| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable coefficient of friction |
| Difficulty | Challenging +1.8 This question requires setting up Newton's second law with variable friction, using the chain rule to convert acceleration (v dv/dx = a), solving a separable differential equation, and finding both maximum speed (dv/dx = 0) and rest position (v = 0). While mechanically demanding with multiple steps, the techniques are standard for M2 variable force problems, though the variable coefficient of friction adds conceptual complexity beyond typical A-level 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/Working | Mark | Guidance |
| \(0.2v\,dv/dx = 0.2g\sin30 - 0.1x^2(0.2g\cos30)\) | M1 | N2L parallel to the slope |
| \(2v\,dv/dx = 10 - (\sqrt{3})\,x^2\) | A1 (AG) | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\int v\,dv = \int(10 - \sqrt{3}x^2)\,dx\) | M1 | Integrates acceleration |
| \(v^2 = 10x - \sqrt{3}x^3/3\) | A1 | No need to show \(c = 0\) |
| \(0 = 10 - \sqrt{3}x^2\) | M1* | Solves accn \(= 0\ (x = 2.4028...)\) |
| \(v^2 = 10 \times 2.4 - \sqrt{3} \times 2.4^3/3\) | dep M1* | Puts solution of accn \(= 0\) in \(v^2(x)\) |
| Max \(v = 4(.002)\) | A1 | Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 10x - \sqrt{3}x^3/3\) | M1 | |
| \(x = 4.16\) | A1 | Total: 2 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.2v\,dv/dx = 0.2g\sin30 - 0.1x^2(0.2g\cos30)$ | M1 | N2L parallel to the slope |
| $2v\,dv/dx = 10 - (\sqrt{3})\,x^2$ | A1 (AG) | **Total: 2** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\int v\,dv = \int(10 - \sqrt{3}x^2)\,dx$ | M1 | Integrates acceleration |
| $v^2 = 10x - \sqrt{3}x^3/3$ | A1 | No need to show $c = 0$ |
| $0 = 10 - \sqrt{3}x^2$ | M1* | Solves accn $= 0\ (x = 2.4028...)$ |
| $v^2 = 10 \times 2.4 - \sqrt{3} \times 2.4^3/3$ | dep M1* | Puts solution of accn $= 0$ in $v^2(x)$ |
| Max $v = 4(.002)$ | A1 | **Total: 5** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 10x - \sqrt{3}x^3/3$ | M1 | |
| $x = 4.16$ | A1 | **Total: 2** |
---6 A particle $P$ of mass 0.2 kg is released from rest at a point $O$ on a plane inclined at $30 ^ { \circ }$ to the horizontal. At time $t \mathrm {~s}$ after its release, $P$ has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and displacement $x \mathrm {~m}$ down the plane from $O$. The coefficient of friction between $P$ and the plane increases as $P$ moves down the plane, and equals $0.1 x ^ { 2 }$.\\
(i) Show that $2 v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 10 - ( \sqrt { } 3 ) x ^ { 2 }$.\\
(ii) Calculate the maximum speed of $P$.\\
(iii) Find the value of $x$ at the point where $P$ comes to rest.
\hfill \mbox{\textit{CAIE M2 2016 Q6 [9]}}