| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2008 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.2 This is a standard M4 mechanics question requiring force analysis to derive a second-order DE with constant coefficients, then solving it using auxiliary equation methods. While it involves multiple forces (spring, friction, air resistance) and requires careful bookkeeping, the mathematical techniques are routine for Further Maths students. The critical damping case makes the solution straightforward once identified. Slightly above average due to the modeling setup and multi-part nature, but follows predictable M4 patterns. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)4.10e Second order non-homogeneous: complementary + particular integral6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(-T - \frac{1}{2}mg - 2mv\sqrt{\frac{g}{l}} = m\ddot{x}\) | M1, A3,2,1,0 | |
| \(\frac{-mgx}{l} - \frac{1}{2}mg - 2m\dot{x}\sqrt{\frac{g}{l}} = m\ddot{x}\) | M1 | |
| \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega\dfrac{\mathrm{d}x}{\mathrm{d}t} + \omega^2 x = -0.5g\) (AG) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(u^2+2\omega u+\omega^2=0 \Rightarrow u=\omega\) (twice) | ||
| CF is \(x = e^{-\omega t}(At+B)\) | B1 | |
| PI is \(x = -\frac{1}{2}l \quad \left(-\frac{g}{2\omega^2}\right)\) | ||
| GS is \(x = e^{-\omega t}(At+B)-\frac{1}{2}l\) | M1 | |
| \(t=0, x=0 \Rightarrow B = \frac{1}{2}l \quad \left(\frac{g}{2\omega^2}\right)\) | M1 | |
| \(\frac{\mathrm{d}x}{\mathrm{d}t} = -\omega e^{-\omega t}(At+B)+Ae^{-\omega t}\) | M1 | |
| \(t=0, \frac{\mathrm{d}x}{\mathrm{d}t}=\sqrt{gl}=\omega l \Rightarrow A = \frac{3}{2}\omega l\left(=\frac{3\sqrt{gl}}{2}\right)\left(=\sqrt{gl}+\frac{0.5g}{\omega}\right)\) | M1 | |
| \(x = e^{-\omega t}\!\left(\tfrac{3}{2}\omega lt + \tfrac{1}{2}l\right)-\tfrac{1}{2}l = \tfrac{1}{2}le^{-\omega t}(3\omega t+1)-\tfrac{1}{2}l\) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\frac{\mathrm{d}x}{\mathrm{d}t}=0 \Rightarrow -\omega e^{-\omega t}(At+B)+Ae^{-\omega t}=0\) | M1 | |
| \(\Rightarrow t = \dfrac{2}{3\omega}\) | M1 A1 | (3), Total: 15 |
## Question 5(a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $-T - \frac{1}{2}mg - 2mv\sqrt{\frac{g}{l}} = m\ddot{x}$ | M1, A3,2,1,0 | |
| $\frac{-mgx}{l} - \frac{1}{2}mg - 2m\dot{x}\sqrt{\frac{g}{l}} = m\ddot{x}$ | M1 | |
| $\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega\dfrac{\mathrm{d}x}{\mathrm{d}t} + \omega^2 x = -0.5g$ (AG) | A1 | **(6)** |
## Question 5(b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $u^2+2\omega u+\omega^2=0 \Rightarrow u=\omega$ (twice) | | |
| CF is $x = e^{-\omega t}(At+B)$ | B1 | |
| PI is $x = -\frac{1}{2}l \quad \left(-\frac{g}{2\omega^2}\right)$ | | |
| GS is $x = e^{-\omega t}(At+B)-\frac{1}{2}l$ | M1 | |
| $t=0, x=0 \Rightarrow B = \frac{1}{2}l \quad \left(\frac{g}{2\omega^2}\right)$ | M1 | |
| $\frac{\mathrm{d}x}{\mathrm{d}t} = -\omega e^{-\omega t}(At+B)+Ae^{-\omega t}$ | M1 | |
| $t=0, \frac{\mathrm{d}x}{\mathrm{d}t}=\sqrt{gl}=\omega l \Rightarrow A = \frac{3}{2}\omega l\left(=\frac{3\sqrt{gl}}{2}\right)\left(=\sqrt{gl}+\frac{0.5g}{\omega}\right)$ | M1 | |
| $x = e^{-\omega t}\!\left(\tfrac{3}{2}\omega lt + \tfrac{1}{2}l\right)-\tfrac{1}{2}l = \tfrac{1}{2}le^{-\omega t}(3\omega t+1)-\tfrac{1}{2}l$ | A1 | **(6)** |
## Question 5(c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\frac{\mathrm{d}x}{\mathrm{d}t}=0 \Rightarrow -\omega e^{-\omega t}(At+B)+Ae^{-\omega t}=0$ | M1 | |
| $\Rightarrow t = \dfrac{2}{3\omega}$ | M1 A1 | **(3), Total: 15** |
---5. A light elastic spring has natural length $l$ and modulus of elasticity $m g$. One end of the spring is fixed to a point $O$ on a rough horizontal table. The other end is attached to a particle $P$ of mass $m$ which is at rest on the table with $O P = l$. At time $t = 0$ the particle is projected with speed $\sqrt { } ( g l )$ along the table in the direction $O P$. At time $t$ the displacement of $P$ from its initial position is $x$ and its speed is $v$. The motion of $P$ is subject to air resistance of magnitude $2 m v \omega$, where $\omega = \sqrt { \frac { g } { l } }$. The coefficient of friction between $P$ and the table is 0.5 .
\begin{enumerate}[label=(\alph*)]
\item Show that, until $P$ first comes to rest,
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + \omega ^ { 2 } x = - 0.5 g$$
\item Find $x$ in terms of $t , l$ and $\omega$.
\item Hence find, in terms of $\omega$, the time taken for $P$ to first come to instantaneous rest.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2008 Q5 [15]}}