I'm trying to only post 3-4 star problem solutions, as these are usually the ones with more nuanced solutions
that are worth typing up (plus I can't solve most of them, so I feel more accomplished if I do manage to get one,
which gives me more motivation to write up my solution!). This is a 2 star problem, but nevertheless, the result of this
problem is pretty cool, so it feels like a good solution to docoument/share.
In this problem, we are asked to consider a very light dustpan of wheels, rolling
down a hill. Spread along the hill is dust of mass density $\sigma \ \text{kg/m}$, along the
slope. The dustpan begins to roll down the hill and collects all of the dust in its path. We are
asked to find the acceleration of the dustpan.
In this particular instance, the mass of the rolling dustpan is increasing, so we use the following
form of Newton's second law:
$$F_{\text{Net}} \ = \ \frac{d p}{dt} \ = \ ma \ + \ \dot{m} v$$
It is clear that $m(x) \ = \ \sigma x$, which implies that $\dot{m} \ = \ \sigma v$. One can also see
that the net force acting on the dustpan at any given point in time is the gravitational force parallel
to the slope, which gives us:
$$\sigma x g \sin \theta \ = \ \sigma x a \ + \ \sigma v^2 \ \Rightarrow \ a \ = \ g \sin \theta \ = \ v \frac{dv}{dx} \ + \ \frac{v^2}{x}$$
This appears to be a linear differential equation, as $\frac{d}{dx} v^2 \ = \ 2v \frac{dv}{dx}$. Let the integrating factor be $\mu$. We assert that:
$$\frac{1}{2} \frac{d}{dx} \Big( \mu v^2 \Big) \ = \ \mu v \frac{dv}{dx} \ + \ \frac{\mu v^2}{x} \ \Rightarrow \ \mu v \frac{dv}{dx} \ + \ \frac{v^2}{2} \frac{d \mu}{dx} \ = \ \mu v \frac{dv}{dx} \ + \ \frac{\mu v^2}{x}$$
$$\Rightarrow \ \frac{d \mu}{dx} \ = \ \frac{2 \mu}{x} \ \Rightarrow \ \mu(x) \ = \ Ax^2$$
where we solved the differential equation on the second line through separation and integration. We thus have:
$$\mu v \frac{dv}{dx} \ + \ \frac{\mu v^2}{x} \ = \ \frac{1}{2} \frac{d}{dx} \Big( x^2 v^2 \Big) \ = \ g \sin \theta x^2$$
$$\Rightarrow \ \frac{1}{2} x^2 v^2 \ = \ \frac{g \sin \theta x^3}{3} \ \Rightarrow \ v \ = \ \sqrt{\frac{2}{3} g \sin \theta x}$$
Thus, we can calculate the acceleration:
$$a \ = \ v \frac{dv}{dx} \ = \ \frac{2}{3} g \sin \theta \sqrt{x} \frac{d}{dx} \sqrt{x} \ = \ \frac{1}{3} g \sin \theta$$
This is an interesting result. Independent of the density of the dust, the acceleration is a constant value, exactly one-third of the acceleration that
the dustpan would have if there was no dust along the slope!