IIT JEE , AIEEE Forum - Physics , Mechanics

INDIAN_ARMY19890

let the origin be at a distance of X1 from the centre of mass of trey and at a distance of X2 from centre of mass of ice

X= MX1+mX2/M+m

when the ice melt the centre of mass will come down by a distance=L/2 (assuming the water will form very thin layer on tray

hence
now X1' =X1
X2'=X2-L/2

new centre of mass
X'= X1*M+(X2-L/2)m /M+m = X-L/2*(m+M)

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