 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/c/solution

s = r_{0} + μ t

. The proposed velocity and its derivative are

v = \frac{g}{4 μ} (s - \frac{r _{0}^{4}}{s ^{3}}), \overset{v}{˙} = \frac{g}{4} (1 + \frac{3 r _{0}^{4}}{s ^{4}}) .

\frac{3 μ}{s} v = \frac{3 g}{4} (1 - \frac{r _{0}^{4}}{s ^{4}}),

so the sum is exactly

g

. Also

v (0) = g (r_{0} - r_{0}) / (4 μ) = 0

, establishing both the equation and its initial condition:

v (t) = \frac{g}{4 μ} [(r_{0} + μ t) - \frac{r _{0}^{4}}{( r _{0} + μ t ) ^{3}}] .

(r_{0} + μ t)^{3}

explains where the expression comes from:

\frac{d}{d t} (s^{3} v) = g s^{3}, s^{3} v = \frac{g}{4 μ} (s^{4} - r_{0}^{4}) .

The result has units of velocity. Its continuous

μ \to 0

limit is

g t

, despite the apparent

1/ μ

in the written expression.

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