| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Standard +0.3 This is a standard M3 variable force question requiring the v dv/dx = f(v) method to establish a differential equation, then integration and application of boundary conditions. Part (i) involves routine separation of variables and finding a terminal velocity limit. Part (ii) is direct substitution. While it requires multiple steps and careful algebra, it follows a well-practiced template for this topic with no novel insights required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v\frac{dv}{dx} = g - 0.0025v^2\) | M1, A1 | For using Newton's 2nd law with \(a = v\frac{dv}{dx}\); 3 terms |
| \(\int \frac{v\,dv}{g - 0.0025v^2} = \int dx\) | M1 | For correctly separating variables and attempting to integrate |
| \(-200\ln(g - 0.0025v^2) = x\,(+A)\) | A1 | |
| \(A = -200\ln g\) | M1* | Attempt to find \(A\) from \(B\ln(C - Dv^2)\) |
| \([g - 0.0025v^2 = ge^{-0.005x}]\) | *M1 | For transposing equation to remove ln |
| \(v^2 = 400g(1 - e^{-0.005x})\) | A1 | |
| \(0 < e^{-0.005x} \leq 1 \Rightarrow v^2\) cannot reach \(400g\), i.e. cannot reach 3920 | B1 | Dependent on getting other 7 marks. Need '\(0 <\)' oe |
| [8] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 400g(1 - e^{-0.5})\) | M1 | For substituting for \(x\) and evaluating \(v\); must have \(v^2 = A + Be^{Cx}\) for (i), but not necessarily in this form |
| Speed of \(P\) is 39.3 ms\(^{-1}\) | A1 | |
| [2] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v\frac{dv}{dx} = g - 0.0025v^2$ | M1, A1 | For using Newton's 2nd law with $a = v\frac{dv}{dx}$; 3 terms |
| $\int \frac{v\,dv}{g - 0.0025v^2} = \int dx$ | M1 | For correctly separating variables and attempting to integrate |
| $-200\ln(g - 0.0025v^2) = x\,(+A)$ | A1 | |
| $A = -200\ln g$ | M1* | Attempt to find $A$ from $B\ln(C - Dv^2)$ |
| $[g - 0.0025v^2 = ge^{-0.005x}]$ | *M1 | For transposing equation to remove ln |
| $v^2 = 400g(1 - e^{-0.005x})$ | A1 | |
| $0 < e^{-0.005x} \leq 1 \Rightarrow v^2$ cannot reach $400g$, i.e. cannot reach 3920 | B1 | Dependent on getting other 7 marks. Need '$0 <$' oe |
| | **[8]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 400g(1 - e^{-0.5})$ | M1 | For substituting for $x$ and evaluating $v$; must have $v^2 = A + Be^{Cx}$ for (i), but not necessarily in this form |
| Speed of $P$ is 39.3 ms$^{-1}$ | A1 | |
| | **[2]** | |
---3 A particle $P$ of mass $m \mathrm {~kg}$ is released from rest and falls vertically. When $P$ has fallen a distance of $x \mathrm {~m}$ it has a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The only forces acting on $P$ are its weight and air resistance of magnitude $\frac { 1 } { 400 } m v ^ { 2 } \mathrm {~N}$.\\
(i) Find $v ^ { 2 }$ in terms of $x$ and show that $v ^ { 2 }$ must be less than 3920 .\\
(ii) Find the speed of $P$ when it has fallen 100 m .
\hfill \mbox{\textit{OCR M3 2012 Q3 [10]}}