| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: equilibrium (find unknowns) |
| Difficulty | Easy -1.2 This is a straightforward vector addition question requiring only basic operations: adding vectors component-wise, calculating magnitude using Pythagoras in 3D, and applying equilibrium condition (forces sum to zero). These are routine M1 skills with no problem-solving or conceptual challenge beyond direct application of standard formulas. |
| Spec | 1.10d Vector operations: addition and scalar multiplication3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Resultant is \(\begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} -6 \\ 4 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}\) | M1 | Adding the vectors. Condone spurious notation. |
| Magnitude is \(\sqrt{(-2)^2 + 3^2 + 6^2} = \sqrt{49} = 7\) N | M1, F1 | Vector must be in proper form (penalise only once in the paper). Accept clear components. Pythagoras on their 3 component vector. Allow e.g. \(2^2\) for \((-2)^2\) even if evaluated as \(-4\). FT their resultant. |
| Sub: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{F} + 2\mathbf{G} + \mathbf{H} = 0\) | M1 | Either \(\mathbf{F} + 2\mathbf{G} + \mathbf{H} = 0\) or \(\mathbf{F} + 2\mathbf{G} = \mathbf{H}\) |
| So \(\mathbf{H} = -2\mathbf{G} - \mathbf{F} = -\begin{pmatrix} -12 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix}\) | 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). |
| Sub: 3 |
### (i)
Resultant is $\begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} -6 \\ 4 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \end{pmatrix}$ | M1 | Adding the vectors. Condone spurious notation.
Magnitude is $\sqrt{(-2)^2 + 3^2 + 6^2} = \sqrt{49} = 7$ N | M1, F1 | Vector must be in proper form (penalise only once in the paper). Accept clear components. Pythagoras on their 3 component vector. Allow e.g. $2^2$ for $(-2)^2$ even if evaluated as $-4$. FT their resultant.
| Sub: 4
### (ii)
$\mathbf{F} + 2\mathbf{G} + \mathbf{H} = 0$ | M1 | Either $\mathbf{F} + 2\mathbf{G} + \mathbf{H} = 0$ or $\mathbf{F} + 2\mathbf{G} = \mathbf{H}$
So $\mathbf{H} = -2\mathbf{G} - \mathbf{F} = -\begin{pmatrix} -12 \\ 4 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ | 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).
| Sub: 3
---Force $\mathbf{F}$ is $\begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}$ N and force $\mathbf{G}$ is $\begin{pmatrix} -6 \\ 2 \\ 4 \end{pmatrix}$ N.
\begin{enumerate}[label=(\roman*)]
\item Find the resultant of $\mathbf{F}$ and $\mathbf{G}$ and calculate its magnitude. [4]
\item Forces $\mathbf{F}$, $2\mathbf{G}$ and $\mathbf{H}$ act on a particle which is in equilibrium. Find $\mathbf{H}$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 2008 Q4 [7]}}