A block A of mass m is lying on a rough horizontal surface. Another block B of mass 2m having velocity 10 m/s just before collision, makes head on collision with stationary block A. The coefficient of restitution between blocks is 0.5, and the coefficient of friction between blocks and ground is 0.3. Find the distance between the blocks when they stop.

Answer

To solve the problem step by step, we will follow the principles of conservation of momentum and the coefficient of restitution, and then calculate the stopping distance due to friction.

### Step 1: Understand the given information
- Mass of block A (m₁) = m
- Mass of block B (m₂) = 2m
- Initial velocity of block A (u₁) = 0 m/s (stationary)
- Initial velocity of block B (u₂) = 10 m/s
- Coefficient of restitution (e) = 0.5
- Coefficient of friction (μ) = 0.3
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Apply conservation of momentum
The total momentum before the collision must equal the total momentum after the collision.

\[ m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ \]

Substituting the values we have:

\[ m(0) + 2m(10) = mv₁ + 2mv₂ \]

This simplifies to:

\[ 20m = mv₁ + 2mv₂ \quad \text{(1)} \]

### Step 3: Apply the coefficient of restitution
The coefficient of restitution relates the relative velocities of separation and approach:

\[ e = \frac{v₂ - v₁}{u₁ - u₂} \]

Substituting the known values:

\[ 0.5 = \frac{v₂ - v₁}{0 - 10} \]

This simplifies to:

\[ 0.5 = \frac{v₂ - v₁}{-10} \]

Rearranging gives:

\[ v₂ - v₁ = -5 \quad \text{(2)} \]

### Step 4: Solve the equations (1) and (2)
From equation (2), we can express \(v₂\) in terms of \(v₁\):

\[ v₂ = v₁ - 5 \]

Substituting this into equation (1):

\[ 20m = mv₁ + 2m(v₁ - 5) \]

Expanding this gives:

\[ 20m = mv₁ + 2mv₁ - 10m \]

Combining like terms:

\[ 20m = 3mv₁ - 10m \]

Adding \(10m\) to both sides:

\[ 30m = 3mv₁ \]

Dividing both sides by \(3m\):

\[ v₁ = 10 \text{ m/s} \]

Now substituting \(v₁\) back into equation (2):

\[ v₂ = 10 - 5 = 5 \text{ m/s} \]

### Step 5: Calculate the acceleration due to friction
The frictional force acting on each block is given by:

\[ F_f = μm₁g = 0.3 \cdot m \cdot 10 = 3m \]

The acceleration \(a\) due to friction for each block is:

\[ a = \frac{F_f}{m} = \frac{3m}{m} = 3 \text{ m/s}² \]

### Step 6: Calculate the distance each block travels before stopping
Using the equation of motion:

\[ v^2 = u^2 + 2as \]

For block A (initial velocity \(u₁ = 10\) m/s):

\[ 0 = (10)^2 - 2(3)s₁ \]

This simplifies to:

\[ 100 = 6s₁ \implies s₁ = \frac{100}{6} \approx 16.67 \text{ m} \]

For block B (initial velocity \(u₂ = 5\) m/s):

\[ 0 = (5)^2 - 2(3)s₂ \]

This simplifies to:

\[ 25 = 6s₂ \implies s₂ = \frac{25}{6} \approx 4.17 \text{ m} \]

### Step 7: Calculate the distance between the blocks when they stop
The distance between the two blocks when they stop is:

\[ d = s₁ - s₂ = 16.67 - 4.17 \approx 12.5 \text{ m} \]

### Final Answer
The distance between the blocks when they stop is approximately **12.5 meters**.

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