| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| 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 elastic string problem requiring equilibrium analysis to find the modulus of elasticity, then energy conservation with careful geometry to find maximum speed. It demands understanding of elastic potential energy, geometric reasoning with Pythagoras, and recognizing that maximum speed occurs at equilibrium position. More challenging than standard mechanics questions due to the combination of equilibrium conditions and energy methods with non-trivial geometry. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2T\cos\theta = 0.28g\) | M1 | Tension component = weight |
| \(2T \times 0.7/2.5 = 2.8,\ T = 5\) | A1 | |
| \(5 = \lambda \times 0.5/2\) | M1 | Hookes Law |
| \(\lambda = 20\) N | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(0.28v^2/2 + 2\times20\times0.5^2/(2\times2) = 0.28g\times0.7 + 2\times20\times0.4^2/(2\times2)\) | M1 A1 | PE/EE/KE conservation with 4 terms |
| \(v = 2.75\) ms\(^{-1}\) | A1 [3] |
## Question 5:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2T\cos\theta = 0.28g$ | M1 | Tension component = weight |
| $2T \times 0.7/2.5 = 2.8,\ T = 5$ | A1 | |
| $5 = \lambda \times 0.5/2$ | M1 | Hookes Law |
| $\lambda = 20$ N | A1 [4] | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $0.28v^2/2 + 2\times20\times0.5^2/(2\times2) = 0.28g\times0.7 + 2\times20\times0.4^2/(2\times2)$ | M1 A1 | PE/EE/KE conservation with 4 terms |
| $v = 2.75$ ms$^{-1}$ | A1 [3] | |
---5 A particle $P$ of mass 0.28 kg is attached to the mid-point of a light elastic string of natural length 4 m . The ends of the string are attached to fixed points $A$ and $B$ which are at the same horizontal level and 4.8 m apart. $P$ is released from rest at the mid-point of $A B$. In the subsequent motion, the acceleration of $P$ is zero when $P$ is at a distance 0.7 m below $A B$.\\
(i) Show that the modulus of elasticity of the string is 20 N .\\
(ii) Calculate the maximum speed of $P$.
\hfill \mbox{\textit{CAIE M2 2010 Q5 [7]}}