| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Newton's second law with vector forces (find acceleration or force) |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question requiring standard vector operations (magnitude using Pythagoras, direction using arctangent, and Newton's second law F=ma). All parts are routine textbook exercises with no problem-solving insight needed, making it slightly easier than average A-level difficulty. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt{(-6)^2 + 13^2} = 14.31782...\) so \(14.3\) N (3 s.f.) | M1 | Accept \(\sqrt{-6^2 + 13^2}\) |
| A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Resultant is \(\begin{pmatrix}-6\\13\end{pmatrix} - \begin{pmatrix}-3\\5\end{pmatrix} = \begin{pmatrix}-3\\8\end{pmatrix}\) | B1 | May not be explicit. If diagram used it must have correct orientation. Give if final angle correct |
| Require \(270 + \arctan\dfrac{8}{3}\) | M1 | Use of \(\arctan\left(\pm\frac{8}{3}\right)\) or \(\arctan\left(\pm\frac{3}{8}\right)\) (\(\pm 20.6°\) or \(\pm 69.4°\)) or equivalent on their resultant |
| so \(339.4439...°\) so \(339°\) | A1 [3] | cao. Do not accept \(-21°\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}-3\\5\end{pmatrix} = 5\mathbf{a}\) | M1 | Use of N2L with acceleration *used* in vector form |
| so \((-0.6\mathbf{i} + \mathbf{j})\) m s\(^{-2}\) | A1 | Any form. Units not required. isw |
| change in velocity is \((-6\mathbf{i} + 10\mathbf{j})\) m s\(^{-1}\) | F1 [3] | \(10\mathbf{a}\) seen. Units not required. Must be a vector. [SC1 for \(a = \sqrt{3^2+5^2}/5 = 1.17\)] |
## Question 6:
### Part (i)
$\sqrt{(-6)^2 + 13^2} = 14.31782...$ so $14.3$ N (3 s.f.) | M1 | Accept $\sqrt{-6^2 + 13^2}$
| A1 [2] |
### Part (ii)
Resultant is $\begin{pmatrix}-6\\13\end{pmatrix} - \begin{pmatrix}-3\\5\end{pmatrix} = \begin{pmatrix}-3\\8\end{pmatrix}$ | B1 | May not be explicit. If diagram used it must have correct orientation. Give if final angle correct
Require $270 + \arctan\dfrac{8}{3}$ | M1 | Use of $\arctan\left(\pm\frac{8}{3}\right)$ or $\arctan\left(\pm\frac{3}{8}\right)$ ($\pm 20.6°$ or $\pm 69.4°$) or equivalent on **their** resultant
so $339.4439...°$ so $339°$ | A1 [3] | cao. Do not accept $-21°$
### Part (iii)
$\begin{pmatrix}-3\\5\end{pmatrix} = 5\mathbf{a}$ | M1 | Use of N2L with acceleration *used* in vector form
so $(-0.6\mathbf{i} + \mathbf{j})$ m s$^{-2}$ | A1 | Any form. Units not required. isw
change in velocity is $(-6\mathbf{i} + 10\mathbf{j})$ m s$^{-1}$ | F1 [3] | $10\mathbf{a}$ seen. Units not required. Must be a vector. [SC1 for $a = \sqrt{3^2+5^2}/5 = 1.17$]6 Force $\mathbf { F } _ { 1 }$ is $\binom { 6 } { 13 } \mathrm {~N}$ and force $\mathbf { F } _ { 2 }$ is $\binom { 3 } { 5 }$, where $\left. \int _ { 0 } \right] _ { \text {and } } \binom { 0 } { 1 }$ are vectors east and north respectively.\\
(i) Calculate the magnitude of $\mathbf { F } _ { 1 }$, correct to three significant figures.\\
(ii) Calculate the direction of the force $\mathbf { F } _ { 1 } - \mathbf { F } _ { 2 }$ as a bearing.
Force $\mathbf { F } _ { 2 }$ is the resultant of all the forces acting on an object of mass 5 kg .\\
(iii) Calculate the acceleration of the object and the change in its velocity after 10 seconds.
\hfill \mbox{\textit{OCR MEI M1 Q6 [8]}}