| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| 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 straightforward application of the cosine rule and sine rule to find the magnitude and direction of a resultant force. The question involves routine two-step calculations with given values and no conceptual challenges—easier than the average A-level question which typically requires more problem-solving or integration of multiple concepts. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ | \mathbf{R}\ | ^2 = 8^2 + 5^2 - 2 \times 8 \times 5\cos130°\) |
| A1 | At most one error e.g. 50 in place of 130 | |
| A1 | Correct unsimplified | |
| \(\ | \mathbf{R}\ | = 11.9\) N (3 SF) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ | \mathbf{R}\ | ^2 = (5 + 8\cos50°)^2 + (8\sin50°)^2\) |
| \(= (10.14^2 + 6.13^2)\) | A1 | At most one error |
| A1 | Correct unsimplified | |
| \(\ | \mathbf{R}\ | = 11.9\) N (3 SF) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\sin\theta}{5} = \frac{\sin130}{11.85}\) | M1 | Independent M1. Use of sine rule or cosine rule with their \(\ |
| A1ft | Follow their \(\ | \mathbf{R}\ |
| \(\sin\theta = \frac{\sin130}{11.85}\) | DM1 | Solve for \(\theta\) |
| \(\theta = 19°\) | A1 | |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \frac{8\sin50°}{5 + 8\cos50°}\) | M1 | Independent M1. Correct use of trig to find direction of \(\mathbf{R}\), or use cosine rule to find \(\alpha\) |
| \((\alpha = 31.1...°)\) | A1ft | Correct unsimplified. Follow their components |
| \(\theta = 50° - \alpha\) | DM1 | Use their \(\alpha\) to solve for \(\theta\) |
| \(\theta = 19°\) | A1 | Alternatively, find \(\beta = 58.8...\) and use \(\theta = \beta - 40\) |
| (4) | ||
| [8] |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|\mathbf{R}\|^2 = 8^2 + 5^2 - 2 \times 8 \times 5\cos130°$ | M1 | Use of cosine rule |
| | A1 | At most one error e.g. 50 in place of 130 |
| | A1 | Correct unsimplified |
| $\|\mathbf{R}\| = 11.9$ N (3 SF) | A1 | 12 or better |
| | **(4)** | |
### Part (a) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\|\mathbf{R}\|^2 = (5 + 8\cos50°)^2 + (8\sin50°)^2$ | M1 | Use of Pythagoras (with usual rules for resolved components) |
| $= (10.14^2 + 6.13^2)$ | A1 | At most one error |
| | A1 | Correct unsimplified |
| $\|\mathbf{R}\| = 11.9$ N (3 SF) | A1 | |
| | **(4)** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\sin\theta}{5} = \frac{\sin130}{11.85}$ | M1 | Independent M1. Use of sine rule or cosine rule with their $\|\mathbf{R}\|$ |
| | A1ft | Follow their $\|\mathbf{R}\|$ |
| $\sin\theta = \frac{\sin130}{11.85}$ | DM1 | Solve for $\theta$ |
| $\theta = 19°$ | A1 | |
| | **(4)** | |
### Part (b) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \frac{8\sin50°}{5 + 8\cos50°}$ | M1 | Independent M1. Correct use of trig to find direction of $\mathbf{R}$, or use cosine rule to find $\alpha$ |
| $(\alpha = 31.1...°)$ | A1ft | Correct unsimplified. Follow their components |
| $\theta = 50° - \alpha$ | DM1 | Use their $\alpha$ to solve for $\theta$ |
| $\theta = 19°$ | A1 | Alternatively, find $\beta = 58.8...$ and use $\theta = \beta - 40$ |
| | **(4)** | |
| | **[8]** | |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5c3869c7-008f-4131-b68d-8ecdd4da3377-22_254_291_251_831}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Two forces, $\mathbf { P }$ and $\mathbf { Q }$, act on a particle. The force $\mathbf { P }$ has magnitude 8 N and the force $\mathbf { Q }$ has magnitude 5 N . The angle between the directions of $\mathbf { P }$ and $\mathbf { Q }$ is $50 ^ { \circ }$, as shown in Figure 3. The resultant of $\mathbf { P }$ and $\mathbf { Q }$ is the force $\mathbf { R }$.
\begin{enumerate}[label=(\alph*)]
\item Find, to 3 significant figures, the magnitude of $\mathbf { R }$.
\item Find, to the nearest degree, the size of the angle between the direction of $\mathbf { P }$ and the direction of $\mathbf { R }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2017 Q7 [8]}}