| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with straight rod/wire components only |
| Difficulty | Standard +0.3 This is a straightforward centre of mass problem requiring basic geometry (finding the height of an isosceles triangle), weighted average calculations for composite bodies, and simple moment equilibrium. The geometry is given clearly, the calculations are routine, and the equilibrium part follows standard methods with no novel insight required. Slightly easier than average due to the symmetric setup and clear structure. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(d(3+3+4) = 3 \times 0.4\sin30 \times 2\) | M1, A1 | Taking moments about AC |
| \(d = 0.12\text{ m}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((3+3+4) \times 0.12 = F \times 0.8\sin30\) | M1 | Taking moments about A, allow candidate's \(d\) |
| \(F = 3\) | A1 | |
| At hinge, \(7\text{ N}\) upwards | B1\(\checkmark\) | \(Ft\ 10 -\) candidate's value \((F)\) (downwards if negative) |
## Question 3:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $d(3+3+4) = 3 \times 0.4\sin30 \times 2$ | M1, A1 | Taking moments about AC |
| $d = 0.12\text{ m}$ | A1 | |
**Total: 3 marks**
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(3+3+4) \times 0.12 = F \times 0.8\sin30$ | M1 | Taking moments about A, allow candidate's $d$ |
| $F = 3$ | A1 | |
| At hinge, $7\text{ N}$ upwards | B1$\checkmark$ | $Ft\ 10 -$ candidate's value $(F)$ (downwards if negative) |
**Total: 3 marks**
---3 A triangular frame $A B C$ consists of two uniform rigid rods each of length 0.8 m and weight 3 N , and a longer uniform rod of weight 4 N . The triangular frame has $A B = B C$, and angle $B A C =$ angle $B C A = 30 ^ { \circ }$.\\
(i) Calculate the distance of the centre of mass of the frame from $A C$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a03ad6c1-b4a3-4007-8d3b-ce289a998a55-2_722_335_1302_904}
The vertex $A$ of the frame is attached to a smooth hinge at a fixed point. The frame is held in equilibrium with $A C$ vertical by a vertical force of magnitude $F \mathrm {~N}$ applied to the frame at $B$ (see diagram).\\
(ii) Calculate $F$, and state the magnitude and direction of the force acting on the frame at the hinge.
\hfill \mbox{\textit{CAIE M2 2015 Q3 [6]}}