Equal forces F act on isolated bodies A and B as shown in figure. The mass of B is three times that of A. The magnitude of the acceleration of A is:

Options

A. three times that of B
B. 1/3 that of B
C. nine times that of B
D. 1/9 that of B

Explanation

To solve the problem, we need to apply 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 Forces and Masses**:
- Let the mass of body A be \( m_A \).
- Since the mass of body B is three times that of A, we have \( m_B = 3m_A \).

2. **Apply Newton’s Second Law**:
- For body A, the force acting on it is \( F \), and according to Newton's second law:
\[ F = m_A \cdot a_A \quad \text{(Equation 1)} \]
- For body B, the force acting on it is also \( F \):
\[ F = m_B \cdot a_B \quad \text{(Equation 2)} \]

3. **Substitute the Mass of B**:
- Substitute \( m_B = 3m_A \) into Equation 2:
\[ F = 3m_A \cdot a_B \]

4. **Set the Two Equations Equal**:
- Since both equations equal \( F \), we can set them equal to each other:
\[ m_A \cdot a_A = 3m_A \cdot a_B \]

5. **Cancel \( m_A \) from Both Sides**:
- Assuming \( m_A \neq 0 \), we can divide both sides by \( m_A \):
\[ a_A = 3a_B \]

6. **Conclusion**:
- The magnitude of the acceleration of A is three times that of B:
\[ a_A = 3 \cdot a_B \]

### Final Answer:
The magnitude of the acceleration of A is **3 times that of B**.

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