| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Conical or hemispherical shell composite |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass problem requiring standard formulas for hemispherical shell (0.5r from centre) and semicircular arc (2r/π from centre), then composite centre of mass calculations and equilibrium analysis. While it involves multiple steps and careful coordinate geometry, the techniques are standard M2 material with no novel insight required—harder than routine but well within typical Further Maths mechanics scope. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{OG (handle)} = 0.3\sin(\pi/2)/(\pi/2)\) | B1 | \(0.6/\pi,\ 0.19098\ldots\) |
| \((8+14)\,d = 14 \times 0.15 - 8 \times 0.6/\pi\) | M1 | |
| \(d = 0.026(0)\text{m}\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((8+14)\,x = 8 \times 0.191\) | M1 | Horizontal distance from bowl axis |
| \(x = 0.0694545\ldots\) | A1 | |
| \((8+14)\,y = 14 \times 0.15\) | M1 | Vertical distance from bowl rim |
| \(y = 0.0954545\ldots\) | A1 | |
| Distance \(= 0.118\) | AG, A1 | [5] Distance\(^2 = 0.06945^2 + 0.095454^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan\theta = 0.0694545/0.0954545\) or \(\sin\theta = 0.0694545/0.118\) or \(\cos\theta = 0.094545/0.118\) | M1 | Any trig ratio correctly using the two answers in (ii) |
| \(\theta = 36(.0)°\) | A1 | [2] Accept 36.1, 35.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan H = 0.15/0.19098\) | M1 | \(H = 38.147°\) |
| \(HG = \dfrac{14}{22} \times \sqrt{(0.15^2 + 0.19098^2)}\) | A1 | \(0.15454\text{ m}\) |
| \(OG^2 = 0.19098^2 + 0.15454^2 - 2 \times 0.19098 \times 0.15454\cos 38.147\) | M1 | Cosine rule on triangle OHG |
| Distance \(= 0.118\) | AG, A1, A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\sin HOG = 0.15454\sin 38.147/0.118\) | M1 | \(54.06°\) |
| Angle \((= 90 - HOG) = 35.9\) | A1 | [2] Accept 36(.0), 36.1 |
## Question 7:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\text{OG (handle)} = 0.3\sin(\pi/2)/(\pi/2)$ | B1 | $0.6/\pi,\ 0.19098\ldots$ |
| $(8+14)\,d = 14 \times 0.15 - 8 \times 0.6/\pi$ | M1 | |
| $d = 0.026(0)\text{m}$ | A1 | [3] |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $(8+14)\,x = 8 \times 0.191$ | M1 | Horizontal distance from bowl axis |
| $x = 0.0694545\ldots$ | A1 | |
| $(8+14)\,y = 14 \times 0.15$ | M1 | Vertical distance from bowl rim |
| $y = 0.0954545\ldots$ | A1 | |
| Distance $= 0.118$ | AG, A1 | [5] Distance$^2 = 0.06945^2 + 0.095454^2$ |
### Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan\theta = 0.0694545/0.0954545$ or $\sin\theta = 0.0694545/0.118$ or $\cos\theta = 0.094545/0.118$ | M1 | Any trig ratio correctly using the two answers in (ii) |
| $\theta = 36(.0)°$ | A1 | [2] Accept 36.1, 35.9 |
### Alternative for 7(ii), 7(iii):
**Part (ii) alternative:**
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan H = 0.15/0.19098$ | M1 | $H = 38.147°$ |
| $HG = \dfrac{14}{22} \times \sqrt{(0.15^2 + 0.19098^2)}$ | A1 | $0.15454\text{ m}$ |
| $OG^2 = 0.19098^2 + 0.15454^2 - 2 \times 0.19098 \times 0.15454\cos 38.147$ | M1 | Cosine rule on triangle OHG |
| Distance $= 0.118$ | AG, A1, A1 | [5] |
**Part (iii) alternative:**
| Working | Mark | Guidance |
|---------|------|----------|
| $\sin HOG = 0.15454\sin 38.147/0.118$ | M1 | $54.06°$ |
| Angle $(= 90 - HOG) = 35.9$ | A1 | [2] Accept 36(.0), 36.1 |7\\
\includegraphics[max width=\textwidth, alt={}, center]{9c82b387-8e5e-48b9-973d-5337b4e56a66-4_553_630_258_753}
The diagram shows a container which consists of a bowl of weight 14 N and a handle of weight 8 N . The bowl of the container is in the form of a uniform hemispherical shell with centre $O$ and radius 0.3 m . The handle is in the form of a uniform semicircular arc of radius 0.3 m and is freely hinged to the bowl at $A$ and $B$, where $A B$ is a diameter of the bowl.\\
(i) Calculate the distance of the centre of mass of the container from $O$ for the position indicated in the diagram, where the handle is perpendicular to the rim of the bowl.\\
(ii) Show that the distance of the centre of mass of the container from $O$ when the handle lies on the rim of the bowl is 0.118 m , correct to 3 significant figures.
In the case when the handle lies on the rim of the bowl, the container rests in equilibrium with the curved surface of the bowl on a horizontal table.\\
(iii) Find the angle which the plane containing the rim of the bowl makes with the horizontal.
\hfill \mbox{\textit{CAIE M2 2014 Q7 [10]}}