Number 10 is tricky. I solved it by using the fact that when you remove mass the center of mass moves away from the mass you removed, so the minimum will occur

Number 10 is tricky. I solved it by using the fact that when you remove mass the center of mass moves away from the mass you removed, so the minimum will occur when the height of soda matches the height of the center of mass. This gives the equation: (1/2 c + (1/2 h) (h f)) / (c + h f) = h Where c is the mass of the can, the can has height 1, h is the height of soda remaining in the can, and f is the mass of the fluid when the can is full. (Incidentally, I wrote this equation wrong like eight times because I kept missing a factor of h in the fluid's center of mass height.) Working the equation: c + h^2 f = 2 h (c + h f) f h^2 + 2 c h - c = 0 Oh hey this will get simpler if we just look at the ratio: let r = c/f h^2 + 2 r h - r = 0 h = -r +- sqrt(r^2 + r) h = sqrt(r^2 + r) - r Well that's kind of gross. I'm actually surprised it's in the right range based on how it looks (but plotting it shows it is in fact in [0, 0.5]). Maybe the values from the problem are nice: r = 1.5/12 = 3/24 h = sqrt((3/24)^2 + 3/24) - 3/24 h = 1/4 I guess that's okay..