A point object of mass m is kept at (a, 0) along x-axis. What mass should be kept at (-3a,0), so that centre of mass lies at origin ?

Options

A. m
B. 2m
C. `(m)/(3)`
D. 3m

Answer

To solve the problem, we need to find the mass \( M \) that should be placed at the point \((-3a, 0)\) such that the center of mass of the system lies at the origin \((0, 0)\).

### Step-by-Step Solution:

1. **Identify the masses and their positions**:
- We have a mass \( m \) located at the point \( (a, 0) \).
- We need to find the mass \( M \) that will be placed at the point \( (-3a, 0) \).

2. **Use the formula for the center of mass**:
The formula for the center of mass \( x_{cm} \) for a system of particles along the x-axis is given by:
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
where \( m_1 \) and \( m_2 \) are the masses, and \( x_1 \) and \( x_2 \) are their respective positions.

3. **Substituting the known values**:
Here, we can substitute \( m_1 = m \), \( x_1 = a \), \( m_2 = M \), and \( x_2 = -3a \):
\[ x_{cm} = \frac{m \cdot a + M \cdot (-3a)}{m + M} \]

4. **Set the center of mass to zero**:
Since we want the center of mass to be at the origin, we set \( x_{cm} = 0 \):
\[ 0 = \frac{m \cdot a - 3M \cdot a}{m + M} \]

5. **Eliminate the denominator**:
To eliminate the denominator, we can multiply both sides by \( m + M \) (assuming \( m + M \neq 0 \)):
\[ 0 = m \cdot a - 3M \cdot a \]

6. **Factor out \( a \)**:
Since \( a \neq 0 \), we can factor it out:
\[ 0 = m - 3M \]

7. **Solve for \( M \)**:
Rearranging gives us:
\[ 3M = m \quad \Rightarrow \quad M = \frac{m}{3} \]

### Final Answer:
The mass \( M \) that should be placed at \((-3a, 0)\) is \( \frac{m}{3} \).