2 mass of block 2 kg & 4 kg are attached to a stiffness spring 100 N/m. 2 kg mass is given velocity 2 m/s & 4 kg to 4 m/s in same direction. Find maximum elongation in the spring during the motion of two blocks.

Answer

To solve the problem of finding the maximum elongation in the spring when two blocks are attached to it, we can follow these steps:

### Step 1: Determine the Initial Momentum of the System
The total momentum of the system before the blocks interact with the spring can be calculated using the formula:
\[ p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot v_2 \]
where:
- \( m_1 = 2 \, \text{kg} \) (mass of the first block)
- \( v_1 = 2 \, \text{m/s} \) (velocity of the first block)
- \( m_2 = 4 \, \text{kg} \) (mass of the second block)
- \( v_2 = 4 \, \text{m/s} \) (velocity of the second block)

Calculating the momentum:
\[ p_{\text{initial}} = (2 \, \text{kg} \cdot 2 \, \text{m/s}) + (4 \, \text{kg} \cdot 4 \, \text{m/s}) = 4 \, \text{kg m/s} + 16 \, \text{kg m/s} = 20 \, \text{kg m/s} \]

### Step 2: Calculate the Center of Mass Velocity
The center of mass (CM) velocity \( V_{\text{cm}} \) can be calculated using the formula:
\[ V_{\text{cm}} = \frac{p_{\text{initial}}}{m_1 + m_2} \]
Substituting the values:
\[ V_{\text{cm}} = \frac{20 \, \text{kg m/s}}{2 \, \text{kg} + 4 \, \text{kg}} = \frac{20 \, \text{kg m/s}}{6 \, \text{kg}} \approx 3.33 \, \text{m/s} \]

### Step 3: Determine the Relative Velocities
Next, we find the relative velocities of each block with respect to the center of mass:
- For the 2 kg block:
\[ v_{1,\text{rel}} = v_1 - V_{\text{cm}} = 2 \, \text{m/s} - 3.33 \, \text{m/s} = -1.33 \, \text{m/s} \]
- For the 4 kg block:
\[ v_{2,\text{rel}} = v_2 - V_{\text{cm}} = 4 \, \text{m/s} - 3.33 \, \text{m/s} = 0.67 \, \text{m/s} \]

### Step 4: Calculate the Kinetic Energy of Each Block
The kinetic energy of each block can be calculated using:
\[ KE = \frac{1}{2} m v^2 \]
- For the 2 kg block:
\[ KE_1 = \frac{1}{2} \cdot 2 \, \text{kg} \cdot (-1.33 \, \text{m/s})^2 = \frac{1}{2} \cdot 2 \cdot 1.7689 \approx 1.77 \, \text{J} \]
- For the 4 kg block:
\[ KE_2 = \frac{1}{2} \cdot 4 \, \text{kg} \cdot (0.67 \, \text{m/s})^2 = \frac{1}{2} \cdot 4 \cdot 0.4489 \approx 0.90 \, \text{J} \]

### Step 5: Total Kinetic Energy
The total kinetic energy of the system is:
\[ KE_{\text{total}} = KE_1 + KE_2 \approx 1.77 \, \text{J} + 0.90 \, \text{J} \approx 2.67 \, \text{J} \]

### Step 6: Relate Kinetic Energy to Spring Potential Energy
At maximum elongation, all kinetic energy will be converted into potential energy stored in the spring:
\[ PE_{\text{spring}} = \frac{1}{2} k x^2 \]
where \( k = 100 \, \text{N/m} \) is the spring constant and \( x \) is the maximum elongation. Setting the kinetic energy equal to the potential energy:
\[ 2.67 \, \text{J} = \frac{1}{2} \cdot 100 \, \text{N/m} \cdot x^2 \]
Solving for \( x \):
\[ 2.67 = 50 x^2 \implies x^2 = \frac{2.67}{50} \implies x^2 = 0.0534 \implies x \approx 0.231 \, \text{m} \]

### Final Answer
The maximum elongation in the spring during the motion of the two blocks is approximately **0.231 m**.

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