| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Standard +0.3 This is a straightforward elastic string equilibrium problem requiring basic trigonometry to find geometry, resolving forces vertically to find tension, and applying Hooke's law. All steps are standard M2 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 3.03b Newton's first law: equilibrium6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Tension is 1.3 N | M1, A1 | For resolving forces vertically |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda = 15.6\) | A1, M1 | For using \(T = \lambda x/l\) |
**(i)** $m g = 2T \cos \alpha$
$0.064 \times 10 = 2(0.08/0.325)$ (or $2T \cos 75.7°$)
Tension is 1.3 N | M1, A1 | For resolving forces vertically
**Alternative:** $1.3 = \lambda (0.0025/0.3)$ (or $\lambda (0.05/0.6)$)
$\lambda = 15.6$ | A1, M1 | For using $T = \lambda x/l$
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\includegraphics[max width=\textwidth, alt={}, center]{ece63d46-5e56-4668-939a-9dbbcfc1a77a-2_248_1267_276_440}
A light elastic string has natural length 0.6 m and modulus of elasticity $\lambda \mathrm { N }$. The ends of the string are attached to fixed points $A$ and $B$, which are at the same horizontal level and 0.63 m apart. A particle $P$ of mass 0.064 kg is attached to the mid-point of the string and hangs in equilibrium at a point 0.08 m below $A B$ (see diagram). Find\\
(i) the tension in the string,\\
(ii) the value of $\lambda$.
\hfill \mbox{\textit{CAIE M2 2006 Q1 [5]}}