| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 15 |
| 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 +0.8 This M3 question combines elastic strings, geometry, energy methods, and dynamics across three parts. Part (a) requires geometric reasoning with Hooke's law; part (b) needs vector resolution of tensions and Newton's second law; part (c) requires energy conservation with elastic potential energy. The multi-step nature, 3D geometry considerations, and integration of multiple mechanics topics makes this moderately challenging but still within standard M3 scope. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(T = \frac{\lambda x}{l} \Rightarrow 240 = \frac{\lambda \times 18}{30}\) | M1A1 | Correct use of Hooke's Law with correct values |
| \(\lambda = 400\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Extension \(= 10\) cm or \(20\) cm (used in (b) or (c)) | B1 | |
| \(T = \frac{400 \times 10}{15} = \left(\frac{800}{3}\right)\) | M1A1ft | Follow through on \(\lambda\) |
| \(R(\uparrow) \quad 2T\cos\theta - 1.5g = (\pm)1.5a\) | M1A1 | Correct resolution vertically with Newton's 2nd Law |
| \(\frac{1600}{3} \times \frac{7}{25} - 1.5 \times 9.8 = (\pm)1.5a\) | ||
| \(a = 89.75...\ \ a = 90 \text{ m s}^{-2}\) or \(89.8\) (positive) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\text{E.P.E.} = \frac{1}{2} \times 400 \times \frac{0.2^2}{0.3}\) | B1ft | Any correct EPE |
| \(1.5g \times 0.07 + \frac{1}{2} \times 1.5v^2 = 200 \times \frac{0.2^2}{0.3} - \frac{200 \times 0.18^2}{0.3}\) | M1A1A1 | Energy equation |
| \(v^2 = \frac{1}{0.75}\left(200 \times \frac{0.2^2}{0.3} - \frac{200 \times 0.18^2}{0.3} - 1.5g \times 0.07\right)\) | M1dep | Dependent method mark |
| \(v = 2.32... = 2.3 \text{ m s}^{-1}\) | A1 |
## Question 7:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $T = \frac{\lambda x}{l} \Rightarrow 240 = \frac{\lambda \times 18}{30}$ | M1A1 | Correct use of Hooke's Law with correct values |
| $\lambda = 400$ | A1 | |
---
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Extension $= 10$ cm or $20$ cm (used in (b) or (c)) | B1 | |
| $T = \frac{400 \times 10}{15} = \left(\frac{800}{3}\right)$ | M1A1ft | Follow through on $\lambda$ |
| $R(\uparrow) \quad 2T\cos\theta - 1.5g = (\pm)1.5a$ | M1A1 | Correct resolution vertically with Newton's 2nd Law |
| $\frac{1600}{3} \times \frac{7}{25} - 1.5 \times 9.8 = (\pm)1.5a$ | | |
| $a = 89.75...\ \ a = 90 \text{ m s}^{-2}$ or $89.8$ (positive) | A1 | |
---
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\text{E.P.E.} = \frac{1}{2} \times 400 \times \frac{0.2^2}{0.3}$ | B1ft | Any correct EPE |
| $1.5g \times 0.07 + \frac{1}{2} \times 1.5v^2 = 200 \times \frac{0.2^2}{0.3} - \frac{200 \times 0.18^2}{0.3}$ | M1A1A1 | Energy equation |
| $v^2 = \frac{1}{0.75}\left(200 \times \frac{0.2^2}{0.3} - \frac{200 \times 0.18^2}{0.3} - 1.5g \times 0.07\right)$ | M1dep | Dependent method mark |
| $v = 2.32... = 2.3 \text{ m s}^{-1}$ | A1 | |7. A particle $P$ of mass 1.5 kg is attached to the mid-point of a light elastic string of natural length 0.30 m and modulus of elasticity $\lambda$ newtons. The ends of the string are attached to two fixed points $A$ and $B$, where $A B$ is horizontal and $A B = 0.48 \mathrm {~m}$. Initially $P$ is held at rest at the mid-point, $M$, of the line $A B$ and the tension in the string is 240 N .
\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = 400$
The particle is now held at rest at the point $C$, where $C$ is 0.07 m vertically below $M$. The particle is released from rest at $C$.
\item Find the magnitude of the initial acceleration of $P$.
\item Find the speed of $P$ as it passes through $M$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q7 [15]}}