| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Range of equilibrium positions |
| Difficulty | Standard +0.3 This is a straightforward mechanics question testing basic moments and equilibrium concepts. Parts (a) and (b) are routine recall/application, part (c) is a simple algebraic expression, and part (d) requires setting up moment equations but follows standard textbook methods with no novel insight required. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The bricks have negligible size so contact force with the plank acts at a point | B1 | Allow "no size" or "size doesn't matter" or "shape is not relevant"; Allow "weight of bricks acts on the plank at point" |
| The mass of plank is evenly spread across its length / the weight of plank acts at centre of plank | B1 | Allow for either statement; Allow the plank is the same throughout or centre of mass at centre; Do not allow "mass acts" at the centre |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When placed at centre, tensions are equal: \(2 \times 75 = 2.3ng + 5g\) | M1 | Using symmetry to establish equation for \(n\) soi; Allow if weight of plank or one tensions missing; Trial and improvement may be used |
| \(n = \left\lfloor\dfrac{101}{2.3g}\right\rfloor = 4.48...\) so 4 bricks | A1 | Final answer must be integer; Allow if 4 seen www |
| Alternative: \(5g \times 0.4 + 2.3gn \times 0.4 = 75 \times 0.8\) | M1 | Allow for missing moment of weight or one of tensions; Every term must be a moment |
| \(n = \left\lfloor\dfrac{40.4}{9.016}\right\rfloor = 4.48...\) so 4 bricks | A1 | Final answer must be integer; Allow if 4 seen www |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2.3gnx\) | B1 | Allow positive or negative \(22.54nx\); Allow in an equation |
| Nm | B1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 4 bricks on the point of breaking, \(x\) m from A; Taking moments about A: \(5g \times 0.4 + 4 \times 2.3gx = 75 \times 0.8\) | M1 | Taking moments about any point to form an equation; FT their \(n\); All forces used in a moment; Allow sign errors; Allow incorrect distance used; Could be an inequality |
| \(9.2gx = 60 - 2g\) | A1 | Fully correct equation FT their \(n\); Allow corresponding inequality; Need not be simplified |
| \(x = 0.448\) [so 44.8 cm from A] | A1 | cao |
| [3] |
# Question 6:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The bricks have negligible size so contact force with the plank acts at a point | B1 | Allow "no size" or "size doesn't matter" or "shape is not relevant"; Allow "weight of bricks acts on the plank at point" |
| The mass of plank is evenly spread across its length / the weight of plank acts at centre of plank | B1 | Allow for either statement; Allow the plank is the same throughout or centre of mass at centre; Do not allow "mass acts" at the centre |
| **[2]** | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When placed at centre, tensions are equal: $2 \times 75 = 2.3ng + 5g$ | M1 | Using symmetry to establish equation for $n$ soi; Allow if weight of plank or one tensions missing; Trial and improvement may be used |
| $n = \left\lfloor\dfrac{101}{2.3g}\right\rfloor = 4.48...$ so 4 bricks | A1 | Final answer must be integer; Allow if 4 seen www |
| Alternative: $5g \times 0.4 + 2.3gn \times 0.4 = 75 \times 0.8$ | M1 | Allow for missing moment of weight or one of tensions; Every term must be a moment |
| $n = \left\lfloor\dfrac{40.4}{9.016}\right\rfloor = 4.48...$ so 4 bricks | A1 | Final answer must be integer; Allow if 4 seen www |
| **[2]** | | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.3gnx$ | B1 | Allow positive or negative $22.54nx$; Allow in an equation |
| Nm | B1 | |
| **[2]** | | |
## Part (d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 4 bricks on the point of breaking, $x$ m from A; Taking moments about A: $5g \times 0.4 + 4 \times 2.3gx = 75 \times 0.8$ | M1 | Taking moments about any point to form an equation; FT their $n$; All forces used in a moment; Allow sign errors; Allow incorrect distance used; Could be an inequality |
| $9.2gx = 60 - 2g$ | A1 | Fully correct equation FT their $n$; Allow corresponding inequality; Need not be simplified |
| $x = 0.448$ [so 44.8 cm from A] | A1 | cao |
| **[3]** | | |
---6 A shelf consists of a horizontal uniform plank AB of length 0.8 m and mass 5 kg with light inextensible vertical strings attached at each end. A stack of bricks each of mass 2.3 kg is placed on the plank as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{9dd6fc6d-b51e-4a73-ace5-d26a7558032c-06_397_734_641_242}
\begin{enumerate}[label=(\alph*)]
\item Explain the meaning of each of the following modelling assumptions.
\begin{itemize}
\item The stack of bricks is modelled as a particle.
\item The plank is modelled as uniform.
\end{itemize}
Either of the strings will break if the tension exceeds 75 N.
\item Find the greatest number of bricks that can be placed at the centre of the plank without breaking the strings.
\item Find an expression for the moment about A of the weight of a stack of $n$ bricks when the stack is at a distance of $x \mathrm {~m}$ from A . State the units for your answer.
\item Calculate the greatest distance from A that the largest stack of bricks can be placed without a string breaking.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q6 [9]}}