| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Toppling and sliding of solids |
| Difficulty | Challenging +1.2 This is a two-part mechanics question requiring centre of mass calculation for a composite body (part i) and then applying equilibrium conditions with limiting friction (part ii). While it involves multiple concepts (moments, friction, resolving forces), the techniques are standard M2 material with straightforward application of formulas. The 30° angle and symmetry simplify the geometry, making this moderately above average but not requiring novel insight. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area of hole \(= \pi r^2\) and Area of original circle \(= 25\pi r^2\) | M1 | |
| Area of cross-section \(= 24\pi r^2\) | A1 | |
| \(\pi r^2(2r) = 24\pi r^2(d)\) | M1 | Take moments about the centre of the cylinder |
| \(d = \frac{r}{12} (= 0.083333...r)\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(2 \times 5r) = W\left(\frac{r}{12}\right)\cos 60\) | M1 | Take moments about the point of contact with the plane |
| \(P = \frac{W\cos 60}{120} = \frac{W}{240} = 0.00417W (= F)\) | A1 | |
| \(\mu = \frac{W\cos 60}{120} / W\) | M1 | Use \(F = \mu R\). Note \(R = W\) by resolving vertically |
| \(\mu = \frac{1}{240} = 0.00417\) | A1 | |
| 4 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area of hole $= \pi r^2$ and Area of original circle $= 25\pi r^2$ | M1 | |
| Area of cross-section $= 24\pi r^2$ | A1 | |
| $\pi r^2(2r) = 24\pi r^2(d)$ | M1 | Take moments about the centre of the cylinder |
| $d = \frac{r}{12} (= 0.083333...r)$ | A1 | |
| | **4** | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(2 \times 5r) = W\left(\frac{r}{12}\right)\cos 60$ | M1 | Take moments about the point of contact with the plane |
| $P = \frac{W\cos 60}{120} = \frac{W}{240} = 0.00417W (= F)$ | A1 | |
| $\mu = \frac{W\cos 60}{120} / W$ | M1 | Use $F = \mu R$. Note $R = W$ by resolving vertically |
| $\mu = \frac{1}{240} = 0.00417$ | A1 | |
| | **4** | |(i) Find, in terms of $r$, the distance of the centre of mass of the prism from the centre of the cylinder.\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b8e52188-f9a6-46fc-90bf-97965c6dd324-11_633_729_258_708}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The prism has weight $W \mathrm {~N}$ and is placed with its curved surface on a rough horizontal plane. The axis of symmetry of the cross-section makes an angle of $30 ^ { \circ }$ with the vertical. A horizontal force of magnitude $P \mathrm {~N}$ acting in the plane of the cross-section through the centre of mass is applied to the cylinder at the highest point of this cross-section (see Fig. 2). The prism rests in limiting equilibrium.\\
(ii) Find the coefficient of friction between the prism and the plane.\\
\hfill \mbox{\textit{CAIE M2 2019 Q6 [8]}}