| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – equilibrium (finding friction, normal reaction, or coefficient of friction) |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring resolution of forces and equilibrium conditions. Part (i) is trivial (R = mg), while parts (ii) and (iii) involve standard resolution of a force at an angle and applying vertical/horizontal equilibrium—routine M1 content with no problem-solving insight needed. |
| Spec | 3.03e Resolve forces: two dimensions3.03f Weight: W=mg3.03i Normal reaction force3.03r Friction: concept and vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = mg\) so \(49\) N | B1 | Equating to weight. Accept \(5g\) (but not \(mg\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram with R upward, \(40°\) force of \(10\) N, weight \(49\) N downward, \(F\) horizontal | B1 | All except \(F\) correct (arrows and labels). Accept \(mg\), \(W\) etc and no angle. Accept components instead of 10N. No extra forces. |
| \(F\) clearly marked and labelled | B1 | \(F\) clearly marked and labelled |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\uparrow \quad R + 10\cos 40 - 49 = 0\) | M1 | Resolve vertically. All forces present and 10N resolved |
| Resolution correct and seen in equation | B1 | Resolution correct and seen in an equation. (Accept \(R = \pm 10\cos 40\) as an equation) |
| \(R = 41.339\ldots\) so \(41.3\) N (3 s.f.) | A1 | |
| \(F = 10\sin 40 = 6.4278\ldots\) so \(6.43\) N (3 s.f.) | B1 | Allow \(-\)ve if consistent with the diagram. |
# Question 3:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = mg$ so $49$ N | B1 | Equating to weight. Accept $5g$ (but not $mg$) |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram with **R** upward, $40°$ force of $10$ N, weight $49$ N downward, $F$ horizontal | B1 | All except $F$ correct (arrows and labels). Accept $mg$, $W$ etc and no angle. Accept components instead of 10N. No extra forces. |
| $F$ clearly marked and labelled | B1 | $F$ clearly marked and labelled |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\uparrow \quad R + 10\cos 40 - 49 = 0$ | M1 | Resolve vertically. All forces present and 10N resolved |
| Resolution correct and seen in equation | B1 | Resolution correct and seen in an equation. (Accept $R = \pm 10\cos 40$ as an equation) |
| $R = 41.339\ldots$ so $41.3$ N (3 s.f.) | A1 | |
| $F = 10\sin 40 = 6.4278\ldots$ so $6.43$ N (3 s.f.) | B1 | Allow $-$ve if consistent with the diagram. |
---3 A box of mass 5 kg is at rest on a rough horizontal floor.\\
(i) Find the value of the normal reaction of the floor on the box.
The box remains at rest on the floor when a force of 10 N is applied to it at an angle of $40 ^ { \circ }$ to the upward vertical, as shown in Fig. 3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{94f23528-931c-47b6-89aa-4b6edd25cc30-2_286_470_1067_803}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(ii) Draw a diagram showing all the forces acting on the box.\\
(iii) Calculate the new value of the normal reaction of the floor on the box and also the frictional force.
\hfill \mbox{\textit{OCR MEI M1 Q3 [7]}}