| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.3 This is a straightforward application of Newton's second law in vector form (F = ma) requiring basic vector arithmetic and angle calculation using dot product. The two-part structure and standard techniques make it slightly easier than average, though it does require competent handling of vectors. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{F} + \begin{pmatrix}-4\\8\end{pmatrix} = 6\begin{pmatrix}2\\3\end{pmatrix}\) | M1 | N2L. \(F = ma\). All forces present |
| B1 | Addition to get resultant. May be implied | |
| B1 | For \(\mathbf{F} \pm \begin{pmatrix}-4\\8\end{pmatrix} = 6\begin{pmatrix}2\\3\end{pmatrix}\) | |
| \(\mathbf{F} = \begin{pmatrix}16\\10\end{pmatrix}\) | A1 | SC4 for \(\mathbf{F} = \begin{pmatrix}16\\10\end{pmatrix}\) WW. If magnitude is given, final mark is lost unless vector answer is clearly intended |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\arctan\left(\frac{16}{10}\right)\) | M1 | Accept equivalent and FT their F only. Do not accept wrong angle. Accept \(360 - \arctan\left(\frac{16}{10}\right)\) |
| \(57.994...\) so \(58.0°\) (3 s.f.) | A1 | cao. Accept \(302°\) (3 s.f.) |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{F} + \begin{pmatrix}-4\\8\end{pmatrix} = 6\begin{pmatrix}2\\3\end{pmatrix}$ | M1 | N2L. $F = ma$. All forces present |
| | B1 | Addition to get resultant. May be implied |
| | B1 | For $\mathbf{F} \pm \begin{pmatrix}-4\\8\end{pmatrix} = 6\begin{pmatrix}2\\3\end{pmatrix}$ |
| $\mathbf{F} = \begin{pmatrix}16\\10\end{pmatrix}$ | A1 | SC4 for $\mathbf{F} = \begin{pmatrix}16\\10\end{pmatrix}$ WW. If magnitude is given, final mark is lost unless vector answer is clearly intended |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\arctan\left(\frac{16}{10}\right)$ | M1 | Accept equivalent and FT their **F** only. Do not accept wrong angle. Accept $360 - \arctan\left(\frac{16}{10}\right)$ |
| $57.994...$ so $58.0°$ (3 s.f.) | A1 | cao. Accept $302°$ (3 s.f.) |
---5 The resultant of the force $\binom { - 4 } { 8 } \mathrm {~N}$ and the force $\mathbf { F }$ gives an object of mass 6 kg an acceleration of $\binom { 2 } { 3 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Calculate $\mathbf { F }$.\\
(ii) Calculate the angle between $\mathbf { F }$ and the vector $\binom { 0 } { 1 }$.
\hfill \mbox{\textit{OCR MEI M1 Q5 [6]}}