| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| 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 +0.8 This M3 question requires applying Newton's second law with variable force, using the v dv/dx form of acceleration, then solving a separable differential equation and working backwards from boundary conditions. While the steps are guided ('show that'), it requires confident handling of calculus in mechanics context and exponential integration, placing it 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 |
| \([0.2v\,dv/dx = -0.4v^2]\) | M1 | For using Newton's second law with \(a = v\,dv/dx\) |
| \((1/v)\,dv/dx = -2\) | A1 | [2 marks total] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\int(1/v)\,dv = \int -2\,dx]\) | M1 | For separating variables and attempting to integrate |
| \(\ln v = -2x \quad (+C)\) | A1 | |
| \([\ln v = -2x + \ln u]\) | M1 | For using \(v(0) = u\) |
| \(v = ue^{-2x}\) | A1 | [4 marks total] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\int e^{2x}\,dx = \int u\,dt]\) | M1 | For using \(v = dx/dt\) and separating variables |
| \(e^{2x}/2 = ut \quad (+C)\) | A1 | |
| \([e^{2x}/2 = ut + \frac{1}{2}]\) | M1 | For using \(x(0) = 0\) |
| \(u = 6.70\) | A1 | [4 marks total] Accept \((e^4-1)/8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\int\frac{1}{v^2}\,dv = -2\int dt \rightarrow -1/v = -2t + A]\), and \(A = -1/u\) | M1 | For using \(a = dv/dt\), separating variables, attempting to integrate and using \(v(0) = u\) |
| M1 | For substituting \(v = ue^{-2x}\) | |
| \(-e^{2x}/u = -2t - 1/u\) | A1 | |
| \(u = 6.70\) | A1 | [4 marks total] Accept \((e^4-1)/8\) |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.2v\,dv/dx = -0.4v^2]$ | M1 | For using Newton's second law with $a = v\,dv/dx$ |
| $(1/v)\,dv/dx = -2$ | A1 | [2 marks total] AG |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\int(1/v)\,dv = \int -2\,dx]$ | M1 | For separating variables and attempting to integrate |
| $\ln v = -2x \quad (+C)$ | A1 | |
| $[\ln v = -2x + \ln u]$ | M1 | For using $v(0) = u$ |
| $v = ue^{-2x}$ | A1 | [4 marks total] AG |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\int e^{2x}\,dx = \int u\,dt]$ | M1 | For using $v = dx/dt$ and separating variables |
| $e^{2x}/2 = ut \quad (+C)$ | A1 | |
| $[e^{2x}/2 = ut + \frac{1}{2}]$ | M1 | For using $x(0) = 0$ |
| $u = 6.70$ | A1 | [4 marks total] Accept $(e^4-1)/8$ |
## Alternative Method for Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\int\frac{1}{v^2}\,dv = -2\int dt \rightarrow -1/v = -2t + A]$, and $A = -1/u$ | M1 | For using $a = dv/dt$, separating variables, attempting to integrate and using $v(0) = u$ |
| | M1 | For substituting $v = ue^{-2x}$ |
| $-e^{2x}/u = -2t - 1/u$ | A1 | |
| $u = 6.70$ | A1 | [4 marks total] Accept $(e^4-1)/8$ |
---3 A particle $P$ of mass 0.2 kg is projected horizontally with speed $u \mathrm {~ms} ^ { - 1 }$ from a fixed point $O$ on a smooth horizontal surface. $P$ moves in a straight line and, at time $t \mathrm {~s}$ after projection, $P$ has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is $x \mathrm {~m}$ from $O$. The only force acting on $P$ has magnitude $0.4 v ^ { 2 } \mathrm {~N}$ and is directed towards $O$.\\
(i) Show that $\frac { 1 } { v } \frac { \mathrm {~d} v } { \mathrm {~d} x } = - 2$.\\
(ii) Hence show that $v = u \mathrm { e } ^ { - 2 x }$.\\
(iii) Find $u$, given that $x = 2$ when $t = 4$.
\hfill \mbox{\textit{OCR M3 2007 Q3 [10]}}