| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Challenging +1.8 This M4 question requires setting up and solving a second-order differential equation with damping for elastic string motion. Part (a) is straightforward kinematics, part (b) requires careful application of Newton's second law with Hooke's law and resistance forces, and part (c) involves solving a non-homogeneous second-order ODE with specific initial conditions. While the individual techniques are standard for M4, the multi-step setup and algebraic manipulation across all three parts, combined with the need to track multiple variables (x, y, extension, displacement), elevates this above average difficulty. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts4.10e Second order non-homogeneous: complementary + particular integral6.02j Conservation with elastics: springs and strings |
| Answer | Marks |
|---|---|
| \(u\dot{t} = y + (\dot{x} + x)\) \(\Rightarrow\) \(u\dot{t} = y + x\) | M1, A1 (2) |
| Answer | Marks |
|---|---|
| \(\Rightarrow\) \(\dot{x} + 2\dot{x} + 5u\dot{x} = 5u\dot{u}\) | M1 A1, M1, B1, B1, M1, A1 (7) |
| Answer | Marks |
|---|---|
| \(x = e^{-u\dot{t}}\left(\frac{3u}{u}\sin 2u\dot{t} - \frac{2u}{5u}\cos 2u\dot{t}\right) + \frac{2u}{5u}\) | B1, M1, B1, B1, M1 A1 V, M1, A1 (8), (19) |
## Part (a)
$u\dot{t} = y + (\dot{x} + x)$ $\Rightarrow$ $u\dot{t} = y + x$ | M1, A1 (2) |
## Part (b)
For particle, $k(\to), T - 2m_0g = ~_3j$ $T = \frac{5m_0^2}{a}$
$u = \dot{y} + \dot{x}; \quad 0 = \dot{y} + \dot{x}^2$
$5m_0^2 - 2m_0(u-\dot{x}) = m_0(\dot{x}-^2)$
$\Rightarrow$ $\dot{x} + 2\dot{x} + 5u\dot{x} = 5u\dot{u}$ | M1 A1, M1, B1, B1, M1, A1 (7) |
## Part (c)
AE: $u^2 + 2u\dot{u} + 5u\dot{x} = 0$ $\Rightarrow$ $(u + \dot{u})^2 = -4u\dot{x}^2$
$\Rightarrow$ $u = -\dot{u} + 2i\dot{u}$
CF: $x = e^{-u\dot{t}}(A\cos 2u\dot{t} + B\sin 2u\dot{t})$
PI: $x = \frac{2u\dot{u}}{5u^2}$
ES: $x = e^{-u\dot{t}}(A\cos 2u\dot{t} + B\sin 2u\dot{t}) + \frac{2u}{5u}$
$u = 0, \dot{x} = 0$: $0 = A + \frac{2u}{5u}$ $\Rightarrow$ $A = -\frac{2u}{5u}$
$\dot{x} = -u\dot{e}^{-u\dot{t}}(A\cos 2u\dot{t} + B\sin 2u\dot{t}) + e^{-u\dot{t}}$
$u = 0, \dot{y} = 0$: $\dot{x} = u$; $u = -\omega\theta + 2i\omega B$ $\Rightarrow$ $B = \frac{1u}{1u\omega}$
$x = e^{-u\dot{t}}\left(\frac{3u}{u}\sin 2u\dot{t} - \frac{2u}{5u}\cos 2u\dot{t}\right) + \frac{2u}{5u}$ | B1, M1, B1, B1, M1 A1 V, M1, A1 (8), (19) |A light elastic string, of natural length $a$ and modulus of elasticity $5ma\omega^2$, lies unstretched along a straight line on a smooth horizontal plane. A particle of mass $m$ is attached to one end of the spring. At time $t = 0$, the other end of the spring starts to move with constant speed $U$ along the line of the spring and away from the particle. As the particle moves along the plane it is subject to a resistance of magnitude $2m\omega v$, where $v$ is its speed. At time $t$, the extension of the spring is $x$ and the displacement of the particle from its initial position is $y$. Show that
\begin{enumerate}[label=(\alph*)]
\item $x + y = Ut$, [2]
\item $\frac{d^2x}{dt^2} + 2\omega \frac{dx}{dt} + 5\omega^2 x = 2\omega U$. [7]
\end{enumerate}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find $x$ in terms of $\omega$, $U$ and $t$. [8]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2005 Q7 [17]}}