| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | January |
| 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 -0.8 This is a straightforward M1 vectors question requiring basic magnitude/angle calculations and simple vector addition with algebraic coefficients. Part (i) uses Pythagoras and inverse tan (standard 2-mark work), while part (ii) involves equating components of 4F + G = H to find constants a and b—routine manipulation with no problem-solving insight needed. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03a Force: vector nature and diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \( | \mathbf{F} | = \sqrt{(-1)^2 + 5^2}\) |
| \(= \sqrt{26} = 5.0990... = 5.10\) (3 s.f.) | A1 | |
| Angle with j is \(\arctan(0.2)\) | M1 | Accept \(\arctan(p)\) where \(p = \pm 0.2\) or \(\pm 5\) o.e. |
| so \(11.309...\) so \(11.3°\) (3 s.f.) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix}-2\\3b\end{pmatrix} = 4\begin{pmatrix}-1\\5\end{pmatrix} + \begin{pmatrix}2a\\a\end{pmatrix}\) | M1 | \(\mathbf{H} = 4\mathbf{F} + \mathbf{G}\) soi |
| M1 | Formulating at least 1 scalar equation from their vector equation soi | |
| \(a = 1, b = 7\) | A1 | \(a\) correct or G follows from their wrong \(a\) |
| so \(\mathbf{G} = \begin{pmatrix}2\\1\end{pmatrix}\) and \(\mathbf{H} = \begin{pmatrix}-2\\21\end{pmatrix}\) or \(\mathbf{G} = 2\mathbf{i} + \mathbf{j}\) and \(\mathbf{H} = -2\mathbf{i} + 21\mathbf{j}\) | A1 | H cao |
# Question 3:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $|\mathbf{F}| = \sqrt{(-1)^2 + 5^2}$ | M1 | Accept $\sqrt{-1^2 + 5^2}$ even if taken to be $\sqrt{24}$ |
| $= \sqrt{26} = 5.0990... = 5.10$ (3 s.f.) | A1 | |
| Angle with **j** is $\arctan(0.2)$ | M1 | Accept $\arctan(p)$ where $p = \pm 0.2$ or $\pm 5$ o.e. |
| so $11.309...$ so $11.3°$ (3 s.f.) | A1 | cao |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}-2\\3b\end{pmatrix} = 4\begin{pmatrix}-1\\5\end{pmatrix} + \begin{pmatrix}2a\\a\end{pmatrix}$ | M1 | $\mathbf{H} = 4\mathbf{F} + \mathbf{G}$ soi |
| | M1 | Formulating at least 1 scalar equation from their vector equation soi |
| $a = 1, b = 7$ | A1 | $a$ correct **or** **G** follows from their wrong $a$ |
| so $\mathbf{G} = \begin{pmatrix}2\\1\end{pmatrix}$ and $\mathbf{H} = \begin{pmatrix}-2\\21\end{pmatrix}$ or $\mathbf{G} = 2\mathbf{i} + \mathbf{j}$ and $\mathbf{H} = -2\mathbf{i} + 21\mathbf{j}$ | A1 | **H** cao |
---3 In this question, $\mathbf { i }$ is a horizontal unit vector and $\mathbf { j }$ is a unit vector pointing vertically upwards.\\
A force $\mathbf { F }$ is $- \mathbf { i } + 5 \mathbf { j }$.\\
(i) Calculate the magnitude of $\mathbf { F }$.
Calculate also the angle between $\mathbf { F }$ and the upward vertical.
Force $\mathbf { G }$ is $2 a \mathbf { i } + a \mathbf { j }$ and force $\mathbf { H }$ is $- 2 \mathbf { i } + 3 b \mathbf { j }$, where $a$ and $b$ are constants. The force $\mathbf { H }$ is the resultant of forces $4 \mathbf { F }$ and $\mathbf { G }$.\\
(ii) Find $\mathbf { G }$ and $\mathbf { H }$.
\hfill \mbox{\textit{OCR MEI M1 2010 Q3 [8]}}