| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance with other powers |
| Difficulty | Challenging +1.2 This is a standard M3 variable force question requiring separation of variables and integration. Part (i) uses F=ma with the chain rule (dv/dt = v·dv/dx), leading to straightforward separation and integration. Part (ii) requires integrating the velocity expression to find x(t) and solving for u. While it involves multiple steps and careful algebraic manipulation, the techniques are routine for M3 students and the question provides significant scaffolding by giving the target result in part (i). |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3v(\mathrm{d}v/\mathrm{d}x) = -1.2v^3\) | M1, A1 | For using Newton's second law and \(a = v(\mathrm{d}v/\mathrm{d}x)\); allow missed sign, stray \(g\), missed \(0.3\) |
| \([-v^{-1} = -4x + A]\) | M1* | For finding \(\mathrm{d}v/\mathrm{d}x\) in terms of \(v\) and attempting to integrate; allow \(A/v = Bx + C\) |
| \([-u^{-1} = 0 + A]\) | *M1 | For using \(v(0) = u\) |
| \(v = \dfrac{u}{4ux + 1}\) | A1 | AG |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int(4ux + 1)\,\mathrm{d}x = \int u\,\mathrm{d}t\) | M1* | For using \(v = \mathrm{d}x/\mathrm{d}t\), separating variables and attempting to integrate one side; \(-1.2v^3 = 0.3\,\mathrm{d}v/\mathrm{d}t\) and attempt to int one side M1*; \(8t = 1/v^2 - 1/u^2\) and subst for \(v\) A1 |
| \(2ux^2 + x = ut + B\) | A1 | |
| \([(2 \times 4 - 9)u = -2]\) | *M1 | For using \(x(0) = 0\) (may be implied by absence of \(B\)) and \(x(9) = 2\), dep on int being done |
| \(u = 2\) | A1 | |
| [4] |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3v(\mathrm{d}v/\mathrm{d}x) = -1.2v^3$ | M1, A1 | For using Newton's second law and $a = v(\mathrm{d}v/\mathrm{d}x)$; allow missed sign, stray $g$, missed $0.3$ |
| $[-v^{-1} = -4x + A]$ | M1* | For finding $\mathrm{d}v/\mathrm{d}x$ in terms of $v$ and attempting to integrate; allow $A/v = Bx + C$ |
| $[-u^{-1} = 0 + A]$ | *M1 | For using $v(0) = u$ |
| $v = \dfrac{u}{4ux + 1}$ | A1 | AG |
| **[5]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int(4ux + 1)\,\mathrm{d}x = \int u\,\mathrm{d}t$ | M1* | For using $v = \mathrm{d}x/\mathrm{d}t$, separating variables and attempting to integrate one side; $-1.2v^3 = 0.3\,\mathrm{d}v/\mathrm{d}t$ and attempt to int one side M1*; $8t = 1/v^2 - 1/u^2$ and subst for $v$ A1 |
| $2ux^2 + x = ut + B$ | A1 | |
| $[(2 \times 4 - 9)u = -2]$ | *M1 | For using $x(0) = 0$ (may be implied by absence of $B$) and $x(9) = 2$, dep on int being done |
| $u = 2$ | A1 | |
| **[4]** | | |
---3 A particle $P$ of mass 0.3 kg is projected horizontally with speed $u \mathrm {~ms} ^ { - 1 }$ from a fixed point $O$ on a smooth horizontal surface. At time $t \mathrm {~s}$ after projection $P$ is $x \mathrm {~m}$ from $O$ and is moving with speed $v \mathrm {~ms} ^ { - 1 }$. There is a force of magnitude $1.2 v ^ { 3 } \mathrm {~N}$ resisting the motion of $P$.\\
(i) Find an expression for $\frac { \mathrm { d } v } { \mathrm {~d} x }$ in terms of $v$ and hence show that $v = \frac { u } { 4 u x + 1 }$.\\
(ii) Given that $x = 2$ when $t = 9$ find the value of $u$.
\hfill \mbox{\textit{OCR M3 2012 Q3 [9]}}