| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics differential equation problem requiring separation of variables with a quadratic resistance term. While it involves partial fractions and integration, the setup is straightforward (F=ma with given forces), and both parts follow a well-practiced procedure. The limiting velocity is a direct deduction from the differential equation. More routine than the average Further Maths question but still requires solid technique. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m\frac{dv}{dt} = 50 - 2v^2\), \(\frac{dv}{dt} = 4(25-v^2)\) | B1 | N2L |
| \(\frac{1}{10}\int\frac{1}{5-v} + \frac{1}{5+v}\,dv = \int 4\,dt\) | M1 | Separate variables and use partial fractions |
| \(\frac{1}{10}(-\ln(5-v) + \ln(5+v)) = 4t + A\) | M1 A1 | Integrate into log terms (Note: formula on MF19) |
| Use \(t=0, v=3\) to give \(A = \frac{1}{10}\ln 4\) | M1 | Use initial condition |
| \(4t = \frac{1}{10}\ln\frac{5+v}{4(5-v)}\) leading to \(\frac{5+v}{20-4v} = e^{40t}\) | M1 | Rearrange to make \(v\) the subject |
| \(v = \frac{5(4-e^{-40t})}{4+e^{-40t}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| As \(t \to \infty, v \to 5\) | B1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m\frac{dv}{dt} = 50 - 2v^2$, $\frac{dv}{dt} = 4(25-v^2)$ | B1 | N2L |
| $\frac{1}{10}\int\frac{1}{5-v} + \frac{1}{5+v}\,dv = \int 4\,dt$ | M1 | Separate variables and use partial fractions |
| $\frac{1}{10}(-\ln(5-v) + \ln(5+v)) = 4t + A$ | M1 A1 | Integrate into log terms (Note: formula on MF19) |
| Use $t=0, v=3$ to give $A = \frac{1}{10}\ln 4$ | M1 | Use initial condition |
| $4t = \frac{1}{10}\ln\frac{5+v}{4(5-v)}$ leading to $\frac{5+v}{20-4v} = e^{40t}$ | M1 | Rearrange to make $v$ the subject |
| $v = \frac{5(4-e^{-40t})}{4+e^{-40t}}$ | A1 | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| As $t \to \infty, v \to 5$ | B1 | |4 A particle of mass 0.5 kg moves along a horizontal straight line. Its velocity is $v \mathrm {~ms} ^ { - 1 }$ at time $t \mathrm {~s}$. The forces acting on the particle are a driving force of magnitude 50 N and a resistance of magnitude $2 v ^ { 2 } \mathrm {~N}$. The initial velocity of the particle is $3 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$.
\item Deduce the limiting value of $v$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q4 [8]}}