A

Step 1: Identify the system. The total mass of the system is \( m_1 + m_2 + M = 50 \, kg + 70 \, kg + 80 \, kg = 200 \, kg \).
Step 2: Calculate the initial center of mass (COM) position. Assuming person A is at position 0 m and person B is at position 2 m, the initial position of the COM is \( \text{COM}_0 = \frac{m_1 \cdot 0 + m_2 \cdot 2 + M \cdot 1}{m_1 + m_2 + M} = \frac{50 \cdot 0 + 70 \cdot 2 + 80 \cdot 1}{200} = \frac{140}{200} = 0.7 \, m \).
Step 3: After exchanging positions, A will be at 2 m and B at 0 m. Calculate the new COM position: \( \text{COM}_1 = \frac{m_1 \cdot 2 + m_2 \cdot 0 + M \cdot 1}{200} = \frac{50 \cdot 2 + 70 \cdot 0 + 80 \cdot 1}{200} = \frac{180}{200} = 0.9 \, m \).
Step 4: The distance moved by the center of mass, \( d_{COM} = \text{COM}_1 - \text{COM}_0 = 0.9 - 0.7 = 0.2 \, m \).
Step 5: Since momentum is conserved in the system, the plank will move towards A (to the left) when they switch positions.
Therefore, the distance moved by the plank will be opposite to A's movement.
Step 6: Calculate the distances moved by A and B with respect to the ground. Since A moves 2 meters towards the right while the plank moves to the left by a distance that conserves momentum and distance, we find: \( d_{plank} = \frac{70}{50 + 70} \cdot d_{A} = \frac{70}{120} \cdot 2 = 1.1667 \, m \approx 1.17 \, m \).
Thus, distances for A and B will be 2 m and \( 1.17 \) m respectively, relating back to ground.
Overall, we conclude all answers based on the derived equations.