| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem using vector notation. Part (i) requires basic Pythagoras and trigonometry, part (ii) is direct recall of weight as a vector, and part (iii) involves resolving forces in equilibrium (sum to zero). All steps are standard M1 techniques with no problem-solving insight required, making it easier than average. |
| 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 |
|---|---|---|
| \(\sqrt{10^2 + 24^2} = 26\) so \(26\) N | B1 | |
| \(\arctan\left(\frac{10}{24}\right)\) | M1 | Using arctan or equivalent. Accept \(\arctan\left(\frac{24}{10}\right)\) or equivalent. Accept \(157.4°\) |
| \(= 22.619...\) so \(22.6°\) (3 s.f.) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{W} = -w\mathbf{j}\) | B1 [1] | Accept \(\begin{pmatrix}0\\-w\end{pmatrix}\) and \(\begin{pmatrix}0\\-w\mathbf{j}\end{pmatrix}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{T}_1 + \mathbf{T}_2 + \mathbf{W} = \mathbf{0}\) | M1 | Accept in any form and recovery from \(\mathbf{W} = w\mathbf{j}\). Award if not explicit and part (ii) and both \(k\) and \(w\) correct |
| \(k = -10\) | B1 | Accept from wrong working |
| \(w = 34\) | B1 [3] | Accept from wrong working but not \(-34\). [Accept \(-10\mathbf{i}\) or \(34\mathbf{j}\) but not both] |
## Question 4:
### Part (i)
$\sqrt{10^2 + 24^2} = 26$ so $26$ N | B1 |
$\arctan\left(\frac{10}{24}\right)$ | M1 | Using arctan or equivalent. Accept $\arctan\left(\frac{24}{10}\right)$ or equivalent. Accept $157.4°$
$= 22.619...$ so $22.6°$ (3 s.f.) | A1 [3] |
### Part (ii)
$\mathbf{W} = -w\mathbf{j}$ | B1 [1] | Accept $\begin{pmatrix}0\\-w\end{pmatrix}$ and $\begin{pmatrix}0\\-w\mathbf{j}\end{pmatrix}$
### Part (iii)
$\mathbf{T}_1 + \mathbf{T}_2 + \mathbf{W} = \mathbf{0}$ | M1 | Accept in any form and recovery from $\mathbf{W} = w\mathbf{j}$. Award if not explicit and part (ii) and **both** $k$ and $w$ correct
$k = -10$ | B1 | Accept from wrong working
$w = 34$ | B1 [3] | Accept from wrong working but not $-34$. [Accept $-10\mathbf{i}$ or $34\mathbf{j}$ but not both]
---4 A small box has weight $\mathbf { W } \mathrm { N }$ and is held in equilibrium by two strings with tensions $\mathbf { T } _ { 1 } \mathrm {~N}$ and $\mathbf { T } _ { 2 } \mathrm {~N}$. This situation is shown in Fig. 2 which also shows the standard unit vectors $\mathbf { i }$ and $\mathbf { j }$ that are horizontal and vertically upwards, respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b80eced6-2fea-4b95-9104-d13339643df0-2_252_631_414_803}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The tension $\mathbf { T } _ { 1 }$ is $10 \mathbf { i } + 24 \mathbf { j }$.\\
(i) Calculate the magnitude of $\mathbf { T } _ { 1 }$ and the angle between $\mathbf { T } _ { 1 }$ and the vertical.
The magnitude of the weight is $w \mathrm {~N}$.\\
(ii) Write down the vector $\mathbf { W }$ in terms of $w$ and $\mathbf { j }$.
The tension $\mathbf { T } _ { 2 }$ is $k \mathbf { i } + 10 \mathbf { j }$, where $k$ is a scalar.\\
(iii) Find the values of $k$ and of $w$.
\hfill \mbox{\textit{OCR MEI M1 Q4 [7]}}