| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Force given resultant and one force |
| Difficulty | Moderate -0.3 This is a standard M1 vector triangle problem requiring application of the cosine and sine rules to find an unknown force. While it involves multiple steps (drawing a diagram, applying trigonometry, finding magnitude and bearing), these are routine techniques for mechanics students with no novel insight required. Slightly easier than average due to being a textbook-style question with straightforward application of standard methods. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03a Force: vector nature and diagrams3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(\mathbf{P} = 7\mathbf{j}\) and \(\mathbf{Q} = \mathbf{R} - \mathbf{P}\) to obtain \(\mathbf{Q} = 5\sqrt{3}\mathbf{i} - 12\mathbf{j}\) | M1 A1 ↓ M1 A1 ↓ M1 A1 | |
| Magnitude \(= \sqrt{(5\sqrt{3})^2 + 12^2} \approx 14.8 \text{ N}\) (AWRT) | M1 A1 | |
| Angle with \(\mathbf{i} = \arctan(12/5\sqrt{3}) \approx 64.2°\) | M1 A1 | |
| Bearing \(\approx 144°\) (AWRT) | A1 | (9 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Vector triangle correct: \(Q^2 = 10^2 + 7^2 + 2 \times 10 \times 7 \cos 60\) | B1 M1 A1 | |
| \(Q \approx 14.8 \text{ N}\) (AWRT) | A1 | |
| \(\frac{14.8}{\sin 120} = \frac{10}{\sin \theta}\) → \(\theta = 35.8°\) → bearing \(144\) (AWRT) | M1 A1 √ ↓ M1 A1, A1 | (9 marks) |
$R = 10\sqrt{3}/2 \mathbf{i} - 5\mathbf{j}$
Using $\mathbf{P} = 7\mathbf{j}$ and $\mathbf{Q} = \mathbf{R} - \mathbf{P}$ to obtain $\mathbf{Q} = 5\sqrt{3}\mathbf{i} - 12\mathbf{j}$ | M1 A1 ↓ M1 A1 ↓ M1 A1 |
Magnitude $= \sqrt{(5\sqrt{3})^2 + 12^2} \approx 14.8 \text{ N}$ (AWRT) | M1 A1 |
Angle with $\mathbf{i} = \arctan(12/5\sqrt{3}) \approx 64.2°$ | M1 A1 |
Bearing $\approx 144°$ (AWRT) | A1 | (9 marks)
### Alternative method
Vector triangle correct: $Q^2 = 10^2 + 7^2 + 2 \times 10 \times 7 \cos 60$ | B1 M1 A1 |
$Q \approx 14.8 \text{ N}$ (AWRT) | A1 |
$\frac{14.8}{\sin 120} = \frac{10}{\sin \theta}$ → $\theta = 35.8°$ → bearing $144$ (AWRT) | M1 A1 √ ↓ M1 A1, A1 | (9 marks)
**Total for Question 4: 9 marks**
---Two forces $\mathbf{P}$ and $\mathbf{Q}$ act on a particle. The force $\mathbf{P}$ has magnitude $7$ N and acts due north. The resultant of $\mathbf{P}$ and $\mathbf{Q}$ is a force of magnitude $10$ N acting in a direction with bearing $120°$. Find
\begin{enumerate}[label=(\roman*)]
\item the magnitude of $\mathbf{Q}$,
\item the direction of $\mathbf{Q}$, giving your answer as a bearing.
\end{enumerate}
[9]
\hfill \mbox{\textit{Edexcel M1 2006 Q4 [9]}}