| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| 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 M2 elastic string problem requiring equilibrium analysis (Hooke's law) followed by energy conservation. Part (i) is routine application of equilibrium conditions, while part (ii) involves setting up and solving an energy equation with elastic PE, gravitational PE, and KE—all standard techniques for this topic with no novel insight required. Slightly above average due to the multi-step energy calculation. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20x/0.4 = 0.5g\sin30\) | M1 | \(\lambda\text{ext/nat length} = \text{comp weight}\) |
| \(x = 0.05\) m | A1 | [2 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20(0.05)^2/(2 \times 0.4) + 0.5 \times 5^2/2 = 20e^2/(2 \times 0.4) + 0.5 \times 2^2/2 - 0.5(e - 0.05)g\sin30\) | M1, A1 | KE/PE/EE balance with 2 KE and 2 EE terms; all terms without \(e\) correct |
| \(2.5e^2 - 2.5e - 5.1875 = 0\) | M1 | Creates/attempts to solve a 3 term quadratic equation |
| \(e = 0.508\) | A1 | [4 marks] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20x/0.4 = 0.5g\sin30$ | M1 | $\lambda\text{ext/nat length} = \text{comp weight}$ |
| $x = 0.05$ m | A1 | **[2 marks]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20(0.05)^2/(2 \times 0.4) + 0.5 \times 5^2/2 = 20e^2/(2 \times 0.4) + 0.5 \times 2^2/2 - 0.5(e - 0.05)g\sin30$ | M1, A1 | KE/PE/EE balance with 2 KE and 2 EE terms; all terms without $e$ correct |
| $2.5e^2 - 2.5e - 5.1875 = 0$ | M1 | Creates/attempts to solve a 3 term quadratic equation |
| $e = 0.508$ | A1 | **[4 marks]** |
---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 $A$ on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. The other end of the string is attached to a particle $P$ of mass 0.5 kg which rests in equilibrium on the plane.\\
(i) Calculate the extension of the string.\\
$P$ is projected down the plane from the equilibrium position with speed $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The extension of the string is $e \mathrm {~m}$ when the speed of the particle is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for the first time.\\
(ii) Find $e$.
\hfill \mbox{\textit{CAIE M2 2015 Q3 [6]}}