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


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Question Number 61205 by necx1 last updated on 30/May/19

$A c u b i c a l b l o c k o f i c e o f m a s s m a n d$ $e d g e L i s p l a c e d i n a l a r g e t r a y o f m a s s$ $M . I f t h e i c e b l o c k m e l t s, h o w f a r d o e s$ $t h e c e n t r e o f m a s s o f t h e s y s t e m ‘ ‘ i c e + {t r a y}^{″}$ $c o m e d o w n ?$ $a) \frac{m l}{m + M} b) \frac{2 m l}{m + M} c) \frac{m l}{2 (m + M)} d) n o n e$
Commented by necx1 last updated on 30/May/19

Commented by mr W last updated on 30/May/19

$c) i s r i g h t .$
Commented by necx1 last updated on 30/May/19

$p l e a s e s h o w w o r k i n g s$
Commented by mr W last updated on 30/May/19

$t r a y i s l a r g e a n d t h i n, h e i g h t o f i t s$ $c e n t e r o f m a s s i s \approx 0 .$ $i c e b l o c k h a s a h e i g h t o f L, t h e h e i g h t$ $i t s c e n t e r o f m a s s i s \frac{L}{2} .$ $t h e h e i g h t o f t h e c o m o f ‘ ‘ i c e + {t r a y}^{″}$ $b e f o r e m e l t i n g i s \frac{0 \times M + \frac{L}{2} \times m}{M + m} = \frac{m L}{2 (M + m)} .$ $a f t e r m e l t i n g i c e b e c o m e s w a t e r w h i c h$ $i s d i s t r i b u t e d o v e r t h e t r a y a s t h i n$ $w a t e r f i l m w h o s e h e i g h t o f c o m i s \approx 0 .$ $t h u s t h e c h a n g e o f c o m o f t h e s y s t e m$ $i c e + t r a y i s f r o m \frac{m L}{2 (M + m)} t o 0 .$
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