Fluid: Archimedes Principle Homework Question

[VincentweZu]

Homework Statement

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.

Homework 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
F_g = mg
F_B = p_watergV

The Attempt at a Solution

Since the apparent weight is given (0.1011N). the actual weight should be defined using:
F_g = 0.1011 + F_B (Buoyancy force)
F_g = 0.1011 + p_watergV

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

After substituting V and I solved for m_silver 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?

 

[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.

 

[gneill]

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.

 

[BruceW]

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

I think this is where you went wrong. This V should actually be V_copper (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.

The buoyant force is mg - .1011 = .0117 N.
V = FB/ρ = .0117/1000 = 1.17 x 10^-5

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!

 

[gneill]

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

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

 

[VincentweZu]

I think this is where you went wrong. This V should actually be V_copper (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!

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

Thanks everyone for helping out! It has been appreciated.

 

[Delphi51]

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