| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: equilibrium (find unknowns) |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem requiring only the basic principle that forces sum to zero, followed by routine magnitude and angle calculations. It involves simple arithmetic with vectors and standard trigonometry, making it easier than average for A-level mechanics questions which typically involve more problem-solving or multi-step reasoning. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.03b Newton's first law: equilibrium3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{R} + \begin{pmatrix}-3\\4\end{pmatrix} + \begin{pmatrix}21\\-7\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\) | M1 | Sum to zero |
| \(\mathbf{R} = \begin{pmatrix}-18\\3\end{pmatrix}\) | A1 | Award if seen here or in (ii) or used in (ii). [SC1 for \(\begin{pmatrix}18\\-3\end{pmatrix}\)] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | \mathbf{R} | = \sqrt{18^2 + 3^2}\) |
| \(= 18.248...\) so \(18.2\) N (3 s.f.) | A1 | Any reasonable accuracy. FT R with 2 non-zero components |
| angle is \(180 - \arctan\left(\frac{3}{18}\right) = 170.53...°\) | M1 | Allow \(\arctan\left(\frac{\pm3}{\pm18}\right)\) or \(\arctan\left(\frac{\pm18}{\pm3}\right)\) |
| so \(171°\) (3 s.f.) | A1 | Any reasonable accuracy. FT R provided angle is obtuse but not \(180°\) |
## Question 3:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{R} + \begin{pmatrix}-3\\4\end{pmatrix} + \begin{pmatrix}21\\-7\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}$ | M1 | Sum to zero |
| $\mathbf{R} = \begin{pmatrix}-18\\3\end{pmatrix}$ | A1 | Award if seen here or in (ii) or used in (ii). [SC1 for $\begin{pmatrix}18\\-3\end{pmatrix}$] |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|\mathbf{R}| = \sqrt{18^2 + 3^2}$ | M1 | Use of Pythagoras |
| $= 18.248...$ so $18.2$ N (3 s.f.) | A1 | Any reasonable accuracy. FT **R** with 2 non-zero components |
| angle is $180 - \arctan\left(\frac{3}{18}\right) = 170.53...°$ | M1 | Allow $\arctan\left(\frac{\pm3}{\pm18}\right)$ or $\arctan\left(\frac{\pm18}{\pm3}\right)$ |
| so $171°$ (3 s.f.) | A1 | Any reasonable accuracy. FT **R** provided angle is obtuse but not $180°$ |
---3 A particle rests on a smooth, horizontal plane. Horizontal unit vectors $\mathbf { i }$ and $\mathbf { j }$ lie in this plane. The particle is in equilibrium under the action of the three forces $( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { N }$ and $( 21 \mathbf { i } - 7 \mathbf { j } ) \mathrm { N }$ and $\mathbf { R N }$.\\
(i) Write down an expression for $\mathbf { R }$ in terms of $\mathbf { i }$ and $\mathbf { j }$.\\
(ii) Find the magnitude of $\mathbf { R }$ and the angle between $\mathbf { R }$ and the $\mathbf { i }$ direction.
\hfill \mbox{\textit{OCR MEI M1 2005 Q3 [6]}}