| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 11 |
| 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 | Standard +0.3 This is a standard M2 mechanics problem involving elastic strings in equilibrium and energy conservation. Parts (i)-(ii) require routine application of Hooke's law and resolving forces vertically, while parts (iii)-(iv) use the standard elastic PE formula and energy conservation. The geometry is straightforward (Pythagoras), and all techniques are textbook applications with no novel insight required. Slightly above average difficulty due to multiple parts and the energy calculation, but well within typical M2 scope. |
| Spec | 3.03b Newton's first law: equilibrium6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Using \(T = \dfrac{\lambda x}{L}\) \(\left(\dfrac{130 \times 3}{10}\text{ or }\dfrac{130 \times 1.5}{5}\right)\) | M1 | |
| Tension is 39 N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolving forces vertically \(\left(mg = 2 \times 39 \times \dfrac{5}{13}\right)\) | M1 | |
| Mass is 3 kg | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Extension \(= 20 - 10\) (or \(10 - 5\)) | B1 | |
| Using EPE \(= \dfrac{\lambda x^2}{2L}\) (\(L\) must be 10 or 5; must be attempt at extension, e.g. \(x=20\) or \(x = 8 - 2.5\) is M0) | M1 | |
| \(\text{EPE} = \dfrac{130 \times 10^2}{2 \times 10}\) or EPE \(= 2 \times \dfrac{130 \times 5^2}{2 \times 5}\) (Allow M1 only for \(x=2\) or \(3\)) | ||
| EPE is 650 J (ft attempted extension in lowest position) | A1 ft |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Change in GPE \(= 3 \times 10 \times 8\) | B1 ft | |
| Using principle of conservation of energy with KE, GPE and EPE all represented | M1 | |
| \(650 = \frac{1}{2}(3)v^2 + 3 \times 10 \times 8 + \dfrac{130 \times 2^2}{2 \times 10}\) | A1 ft | |
| Speed is 16 ms\(^{-1}\) | A1 |
# Question 7:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Using $T = \dfrac{\lambda x}{L}$ $\left(\dfrac{130 \times 3}{10}\text{ or }\dfrac{130 \times 1.5}{5}\right)$ | M1 | |
| Tension is 39 N | A1 | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolving forces vertically $\left(mg = 2 \times 39 \times \dfrac{5}{13}\right)$ | M1 | |
| Mass is 3 kg | A1 | |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Extension $= 20 - 10$ (or $10 - 5$) | B1 | |
| Using EPE $= \dfrac{\lambda x^2}{2L}$ ($L$ must be 10 or 5; must be attempt at extension, e.g. $x=20$ or $x = 8 - 2.5$ is M0) | M1 | |
| $\text{EPE} = \dfrac{130 \times 10^2}{2 \times 10}$ or EPE $= 2 \times \dfrac{130 \times 5^2}{2 \times 5}$ (Allow M1 only for $x=2$ or $3$) | | |
| EPE is 650 J (ft attempted extension in lowest position) | A1 ft | |
## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| Change in GPE $= 3 \times 10 \times 8$ | B1 ft | |
| Using principle of conservation of energy with KE, GPE and EPE all represented | M1 | |
| $650 = \frac{1}{2}(3)v^2 + 3 \times 10 \times 8 + \dfrac{130 \times 2^2}{2 \times 10}$ | A1 ft | |
| Speed is 16 ms$^{-1}$ | A1 | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{7f8646df-a7d8-4ca1-a6ee-3ceab6bb83af-4_232_905_762_621}
A light elastic string has natural length 10 m and modulus of elasticity 130 N . The ends of the string are attached to fixed points $A$ and $B$, which are at the same horizontal level. A small stone is attached to the mid-point of the string and hangs in equilibrium at a point 2.5 m below $A B$, as shown in the diagram. With the stone in this position the length of the string is 13 m .\\
(i) Find the tension in the string.\\
(ii) Show that the mass of the stone is 3 kg .
The stone is now held at rest at a point 8 m vertically below the mid-point of $A B$.\\
(iii) Find the elastic potential energy of the string in this position.\\
(iv) The stone is now released. Find the speed with which it passes through the mid-point of $A B$.
\hfill \mbox{\textit{CAIE M2 2003 Q7 [11]}}