| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - inclined plane motion |
| Difficulty | Challenging +1.2 This is a standard M2 variable force question requiring Newton's second law with v dv/dx substitution and separation of variables. Part (i) is straightforward force resolution; part (ii) requires integrating a separable differential equation with partial fractions, which is routine for M2 students but involves multiple technical steps beyond basic mechanics. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4v\frac{dv}{dx} = 0.4g\sin 30 - 0.2v^2\) | M1 | Uses Newton's Second Law down the plane. Allow \(a\) for \(v\frac{dv}{dx}\). |
| \(v\frac{dv}{dx} = 5 - 0.5v^2\) | A1 | AG |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \frac{v}{5 - 0.5v^2} \, dv = \int x \, dx\) | M1 | Separates the variables and attempts to integrate. |
| \(-\ln(5 - 0.5v^2) = x (+c)\) | A1 | |
| \(c = -\ln 5 \; [5 - 0.5v^2 = 5e^{-x}]\) | M1 | Puts \(x=0\), \(v=0\) to find \(c\) and attempts to solve for \(v\). |
| \(v = \sqrt{(10 - 10e^{-x})}\) | A1 | |
| Total | 4 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4v\frac{dv}{dx} = 0.4g\sin 30 - 0.2v^2$ | M1 | Uses Newton's Second Law down the plane. Allow $a$ for $v\frac{dv}{dx}$. |
| $v\frac{dv}{dx} = 5 - 0.5v^2$ | A1 | AG |
| **Total** | **2** | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \frac{v}{5 - 0.5v^2} \, dv = \int x \, dx$ | M1 | Separates the variables and attempts to integrate. |
| $-\ln(5 - 0.5v^2) = x (+c)$ | A1 | |
| $c = -\ln 5 \; [5 - 0.5v^2 = 5e^{-x}]$ | M1 | Puts $x=0$, $v=0$ to find $c$ and attempts to solve for $v$. |
| $v = \sqrt{(10 - 10e^{-x})}$ | A1 | |
| **Total** | **4** | |
---3 A particle $P$ of mass 0.4 kg is released from rest at a point $O$ on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. $P$ moves down the line of greatest slope through $O$. The velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when its displacement from $O$ is $x \mathrm {~m}$. A retarding force of magnitude $0.2 v ^ { 2 } \mathrm {~N}$ acts on $P$ in the direction $P O$.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 5 - 0.5 v ^ { 2 }$.\\
(ii) Express $v$ in terms of $x$.\\
\hfill \mbox{\textit{CAIE M2 2017 Q3 [6]}}