| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Applied force in addition to weights |
| Difficulty | Standard +0.3 This is a standard M1 pulley problem with connected particles on inclined planes. Part (i) requires setting up Newton's second law for both particles and solving simultaneous equations—routine mechanics. Part (ii) adds friction at limiting equilibrium, requiring careful force resolution but following standard procedures. The multi-part structure and inclusion of friction elevates it slightly above average, but it remains a textbook-style question testing standard techniques without requiring novel insight. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution3.03m Equilibrium: sum of resolved forces = 03.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | For applying Newton's 2nd law to either particle (correct number of terms) | |
| \(T - 0.9g\sin 15 = 0.9a\) | A1 | |
| \(2.5 + 0.4g\sin 25 - T = 0.4a\) | A1 | |
| \(1.3a = 1.86\ldots\) | M1 | Solving simultaneously for \(a\) |
| \(a = 1.43\) m s\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F = 0.8 \times 0.4g\cos 25\) | B1 | |
| \(2.5 + 0.4g\sin 25 - T - F = 0\) | M1 | For using equilibrium of forces acting on particle \(B\) with 4 terms |
| \(T - 0.9g\sin\theta = 0\) | M1 | For using equilibrium of forces acting on particle \(A\) with 2 terms |
| M1 | For solving for \(\theta\) | |
| \(\theta = 8.2°\) | A1 | |
| Total | 5 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | For applying Newton's 2nd law to either particle (correct number of terms) |
| $T - 0.9g\sin 15 = 0.9a$ | A1 | |
| $2.5 + 0.4g\sin 25 - T = 0.4a$ | A1 | |
| $1.3a = 1.86\ldots$ | M1 | Solving simultaneously for $a$ |
| $a = 1.43$ m s$^{-2}$ | A1 | |
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $F = 0.8 \times 0.4g\cos 25$ | **B1** | |
| $2.5 + 0.4g\sin 25 - T - F = 0$ | **M1** | For using equilibrium of forces acting on particle $B$ with 4 terms |
| $T - 0.9g\sin\theta = 0$ | **M1** | For using equilibrium of forces acting on particle $A$ with 2 terms |
| | **M1** | For solving for $\theta$ |
| $\theta = 8.2°$ | **A1** | |
| **Total** | **5** | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{db1b5f31-1a41-44dd-ae9a-0c67336997eb-10_212_1029_255_557}
Two particles $A$ and $B$ of masses 0.9 kg and 0.4 kg respectively are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to the top of two inclined planes. The particles are initially at rest with $A$ on a smooth plane inclined at angle $\theta ^ { \circ }$ to the horizontal and $B$ on a plane inclined at angle $25 ^ { \circ }$ to the horizontal. The string is taut and the particles can move on lines of greatest slope of the two planes. A force of magnitude 2.5 N is applied to $B$ acting down the plane (see diagram).\\
(i) For the case where $\theta = 15$ and the plane on which $B$ rests is smooth, find the acceleration of $B$.\\
(ii) For a different value of $\theta$, the plane on which $B$ rests is rough with coefficient of friction between the plane and $B$ of 0.8 . The system is in limiting equilibrium with $B$ on the point of moving in the direction of the 2.5 N force. Find the value of $\theta$.\\
\hfill \mbox{\textit{CAIE M1 2017 Q7 [10]}}