| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.3 This is a standard M2 differential equations question involving air resistance proportional to velocity. Part (i) requires setting up Newton's second law with resistance, part (ii) involves separating variables and integrating (a routine technique), and part (iii) is straightforward substitution at t=0. The 'show that' structure and the ln 3 validity check add slight complexity, but this follows a well-established template for this topic, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(0.1\,dv/dt = -0.1\,g - 0.1v \Rightarrow \dfrac{1}{v+10}\dfrac{dv}{dt} = -1\) | M1 A1 | For using Newton's second law with \(a = dv/dt\) AG [2] |
| (ii) \(\left[\int\dfrac{1}{v+10}\,dv = -\int dt\right]\) | M1 | For separating variables and attempting to integrate |
| \(\ln(v+10) = -t\ (+A)\) | A1 | |
| \(A = \ln 30\) | B1\(\sqrt{}\) | |
| \([\ln\{(v+10)/30\} = -t \Rightarrow (v+10)/30 = e^{-t}]\) | M1 | For transposing to eliminate ln |
| \(v = 30e^{-t} - 10\) | A1 | |
| Until \(0 = 30e^{-t} \Rightarrow\) valid for \(0 \leq t \leq \ln 3\) | B1 | [6] |
| (iii) | M1 | For substituting \(v = 20\) into \(a = -(v+10)\) or \(t = 0\) into \(a = -30e^{-t}\) |
| Acceleration is \(-30\text{ m s}^{-2}\) | A1 | [2] |
## Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $0.1\,dv/dt = -0.1\,g - 0.1v \Rightarrow \dfrac{1}{v+10}\dfrac{dv}{dt} = -1$ | M1 A1 | For using Newton's second law with $a = dv/dt$ AG **[2]** |
| **(ii)** $\left[\int\dfrac{1}{v+10}\,dv = -\int dt\right]$ | M1 | For separating variables and attempting to integrate |
| $\ln(v+10) = -t\ (+A)$ | A1 | |
| $A = \ln 30$ | B1$\sqrt{}$ | |
| $[\ln\{(v+10)/30\} = -t \Rightarrow (v+10)/30 = e^{-t}]$ | M1 | For transposing to eliminate ln |
| $v = 30e^{-t} - 10$ | A1 | |
| Until $0 = 30e^{-t} \Rightarrow$ valid for $0 \leq t \leq \ln 3$ | B1 | **[6]** |
| **(iii)** | M1 | For substituting $v = 20$ into $a = -(v+10)$ or $t = 0$ into $a = -30e^{-t}$ |
| Acceleration is $-30\text{ m s}^{-2}$ | A1 | **[2]** |7 A particle $P$ of mass 0.1 kg is projected vertically upwards from a point $O$ with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Air resistance of magnitude $0.1 v \mathrm {~N}$ opposes the motion, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of $P$ at time $t \mathrm {~s}$ after projection.\\
(i) Show that, while $P$ is moving upwards, $\frac { 1 } { v + 10 } \frac { \mathrm {~d} v } { \mathrm {~d} t } = - 1$.\\
(ii) Hence find an expression for $v$ in terms of $t$, and explain why it is valid only for $0 \leqslant t \leqslant \ln 3$.\\
(iii) Find the initial acceleration of $P$.
\hfill \mbox{\textit{CAIE M2 2009 Q7 [10]}}