A particle A of unit mass travels horizontally through a viscous medium - HSC - SSCE Mathematics Extension 2 - Question 15 - 2015 - Paper 1

Considering particle B, which is projected vertically, we have two forces acting on it; the gravitational force mg downward and the resistive force $k w^2$ upward. The differential equation governing the motion of particle B can be expressed as:

$m\frac{dw}{dt} = -mg - kw^2$

Rearranging gives:

$\frac{dw}{dt} = -g - \frac{k}{m} w^2$

Letting $g = 1$ for simplicity, the solution of this equation leads us to:

$t = \frac{1}{g} \left( \sqrt{k} \tan^{-1}\left( \frac{w}{\sqrt{k}} \right) \right).$

When particle B is at rest, its velocity w approaches 0, leading to:

$t = \frac{1}{g} \left( \sqrt{k} \tan^{-1}\left( \frac{0}{\sqrt{k}} \right) \right) = 0.$

Now, substituting back into the equation we derived in part (i) for particle A,

As particle B comes to a stop, we want the time taken which is derived from the above substitution. Hence, we simplify to find V as:

$\frac{1}{V} = \frac{1}{u} + \frac{k}{u^2} \tan^{-1}\left(\frac{\sqrt{k}}{g}\right).$

For large values of u, the term $\frac{1}{u}$ approaches 0. Thus, the behavior of V simplifies to:

$V \approx \frac{2}{\pi} \sqrt{k}$

This implies that as the initial speed increases, the velocity V of particle A approaches a constant, diminishing the effects of resistance, resulting in the final approximation for V.