| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: projected from equilibrium or other point |
| Difficulty | Challenging +1.2 This is a standard energy conservation problem in elastic strings requiring students to find equilibrium position using Hooke's law, then apply conservation of energy with elastic potential energy. While it involves multiple steps (equilibrium, energy equation, solving), the approach is methodical and well-practiced in Further Mechanics syllabi. The 6 marks reflect routine application rather than novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks |
|---|---|
| 3(a) | 2mg |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | Equilibrium position. |
| Extension = 1 m | A1 |
| Answer | Marks |
|---|---|
| 2 2 | B1 |
| Answer | Marks |
|---|---|
| Gain in GPE = | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | |
| d = 5 | A1 | SC: 3 marks for final answer of 5+1 . |
| Answer | Marks |
|---|---|
| 3(b) | 1 1 2mg |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 2 | M1 | GPE, KE, EPE terms. |
| V2 =g ( d2 −1 ) V = 40 =2 10 | A1 |
| Answer | Marks |
|---|---|
| 2 | M1 |
| V2 =4g V = 40 =2 10 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(a) ---
3(a) | 2mg
Hooke’s law: T = extension and T =mg
2 | M1 | Equilibrium position.
Extension = 1 m | A1
1 2mg
EPE loss = (1+d)2
2 2 | B1
mg(2+1+d)
Gain in GPE = | B1
1
Equate: mg(1+d)2 =mg(3+d)
2 | M1
d = 5 | A1 | SC: 3 marks for final answer of 5+1 .
SC: 2 marks for final answer of 5+k , k 1 .
6
--- 3(b) ---
3(b) | 1 1 2mg
Energy equation: mV2 +mg(1+d)= (1+d)2
2 2 2 | M1 | GPE, KE, EPE terms.
V2 =g ( d2 −1 ) V = 40 =2 10 | A1
Alternatively:
Using KE and GPE from 2 m below O to point O
1
mV2 =2mg
2 | M1
V2 =4g V = 40 =2 10 | A1
2
Question | Answer | Marks | GuidanceA particle $P$ of mass $m\text{kg}$ is attached to one end of a light elastic string of natural length $2\text{m}$ and modulus of elasticity $2mg\text{N}$. The other end of the string is attached to a fixed point $O$. The particle $P$ hangs in equilibrium vertically below $O$. The particle $P$ is pulled down vertically a distance $d\text{m}$ below its equilibrium position and released from rest.
\begin{enumerate}[label=(\alph*)]
\item Given that the particle just reaches $O$ in the subsequent motion, find the value of $d$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q3 [6]}}