| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with elastic string or spring support |
| Difficulty | Standard +0.3 This is a straightforward equilibrium problem requiring moments about the hinge and Hooke's law. The geometry is given explicitly, and students need only take moments about B (eliminating the reaction force), apply T = λx/l, and solve two simple equations. While it combines topics (moments + elasticity), each step is standard with no conceptual subtlety. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Takes moments about \(B\) | M1 | |
| \([T\cos 60° \times 2 = 10g \times 1]\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Uses Hooke's Law | M1 | |
| Obtains \(100 = 200(3-L)/L\) or \(100 = 200x/(3-x)\) | A1 ft | |
| Obtains natural length as 2m | A1 | 3 marks |
**(i)**
Takes moments about $B$ | M1 |
$[T\cos 60° \times 2 = 10g \times 1]$ | A1 | 2 marks |
**(ii)**
Uses Hooke's Law | M1 |
Obtains $100 = 200(3-L)/L$ or $100 = 200x/(3-x)$ | A1 ft |
Obtains natural length as 2m | A1 | 3 marks |2\\
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A uniform rod $A B$, of length 2 m and mass 10 kg , is freely hinged to a fixed point at the end $B$. A light elastic string, of modulus of elasticity 200 N , has one end attached to the end $A$ of the rod and the other end attached to a fixed point $O$, which is in the same vertical plane as the rod. The rod is horizontal and in equilibrium, with $O A = 3 \mathrm {~m}$ and angle $O A B = 150 ^ { \circ }$ (see diagram). Find\\
(i) the tension in the string,\\
(ii) the natural length of the string.
\hfill \mbox{\textit{CAIE M2 2002 Q2 [5]}}