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

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(B) Since the man is in a gravity-free space,the net external force on the man-stone system is zero. Therefore,the center of mass of the system remain

Vedclass · Accepted Answer

$10.1$

Vedclass · Answer

(B) Since the man is in a gravity-free space,the net external force on the man-stone system is zero. Therefore,the center of mass of the system remains at rest.Let the man move upwards by a distance $x$ when the stone reaches the floor.According to the principle of center of mass,the displacement of the center of mass is zero:$M_{man} \cdot \Delta x_{man} + M_{stone} \cdot \Delta x_{stone} = 0$Here,the stone moves downwards by $10\, m$ (so $\Delta x_{stone} = -10\, m$) and the man moves upwards by $x$ (so $\Delta x_{man} = x$).$50 \cdot x + 0.5 \cdot (-10) = 0$$50x = 5$$x = \frac{5}{50} = 0.1\, m$Therefore,the final 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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