| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question on variable force with resistance proportional to v². Part (a) requires setting up F=ma with the given substitution and separating variables to integrate—a routine technique for this topic. Parts (b)(i-ii) involve straightforward substitution and algebraic manipulation. While it requires familiarity with differential equations and exponential functions, the method is well-practiced in Further Maths syllabi with no novel insight needed. |
| Spec | 3.02h Motion under gravity: vector form6.06a Variable force: dv/dt or v*dv/dx methods |
A particle $P$ of mass $mk$ falls from rest due to gravity. There is a resistance force of magnitude $mkv^2$ N, where $v$ ms$^{-1}$ is the speed of $P$ after it has fallen a distance $x$ m and $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item By using $v \frac{dv}{dx} = \frac{dv}{dt}$ and appropriate differential equation, show that
$$v^2 = \frac{g}{k}(1 - e^{-2kx}).$$ [7]
It is given that $k = 0.01$. The speed of $P$ when $x = 0.2$ comes to approximately $v$ ms$^{-1}$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $V$ correct to 2 decimal places. [1]
\item Hence find how far $P$ has fallen when its speed is $\frac{1}{2}V$ ms$^{-1}$. [2]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q3 [10]}}