| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Standard +0.8 This is a non-uniform body (cone) equilibrium problem requiring moment calculations about a point, resolution of forces in two directions, and knowledge that the center of mass of a cone is 3h/4 from vertex. It involves multiple steps with trigonometry and simultaneous equations, making it moderately challenging but still within standard M2 scope. |
| Spec | 3.03n Equilibrium in 2D: particle under forces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moment of \(T_O\) about \(P = T_O\,h\cos20°\) | B1 | |
| Moment of \(W\) about \(P = 5g \times 0.75h\) | B1 | |
| \([T_O\,h\cos20° = 37.5h]\) | M1 | For taking moments about \(P\) |
| Tension in string at \(O\) is 39.9 N | A1 | Subtotal: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving forces horizontally or vertically or for taking moments | |
| \(T_P\sin\theta = 39.9\sin20°\) | A1ft | ft incorrect \(T_O\) |
| \(T_P\cos\theta + 39.9\cos20° = 5g\) or \((T_P\cos\theta)h = \frac{1}{4}h \times 50\) or \((T_P\cos\theta)\frac{3}{4}h = (T_O\cos20°)\frac{1}{4}h\) | A1ft | ft incorrect \(T_O\) |
| M1 | For eliminating \(T_P\) | |
| \(\theta = 47.5\) | A1 | |
| Tension in string at \(P\) is 18.5 N | A1 | Subtotal: 6 |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment of $T_O$ about $P = T_O\,h\cos20°$ | B1 | |
| Moment of $W$ about $P = 5g \times 0.75h$ | B1 | |
| $[T_O\,h\cos20° = 37.5h]$ | M1 | For taking moments about $P$ |
| Tension in string at $O$ is 39.9 N | A1 | **Subtotal: 4** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces horizontally or vertically or for taking moments |
| $T_P\sin\theta = 39.9\sin20°$ | A1ft | ft incorrect $T_O$ |
| $T_P\cos\theta + 39.9\cos20° = 5g$ or $(T_P\cos\theta)h = \frac{1}{4}h \times 50$ or $(T_P\cos\theta)\frac{3}{4}h = (T_O\cos20°)\frac{1}{4}h$ | A1ft | ft incorrect $T_O$ |
| | M1 | For eliminating $T_P$ |
| $\theta = 47.5$ | A1 | |
| Tension in string at $P$ is 18.5 N | A1 | **Subtotal: 6** | **Total: 10** |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{fe5c198d-5d05-4241-98f5-894ba92f7afe-4_447_736_269_701}\\
$P$ is the vertex of a uniform solid cone of mass 5 kg , and $O$ is the centre of its base. Strings are attached to the cone at $P$ and at $O$. The cone hangs in equilibrium with $P O$ horizontal and the strings taut. The strings attached at $P$ and $O$ make angles of $\theta ^ { \circ }$ and $20 ^ { \circ }$, respectively, with the vertical (see diagram, which shows a cross-section).\\
(i) By taking moments about $P$ for the cone, find the tension in the string attached at $O$.\\
(ii) Find the value of $\theta$ and the tension in the string attached at $P$.
\hfill \mbox{\textit{CAIE M2 2009 Q6 [10]}}