A man weighing 80, kg is standing at the cent | vedclass

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(C) Let $m_1 = 80\, kg$ (mass of the man) and $m_2 = 200\, kg$ (mass of the boat).Since there is no external horizontal force on the system,the centre

Vedclass · Accepted Answer

$14.3$

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(C) Let $m_1 = 80\, kg$ (mass of the man) and $m_2 = 200\, kg$ (mass of the boat).Since there is no external horizontal force on the system,the centre of mass remains stationary: $\Delta x_{cm} = 0$.Let $x$ be the displacement of the boat away from the shore. The displacement of the man relative to the shore is $(8 - x)$.Using the centre of mass formula: $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.$80(8 - x) + 200(-x) = 0$.$640 - 80x - 200x = 0$.$280x = 640$.$x = \frac{640}{280} \approx 2.286\, m$.The man's initial distance from the shore was $20\, m$. After walking $8\, m$ towards the shore and the boat moving $x$ away from the shore,his new distance is $20 - 8 + x = 12 + 2.286 = 14.286\, m \approx 14.3\, m$.

$A$ man weighing $80\, kg$ is standing at the centre of a flat boat and he is $20\, m$ from the shore. He walks $8\, m$ on the boat towards the shore and then halts. The boat weighs $200\, kg$. How far is he from the shore at the end of this time?

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