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 need to analyze the situation involving two blocks, A and B, during and after their collision. Here's the detailed solution:

### Step 1: Understand the initial conditions
- Block A (mass = m) is stationary.
- Block B (mass = 2m) is moving with a velocity of 10 m/s before the collision.
- Coefficient of restitution (e) = 0.5.
- Coefficient of friction (μ) = 0.3.

### Step 2: Apply conservation of momentum
Before the collision, the total momentum of the system is:
\[ \text{Initial momentum} = (2m \cdot 10) + (m \cdot 0) = 20m \]
After the collision, let the velocities of blocks A and B be \(v_1\) and \(v_2\) respectively. The total momentum after the collision is:
\[ \text{Final momentum} = m v_1 + 2m v_2 \]
Setting the initial momentum equal to the final momentum:
\[ 20m = mv_1 + 2mv_2 \]
Dividing through by \(m\):
\[ 20 = v_1 + 2v_2 \quad \text{(Equation 1)} \]

### Step 3: Apply the coefficient of restitution
The coefficient of restitution relates the velocities before and after the collision:
\[ e = \frac{v_1 - v_2}{10 - 0} \]
Substituting the value of e:
\[ 0.5 = \frac{v_1 - v_2}{10} \]
Multiplying through by 10:
\[ v_1 - v_2 = 5 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
We have two equations:
1. \(20 = v_1 + 2v_2\)
2. \(v_1 - v_2 = 5\)

From Equation 2, we can express \(v_1\) in terms of \(v_2\):
\[ v_1 = v_2 + 5 \]
Substituting this into Equation 1:
\[ 20 = (v_2 + 5) + 2v_2 \]
Simplifying:
\[ 20 = 3v_2 + 5 \]
\[ 15 = 3v_2 \]
\[ v_2 = 5 \, \text{m/s} \]
Now substituting \(v_2\) back to find \(v_1\):
\[ v_1 = 5 + 5 = 10 \, \text{m/s} \]

### Step 5: Determine the deceleration due to friction
The frictional force acting on each block is given by:
\[ f = \mu \cdot m \cdot g \]
Where \(g \approx 10 \, \text{m/s}^2\). Thus, for block A (mass = m):
\[ f_A = 0.3 \cdot m \cdot 10 = 3m \]
And for block B (mass = 2m):
\[ f_B = 0.3 \cdot 2m \cdot 10 = 6m \]
The deceleration (a) for both blocks due to friction is:
\[ a = \frac{f}{m} = \frac{3m}{m} = 3 \, \text{m/s}^2 \quad \text{(for A)} \]
\[ a = \frac{6m}{2m} = 3 \, \text{m/s}^2 \quad \text{(for B)} \]

### Step 6: Calculate the stopping distances
Using the equation of motion:
\[ v^2 = u^2 + 2as \]
Where \(v = 0\) (final velocity), \(u\) is the initial velocity, \(a\) is the deceleration, and \(s\) is the distance traveled.

For block A (initial velocity \(u_1 = 10 \, \text{m/s}\)):
\[ 0 = (10)^2 - 2 \cdot 3 \cdot s_1 \]
\[ 100 = 6s_1 \implies s_1 = \frac{100}{6} \approx 16.67 \, \text{m} \]

For block B (initial velocity \(u_2 = 5 \, \text{m/s}\)):
\[ 0 = (5)^2 - 2 \cdot 3 \cdot s_2 \]
\[ 25 = 6s_2 \implies s_2 = \frac{25}{6} \approx 4.17 \, \text{m} \]

### Step 7: Find the distance between the blocks when they stop
The distance \(x\) between the two blocks when they stop is:
\[ x = s_1 - s_2 = 16.67 - 4.17 \approx 12.5 \, \text{m} \]

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

---