| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of three coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question requiring resolution of forces into components (using bearing 240° and South), vector addition, then finding magnitude using Pythagoras and direction using inverse tan. It's slightly easier than average A-level because it's a routine two-force resultant problem with straightforward trigonometry, though bearing conversions add minor complexity compared to pure Cartesian problems. |
| Spec | 1.10c Magnitude and direction: of vectors3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{R} = \mathbf{F} + \mathbf{G}\); \(R^2 = 8^2 + 10^2 - 2 \times 8 \times 10\cos 120°\) (or 244) | M1A1 | M1 for equation in \(R\) only. M0 for \(R^2 = 8^2 + 10^2 - 2\times8\times10\cos60°\) or clear misquote of cosine rule. Condone sin/cos confusion and sign errors |
| \(R = \sqrt{244} = 15.620499...\) N | A1 | \(\sqrt{244}\) or 16 or better (N) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{\sin\alpha}{8} = \frac{\sin 120°}{\sqrt{244}}\) or \(\frac{\sin\beta}{10} = \frac{\sin 120°}{\sqrt{244}}\) (allow sin 60°) | M1A1 | M1 for equation in relevant angle only using their \(R\). For SOHCAHTOA alternatives allow sin/cos confusion and sign errors |
| \(\alpha = 26.3°\) OR \(\beta = 33.67°\) (accept 34) | A1 | Correct angle to nearest degree |
| Bearing is \(206°\) (nearest degree) | A1 | cao |
# Question 5:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{R} = \mathbf{F} + \mathbf{G}$; $R^2 = 8^2 + 10^2 - 2 \times 8 \times 10\cos 120°$ (or 244) | M1A1 | M1 for equation in $R$ only. M0 for $R^2 = 8^2 + 10^2 - 2\times8\times10\cos60°$ or clear misquote of cosine rule. Condone sin/cos confusion and sign errors |
| $R = \sqrt{244} = 15.620499...$ N | A1 | $\sqrt{244}$ or 16 or better (N) |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\sin\alpha}{8} = \frac{\sin 120°}{\sqrt{244}}$ or $\frac{\sin\beta}{10} = \frac{\sin 120°}{\sqrt{244}}$ (allow sin 60°) | M1A1 | M1 for equation in relevant angle only using their $R$. For SOHCAHTOA alternatives allow sin/cos confusion and sign errors |
| $\alpha = 26.3°$ **OR** $\beta = 33.67°$ (accept 34) | A1 | Correct angle to nearest degree |
| Bearing is $206°$ (nearest degree) | A1 | cao |5. A particle is acted upon by two forces $\mathbf { F }$ and $\mathbf { G }$. The force $\mathbf { F }$ has magnitude 8 N and acts in a direction with a bearing of $240 ^ { \circ }$. The force $\mathbf { G }$ has magnitude 10 N and acts due South.
Given that $\mathbf { R } = \mathbf { F } + \mathbf { G }$, find\\
(i) the magnitude of $\mathbf { R }$,\\
(ii) the direction of $\mathbf { R }$, giving your answer as a bearing to the nearest degree. in a direction with a bearing of $240 ^ { \circ }$. The force $\mathbf { G }$ has magnitude 10 N and acts due South. Given that $\mathbf { R } = \mathbf { F } + \mathbf { G }$, find
\hfill \mbox{\textit{Edexcel M1 2021 Q5 [7]}}