| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Motion with exponential force |
| Difficulty | Standard +0.8 This question requires applying F=ma with variable force, integrating to find v(x), then separating variables and integrating again for time. While it involves exponential functions and multiple integration steps, the setup is straightforward and the algebra is manageable. It's moderately harder than average due to the multi-step calculus with exponentials, but follows a standard variable-force framework taught in M2. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5a = 0.16e^x\) | M1 | N2L, single force |
| \(a = 0.32e^x\) | A1 | |
| \(\int v\,dv = \int 0.32e^x\,dx\) | M1 | Forms integral from \(v\,dv/dx = a\) |
| \(v^2/2 = 0.32e^x\ (+c)\) | A1 | Award if \(c\) omitted |
| \(x=0,\ v=0.8\) hence \(c=0\), so \(v^2 = 0.64e^x\) | M1 | Trying to find the value of \(c\) |
| \(v = 0.8e^{x/2}\) | A1 (AG) [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(dv/dt = 0.8e^{x/2}\cdot x\,dx/dt\) | M1 | Uses chain rule on given answer |
| \(dv/dt = 0.4e^{x/2}\cdot v\) | A1 | Maybe implied by later work |
| \(x=0,\ v=0.8e^0\) | M1 | Finding speed where \(x=0\) |
| \(x=0,\ v=0.8\) | A1 | |
| \(0.5\,dv/dt = (0.2e^{x/2})(0.8e^{x/2})\) | M1 | Expresses "ma" in terms of \(x\) |
| \(0.5\,\text{acc}^n = 0.16e^x\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int e^{-x/2}\,dx = \int 0.8\,dt\) | M1 | Forms integral from \(dx/dt = 0.8e^{x/2}\) |
| \(e^{-x/2}/(-1/2) = 0.8t\ (+c)\) | A1 | Award if \(c\) omitted |
| \(x=0,\ t=0\), hence \(c=-2\) and \(-2e^{-1.4/2} = 0.8t - 2\) | M1 | Finding \(c\) and using \(x=1.4\) or \(\left[e^{-x/2}/(-1/2)\right]_0^{1.4} = 0.8t\) |
| \(t = 1.26\) s | A1 [4] [10] | \(1.2585...\) |
## Question 7:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5a = 0.16e^x$ | M1 | N2L, single force |
| $a = 0.32e^x$ | A1 | |
| $\int v\,dv = \int 0.32e^x\,dx$ | M1 | Forms integral from $v\,dv/dx = a$ |
| $v^2/2 = 0.32e^x\ (+c)$ | A1 | Award if $c$ omitted |
| $x=0,\ v=0.8$ hence $c=0$, so $v^2 = 0.64e^x$ | M1 | Trying to find the value of $c$ |
| $v = 0.8e^{x/2}$ | A1 (AG) [6] | |
**OR:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $dv/dt = 0.8e^{x/2}\cdot x\,dx/dt$ | M1 | Uses chain rule on given answer |
| $dv/dt = 0.4e^{x/2}\cdot v$ | A1 | Maybe implied by later work |
| $x=0,\ v=0.8e^0$ | M1 | Finding speed where $x=0$ |
| $x=0,\ v=0.8$ | A1 | |
| $0.5\,dv/dt = (0.2e^{x/2})(0.8e^{x/2})$ | M1 | Expresses "ma" in terms of $x$ |
| $0.5\,\text{acc}^n = 0.16e^x$ | A1 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int e^{-x/2}\,dx = \int 0.8\,dt$ | M1 | Forms integral from $dx/dt = 0.8e^{x/2}$ |
| $e^{-x/2}/(-1/2) = 0.8t\ (+c)$ | A1 | Award if $c$ omitted |
| $x=0,\ t=0$, hence $c=-2$ and $-2e^{-1.4/2} = 0.8t - 2$ | M1 | Finding $c$ and using $x=1.4$ or $\left[e^{-x/2}/(-1/2)\right]_0^{1.4} = 0.8t$ |
| $t = 1.26$ s | A1 [4] [10] | $1.2585...$ |7 A particle $P$ of mass 0.5 kg moves in a straight line on a smooth horizontal surface. The velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when the displacement of $P$ from $O$ is $x \mathrm {~m}$. A single horizontal force of magnitude $0.16 \mathrm { e } ^ { x } \mathrm {~N}$ acts on $P$ in the direction $O P$. The velocity of $P$ when it is at $O$ is $0.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $v = 0.8 \mathrm { e } ^ { \frac { 1 } { 2 } x }$.\\
(ii) Find the time taken by $P$ to travel 1.4 m from $O$.
\hfill \mbox{\textit{CAIE M2 2013 Q7 [10]}}