| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question requiring geometric decomposition and trigonometric manipulation. Part (a) involves standard coordinate geometry to find vertical height components, while part (b) requires solving a trigonometric equation. The geometry is moderately complex with the bent rod and 120° angle, but the techniques are standard for Further Maths Mechanics. More challenging than basic equilibrium problems but doesn't require novel insight. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(BAC = 360 - 120 - 90 - (90 - \theta) = \theta + 60\) | B1 | 3.1a |
| \(\Rightarrow BC = 2\sin(\theta + 60)\) | M1 | 1.1 |
| \(CD = AE = \sin\theta\) | ||
| \(\Rightarrow h = CD + BC = \sin\theta + 2\sin(\theta + 60°)\) | E1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(h = \sin\theta + 2\sin(\theta + 60°)\) | ||
| \(= \sin\theta + 2(\sin\theta\cos 60 + \cos\theta\sin 60)\) | M1 | 3.1a |
| \(= \sin\theta + \sin\theta + \sqrt{3}\cos\theta = 2\sin\theta + \sqrt{3}\cos\theta\) | A1 | 2.1 |
| \(h = 0 \Rightarrow 2\sin\theta + \sqrt{3}\cos\theta = 0\) | M1 | 1.1 |
| \(\Rightarrow \tan\theta = -\frac{\sqrt{3}}{2}\) | M1 | 1.1 |
| \(\Rightarrow \theta = -40.9°\) [so \(40.9°\) below the horizontal] | A1 | 1.1 |
| Alternative method: diagram with \(h=0\); \(a^2 = 1^2 + 2^2 - 4\cos 120°\); \(a = \sqrt{7}\); \(\sin\theta = \frac{2\sin 120°}{\sqrt{7}} = \sqrt{\frac{3}{7}}\); \(\theta = -40.9°\) | M1, M1, A1, M1, A1 | 3.1a, 2.1, 1.1, 1.1, 1.1 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $BAC = 360 - 120 - 90 - (90 - \theta) = \theta + 60$ | B1 | 3.1a |
| $\Rightarrow BC = 2\sin(\theta + 60)$ | M1 | 1.1 |
| $CD = AE = \sin\theta$ | | |
| $\Rightarrow h = CD + BC = \sin\theta + 2\sin(\theta + 60°)$ | E1 | 2.1 | AG **[3]** |
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## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $h = \sin\theta + 2\sin(\theta + 60°)$ | | |
| $= \sin\theta + 2(\sin\theta\cos 60 + \cos\theta\sin 60)$ | M1 | 3.1a | use of compound angle formula |
| $= \sin\theta + \sin\theta + \sqrt{3}\cos\theta = 2\sin\theta + \sqrt{3}\cos\theta$ | A1 | 2.1 |
| $h = 0 \Rightarrow 2\sin\theta + \sqrt{3}\cos\theta = 0$ | M1 | 1.1 | $h=0$ soi |
| $\Rightarrow \tan\theta = -\frac{\sqrt{3}}{2}$ | M1 | 1.1 | Use of $\frac{\sin}{\cos} = \tan$ |
| $\Rightarrow \theta = -40.9°$ [so $40.9°$ below the horizontal] | A1 | 1.1 | or $319.1°$ or $139.1°$ |
| **Alternative method:** diagram with $h=0$; $a^2 = 1^2 + 2^2 - 4\cos 120°$; $a = \sqrt{7}$; $\sin\theta = \frac{2\sin 120°}{\sqrt{7}} = \sqrt{\frac{3}{7}}$; $\theta = -40.9°$ | M1, M1, A1, M1, A1 | 3.1a, 2.1, 1.1, 1.1, 1.1 | For final mark, $\theta$ shown below horizontal in diagram together with $40.9°$ is acceptable **[5]** |
---8 In Fig. 8, OAB is a thin bent rod, with $\mathrm { OA } = 1 \mathrm {~m} , \mathrm { AB } = 2 \mathrm {~m}$ and angle $\mathrm { OAB } = 120 ^ { \circ }$. Angles $\theta , \phi$ and $h$ are as shown in Fig. 8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b4e10fd2-4144-4019-bf00-070f93a2b05d-07_949_949_429_214}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that $h = \sin \theta + 2 \sin \left( \theta + 60 ^ { \circ } \right)$.
The rod is free to rotate about the origin so that $\theta$ and $\phi$ vary. You may assume that the result for $h$ in part (a) holds for all values of $\theta$.
\item Find an angle $\theta$ for which $h = 0$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 Q8 [8]}}