| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Resultant force on lamina |
| Difficulty | Moderate -0.3 This is a straightforward moments problem requiring students to take moments about the hinge P to find the horizontal force F, then describe qualitative behavior. The setup is standard (uniform lamina, single hinge, one unknown force), requiring only basic moment equilibrium and understanding that removing the force causes rotation. Slightly easier than average due to the simple geometry and minimal problem-solving beyond applying the standard moments formula. |
| Spec | 3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Weight acts through the centre of the ruler, 15cm from P | B1 | Uses uniformity of the ruler; distance shown on diagram or 15 used in calculation |
| Take moments about P: \(0.02g \times 15 = F \times 4\) | M1 | Taking moments about any suitable point with all relevant forces and distances attempted. Must include weight (not mass only) term. Allow one incorrect distance. Allow 20g for weight or error converting mass to kg |
| \(F = \frac{0.02g \times 15}{4} = \frac{3g}{40} = 0.735\) | A1 | Allow \(0.075g\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Without \(F\) the ruler would rotate clockwise [about P] | E1 | Must indicate the turning effect of the weight and the direction. "Clockwise" implies rotation |
## Question 4:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Weight acts through the centre of the ruler, 15cm from P | B1 | Uses uniformity of the ruler; distance shown on diagram or 15 used in calculation |
| Take moments about P: $0.02g \times 15 = F \times 4$ | M1 | Taking moments about any suitable point with all relevant forces and distances attempted. Must include weight (not mass only) term. Allow one incorrect distance. Allow 20g for weight or error converting mass to kg |
| $F = \frac{0.02g \times 15}{4} = \frac{3g}{40} = 0.735$ | A1 | Allow $0.075g$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Without $F$ the ruler would rotate clockwise [about P] | E1 | Must indicate the turning effect of the weight and the direction. "Clockwise" implies rotation |
---4 A ruler PQRS is a uniform rectangular lamina with mass 20 grams. The length of PQ is 30 cm and the length of PS is 4 cm . The ruler is attached at P to a smooth hinge and held with S vertically below P by a horizontal force of magnitude $F \mathrm {~N}$ as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{8eeff88d-8b05-43c6-86a5-bd82221c0bea-04_303_1495_1363_239}
\begin{enumerate}[label=(\alph*)]
\item Calculate the value of $F$.
\item Explain what would happen to the lamina if the force at S were removed.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q4 [4]}}