| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2004 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string horizontal surface projection |
| Difficulty | Challenging +1.2 This is a multi-part M3 question involving elastic strings, energy methods, and contact forces. Part (a) requires calculating extension and elastic PE (straightforward). Part (b) uses energy conservation with both elastic and gravitational PE (standard technique). Part (c) requires resolving tension and showing the normal reaction remains positive throughout motion (more demanding but follows a clear method). The trigonometry with tan β = 3/4 is routine. Overall, this is above-average difficulty due to the combination of topics and the verification in part (c), but follows standard M3 procedures without requiring novel insight. |
| Spec | 6.02g 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 | Guidance |
|---|---|---|
| (a) Length of string \(= \frac{10}{3}a\) | B1 | |
| \(\frac{1}{2}mg\) | ||
| EPE \(= \frac{2}{2a}\frac{1}{2}mg(L-a)^2\) | M1 | |
| \(= \frac{49}{36}mga\) | A1 (3) | |
| (b) Energy equation: \(\frac{1}{2}mv^2 + \frac{2}{2a}\frac{1}{2}mga^2 = \left(\frac{39}{36}mga\right)_C\) | M1A1 ⋈ | |
| \(v = \frac{2}{3}\sqrt{5ga}\) or equivalent | A1 (3) | |
| (c) When string at angle \(\theta\) to horizontal, length of string \(= \frac{2a}{\sin\theta}\) | M1A1 | |
| \(\Rightarrow\) Vert. Comp. of \(T\), \(T_V = T\sin\theta = \frac{mg}{2a}\left(\frac{2a}{\sin\theta} - a\right)\sin\theta\) | M1A1 | |
| \(= \frac{mg}{2}(2 - \sin\theta)\) | ||
| (\(\Downarrow\)) \(R + T_V = mg\) and find \(R = \ldots\) | M1 | |
| \(R = mg - \frac{mg}{2}(2 - \sin\theta) = \frac{mg}{2}\sin\theta\) | A1 | |
| \(\Rightarrow R > 0\) (as \(\sin\theta > 0\)), so stays on table | A1 (5) | |
| [Alternative final 3 marks: As \(\theta\) increases so \(T_V\) decreases | M1 | |
| Initial \(T_V\) (string at \(\beta\) to hor.) \(= \frac{7}{10}mg\) | A1 | |
| \(\Rightarrow T_V \leq \frac{7}{10}mg < mg\), so stays on table | A1 | ] |
(a) Length of string $= \frac{10}{3}a$ | B1 |
$\frac{1}{2}mg$ | |
EPE $= \frac{2}{2a}\frac{1}{2}mg(L-a)^2$ | M1 |
$= \frac{49}{36}mga$ | A1 (3) |
(b) Energy equation: $\frac{1}{2}mv^2 + \frac{2}{2a}\frac{1}{2}mga^2 = \left(\frac{39}{36}mga\right)_C$ | M1A1 ⋈ |
$v = \frac{2}{3}\sqrt{5ga}$ or equivalent | A1 (3) |
(c) When string at angle $\theta$ to horizontal, length of string $= \frac{2a}{\sin\theta}$ | M1A1 |
$\Rightarrow$ Vert. Comp. of $T$, $T_V = T\sin\theta = \frac{mg}{2a}\left(\frac{2a}{\sin\theta} - a\right)\sin\theta$ | M1A1 |
$= \frac{mg}{2}(2 - \sin\theta)$ | |
($\Downarrow$) $R + T_V = mg$ and find $R = \ldots$ | M1 |
$R = mg - \frac{mg}{2}(2 - \sin\theta) = \frac{mg}{2}\sin\theta$ | A1 |
$\Rightarrow R > 0$ (as $\sin\theta > 0$), so stays on table | A1 (5) |
[Alternative final 3 marks: As $\theta$ increases so $T_V$ decreases | M1 |
Initial $T_V$ (string at $\beta$ to hor.) $= \frac{7}{10}mg$ | A1 |
$\Rightarrow T_V \leq \frac{7}{10}mg < mg$, so stays on table | A1 |] | [11] |
---4. A particle $P$ of mass $m$ is attached to one end of a light elastic string of length $a$ and modulus of elasticity $\frac { 1 } { 2 } m g$. The other end of the string is fixed at the point $A$ which is at a height $2 a$ above a smooth horizontal table. The particle is held on the table with the string making an angle $\beta$ with the horizontal, where $\tan \beta = \frac { 3 } { 4 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the elastic energy stored in the string in this position.
The particle is now released. Assuming that $P$ remains on the table,
\item find the speed of $P$ when the string is vertical.
By finding the vertical component of the tension in the string when $P$ is on the table and $A P$ makes an angle $\theta$ with the horizontal,
\item show that the assumption that $P$ remains in contact with the table is justified.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2004 Q4 [11]}}