The position Vectors of two identical particles with respect to the origin in the three-dimensional coordinator system are `r_1and r_2` The position of the centre of mass of the system is given by

Options

A. `r_1 +r_2`
B. `2 (r_1+r_2)`
C. `r_1 -r_2`
D. `(r_1+r_2)/2`

Answer

To find the position of the center of mass of two identical particles with position vectors \( \mathbf{r_1} \) and \( \mathbf{r_2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have two identical particles, which means they have the same mass. Let's denote the mass of each particle as \( m \). The position vectors of the particles with respect to the origin are given as \( \mathbf{r_1} \) and \( \mathbf{r_2} \).

2. **Formula for Center of Mass**: The formula for the center of mass \( \mathbf{R}_{CM} \) for a system of particles is given by:
\[ \mathbf{R}_{CM} = \frac{1}{M} \sum_{i=1}^{n} m_i \mathbf{r_i} \]
where \( M \) is the total mass of the system, \( m_i \) is the mass of each particle, and \( \mathbf{r_i} \) is the position vector of each particle.

3. **Applying the Formula**: For our case with two identical particles:
- The total mass \( M = m_1 + m_2 = m + m = 2m \).
- The position vectors are \( \mathbf{r_1} \) and \( \mathbf{r_2} \).
- Thus, the center of mass can be calculated as:
\[ \mathbf{R}_{CM} = \frac{1}{2m} (m \mathbf{r_1} + m \mathbf{r_2}) \]

4. **Simplifying the Expression**: Since both particles have the same mass \( m \), we can factor out \( m \):
\[ \mathbf{R}_{CM} = \frac{1}{2m} m (\mathbf{r_1} + \mathbf{r_2}) = \frac{\mathbf{r_1} + \mathbf{r_2}}{2} \]

5. **Final Result**: Therefore, the position of the center of mass of the system is:
\[ \mathbf{R}_{CM} = \frac{\mathbf{r_1} + \mathbf{r_2}}{2} \]