| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 13 |
| 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 requires multiple steps: finding extensions using Pythagoras, resolving forces in equilibrium with symmetry, then applying energy conservation with elastic potential energy. Part (a) involves careful geometry and force resolution to prove k=10. Part (b) requires setting up and solving an energy equation with multiple elastic strings. While systematic, it demands strong spatial reasoning, accurate handling of elastic energy formulae, and algebraic manipulation across 13 marks total—significantly above average difficulty for A-level. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| \(\text{length } AP = \text{length } BP = \frac{5}{4}l\) | B1 |
| \(T_a = T_b = \frac{kmg(\frac{1}{4}l)}{l} = \frac{1}{4}kmg\) (or \(T = ...\)) | M1A1 |
| \(R(\uparrow): T_a\cos\theta + T_b\cos\theta = 3mg\) (or \(2T\cos\theta = 3mg\)) | M1A1 |
| \(\frac{1}{4}kmg \times \frac{3}{5} + \frac{1}{4}kmg \times \frac{3}{5} = 3mg\) | |
| (or \(\frac{1}{2}kmg \times \frac{3}{5} = 3mg\)) | |
| \(\frac{3}{10}kmg = 3mg\) | A1 |
| \(k = 10\) * | A1 |
| Answer | Marks |
|---|---|
| \(\text{initial extn} = \frac{13}{5}l - l = \frac{8}{5}l\) | B1 |
| \(\text{E.P.E. lost} = 2 \times \frac{\lambda x^2}{2l} = 2 \times \frac{10mg(\frac{8l}{5})^2}{2l} = \frac{128mgl}{5}\) | M1A1 |
| \(\text{P.E. gained} = 3mg \times \frac{12l}{5} = \frac{36mgl}{5}\) | |
| \(\frac{1}{2} \times 3mv^2 + \frac{36mgl}{5} = \frac{128mgl}{5}\) | M1A1 |
| \(v^2 = \frac{256gl}{15} - \frac{72gl}{15}\) | |
| \(v = \sqrt{(\frac{184}{15}gl)}\) | A1 |
## Part (a):
$\text{length } AP = \text{length } BP = \frac{5}{4}l$ | B1 |
$T_a = T_b = \frac{kmg(\frac{1}{4}l)}{l} = \frac{1}{4}kmg$ (or $T = ...$) | M1A1 |
$R(\uparrow): T_a\cos\theta + T_b\cos\theta = 3mg$ (or $2T\cos\theta = 3mg$) | M1A1 |
$\frac{1}{4}kmg \times \frac{3}{5} + \frac{1}{4}kmg \times \frac{3}{5} = 3mg$ | |
(or $\frac{1}{2}kmg \times \frac{3}{5} = 3mg$) | |
$\frac{3}{10}kmg = 3mg$ | A1 |
$k = 10$ * | A1 |
## Part (b):
$\text{initial extn} = \frac{13}{5}l - l = \frac{8}{5}l$ | B1 |
$\text{E.P.E. lost} = 2 \times \frac{\lambda x^2}{2l} = 2 \times \frac{10mg(\frac{8l}{5})^2}{2l} = \frac{128mgl}{5}$ | M1A1 |
$\text{P.E. gained} = 3mg \times \frac{12l}{5} = \frac{36mgl}{5}$ | |
$\frac{1}{2} \times 3mv^2 + \frac{36mgl}{5} = \frac{128mgl}{5}$ | M1A1 |
$v^2 = \frac{256gl}{15} - \frac{72gl}{15}$ | |
$v = \sqrt{(\frac{184}{15}gl)}$ | A1 |\includegraphics{figure_4}
A small ball of mass $3m$ is attached to the ends of two light elastic strings $AP$ and $BP$, each of natural length $l$ and modulus of elasticity $kmg$. The ends $A$ and $B$ of the strings are attached to fixed points on the same horizontal level, with $AB = 2l$. The mid-point of $AB$ is $C$. The ball hangs in equilibrium at a distance $\frac{3}{4}l$ vertically below $C$ as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = 10$
[7]
\end{enumerate}
The ball is now pulled vertically downwards until it is at a distance $\frac{15}{8}l$ below $C$. The ball is released from rest.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the speed of the ball as it reaches $C$.
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2011 Q6 [13]}}