| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Challenging +1.2 This is a multi-step energy problem requiring work-energy principle with elastic potential energy, gravitational potential energy, and friction work. While it involves several components (spring, incline, friction), the approach is methodical and standard for M3: identify initial/final states, apply energy conservation with work done against friction. The calculation is straightforward once set up correctly, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02g 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 principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| EPE lost \(= \dfrac{\lambda \times 0.6^2}{2\times 0.9} - \dfrac{\lambda \times 0.1^2}{2\times 0.9} \left(= \dfrac{7}{36}\lambda\right)\) | M1 A1 | |
| \(R(\uparrow)\quad R = mg\cos\theta = 0.5g \times \dfrac{4}{5} = 0.4g\) | M1 | |
| \(F = \mu R = 0.15 \times 0.4g\) | M1 A1 | |
| P.E. gained = E.P.E. lost \(-\) work done against friction | ||
| \(0.5g \times 0.7\sin\theta = \dfrac{\lambda\times 0.6^2}{2\times 0.9} - \dfrac{\lambda\times 0.1^2}{2\times 0.9} - 0.15\times 0.4g\times 0.7\) | M1 A1 A1 | |
| \(0.1944\lambda = 0.5\times 9.8\times 0.7\times\dfrac{3}{5} + 0.15\times 0.4\times 9.8\times 0.7\) | ||
| \(\lambda = 12.70\ldots\) | ||
| \(\lambda = 13\ \text{N}\) or \(12.7\) | A1 |
# Question 3:
| Working | Marks | Notes |
|---------|-------|-------|
| EPE lost $= \dfrac{\lambda \times 0.6^2}{2\times 0.9} - \dfrac{\lambda \times 0.1^2}{2\times 0.9} \left(= \dfrac{7}{36}\lambda\right)$ | M1 A1 | |
| $R(\uparrow)\quad R = mg\cos\theta = 0.5g \times \dfrac{4}{5} = 0.4g$ | M1 | |
| $F = \mu R = 0.15 \times 0.4g$ | M1 A1 | |
| P.E. gained = E.P.E. lost $-$ work done against friction | | |
| $0.5g \times 0.7\sin\theta = \dfrac{\lambda\times 0.6^2}{2\times 0.9} - \dfrac{\lambda\times 0.1^2}{2\times 0.9} - 0.15\times 0.4g\times 0.7$ | M1 A1 A1 | |
| $0.1944\lambda = 0.5\times 9.8\times 0.7\times\dfrac{3}{5} + 0.15\times 0.4\times 9.8\times 0.7$ | | |
| $\lambda = 12.70\ldots$ | | |
| $\lambda = 13\ \text{N}$ or $12.7$ | A1 | |
**Total: [9]**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{049ea68c-d15f-41f8-860e-0816d36a2748-05_342_718_255_610}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle of mass 0.5 kg is attached to one end of a light elastic spring of natural length 0.9 m and modulus of elasticity $\lambda$ newtons. The other end of the spring is attached to a fixed point $O$ on a rough plane which is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 3 } { 5 }$. The coefficient of friction between the particle and the plane is 0.15 . The particle is held on the plane at a point which is 1.5 m down the line of greatest slope from $O$, as shown in Figure 2. The particle is released from rest and first comes to rest again after moving 0.7 m up the plane.
Find the value of $\lambda$.\\
\hfill \mbox{\textit{Edexcel M3 2010 Q3 [9]}}