| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Challenging +1.2 This is a standard M4 elastic spring SHM problem requiring derivation of the differential equation from Hooke's law and Newton's second law, then solving a non-homogeneous second-order ODE with particular integral. While it involves multiple steps and requires knowledge of complementary function/particular integral methods, it follows a well-established template for this topic with no novel insights required. The multi-part structure and calculus make it above average difficulty, but it's routine for M4 students. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Position of end \(B\) at time \(t\) is \(a + Ut\) | M1 | Considering positions of \(A\) and \(B\) |
| Length of string \(= a + x\), so position of \(A\) is \(y\), giving \(y + (a+x) - a = Ut\), hence \(x + y = Ut\) | A1 | Correct argument shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Tension in string \(= \frac{9ma \cdot x}{a} = 9mx\) | B1 | Correct tension |
| Equation of motion: \(m\ddot{y} = -9mx - 6m\dot{y}\) | M1 | Newton's second law applied |
| \(\ddot{y} = -9x - 6\dot{y}\) | A1 | Correct equation |
| From (a): \(\dot{y} = U - \dot{x}\), \(\ddot{y} = -\ddot{x}\) | M1 | Differentiating \(x+y=Ut\) |
| Substituting: \(-\ddot{x} = -9x - 6(U-\dot{x})\), giving \(\ddot{x} + 6\dot{x} + 9x = 6U\) | A1 | Correct equation shown |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At \(t=0\): \(x=0\) (string at natural length, no extension) | B1 | Correct initial condition |
| \(0 = A \cdot U \cdot 1 + \frac{2U}{3}\), so \(A = -\frac{2}{3}\) | M1 A1 | Using \(x=0\) at \(t=0\) |
| \(\dot{x} = (B - 3A - 3Bt)Ue^{-3t}\) | M1 | Differentiating \(x\) |
| At \(t=0\): \(\dot{x}=0\) (particle initially at rest), so \(B - 3A = 0\), giving \(B = 3A = -2\) | A1 | Using \(\dot{x}=0\) at \(t=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Speed of \(P\) = \(\dot{y} = U - \dot{x}\) | M1 | Using result from (a) |
| \(\dot{x} = (-2 - 3(-2)t + 6)Ue^{-3t} \cdot (-3) + ...\); speed \(= U(1+2+6t)e^{-3t} \cdot ... \) | ||
| Speed \(= U(1-(6t+2)e^{-3t}\cdot(-3) + 6e^{-3t})\); \(\dot{y} = U(1-(6t+4)e^{-3t})\) | A1 | Accept equivalent forms |
# Question 6:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Position of end $B$ at time $t$ is $a + Ut$ | M1 | Considering positions of $A$ and $B$ |
| Length of string $= a + x$, so position of $A$ is $y$, giving $y + (a+x) - a = Ut$, hence $x + y = Ut$ | A1 | Correct argument shown |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Tension in string $= \frac{9ma \cdot x}{a} = 9mx$ | B1 | Correct tension |
| Equation of motion: $m\ddot{y} = -9mx - 6m\dot{y}$ | M1 | Newton's second law applied |
| $\ddot{y} = -9x - 6\dot{y}$ | A1 | Correct equation |
| From (a): $\dot{y} = U - \dot{x}$, $\ddot{y} = -\ddot{x}$ | M1 | Differentiating $x+y=Ut$ |
| Substituting: $-\ddot{x} = -9x - 6(U-\dot{x})$, giving $\ddot{x} + 6\dot{x} + 9x = 6U$ | A1 | Correct equation shown |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| At $t=0$: $x=0$ (string at natural length, no extension) | B1 | Correct initial condition |
| $0 = A \cdot U \cdot 1 + \frac{2U}{3}$, so $A = -\frac{2}{3}$ | M1 A1 | Using $x=0$ at $t=0$ |
| $\dot{x} = (B - 3A - 3Bt)Ue^{-3t}$ | M1 | Differentiating $x$ |
| At $t=0$: $\dot{x}=0$ (particle initially at rest), so $B - 3A = 0$, giving $B = 3A = -2$ | A1 | Using $\dot{x}=0$ at $t=0$ |
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Speed of $P$ = $\dot{y} = U - \dot{x}$ | M1 | Using result from (a) |
| $\dot{x} = (-2 - 3(-2)t + 6)Ue^{-3t} \cdot (-3) + ...$; speed $= U(1+2+6t)e^{-3t} \cdot ... $ | | |
| Speed $= U(1-(6t+2)e^{-3t}\cdot(-3) + 6e^{-3t})$; $\dot{y} = U(1-(6t+4)e^{-3t})$ | A1 | Accept equivalent forms |
---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 $x$.\\
(a) Show that
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = f .$$
(b) Find $x$ in terms of $n , f$ and $t$.
Hence find\\
(c) the maximum extension of the spring,\\
(d) the speed of $P$ when the spring first reaches its maximum extension.\\
\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are unit vectors due east and due north respectively]
\end{enumerate}
A man cycles at a constant speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on level ground and finds that when his velocity is $u \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the velocity of the wind appears to be $v ( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, where $v$ is a positive constant.
When the man cycles with velocity $\frac { 1 } { 5 } u ( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, the velocity of the wind appears to be $w \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $w$ is a positive constant.
Find, in terms of $u$, the true velocity of the wind.\\
\hfill \mbox{\textit{Edexcel M4 Q6 [14]}}