| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Equilibrium with applied force |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass problem requiring composite shapes (rectangle + triangle), then equilibrium analysis with toppling condition. The steps are routine: find centroid by splitting into standard shapes, apply moments about pivot point when toppling occurs. While multi-part, each step follows textbook methods without requiring novel insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0.9a + 0.9a/2)Y = 0.9a \times 0.45 + 0.45a \times 0.9 \times 2/3\) | M1 | \(1.5Y = 1 \times 0.45 + 0.5 \times 0.6\), Moments about AD |
| \(Y = 0.5\) m | A1 | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0.9a + 0.9a/2)X = 0.9a \times a/2 + 0.45a \times (a + a/3)\) | M1 | \(1.5X = 1 \times a/2 + 0.5 \times 4a/3\) |
| \(X = 7a/9\) | A1 | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5 \times 6 = (a - 7a/9) \times 18\) | M1, A1\(\checkmark\) | Ft [\(Yi\) and \((a - Xii)\)] |
| \(a = 0.75\) | A1 | Total: 3 marks |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0.9a + 0.9a/2)Y = 0.9a \times 0.45 + 0.45a \times 0.9 \times 2/3$ | M1 | $1.5Y = 1 \times 0.45 + 0.5 \times 0.6$, Moments about AD |
| $Y = 0.5$ m | A1 | Total: 2 marks |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0.9a + 0.9a/2)X = 0.9a \times a/2 + 0.45a \times (a + a/3)$ | M1 | $1.5X = 1 \times a/2 + 0.5 \times 4a/3$ |
| $X = 7a/9$ | A1 | Total: 2 marks |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5 \times 6 = (a - 7a/9) \times 18$ | M1, A1$\checkmark$ | Ft [$Yi$ and $(a - Xii)$] |
| $a = 0.75$ | A1 | Total: 3 marks |
---4\\
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The diagram shows the cross-section $A B C D$ through the centre of mass of a uniform solid prism. $A B = 0.9 \mathrm {~m} , B C = 2 a \mathrm {~m} , A D = a \mathrm {~m}$ and angle $A B C =$ angle $B A D = 90 ^ { \circ }$.\\
(i) Calculate the distance of the centre of mass of the prism from $A D$.\\
(ii) Express the distance of the centre of mass of the prism from $A B$ in terms of $a$.
The prism has weight 18 N and rests in equilibrium on a rough horizontal surface, with $A D$ in contact with the surface. A horizontal force of magnitude 6 N is applied to the prism. This force acts through the centre of mass in the direction $B C$.\\
(iii) Given that the prism is on the point of toppling, calculate $a$.
\hfill \mbox{\textit{CAIE M2 2016 Q4 [7]}}