| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 6 |
| 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 standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions to find two unknowns. It's a routine textbook exercise with straightforward application of ΣFₓ=0 and ΣFᵧ=0, requiring only basic trigonometry and algebraic manipulation—easier than average A-level questions which typically involve more steps or conceptual depth. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(F = \sqrt{6^2 + 5^2}\) | M1 A1 | Obtaining an equation for \(F\) with square or root. Correct equation |
| \(= \sqrt{61} = 7.81\) | A1 | Total: 3 — Correct force |
| Alt: \(\alpha = \tan^{-1}\left(\frac{5}{6}\right) = 39.8°\) | ||
| \(F = \frac{6}{\cos 39.8} = 7.81\) or \(F = \frac{5}{\sin 39.8} = 7.81\) | (M1 A1)(A1) | Equation for \(F\) with a value for \(\alpha\). Correct equation. Correct force |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\alpha = \tan^{-1}\left(\frac{5}{6}\right)\) or \(\cos^{-1}\left(\frac{6}{7.81}\right)\) or \(\sin^{-1}\left(\frac{6}{7.81}\right)\) | M1 | Obtaining an equation for \(\alpha\) using trigonometry. Correct equation (using their \(F\)) |
| A1 | ||
| \(= 39.8°\) | A1 | Total: 3 — Correct angle. Accept values between 39.7 and 39.9 |
| Alt: \(\frac{\sin\alpha}{5} = \frac{\sin 90°}{\sqrt{61}}\), \(\alpha = 39.8°\) |
## Question 3:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $F = \sqrt{6^2 + 5^2}$ | M1 A1 | Obtaining an equation for $F$ with square or root. Correct equation |
| $= \sqrt{61} = 7.81$ | A1 | **Total: 3** — Correct force |
| **Alt:** $\alpha = \tan^{-1}\left(\frac{5}{6}\right) = 39.8°$ | | |
| $F = \frac{6}{\cos 39.8} = 7.81$ or $F = \frac{5}{\sin 39.8} = 7.81$ | (M1 A1)(A1) | Equation for $F$ with a value for $\alpha$. Correct equation. Correct force |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\alpha = \tan^{-1}\left(\frac{5}{6}\right)$ or $\cos^{-1}\left(\frac{6}{7.81}\right)$ or $\sin^{-1}\left(\frac{6}{7.81}\right)$ | M1 | Obtaining an equation for $\alpha$ using trigonometry. Correct equation (using their $F$) |
| | A1 | |
| $= 39.8°$ | A1 | **Total: 3** — Correct angle. Accept values between 39.7 and 39.9 |
| **Alt:** $\frac{\sin\alpha}{5} = \frac{\sin 90°}{\sqrt{61}}$, $\alpha = 39.8°$ | | |
---3 The diagram shows three forces which act in the same plane and are in equilibrium.\\
\includegraphics[max width=\textwidth, alt={}, center]{965a176a-848c-478d-a748-80fc9dfe4399-3_419_516_383_761}
\begin{enumerate}[label=(\alph*)]
\item Find $F$.
\item Find $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q3 [6]}}