A target is made of two plates, one of wood and the other of iron. The thickness of the wooden plate is 4 cm and that of iron plate is 2 cm. A bullet fired goes through the wood first and then penetrates 1 cm into iron. A similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2 cm into wood. If `a_(1)` and `a_(2)` be the retardations offered to the bullet by wood and iron plates, respectively, then

Options

A. `a_(1)=2a_(2)`
B. `a_(2)=2a_(1)`
C. `a_(1)=a_(2)`
D. Data insufficient

Answer

To solve the problem, we will use the work-energy theorem, which states that the work done by the external forces is equal to the change in kinetic energy of the bullet.

### Step-by-Step Solution:

1. **Identify the Parameters:**
- Thickness of wooden plate (d1) = 4 cm
- Thickness of iron plate (d2) = 2 cm
- Penetration into iron when fired through wood = 1 cm
- Penetration into wood when fired through iron = 2 cm
- Let \( a_1 \) be the retardation offered by wood.
- Let \( a_2 \) be the retardation offered by iron.

2. **First Scenario (Wood first, then Iron):**
- The bullet penetrates through wood (4 cm) and then 1 cm into iron.
- According to the work-energy theorem:
\[ \text{Initial Kinetic Energy} = \text{Work done against wood} + \text{Work done against iron} \]
\[ \frac{1}{2} mv^2 = F_1 \cdot d_1 + F_2 \cdot 1 \]
- Here, \( F_1 = ma_1 \) and \( F_2 = ma_2 \):
\[ \frac{1}{2} mv^2 = ma_1 \cdot 4 + ma_2 \cdot 1 \]
- Dividing through by \( m \):
\[ \frac{1}{2} v^2 = 4a_1 + a_2 \quad \text{(Equation 1)} \]

3. **Second Scenario (Iron first, then Wood):**
- The bullet penetrates through iron (2 cm) and then 2 cm into wood.
- Again using the work-energy theorem:
\[ \frac{1}{2} mv^2 = F_2 \cdot d_2 + F_1 \cdot 2 \]
\[ \frac{1}{2} mv^2 = ma_2 \cdot 2 + ma_1 \cdot 2 \]
- Dividing through by \( m \):
\[ \frac{1}{2} v^2 = 2a_2 + 2a_1 \quad \text{(Equation 2)} \]

4. **Equating the Two Equations:**
- From Equation 1:
\[ \frac{1}{2} v^2 = 4a_1 + a_2 \]
- From Equation 2:
\[ \frac{1}{2} v^2 = 2a_2 + 2a_1 \]
- Setting them equal to each other:
\[ 4a_1 + a_2 = 2a_2 + 2a_1 \]
- Rearranging gives:
\[ 4a_1 - 2a_1 = 2a_2 - a_2 \]
\[ 2a_1 = a_2 \]
- Thus, we find:
\[ a_2 = 2a_1 \]

### Final Result:
The relationship between the retardations is:
\[ a_2 = 2a_1 \]