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

For particle B, which is projected vertically and experiences gravitational force as well as viscous resistance, similar force evaluations can be done. The forces on particle B can be expressed as:

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

Substituting m = 1 gives us:

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

Rearranging yields:

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

This expression needs to be integrated. To do this, we utilize a substitution where (k/g) terms are introduced:

Using partial fractions and integration, we arrive at:

$t = \frac{1}{gk} \left[ \ln(\sqrt{\frac{k}{g}} + \tan(\sqrt{\frac{g}{k}}w) \right] + C'$

At t=0, the initial conditions will help determine C'. Finalizing shows:

$t = \frac{1}{gk} [\ln(\sqrt{k/g} + \tan(\sqrt{g/k}w)].$

For this part, we know that particle B will experience a certain time duration until it reaches its maximum height and then comes to rest, allowing us to equate conditions at that state. The time taken for particle B to rest can be deduced from the previous step's derivation. Thus, substituting this time back into the original equation for particle A:

At rest, the dynamics imply:

$t_{B} = \frac{1}{gk} [\ln(\sqrt{k/g} + \tan(\sqrt{g/k}(w_{rest})))] \implies V = 1/(1/u + kt_{b}).$

Using the values leads to:

$\frac{1}{V} = \frac{1}{u} + k\sqrt{\frac{g}{k}}$

When the initial value u is considered very large, in practical terms, it suggests that the impact of k over time becomes a negligible fraction of velocity, particularly the logarithmic terms become stable and proportional. This brings a ratio as:

V \approx $c * constant - limit.

From physical approximations and empirical observations with behavior in viscous mediums:

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

where the behavior resembles a scenario of critical damping effects leading into resonance of motion halted from the vertical thrust.