| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - horizontal motion |
| Difficulty | Standard +0.3 This is a standard M2 variable force question involving air resistance proportional to velocity. Part (i) requires applying F=ma with v(dv/dx) and simple integration. Part (ii) adds friction but follows the same template: set up the differential equation, separate variables, and integrate. The question is structured with clear 'show that' scaffolding and uses routine techniques for this topic, making it slightly easier than average overall. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Uses Newton's 2nd Law with \(a = \frac{dv}{dt}\) (error flag) | B1, M1 | |
| Integrates and uses v(0) = 2 \([v = -x/4 + 2]\) | M1 | |
| Obtains the distance as 8 m | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains \(F = 3/40 \times 0.4g\) \([=0.3]\) | B1 | |
| Uses Newton's 2nd Law and \(a = \frac{dv}{dt}\) with \(F\) \([0.4 \frac{dv}{dt} = -0.1v - F]\) | M1 | |
| Obtains the given equation \(\frac{dv}{dt} = -(v + 3)\) correctly | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains \(t = -4 \ln(v + 3)\) \((+C)\) | B1 | |
| Uses v(0) = 2 to find C (or evaluates \(\int \frac{dv}{v + 3}\) (or evaluates \(\int \frac{dv}{t + f(v) \text{ is a ln function}}\)) | M1 | |
| \(t = 4 \ln 5/3 (= 2.04)\) | A1 | 3 marks |
**(i)**
| Uses Newton's 2nd Law with $a = \frac{dv}{dt}$ (error flag) | B1, M1 |
| Integrates and uses v(0) = 2 $[v = -x/4 + 2]$ | M1 |
| Obtains the distance as 8 m | A1 | 4 marks |
**(ii)(a)**
| Obtains $F = 3/40 \times 0.4g$ $[=0.3]$ | B1 |
| Uses Newton's 2nd Law and $a = \frac{dv}{dt}$ with $F$ $[0.4 \frac{dv}{dt} = -0.1v - F]$ | M1 |
| Obtains the given equation $\frac{dv}{dt} = -(v + 3)$ correctly | A1 | 3 marks |
**(b)**
| Obtains $t = -4 \ln(v + 3)$ $(+C)$ | B1 |
| Uses v(0) = 2 to find C (or evaluates $\int \frac{dv}{v + 3}$ (or evaluates $\int \frac{dv}{t + f(v) \text{ is a ln function}}$) | M1 |
| $t = 4 \ln 5/3 (= 2.04)$ | A1 | 3 marks |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{3e7472a8-df1e-45c4-81fb-e4397bddf5ad-4_182_844_264_653}
A particle $P$ of mass 0.4 kg travels on a horizontal surface along the line $O A$ in the direction from $O$ to $A$. 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 it passes through the fixed point $O$ (see diagram). The speed of $P$ at $O$ is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Assume that the horizontal surface is smooth. Show that $\frac { \mathrm { d } v } { \mathrm {~d} x } = - \frac { 1 } { 4 }$, where $x \mathrm {~m}$ is the distance of $P$ from $O$ at time $t \mathrm {~s}$, and hence find the distance from $O$ at which the speed of $P$ is zero.\\
(ii) Assume instead that the horizontal surface is not smooth and that the coefficient of friction between $P$ and the surface is $\frac { 3 } { 40 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $4 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - ( v + 3 )$.
\item Hence find the value of $t$ for which the speed of $P$ is zero.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2002 Q6 [10]}}