| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | January |
| 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 routine two-part calculation with no problem-solving insight needed, making it easier than average but not trivial since students must correctly apply trigonometry with the supplementary angle (60°). |
| Spec | 1.10c Magnitude and direction: of vectors3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Horizontal component \(= 6 - 7\cos 60°\) (N) | M1A1 | First M1 for attempt, allow sin/cos confusion, to find component parallel to P. First A1 for a correct expression. |
| Vertical component (N) \(= 7\cos 30°\) | M1A1 | Second M1 for attempt, allow sin/cos confusion, to find component perp to P. First A1 for a correct expression. |
| Use Pythagoras: \(\sqrt{2.5^2 + 6.06^2} = \sqrt{43} = 6.6\) (N) or better | M1A1 | Third M1 for using Pythagoras to find magnitude of R. Third A1 for \(\sqrt{43}\), 6.6 (N) or better. |
| Use trig: angle \(= \tan^{-1}\!\left(\dfrac{7\cos 30°}{2.5}\right) = 68°\) (below P) or better. Also allow \(112°\), \(292°\) or \(248°\) | M1A1 | Fourth M1 for complete method to find angle (M0 if 6 used for 'horiz' cpt). Fourth A1 for \(68°\) or better (67.589089...) \(112°\) or \(292°\) or \(248°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Cosine rule to find \( | \mathbf{R} | \): \(R^2 = 36 + 49 - 2 \times 6 \times 7 \times \cos 60°\ (= 43)\) |
| \(R = 6.6\) (N) or better | M1A1 | Third M1 for solving for \(R\). Third A1 for \(\sqrt{43}\), 6.6 (N) or better. |
| Sine rule for \(\theta\): \(\sin^{-1}\!\left(\dfrac{7\sin 60°}{R}\right) = 68°\) or better. Also allow \(112°\) or \(292°\) or \(248°\) | M1A1 | Fourth M1 for complete method (e.g. sine rule) to find angle between their R and P. Fourth A1 for \(68°\) or better. |
# Question 3:
## Method 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Horizontal component $= 6 - 7\cos 60°$ (N) | M1A1 | First M1 for attempt, allow sin/cos confusion, to find component parallel to **P**. First A1 for a correct expression. |
| Vertical component (N) $= 7\cos 30°$ | M1A1 | Second M1 for attempt, allow sin/cos confusion, to find component perp to **P**. First A1 for a correct expression. |
| Use Pythagoras: $\sqrt{2.5^2 + 6.06^2} = \sqrt{43} = 6.6$ (N) or better | M1A1 | Third M1 for using Pythagoras to find magnitude of **R**. Third A1 for $\sqrt{43}$, 6.6 (N) or better. |
| Use trig: angle $= \tan^{-1}\!\left(\dfrac{7\cos 30°}{2.5}\right) = 68°$ (below **P**) or better. Also allow $112°$, $292°$ or $248°$ | M1A1 | Fourth M1 for complete method to find angle (M0 if 6 used for 'horiz' cpt). Fourth A1 for $68°$ or better (67.589089...) $112°$ or $292°$ or $248°$ |
## Alternative Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Cosine rule to find $|\mathbf{R}|$: $R^2 = 36 + 49 - 2 \times 6 \times 7 \times \cos 60°\ (= 43)$ | M2A2 | First M2 for use of cosine rule with correct structure but allow $\cos 120°$ and allow $R^2$. First A2 for a correct equation (A0 if $120°$ used). |
| $R = 6.6$ (N) or better | M1A1 | Third M1 for solving for $R$. Third A1 for $\sqrt{43}$, 6.6 (N) or better. |
| Sine rule for $\theta$: $\sin^{-1}\!\left(\dfrac{7\sin 60°}{R}\right) = 68°$ or better. Also allow $112°$ or $292°$ or $248°$ | M1A1 | Fourth M1 for complete method (e.g. sine rule) to find angle between their **R** and **P**. Fourth A1 for $68°$ or better. |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ba698f74-a51c-409a-a9d9-e9080fc87be2-05_520_730_264_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Two forces $\mathbf { P }$ and $\mathbf { Q }$ act on a particle at a point $O$. Force $\mathbf { P }$ has magnitude 6 N and force $\mathbf { Q }$ has magnitude 7 N . The angle between the line of action of $\mathbf { P }$ and the line of action of $\mathbf { Q }$ is $120 ^ { \circ }$, as shown in Figure 1.
The resultant of $\mathbf { P }$ and $\mathbf { Q }$ is $\mathbf { R }$.
Find\\
(i) the magnitude of $\mathbf { R }$,\\
(ii) the angle between the line of action of $\mathbf { R }$ and the line of action of $\mathbf { P }$.\\
\hfill \mbox{\textit{Edexcel M1 2017 Q3 [8]}}