In nuclear disintegration, the conservation of momentum plays a crucial role in determining the motion of the resulting fragments. Since the original nucleus is at rest, its total initial momentum is zero. After the disintegration, the total momentum of the system must still be zero to conserve momentum.

Given the mass ratio of the resulting two smaller nuclei is $2:1$, let's denote the masses of the two nuclei as $2m$ and $m$, respectively.

Applying Conservation of Momentum

For the nucleus with mass $2m$ and velocity $v_1$ and for the nucleus with mass $m$ and velocity $v_2$, the conservation of momentum equation is:

$ 2m \cdot v_1 + m \cdot v_2 = 0 $

Given that momentum is a vector quantity, and the total initial momentum was zero, the nuclei must move in opposite directions for their momenta to cancel each other out. Therefore, we rearrange the equation:

$ 2m \cdot v_1 = -m \cdot v_2 $

$ 2 \cdot v_1 = -v_2 $

$ v_2 = -2 \cdot v_1 $

The negative sign indicates that $v_1$ and $v_2$ are in opposite directions. Taking magnitudes and considering the ratio:

$ \left| v_2 \right| = 2 \cdot \left| v_1 \right| $

This equation shows that the velocity of the lighter fragment (mass $m$) is twice the velocity of the heavier fragment (mass $2m$). Since they must move in opposite directions for momentum conservation, this confirms:

Correct Answer:

Option A - in opposite directions with speed in the ratio of $1:2$ respectively.