- #1
kiloNewton
- 5
- 1
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.
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.