| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| 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 | Challenging +1.2 This is a multi-part elastic strings problem requiring energy conservation for part (a) and equilibrium resolution for part (b). While it involves several steps (geometry, EPE calculations, energy methods, force resolution), the techniques are standard M3 fare with no novel insights required. The algebra in part (b) is moderately involved but follows a predictable pattern. Slightly above average due to the geometric setup and algebraic manipulation needed. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 06.02g 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 |
|---|---|
| \(AP = \sqrt{(0.75^2 + 1^2)} = 1.25\) | M1 A1 |
| Conservation of energy: \(\frac{1}{2} \times 2 \times v^2 + 2 \times \frac{49 \times 0.5^2}{2 \times 0.75} = 2g \times 1\) | M1 A2 (1, 0) |
| Leading to \(v \approx 1.8\) (m s\(^{-1}\)) accept 1.81 | A1 |
| (6) |
| Answer | Marks |
|---|---|
| \(R(\uparrow)\) \(2T\cos\alpha = 2g\) | M1 A1 |
| \(y = \frac{0.75}{\sin\alpha}\) | M1 A1 |
| Hooke's Law: \(T = \frac{49}{0.75}\left(\frac{0.75}{\sin\alpha} - 0.75\right)\) | M1 A1 |
| \(= 49\left(\frac{1}{\sin\alpha} - 1\right)\) | |
| \(\frac{9.8}{\cos\alpha} = 49\left(\frac{1}{\sin\alpha} - 1\right)\) | M1 |
| \(\tan\alpha = 5(1 - \sin\alpha)\) | |
| \(5 = \tan\alpha + 5\sin\alpha\) ✱ | A1 |
| Total: [12] |
**(a)**
| $AP = \sqrt{(0.75^2 + 1^2)} = 1.25$ | M1 A1 |
| Conservation of energy: $\frac{1}{2} \times 2 \times v^2 + 2 \times \frac{49 \times 0.5^2}{2 \times 0.75} = 2g \times 1$ | M1 A2 (1, 0) |
| Leading to $v \approx 1.8$ (m s$^{-1}$) accept 1.81 | A1 |
| | (6) |
**Note:** −1 for each incorrect term
**(b)**
| $R(\uparrow)$ $2T\cos\alpha = 2g$ | M1 A1 |
| $y = \frac{0.75}{\sin\alpha}$ | M1 A1 |
| Hooke's Law: $T = \frac{49}{0.75}\left(\frac{0.75}{\sin\alpha} - 0.75\right)$ | M1 A1 |
| $= 49\left(\frac{1}{\sin\alpha} - 1\right)$ | |
| $\frac{9.8}{\cos\alpha} = 49\left(\frac{1}{\sin\alpha} - 1\right)$ | M1 |
| $\tan\alpha = 5(1 - \sin\alpha)$ | |
| $5 = \tan\alpha + 5\sin\alpha$ ✱ | A1 |
| **Total:** [12] |
**Note:** Eliminating $T$; cso
---Two light elastic strings each have natural length $0.75$ m and modulus of elasticity $49$ N. A particle $P$ of mass $2$ kg is attached to one end of each string. The other ends of the strings are attached to fixed points $A$ and $B$, where $AB$ is horizontal and $AB = 1.5$ m.
\includegraphics{figure_2}
The particle is held at the mid-point of $AB$. The particle is released from rest, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ when it has fallen a distance of $1$ m.
[6]
\end{enumerate}
Given instead that $P$ hangs in equilibrium vertically below the mid-point of $AB$, with $\angle APB = 2\alpha$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that $\tan \alpha + 5 \sin \alpha = 5$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2006 Q5 [12]}}