| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions. Students apply standard M1 techniques (resolving horizontally and vertically) with given angles to find two unknowns. The 'show that' part guides them to one answer, making it easier than average but still requiring correct trigonometric setup and algebraic manipulation. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(P = 5 + 8\cos 60°\) | M1 | Both relevant forces, component of 8N attempted |
| A1 | All correct | |
| \(P = 9\) | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(Q = 8\cos 30°\) | M1 | Component of 8N attempted |
| \(Q = 6.93\) or \(4\sqrt{3}\) | A1 | AWRT 6.93 |
## Question 2:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $P = 5 + 8\cos 60°$ | M1 | Both relevant forces, component of 8N attempted |
| | A1 | All correct |
| $P = 9$ | A1 | CAO | **Total: 3** |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $Q = 8\cos 30°$ | M1 | Component of 8N attempted |
| $Q = 6.93$ or $4\sqrt{3}$ | A1 | AWRT 6.93 | **Total: 2** |
---2 A particle is in equilibrium under the action of four horizontal forces of magnitudes 5 newtons, 8 newtons, $P$ newtons and $Q$ newtons, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{cfe0bdbc-35e3-485f-a922-b652a72f4c95-2_355_357_1146_852}
\begin{enumerate}[label=(\alph*)]
\item Show that $P = 9$.
\item Find the value of $Q$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2006 Q2 [5]}}