| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on smooth inclined plane |
| Difficulty | Standard +0.3 This is a standard M3 elastic string energy problem requiring energy conservation and equilibrium concepts. While it involves multiple steps (finding extension at rest, using energy equation, then finding maximum speed position), the techniques are routine for this module with no novel insight required. The inclined plane adds mild complexity but follows textbook methodology. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| GPE lost = EPE gained: \(mgx\sin 30° = \frac{6mgx^2}{2a}\) | M1 A1=A1 | |
| \(x = \frac{a}{6}\) | M1 | |
| \(AC = \frac{7a}{6}\) | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Greatest speed when acceleration of \(B\) is zero, i.e. where forces on \(B\) are equal | ||
| \((\nwarrow)\quad T = mg\sin 30° = \frac{6mge}{a}\) | M1 | |
| \(e = \frac{a}{12}\) | A1 | |
| CE: \(\frac{1}{2}mv^2 + \frac{6mg}{2a}\left(\frac{a}{12}\right)^2 = mg\frac{a}{12}\sin 30°\) | M1 A1=A1 | |
| \(v = \sqrt{\left(\frac{ga}{24}\right)} = \frac{\sqrt{6ga}}{12}\) | M1 A1 | (7) [12] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| CE: \(\frac{1}{2}mv^2 + \frac{6mg}{2a}x^2 = mgx\sin 30°\) | M1 A1=A1 | |
| \(v^2 = gx - \frac{6g}{a}x^2\) | ||
| \(\frac{d}{dx}(v^2) = 2v\frac{dv}{dx} = g - \frac{12g}{a}x = 0\) | M1 A1 | |
| \(x = \frac{a}{12}\) | ||
| \(v^2 = g\left(\frac{a}{12}\right) - \frac{6g}{a}\left(\frac{a}{12}\right)^2 = \frac{ga}{24}\) | M1 | |
| \(v = \sqrt{\left(\frac{ga}{24}\right)}\) | A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Centre of oscillation when extension is \(\frac{a}{12}\) | M1 A1 | |
| N2L: \(mg\sin 30° - T = m\ddot{x}\); \(\frac{1}{2}mg - \frac{6mg\left(\frac{a}{12}+x\right)}{a} = m\ddot{x}\) | M1 A1 | |
| \(\ddot{x} = -\frac{6g}{a}x \Rightarrow \omega^2 = \frac{6g}{a}\) | A1 | |
| \(v_{\max} = \omega a = \sqrt{\left(\frac{6g}{a}\right)} \times \frac{a}{12} = \sqrt{\left(\frac{ga}{24}\right)}\) | M1 A1 | (7) |
## Question 5:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| GPE lost = EPE gained: $mgx\sin 30° = \frac{6mgx^2}{2a}$ | M1 A1=A1 | |
| $x = \frac{a}{6}$ | M1 | |
| $AC = \frac{7a}{6}$ | A1 | **(5)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Greatest speed when acceleration of $B$ is zero, i.e. where forces on $B$ are equal | | |
| $(\nwarrow)\quad T = mg\sin 30° = \frac{6mge}{a}$ | M1 | |
| $e = \frac{a}{12}$ | A1 | |
| CE: $\frac{1}{2}mv^2 + \frac{6mg}{2a}\left(\frac{a}{12}\right)^2 = mg\frac{a}{12}\sin 30°$ | M1 A1=A1 | |
| $v = \sqrt{\left(\frac{ga}{24}\right)} = \frac{\sqrt{6ga}}{12}$ | M1 A1 | **(7) [12]** |
#### Alternative (b) – calculus with energy:
| Working/Answer | Marks | Guidance |
|---|---|---|
| CE: $\frac{1}{2}mv^2 + \frac{6mg}{2a}x^2 = mgx\sin 30°$ | M1 A1=A1 | |
| $v^2 = gx - \frac{6g}{a}x^2$ | | |
| $\frac{d}{dx}(v^2) = 2v\frac{dv}{dx} = g - \frac{12g}{a}x = 0$ | M1 A1 | |
| $x = \frac{a}{12}$ | | |
| $v^2 = g\left(\frac{a}{12}\right) - \frac{6g}{a}\left(\frac{a}{12}\right)^2 = \frac{ga}{24}$ | M1 | |
| $v = \sqrt{\left(\frac{ga}{24}\right)}$ | A1 | **(7)** |
#### Alternative (b) – calculus with Newton's second law:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Centre of oscillation when extension is $\frac{a}{12}$ | M1 A1 | |
| N2L: $mg\sin 30° - T = m\ddot{x}$; $\frac{1}{2}mg - \frac{6mg\left(\frac{a}{12}+x\right)}{a} = m\ddot{x}$ | M1 A1 | |
| $\ddot{x} = -\frac{6g}{a}x \Rightarrow \omega^2 = \frac{6g}{a}$ | A1 | |
| $v_{\max} = \omega a = \sqrt{\left(\frac{6g}{a}\right)} \times \frac{a}{12} = \sqrt{\left(\frac{ga}{24}\right)}$ | M1 A1 | **(7)** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8374fa0f-cb28-497f-8696-877d7d0762f1-07_311_716_249_612}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
One end $A$ of a light elastic string, of natural length $a$ and modulus of elasticity $6 m g$, is fixed at a point on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. A small ball $B$ of mass $m$ is attached to the other end of the string. Initially $B$ is held at rest with the string lying along a line of greatest slope of the plane, with $B$ below $A$ and $A B = a$. The ball is released and comes to instantaneous rest at a point $C$ on the plane, as shown in Figure 2. Find
\begin{enumerate}[label=(\alph*)]
\item the length $A C$,
\item the greatest speed attained by $B$ as it moves from its initial position to $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2009 Q5 [12]}}