 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/section-b/7/d/solution

τ = μ t / r_{0}

and

w = μv / (g r_{0})

. Then

w (τ) = \frac{1}{4} [(1 + τ) - (1 + τ)^{- 3}] .

t = 0

gives

v (t) = g t - \frac{3 gμ}{2 r _{0}} t^{2} + \frac{5 g μ ^{2}}{2 r _{0}^{2}} t^{3} + O (t^{4}), t ≪ r_{0} / μ .

Thus the initial slope is

g

, as for free fall. At late times,

v (t) = \frac{g t}{4} + \frac{g r _{0}}{4 μ} - \frac{g r _{0}^{4}}{4 μ ( r _{0} + μ t ) ^{3}} \sim \frac{g t}{4}, t ≫ r_{0} / μ .

The oblique asymptote is

v = g t /4 + g r_{0} / (4 μ)

, approached from below. Direct differentiation gives

\overset{v}{˙} = \frac{g}{4} [1 + 3 (1 + τ)^{- 4}] > 0, \overset{v}{¨} = - \frac{3 gμ}{r _{0}} (1 + τ)^{- 5} < 0.

The speed increases monotonically with decreasing slope, from

g

towards

g /4

. It does not reach a terminal velocity in this drag-free model. Without mass accumulation, and still without drag, a drop released from rest has

v = g t

: its acceleration remains

g

, whereas the growing drop continually shares its momentum with newly acquired material.

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