Three men A, B and C of masses 40,kg, 50,kg a | vedclass

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(B) Since the system is on a smooth horizontal plane,there is no external horizontal force acting on the system. Therefore,the centre of mass of the s

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$1/3\,m$ towards right

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(B) Since the system is on a smooth horizontal plane,there is no external horizontal force acting on the system. Therefore,the centre of mass of the system remains stationary,i.e.,$\Delta X_{cm} = 0$.Let the plank move by a distance $x$ towards the right when $A$ and $C$ interchange their positions.Let the initial positions of $A, B, C$ and the plank be $x_A = -2\,m, x_B = 0\,m, x_C = 2\,m$ and $x_P = 0\,m$ respectively.When $A$ and $C$ interchange positions,their new positions relative to the plank are $x'_A = 2\,m$ and $x'_C = -2\,m$. The plank has shifted by $x$ towards the right.The displacement of each mass relative to the ground is:$\Delta x_A = (2 + x) - (-2) = 4 + x$$\Delta x_B = x - 0 = x$$\Delta x_C = (-2 + x) - 2 = -4 + x$$\Delta x_P = x - 0 = x$Using the formula $m_A \Delta x_A + m_B \Delta x_B + m_C \Delta x_C + m_P \Delta x_P = 0$:$40(4 + x) + 50(x) + 60(-4 + x) + 90(x) = 0$$160 + 40x + 50x - 240 + 60x + 90x = 0$$240x - 80 = 0$$x = 80/240 = 1/3\,m$.Since $x$ is positive,the plank (and thus $B$ which is on the plank) shifts $1/3\,m$ towards the right.

Three men $A, B$ and $C$ of masses $40\,kg, 50\,kg$ and $60\,kg$ are standing on a plank of mass $90\,kg$,which is kept on a smooth horizontal plane. If $A$ and $C$ exchange their positions,then mass $B$ will shift:

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