A block of mass `1kg` is at `x=10m` and moving towards negative `x-axis` with velocity `6m//s`. Another block of mass `2kg` is at `x=12m` and moving towards positive `x-`axis with velocity `4m//s` at the same instant. Find position of their centre of mass after `2s`.

Explanation

To find the position of the center of mass after 2 seconds for the two blocks, we can follow these steps:

### Step 1: Identify the initial positions and velocities of the blocks
- Block 1 (mass \( m_1 = 1 \, \text{kg} \)):
- Initial position \( x_1 = 10 \, \text{m} \)
- Velocity \( v_1 = -6 \, \text{m/s} \) (moving towards the negative x-axis)

- Block 2 (mass \( m_2 = 2 \, \text{kg} \)):
- Initial position \( x_2 = 12 \, \text{m} \)
- Velocity \( v_2 = 4 \, \text{m/s} \) (moving towards the positive x-axis)

### Step 2: Calculate the positions of the blocks after 2 seconds
- For Block 1:
\[ x_1' = x_1 + v_1 \cdot t = 10 \, \text{m} + (-6 \, \text{m/s}) \cdot 2 \, \text{s} \]
\[ x_1' = 10 \, \text{m} - 12 \, \text{m} = -2 \, \text{m} \]

- For Block 2:
\[ x_2' = x_2 + v_2 \cdot t = 12 \, \text{m} + (4 \, \text{m/s}) \cdot 2 \, \text{s} \]
\[ x_2' = 12 \, \text{m} + 8 \, \text{m} = 20 \, \text{m} \]

### Step 3: Calculate the position of the center of mass
The formula for the center of mass \( x_{\text{cm}} \) is given by:
\[ x_{\text{cm}} = \frac{m_1 \cdot x_1' + m_2 \cdot x_2'}{m_1 + m_2} \]
Substituting the values:
\[ x_{\text{cm}} = \frac{(1 \, \text{kg}) \cdot (-2 \, \text{m}) + (2 \, \text{kg}) \cdot (20 \, \text{m})}{1 \, \text{kg} + 2 \, \text{kg}} \]
\[ x_{\text{cm}} = \frac{-2 \, \text{kg m} + 40 \, \text{kg m}}{3 \, \text{kg}} = \frac{38 \, \text{kg m}}{3 \, \text{kg}} \approx 12.67 \, \text{m} \]

### Final Answer
The position of the center of mass after 2 seconds is approximately \( 12.67 \, \text{m} \).

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