1. Use conservation of momentum

Since the nucleus is initially at rest, its initial momentum is zero.

So after disintegration, the total momentum must still be zero:

$\vec p_1 + \vec p_2 = 0$

This means the two smaller nuclei must move in opposite directions with equal magnitudes of momentum:

$p_1 = p_2$

2. Let the masses be in the ratio $2:1$

Take

$m_1 : m_2 = 2:1$

So we can write

$m_1 = 2m, \qquad m_2 = m$

Let their speeds be $v_1$ and $v_2$ respectively.

3. Apply momentum equality

Since magnitudes of momenta are equal,

$m_1 v_1 = m_2 v_2$

Substitute the masses:

$(2m)v_1 = (m)v_2$

$2v_1 = v_2$

$v_1 : v_2 = 1:2$

Thus, the heavier fragment moves slower, and the lighter fragment moves faster.

4. Interpret the result

• They move in opposite directions
• Their speeds are in the ratio $1:2$ respectively

5. Check options

• A: in opposite directions with speed in the ratio of $1:2$ respectively. ✅
• B: in the same direction with same speed. ❌
• C: in opposite directions with speed in the ratio of $2:1$ respectively. ❌
• D: in opposite directions with the same speed. ❌

Therefore, the correct option is A.