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Question Number 61205 by necx1 last updated on 30/May/19
A cubical block of ice of mass m and  edge L is placed in a large tray of mass  M.If the ice block melts,how far does  the centre of mass of the system “ice + tray”  come down ?    a)((ml)/(m+M))  b)((2ml)/(m+M))  c)((ml)/(2(m+M)))  d)none
AcubicalblockoficeofmassmandedgeLisplacedinalargetrayofmassM.Iftheiceblockmelts,howfardoesthecentreofmassofthesystemice+traycomedown?a)mlm+Mb)2mlm+Mc)ml2(m+M)d)none
Commented by necx1 last updated on 30/May/19
Commented by mr W last updated on 30/May/19
c) is right.
c)isright.
Commented by necx1 last updated on 30/May/19
please show workings
pleaseshowworkings
Commented by mr W last updated on 30/May/19
tray is large and thin, height of its  center of mass is ≈0.  ice block has a height of L, the height  its center of mass is (L/2).  the height of the com of  “ice+tray”  before melting is ((0×M+(L/2)×m)/(M+m))=((mL)/(2(M+m))).  after melting ice becomes water which  is distributed over the tray as thin  water film whose height of com is ≈0.  thus the change of com of the system  ice+tray is from ((mL)/(2(M+m))) to 0.
trayislargeandthin,heightofitscenterofmassis0.iceblockhasaheightofL,theheightitscenterofmassisL2.theheightofthecomofice+traybeforemeltingis0×M+L2×mM+m=mL2(M+m).aftermeltingicebecomeswaterwhichisdistributedoverthetrayasthinwaterfilmwhoseheightofcomis0.thusthechangeofcomofthesystemice+trayisfrommL2(M+m)to0.

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