Question

A small diamond of mass 15.9 g drops from a swimmer's earring and falls through the water, reaching a terminal velocity of 1.9 m/s.

(a) Assuming the frictional force on the diamond obeys f = −bv, what is b (in kg/s)? (Round your answer to at least four decimal places.)

(b) How far (in m) does the diamond fall before it reaches 90 percent of its terminal speed?

Answer 1

Part A.

Using Force balance on diamond

Frictional force at terminal velocity = Weight of diamond

f = Fg

b*v = m*g

b = m*g/v

m = mass of diamond = 15.9 gm = 15.9*10^-3 kg

v = 1.9 m/sec = terminal velocity

So,

b = 15.9*10^-3*9.8/1.9

b = 0.0820 kg/sec

Part B

I'm assuming these equations are solved in reference book, let me know if you have any query regarding these equations:

From force balance on diamond before reaching terminal velocity

mg - bv = m*dv/dt

v(t) = (mg/b)*(1 - exp(-bt/m))

y = mgt/b + (m^2g/b^2)*(exp(-bt/m) - 1)

from 2nd equation

1 - exp (-bt/m) = v*b/(mg)

t = (-m/b)*ln (1 - bv/mg)

Using given info

v = 90% of terminal velocity = 0.9*1.9

t = (-15.9*10^-3/0.0820)*ln (1 - 0.0820*0.9*1.9/(15.9*10^-3*9.8))

t = 0.446 sec

Now using this value of time in 3rd equation

y = mgt/b + (m^2g/b^2)*(exp(-bt/m) - 1)

y = 15.9*10^-3*9.8*0.446/0.0820 + (15.9^2*10^-6*9.8/0.082^2)*(exp(-0.0820*0.446/(15.9*10^-3)) - 1)

y = 0.516 m = distance traveled by diamond before reach 90% of terminal speed

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