| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2001 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of two forces (triangle/parallelogram law) |
| Difficulty | Moderate -0.8 This is a standard M1 mechanics question requiring direct application of the cosine rule and sine rule to find resultant force magnitude and direction. It's a textbook exercise with straightforward calculation steps, making it easier than average A-level questions which typically require more problem-solving or multi-step reasoning. |
| Spec | 3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(F^2 = 5^2 + 3^2 - 2(5)(3)\cos 140°\) | M1 A1 | Vector triangle attempt, correct |
| \(F \approx 7.55 \text{ N}\) | M1 A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{F}{\sin 140°} = \frac{3}{\sin \theta} \Rightarrow \theta \approx 14.8°\) | M1 A1 A1 | (3 marks) Total: 8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F^2 = (5 + 3\cos 40°)^2 + (3\sin 40°)^2\) | M1 A1 M1 A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{3\sin 40°}{5 + 3\cos 40°}, \quad \theta \approx 14.8°\) | M1 A1 A1 | (3 marks) Total: 8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\underline{P} = \begin{pmatrix}5\\0\end{pmatrix}\), \(\underline{Q} = \begin{pmatrix}3\cos 40°\\3\sin 40°\end{pmatrix}\) | M1 | |
| \(\Rightarrow \underline{F} = \begin{pmatrix}5+3\cos 40°\\3\sin 40°\end{pmatrix}\) | A1 | |
| \( | \underline{F} | = \sqrt{(5+3\cos 40°)^2 + (3\sin 40°)^2} \approx 7.55 \text{ N}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{3\sin 40°}{5 + 3\cos 40°} \approx 14.8°\) | M1 A1 A1 | (3 marks) Total: 8 |
## Question 2:
### Part (a) — EITHER method (Cosine Rule):
$F^2 = 5^2 + 3^2 - 2(5)(3)\cos 140°$ | M1 A1 | Vector triangle attempt, correct
$F \approx 7.55 \text{ N}$ | M1 A1 | (5 marks)
### Part (b):
$\frac{F}{\sin 140°} = \frac{3}{\sin \theta} \Rightarrow \theta \approx 14.8°$ | M1 A1 A1 | (3 marks) **Total: 8**
### Part (a) — OR method (Components):
$F^2 = (5 + 3\cos 40°)^2 + (3\sin 40°)^2$ | M1 A1 M1 A1 | (5 marks)
$F \approx 7.55 \text{ N}$
### Part (b):
$\tan\theta = \frac{3\sin 40°}{5 + 3\cos 40°}, \quad \theta \approx 14.8°$ | M1 A1 A1 | (3 marks) **Total: 8**
### Part (a) — OR method (Column vectors):
$\underline{P} = \begin{pmatrix}5\\0\end{pmatrix}$, $\underline{Q} = \begin{pmatrix}3\cos 40°\\3\sin 40°\end{pmatrix}$ | M1 |
$\Rightarrow \underline{F} = \begin{pmatrix}5+3\cos 40°\\3\sin 40°\end{pmatrix}$ | A1 |
$|\underline{F}| = \sqrt{(5+3\cos 40°)^2 + (3\sin 40°)^2} \approx 7.55 \text{ N}$ | M1 A1 | (5 marks)
### Part (b):
$\tan\theta = \frac{3\sin 40°}{5 + 3\cos 40°} \approx 14.8°$ | M1 A1 A1 | (3 marks) **Total: 8**
---2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{218383c1-0875-46f2-9416-8e827065a7a6-2_272_592_1239_648}
\end{center}
\end{figure}
Two forces $\mathbf { P }$ and $\mathbf { Q }$, act on a particle. The force $\mathbf { P }$ has magnitude 5 N and the force $\mathbf { Q }$ has magnitude 3 N . The angle between the directions of $\mathbf { P }$ and $\mathbf { Q }$ is $40 ^ { \circ }$, as shown in Fig. 1. The resultant of $\mathbf { P }$ and $\mathbf { Q }$ is $\mathbf { F }$.
\begin{enumerate}[label=(\alph*)]
\item Find, to 3 significant figures, the magnitude of $\mathbf { F }$.
\item Find, in degrees to 1 decimal place, the angle between the directions of $\mathbf { F }$ and $\mathbf { P }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2001 Q2 [8]}}