| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - vertical motion |
| Difficulty | Standard +0.8 This is a two-part variable force question requiring setting up and solving a differential equation with air resistance, then integrating velocity to find distance. While the integration is guided ('show that'), part (ii) requires integrating an exponential function over a specific interval, which is more demanding than standard M3 questions. The multi-step nature and need to handle both the differential equation setup and subsequent integration places this moderately above average difficulty. |
| 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 | Mark | Guidance |
| Initial speed in medium is \(\sqrt{2g\times10}\ (=14)\) | B1 | |
| \([0.125\ dv/dt = 0.125g - 0.025v]\) | M1 | For using Newton's second law with \(a = dv/dt\) (3 terms required) |
| \(\int \frac{5\,dv}{5g - v} = \int dt\) | M1 | For separating variables and attempting to integrate |
| \(-5\ln(5g - v) = t\ (+A)\) | A1 | |
| \([-5\ln 35 = A]\) | M1 | For using \(v(0) = 14\) |
| \(t = 5\ln\{35/(49-v)\}\) | A1 | |
| M1 | For method of transposition | |
| \(v = 49 - 35e^{-0.2t}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 49t + 175e^{-0.2t}\ (+B)\) | M1 | For integrating to find \(x(t)\) |
| A1 | ||
| \([x(3) = (49\times3 + 175e^{-0.6}) - (0 + 175)]\) | M1 | For using limits 0 to 3 or for using \(x(0)=0\) and evaluating \(x(3)\) |
| Distance is \(68.0\)m | A1 |
## Question 6:
**Part (i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Initial speed in medium is $\sqrt{2g\times10}\ (=14)$ | B1 | |
| $[0.125\ dv/dt = 0.125g - 0.025v]$ | M1 | For using Newton's second law with $a = dv/dt$ (3 terms required) |
| $\int \frac{5\,dv}{5g - v} = \int dt$ | M1 | For separating variables and attempting to integrate |
| $-5\ln(5g - v) = t\ (+A)$ | A1 | |
| $[-5\ln 35 = A]$ | M1 | For using $v(0) = 14$ |
| $t = 5\ln\{35/(49-v)\}$ | A1 | |
| | M1 | For method of transposition |
| $v = 49 - 35e^{-0.2t}$ | A1 | AG |
**Part (ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 49t + 175e^{-0.2t}\ (+B)$ | M1 | For integrating to find $x(t)$ |
| | A1 | |
| $[x(3) = (49\times3 + 175e^{-0.6}) - (0 + 175)]$ | M1 | For using limits 0 to 3 or for using $x(0)=0$ and evaluating $x(3)$ |
| Distance is $68.0$m | A1 | |
---6 A stone of mass 0.125 kg falls freely under gravity, from rest, until it has travelled a distance of 10 m . The stone then continues to fall in a medium which exerts an upward resisting force of $0.025 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the stone $t \mathrm {~s}$ after the instant that it enters the resisting medium.\\
(i) Show by integration that $v = 49 - 35 \mathrm { e } ^ { - 0.2 t }$.\\
(ii) Find how far the stone travels during the first 3 seconds in the medium.
\hfill \mbox{\textit{OCR M3 2009 Q6 [12]}}