| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Equilibrium on slope with force parallel to slope |
| Difficulty | Moderate -0.3 This is a standard equilibrium problem on a slope requiring resolution of forces in two perpendicular directions. While it involves multiple forces including a vertical string (slightly less routine), the method is straightforward: draw force diagram, resolve parallel and perpendicular to slope, solve simultaneous equations. The friction force is given rather than needing to be calculated from μR, simplifying the problem. Slightly easier than average due to clear structure and given values. |
| Spec | 3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03i Normal reaction force3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram showing \(10\) N, \(T\) N, \(R\) N, \(4g\) N at \(60°\) | B1 | All forces present. No extras. Accept \(mg\), \(w\) etc. All labelled with arrows. Accept resolved parts only if clearly additional. Accept no angles. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve parallel to plane: \(10 + T\cos 30 = 4g\cos 30\) | M1 | All terms present. Must be resolution in at least 1 term. Accept \(\sin \leftrightarrow \cos\). If resolution in another direction there must be an equation only in \(T\) with no forces omitted. No extra forces. |
| \(T = 27.65299...\) so \(27.7\) N (3 s.f.) | A1 A1 | All correct. Any reasonable accuracy. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve perpendicular to plane: \(R + 0.5T = 2g\) | M1 | At least one resolution correct. Accept resolution horizontal or vertical if at least 1 resolution correct. All forces present. No extra forces. |
| \(R = 5.7735...\) so \(5.77\) N (3 s.f.) | A1 A1 | Correct. FT \(T\) if evaluated. Any reasonable accuracy. cao. |
## Question 7:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram showing $10$ N, $T$ N, $R$ N, $4g$ N at $60°$ | B1 | All forces present. No extras. Accept $mg$, $w$ etc. All labelled with arrows. Accept resolved parts only if clearly additional. Accept no angles. |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve parallel to plane: $10 + T\cos 30 = 4g\cos 30$ | M1 | All terms present. Must be resolution in at least 1 term. Accept $\sin \leftrightarrow \cos$. If resolution in another direction there must be an equation only in $T$ with no forces omitted. No extra forces. |
| $T = 27.65299...$ so $27.7$ N (3 s.f.) | A1 A1 | All correct. Any reasonable accuracy. |
**(iii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to plane: $R + 0.5T = 2g$ | M1 | At least one resolution correct. Accept resolution horizontal or vertical if at least 1 resolution correct. All forces present. No extra forces. |
| $R = 5.7735...$ so $5.77$ N (3 s.f.) | A1 A1 | Correct. FT $T$ if evaluated. Any reasonable accuracy. cao. |7 A block of mass 4 kg is in equilibrium on a rough plane inclined at $60 ^ { \circ }$ to the horizontal, as shown in Fig. 4. A frictional force of 10 N acts up the plane and a vertical string AB attached to the block is in tension.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-5_492_347_1545_870}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Draw a diagram showing the four forces acting on the block.\\
(ii) By considering the components of the forces parallel to the slope, calculate the tension in the string.\\
(iii) Calculate the normal reaction of the plane on the block.
\hfill \mbox{\textit{OCR MEI M1 Q7 [7]}}