| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - horizontal motion |
| Difficulty | Standard +0.3 This is a standard M2 variable force question with air resistance proportional to velocity. It requires setting up Newton's second law with friction and air resistance, separating variables to integrate a simple differential equation, then integrating again for displacement. The 'show that' structure guides students through each step, making it slightly easier than average for M2 level. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)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 |
| \(0.5a = -(0.2v + 0.08 \times 0.5g)\) | M1 | For using Newton's second law and \(F = \mu R\) |
| \(2.5dv/dt = -(v + 2)\) | A1 | AG |
| M1 | For separating variables and integrating | |
| \(2.5\ln(v + 2) = -t\ (+c)\) | A1 | |
| M1 | For using \(v(0) = 4\) | |
| \(t = 2.5\ln[6/(v+2)]\) | A1 | |
| \(t = 2.75\) | A1 | Subtotal: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(e^{0.4t} = 6/(v+2) \Rightarrow dx/dt = 6e^{-0.4t} - 2\) | B1 | |
| M1 | For integrating | |
| \(x = -15e^{-0.4t} - 2t\ (+k)\) | A1 | |
| \(\left[x = 15(1 - e^{-0.4t}) - 2t\right]\) | M1 | For using \(x(0) = 0\) i.e. \(x = 0,\ t = 0\) |
| Distance is 4.51m | A1 | Subtotal: 5. Total: 12 |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5a = -(0.2v + 0.08 \times 0.5g)$ | M1 | For using Newton's second law and $F = \mu R$ |
| $2.5dv/dt = -(v + 2)$ | A1 | AG |
| | M1 | For separating variables and integrating |
| $2.5\ln(v + 2) = -t\ (+c)$ | A1 | |
| | M1 | For using $v(0) = 4$ |
| $t = 2.5\ln[6/(v+2)]$ | A1 | |
| $t = 2.75$ | A1 | **Subtotal: 7** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $e^{0.4t} = 6/(v+2) \Rightarrow dx/dt = 6e^{-0.4t} - 2$ | B1 | |
| | M1 | For integrating |
| $x = -15e^{-0.4t} - 2t\ (+k)$ | A1 | |
| $\left[x = 15(1 - e^{-0.4t}) - 2t\right]$ | M1 | For using $x(0) = 0$ i.e. $x = 0,\ t = 0$ |
| Distance is 4.51m | A1 | **Subtotal: 5. Total: 12** |7 A particle $P$ of mass 0.5 kg moves on a horizontal surface along the straight line $O A$, in the direction from $O$ to $A$. The coefficient of friction between $P$ and the surface is 0.08 . Air resistance of magnitude $0.2 v \mathrm {~N}$ opposes the motion, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of $P$ at time $t \mathrm {~s}$. The particle passes through $O$ with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $t = 0$.\\
(i) Show that $2.5 \frac { \mathrm {~d} v } { \mathrm {~d} t } = - ( v + 2 )$ and hence find the value of $t$ when $v = 0$.\\
(ii) Show that $\frac { \mathrm { d } x } { \mathrm {~d} t } = 6 \mathrm { e } ^ { - 0.4 t } - 2$, where $x \mathrm {~m}$ is the displacement of $P$ from $O$ at time $t \mathrm {~s}$, and hence find the distance $O P$ when $v = 0$.
\hfill \mbox{\textit{CAIE M2 2008 Q7 [12]}}