| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Second order differential equations |
| Type | Particular solution with initial conditions |
| Difficulty | Standard +0.8 This is a mechanics-based second order differential equation requiring force analysis to derive the DE (part a), then applying initial conditions to find constants (part b), solving for when velocity is zero (part c), and proving an inequality involving the physical constraint that string remains taut (part d). While the general solution is given, the multi-step nature, combination of calculus and mechanics reasoning, and the non-trivial inequality proof make this significantly harder than average A-level questions. |
| Spec | 4.10d Second order homogeneous: auxiliary equation method4.10e 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 |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(R(\downarrow),\ mg - 2k\dot{x}m - T = m\ddot{x}\) | ||
| \(g - 2k\dot{x} - \frac{2ak^2x}{a} = \ddot{x}\) | M1 | |
| \(\Rightarrow \ddot{x} + 2k\dot{x} + 2k^2x = g\) | A1 c.s.o | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(t=0,\ x=a\): \(\quad a = D + \frac{g}{2k^2} \Rightarrow D = a - \frac{g}{2k^2}\) | M1 A1 | |
| \(\dot{x} = -ke^{-kt}(C\sin kt + D\cos kt) + -ke^{-kt}(C\cos kt - D\sin kt)\) | M1 A1 | |
| \(t=0,\ \dot{x}=0\): \(\quad 0 = -kD + kC \Rightarrow C = D\) | ||
| \(\Rightarrow C = a - \frac{g}{2k^2}\) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\dot{x} = 0 \Rightarrow C(\sin kt + \cos kt) + C(\cos kt - \sin kt)\) | M1 | |
| \(\Rightarrow \sin kt = 0\) | A1 ft | |
| \(\Rightarrow kt = \pi\) | ||
| \(\Rightarrow t = \frac{\pi}{k}\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| When \(t = \frac{\pi}{k}\), \(x = -De^{-\pi} + \frac{g}{2k^2}\) | M1 | |
| \(= \frac{g}{2k^2} - e^{-\pi}\!\left(a - \frac{g}{2k^2}\right)\) | A1 ft | |
| \(\Rightarrow xe^{\pi} = \frac{g}{2k^2}(e^{\pi}+1) - a\) | ||
| \(> 0\) (given) | M1 | |
| \(\Rightarrow g(e^{\pi}+1) > 2k^2a\) | A1 c.s.o | (4 marks) |
# Question 5:
## Part (a)
| Answer/Working | Marks | Notes |
|---|---|---|
| $R(\downarrow),\ mg - 2k\dot{x}m - T = m\ddot{x}$ | | |
| $g - 2k\dot{x} - \frac{2ak^2x}{a} = \ddot{x}$ | M1 | |
| $\Rightarrow \ddot{x} + 2k\dot{x} + 2k^2x = g$ | A1 c.s.o | (4 marks) |
## Part (b)
| Answer/Working | Marks | Notes |
|---|---|---|
| $t=0,\ x=a$: $\quad a = D + \frac{g}{2k^2} \Rightarrow D = a - \frac{g}{2k^2}$ | M1 A1 | |
| $\dot{x} = -ke^{-kt}(C\sin kt + D\cos kt) + -ke^{-kt}(C\cos kt - D\sin kt)$ | M1 A1 | |
| $t=0,\ \dot{x}=0$: $\quad 0 = -kD + kC \Rightarrow C = D$ | | |
| $\Rightarrow C = a - \frac{g}{2k^2}$ | A1 | (5 marks) |
## Part (c)
| Answer/Working | Marks | Notes |
|---|---|---|
| $\dot{x} = 0 \Rightarrow C(\sin kt + \cos kt) + C(\cos kt - \sin kt)$ | M1 | |
| $\Rightarrow \sin kt = 0$ | A1 ft | |
| $\Rightarrow kt = \pi$ | | |
| $\Rightarrow t = \frac{\pi}{k}$ | A1 | (3 marks) |
## Part (d)
| Answer/Working | Marks | Notes |
|---|---|---|
| When $t = \frac{\pi}{k}$, $x = -De^{-\pi} + \frac{g}{2k^2}$ | M1 | |
| $= \frac{g}{2k^2} - e^{-\pi}\!\left(a - \frac{g}{2k^2}\right)$ | A1 ft | |
| $\Rightarrow xe^{\pi} = \frac{g}{2k^2}(e^{\pi}+1) - a$ | | |
| $> 0$ (given) | M1 | |
| $\Rightarrow g(e^{\pi}+1) > 2k^2a$ | A1 c.s.o | (4 marks) |
**Total: 16 marks**
---5. A particle $P$ of mass $m$ is attached to one end of a light elastic string, of natural length $a$ and modulus of elasticity $2 m a k ^ { 2 }$, where $k$ is a positive constant. The other end of the string is attached to a fixed point $A$. At time $t = 0 , P$ is released from rest from a point which is a distance $2 a$ vertically below $A$. When $P$ is moving with speed $v$, the air resistance has magnitude $2 m k v$. At time $t$, the extension of the string is $x$.
\begin{enumerate}[label=(\alph*)]
\item Show that, while the string is taut,
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = g$$
You are given that the general solution of this differential equation is
$$x = \mathrm { e } ^ { - k t } ( C \sin k t + D \cos k t ) + \frac { g } { 2 k ^ { 2 } } , \quad \text { where } C \text { and } D \text { are constants. }$$
\item Find the value of $C$ and the value of $D$.
Assuming that the string remains taut,
\item find the value of $t$ when $P$ first comes to rest,
\item show that $2 k ^ { 2 } a < g \left( 1 + \mathrm { e } ^ { \pi } \right)$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2004 Q5 [16]}}