A force produces an acceleration of `4ms^(-2)` in a body of mass `m_(1)` and the same force produces an acceleration of `6ms^(-2)` in another body oa mass `m_(2)`. If the same force is applied to `(m_(1)+m_(2))`, then the acceleration will be

Options

A. `1.6ma^(-2)`
B. `2ms^(-2)`
C. `2.4ms^(-2)`
D. `3.2ms^(-2)`

Answer

To solve the problem, we will use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The acceleration produced by the force \( F \) on mass \( m_1 \) is \( 4 \, \text{m/s}^2 \).
- The acceleration produced by the same force \( F \) on mass \( m_2 \) is \( 6 \, \text{m/s}^2 \).

2. **Express the Force in Terms of Mass and Acceleration:**
- For mass \( m_1 \):
\[ F = m_1 \cdot 4 \]
- For mass \( m_2 \):
\[ F = m_2 \cdot 6 \]

3. **Set the Two Expressions for Force Equal to Each Other:**
- Since both expressions equal \( F \), we can write:
\[ m_1 \cdot 4 = m_2 \cdot 6 \]

4. **Rearranging the Equation:**
- From the equation \( m_1 \cdot 4 = m_2 \cdot 6 \), we can express \( m_1 \) in terms of \( m_2 \):
\[ m_1 = \frac{6}{4} m_2 = \frac{3}{2} m_2 \]

5. **Calculate the Total Mass \( m_1 + m_2 \):**
- Substitute \( m_1 \) in terms of \( m_2 \):
\[ m_1 + m_2 = \frac{3}{2} m_2 + m_2 = \frac{3}{2} m_2 + \frac{2}{2} m_2 = \frac{5}{2} m_2 \]

6. **Express the Force in Terms of Total Mass and Acceleration:**
- When the same force \( F \) is applied to the total mass \( m_1 + m_2 \), we have:
\[ F = (m_1 + m_2) \cdot a \]
- Substituting \( m_1 + m_2 \):
\[ F = \left(\frac{5}{2} m_2\right) \cdot a \]

7. **Substituting the Value of \( F \) from One of the Previous Equations:**
- We can use \( F = m_2 \cdot 6 \):
\[ m_2 \cdot 6 = \left(\frac{5}{2} m_2\right) \cdot a \]

8. **Canceling \( m_2 \) (assuming \( m_2 \neq 0 \)):**
- Dividing both sides by \( m_2 \):
\[ 6 = \frac{5}{2} a \]

9. **Solving for \( a \):**
- Multiply both sides by \( \frac{2}{5} \):
\[ a = 6 \cdot \frac{2}{5} = \frac{12}{5} = 2.4 \, \text{m/s}^2 \]

### Final Answer:
The acceleration when the same force is applied to \( m_1 + m_2 \) is \( 2.4 \, \text{m/s}^2 \).