| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - projected vertically upward |
| Difficulty | Challenging +1.8 This is a Further Maths M4 question requiring separation of variables for non-linear differential equations with air resistance proportional to v². Part (a) involves setting up F=ma with resistance, separating variables, and integrating to a given time expression. Part (b) requires a second integration or energy methods. While the topic is advanced (M4), the solution follows standard techniques for this question type with clear guidance toward the answer. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration4.10a General/particular solutions: of differential equations4.10e Second order non-homogeneous: complementary + particular integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-mg\!\left(1+\dfrac{v^2}{k^2}\right) = m\dfrac{dv}{dt}\) | M1 A1 | |
| \(g\displaystyle\int_0^T dt = \int_U^0 \dfrac{-k^2\,dv}{(k^2+v^2)}\) | DM1 | Separating variables and integrating |
| \(T = \dfrac{k}{g}\!\left[\tan^{-1}\dfrac{v}{k}\right]_0^U\) | A1 | |
| \(= \dfrac{k}{g}\tan^{-1}\dfrac{U}{k}\) | DM1 A1 | Total: 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-mg\!\left(1+\dfrac{v^2}{k^2}\right) = mv\dfrac{dv}{dx}\) | M1 A1 | |
| \(g\displaystyle\int_0^H dx = \int_U^0 \dfrac{-k^2 v\,dv}{(k^2+v^2)}\) | DM1 | |
| \(H = \dfrac{k^2}{2g}\!\left[\ln(k^2+v^2)\right]_0^U\) | A1 | |
| \(H = \dfrac{k^2}{2g}\ln\dfrac{(k^2+U^2)}{k^2}\) | DM1 A1 | Total: 6 marks |
## Question 4:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-mg\!\left(1+\dfrac{v^2}{k^2}\right) = m\dfrac{dv}{dt}$ | M1 A1 | |
| $g\displaystyle\int_0^T dt = \int_U^0 \dfrac{-k^2\,dv}{(k^2+v^2)}$ | DM1 | Separating variables and integrating |
| $T = \dfrac{k}{g}\!\left[\tan^{-1}\dfrac{v}{k}\right]_0^U$ | A1 | |
| $= \dfrac{k}{g}\tan^{-1}\dfrac{U}{k}$ | DM1 A1 | Total: 6 marks |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-mg\!\left(1+\dfrac{v^2}{k^2}\right) = mv\dfrac{dv}{dx}$ | M1 A1 | |
| $g\displaystyle\int_0^H dx = \int_U^0 \dfrac{-k^2 v\,dv}{(k^2+v^2)}$ | DM1 | |
| $H = \dfrac{k^2}{2g}\!\left[\ln(k^2+v^2)\right]_0^U$ | A1 | |
| $H = \dfrac{k^2}{2g}\ln\dfrac{(k^2+U^2)}{k^2}$ | DM1 A1 | Total: 6 marks |
---\begin{enumerate}
\item A particle of mass $m$ is projected vertically upwards, at time $t = 0$, with speed $U$. The particle is subject to air resistance of magnitude $\frac { m g v ^ { 2 } } { k ^ { 2 } }$, where $v$ is the speed of the particle at time $t$ and $k$ is a positive constant.\\
(a) Show that the particle reaches its greatest height above the point of projection at time
\end{enumerate}
$$\frac { k } { g } \tan ^ { - 1 } \left( \frac { U } { k } \right)$$
(b) Find the greatest height above the point of projection attained by the particle.
\hfill \mbox{\textit{Edexcel M4 2010 Q4 [12]}}