| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Elastic string with variable force |
| Difficulty | Challenging +1.8 This is a challenging M2 mechanics problem requiring multiple sophisticated techniques: deriving equations of motion with variable forces, using v dv/dx = a, integrating exponential functions, and applying Hooke's law with proportionality conditions. The multi-stage nature (three distinct phases of motion) and the need to handle both exponential decay and polynomial terms elevate this significantly above standard mechanics questions, though the individual calculus steps are methodical once set up correctly. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.02g Hooke's law: T = k*x or T = lambda*x/l6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.4a = 0.4v\frac{dv}{dx} = 8x - (2e^{-x} + 4)\) | M1 | Use Newton's Second Law horizontally |
| \(v\frac{dv}{dx} = 20x - 10 - 5e^{-x}\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int v\,dv = \int(20x - 10 - 5e^{-x})\,dx\) | M1 | Attempt to integrate the equation from part (i) |
| \(\left[v^2/2\right]_6^v = \left[10x^2 - 10x + 5e^{-x}\right]_0^{0.5}\) | M1 | Use correct limits in the integration |
| \(v = 5.2(0)\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(8x - 4 = T = \lambda(x - 0.5)/0.5\ (= \lambda x/0.5 - \lambda)\) | M1 | Use \(T = \lambda x/L\) |
| \(\lambda = 4\) N | A1 | |
| 2 |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.4a = 0.4v\frac{dv}{dx} = 8x - (2e^{-x} + 4)$ | M1 | Use Newton's Second Law horizontally |
| $v\frac{dv}{dx} = 20x - 10 - 5e^{-x}$ | A1 | |
| | **2** | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int v\,dv = \int(20x - 10 - 5e^{-x})\,dx$ | M1 | Attempt to integrate the equation from part (i) |
| $\left[v^2/2\right]_6^v = \left[10x^2 - 10x + 5e^{-x}\right]_0^{0.5}$ | M1 | Use correct limits in the integration |
| $v = 5.2(0)$ | A1 | |
| | **3** | |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $8x - 4 = T = \lambda(x - 0.5)/0.5\ (= \lambda x/0.5 - \lambda)$ | M1 | Use $T = \lambda x/L$ |
| $\lambda = 4$ N | A1 | |
| | **2** | |
---3 A particle $P$ of mass 0.4 kg is projected horizontally along a smooth horizontal plane from a point $O$. After projection the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its displacement from $O$ is $x \mathrm {~m}$. A force of magnitude $8 x \mathrm {~N}$ directed away from $O$ acts on $P$ and a force of magnitude ( $2 \mathrm { e } ^ { - x } + 4$ ) N opposes the motion of $P$. One end of a light elastic string of natural length 0.5 m is attached to $O$ and the other end of the string is attached to $P$.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 20 x - 10 - 5 \mathrm { e } ^ { - x }$ before the elastic string becomes taut.\\
(ii) Given that the initial velocity of $P$ is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find $v$ when the string first becomes taut.\\
When the string is taut, the acceleration of $P$ is proportional to $\mathrm { e } ^ { - x }$.\\
(iii) Find the modulus of elasticity of the string.\\
\hfill \mbox{\textit{CAIE M2 2018 Q3 [7]}}