| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem with vertical string geometry. Part (a) requires resolving forces and recognizing the limiting case when tension becomes zero (straightforward inequality derivation). Part (b) applies Hooke's law and force resolution with given angle—routine calculations with no novel insight required. Slightly easier than average due to clear setup and standard techniques. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=16.02g Hooke's law: T = k*x or T = lambda*x/l6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| \(T\cos\theta + N = Mg\) (1) | M1A1 |
| \(T\sin\theta = mro^2\) (2) | M1A1 |
| \(\sin\theta = \frac{r}{l}\) from (2), \(T = ml\omega^2\) | M1 |
| sub into (1) \(ml\cos\theta\omega^2 + N = mg\) | A1 |
| \(N = mg - mh\omega^2\) | |
| Since in contact with table \(N \geq 0 \therefore \omega^2 \leq \frac{g}{h}\) | M1A1cso (8) |
| Answer | Marks |
|---|---|
| \(r : h : l = 3 : 4 : 5 \therefore \text{extension} = \frac{h}{4}\) | B1 |
| \(T = \frac{2mg}{h} \times \frac{h}{4} = \frac{mg}{2}\) | M1A1 |
| \(T = ml\omega^2 = \frac{5mh}{4}\omega^2 \omega = \sqrt{\frac{2g}{5h}}\) | M1A1 (5) |
**Part (a):**
$T\cos\theta + N = Mg$ (1) | M1A1 |
$T\sin\theta = mro^2$ (2) | M1A1 |
$\sin\theta = \frac{r}{l}$ from (2), $T = ml\omega^2$ | M1 |
sub into (1) $ml\cos\theta\omega^2 + N = mg$ | A1 |
$N = mg - mh\omega^2$ | |
Since in contact with table $N \geq 0 \therefore \omega^2 \leq \frac{g}{h}$ | M1A1cso (8) |
**Part (b):**
$r : h : l = 3 : 4 : 5 \therefore \text{extension} = \frac{h}{4}$ | B1 |
$T = \frac{2mg}{h} \times \frac{h}{4} = \frac{mg}{2}$ | M1A1 |
$T = ml\omega^2 = \frac{5mh}{4}\omega^2 \omega = \sqrt{\frac{2g}{5h}}$ | M1A1 (5) |
**Total: 13 marks**
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f07b8a65-ccb5-423f-96cc-b303bd05ad1f-05_495_972_239_484}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a particle $B$, of mass $m$, attached to one end of a light elastic string. The other end of the string is attached to a fixed point $A$, at a distance $h$ vertically above a smooth horizontal table. The particle moves on the table in a horizontal circle with centre $O$, where $O$ is vertically below $A$. The string makes a constant angle $\theta$ with the downward vertical and $B$ moves with constant angular speed $\omega$ about $O A$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\omega ^ { 2 } \leqslant \frac { g } { h }$.
The elastic string has natural length $h$ and modulus of elasticity $2 m g$.\\
Given that $\tan \theta = \frac { 3 } { 4 }$,
\item find $\omega$ in terms of $g$ and $h$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2008 Q3 [13]}}