# Homework Help: Fluid: Archimedes Principle Homework Question

1. Dec 2, 2011

### VincentweZu

1. The problem statement, all variables and given/known data
A Kennedy half-dollar has a mass that is 1.150 x 10-2 kg. The coin is a mixture of silver and copper, and in water weighs 0.1011 N. Determine the mass of silver in the coin.

2. Relevant equations
Explicit givens: m = 1.150 x 10-2kg
apparent weight = 0.1011N
Implicit givens: density of water = 1 x 103kg/m3
density of copper = 8890 kg/m3
density of silver = 10500 kg/m3
Fg = mg
FB = pwatergV

3. The attempt at a solution

Since the apparent weight is given (0.1011N). the actual weight should be defined using:
Fg = 0.1011 + FB (Buoyancy force)
Fg = 0.1011 + pwatergV

the mass of the coin should be the sum of the mass of silver and copper therefore:
m = msilver + mcopper
m = msilver + pcopperV
I isolated for V since it is an unknown variable in both equations
V = (m - msilver)/pcopper

After substituting V and I solved for msilver and got 9.77 x 10-4kg however the answer which is provided to me is 6.3 x 10-3kg

I don't understand what I did wrong for this question! I've triple checked my work, may someone please point out where I have made a mistake?

2. Dec 2, 2011

### Delphi51

I think there is an error in the given numbers.
The buoyant force is mg - .1011 = .0117 N.
V = FB/ρ = .0117/1000 = 1.17 x 10^-5
density of coin: .0116/(1.17 x 10^-5) = 983.
But the coin has to be more dense than water, so this makes no sense.
It would be worth checking the questions to make sure the numbers were copied correctly.

3. Dec 2, 2011

### Staff: Mentor

I think the numbers will work out.

You were given the mass of the coin in the question statement so you automatically have its weight (mg). The apparent weight in water is also given. The difference in these weights should be the weight of water displaced. From there it's a small hop to the volume. So no need to solve a pair of equations to determine the volume!

So you have total mass and total volume.

To determine the masses of each element present you should be able to set up two equations, one for mass and one for volume, to solve for the individual masses.

Note that the results will be very sensitive to the value used for g. Use g = 9.8m/s2 to get the book result.

4. Dec 2, 2011

### BruceW

I think this is where you went wrong. This V should actually be Vcopper (in other words, the proportionate volume of the coin which the copper takes up). And this is not equal to V the total volume of the coin.

That's right for the buoyant force, but the equation for V should be: V = FB/ρg.

Edit: wow, 3 homework helpers! this guy is getting a lot of attention!

5. Dec 2, 2011

### Staff: Mentor

Hey, it's Friday night and things are slow...

6. Dec 3, 2011

### VincentweZu

I never caught that mistake haha! Thank you very much!

Thanks everyone for helping out! It has been appreciated.

7. Dec 3, 2011

### Delphi51

Darn, I didn't even check that formula! Thanks, Bruce.