| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: vertical motion |
| Difficulty | Standard +0.8 This is a multi-step energy conservation problem involving elastic strings in 2D geometry. Students must handle the geometric configuration (finding stretched lengths using Pythagoras), apply Hooke's law for elastic potential energy with two string segments, and use energy conservation twice. The geometry and dual-string setup elevate this above routine mechanics questions, but it follows a standard framework once the setup is understood. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(e = \sqrt{(0.6^2 + 0.32^2)} - 0.4 (= 0.28)\), \(0.3g \times 0.32 = 2[\lambda(0.28^2 - 0.2^2)] / (2 \times 0.4)\), \(\lambda = 10\) | B1, M1, A1, A1 | Extension of half string = 0.28 m; PE loss = EE gain [4] |
| (ii) \(e = \sqrt{(0.6^2 + 0.25^2)} - 0.4\), \(0.3g \times 0.25 = 0.3v^2 / 2 +\) \(2[10(0.25^2 - 0.2^2) / (2 \times 0.4)]\), \(v = 1.12\) | B1, M1, A1ft, A1 | Extension of half string = 0.25 m; PE loss = KE gain + EE gain; N.B. 0.25 is extension of half string ft on candidates' \(\lambda\) only [4] |
**(i)** $e = \sqrt{(0.6^2 + 0.32^2)} - 0.4 (= 0.28)$, $0.3g \times 0.32 = 2[\lambda(0.28^2 - 0.2^2)] / (2 \times 0.4)$, $\lambda = 10$ | B1, M1, A1, A1 | Extension of half string = 0.28 m; PE loss = EE gain [4]
**(ii)** $e = \sqrt{(0.6^2 + 0.25^2)} - 0.4$, $0.3g \times 0.25 = 0.3v^2 / 2 +$ $2[10(0.25^2 - 0.2^2) / (2 \times 0.4)]$, $v = 1.12$ | B1, M1, A1ft, A1 | Extension of half string = 0.25 m; PE loss = KE gain + EE gain; N.B. 0.25 is extension of half string ft on candidates' $\lambda$ only [4]
---4 The ends of a light elastic string of natural length 0.8 m and modulus of elasticity $\lambda \mathrm { N }$ are attached to fixed points $A$ and $B$ which are 1.2 m apart at the same horizontal level. A particle of mass 0.3 kg is attached to the centre of the string, and released from rest at the mid-point of $A B$. The particle descends 0.32 m vertically before coming to instantaneous rest.\\
(i) Calculate $\lambda$.\\
(ii) Calculate the speed of the particle when it is 0.25 m below $A B$.
\hfill \mbox{\textit{CAIE M2 2011 Q4 [8]}}