| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Particle on rough incline connected to particle on horizontal surface or other incline |
| Difficulty | Standard +0.3 This is a standard M1 pulley-on-wedge problem requiring systematic application of Newton's second law and friction concepts. While it involves multiple components (inclined planes, friction, pulley forces), each part follows routine mechanics procedures without requiring novel insight. The given acceleration simplifies the problem significantly, making it slightly easier than average for M1. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| ⟹ \(T = \frac{6}{5}mg\) | M1 A1 A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F: R(\text{perp}): R = mg \cos 30°\) | M1 A1 | |
| \(R(\parallel): T - mg \sin 30° - F = m \cdot \frac{1}{10}g\) | M1 A2, 1, 0 | |
| Using \(F = \mu R\) | M1 ↓↓↓ | |
| \(\frac{6}{5}mg - \frac{1}{2}mg - \mu mg \frac{\sqrt{3}}{2} = \frac{1}{10}mg\) | M1 | |
| ⟹ \(\mu = 0.693\) or \(0.69\) or \(\frac{2\sqrt{3}}{5}\) | A1 | (8 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Magn of force on pulley \(= 2T \cos 60° = \frac{6}{5}mg\) | M1 A1 √ | |
| Direction is vertically downwards | B1 (cso) | (3 marks) |
## Part (a)
$A: 3mg \sin 30° - T = 3m \cdot \frac{1}{10}g$
⟹ $T = \frac{6}{5}mg$ | M1 A1 A1 | (3 marks)
## Part (b)
$F: R(\text{perp}): R = mg \cos 30°$ | M1 A1 |
$R(\parallel): T - mg \sin 30° - F = m \cdot \frac{1}{10}g$ | M1 A2, 1, 0 |
Using $F = \mu R$ | M1 ↓↓↓ |
$\frac{6}{5}mg - \frac{1}{2}mg - \mu mg \frac{\sqrt{3}}{2} = \frac{1}{10}mg$ | M1 |
⟹ $\mu = 0.693$ or $0.69$ or $\frac{2\sqrt{3}}{5}$ | A1 | (8 marks)
## Part (c)
Magn of force on pulley $= 2T \cos 60° = \frac{6}{5}mg$ | M1 A1 √ |
Direction is vertically downwards | B1 (cso) | (3 marks)
**Total for Question 7: 14 marks**\includegraphics{figure_3}
A fixed wedge has two plane faces, each inclined at $30°$ to the horizontal. Two particles $A$ and $B$, of mass $3m$ and $m$ respectively, are attached to the ends of a light inextensible string. Each particle moves on one of the plane faces of the wedge. The string passes over a small smooth light pulley fixed at the top of the wedge. The face on which $A$ moves is smooth. The face on which $B$ moves is rough. The coefficient of friction between $B$ and this face is $\mu$. Particle $A$ is held at rest with the string taut. The string lies in the same vertical plane as lines of greatest slope on each plane face of the wedge, as shown in Figure 3.
The particles are released from rest and start to move. Particle $A$ moves downwards and $B$ moves upwards. The accelerations of $A$ and $B$ each have magnitude $\frac{1}{10}g$.
\begin{enumerate}[label=(\alph*)]
\item By considering the motion of $A$, find, in terms of $m$ and $g$, the tension in the string. [3]
\item By considering the motion of $B$, find the value of $\mu$. [8]
\item Find the resultant force exerted by the string on the pulley, giving its magnitude and direction. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2006 Q7 [14]}}