Two blocks A and B of masses `m_A` and `m_B` are connected together by means of a spring and are resting on a horizontal frictionless table. The blocks are then pulled apart so as to stretch the spring and then released. Show that the kinetic energies of the blocks are, at any instant inversely proportional to their masses.

Explanation

To show that the kinetic energies of the blocks A and B are inversely proportional to their masses, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Masses**: Let the mass of block A be \( m_A \) and the mass of block B be \( m_B \).

2. **Conservation of Momentum**: Since there are no external forces acting on the system, the total momentum of the system must be conserved. If we denote the momentum of block A as \( p_A \) and the momentum of block B as \( p_B \), we have:
\[ p_A + p_B = 0 \]
This implies:
\[ p_A = -p_B \]

3. **Kinetic Energy Formula**: The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{p^2}{2m} \]
where \( p \) is the momentum and \( m \) is the mass of the object.

4. **Kinetic Energy of Each Block**:
- For block A:
\[ KE_A = \frac{p_A^2}{2m_A} \]
- For block B:
\[ KE_B = \frac{p_B^2}{2m_B} \]

5. **Substituting Momentum**: Since \( p_A = -p_B \), we can denote the magnitude of the momentum as \( p \):
\[ |p_A| = |p_B| = p \]
Therefore, we can rewrite the kinetic energies as:
- For block A:
\[ KE_A = \frac{p^2}{2m_A} \]
- For block B:
\[ KE_B = \frac{p^2}{2m_B} \]

6. **Ratio of Kinetic Energies**: To show the relationship between the kinetic energies, we can take the ratio of \( KE_A \) to \( KE_B \):
\[ \frac{KE_A}{KE_B} = \frac{\frac{p^2}{2m_A}}{\frac{p^2}{2m_B}} = \frac{m_B}{m_A} \]

7. **Conclusion**: This implies:
\[ KE_A \propto \frac{1}{m_A} \quad \text{and} \quad KE_B \propto \frac{1}{m_B} \]
Thus, the kinetic energies of the blocks are inversely proportional to their masses.