A

Step 1: Define the initial and final positions of A and B.
Initial positions: A at 0 m (left) and B at 2 m (right).
After exchange, A moves to 2 m and B moves to 0 m.
Step 2: Calculate the center of mass (COM) before and after the exchange.
The total mass of the system is \( m_1 + m_2 + M = 50 + 70 + 80 = 200 \) kg.
Initial COM, \( x_{COM_i} = \frac{m_1 \cdot x_A + m_2 \cdot x_B + M \cdot x_{M}}{m_1 + m_2 + M} = \frac{50 \cdot 0 + 70 \cdot 2 + 80 \cdot 1}{200} = \frac{140 + 80}{200} = 1.1 \text{ m} \)
Final COM, \( x_{COM_f} = \frac{m_1 \cdot x_{A_f} + m_2 \cdot x_{B_f} + M \cdot x_M}{200} = \frac{50 \cdot 2 + 70 \cdot 0 + 80 \cdot 1}{200} = \frac{100 + 80}{200} = 0.9 \text{ m} \)
Therefore, the distance moved by the COM is \( |x_{COM_f} - x_{COM_i}| = |0.9 - 1.1| = 0.2 \text{ m} \).
Step 3: The plank moves towards B since it is lighter (having A at the heavier side).
Step 4: The distance moved by the plank is the same as the distance moved by the COM, which is 0.2 m.
Step 5: Distance moved by A with respect to the ground: \( 2 \text{ m} - 0.2 \text{ m} = 1.8 \text{ m} \).
Step 6: Distance moved by B with respect to the ground: \( 0 ext{ m} + 0.2 ext{ m} = 0.2 \text{ m} \).
Therefore, A moves to 1.8 m, B to 0.2 m, and the COM moves 0.2 m.