| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| 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 | Challenging +1.2 This is a multi-step energy conservation problem requiring geometric reasoning to find string extensions, application of Hooke's law, and force resolution. While it involves several techniques (energy conservation, Pythagoras, elastic forces, Newton's second law), the approach is methodical and follows standard M2 procedures without requiring novel insight. The geometry and algebra are moderately involved but straightforward for this level. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(EE = 21(\sqrt{1.2^2+1.6^2}-1.2)^2/(2\times1.2)\) | B1 | Use of EE formula \((= 5.6\text{ J})\) |
| \(m12^2/2 = mg\times1.6 + 2\times21(\sqrt{1.2^2+1.6^2}-1.2)^2/(2\times1.2)\) | M1, A1 | KE/EE/PE conservation |
| \(m = 0.2\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = 21(\sqrt{1.2^2+0.5^2}-1.2)/1.2\) | B1 | |
| \(ma = 2\times21(\sqrt{1.2^2+0.5^2}-1.2)/1.2 \times \frac{0.5}{1.3} - mg\) | M1, A1 | Newton's Second Law with component of T or reversed signs |
| \(a = (-)3.27\text{ ms}^{-2}\) | A1 [4] |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $EE = 21(\sqrt{1.2^2+1.6^2}-1.2)^2/(2\times1.2)$ | B1 | Use of EE formula $(= 5.6\text{ J})$ |
| $m12^2/2 = mg\times1.6 + 2\times21(\sqrt{1.2^2+1.6^2}-1.2)^2/(2\times1.2)$ | M1, A1 | KE/EE/PE conservation |
| $m = 0.2$ | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = 21(\sqrt{1.2^2+0.5^2}-1.2)/1.2$ | B1 | |
| $ma = 2\times21(\sqrt{1.2^2+0.5^2}-1.2)/1.2 \times \frac{0.5}{1.3} - mg$ | M1, A1 | Newton's Second Law with component of T or reversed signs |
| $a = (-)3.27\text{ ms}^{-2}$ | A1 [4] | |
---4 A light elastic string has natural length 2.4 m and modulus of elasticity 21 N . A particle $P$ of mass $m \mathrm {~kg}$ is attached to the mid-point of the string. The ends of the string are attached to fixed points $A$ and $B$ which are 2.4 m apart at the same horizontal level. $P$ is projected vertically upwards with velocity $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the mid-point of $A B$. In the subsequent motion $P$ is at instantaneous rest at a point 1.6 m above $A B$.\\
(i) Find $m$.\\
(ii) Calculate the acceleration of $P$ when it first passes through a point 0.5 m below $A B$.
\hfill \mbox{\textit{CAIE M2 2012 Q4 [8]}}