Two solids A and B of mass 1 kg and 2 kg respectively are moving with equal linear momentum. The ratio of their kinetic energies `(K.E.)_A : (K.E.)_B` will be `A/1` , so the value of A will be _____ .

Answer

To solve the problem, we need to find the ratio of the kinetic energies of two solids A and B, given that they have equal linear momentum.

### Step-by-step Solution:

1. **Define the given information**:
- Mass of solid A, \( m_A = 1 \, \text{kg} \)
- Mass of solid B, \( m_B = 2 \, \text{kg} \)
- Both solids have equal linear momentum, \( P_A = P_B \).

2. **Express linear momentum**:
- The linear momentum of solid A is given by:
\[ P_A = m_A \cdot v_A = 1 \cdot v_A = v_A \]
- The linear momentum of solid B is given by:
\[ P_B = m_B \cdot v_B = 2 \cdot v_B \]

3. **Set the momenta equal**:
- Since \( P_A = P_B \):
\[ v_A = 2 \cdot v_B \]

4. **Calculate kinetic energies**:
- The kinetic energy of solid A is:
\[ K.E._A = \frac{1}{2} m_A v_A^2 = \frac{1}{2} \cdot 1 \cdot v_A^2 = \frac{1}{2} v_A^2 \]
- The kinetic energy of solid B is:
\[ K.E._B = \frac{1}{2} m_B v_B^2 = \frac{1}{2} \cdot 2 \cdot v_B^2 = v_B^2 \]

5. **Substitute \( v_A \) in terms of \( v_B \)**:
- From step 3, we have \( v_A = 2 v_B \). Now substitute this into the kinetic energy of A:
\[ K.E._A = \frac{1}{2} (2 v_B)^2 = \frac{1}{2} \cdot 4 v_B^2 = 2 v_B^2 \]

6. **Find the ratio of kinetic energies**:
- Now we can find the ratio of the kinetic energies:
\[ \frac{K.E._A}{K.E._B} = \frac{2 v_B^2}{v_B^2} = 2 \]

7. **Express the ratio in the required form**:
- The problem states that the ratio of their kinetic energies is \( \frac{K.E._A}{K.E._B} = \frac{A}{1} \).
- From our calculation, we found that \( \frac{K.E._A}{K.E._B} = 2 \), which means \( A = 2 \).

### Final Answer:
The value of \( A \) is \( 2 \).