| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance with other powers |
| Difficulty | Standard +0.8 This question requires setting up and solving a non-standard differential equation with a force proportional to √v, then using the result in a second integration. The √v resistance creates a fractional power that makes the separation of variables less routine than standard exponential resistance models, and part (ii) requires careful manipulation of the implicit result to find x(t). |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.5v\frac{dv}{dx} = -3v^{1/2}\) | M1 | Newton's 2nd Law with \(a = v\frac{dv}{dx}\) |
| \(\int v^{1/2}\,dv = -\int 6\,dx\) | M1 | Separates variables and integrates |
| \(\frac{v^{3/2}}{(3/2)} = -6x \,(+\,c)\) | A1 | |
| \(x=0,\; v=9\) hence \(c=18\) | M1 | Or uses limits |
| \(v^{3/2} = \frac{3(18-6x)}{2}\) | A1 | |
| \(v = (27 - 9x)^{2/3}\) | A1 AG [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = (27-9x)^{2/3}\) | M1 | \(0.5\frac{dv}{dt} = -3v^{1/2}\); \(\int v^{-1/2}\,dv = -\int 6\,dt\) |
| \(\int(27-9x)^{-2/3}\,dx = \int dt\) | \(v^{1/2} = -3t + c\) | |
| \(\frac{(27-9x)^{1/3}}{-3} = t\,(+\,c)\) | A1ft | \(t=0,\;v=9\) hence \(c=3\) and \(t=0.5\), giving \(v=2.25\) |
| \(t=0,\;x=0\) hence \(c=-1\) | M1 | |
| \(t = 0.5,\; x = 2.625\) | A1 [4] | \(v = 2.25,\; x = 2.625\) |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.5v\frac{dv}{dx} = -3v^{1/2}$ | M1 | Newton's 2nd Law with $a = v\frac{dv}{dx}$ |
| $\int v^{1/2}\,dv = -\int 6\,dx$ | M1 | Separates variables and integrates |
| $\frac{v^{3/2}}{(3/2)} = -6x \,(+\,c)$ | A1 | |
| $x=0,\; v=9$ hence $c=18$ | M1 | Or uses limits |
| $v^{3/2} = \frac{3(18-6x)}{2}$ | A1 | |
| $v = (27 - 9x)^{2/3}$ | A1 AG **[5]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = (27-9x)^{2/3}$ | M1 | $0.5\frac{dv}{dt} = -3v^{1/2}$; $\int v^{-1/2}\,dv = -\int 6\,dt$ |
| $\int(27-9x)^{-2/3}\,dx = \int dt$ | | $v^{1/2} = -3t + c$ |
| $\frac{(27-9x)^{1/3}}{-3} = t\,(+\,c)$ | A1ft | $t=0,\;v=9$ hence $c=3$ and $t=0.5$, giving $v=2.25$ |
| $t=0,\;x=0$ hence $c=-1$ | M1 | |
| $t = 0.5,\; x = 2.625$ | A1 **[4]** | $v = 2.25,\; x = 2.625$ |
---6 A particle $P$ of mass 0.5 kg moves in a straight line on a smooth horizontal surface. At time $t \mathrm {~s}$, the displacement of $P$ from a fixed point on the line is $x \mathrm {~m}$ and the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It is given that when $t = 0 , x = 0$ and $v = 9$. The motion of $P$ is opposed by a force of magnitude $3 \sqrt { } v \mathrm {~N}$.\\
(i) By solving an appropriate differential equation, show that $v = ( 27 - 9 x ) ^ { \frac { 2 } { 3 } }$.\\
(ii) Calculate the value of $x$ when $t = 0.5$.
\hfill \mbox{\textit{CAIE M2 2010 Q6 [9]}}