| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of three coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question requiring resolution of forces into perpendicular components and finding the resultant magnitude using Pythagoras. The symmetry (PQ bisects the angle between equal forces) simplifies the calculation significantly. While it requires multiple steps and careful angle work, it's a routine textbook-style problem with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks |
|---|---|
| For resolving in the direction \(PQ\) | M1 |
| Component is \(2 \times 10\cos30° - 6\cos60°\) or \(14.3\) N or \(10\sqrt{3} - 3\) N | A1 |
| Answer | Marks |
|---|---|
| Component is \(\pm 6\cos30° - 6\cos60°\) or \(\pm 5.20\) N or \(\pm 3\sqrt{3}\) N | B1 |
| Answer | Marks |
|---|---|
| For resolving in both directions | M1 |
| For \(X = 6 - 10\cos30°\) or \(-2.66\) N and \(Y = 10 + 10\sin30°\) or 15 N | A1 |
| Answer | Marks |
|---|---|
| For resolving in both directions | M1 |
| For \((6\cos30°)\mathbf{i} + (2 \times 10\cos30° - 6\cos60°)\mathbf{j}\) or any vector equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| For using Magnitude \(= \sqrt{ans(i)^2 + ans(ii)^2}\) | M1 | |
| Magnitude is 15.2 N | A1ft | ft only following sin/cos mix and for answer 5.66 N |
# Question 2:
## Part (i)(a)
| For resolving in the direction $PQ$ | M1 | |
| Component is $2 \times 10\cos30° - 6\cos60°$ or $14.3$ N or $10\sqrt{3} - 3$ N | A1 | |
## Part (i)(b)
| Component is $\pm 6\cos30° - 6\cos60°$ or $\pm 5.20$ N or $\pm 3\sqrt{3}$ N | B1 | |
**SR** (for candidates who resolve parallel to and perpendicular to the force of magnitude 6 N) (Max 2 out of 3):
| For resolving in both directions | M1 | |
| For $X = 6 - 10\cos30°$ or $-2.66$ N and $Y = 10 + 10\sin30°$ or 15 N | A1 | |
**SR** (for candidates who give a combined answer for (a) and (b)) (Max 2 out of 3):
| For resolving in both directions | M1 | |
| For $(6\cos30°)\mathbf{i} + (2 \times 10\cos30° - 6\cos60°)\mathbf{j}$ or any vector equivalent | A1 | |
## Part (ii)
| For using Magnitude $= \sqrt{ans(i)^2 + ans(ii)^2}$ | M1 | |
| Magnitude is 15.2 N | A1ft | ft only following sin/cos mix and for answer 5.66 N |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{cb04a09c-af23-4e9d-b3da-da9e351fe879-2_405_384_550_884}
Three coplanar forces of magnitudes $10 \mathrm {~N} , 10 \mathrm {~N}$ and 6 N act at a point $P$ in the directions shown in the diagram. $P Q$ is the bisector of the angle between the two forces of magnitude 10 N .\\
(i) Find the component of the resultant of the three forces
\begin{enumerate}[label=(\alph*)]
\item in the direction of $P Q$,
\item in the direction perpendicular to $P Q$.\\
(ii) Find the magnitude of the resultant of the three forces.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2003 Q2 [5]}}