| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Forces as vectors |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic vector addition and the equilibrium condition. Part (i) involves simple component-wise addition and magnitude calculation using Pythagoras. Part (ii) applies the standard equilibrium principle (sum of forces = 0) with minimal algebraic manipulation. Both parts are routine textbook exercises with no problem-solving insight required. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03b Newton's first law: equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resultant is \(\begin{pmatrix}4\\1\\2\end{pmatrix} + \begin{pmatrix}-6\\2\\4\end{pmatrix} = \begin{pmatrix}-2\\3\\6\end{pmatrix}\) | M1 | Adding the vectors. Condone spurious notation. |
| Resultant vector \(\begin{pmatrix}-2\\3\\6\end{pmatrix}\) | A1 | Vector must be in proper form (penalise only once in the paper). Accept clear components. |
| Magnitude is \(\sqrt{(-2)^2+3^2+6^2} = \sqrt{49} = 7\) N | M1 | Pythagoras on their 3 component vector. Allow e.g. \(-2^2\) for \((-2)^2\) even if evaluated as \(-4\). |
| \(= 7\) N | F1 | FT their resultant. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{F} + 2\mathbf{G} + \mathbf{H} = \mathbf{0}\) | M1 | Either \(\mathbf{F} + 2\mathbf{G} + \mathbf{H} = \mathbf{0}\) or \(\mathbf{F} + 2\mathbf{G} = \mathbf{H}\) |
| So \(\mathbf{H} = -2\mathbf{G} - \mathbf{F} = -\begin{pmatrix}-12\\4\\8\end{pmatrix} - \begin{pmatrix}4\\1\\2\end{pmatrix}\) | A1 | Must see attempt at \(\mathbf{H} = -2\mathbf{G} - \mathbf{F}\) |
| \(= \begin{pmatrix}8\\-5\\-10\end{pmatrix}\) | A1 | cao. Vector must be in proper form (penalise only once in the paper). |
# Question 2:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resultant is $\begin{pmatrix}4\\1\\2\end{pmatrix} + \begin{pmatrix}-6\\2\\4\end{pmatrix} = \begin{pmatrix}-2\\3\\6\end{pmatrix}$ | M1 | Adding the vectors. Condone spurious notation. |
| Resultant vector $\begin{pmatrix}-2\\3\\6\end{pmatrix}$ | A1 | Vector must be in proper form (penalise only once in the paper). Accept clear components. |
| Magnitude is $\sqrt{(-2)^2+3^2+6^2} = \sqrt{49} = 7$ N | M1 | Pythagoras on **their** 3 component vector. Allow e.g. $-2^2$ for $(-2)^2$ even if evaluated as $-4$. |
| $= 7$ N | F1 | FT **their** resultant. |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{F} + 2\mathbf{G} + \mathbf{H} = \mathbf{0}$ | M1 | Either $\mathbf{F} + 2\mathbf{G} + \mathbf{H} = \mathbf{0}$ or $\mathbf{F} + 2\mathbf{G} = \mathbf{H}$ |
| So $\mathbf{H} = -2\mathbf{G} - \mathbf{F} = -\begin{pmatrix}-12\\4\\8\end{pmatrix} - \begin{pmatrix}4\\1\\2\end{pmatrix}$ | A1 | Must see attempt at $\mathbf{H} = -2\mathbf{G} - \mathbf{F}$ |
| $= \begin{pmatrix}8\\-5\\-10\end{pmatrix}$ | A1 | cao. Vector must be in proper form (penalise only once in the paper). |
---2 Force $\mathbf { F }$ is $\left( \begin{array} { l } 4 \\ 1 \\ 2 \end{array} \right) \mathrm { N }$ and force $\mathbf { G }$ is $\left( \begin{array} { r } - 6 \\ 2 \\ 4 \end{array} \right) \mathrm { N }$.\\
(i) Find the resultant of $\mathbf { F }$ and $\mathbf { G }$ and calculate its magnitude.\\
(ii) Forces $\mathbf { F }$, $2 \mathbf { G }$ and $\mathbf { H }$ act on a particle which is in equilibrium. Find $\mathbf { H }$.
\hfill \mbox{\textit{OCR MEI M1 Q2 [7]}}