Question Number 61205 by necx1 last updated on 30/May/19
$${A}\:{cubical}\:{block}\:{of}\:{ice}\:{of}\:{mass}\:{m}\:{and} \\ $$$${edge}\:{L}\:{is}\:{placed}\:{in}\:{a}\:{large}\:{tray}\:{of}\:{mass} \\ $$$${M}.{If}\:{the}\:{ice}\:{block}\:{melts},{how}\:{far}\:{does} \\ $$$${the}\:{centre}\:{of}\:{mass}\:{of}\:{the}\:{system}\:“{ice}\:+\:{tray}'' \\ $$$${come}\:{down}\:? \\ $$$$ \\ $$$$\left.{a}\left.\right)\left.\frac{{ml}}{{m}+{M}}\left.\:\:{b}\right)\frac{\mathrm{2}{ml}}{{m}+{M}}\:\:{c}\right)\frac{{ml}}{\mathrm{2}\left({m}+{M}\right)}\:\:{d}\right){none} \\ $$
Commented by necx1 last updated on 30/May/19
Commented by mr W last updated on 30/May/19
$$\left.{c}\right)\:{is}\:{right}. \\ $$
Commented by necx1 last updated on 30/May/19
$${please}\:{show}\:{workings} \\ $$
Commented by mr W last updated on 30/May/19
$${tray}\:{is}\:{large}\:{and}\:{thin},\:{height}\:{of}\:{its} \\ $$$${center}\:{of}\:{mass}\:{is}\:\approx\mathrm{0}. \\ $$$${ice}\:{block}\:{has}\:{a}\:{height}\:{of}\:{L},\:{the}\:{height} \\ $$$${its}\:{center}\:{of}\:{mass}\:{is}\:\frac{{L}}{\mathrm{2}}. \\ $$$${the}\:{height}\:{of}\:{the}\:{com}\:{of}\:\:“{ice}+{tray}'' \\ $$$${before}\:{melting}\:{is}\:\frac{\mathrm{0}×{M}+\frac{{L}}{\mathrm{2}}×{m}}{{M}+{m}}=\frac{{mL}}{\mathrm{2}\left({M}+{m}\right)}. \\ $$$${after}\:{melting}\:{ice}\:{becomes}\:{water}\:{which} \\ $$$${is}\:{distributed}\:{over}\:{the}\:{tray}\:{as}\:{thin} \\ $$$${water}\:{film}\:{whose}\:{height}\:{of}\:{com}\:{is}\:\approx\mathrm{0}. \\ $$$${thus}\:{the}\:{change}\:{of}\:{com}\:{of}\:{the}\:{system} \\ $$$${ice}+{tray}\:{is}\:{from}\:\frac{{mL}}{\mathrm{2}\left({M}+{m}\right)}\:{to}\:\mathrm{0}. \\ $$