| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (velocity v) - use v dv/dx |
| Difficulty | Standard +0.3 This is a standard M3 variable force question using v dv/dx = a. Part (i) requires applying F=ma with the given resistance force, separating variables, and integrating—a routine technique for this module. Part (ii) involves separating variables again (dt = dx/v) and integrating with limits. While it requires multiple steps and careful algebra, it follows a well-established method pattern that M3 students practice extensively. Slightly easier than average due to the straightforward setup and clear path to solution. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use \(F = mv\frac{dv}{dx}\): \(-4v = \frac{dv}{dx}\) | M1, A1 | Expression for \(\frac{dv}{dx}\) required; allow sign error, missing \(m\) or \(g\) inc |
| \(-4x = \ln v + c\); \(0 = \ln 2 + c\) | M1, M1 | Get \((+/-) Ax = \ln v + c\); valid attempt to find \(c\); need a step leading to given answer |
| \(\ln\frac{v}{2} = -4x\); \(v = 2e^{-4x}\) | A1 [5] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{4x}\ dx = 2\ dt\); \(\frac{1}{4}e^{4x} = 2t + c\) | M1*, A1 | Write \(v\) as \(\frac{dx}{dt}\) and separate variables; must have \(c\) or use limits; \(dv/4v^2 = -dt\); \(\frac{1}{v} = 4t + \frac{1}{2}\) |
| \(\frac{1}{4} = 0 + c\) | *M1 | Valid attempt to find \(c\) or subst limits; \(\frac{dx}{dt} = \frac{2}{8t+1}\) OR \(t=0.5\) gives \(v=0.4\) |
| \(e^{4x} = 4\left(1+\frac{1}{4}\right)\) | *M1 | Find \(x\) when \(t = 0.5\); need to remove exp; allow even if no \(c\); \(x = \frac{1}{4}\ln(8t+1)+c\) OR \(-4x = \ln 0.2\) |
| \(x = \frac{1}{4}\ln 5\) | A1 [5] | Accept \(0.402(359...)\) |
## Question 4:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use $F = mv\frac{dv}{dx}$: $-4v = \frac{dv}{dx}$ | M1, A1 | Expression for $\frac{dv}{dx}$ required; allow sign error, missing $m$ or $g$ inc |
| $-4x = \ln v + c$; $0 = \ln 2 + c$ | M1, M1 | Get $(+/-) Ax = \ln v + c$; valid attempt to find $c$; need a step leading to given answer |
| $\ln\frac{v}{2} = -4x$; $v = 2e^{-4x}$ | A1 [5] | AG |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{4x}\ dx = 2\ dt$; $\frac{1}{4}e^{4x} = 2t + c$ | M1*, A1 | Write $v$ as $\frac{dx}{dt}$ and separate variables; must have $c$ or use limits; $dv/4v^2 = -dt$; $\frac{1}{v} = 4t + \frac{1}{2}$ |
| $\frac{1}{4} = 0 + c$ | *M1 | Valid attempt to find $c$ or subst limits; $\frac{dx}{dt} = \frac{2}{8t+1}$ OR $t=0.5$ gives $v=0.4$ |
| $e^{4x} = 4\left(1+\frac{1}{4}\right)$ | *M1 | Find $x$ when $t = 0.5$; need to remove exp; allow even if no $c$; $x = \frac{1}{4}\ln(8t+1)+c$ OR $-4x = \ln 0.2$ |
| $x = \frac{1}{4}\ln 5$ | A1 [5] | Accept $0.402(359...)$ |
---4 A particle $P$ of mass 0.4 kg is projected horizontally with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ 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 away from $O$ with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. There is a force of magnitude $1.6 v ^ { 2 } \mathrm {~N}$ 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 $v = 2 \mathrm { e } ^ { - 4 x }$.\\
(ii) Find the distance travelled by $P$ in the 0.5 seconds after it leaves $O$.
\hfill \mbox{\textit{OCR M3 2014 Q4 [10]}}