| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on smooth inclined plane |
| Difficulty | Standard +0.8 This M3 question requires setting up energy conservation with elastic potential energy on an inclined plane, finding equilibrium conditions, and tracking energy through two different configurations. It demands careful bookkeeping of gravitational PE, elastic PE, and KE across multiple states, plus understanding when the string becomes slack—more sophisticated than standard M1/M2 problems but follows established M3 techniques. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Energy equation with at least three terms, including KE term \(\frac{1}{2}mV^2 + \ldots\) | M1 | |
| \(+\frac{1}{2}\cdot\frac{2mg}{a}\cdot\frac{a^2}{16} + mg\cdot\frac{1}{2}a\sin 30 = \frac{1}{2}\cdot\frac{2mg}{a}\cdot\frac{9a^2}{16}\) | A1, A1, A1 | |
| \(\Rightarrow V = \sqrt{\frac{ga}{2}}\) | dM1 A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using point where velocity is zero and point where string becomes slack: \(\frac{1}{2}mw^2 =\) | M1 | |
| \(\frac{1}{2}\cdot\frac{2mg}{a}\cdot\frac{9a^2}{16} - mg\cdot\frac{3a}{4}\cdot\sin 30\) | A1, A1 | |
| \(\Rightarrow w = \sqrt{\frac{3ag}{8}}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mw^2 - \frac{1}{2}mV_1^2 = \frac{mga}{16} - \frac{mga}{8}\) | M1, A1 A1 | |
| So \(w = \sqrt{\frac{3ag}{8}}\) | A1 |
| Answer | Marks |
|---|---|
| Using \(v^2 = \omega^2(a^2 - x^2)\) with \(\omega^2 = \frac{2g}{a}\) and \(x = \pm\frac{a}{4}\), using \('a' = \frac{a}{2}\) to give \(w = \sqrt{\frac{3ag}{8}}\) | M1 A1 A1 / A1 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy equation with at least three terms, including KE term $\frac{1}{2}mV^2 + \ldots$ | M1 | |
| $+\frac{1}{2}\cdot\frac{2mg}{a}\cdot\frac{a^2}{16} + mg\cdot\frac{1}{2}a\sin 30 = \frac{1}{2}\cdot\frac{2mg}{a}\cdot\frac{9a^2}{16}$ | A1, A1, A1 | |
| $\Rightarrow V = \sqrt{\frac{ga}{2}}$ | dM1 A1 | (6) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using point where velocity is zero and point where string becomes slack: $\frac{1}{2}mw^2 =$ | M1 | |
| $\frac{1}{2}\cdot\frac{2mg}{a}\cdot\frac{9a^2}{16} - mg\cdot\frac{3a}{4}\cdot\sin 30$ | A1, A1 | |
| $\Rightarrow w = \sqrt{\frac{3ag}{8}}$ | A1 | (4) |
**Alternative** (using point of projection and point where string becomes slack):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mw^2 - \frac{1}{2}mV_1^2 = \frac{mga}{16} - \frac{mga}{8}$ | M1, A1 A1 | |
| So $w = \sqrt{\frac{3ag}{8}}$ | A1 | |
**Notes:** DM1 in part (a) requires EE, PE and KE included. If sign errors lead to $V^2 = -\frac{ga}{2}$, last two marks are M0 A0. In part (b) for M1 need **one** KE term in energy equation of at least **3 terms** with distance $\frac{3a}{4}$ to indicate first method, and **two** KE terms in energy equation of at least **4 terms** with distance $\frac{a}{4}$ to indicate second method.
**SHM approach in part (b): Condone this method only if SHM is proved.**
| Using $v^2 = \omega^2(a^2 - x^2)$ with $\omega^2 = \frac{2g}{a}$ and $x = \pm\frac{a}{4}$, using $'a' = \frac{a}{2}$ to give $w = \sqrt{\frac{3ag}{8}}$ | M1 A1 A1 / A1 | |
**Total: 10 marks**
---\begin{enumerate}
\item A particle $P$ of mass $m$ lies on a smooth plane inclined at an angle $30 ^ { \circ }$ to the horizontal. The particle is attached to one end of a light elastic string, of natural length $a$ and modulus of elasticity $2 m g$. The other end of the string is attached to a fixed point $O$ on the plane. The particle $P$ is in equilibrium at the point $A$ on the plane and the extension of the string is $\frac { 1 } { 4 } a$. The particle $P$ is now projected from $A$ down a line of greatest slope of the plane with speed $V$. It comes to instantaneous rest after moving a distance $\frac { 1 } { 2 } a$.
\end{enumerate}
By using the principle of conservation of energy,\\
(a) find $V$ in terms of $a$ and $g$,\\
(b) find, in terms of $a$ and $g$, the speed of $P$ when the string first becomes slack.\\
\hfill \mbox{\textit{Edexcel M3 2008 Q4 [10]}}