| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| 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 using the triangle of forces method, which is explicitly prompted. Students need only draw a triangle with weight (3g N), tension T, and horizontal force F, then use basic trigonometry (tan 25° = F/3g or sin/cos ratios) to find T and F. It requires standard recall of the triangle of forces technique with minimal problem-solving, making it easier than average but not trivial since it involves setting up the geometry correctly. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Force triangle with \(F\) N, Tension, Weight correctly arranged with 25° angle | B1 | Arrows making a closed loop in roughly the right directions |
| Labels: Tension, Weight, \(F\) correctly labelled with 25° (or 65°) | B1 | Tension, weight and \(F\) labelled on their triangle and 25° (or 65°) correctly labelled. May be given as a suitable angle outside the triangle |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using the triangle of forces: \(F = 3g\tan 25°\) or Tension \(= \dfrac{3g}{\cos 25°}\) | M1 | Allow sin/cos or 25°/65° interchange to find \(F\) or \(T\) |
| \(F = 13.7\) | A1 | cao |
| Tension \(= 32.4\) N | A1 | cao |
| Alternative method: | ||
| Resolve vertically \(T\cos 25° = 3g\) | M1 | Allow sin/cos interchange or 25°/65° interchange |
| \(T = 32.4\) N | A1 | cao |
| Resolve horizontally \(F = T\sin 25° = 13.7\) | A1 | cao |
| [3] |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Force triangle with $F$ N, Tension, Weight correctly arranged with 25° angle | B1 | Arrows making a closed loop in roughly the right directions |
| Labels: Tension, Weight, $F$ correctly labelled with 25° (or 65°) | B1 | Tension, weight and $F$ labelled on their triangle and 25° (or 65°) correctly labelled. May be given as a suitable angle outside the triangle |
| **[2]** | | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using the triangle of forces: $F = 3g\tan 25°$ or Tension $= \dfrac{3g}{\cos 25°}$ | M1 | Allow sin/cos or 25°/65° interchange to find $F$ or $T$ |
| $F = 13.7$ | A1 | cao |
| Tension $= 32.4$ N | A1 | cao |
| **Alternative method:** | | |
| Resolve vertically $T\cos 25° = 3g$ | M1 | Allow sin/cos interchange or 25°/65° interchange |
| $T = 32.4$ N | A1 | cao |
| Resolve horizontally $F = T\sin 25° = 13.7$ | A1 | cao |
| **[3]** | | |5 A sphere of mass 3 kg hangs on a string. A horizontal force of magnitude $F \mathrm {~N}$ acts on the sphere so that it hangs in equilibrium with the string making an angle of $25 ^ { \circ }$ to the vertical. The force diagram for the sphere is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{9dd6fc6d-b51e-4a73-ace5-d26a7558032c-05_502_513_408_244}
\begin{enumerate}[label=(\alph*)]
\item Sketch the triangle of forces for these forces.
\item Hence or otherwise determine each of the following:
\begin{itemize}
\item the tension in the string
\item the value of $F$.
\end{itemize}
Answer all the questions.\\
Section B (76 marks)
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q5 [5]}}