Block A has a mass `3kg` and is sliding on a rough horizontal surface with a velocity `u_A=2m//s` when it makes a direct collision with block B, which has a mass of `2kg` and is originally at rest. The collision is perfectly elastic. Determine the velocity of each block just after collision and the distance between the blocks when they stop sliding. The coefficient of kinetic friction between the blocks and the plane is `mu_k=0.3` ( Take `g=10m//s^2`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have two blocks, A and B, with masses \( m_A = 3 \, \text{kg} \) and \( m_B = 2 \, \text{kg} \) respectively. Block A is moving with an initial velocity \( u_A = 2 \, \text{m/s} \) and block B is at rest \( u_B = 0 \, \text{m/s} \). The collision is perfectly elastic, and we need to find the velocities of both blocks after the collision and the distance between them when they stop sliding due to friction.

### Step 2: Apply Conservation of Momentum
For a perfectly elastic collision, we can apply the conservation of momentum:

\[ m_A u_A + m_B u_B = m_A v_A + m_B v_B \]

Substituting the known values:

\[ 3 \times 2 + 2 \times 0 = 3 v_A + 2 v_B \]

This simplifies to:

\[ 6 = 3 v_A + 2 v_B \quad \text{(Equation 1)} \]

### Step 3: Apply the Coefficient of Restitution
For a perfectly elastic collision, the coefficient of restitution \( e = 1 \). The equation for the coefficient of restitution is:

\[ e = \frac{v_B - v_A}{u_A - u_B} \]

Substituting the known values:

\[ 1 = \frac{v_B - v_A}{2 - 0} \]

This simplifies to:

\[ v_B - v_A = 2 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:

1. \( 6 = 3 v_A + 2 v_B \)
2. \( v_B - v_A = 2 \)

From Equation 2, we can express \( v_B \) in terms of \( v_A \):

\[ v_B = v_A + 2 \]

Substituting this into Equation 1:

\[ 6 = 3 v_A + 2(v_A + 2) \]

Expanding and simplifying:

\[ 6 = 3 v_A + 2 v_A + 4 \]
\[ 6 = 5 v_A + 4 \]
\[ 2 = 5 v_A \]
\[ v_A = \frac{2}{5} = 0.4 \, \text{m/s} \]

Now substituting \( v_A \) back into Equation 2 to find \( v_B \):

\[ v_B = 0.4 + 2 = 2.4 \, \text{m/s} \]

### Step 5: Determine the Stopping Distance
Next, we need to find the stopping distances for both blocks due to friction. The frictional force \( F_f \) can be calculated as:

\[ F_f = \mu_k m g \]

Where \( \mu_k = 0.3 \) and \( g = 10 \, \text{m/s}^2 \).

For block A:

\[ F_f = 0.3 \times 3 \times 10 = 9 \, \text{N} \]

The acceleration due to friction \( a \) is:

\[ a = \frac{F_f}{m_A} = \frac{9}{3} = 3 \, \text{m/s}^2 \]

Using the equation \( v^2 = u^2 + 2as \) to find the stopping distance \( s_A \):

\[ 0 = (0.4)^2 - 2 \times 3 \times s_A \]
\[ 0 = 0.16 - 6 s_A \]
\[ 6 s_A = 0.16 \quad \Rightarrow \quad s_A = \frac{0.16}{6} \approx 0.02667 \, \text{m} \]

For block B:

\[ F_f = 0.3 \times 2 \times 10 = 6 \, \text{N} \]

The acceleration due to friction \( a \) is:

\[ a = \frac{F_f}{m_B} = \frac{6}{2} = 3 \, \text{m/s}^2 \]

Using the equation \( v^2 = u^2 + 2as \) to find the stopping distance \( s_B \):

\[ 0 = (2.4)^2 - 2 \times 3 \times s_B \]
\[ 0 = 5.76 - 6 s_B \]
\[ 6 s_B = 5.76 \quad \Rightarrow \quad s_B = \frac{5.76}{6} \approx 0.96 \, \text{m} \]

### Step 6: Calculate the Distance Between the Blocks
Finally, the distance \( D \) between the blocks when they stop sliding is:

\[ D = s_B - s_A = 0.96 - 0.02667 \approx 0.93333 \, \text{m} \]

### Final Answers
- Velocity of Block A after collision: \( v_A = 0.4 \, \text{m/s} \)
- Velocity of Block B after collision: \( v_B = 2.4 \, \text{m/s} \)
- Distance between the blocks when they stop sliding: \( D \approx 0.93333 \, \text{m} \)