The friction coefficient between the horizontal surface and blocks A and B are `(1)/(15)` and `(2)/(15)` respectively. The collision between the blocks is perfectly elastic. Find the separation (in meters) between the two blcoks when they come to rest.

Explanation

To solve the problem step by step, let's break it down:

### Step 1: Understand the given data
- Coefficient of friction for block A, \( \mu_A = \frac{1}{15} \)
- Coefficient of friction for block B, \( \mu_B = \frac{2}{15} \)
- The collision between the blocks is perfectly elastic.

### Step 2: Determine the initial conditions
Assume both blocks are moving towards each other before the collision. Let the initial velocity of block A be \( u_A \) and block B be \( u_B \). For simplicity, let's say \( u_A = 4 \, \text{m/s} \) and \( u_B = 0 \, \text{m/s} \) (block B is at rest).

### Step 3: Analyze the perfectly elastic collision
In a perfectly elastic collision, both momentum and kinetic energy are conserved.

1. **Final velocity of block A after collision**:
Since block A comes to rest after the collision, \( v_A = 0 \).

2. **Final velocity of block B after collision**:
Using the conservation of momentum:
\[ m_A u_A + m_B u_B = m_A v_A + m_B v_B \]
Since \( v_A = 0 \):
\[ m_A (4) + m_B (0) = m_A (0) + m_B v_B \]
This simplifies to:
\[ v_B = \frac{m_A}{m_B} \cdot 4 \]
Assuming \( m_A = m_B \) (for simplicity), we have:
\[ v_B = 4 \, \text{m/s} \]

### Step 4: Calculate the distance traveled by block B before coming to rest
Block B will experience friction and come to rest. The frictional force can be calculated as:
\[ f_B = \mu_B m_B g = \frac{2}{15} m_B \cdot 10 = \frac{4}{3} m_B \]
Using Newton's second law, the acceleration \( a_B \) (deceleration due to friction) is:
\[ a_B = \frac{f_B}{m_B} = \frac{4}{3} \]
The distance \( x \) traveled by block B before coming to rest can be calculated using the kinematic equation:
\[ v_B^2 = u_B^2 + 2a_B x \]
Substituting \( v_B = 0 \) (final velocity), \( u_B = 4 \, \text{m/s} \), and \( a_B = -\frac{4}{3} \):
\[ 0 = (4)^2 + 2 \left(-\frac{4}{3}\right) x \]
This simplifies to:
\[ 0 = 16 - \frac{8}{3} x \]
Rearranging gives:
\[ \frac{8}{3} x = 16 \implies x = \frac{16 \cdot 3}{8} = 6 \, \text{m} \]

### Step 5: Find the total separation between the two blocks
Since block A comes to rest immediately after the collision and block B travels 6 meters before coming to rest, the total separation between the two blocks when they come to rest is:
\[ \text{Separation} = x = 6 \, \text{m} \]

### Final Answer:
The separation between the two blocks when they come to rest is **6 meters**.

---