| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on smooth inclined plane |
| Difficulty | Standard +0.3 This is a standard energy conservation problem with elastic strings on an inclined plane. While it requires setting up energy equations correctly (EPE, GPE, KE) and some algebraic manipulation, the method is routine for M2 students and the given result provides scaffolding for parts (ii) and (iii), which become straightforward optimization exercises. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EE gain \(= 2x^2/(2\times4)\) | B1 | |
| PE loss \(= 0.1g(4+x)\sin30^\circ\) | B1 | |
| \(\frac{1}{2}(0.1)v^2 = 2 + 0.5x - 0.25x^2\) | M1 | For using KE gain = PE loss \(-\) EE gain |
| \(v^2 = 40 + 10x - 5x^2 = 45-(5x^2-10x+5)\) | M1 | For attempting to express \(v^2\) in required form |
| \(v^2 = 45 - 5(x-1)^2\) | A1 | AG [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5(x-1)^2 = 45\) | M1 | For substituting \(v = 0\) |
| Distance is \(8\text{ m}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(x = 1\) for maximum \(v\) | |
| Maximum speed is \(6.71 \text{ ms}^{-1}\) | A1 | [2] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| EE gain $= 2x^2/(2\times4)$ | B1 | |
| PE loss $= 0.1g(4+x)\sin30^\circ$ | B1 | |
| $\frac{1}{2}(0.1)v^2 = 2 + 0.5x - 0.25x^2$ | M1 | For using KE gain = PE loss $-$ EE gain |
| $v^2 = 40 + 10x - 5x^2 = 45-(5x^2-10x+5)$ | M1 | For attempting to express $v^2$ in required form |
| $v^2 = 45 - 5(x-1)^2$ | A1 | AG **[5]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5(x-1)^2 = 45$ | M1 | For substituting $v = 0$ |
| Distance is $8\text{ m}$ | A1 | **[2]** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $x = 1$ for maximum $v$ |
| Maximum speed is $6.71 \text{ ms}^{-1}$ | A1 | **[2]** |
---6 A light elastic string has natural length 4 m and modulus of elasticity 2 N . One end of the string is attached to a fixed point $O$ of a smooth plane which is inclined at $30 ^ { \circ }$ to the horizontal. The other end of the string is attached to a particle $P$ of mass $0.1 \mathrm {~kg} . P$ is held at rest at $O$ and then released. The speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when the extension of the string is $x \mathrm {~m}$.\\
(i) Show that $v ^ { 2 } = 45 - 5 ( x - 1 ) ^ { 2 }$.
Hence find\\
(ii) the distance of $P$ from $O$ when $P$ is at its lowest point,\\
(iii) the maximum speed of $P$.
\hfill \mbox{\textit{CAIE M2 2008 Q6 [9]}}