| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Potential energy with inextensible strings or gravity only |
| Difficulty | Challenging +1.8 This is an M4 (Further Maths Mechanics) question requiring potential energy formulation from geometry, finding equilibrium via differentiation, and stability analysis via second derivative test. While systematic, it demands careful coordinate geometry, trigonometric manipulation across multiple parts, and understanding of energy methods—significantly above standard A-level but routine for M4 students. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AR = 2r\cos\theta\) | B1 | |
| For \(P\): \(-mg(L - 2r\cos\theta)\) | B1 | |
| For \(R\): \(-mg\cdot 2r\cos^2\theta\) | M1 A1 | |
| \(V = 2mgr(\cos\theta - \cos^2\theta) - mgL\) (*) | M1 A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 2mgr(-\sin\theta + 2\cos\theta\sin\theta)\) | M1 A1 | |
| \(= 2mgr\sin\theta(2\cos\theta - 1)\) | A1 | |
| \(0 = 2mgr\sin\theta(2\cos\theta - 1)\) | M1 | |
| \(\sin\theta = 0\) or \(\cos\theta = \tfrac{1}{2}\) | ||
| \(\theta = 0\) or \(\theta = \dfrac{\pi}{3}\) | A1 A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 2mgr(-\cos\theta + 2\cos 2\theta)\) | M1 A1 | |
| \(\theta = 0\): \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 2mgr > 0 \Rightarrow\) STABLE | M1 A1 | |
| \(\theta = \dfrac{\pi}{3}\): \(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = -3mgr < 0 \Rightarrow\) UNSTABLE | A1 | (5), (17) |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AR = 2r\cos\theta$ | B1 | |
| For $P$: $-mg(L - 2r\cos\theta)$ | B1 | |
| For $R$: $-mg\cdot 2r\cos^2\theta$ | M1 A1 | |
| $V = 2mgr(\cos\theta - \cos^2\theta) - mgL$ (*) | M1 A1 | **(6)** |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 2mgr(-\sin\theta + 2\cos\theta\sin\theta)$ | M1 A1 | |
| $= 2mgr\sin\theta(2\cos\theta - 1)$ | A1 | |
| $0 = 2mgr\sin\theta(2\cos\theta - 1)$ | M1 | |
| $\sin\theta = 0$ or $\cos\theta = \tfrac{1}{2}$ | | |
| $\theta = 0$ or $\theta = \dfrac{\pi}{3}$ | A1 A1 | **(6)** |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 2mgr(-\cos\theta + 2\cos 2\theta)$ | M1 A1 | |
| $\theta = 0$: $\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 2mgr > 0 \Rightarrow$ STABLE | M1 A1 | |
| $\theta = \dfrac{\pi}{3}$: $\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = -3mgr < 0 \Rightarrow$ UNSTABLE | A1 | **(5), (17)** |7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{d57ea92a-4d6a-46bf-a6aa-bbd5083e8726-5_955_855_349_573}
\end{center}
\end{figure}
A smooth wire $A B$, in the shape of a circle of radius $r$, is fixed in a vertical plane with $A B$ vertical. A small smooth ring $R$ of mass $m$ is threaded on the wire and is connected by a light inextensible string to a particle $P$ of mass $m$. The length of the string is greater than the diameter of the circle. The string passes over a small smooth pulley which is fixed at the highest point $A$ of the wire and angle $R \hat { A } P = \theta$, as shown in Fig. 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the potential energy of the system is given by
$$2 m g r \left( \cos \theta - \cos ^ { 2 } \theta \right) + \text { constant. }$$
\item Hence determine the values of $\theta , \theta \geq 0$, for which the system is in equilibrium. (6 marks)
\item Determine the stability of each position of equilibrium.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 Q7 [17]}}