| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Standard +0.8 This is a multi-part M3 question requiring derivation of differential equations with air resistance proportional to v², solving by separation of variables, finding limiting velocity, and applying given conditions. While the techniques are standard for M3 (Newton's second law with resistance, chain rule for dv/dx, integration), it requires careful algebraic manipulation and understanding of terminal velocity concepts. The three parts build systematically but demand sustained accuracy across multiple steps, placing it moderately above average difficulty. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \([mg - mkv^2 = ma]\) | M1 | For using Newton's second law |
| \((v\ dv/dx)/(g-kv^2) = 1\) | A1 | 2 AG |
| (ii) \([-\frac{1}{2}\ln(g-kv^2)]/k = x\ (+C)]\) | M1 | For separating variables and attempting to integrate |
| \([-(\ln g)/2k = C]\) | M1 | For using \(v(0)=0\) to find C |
| \(x = [-\frac{1}{2}\ln\{(g-kv^2)/g\}]/k\) | A1 | Any equivalent expression for x |
| \([\ln\{(g-kv^2)/g\} = \ln(e^{-2kx})]\) | M1 | For expressing in the form \(\ln f(v^2) = \ln g(x)\) or equivalent |
| \(v^2 = (1-e^{-2kx})g/k\) | A1 | |
| M1 | For using \(e^{-Ax} \to 0\) for \(+ve\ A\) | |
| Limiting value is \(\sqrt{g/k}\) | A1ft | 7 AG |
| (iii) \([1-e^{-600k} = 0.81]\) | M1 | For using \(v^2(300) = 0.9^2 g/k\) |
| \([-600k = \ln(0.19)]\) | M1 | For using logarithms to solve for k |
| \(k = 0.00277\) | A1 | 3 |
# Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $[mg - mkv^2 = ma]$ | M1 | For using Newton's second law |
| $(v\ dv/dx)/(g-kv^2) = 1$ | A1 | **2** AG |
| **(ii)** $[-\frac{1}{2}\ln(g-kv^2)]/k = x\ (+C)]$ | M1 | For separating variables and attempting to integrate |
| $[-(\ln g)/2k = C]$ | M1 | For using $v(0)=0$ to find C |
| $x = [-\frac{1}{2}\ln\{(g-kv^2)/g\}]/k$ | A1 | Any equivalent expression for x |
| $[\ln\{(g-kv^2)/g\} = \ln(e^{-2kx})]$ | M1 | For expressing in the form $\ln f(v^2) = \ln g(x)$ or equivalent |
| $v^2 = (1-e^{-2kx})g/k$ | A1 | |
| | M1 | For using $e^{-Ax} \to 0$ for $+ve\ A$ |
| Limiting value is $\sqrt{g/k}$ | A1ft | **7** AG |
| **(iii)** $[1-e^{-600k} = 0.81]$ | M1 | For using $v^2(300) = 0.9^2 g/k$ |
| $[-600k = \ln(0.19)]$ | M1 | For using logarithms to solve for k |
| $k = 0.00277$ | A1 | **3** |
---5 The pilot of a hot air balloon keeps it at a fixed altitude by dropping sand from the balloon. Each grain of sand has mass $m \mathrm {~kg}$ and is released from rest. When a grain has fallen a distance $x \mathrm {~m}$, it has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Each grain falls vertically and the only forces acting on it are its weight and air resistance of magnitude $m k v ^ { 2 } \mathrm {~N}$, where $k$ is a positive constant.\\
(i) Show that $\left( \frac { v } { g - k v ^ { 2 } } \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1$.\\
(ii) Find $v ^ { 2 }$ in terms of $k , g$ and $x$. Hence show that, as $x$ becomes large, the limiting value of $v$ is $\sqrt { \frac { g } { k } }$.\\
(iii) Given that the altitude of the balloon is 300 m and that each grain strikes the ground at $90 \%$ of its limiting velocity, find $k$.
\hfill \mbox{\textit{OCR M3 2007 Q5 [12]}}