| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle attached to two separate elastic strings |
| Difficulty | Challenging +1.8 This is a challenging Further Maths FM2 question requiring: (1) proving SHM by showing restoring force ∝ -x (non-trivial with two strings), (2) finding equilibrium position and ω, (3) using energy methods for maximum speed, and (4) solving a time-in-region problem requiring inverse trig. Multi-step with several conceptually demanding elements, but follows standard FM2 SHM framework. |
| 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 |
| \(T_A = \frac{20e}{2},\ T_B = \frac{50(2-e)}{2}\ e\) | M1 | Use of \(T = \frac{\lambda x}{a}\) |
| In equilibrium \(T_A = T_B,\ 10e = 25(2-e)\) | M1 | Dependent on preceding M1. Equate their tensions |
| \((35e = 50),\ e = \frac{10}{7}\) | A1 | cao |
| Equation of motion for \(P\) when distance \(x\) from equilibrium position towards \(B\): | M1 | Condone sign error |
| \(3.5\ddot{x} = T_B - T_A = \frac{50(2-e-x)}{2} - \frac{20(e+x)}{2}\) | A1 A1 | Correct unsimplified equation in \(e\) and \(x\). Equation with one error A1A0 |
| \(= \frac{50\left(\frac{4}{7}-x\right)}{2} - \frac{20\left(\frac{10}{7}+x\right)}{2}\) | ||
| \(\Rightarrow 3.5\ddot{x} = -35x,\ \ddot{x} = -10x\) and hence SHM about the equilibrium position | A1 | Full working to justify conclusion that it is SHM about the equilibrium position |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Amplitude \(= 2 - \frac{10}{7} = \frac{4}{7}\) | B1 ft | Seen or implied. Follow their \(e\) |
| Use of max speed \(= a\omega\) | M1 | Correct method for max. speed |
| \(= \frac{4}{7}\sqrt{10} = 1.81\ (\text{m s}^{-1})\) | A1 ft | 1.81 or better. Follow their \(a, \omega\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Nearer to \(A\) than to \(B\): \(x < -\frac{3}{7}\) | B1 | Seen or implied |
| Solve for \(\sqrt{10}t\): \(\cos\sqrt{10}t = -\frac{3}{4},\ \sqrt{10}t = 2.418\ldots\) | M1 | Use of \(x = a\cos\omega t\) |
| Length of time: \(\frac{2}{\sqrt{10}}(\pi - 2.418\ldots)\) | M1 | Correct strategy for the required interval |
| \(0.457\) (seconds) | A1 | 0.457 or better |
# Question 7:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T_A = \frac{20e}{2},\ T_B = \frac{50(2-e)}{2}\ e$ | M1 | Use of $T = \frac{\lambda x}{a}$ |
| In equilibrium $T_A = T_B,\ 10e = 25(2-e)$ | M1 | Dependent on preceding M1. Equate their tensions |
| $(35e = 50),\ e = \frac{10}{7}$ | A1 | cao |
| Equation of motion for $P$ when distance $x$ from equilibrium position towards $B$: | M1 | Condone sign error |
| $3.5\ddot{x} = T_B - T_A = \frac{50(2-e-x)}{2} - \frac{20(e+x)}{2}$ | A1 A1 | Correct unsimplified equation in $e$ and $x$. Equation with one error A1A0 |
| $= \frac{50\left(\frac{4}{7}-x\right)}{2} - \frac{20\left(\frac{10}{7}+x\right)}{2}$ | | |
| $\Rightarrow 3.5\ddot{x} = -35x,\ \ddot{x} = -10x$ and hence SHM about the equilibrium position | A1 | Full working to justify conclusion that it is SHM about the equilibrium position |
**(7 marks)**
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Amplitude $= 2 - \frac{10}{7} = \frac{4}{7}$ | B1 ft | Seen or implied. Follow their $e$ |
| Use of max speed $= a\omega$ | M1 | Correct method for max. speed |
| $= \frac{4}{7}\sqrt{10} = 1.81\ (\text{m s}^{-1})$ | A1 ft | 1.81 or better. Follow their $a, \omega$ |
**(3 marks)**
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Nearer to $A$ than to $B$: $x < -\frac{3}{7}$ | B1 | Seen or implied |
| Solve for $\sqrt{10}t$: $\cos\sqrt{10}t = -\frac{3}{4},\ \sqrt{10}t = 2.418\ldots$ | M1 | Use of $x = a\cos\omega t$ |
| Length of time: $\frac{2}{\sqrt{10}}(\pi - 2.418\ldots)$ | M1 | Correct strategy for the required interval |
| $0.457$ (seconds) | A1 | 0.457 or better |
**(4 marks) (14 marks total)**
The image appears to be a blank page (page 286) from Pearson Edexcel Level 3 Advanced GCE in Further Mathematics Sample Assessment Materials, July 2017. There is no mark scheme content visible on this page to extract.\begin{enumerate}
\item Two points $A$ and $B$ are 6 m apart on a smooth horizontal surface.
\end{enumerate}
A light elastic string of natural length 2 m and modulus of elasticity 20 N , has one end attached to the point $A$.
A second light elastic string of natural length 2 m and modulus of elasticity 50 N , has one end attached to the point $B$.
A particle $P$ of mass 3.5 kg is attached to the free end of each string.\\
The particle $P$ is held at the point on $A B$ which is 2 m from $B$ and then released from rest.\\
In the subsequent motion both strings remain taut.\\
(a) Show that $P$ moves with simple harmonic motion about its equilibrium position.\\
(b) Find the maximum speed of $P$.\\
(c) Find the length of time within each oscillation for which $P$ is closer to $A$ than to $B$.
\hfill \mbox{\textit{Edexcel FM2 Q7 [14]}}