| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| 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 two-part elastic string problem requiring (i) equilibrium resolution with Hooke's law and Pythagoras to find mass, and (ii) energy conservation with elastic PE. While it involves multiple steps and careful geometry, the techniques are standard M2 content with no novel insight required—moderately above average difficulty. |
| 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 principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e = \sqrt{(0.5^2 + 1.2^2)} - 1 = 0.3\) | B1 | |
| \(T = 39 \times 0.3/1\) | M1 | Uses \(T = \lambda x / L\). |
| \(mg = 2 \times (39 \times 0.3/1) \times 0.5/1.3\) | M1 | Resolves vertically. |
| \(m = 0.9\) | A1 | AG |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E = \sqrt{(1.6^2 + 1.2^2)} - 1 = 1\text{ m}\) | B1 | \(E\) = extension when the particle comes to instantaneous rest. |
| \(EE = 39 \times 1^2/(2 \times 1)\) or \(39 \times 0.3^2/(2 \times 1)\) | B1 | |
| \(0.9v^2/2 + 0.9g(1.6 - 0.5) = 2[39 \times 1^2/(2 \times 1) - 39 \times 0.3^2/(2 \times 1)]\) | M1A1 | Set up a 4 term energy equation involving \(EE\), \(KE\) and \(PE\). |
| \(v = 7.54\text{ m s}^{-1}\) | A1 | |
| Total | 5 |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e = \sqrt{(0.5^2 + 1.2^2)} - 1 = 0.3$ | B1 | |
| $T = 39 \times 0.3/1$ | M1 | Uses $T = \lambda x / L$. |
| $mg = 2 \times (39 \times 0.3/1) \times 0.5/1.3$ | M1 | Resolves vertically. |
| $m = 0.9$ | A1 | AG |
| **Total** | **4** | |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E = \sqrt{(1.6^2 + 1.2^2)} - 1 = 1\text{ m}$ | B1 | $E$ = extension when the particle comes to instantaneous rest. |
| $EE = 39 \times 1^2/(2 \times 1)$ or $39 \times 0.3^2/(2 \times 1)$ | B1 | |
| $0.9v^2/2 + 0.9g(1.6 - 0.5) = 2[39 \times 1^2/(2 \times 1) - 39 \times 0.3^2/(2 \times 1)]$ | M1A1 | Set up a 4 term energy equation involving $EE$, $KE$ and $PE$. |
| $v = 7.54\text{ m s}^{-1}$ | A1 | |
| **Total** | **5** | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{6b220343-1d64-4dbc-a42d-77967eef9c6d-06_264_839_260_653}
A light elastic string has natural length 2 m and modulus of elasticity 39 N . The ends of the string are attached to fixed points $A$ and $B$ which are at the same horizontal level and 2.4 m apart. A particle $P$ of mass $m \mathrm {~kg}$ is attached to the mid-point of the string and hangs in equilibrium at a point 0.5 m below $A B$ (see diagram).\\
(i) Show that $m = 0.9$.\\
$P$ is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point 1.6 m below $A B$.\\
(ii) Calculate the speed of projection of $P$.\\
\hfill \mbox{\textit{CAIE M2 2017 Q4 [9]}}