| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2004 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Challenging +1.3 This is a vector differential equation requiring auxiliary equation solution with complex roots, application of initial conditions, and kinetic energy calculation. While it involves multiple steps and vector manipulation, the techniques are standard for M5 level—solving the characteristic equation, using Euler's form for complex exponentials, and applying given initial conditions systematically. The structure is methodical rather than requiring novel insight, making it moderately harder than average but well within reach of prepared students. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation4.10d Second order homogeneous: auxiliary equation method4.10e Second order non-homogeneous: complementary + particular integral6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Aux. equ. \(m^2 + 2m + 2 = 0 \Rightarrow m = -1 \pm \mathbf{i}\) | M1 A1 | |
| G. soln.: \(\mathbf{r} = e^{-t}(\mathbf{A}\cos t + \mathbf{B}\sin t)\) | A1 ft | |
| \(t=0, \mathbf{r} = \mathbf{i} \Rightarrow \mathbf{A} = \mathbf{i}\) | M1 A1 | |
| \(\dot{\mathbf{r}} = -e^{-t}(\mathbf{A}\cos t + \mathbf{B}\sin t) + e^{-t}(-\mathbf{A}\sin t + \mathbf{B}\cos t)\) | M1 | |
| \(t=0,\; \dot{\mathbf{r}} = (-\mathbf{i}+\mathbf{j}) \Rightarrow -\mathbf{i}+\mathbf{j} = -\mathbf{i}+\mathbf{B} \Rightarrow \mathbf{B} = \mathbf{j}\) | M1 | |
| \(\therefore \mathbf{r} = e^{-t}(\cos t\,\mathbf{i} + \sin t\,\mathbf{j})\) | A1 | (8 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dot{\mathbf{r}} = -e^{-t}(\cos t\,\mathbf{i}+\sin t\,\mathbf{j}) + e^{-t}(-\sin t\,\mathbf{i}+\cos t\,\mathbf{j})\) | M1 | |
| \(= e^{-t}\{-(\cos t+\sin t)\mathbf{i} + (\cos t - \sin t)\mathbf{j}\}\) | ||
| Speed \(= e^{-t}\sqrt{(\cos t+\sin t)^2 + (\cos t - \sin t)^2}\) | M1 A1 | |
| \(= e^{-t}\sqrt{1 + 2\cos t\sin t + 1 - 2\cos t\sin t}\) | M1 | |
| \(= e^{-t}\sqrt{2}\) (*) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Loss of \(KE = \frac{1}{2} \times 2 \times 2(1 - e^{-2})\) | M1 | |
| \(= 2\!\left(1 - \frac{1}{e^2}\right) \approx 1.73\) (AWRT) | A1 | (2 marks) |
## Question 6:
### Part (a)
Aux. equ. $m^2 + 2m + 2 = 0 \Rightarrow m = -1 \pm \mathbf{i}$ | M1 A1 |
G. soln.: $\mathbf{r} = e^{-t}(\mathbf{A}\cos t + \mathbf{B}\sin t)$ | A1 ft |
$t=0, \mathbf{r} = \mathbf{i} \Rightarrow \mathbf{A} = \mathbf{i}$ | M1 A1 |
$\dot{\mathbf{r}} = -e^{-t}(\mathbf{A}\cos t + \mathbf{B}\sin t) + e^{-t}(-\mathbf{A}\sin t + \mathbf{B}\cos t)$ | M1 |
$t=0,\; \dot{\mathbf{r}} = (-\mathbf{i}+\mathbf{j}) \Rightarrow -\mathbf{i}+\mathbf{j} = -\mathbf{i}+\mathbf{B} \Rightarrow \mathbf{B} = \mathbf{j}$ | M1 |
$\therefore \mathbf{r} = e^{-t}(\cos t\,\mathbf{i} + \sin t\,\mathbf{j})$ | A1 | (8 marks)
### Part (b)
$\dot{\mathbf{r}} = -e^{-t}(\cos t\,\mathbf{i}+\sin t\,\mathbf{j}) + e^{-t}(-\sin t\,\mathbf{i}+\cos t\,\mathbf{j})$ | M1 |
$= e^{-t}\{-(\cos t+\sin t)\mathbf{i} + (\cos t - \sin t)\mathbf{j}\}$ | |
Speed $= e^{-t}\sqrt{(\cos t+\sin t)^2 + (\cos t - \sin t)^2}$ | M1 A1 |
$= e^{-t}\sqrt{1 + 2\cos t\sin t + 1 - 2\cos t\sin t}$ | M1 |
$= e^{-t}\sqrt{2}$ (*) | A1 | (5 marks)
### Part (c)
Loss of $KE = \frac{1}{2} \times 2 \times 2(1 - e^{-2})$ | M1 |
$= 2\!\left(1 - \frac{1}{e^2}\right) \approx 1.73$ (AWRT) | A1 | (2 marks)
---6. A particle $P$ of mass 2 kg moves in the $x - y$ plane. At time $t$ seconds its position vector is $\mathbf { r }$ metres. When $t = 0$, the position vector of $P$ is $\mathbf { i }$ metres and the velocity of $P$ is ( $- \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
The vector $\mathbf { r }$ satisfies the differential equation
$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = \mathbf { 0 }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { r }$ in terms of $t$.
\item Show that the speed of $P$ at time $t$ is $\mathrm { e } ^ { - t } \sqrt { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Find, in terms of e, the loss of kinetic energy of $P$ in the interval $t = 0$ to $t = 1$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2004 Q6 [15]}}