| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Standard +0.3 This is a straightforward elastic string problem requiring Hooke's law for part (i) and resolving forces with symmetry for part (ii). The geometry is simple (symmetric configuration with clear right triangles), and the solution follows standard textbook methods without requiring novel insight or complex multi-step reasoning. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l |
| Answer | Marks | Guidance |
|---|---|---|
| (i) For using \(T = \frac{\lambda x}{L}\) | M1 | (\(T = 60(0.5)/1.5\)) |
| Tension is 20 N | A1 | 2 marks |
| (ii) For using extension 0.5 m (or 1 m) in Hooke's Law | M1 | |
| \(T = 60(0.5)/0.75\) or \(T = 60(1)/1.5\) | A1 | (\(T = 40\)) |
| For resolving forces vertically | M1 | (2 components of \(T\) necessary) |
| \(W = 2T\left(\frac{3}{5}\right)\) | M1 | |
| \(W = 48\) N | A1 | 4 marks |
**(i)** For using $T = \frac{\lambda x}{L}$ | M1 | ($T = 60(0.5)/1.5$)
Tension is 20 N | A1 | 2 marks
**(ii)** For using extension 0.5 m (or 1 m) in Hooke's Law | M1 |
$T = 60(0.5)/0.75$ or $T = 60(1)/1.5$ | A1 | ($T = 40$)
For resolving forces vertically | M1 | (2 components of $T$ necessary)
$W = 2T\left(\frac{3}{5}\right)$ | M1 |
$W = 48$ N | A1 | 4 marks
---1 A light elastic string has natural length 1.5 m and modulus of elasticity 60 N . The string is stretched between two fixed points $A$ and $B$, which are at the same horizontal level and 2 m apart.\\
(i) Find the tension in the string.
A particle of weight $W \mathrm {~N}$ is now attached to the mid-point of the string and the particle is in equilibrium at a point 0.75 m vertically below the mid-point of $A B$.\\
(ii) Find the value of $W$.
\hfill \mbox{\textit{CAIE M2 2004 Q1 [6]}}