| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with elastic strings. Students need to equate gravitational PE lost to elastic PE gained, using standard formulas. It's a single-step calculation with clearly defined values, making it slightly easier than average for M3 level. |
| Spec | 6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{EPE} = \dfrac{\lambda \times 0.5^2}{1.2}\) | B1 | Correct expression for EPE |
| \(\text{GPE lost} = \text{EPE gained}\) | M1 (used) | Energy conservation equation used |
| \(0.8 \times 9.8 \times 1.1 = \dfrac{\lambda \times 0.5^2}{1.2}\) | A1ft | Follow through on EPE expression |
| \(\lambda = 41.4 \text{ N or } 41 \text{ N}\) | A1 | Correct final answer |
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{EPE} = \dfrac{\lambda \times 0.5^2}{1.2}$ | B1 | Correct expression for EPE |
| $\text{GPE lost} = \text{EPE gained}$ | M1 (used) | Energy conservation equation used |
| $0.8 \times 9.8 \times 1.1 = \dfrac{\lambda \times 0.5^2}{1.2}$ | A1ft | Follow through on EPE expression |
| $\lambda = 41.4 \text{ N or } 41 \text{ N}$ | A1 | Correct final answer |
**Total: 4 marks**\begin{enumerate}
\item A particle of mass 0.8 kg is attached to one end of a light elastic string of natural length 0.6 m . The other end of the string is attached to a fixed point $A$. The particle is released from rest at $A$ and comes to instantaneous rest 1.1 m below $A$.
\end{enumerate}
Find the modulus of elasticity of the string.\\
\hfill \mbox{\textit{Edexcel M3 2012 Q1 [4]}}