| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| 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 M1 equilibrium problem requiring resolution of forces in two perpendicular directions. While it involves a vertical string (slightly less routine than horizontal forces), the method is straightforward: resolve parallel and perpendicular to the slope, apply equilibrium conditions. 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 friction value. |
| Spec | 3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram with \(T\) N, \(R\) N, \(10\) 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 the plane: \(10 + T\cos30 = 4g\cos30\) | 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. |
| A1 | All correct | |
| \(T = 27.65299...\) so \(27.7\) N (3 s.f.) | A1 | 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 horiz or vert if at least 1 resolution correct. All forces present. No extra forces. |
| A1 | Correct. FT \(T\) if evaluated. | |
| \(R = 5.7735...\) so \(5.77\) N (3 s.f.) | A1 | Any reasonable accuracy. cao. |
## Question 4:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram with $T$ N, $R$ N, $10$ 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 |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve parallel to the plane: $10 + T\cos30 = 4g\cos30$ | 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. |
| | A1 | All correct |
| $T = 27.65299...$ so $27.7$ N (3 s.f.) | A1 | Any reasonable accuracy |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve perpendicular to plane: $R + 0.5T = 2g$ | M1 | At least one resolution correct. Accept resolution horiz or vert if at least 1 resolution correct. All forces present. No extra forces. |
| | A1 | Correct. FT $T$ if evaluated. |
| $R = 5.7735...$ so $5.77$ N (3 s.f.) | A1 | Any reasonable accuracy. cao. |
---4 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]{04848aba-9e64-4265-a4a5-e9336b958a05-3_533_378_852_831}
\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 2005 Q4 [7]}}