Block 1, with mass `m_1` and speed 3.0 m/s, slides along an x axis on a frictionless floor and then undergoes a one dimensional elastic collision with stationary block 2, with mass `m_2 = 0.40m_1`. The two blocks then slide into a region where the coefficient of kinetic friction is 0.50, there they stop. How far into that region do (a) block 1 and (b) block 2 slide?

Answer

To solve the problem step by step, we need to analyze the elastic collision and the motion of the blocks after the collision.

### Step 1: Understand the initial conditions
- Block 1 has mass \( m_1 \) and initial speed \( u_1 = 3.0 \, \text{m/s} \).
- Block 2 has mass \( m_2 = 0.40 m_1 \) and is initially at rest, so \( u_2 = 0 \, \text{m/s} \).

### Step 2: Apply conservation of momentum
For a one-dimensional elastic collision, the conservation of momentum states:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ m_1 (3.0) + 0.40 m_1 (0) = m_1 v_1 + 0.40 m_1 v_2 \]
This simplifies to:
\[ 3.0 m_1 = m_1 v_1 + 0.40 m_1 v_2 \]
Dividing through by \( m_1 \):
\[ 3.0 = v_1 + 0.40 v_2 \quad (1) \]

### Step 3: Apply conservation of kinetic energy
For elastic collisions, kinetic energy is also conserved:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the known values:
\[ \frac{1}{2} m_1 (3.0)^2 + 0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} (0.40 m_1) v_2^2 \]
This simplifies to:
\[ 4.5 m_1 = \frac{1}{2} m_1 v_1^2 + 0.20 m_1 v_2^2 \]
Dividing through by \( m_1 \):
\[ 4.5 = \frac{1}{2} v_1^2 + 0.20 v_2^2 \quad (2) \]

### Step 4: Solve the system of equations (1) and (2)
From equation (1):
\[ v_1 = 3.0 - 0.40 v_2 \]
Substituting \( v_1 \) into equation (2):
\[ 4.5 = \frac{1}{2} (3.0 - 0.40 v_2)^2 + 0.20 v_2^2 \]
Expanding and simplifying:
\[ 4.5 = \frac{1}{2} (9 - 2.4 v_2 + 0.16 v_2^2) + 0.20 v_2^2 \]
\[ 4.5 = 4.5 - 1.2 v_2 + 0.08 v_2^2 + 0.20 v_2^2 \]
\[ 0 = -1.2 v_2 + 0.28 v_2^2 \]
Rearranging gives:
\[ 0.28 v_2^2 - 1.2 v_2 = 0 \]
Factoring out \( v_2 \):
\[ v_2 (0.28 v_2 - 1.2) = 0 \]
Thus, \( v_2 = 0 \) or \( v_2 = \frac{1.2}{0.28} \approx 4.29 \, \text{m/s} \).

### Step 5: Calculate \( v_1 \)
Substituting \( v_2 \) back into equation (1):
\[ v_1 = 3.0 - 0.40(4.29) \approx 1.28 \, \text{m/s} \]

### Step 6: Calculate the distance each block slides
The frictional force acting on each block is given by:
\[ F_f = \mu m g \]
Where \( \mu = 0.5 \) and \( g \approx 9.8 \, \text{m/s}^2 \). Thus:
\[ F_f = 0.5 m_1 g \]
The deceleration \( a \) due to friction is:
\[ a = \frac{F_f}{m} = \mu g = 0.5 \times 9.8 \approx 4.9 \, \text{m/s}^2 \]

Using the equation of motion \( v^2 = u^2 + 2as \), where \( v = 0 \):
1. For Block 1:
\[ 0 = (1.28)^2 - 2(4.9)s_1 \implies s_1 = \frac{(1.28)^2}{2 \times 4.9} \approx 0.083 \, \text{m} \]

2. For Block 2:
\[ 0 = (4.29)^2 - 2(4.9)s_2 \implies s_2 = \frac{(4.29)^2}{2 \times 4.9} \approx 1.87 \, \text{m} \]

### Final Answers:
- (a) Block 1 slides approximately \( 0.083 \, \text{m} \).
- (b) Block 2 slides approximately \( 1.87 \, \text{m} \).