A man of 50, kg mass is standing in a gravity | vedclass

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Question

(C) In a gravity-free space,there is no external force acting on the system (man + stone). Therefore,the center of mass of the system remains at rest.

Vedclass · Accepted Answer

$10.1$

Vedclass · Answer

(C) In a gravity-free space,there is no external force acting on the system (man + stone). Therefore,the center of mass of the system remains at rest.Let the initial position of the center of mass be $Y_{cm} = 10\, m$ from the floor.Let the man move upwards by a distance $x$ and the stone move downwards by a distance $y$ when the stone hits the floor.The stone travels $10\, m$ downwards,so $y = 10\, m$.Since the center of mass remains stationary,the displacement of the center of mass is zero: $m_1 \Delta y_1 + m_2 \Delta y_2 = 0$.Here,$m_1 = 50\, kg$,$\Delta y_1 = x$ (upwards),$m_2 = 0.5\, kg$,$\Delta y_2 = -10\, m$ (downwards).$50(x) + 0.5(-10) = 0$.$50x = 5$.$x = 0.1\, m$.The new height of the man above the floor is $10 + x = 10 + 0.1 = 10.1\, m$.

$A$ man of $50\, kg$ mass is standing in a gravity-free space at a height of $10\, m$ above the floor. He throws a stone of $0.5\, kg$ mass downwards with a speed of $2\, m/s$. When the stone reaches the floor,the distance of the man above the floor will be ........ $m.$

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