| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (position x) - find velocity |
| Difficulty | Challenging +1.2 This is a standard M3 variable force question using v dv/dx = F/m. Part (i) requires setting up and integrating a separable differential equation with given initial conditions - a routine technique for this module. Part (ii) requires separating variables again using v = dx/dt. While it involves multiple steps and careful algebra, the method is well-practiced in M3 and follows a predictable pattern. Slightly above average difficulty due to the two-part integration process and algebraic manipulation required. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(F = ma\), using \(v\frac{dv}{dx}\) | M1* | Allow sign error / 0.3 omitted |
| \(0.3v\frac{dv}{dx} = 1.5x\) | A1 | |
| Attempt to rearrange and integrate | *M1 | \(0.3v^2 = 1.5x^2 (+c)\); No need for \(c\); at least one side integrated correctly |
| \(v = \sqrt{5x}\) AG | A1 [4] | Correct derivation WWW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate to find \(x\) in terms of \(t\) | M1 | \(dx/x = \sqrt{5}\,dt\) and int 1 side correctly; Need to separate variables |
| \(\ln x = \sqrt{5}t + c\) | A1 | No need for \(c\) for first 2 marks |
| \(x = e^{\sqrt{5}t}\) | A1 | Must include showing \(c = 0\) |
| \(v = \sqrt{5}\,e^{\sqrt{5}t}\) | A1 [4] | CAO |
| OR: Integrate to find \(v\) in terms of \(t\) | M1 | Use \(0.3\frac{dv}{dt} = 1.5x\) and int 1 side correctly; No need for \(c\) for first 2 marks |
| \(\frac{dv}{v} = \sqrt{5}\,dt\) | ||
| \(\ln v = \sqrt{5}t + c\) | A1 | |
| \(\ln v = \sqrt{5}t + \ln(\sqrt{5})\) | A1 | Must include showing \(c = \ln(\sqrt{5})\) |
| \(v = \sqrt{5}\,e^{\sqrt{5}t}\) | A1 | CAO |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $F = ma$, using $v\frac{dv}{dx}$ | M1* | Allow sign error / 0.3 omitted |
| $0.3v\frac{dv}{dx} = 1.5x$ | A1 | |
| Attempt to rearrange and integrate | *M1 | $0.3v^2 = 1.5x^2 (+c)$; No need for $c$; at least one side integrated correctly |
| $v = \sqrt{5x}$ **AG** | A1 **[4]** | Correct derivation WWW |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to find $x$ in terms of $t$ | M1 | $dx/x = \sqrt{5}\,dt$ and int 1 side correctly; Need to separate variables |
| $\ln x = \sqrt{5}t + c$ | A1 | No need for $c$ for first 2 marks |
| $x = e^{\sqrt{5}t}$ | A1 | Must include showing $c = 0$ |
| $v = \sqrt{5}\,e^{\sqrt{5}t}$ | A1 **[4]** | CAO |
| OR: Integrate to find $v$ in terms of $t$ | M1 | Use $0.3\frac{dv}{dt} = 1.5x$ and int 1 side correctly; No need for $c$ for first 2 marks |
| $\frac{dv}{v} = \sqrt{5}\,dt$ | | |
| $\ln v = \sqrt{5}t + c$ | A1 | |
| $\ln v = \sqrt{5}t + \ln(\sqrt{5})$ | A1 | Must include showing $c = \ln(\sqrt{5})$ |
| $v = \sqrt{5}\,e^{\sqrt{5}t}$ | A1 | CAO |
---3 At time $t = 0 \mathrm {~s}$ a particle $P$, of mass 0.3 kg , is 1 m away from a point $O$ on a smooth horizontal plane and is moving away from $O$ with speed $\sqrt { 5 } \mathrm {~ms} ^ { - 1 }$. The only horizontal force acting on $P$ has magnitude $1.5 x \mathrm {~N}$, where $x$ is the distance $O P$, and acts away from $O$.\\
(i) Show that the speed of $P , v \mathrm {~ms} ^ { - 1 }$, is given by $v = \sqrt { 5 } x$.\\
(ii) Find an expression for $v$ in terms of $t$.
\hfill \mbox{\textit{OCR M3 2013 Q3 [8]}}