| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| 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 equilibrium problem requiring resolution of forces in two perpendicular directions. The steps are routine: draw force diagram, resolve horizontally to find friction (25N given), resolve vertically to find normal reaction, then use Pythagoras. No problem-solving insight needed, just standard M1 technique with simple arithmetic. |
| 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 |
|---|---|
| B2 | Diagram showing weight \(mg\) and normal reaction \(N\) of 50 N with arrows and labels |
| Answer | Marks |
|---|---|
| M1 | Horizontal equilibrium: \(R = 50\sin 30°\) |
| A1 | \(R = 25\) |
| Answer | Marks |
|---|---|
| M1 | Vertical equilibrium: \(N + 50\cos 30° = 10g\) |
| A1 | \(N = 54.7\) to 3 s.f. |
| Answer | Marks |
|---|---|
| M1 | Resultant \(= \sqrt{25^2 + 54.7^2}\) |
| A1 | Resultant is 60.1 N |
# Question 1
## (i)
B2 | Diagram showing weight $mg$ and normal reaction $N$ of 50 N with arrows and labels
Subtract one mark for each error, omission or addition down to a minimum of zero. Each force must have a label and an arrow. Accept T for 50 N. Units not required.
If a candidate gives the tension in components: Accept if the components are a replacement for the tension. Treat as an error if the components duplicate the tension. However, accept dotted lines for the components as not being duplication.
[2]
## (ii)
M1 | Horizontal equilibrium: $R = 50\sin 30°$
A1 | $R = 25$
May be implied. Allow sin-cos interchange for this mark only. Award both marks for a correct answer after a mistake in part (i) (e.g. omission of $R$).
[2]
## (iii)
M1 | Vertical equilibrium: $N + 50\cos 30° = 10g$
A1 | $N = 54.7$ to 3 s.f.
Relationship must be seen and involve all 3 elements. No credit given in the case of sin-cos interchange. CAO.
[2]
## (iv)
M1 | Resultant $= \sqrt{25^2 + 54.7^2}$
A1 | Resultant is 60.1 N
Use of Pythagoras. Components must be correct but allow follow-through from both (ii) and (iii) for M1 only. CAO.
[2]1 Fig. 5 shows a block of mass 10 kg at rest on a rough horizontal floor. A light string, at an angle of $30 ^ { \circ }$ to the vertical, is attached to the block. The tension in the string is 50 N .
The block is in equilibrium.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a1895e1-abe3-4739-876a-f19458f0f6ed-1_409_585_472_768}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(i) Show all the forces acting on the block.\\
(ii) Show that the frictional force acting on the block is 25 N .\\
(iii) Calculate the normal reaction of the floor on the block.\\
(iv) Calculate the magnitude of the total force the floor is exerting on the block.
\hfill \mbox{\textit{OCR MEI M1 Q1 [8]}}