| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Energy considerations in circular motion |
| Difficulty | Challenging +1.2 This is a multi-part Further Mechanics question requiring equilibrium analysis, SHM verification, and energy/kinematics calculations. While it involves several steps and FM2 content (elastic strings, SHM), the techniques are standard: Hooke's law for part (a), showing resultant force ∝ -x for part (b), and using SHM formulas for parts (c)-(d). The problem-solving is methodical rather than requiring novel insight, making it moderately above average difficulty for A-level but routine for FM2 students. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| In equilibrium: \(T_A = T_B\) | M1 | Form equation for equilibrium of forces acting on \(P\). Dimensionally correct. |
| \(\dfrac{40e}{2} = \dfrac{20(2-e)}{2}\) | A1 | Correct unsimplified equation in one unknown. |
| \(2e = 2-e \Rightarrow e = \dfrac{2}{3} \Rightarrow EB = \dfrac{8}{3}\) m * | A1* | Obtain given answer from correct working. Condone missing units. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion for \(P\) when displaced \(x\) from equilibrium in direction of \(A\) | M1 | Equation of motion about the equilibrium. Need all terms and dimensionally correct. Condone sign errors. |
| \(\dfrac{40(e+x)}{2} - \dfrac{20(2-e-x)}{2} = -0.3\ddot{x}\) | A1, A1 | Unsimplified equation with \(e\) or given \(e\) with at most one error. / Correct unsimplified equation with \(e\) or given \(e\). |
| \(\ddot{x} = -100x\), of the form \(\ddot{x} = -\omega^2 x\) hence SHM* | A1* | Reach given conclusion in correct form from correct working. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Periodic time \(= \dfrac{2\pi}{\omega} = \dfrac{\pi}{5}\) (s) | M1 | Use the model to find the periodic time for \(\omega\) obtained correctly. |
| \(\Rightarrow S = \dfrac{1}{4}\times\dfrac{2\pi}{\omega}\) | M1 | Correct method to find the required time for \(\omega\) obtained correctly. |
| \(S = \dfrac{\pi}{20}\) | A1 | Correct only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(x = a\cos(\omega t)\) or \(x = a\sin(\omega t)\) | M1 | |
| \(0 = \frac{1}{3}\cos(\omega)S\) or \(\frac{1}{3} = \frac{1}{3}\sin(\omega S)\) | M1 | |
| \(S = \dfrac{\pi}{20}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(v = a\omega\cos\omega t\) or \(v = a\omega\sin\omega t\) | M1 | Use a correct model for the speed or displacement of \(P\) for \(\omega\) obtained correctly. |
| \(2 = \dfrac{\omega}{3}\cos\omega t\) \((t=0.0927...)\) or \(2 = \dfrac{\omega}{3}\sin\omega t\) \((t=0.06435...)\) | A1ft | Follow their \(\omega\). NB \(v=2\) when \(x = \dfrac{4}{15}\) |
| time \(= 4\times 0.0927...\) or time \(= \dfrac{\pi}{5} - 4\times 0.06435...\) | DM1 | Complete method to find the required time. |
| time \(= 0.37\) s \((0.371\) s\()\) | A1 | 0.37 or better \((0.370918....)\) |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| In equilibrium: $T_A = T_B$ | M1 | Form equation for equilibrium of forces acting on $P$. Dimensionally correct. |
| $\dfrac{40e}{2} = \dfrac{20(2-e)}{2}$ | A1 | Correct unsimplified equation in one unknown. |
| $2e = 2-e \Rightarrow e = \dfrac{2}{3} \Rightarrow EB = \dfrac{8}{3}$ m * | A1* | Obtain given answer from correct working. Condone missing units. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion for $P$ when displaced $x$ from equilibrium in direction of $A$ | M1 | Equation of motion about the equilibrium. Need all terms and dimensionally correct. Condone sign errors. |
| $\dfrac{40(e+x)}{2} - \dfrac{20(2-e-x)}{2} = -0.3\ddot{x}$ | A1, A1 | Unsimplified equation with $e$ or given $e$ with at most one error. / Correct unsimplified equation with $e$ or given $e$. |
| $\ddot{x} = -100x$, of the form $\ddot{x} = -\omega^2 x$ hence SHM* | A1* | Reach given conclusion in correct form from correct working. |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Periodic time $= \dfrac{2\pi}{\omega} = \dfrac{\pi}{5}$ (s) | M1 | Use the model to find the periodic time for $\omega$ obtained correctly. |
| $\Rightarrow S = \dfrac{1}{4}\times\dfrac{2\pi}{\omega}$ | M1 | Correct method to find the required time for $\omega$ obtained correctly. |
| $S = \dfrac{\pi}{20}$ | A1 | Correct only. |
**Alternative (8c alt):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $x = a\cos(\omega t)$ or $x = a\sin(\omega t)$ | M1 | |
| $0 = \frac{1}{3}\cos(\omega)S$ or $\frac{1}{3} = \frac{1}{3}\sin(\omega S)$ | M1 | |
| $S = \dfrac{\pi}{20}$ | A1 | |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $v = a\omega\cos\omega t$ or $v = a\omega\sin\omega t$ | M1 | Use a correct model for the speed or displacement of $P$ for $\omega$ obtained correctly. |
| $2 = \dfrac{\omega}{3}\cos\omega t$ $(t=0.0927...)$ or $2 = \dfrac{\omega}{3}\sin\omega t$ $(t=0.06435...)$ | A1ft | Follow their $\omega$. NB $v=2$ when $x = \dfrac{4}{15}$ |
| time $= 4\times 0.0927...$ or time $= \dfrac{\pi}{5} - 4\times 0.06435...$ | DM1 | Complete method to find the required time. |
| time $= 0.37$ s $(0.371$ s$)$ | A1 | 0.37 or better $(0.370918....)$ |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b070338-1de4-4c33-be29-d37ac06c9fed-28_200_1086_214_552}
\captionsetup{labelformat=empty}
\caption{Figure 7}
\end{center}
\end{figure}
The fixed points $A$ and $B$ lie on a smooth horizontal surface with $A B = 6 \mathrm {~m}$.\\
A particle $P$ has mass 0.3 kg .\\
One end of a light elastic string, of natural length 2 m and modulus of elasticity 20 N , is attached to $P$, and the other end is attached to $A$.
One end of another light elastic string, of natural length 2 m and modulus of elasticity 40 N , is attached to $P$ and the other end is attached to $B$.
The particle $P$ is at rest in equilibrium at the point $E$ on the surface, as shown in Figure 7.
\begin{enumerate}[label=(\alph*)]
\item Show that $E B = \frac { 8 } { 3 } \mathrm {~m}$.
The particle $P$ is now held at the midpoint of $A B$ and released from rest.
\item Show that $P$ oscillates with simple harmonic motion about the point $E$.
The time between the instant when $P$ is released and the instant when it first returns to the point $E$ is $S$ seconds.
\item Find the exact value of $S$.
\item Find the length of time during one oscillation for which the speed of $P$ is more than $2 \mathrm {~ms} ^ { - 1 }$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2023 Q8 [14]}}