| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Limiting equilibrium on incline |
| Difficulty | Challenging +1.2 This is a standard two-particle pulley problem requiring resolution of forces, tension analysis using the resultant force formula, and friction at limiting equilibrium. While it involves multiple steps (finding tension from resultant, then applying equilibrium with friction on an incline), the techniques are all routine M1 content with no novel insight required. The geometry is straightforward (30-60-90 triangle) and the friction calculation is standard. Slightly above average difficulty due to the multi-stage nature and the resultant force calculation, but well within typical M1 examination scope. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([2T\cos30° = 3\sqrt{3}\) or \(T/\sin30° = 3\sqrt{3}/\sin120°\) or \(T^2 = T^2 + (3\sqrt{3})^2 - 2T(3\sqrt{3})\cos30°\) or \(\sqrt{[(Tcos30°)^2 + (T + Tcos60°)^2]} = 3\sqrt{3}]\) Tension is \(3 \text{ N}\) | M1, A1 | For expressing resultant in terms of \(T\) and equating with value or for using sine rule or for using cosine rule or for finding \(R_x\) and \(R_y\) and equating resultant to \(3\sqrt{3}\) |
| (ii) \([T = F + mg\sin30°]\) \(R = mg\cos30°\) or \(3 = 0.75(10\cos30°)m + 10m\sin30°\) \(3 = 0.75(10\cos30°)m + 10m\sin30°\) Mass is \(0.261 \text{ kg}\) | M1, B1, M1, A1, A1 | For resolving forces on Q parallel to AC or For using \(F = \mu R\) |
(i) $[2T\cos30° = 3\sqrt{3}$ or $T/\sin30° = 3\sqrt{3}/\sin120°$ or $T^2 = T^2 + (3\sqrt{3})^2 - 2T(3\sqrt{3})\cos30°$ or $\sqrt{[(Tcos30°)^2 + (T + Tcos60°)^2]} = 3\sqrt{3}]$ Tension is $3 \text{ N}$ | M1, A1 | For expressing resultant in terms of $T$ and equating with value or for using sine rule or for using cosine rule or for finding $R_x$ and $R_y$ and equating resultant to $3\sqrt{3}$ | AG | [2]
(ii) $[T = F + mg\sin30°]$ $R = mg\cos30°$ or $3 = 0.75(10\cos30°)m + 10m\sin30°$ $3 = 0.75(10\cos30°)m + 10m\sin30°$ Mass is $0.261 \text{ kg}$ | M1, B1, M1, A1, A1 | For resolving forces on Q parallel to AC or For using $F = \mu R$ | [5]
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\includegraphics[max width=\textwidth, alt={}, center]{f0200d12-4ab0-4395-804c-e693f7f26507-2_368_853_1503_644}
A small smooth pulley is fixed at the highest point $A$ of a cross-section $A B C$ of a triangular prism. Angle $A B C = 90 ^ { \circ }$ and angle $B C A = 30 ^ { \circ }$. The prism is fixed with the face containing $B C$ in contact with a horizontal surface. Particles $P$ and $Q$ are attached to opposite ends of a light inextensible string, which passes over the pulley. The particles are in equilibrium with $P$ hanging vertically below the pulley and $Q$ in contact with $A C$. The resultant force exerted on the pulley by the string is $3 \sqrt { } 3 \mathrm {~N}$ (see diagram).\\
(i) Show that the tension in the string is 3 N .
The coefficient of friction between $Q$ and the prism is 0.75 .\\
(ii) Given that $Q$ is in limiting equilibrium and on the point of moving upwards, find its mass.
\hfill \mbox{\textit{CAIE M1 2010 Q3 [7]}}