| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Energy methods in SHM |
| Difficulty | Challenging +1.2 This is a multi-step SHM problem requiring geometric setup, equilibrium analysis with Hooke's law, and energy conservation. Part (a) involves resolving forces with Pythagoras (finding string lengths/extensions) and symmetry arguments—standard but requires careful algebra. Part (b) uses energy methods to find the upper turning point. While requiring several techniques and careful bookkeeping across 15 marks, it follows predictable patterns for M3 SHM questions without requiring novel insight or proof techniques beyond the syllabus. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(AP = \sqrt{(1.5l)^2 + (2l)^2} = 2.5l\) | M1 A1 | |
| \(\cos\alpha = \frac{4}{5}\) | B1 | |
| Hooke's Law: \(T = \frac{\lambda(2.5l - 1.5l)}{1.5l} = \left(\frac{2\lambda}{3}\right)\) | M1 A1 | |
| \(\uparrow\) \(2T\cos\alpha = mg\) \(\left(T = \frac{5mg}{8}\right)\) | M1 A1 | |
| \(2 \times \frac{2\lambda}{3} \times \frac{4}{5} = mg\) \(\left(\frac{2\lambda}{3} = \frac{5mg}{8}\right)\) | M1 | |
| \(\lambda = \frac{15mg}{16}\) | A1 (cso) | 9 marks |
| (b) \(h = \sqrt{(3.9l)^2 - (1.5l)^2} = 3.6l\) | M1 A1 | |
| Energy: \(\frac{1}{2}mv^2 + mg \times h = 2 \times \frac{15mg}{16} \times \frac{(2.4l)^2}{2 \times 1.5l}\) ft their \(h\) | M1 A1 ft | |
| Leading to \(v = 0\) | A1 (cso) | 6 marks [15] |
**(a)** $AP = \sqrt{(1.5l)^2 + (2l)^2} = 2.5l$ | M1 A1 |
$\cos\alpha = \frac{4}{5}$ | B1 |
Hooke's Law: $T = \frac{\lambda(2.5l - 1.5l)}{1.5l} = \left(\frac{2\lambda}{3}\right)$ | M1 A1 |
$\uparrow$ $2T\cos\alpha = mg$ $\left(T = \frac{5mg}{8}\right)$ | M1 A1 |
$2 \times \frac{2\lambda}{3} \times \frac{4}{5} = mg$ $\left(\frac{2\lambda}{3} = \frac{5mg}{8}\right)$ | M1 |
$\lambda = \frac{15mg}{16}$ | A1 (cso) | 9 marks
**(b)** $h = \sqrt{(3.9l)^2 - (1.5l)^2} = 3.6l$ | M1 A1 |
Energy: $\frac{1}{2}mv^2 + mg \times h = 2 \times \frac{15mg}{16} \times \frac{(2.4l)^2}{2 \times 1.5l}$ ft their $h$ | M1 A1 ft |
Leading to $v = 0$ | A1 (cso) | 6 marks [15]\includegraphics{figure_1}
A light elastic string, of natural length $3l$ and modulus of elasticity $\lambda$, has its ends attached to two points $A$ and $B$, where $AB = 3l$ and $AB$ is horizontal. A particle $P$ of mass $m$ is attached to the mid-point of the string. Given that $P$ rests in equilibrium at a distance $2l$ below $AB$, as shown in Figure 1,
\begin{enumerate}[label=(\alph*)]
\item show that $\lambda = \frac{15mg}{16}$ [9]
\end{enumerate}
The particle is pulled vertically downwards from its equilibrium position until the total length of the elastic string is $7.8l$. The particle is released from rest.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $P$ comes to instantaneous rest on the line $AB$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2007 Q7 [15]}}