| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Standard +0.3 This is a standard M2 variable force question requiring separation of variables to solve a differential equation with resistance proportional to v², followed by integration to find distance. The 'show that' structure guides students through the solution, and the techniques are routine for this module, making it slightly easier than average overall. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.48v\delta t = 0.2t^2\) | M1 | Newton's Second Law with \(a = \delta v/\delta t\) |
| \(\int v^{-2}\delta v = -0.5\int \delta t\) | A1 | |
| \(-v^{-1} = -0.5(t + c)\) | ||
| \(t = 0, v = 8\), hence \(c = -0.125\) | M1 | |
| \(v = 1/(0.125 + 0.5t) = 8/(1 + 4t)\) AG | A1 | [4] |
| (ii) \(\delta x/\delta t = 8/(1 + 4t)\) | M1* | |
| \(x = 8\int \delta t/(1 + 4t)\) | ||
| \(x = \frac{8}{4}\ln(1 + 4t) (+ c)\) | A1 | Accept \(c = 0\) assumed |
| \(t = 1.5, x = \frac{8}{4} \times \ln(1 + 4 \times 1.5)\) | D*M1 | Or limits used \(\frac{8}{4}[\ln(1 + 4t)]_0^{1.5}\) |
| \(OP = 3.89 \text{ m}\) | A1 | 4 |
**(i)** $0.48v\delta t = 0.2t^2$ | M1 | Newton's Second Law with $a = \delta v/\delta t$
$\int v^{-2}\delta v = -0.5\int \delta t$ | A1 |
$-v^{-1} = -0.5(t + c)$ | |
$t = 0, v = 8$, hence $c = -0.125$ | M1 |
$v = 1/(0.125 + 0.5t) = 8/(1 + 4t)$ AG | A1 | [4]
**(ii)** $\delta x/\delta t = 8/(1 + 4t)$ | M1* |
$x = 8\int \delta t/(1 + 4t)$ | |
$x = \frac{8}{4}\ln(1 + 4t) (+ c)$ | A1 | Accept $c = 0$ assumed
$t = 1.5, x = \frac{8}{4} \times \ln(1 + 4 \times 1.5)$ | D*M1 | Or limits used $\frac{8}{4}[\ln(1 + 4t)]_0^{1.5}$
$OP = 3.89 \text{ m}$ | A1 | 44 A particle $P$ of mass 0.4 kg is projected horizontally with velocity $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $O$ on a smooth horizontal surface. The motion of $P$ is opposed by a resisting force of magnitude $0.2 v ^ { 2 } \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of $P$ at time $t \mathrm {~s}$ after projection.\\
(i) Show that $v = \frac { 8 } { 1 + 4 t }$.\\
(ii) Calculate the distance $O P$ when $t = 1.5$.
\hfill \mbox{\textit{CAIE M2 2011 Q4 [8]}}