A cubical block of ice of mass m and edge L is placed in a large tray of mass M. If the ice melts, how for does the center of mass of the system "ice plus tray" come down?

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Solution

COM of system Before melting $= \frac{M (0) + m (\frac{L}{2})}{M + m} = \frac{m L}{2 (M + m)}$
COM of system After melting $= \frac{M (0) + m (0)}{M + m} = 0$
So, Change in the position of centre of mass $= \frac{m L}{2 (M + m)}$

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A cubical block of ice of mass m and edge L is placed in a large tray of mass M. If the ice melts, how for does the center of mass of the system "ice plus tray" come down?

COM of system Before melting =M(0)+m(L2)M+m=mL2(M+m)COM of system After melting =M(0)+m(0)M+m=0So, Change in the position of centre of mass =mL2(M+m)

COM of system Before melting $= \frac{M (0) + m (\frac{L}{2})}{M + m} = \frac{m L}{2 (M + m)}$
COM of system After melting $= \frac{M (0) + m (0)}{M + m} = 0$
So, Change in the position of centre of mass $= \frac{m L}{2 (M + m)}$