A particle of mass $$m$$ is moving with speed $$2v$$ and collides with a mass $$2m$$ moving with speed $$v$$ in the same direction. After the collision, the first mass is stopped completely while the second one splits into two particles each of mass $$m$$, which move at an angle 45° with respect to the original direction. The speed of each of the moving particle will be:

We have two bodies moving along the same straight line (take this as the positive x-axis). Initially, the momentum of the system is the vector sum of the individual momenta.

The first body has mass $$m$$ and speed $$2v$$ in the +x direction, so its momentum is $$ p_{1i}=m\,(2v)=2mv. $$

The second body has mass $$2m$$ and speed $$v$$, also in the +x direction, so its momentum is $$ p_{2i}=2m\,(v)=2mv. $$

Hence the total initial momentum is $$ p_{\text{initial}} = p_{1i}+p_{2i}=2mv+2mv = 4mv. $$

After the collision the first mass stops completely, so its final momentum is zero. The second body breaks into two equal parts, each of mass $$m$$. Let the speed of each fragment be $$u$$ and let both fragments make an angle $$45^\circ$$ with the original +x-axis, one above and one below, as described.

Because the angles are symmetric, the components of their momenta perpendicular to the +x-axis cancel each other, leaving only the components along the +x-direction.

For one fragment, the momentum component along +x is $$ p_{x} = m\,u\cos 45^\circ. $$

Since $$\cos 45^\circ=\dfrac{1}{\sqrt 2}$$, this becomes $$ p_{x}=m\,u\frac{1}{\sqrt 2}. $$

There are two such fragments, so the total final momentum along the +x-direction is $$ p_{\text{final}} = 2\left(m\,u\frac{1}{\sqrt 2}\right)=\frac{2mu}{\sqrt 2}=\sqrt 2\,m\,u. $$

By the law of conservation of linear momentum, the initial and final momenta along the +x-axis must be equal (no external force acts horizontally), hence $$ p_{\text{initial}} = p_{\text{final}}. $$

Substituting the expressions obtained: $$ 4mv = \sqrt 2\,m\,u. $$

Now divide both sides by $$m$$ to remove the common factor: $$ 4v = \sqrt 2\,u. $$

Finally, isolate $$u$$ by dividing both sides by $$\sqrt 2$$: $$ u = \frac{4v}{\sqrt 2} = 2\sqrt 2\,v. $$

So, the speed of each fragment is $$2\sqrt 2\,v$$, which corresponds to Option C.

Hence, the correct answer is Option C.