Two men,of masses 60 , kg and 80 , kg,are sit | vedclass

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(B) Let the mass of the boat be $M = 60 \, kg$,length $L = 4 \, m$,and the masses of the two men be $m_1 = 60 \, kg$ and $m_2 = 80 \, kg$.Since there

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the boat moves by $0.4 \, m$

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(B) Let the mass of the boat be $M = 60 \, kg$,length $L = 4 \, m$,and the masses of the two men be $m_1 = 60 \, kg$ and $m_2 = 80 \, kg$.Since there is no external horizontal force acting on the system (boat + men),the centre of mass of the system remains stationary.Let the displacement of the boat be $x$. When the men exchange positions,the displacement of the men relative to the ground is $(L - x)$ and $(-L - x)$ respectively.Using the principle of conservation of the centre of mass: $M(x) + m_1(x - L) + m_2(x + L) = 0$.Substituting the values: $60x + 60(x - 4) + 80(x + 4) = 0$.$60x + 60x - 240 + 80x + 320 = 0$.$200x + 80 = 0$.$200x = -80$.$x = -80 / 200 = -0.4 \, m$.The negative sign indicates that the boat moves by $0.4 \, m$ in the opposite direction of the heavier man's movement.

Two men,of masses $60 \, kg$ and $80 \, kg$,are sitting at the ends of a boat of mass $60 \, kg$ and length $4 \, m$. The boat is stationary. If the men now exchange their positions,then:

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