Motion of Center of Mass: Practice Problem & Solution
A man of $50\text{ kg}$ mass is standing in a gravity free space at a height of $10\text{ m}$ above the floor. He throws a stone of $0.5\text{ kg}$ mass downwards with a speed $2\text{ m/s}$. When the stone reaches the floor, the distance of the man above the floor will be : (2010 Pre)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Motion of Center of Mass. Understanding the underlying formula is key to arriving at the correct answer below:
In gravity-free space, no external force acts, so the centre of mass remains at its initial height of $10\text{ m}$. Using COM conservation: $M_m h_m + M_s h_s = (M_m + M_s) Y_{cm} \implies 50(h) + 0.5(0) = (50 + 0.5)(10) \implies h = 10.1\text{ m}$. Option (b) is correct.
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