1. The problem statement Peter threw 10 (same) coins into an empty container (a tube/drum), which was floating in A LIQUID. The diameter of the container was 2,257 cm, and so because the container is a tube the surface of the tube was equal to π times the diameter squared. He measured, after 4 to up to 10 coins, how much the tube sank deeper into the water, and displayed them on this chart (n = number of coins, Δh = change of height in centimeters). The issue I had here was mostly because it wasn't stated that the container was floating in water, but I assumed it did. n | 4 | 5 | 6 | 7 | 8 | 9 | 10 Δh | 9,4 | 10,3 | 12 |13,49 | 15 | 16,9 | 17,6 Q1: What is the mass of one coin? Q2: What is the mass of the container? 2. The attempt at solution I know that the volume of water moved = the volume of coins we put in, but to calculate the mass wouldn't we have to know the density of the coins? ρ = m/V Now I am guessing that the denisty of the water moved = to the density of the coins moved so; ρ(H2O) = ρ(coins) ρ(H2O) = 1000 kg/m^3 Now the second issue: I have no idea how to proceed with this, since from the change of height it seems like we are using different coins, since the height doesn't increase linearly, but I used the first 2 coins measured. I used the change of height between the 5th and 4th coin; h(coin) = 10,3 cm - 9,4 cm = 0,9 cm (not the same with every coin) The volume of the coin equals the volume of the pushed water so it equals to the height times the surface; V(coin) = π*r^2 * h = π*2,257cm^2 * 0,9cm = 14,4 cm^3 = 1,44 * 10^-5 m^3 And so we use the density from above to calculate the mass of one coin: ρ(coins) = m(coin) / V(coin) m(coin) = ρ(coins) * V(coin) = 1000 kg / m^3 * 1,44 * 10^-5 m^3 = 0,0144 kg = 14,4g This is where I finished, I do not know how to calculate the mass of the container. Feel free to also correct me on my way to this point, as I am sure I have went wrong somewhere.