Question

A 1967 Kennedy half-dollar has a mass of 1.150 x 10-2 kg. The coin is a mixture of silver and copper, and in water weighs 0.1012 N. Determine the mass of silver in the coin. Give your answer to two significant figures.

Answer 1

Here we can find the density of the coin (ρ). This density will be a weighted average of the densities of the 2 metals, silver (ρs) and copper (ρc). Suppose (ρw) represents the density of water.

(weight of coin in water) = (weight of coin in air) - (weight of water displaced by coin)

(weight of coin in water) = mg - (ρw)Vg

(weight of coin in water) = g[m - (ρw)V]

0.1012 N = (9.81 m/s^2)[(1.15e-2 kg) - (1000 kg/m^3)(V)]

V = 1.184e-6 m^3

ρ = m/V

ρ = (1.150e-2 kg)/(1.184e-6 m^3)

ρ = 9.713e3 kg/m^3 or 9.713 g/cm^3

ρ = x(ρs) + (1 - x)(ρc)

9.713 = x(10.49) + (1 - x)(8.96)

0.753 = 1.53x

x = 0.4922 (coin is 49.22% silver)

(0.4922)m = (0.4922)(1.150e-2 kg) = 5.66e-3 kg