| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Limiting equilibrium on incline |
| Difficulty | Moderate -0.3 This is a standard M1 pulley problem with straightforward resolution of forces. Part (a) is trivial equilibrium, parts (b-c) involve routine resolution perpendicular to the surface, and part (d) requires the standard limiting friction formula μ = F/R. The angled string adds minor complexity but follows textbook methods with no novel insight required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| For particle \(A\): \(T = 2g\) | M1 | Resolving vertically for \(A\) |
| \(T = 19.6\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram showing on \(B\): Weight \(4g\) downward, Normal reaction \(R\) upward, Tension \(T\) along string (at 30° to vertical, i.e. toward peg), Friction force \(F\) horizontal | B1 B1 | B1 for weight and normal reaction; B1 for tension and friction in correct directions |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving vertically for \(B\): \(R + T\cos 30° = 4g\) | M1 | |
| \(R = 4g - 19.6\cos 30°\) | A1 | |
| \(R = 39.2 - 19.6 \times \frac{\sqrt{3}}{2} = 39.2 - 16.97...\) | A1 | |
| \(R = 22.2\) N (to 3 s.f.) | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving horizontally for \(B\): \(F = T\sin 30°\) | M1 | |
| \(F = 19.6 \times 0.5 = 9.8\) N | A1 | |
| \(\mu \geq \frac{F}{R} = \frac{9.8}{22.2}\) | M1 | Using \(F \leq \mu R\) |
| \(\mu \geq 0.441\) | A1 | Accept \(\frac{9.8}{22.2}\) or equivalent |
# Question 6:
## Part (a):
| For particle $A$: $T = 2g$ | M1 | Resolving vertically for $A$ |
|---|---|---|
| $T = 19.6$ N | A1 | |
## Part (b):
| Diagram showing on $B$: Weight $4g$ downward, Normal reaction $R$ upward, Tension $T$ along string (at 30° to vertical, i.e. toward peg), Friction force $F$ horizontal | B1 B1 | B1 for weight and normal reaction; B1 for tension and friction in correct directions |
|---|---|---|
## Part (c):
| Resolving vertically for $B$: $R + T\cos 30° = 4g$ | M1 | |
|---|---|---|
| $R = 4g - 19.6\cos 30°$ | A1 | |
| $R = 39.2 - 19.6 \times \frac{\sqrt{3}}{2} = 39.2 - 16.97...$ | A1 | |
| $R = 22.2$ N (to 3 s.f.) | | Shown |
## Part (d):
| Resolving horizontally for $B$: $F = T\sin 30°$ | M1 | |
|---|---|---|
| $F = 19.6 \times 0.5 = 9.8$ N | A1 | |
| $\mu \geq \frac{F}{R} = \frac{9.8}{22.2}$ | M1 | Using $F \leq \mu R$ |
| $\mu \geq 0.441$ | A1 | Accept $\frac{9.8}{22.2}$ or equivalent |6 Two particles, $A$ and $B$, are connected by a light inextensible string which passes over a smooth peg. Particle $A$ has mass 2 kg and particle $B$ has mass 4 kg . Particle $A$ hangs freely with the string vertical. Particle $B$ is at rest in equilibrium on a rough horizontal surface with the string at an angle of $30 ^ { \circ }$ to the vertical. The particles, peg and string are shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-14_419_953_571_541}
\begin{enumerate}[label=(\alph*)]
\item By considering particle $A$, find the tension in the string.
\item Draw a diagram to show the forces acting on particle $B$.
\item Show that the magnitude of the normal reaction force acting on particle $B$ is 22.2 newtons, correct to three significant figures.
\item Find the least possible value of the coefficient of friction between particle $B$ and the surface.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f30b02da-a41e-44cb-b45f-9e6a3a9d0528-16_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2011 Q6 [11]}}