| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Coplanar forces in equilibrium |
| Difficulty | Standard +0.3 This is a standard coplanar forces problem requiring resolution in two perpendicular directions and basic trigonometry. Part (a) uses equilibrium conditions (ΣFx=0, ΣFy=0) to find two unknowns, while part (b) requires understanding that perpendicularity means the component parallel to the 10N force is zero. Both parts are routine applications of A-level mechanics techniques with straightforward algebra, making this slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempt to resolve vertically or horizontally | M1 | Correct number of terms |
| \(G\sin 60° + 2F\sin 40° - 10 = 0\) | A1 | Correct resolution vertically |
| \(F + G\cos 60° - 2F\cos 40° = 0\) | A1 | Correct resolution horizontally |
| Attempt to solve simultaneously for \(F\) or \(G\) | M1 | From equations with 3 relevant terms in each |
| \(F = 4.53,\; G = 4.82\) | A1 | For both correct |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(G\sin 60° + 2 \times 3\sin 40° - 10 = 0\) | M1 | Resolve forces parallel to the 10 N force and equate this expression to zero, 3 terms |
| \(G = 7.09\) to 3 sf | A1 | |
| 2 |
## Question 5:
### Part 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempt to resolve vertically or horizontally | M1 | Correct number of terms |
| $G\sin 60° + 2F\sin 40° - 10 = 0$ | A1 | Correct resolution vertically |
| $F + G\cos 60° - 2F\cos 40° = 0$ | A1 | Correct resolution horizontally |
| Attempt to solve simultaneously for $F$ or $G$ | M1 | From equations with 3 relevant terms in each |
| $F = 4.53,\; G = 4.82$ | A1 | For both correct |
| | **5** | |
### Part 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $G\sin 60° + 2 \times 3\sin 40° - 10 = 0$ | M1 | Resolve forces parallel to the 10 N force and equate this expression to zero, 3 terms |
| $G = 7.09$ to 3 sf | A1 | |
| | **2** | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{19a41291-2692-48f4-86af-bb4930353959-08_645_611_258_767}
Four coplanar forces act at a point. The magnitudes of the forces are $10 \mathrm {~N} , F \mathrm {~N} , G \mathrm {~N}$ and $2 F \mathrm {~N}$. The directions of the forces are as shown in the diagram.
\begin{enumerate}[label=(\alph*)]
\item Given that the forces are in equilibrium, find the values of $F$ and $G$.
\item Given instead that $F = 3$, find the value of $G$ for which the resultant of the forces is perpendicular to the 10 N force.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q5 [7]}}