To solve the problem step by step, we will analyze the situation involving two blocks, A and B, and their elastic collision, while considering the effects of friction.
### Step 1: Identify the given data
- Mass of block A, \( m_A = 2 \, \text{kg} \)
- Mass of block B, \( m_B = 1 \, \text{kg} \)
- Initial speed of block B, \( u_B = 1 \, \text{m/s} \)
- Initial speed of block A, \( u_A = 0 \, \text{m/s} \)
- Distance between the blocks before collision, \( d = 16 \, \text{cm} = 0.16 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
### Step 2: Calculate the frictional force acting on block B
The frictional force \( F_f \) acting on block B can be calculated using the formula:
\[
F_f = \mu \cdot m_B \cdot g
\]
Substituting the values:
\[
F_f = 0.2 \cdot 1 \cdot 10 = 2 \, \text{N}
\]
### Step 3: Calculate the acceleration due to friction
The acceleration \( a \) caused by the frictional force on block B is given by:
\[
a = \frac{F_f}{m_B} = \frac{2}{1} = 2 \, \text{m/s}^2
\]
Since the friction acts opposite to the direction of motion, the acceleration will be negative:
\[
a = -2 \, \text{m/s}^2
\]
### Step 4: Calculate the final velocity of block B before collision
Using the kinematic equation:
\[
v_B^2 = u_B^2 + 2as
\]
where \( s = 0.16 \, \text{m} \):
\[
v_B^2 = 1^2 + 2 \cdot (-2) \cdot 0.16
\]
Calculating:
\[
v_B^2 = 1 - 0.64 = 0.36
\]
Thus,
\[
v_B = \sqrt{0.36} = 0.6 \, \text{m/s}
\]
### Step 5: Apply conservation of momentum
For an elastic collision, the law of conservation of momentum states:
\[
m_B \cdot v_B = m_A \cdot v_A + m_B \cdot v_B'
\]
Substituting the known values:
\[
1 \cdot 0.6 = 2 \cdot v_A + 1 \cdot v_B'
\]
This simplifies to:
\[
0.6 = 2v_A + v_B' \quad \text{(1)}
\]
### Step 6: Apply the coefficient of restitution
For elastic collisions, the coefficient of restitution \( e \) is defined as:
\[
e = \frac{v_B' - v_A}{u_B - u_A}
\]
Since \( e = 1 \) for elastic collisions:
\[
v_B' - v_A = 0.6 \quad \text{(2)}
\]
### Step 7: Solve the equations
From equation (2), we can express \( v_B' \) in terms of \( v_A \):
\[
v_B' = v_A + 0.6
\]
Substituting this into equation (1):
\[
0.6 = 2v_A + (v_A + 0.6)
\]
This simplifies to:
\[
0.6 = 3v_A + 0.6
\]
Subtracting 0.6 from both sides:
\[
0 = 3v_A \implies v_A = 0
\]
Now substituting \( v_A = 0 \) back into equation (2):
\[
v_B' = 0 + 0.6 = 0.6 \, \text{m/s}
\]
### Step 8: Calculate the distance traveled after the collision
Both blocks will experience friction after the collision. The frictional force will decelerate both blocks.
For block B:
Using \( a = -2 \, \text{m/s}^2 \):
\[
0 = (0.6)^2 + 2 \cdot (-2) \cdot s_B
\]
\[
0.36 = 4s_B \implies s_B = 0.09 \, \text{m}
\]
For block A:
Using \( a = -2 \, \text{m/s}^2 \):
\[
0 = (0)^2 + 2 \cdot (-2) \cdot s_A
\]
\[
0 = 0 \implies s_A = 0
\]
### Step 9: Calculate the final separation
The final separation between the blocks after they have come to rest due to friction is:
\[
\text{Final Separation} = s_B + s_A = 0.09 + 0 = 0.09 \, \text{m} = 9 \, \text{cm}
\]
### Final Answer
The final separation between the blocks after the collision is **9 cm**.
