| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| 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 mechanics question applying the cosine rule to find an unknown force magnitude and then the sine rule for an angle. It requires straightforward application of triangle laws with given numerical values, making it slightly easier than average A-level difficulty but still requiring proper method and calculation accuracy. |
| Spec | 1.05c Area of triangle: using 1/2 ab sin(C)3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using cosine rule: \(129 = X^2 + (5\sqrt{3})^2 - 2X\times5\sqrt{3}\cos30°\) | M1 A1 | Complete method for equation in \(X\) only |
| OR using sine rule to find \(\alpha\): \(\frac{\sqrt{129}}{\sin30°}=\frac{5\sqrt{3}}{\sin\alpha}\) then further sine/cosine rule application | ||
| OR using components: \(\sqrt{129}=\sqrt{(X\cos30°-5\sqrt{3})^2+(X\sin30°)^2}\) | ||
| Solves equation (if quadratic, must include \(X\) term) | M1 | |
| \(X = 18\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER finds equation in \(\beta\) only using their \(X\): \(\frac{18}{\sin\beta}=\frac{\sqrt{129}}{\sin30°}\) | M2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Complete method to form an equation in \(\beta\) only | M2 | Using cosine rule and/or sine rule or components |
| Correct equation e.g. \(18^2 = (5\sqrt{3})^2 + 129 - 2 \times 5\sqrt{3} \times \sqrt{129}\cos\beta\) | A1 | |
| \(\beta = 128°\) to nearest degree | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Complete method to find a relevant angle | M1 | |
| Correct relevant angle: \(\alpha = \sin^{-1}\left(\frac{\sin 30° \times 5\sqrt{3}}{\sqrt{129}}\right) = 22.4109°\) or \(37.589°\) or \(52.411°\) | A1 | |
| Completes method to find required angle e.g. \(150°-\alpha\), \(210°+\alpha\), \(180°-52.411°\), \(90°+37.859°\) | M1 | |
| \(\beta = 128°\) to nearest degree, accept \(232°\) | A1 | cao |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using cosine rule: $129 = X^2 + (5\sqrt{3})^2 - 2X\times5\sqrt{3}\cos30°$ | M1 A1 | Complete method for equation in $X$ only |
| OR using sine rule to find $\alpha$: $\frac{\sqrt{129}}{\sin30°}=\frac{5\sqrt{3}}{\sin\alpha}$ then further sine/cosine rule application | | |
| OR using components: $\sqrt{129}=\sqrt{(X\cos30°-5\sqrt{3})^2+(X\sin30°)^2}$ | | |
| Solves equation (if quadratic, must include $X$ term) | M1 | |
| $X = 18$ | A1 | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER** finds equation in $\beta$ only using their $X$: $\frac{18}{\sin\beta}=\frac{\sqrt{129}}{\sin30°}$ | M2 | |
## Question 4(ii):
**EITHER method:**
| Working | Mark | Guidance |
|---------|------|----------|
| Complete method to form an equation in $\beta$ only | M2 | Using cosine rule and/or sine rule or components |
| Correct equation e.g. $18^2 = (5\sqrt{3})^2 + 129 - 2 \times 5\sqrt{3} \times \sqrt{129}\cos\beta$ | A1 | |
| $\beta = 128°$ to nearest degree | A1 | cao |
**OR method:**
| Working | Mark | Guidance |
|---------|------|----------|
| Complete method to find a relevant angle | M1 | |
| Correct relevant angle: $\alpha = \sin^{-1}\left(\frac{\sin 30° \times 5\sqrt{3}}{\sqrt{129}}\right) = 22.4109°$ or $37.589°$ or $52.411°$ | A1 | |
| Completes method to find required angle e.g. $150°-\alpha$, $210°+\alpha$, $180°-52.411°$, $90°+37.859°$ | M1 | |
| $\beta = 128°$ to nearest degree, accept $232°$ | A1 | cao |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e59a66b8-c2ad-41fd-9959-9d21e9455c37-08_399_889_246_587}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows two horizontal forces $\mathbf { P }$ and $\mathbf { Q }$ acting on a particle.\\
The angle between the direction of $\mathbf { P }$ and the direction of $\mathbf { Q }$ is $150 ^ { \circ }$\\
Force $\mathbf { P }$ has magnitude $X$ newtons.\\
Force $\mathbf { Q }$ has magnitude $5 \sqrt { 3 } \mathrm {~N}$.\\
The resultant of $\mathbf { P }$ and $\mathbf { Q }$ has magnitude $\sqrt { 129 } \mathrm {~N}$.\\
Find\\
(i) the value of $X$.\\
(ii) the angle between $\mathbf { Q }$ and the resultant, giving your answer to the nearest degree.
\hfill \mbox{\textit{Edexcel M1 2024 Q4 [8]}}