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(C) Let the mass of man $A$ be $m_A = 40 \, kg$,mass of man $B$ be $m_B = 60 \, kg$,and mass of the plank be $m_P = 40 \, kg$. The length of the plank

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the right end of the plank

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(C) Let the mass of man $A$ be $m_A = 40 \, kg$,mass of man $B$ be $m_B = 60 \, kg$,and mass of the plank be $m_P = 40 \, kg$. The length of the plank is $L = 120 \, cm$.Since the bottom surface is smooth and there are no external horizontal forces,the position of the center of mass of the system remains fixed.Let the initial position of the center of the plank be the origin $(0, 0)$.Initially,man $A$ is at $x_A = -60 \, cm$,man $B$ is at $x_B = 0 \, cm$,and the center of the plank is at $x_P = 0 \, cm$.The center of mass of the system is $X_{CM} = \frac{m_A x_A + m_B x_B + m_P x_P}{m_A + m_B + m_P} = \frac{40(-60) + 60(0) + 40(0)}{40 + 60 + 40} = \frac{-2400}{140} = -\frac{120}{7} \, cm$.When $A$ meets $B$,let the plank have shifted by a displacement $d$ to the right. The new position of the plank's center is $x_P' = d$.Since $B$ remains fixed with respect to the ground,$B$ is at $x_B' = 0$. Since $B$ was at the center of the plank,and the plank shifted by $d$,$B$ is now at the right edge of the plank (relative to the plank's frame). Thus,$A$ must also be at the right edge of the plank to meet $B$.Using the conservation of center of mass: $m_A x_A' + m_B x_B' + m_P x_P' = m_A x_A + m_B x_B + m_P x_P$.Since $A$ and $B$ meet at the right end of the plank,$x_A' = x_B' = d + 60$.$40(d + 60) + 60(0) + 40(d) = -2400$.$40d + 2400 + 40d = -2400 \Rightarrow 80d = -4800 \Rightarrow d = -60 \, cm$.The meeting point is $x_B' = 0$. The right end of the plank is at $x_P' + 60 = -60 + 60 = 0$. Thus,they meet at the right end of the plank.

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