Question

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 far does the centre of mass of the system "ice plus tray” come down with reference to the point O as shown in figure

• A. $\frac{mL}{m+M}$
• B. $\frac{mL}{m+M}$
• C. $\frac{(M+m)L}{m}$
• D. $\frac{(M+m)L}{M}$

Answer 1

The correct option is B

$\frac{mL}{m+M}$

Cosider figure . Suppose the centre of mass of the tray is a distance $x_1$ above the origin and that of the ice is a distance $x_2$ above the origin. The hieght of the centre of mass of the ice - tray system is
$x=\frac{m{x}_2+M{x}_1}{m+M}$

When the ice melts, the water of mass m spreads on the surface of the tray . As the tray is large , the height of water is negligible. The centre of mass of the water is then on the surface of the tray and is at a distance $x_2$ - $\frac{L}{2}$ above the origin. The new centre of mass of the ice - tray system will be at the height.
x' = $\frac{m(x_2-\frac{L}{2})+M{x}_1}{m+M}$
The shift in the centre of mass = x - x ' = $\frac{mL}{2(m+M)}$