Two spheres of equal terminal falling velocity settle in water starting from res

Question

Two spheres of equal terminal falling velocity settle in water starting from rest at the
same horizontal level. How far apart vertically will the particles be when they have both
reached 99 per cent of their terminal falling velocities? It may be assumed that Stokes’
law is valid and this assumption should be checked.
The diameter of one sphere is 40 µm and its density is 1500 kg/m3 and the density of
the second sphere is 3000 kg/m3. The density and viscosity of water are 1000 kg/m3 and
1 mNs/m2 respectively.

Explanation / Answer

Assuming Stokes’ law is valid, the terminal velocity is given by equation 3.24 as:
u0 = (d2g/18µ)(?s - ?)
For particle 1:
u0 = {[(40 × 10-6)2 × 9.81]/(18 × 1 × 10-3)}(1500 - 1000)
= 4.36 × 10-4 m/s

Since particle 2 has the same terminal velocity:
4.36 × 10-4 = [(d2
2
× 9.81)/(18 × 1 × 10-3)](3000 - 1000)
From which: d2 = (2 × 10-5)m or 20µm
From equation 3.83: a = 18µ/d2?s
For particle 1: a1 = (18 × 1 × 10-3)/((40 × 10-6)2 × 1500) = 7.5 × 103 s-1
and for particle 2: a2 = (18 × 1 × 10-3)/((20 × 10-6)2 × 3000) = 1.5 × 104 s-1
From equation 3.90: b = (1 - ?/?s)g
For particle 1: b1 = (1 - 1000/1500)9.81 = 3.27 m/s2
and for particle 2: b2 = (1 - 1000/3000)9.81 = 6.54 m/s2
The initial velocity of both particles, v = 0,
get expression for y
differentiate y

and then
When ?y = u0, the terminal velocity, it is not possible to solve for t and hence ?y will be
taken as 0.99u0.
For particle 1:
(0.99 × 4.36 × 10-4) = (4.36 × 10-4)[1 - exp(-7.5 × 103t)]
and: t = 6.14 × 10-4 s
The distance travelled in this time is given by equation 3.88:
y = (3.27/7.5 × 103)6.14 × 10-4 - [3.27/(7.5 × 103)2]
[1 - exp(-7.5 × 103 × 6.14 × 10-4)] = 2.10 × 10-7 m

you can find out for particle 2 ..
rest i think ..m is the difficult part ..

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