum along the original direction.

3. Write final momentum along the original direction

Momentum of each fragment along the original direction:

$m u \cos 45^\circ$

Since there are two such fragments,

$p_f = 2mu\cos 45^\circ$

Using $\cos 45^\circ = \frac{1}{\sqrt2}$,

$p_f = 2mu\left(\frac{1}{\sqrt2}\right) = \sqrt2 \, m u$

4. Apply conservation of linear momentum

$p_i = p_f$

$4mv = \sqrt2 \, m u$

Cancel $m$:

$4v = \sqrt2 \, u$

$u = \frac{4v}{\sqrt2} = 2\sqrt2\,v$

5. Match with the options

The speed of each fragment is

$\boxed{2\sqrt2\,v}$

So the correct option is:

$\boxed{\text{A}}$

6. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They match.