A nucleus moving with velocity `bar(v)` emits an `alpha`-particle. Let the velocities of the `alpha`-particle and the remaining nucleus be `bar(v)_(1)` and `bar(v)_(2)` and their masses be `m_(1)` and `(m_2)` then,

Options

A. `vecv , vecv_1 and vecv_2` must be parallel to each other
B. None of the two of `vecv, vecv_1 and vecv_2` should be parallel to each other
C. `vecv_1 + vec_2` must be parallel to `vecv`
D. `m_1 vecv_1 + m_2vecv_2` must be parallel to `vecv`

Explanation

To solve the problem, we need to apply the principle of conservation of linear momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a nucleus moving with an initial velocity \( \bar{v} \). When it emits an alpha particle, we denote the masses and velocities of the alpha particle and the remaining nucleus as follows:
- Mass of the alpha particle: \( m_1 \)
- Velocity of the alpha particle: \( \bar{v}_1 \)
- Mass of the remaining nucleus: \( m_2 \)
- Velocity of the remaining nucleus: \( \bar{v}_2 \)

### Step 2: Write the Initial Momentum
The initial momentum of the system (the moving nucleus before emission) can be expressed as:
\[ \text{Initial Momentum} = m \bar{v} \]
where \( m \) is the mass of the original nucleus.

### Step 3: Write the Final Momentum
After the emission of the alpha particle, the total momentum of the system can be expressed as:
\[ \text{Final Momentum} = m_1 \bar{v}_1 + m_2 \bar{v}_2 \]

### Step 4: Apply Conservation of Momentum
According to the law of conservation of momentum, the initial momentum must equal the final momentum:
\[ m \bar{v} = m_1 \bar{v}_1 + m_2 \bar{v}_2 \]

### Step 5: Analyze the Directions
Since the alpha particle is emitted, the directions of the velocities must be considered. The momentum vectors must be in the same direction for the conservation to hold true. Therefore, we can say:
\[ \bar{v}_1 \text{ and } \bar{v}_2 \text{ must be such that } m_1 \bar{v}_1 + m_2 \bar{v}_2 \text{ is parallel to } \bar{v} \]

### Step 6: Conclusion
Thus, the final conclusion is that the momentum after the emission of the alpha particle will maintain the same direction as the initial momentum of the nucleus. Hence, the correct option is that the momentum vectors of the alpha particle and the remaining nucleus must be parallel to the initial velocity \( \bar{v} \).

### Final Answer
The correct conclusion is:
- The momentum of the alpha particle and the remaining nucleus must be parallel to the initial momentum of the nucleus.