| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces and application of F=mrω². Part (a) involves setting up two equations (vertical equilibrium and horizontal centripetal force) with straightforward geometry (right angle gives r=l/√2). Part (b) requires recognizing that one tension must be non-negative. While it has multiple steps and careful algebra, it follows a well-established template for conical pendulum problems with no novel insight required, making it slightly easier than average. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| \(r = \frac{l}{\sqrt{2}}\) | B1 |
| \(R(\uparrow): T_a\cos 45° = T_b\cos 45° + mg\) | M1A1 |
| \(T_a - T_b = mg\sqrt{2}\) | (1) |
| \(R(\to): T_a\cos 45° + T_b\cos 45° = mr\omega^2\) | M1A1 |
| \(T_a \times \frac{1}{\sqrt{2}} + T_b \times \frac{1}{\sqrt{2}} = m\frac{l}{\sqrt{2}}\omega^2\) | |
| \(T_a + T_b = ml\omega^2\) | (2) |
| \(T_a - T_b = mg\sqrt{2}\) | (1) |
| \(2T_a = m(l\omega^2 + g\sqrt{2})\) | M1 |
| \(T_a = \frac{1}{2}m(l\omega^2 + g\sqrt{2})\) | A1 |
| \(T_b = ml\omega^2 - T_a\) | A1 |
| \(= \frac{1}{2}m(l\omega^2 - g\sqrt{2})\) |
| Answer | Marks |
|---|---|
| \(T_b > 0\): \(\frac{1}{2}m(l\omega^2 - g\sqrt{2}) > 0\) | M1 |
| \(\omega^2 > \frac{g\sqrt{2}}{l}\) | A1 |
| \(\omega > \sqrt{\frac{g\sqrt{2}}{l}}\) * |
## Part (a):
$r = \frac{l}{\sqrt{2}}$ | B1 |
$R(\uparrow): T_a\cos 45° = T_b\cos 45° + mg$ | M1A1 |
$T_a - T_b = mg\sqrt{2}$ | (1) |
$R(\to): T_a\cos 45° + T_b\cos 45° = mr\omega^2$ | M1A1 |
$T_a \times \frac{1}{\sqrt{2}} + T_b \times \frac{1}{\sqrt{2}} = m\frac{l}{\sqrt{2}}\omega^2$ | |
$T_a + T_b = ml\omega^2$ | (2) |
$T_a - T_b = mg\sqrt{2}$ | (1) |
$2T_a = m(l\omega^2 + g\sqrt{2})$ | M1 |
$T_a = \frac{1}{2}m(l\omega^2 + g\sqrt{2})$ | A1 |
$T_b = ml\omega^2 - T_a$ | A1 |
$= \frac{1}{2}m(l\omega^2 - g\sqrt{2})$ | |
## Part (b):
$T_b > 0$: $\frac{1}{2}m(l\omega^2 - g\sqrt{2}) > 0$ | M1 |
$\omega^2 > \frac{g\sqrt{2}}{l}$ | A1 |
$\omega > \sqrt{\frac{g\sqrt{2}}{l}}$ * | |\includegraphics{figure_3}
A small ball $P$ of mass $m$ is attached to the ends of two light inextensible strings of length $l$. The other ends of the strings are attached to fixed points $A$ and $B$, where $A$ is vertically above $B$. Both strings are taut and $AP$ is perpendicular to $BP$ as shown in Figure 3. The system rotates about the line $AB$ with constant angular speed $\omega$. The ball moves in a horizontal circle.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$, $g$, $l$ and $\omega$, the tension in $AP$ and the tension in $BP$.
[8]
\item Show that $\omega^2 \geq \frac{g\sqrt{2}}{l}$.
[2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2011 Q5 [10]}}