| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| 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 straightforward resolution of forces in two perpendicular directions. Students apply the routine method of resolving horizontally to find θ, then vertically to find m, with no conceptual challenges beyond basic trigonometry and the equilibrium condition ΣF=0. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolve horizontally: \(50\cos\theta = 60\cos48°\) | M1 | Correct equation resolving horizontally |
| \(\cos\theta = \frac{60\cos48°}{50}\) | A1 | Correct expression |
| \(\cos\theta = 0.8034...\) | - | |
| \(\theta = 36.5°\) (or \(36°\) to nearest degree) | A1 A1 | A1 for correct method, A1 for answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolve vertically: \(mg = 50\sin\theta + 60\sin48°\) | M1 | Correct vertical resolution equation |
| \(mg = 50\sin(36.5°) + 60\sin48°\) | A1 | Substitution of their \(\theta\) |
| \(m = \frac{50\sin(36.5°) + 60\sin48°}{9.8}\) | - | |
| \(m \approx 7.63\) kg | A1 | cao, follow through on their \(\theta\) |
## Question 4:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve horizontally: $50\cos\theta = 60\cos48°$ | M1 | Correct equation resolving horizontally |
| $\cos\theta = \frac{60\cos48°}{50}$ | A1 | Correct expression |
| $\cos\theta = 0.8034...$ | - | |
| $\theta = 36.5°$ (or $36°$ to nearest degree) | A1 A1 | A1 for correct method, A1 for answer |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve vertically: $mg = 50\sin\theta + 60\sin48°$ | M1 | Correct vertical resolution equation |
| $mg = 50\sin(36.5°) + 60\sin48°$ | A1 | Substitution of their $\theta$ |
| $m = \frac{50\sin(36.5°) + 60\sin48°}{9.8}$ | - | |
| $m \approx 7.63$ kg | A1 | cao, follow through on their $\theta$ |
---4 A particle, of mass $m \mathrm {~kg}$, remains in equilibrium under the action of three forces, which act in a vertical plane, as shown in the diagram. The force with magnitude 60 N acts at $48 ^ { \circ }$ above the horizontal and the force with magnitude 50 N acts at an angle $\theta$ above the horizontal.\\
\includegraphics[max width=\textwidth, alt={}, center]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-08_576_647_548_701}
\begin{enumerate}[label=(\alph*)]
\item By resolving horizontally, find $\theta$.
\item Find $m$.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5d474771-fe32-47c6-8bf3-60ff7a25dd12-09_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2010 Q4 [7]}}