| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Moderate -0.3 This is a straightforward two-part elastic string problem requiring standard application of Hooke's Law and equilibrium conditions. Part (a) involves simple substitution into F = (λx)/l with vertical forces only. Part (b) requires resolving forces in two directions and using Pythagoras, but follows a well-practiced routine with no novel insight needed. Slightly easier than average due to clear setup and standard method. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\) or \(\frac{\lambda \times e}{l} = mg\) | M1 | even \(T=m\) is M1, A0, A0 special case |
| \(\frac{\lambda \times 0.16}{0.4} = 2g\) | A1 | |
| \(\Rightarrow \lambda = 49\text{ N}\) or \(5g\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R(\uparrow)\ T\cos\theta = mg\) or \(\cos\theta = \frac{mg}{T}\) | M1 | Special case \(T\sin\theta = mg\) giving \(\theta=30°\) is M1 A0 A0 unless evidence they think \(\theta\) is with horizontal — then M1 A1 A0 |
| \(49\cdot\frac{0.32}{0.4}\cdot\cos\theta = 19.6\) or \(4g\cos\theta = 2g\) or \(2mg\cos\theta = mg\) | A1ft | ft on their \(\lambda\) |
| \(\Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = 60°\) (or \(\frac{\pi}{3}\) radians) | A1 | (3) |
## Question 1:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T$ or $\frac{\lambda \times e}{l} = mg$ | M1 | even $T=m$ is M1, A0, A0 special case |
| $\frac{\lambda \times 0.16}{0.4} = 2g$ | A1 | |
| $\Rightarrow \lambda = 49\text{ N}$ or $5g$ | A1 | (3) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R(\uparrow)\ T\cos\theta = mg$ or $\cos\theta = \frac{mg}{T}$ | M1 | Special case $T\sin\theta = mg$ giving $\theta=30°$ is M1 A0 A0 unless evidence they think $\theta$ is with horizontal — then M1 A1 A0 |
| $49\cdot\frac{0.32}{0.4}\cdot\cos\theta = 19.6$ or $4g\cos\theta = 2g$ or $2mg\cos\theta = mg$ | A1ft | ft on their $\lambda$ |
| $\Rightarrow \cos\theta = \frac{1}{2} \Rightarrow \theta = 60°$ (or $\frac{\pi}{3}$ radians) | A1 | (3) |
**Total: 6 marks**
---\begin{enumerate}
\item A light elastic string of natural length 0.4 m has one end $A$ attached to a fixed point. The other end of the string is attached to a particle $P$ of mass 2 kg . When $P$ hangs in equilibrium vertically below $A$, the length of the string is 0.56 m .\\
(a) Find the modulus of elasticity of the string.
\end{enumerate}
A horizontal force is applied to $P$ so that it is held in equilibrium with the string making an angle $\theta$ with the downward vertical. The length of the string is now 0.72 m .\\
(b) Find the angle $\theta$.\\
\hfill \mbox{\textit{Edexcel M3 2008 Q1 [6]}}