| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable coefficient of friction |
| Difficulty | Challenging +1.2 This is a multi-part variable force mechanics problem requiring integration and careful consideration of friction direction changes. While it involves several steps (finding mass from limiting friction, setting up and solving differential equations, integrating for displacement), the techniques are standard M2 content with clear signposting. The 'show that' parts provide targets that guide the solution method, reducing problem-solving demand compared to open-ended questions. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2mg = 0.06 \times 8\) | M1 | Resolve along the plane |
| \(m = 0.24\text{ kg}\) | A1 AG | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(m\frac{dv}{dt} = 0.06t - 0.2mg\) or \(0.24\frac{dv}{dt} = 0.06t - 0.2 \times 0.24g\) | M1 | Use N2L along the plane |
| \(\frac{dv}{dt} = 0.25t - 2\) | A1 AG | |
| \(\int dv = \int(0.25t - 2)dt\) | M1 | Attempt to integrate |
| \(v = \frac{0.25t^2}{2} - 2t + c\), Put \(v = 0\) and \(t = 4\) (leads to \(c = 6\)) | M1 | Attempt to find \(c\) |
| Initial velocity \(= 6 \text{ m s}^{-1}\) | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = \int\left(\frac{0.25t^2}{2} - 2t + 6\right)dt\) | M1 | Attempt to integrate |
| \(x = \frac{0.25t^3}{6} - t^2 + 6t\ (+k)\) | A1ft | ft candidates \(c\) from part (ii) |
| Finds or assumes \(k = 0\) and substitutes \(t = 4\) OR uses limits of \(0\) and \(4\) | M1 | |
| \(OP = \frac{32}{3} = 10\frac{2}{3} = 10.7\text{ m}\) | A1 | |
| 4 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2mg = 0.06 \times 8$ | M1 | Resolve along the plane |
| $m = 0.24\text{ kg}$ | A1 AG | |
| | **2** | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $m\frac{dv}{dt} = 0.06t - 0.2mg$ or $0.24\frac{dv}{dt} = 0.06t - 0.2 \times 0.24g$ | M1 | Use N2L along the plane |
| $\frac{dv}{dt} = 0.25t - 2$ | A1 AG | |
| $\int dv = \int(0.25t - 2)dt$ | M1 | Attempt to integrate |
| $v = \frac{0.25t^2}{2} - 2t + c$, Put $v = 0$ and $t = 4$ (leads to $c = 6$) | M1 | Attempt to find $c$ |
| Initial velocity $= 6 \text{ m s}^{-1}$ | A1 | |
| | **5** | |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = \int\left(\frac{0.25t^2}{2} - 2t + 6\right)dt$ | M1 | Attempt to integrate |
| $x = \frac{0.25t^3}{6} - t^2 + 6t\ (+k)$ | A1ft | ft candidates $c$ from part **(ii)** |
| Finds or assumes $k = 0$ and substitutes $t = 4$ OR uses limits of $0$ and $4$ | M1 | |
| $OP = \frac{32}{3} = 10\frac{2}{3} = 10.7\text{ m}$ | A1 | |
| | **4** | |7 A particle $P$ is projected horizontally from a point $O$ on a rough horizontal surface. The coefficient of friction between the particle and the surface is 0.2 . A horizontal force of magnitude $0.06 t \mathrm {~N}$ directed away from $O$ acts on $P$, where $t \mathrm {~s}$ is the time after projection. $P$ comes to rest when $t = 4$.\\
(i) The particle begins to move again when $t = 8$. Show that the mass of $P$ is 0.24 kg .\\
(ii) Show that, for $0 \leqslant t \leqslant 4 , \frac { \mathrm {~d} v } { \mathrm {~d} t } = 0.25 t - 2$, and find the speed of projection of $P$.\\
(iii) Find the distance from $O$ at which $P$ comes to rest.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M2 2019 Q7 [11]}}