| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Motion with exponential force |
| Difficulty | Standard +0.3 This is a standard M2 variable force question requiring Newton's second law with an exponential force, followed by routine integration twice. The 'show that' part guides students through the setup, and finding maximum velocity is straightforward calculus. While it involves multiple steps, each technique is standard for this topic with no novel problem-solving required. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.4\,dv/dt = 3e^{-t} - 0.4g\sin 30\) | M1 | |
| \(dv/dt = 7.5e^{-t} - 5\) | A1 AG | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| M1 | Integrates acceleration | |
| \(\int_0^v dv = \int_0^t (7.5e^{-t} - 5)\,dt\) | M1 | Limits or finds integration constant |
| \(v = 7.5 - 7.5e^{-t} - 5t\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Solves \(dv/dt = 0\) | M1 | \(t = 0.405(46\ldots)\) |
| \(\int_0^x dx = \int_0^{0.405}(7.5 - 7.5e^{-t} - 5t)\,dt\) | M1 A1 | Integrates expression for \(v\) and uses \(t = 0.405\) |
| \(x = 0.13(0)\) m | ||
| [3] |
## Question 5:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.4\,dv/dt = 3e^{-t} - 0.4g\sin 30$ | M1 | |
| $dv/dt = 7.5e^{-t} - 5$ | A1 AG | |
| | [2] | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| | M1 | Integrates acceleration |
| $\int_0^v dv = \int_0^t (7.5e^{-t} - 5)\,dt$ | M1 | Limits or finds integration constant |
| $v = 7.5 - 7.5e^{-t} - 5t$ | A1 | |
| | [3] | |
### Part (iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Solves $dv/dt = 0$ | M1 | $t = 0.405(46\ldots)$ |
| $\int_0^x dx = \int_0^{0.405}(7.5 - 7.5e^{-t} - 5t)\,dt$ | M1 A1 | Integrates expression for $v$ and uses $t = 0.405$ |
| $x = 0.13(0)$ m | | |
| | [3] | |
---5 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. A force of magnitude $3 \mathrm { e } ^ { - t } \mathrm {~N}$ directed up a line of greatest slope acts on $P$, where $t \mathrm {~s}$ is the time after release.\\
(i) Show that $\frac { \mathrm { d } v } { \mathrm {~d} t } = 7.5 \mathrm { e } ^ { - t } - 5$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of $P$ up the plane at time $t \mathrm {~s}$.\\
(ii) Express $v$ in terms of $t$.\\
(iii) Find the distance of $P$ from $O$ when $v$ has its maximum value.
\hfill \mbox{\textit{CAIE M2 2016 Q5 [8]}}