| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Challenging +1.2 This is a standard M3 variable force question with air resistance proportional to v². Part (i) requires setting up F=ma with the resistance term (routine for M3), part (ii) involves separating variables and integrating to find v² (standard technique), and part (iii) applies work-energy principles. While it requires multiple steps and careful algebra, these are well-practiced M3 techniques without requiring novel insight—moderately above average difficulty due to the algebraic manipulation and multi-part nature. |
| Spec | 6.02a Work done: concept and definition6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.2g - v^2/2000 = 0.2v(dv/dx)\) | M1 | For using Newton's second law with \(a = v(dv/dx)\) |
| \(\frac{400v}{3920 - v^2}\frac{dv}{dx} = 1\) | A1 | AG Convincing, with no slips. |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(-200\ln(3920 - v^2) = x + (A)\) | M1 | For separating variables and integrating |
| \(-200\ln(3920) = A\) | M1 | For using \(v(0) = 0\) |
| \(x = 200\ln\left(\frac{3920}{3920 - v^2}\right)\) | A1 | |
| \(e^{x/200} = 3920/(3920 - v^2)\) | M1 | For using inverse ln process |
| \(v^2 = 3920(1 - e^{-x/200})\) | A1 | |
| \(0 < x/200 \Rightarrow v^2 < 3920\) | B1 | AG Convincingly – dep on correct answer |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(0.2g - v^2/2000 = 0.2a\) | M1 | |
| A1 | ||
| \(v = 40\) | B1ft | |
| Gain in KE = \(\frac{1}{2} \times 0.2 \times 1600\) | B1ft | (=160J) |
| \(x = 200\ln\left(\frac{3920}{3920 - 1600}\right)\) (= 104.90) | B1ft | |
| \(0.2g \times (104.9) - 160\) | M1 | For using WD = loss of PE – gain in KE |
| Work done is \(45.6\) J | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(0.2g - v^2/2000 = 0.2a\) | M1 | |
| A1 | ||
| \(v = 40\) | B1ft | |
| \(x = 200\ln\left(\frac{3920}{3920 - 1600}\right)\) (= 104.90...) | B1ft | |
| \(WD = \int F dx\) and subst for \(v^2\) | M1 | Use of WD = \(\int Fdx\) and subst for \(v^2\) |
| \(= \int_{2000}^{3920} \frac{1}{2000}(1 - e^{-v/200})dx\) | A1 | |
| \(= 3920/2000(x + 200e^{-t/200}) - 392\) | A1 | |
| [6] |
**Part (i)**
$0.2g - v^2/2000 = 0.2v(dv/dx)$ | M1 | For using Newton's second law with $a = v(dv/dx)$
$\frac{400v}{3920 - v^2}\frac{dv}{dx} = 1$ | A1 | AG Convincing, with no slips.
| | [2] |
**Part (ii)**
$-200\ln(3920 - v^2) = x + (A)$ | M1 | For separating variables and integrating
$-200\ln(3920) = A$ | M1 | For using $v(0) = 0$
$x = 200\ln\left(\frac{3920}{3920 - v^2}\right)$ | A1 |
$e^{x/200} = 3920/(3920 - v^2)$ | M1 | For using inverse ln process
$v^2 = 3920(1 - e^{-x/200})$ | A1 |
$0 < x/200 \Rightarrow v^2 < 3920$ | B1 | AG Convincingly – dep on correct answer
| | [7] |
**Part (iii)**
Using $0.2g - v^2/2000 = 0.2a$ | M1 |
| | A1 |
$v = 40$ | B1ft |
Gain in KE = $\frac{1}{2} \times 0.2 \times 1600$ | B1ft | (=160J)
$x = 200\ln\left(\frac{3920}{3920 - 1600}\right)$ (= 104.90) | B1ft |
$0.2g \times (104.9) - 160$ | M1 | For using WD = loss of PE – gain in KE
Work done is $45.6$ J | A1 |
| | [6] |
**OR alternative method:**
Using $0.2g - v^2/2000 = 0.2a$ | M1 |
| | A1 |
$v = 40$ | B1ft |
$x = 200\ln\left(\frac{3920}{3920 - 1600}\right)$ (= 104.90...) | B1ft |
$WD = \int F dx$ and subst for $v^2$ | M1 | Use of WD = $\int Fdx$ and subst for $v^2$
$= \int_{2000}^{3920} \frac{1}{2000}(1 - e^{-v/200})dx$ | A1 |
$= 3920/2000(x + 200e^{-t/200}) - 392$ | A1 |
| | [6] |7 A particle $P$ of mass 0.2 kg is released from rest at a point $O$ and falls vertically. Air resistance of magnitude $\frac { v ^ { 2 } } { 2000 } \mathrm {~N}$ acts upwards on $P$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the velocity of $P$ when it has fallen a distance of $x \mathrm {~m}$.\\
(i) Show that $\left( \frac { 400 v } { 3920 - v ^ { 2 } } \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1$.\\
(ii) Find $v ^ { 2 }$ in terms of $x$ and hence show that $v ^ { 2 } < 3920$ for all values of $x$.\\
(iii) Find the work done against the air resistance while $P$ is falling, from $O$, to the point where its downward acceleration is $5.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
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\hfill \mbox{\textit{OCR M3 2011 Q7 [15]}}