| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2009 |
| Session | June |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Modeling context with interpretation |
| Difficulty | Standard +0.8 This is a mechanics-based second order DE problem requiring physical modeling (Newton's second law with Hooke's law), solving a non-homogeneous DE with particular integral, and interpreting the solution for maximum extension and speed. While the DE solving is standard A-level technique, the physical setup, derivation from first principles, and multi-part interpretation elevate it above average difficulty. |
| 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 |
|---|---|
| M1 B1 B2 DM1 A1 |
| Answer | Marks |
|---|---|
| B1 B1 M1 M1 A1ft M1 A1 A1 |
| Answer | Marks |
|---|---|
| M1 M1 A1 |
| Answer | Marks |
|---|---|
| M1 A1 |
**Part (a)**
$(\to), T = m\ddot{y}$
Hooke's Law: $T = \frac{2mn^2ax}{2a} = mn^2x$
$x + y = \frac{1}{2}ft^2$ ⟹ $\dot{x} + \dot{y} = ft$ ⟹ $\ddot{x} + \ddot{y} = f$ ⟹ $\ddot{x} = f$
so, $(\to), mn^2x = m\dot{y} = m(f - \ddot{x})$
$\ddot{x} + n^2x = f$ **
| M1 B1 B2 DM1 A1 |
**Part (b)**
C.F.: $x = A\cos nt + B\sin nt$
P.I.: $x = \frac{f}{n^2}$
Gen solution: $x = A\cos nt + B\sin nt + \frac{f}{n^2}$
$\dot{x} = -An\sin nt + Bn\cos nt$ follow their PI
$t = 0, x = 0 \Rightarrow A = -\frac{f}{n^2}$ ⟹
$t = 0, \dot{x} = 0 \Rightarrow B = 0$
$x = \frac{f}{n^2}(1 - \cos nt)$
| B1 B1 M1 M1 A1ft M1 A1 A1 |
**Part (c)**
$\dot{x} = 0 \Rightarrow nt = \pi$
$x_{\max} = \frac{f}{n^2}(1-1) = \frac{2f}{n^2}$
| M1 M1 A1 |
**Part (d)**
$\dot{y} = ft - \dot{x}$
$= f\frac{\pi}{n} - 0 = \frac{f\pi}{n}$
| M1 A1 |
**Total: [19]**6. A light elastic spring $A B$ has natural length $2 a$ and modulus of elasticity $2 m n ^ { 2 } a$, where $n$ is a constant. A particle $P$ of mass $m$ is attached to the end $A$ of the spring. At time $t = 0$, the spring, with $P$ attached, lies at rest and unstretched on a smooth horizontal plane. The other end $B$ of the spring is then pulled along the plane in the direction $A B$ with constant acceleration $f$. At time $t$ the extension of the spring is $x$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = f .$$
\item Find $x$ in terms of $n , f$ and $t$.
Hence find
\item the maximum extension of the spring,
\item the speed of $P$ when the spring first reaches its maximum extension.\\
\section*{June 2009}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2009 Q6 [19]}}