| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle with string at angle to wall |
| Difficulty | Moderate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions. Students need to apply T cos(20°) = mg and T sin(20°) = F with given T = 12N, involving only basic trigonometry and no problem-solving insight beyond standard method application. |
| Spec | 3.03f Weight: W=mg3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolve horizontally \(F = 12\sin 20°\) | M1 | Resolving horizontally – allow sin/cos interchange. Allow if \(F = T\sin 20°\) or similar seen |
| \(F = 4.10\) | A1 | www |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolve vertically \([mg =] 12\cos 20°\) | M1 | Resolve to find vertical component of tension. Allow sin/cos interchange if consistent with (a). If triangle of forces used, allow attempt to find by Pythagoras |
| \(m = \frac{12\cos 20°}{g} = 1.15\text{ kg}\) | M1 | Equating weight to their component of tension and dividing their weight by \(g\) soi |
| \(m = 1.15\text{ kg}\) | A1 | cao |
| [3] |
## Question 3:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve horizontally $F = 12\sin 20°$ | M1 | Resolving horizontally – allow sin/cos interchange. Allow if $F = T\sin 20°$ or similar seen |
| $F = 4.10$ | A1 | www |
| [2] | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve vertically $[mg =] 12\cos 20°$ | M1 | Resolve to find vertical component of tension. Allow sin/cos interchange if consistent with (a). If triangle of forces used, allow attempt to find by Pythagoras |
| $m = \frac{12\cos 20°}{g} = 1.15\text{ kg}$ | M1 | Equating weight to their component of tension and dividing their weight by $g$ soi |
| $m = 1.15\text{ kg}$ | A1 | cao |
| [3] | | |
---3 A particle hangs at the end of a string. A horizontal force of magnitude $F \mathrm {~N}$ acting on the particle holds it in equilibrium so that the string makes an angle of $20 ^ { \circ }$ with the vertical, as shown in the diagram. The tension in the string is 12 N .\\
\includegraphics[max width=\textwidth, alt={}, center]{1d0ca3d5-6529-435f-a0b8-50ea4859adde-04_357_374_1409_239}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $F$.
\item Find the mass of the particle.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q3 [5]}}