| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Standard +0.3 This is a standard three-part elastic string problem requiring Hooke's law for equilibrium, energy conservation for speed, and energy methods for maximum extension. While it involves multiple steps and energy principles, the approach is methodical and follows textbook procedures without requiring novel insight. Slightly above average difficulty due to the energy conservation setup and careful tracking of gravitational PE, elastic PE, and KE across three related parts. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.25g = 20e/0.4\) | M1 | Uses \(T = 2\pi/L\) |
| \(\text{OP} ( = 0.05 + 0.4) = 0.45 \text{ m}\) | A1 | [2] |
| (ii) \(20 \times 0.05^2/(2 \times 0.4) + 0.25v^2/2 = 0.25g \times 0.45\) | M1 | |
| A1 | ||
| \(v = 2.92 \text{ ms}^{-1}\) | A1 | [3] |
| (iii) \(20(d - 0.4)^2/(2 \times 0.4) = 0.25gd\) | M1 | Hence \(d^2 - (0.8 + 0.1)d + 0.16 = 0\) |
| \(d = [0.9 \pm \sqrt{(0.9^2 - 4 \times 0.16)}]/2\) | M1 | Solves a 3 term quadratic equation |
| \(d = 0.656\) | A1 | [3] Ignore \(d = 0.244\) if seen |
**(i)** $0.25g = 20e/0.4$ | M1 | Uses $T = 2\pi/L$
$\text{OP} ( = 0.05 + 0.4) = 0.45 \text{ m}$ | A1 | [2]
**(ii)** $20 \times 0.05^2/(2 \times 0.4) + 0.25v^2/2 = 0.25g \times 0.45$ | M1
| A1
$v = 2.92 \text{ ms}^{-1}$ | A1 | [3]
**(iii)** $20(d - 0.4)^2/(2 \times 0.4) = 0.25gd$ | M1 | Hence $d^2 - (0.8 + 0.1)d + 0.16 = 0$
$d = [0.9 \pm \sqrt{(0.9^2 - 4 \times 0.16)}]/2$ | M1 | Solves a 3 term quadratic equation
$d = 0.656$ | A1 | [3] Ignore $d = 0.244$ if seen
---3 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 20 N is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $0.25 \mathrm {~kg} . P$ hangs in equilibrium below $O$.\\
(i) Calculate the distance $O P$.
The particle $P$ is raised, and is released from rest at $O$.\\
(ii) Calculate the speed of $P$ when it passes through the equilibrium position.\\
(iii) Calculate the greatest value of the distance $O P$ in the subsequent motion.
\hfill \mbox{\textit{CAIE M2 2011 Q3 [8]}}