| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2003 |
| Session | June |
| Marks | 11 |
| 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. The geometry is given, making it straightforward to find angles and apply Newton's second law. Part (a) involves routine resolution in two directions, part (b) is immediate from similar working, and part (c) requires the simple insight that tension must be non-negative. While multi-step, it follows a well-established template for conical pendulum problems with no novel reasoning required. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \((↓) (T-S)\cos\theta = mg\) | M1 A1 | |
| \((↔) (T+S)\sin\theta = mr\omega^2 = m(l\sin\theta)\omega^2\) | M1 A1 ft | |
| Finding \(T\) in terms of \(l, m, \omega^2\) and \(g\) | M1 | |
| \(T = \frac{1}{6}m(3l\omega^2 + 4g)\) (*) | A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(S = \frac{1}{6}m(3l\omega^2 - 4g)\) | any correct form | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Setting \(S \geq 0\); \(\omega^2 \geq \frac{4g}{3l}\) (*) | (no wrong working seen) | M1 A1 |
## (a)
$(↓) (T-S)\cos\theta = mg$ | M1 A1 |
$(↔) (T+S)\sin\theta = mr\omega^2 = m(l\sin\theta)\omega^2$ | M1 A1 ft |
Finding $T$ in terms of $l, m, \omega^2$ and $g$ | M1 |
$T = \frac{1}{6}m(3l\omega^2 + 4g)$ (*) | A1 | (7 marks)
## (b)
$S = \frac{1}{6}m(3l\omega^2 - 4g)$ | any correct form | M1 A1 | (2 marks)
## (c)
Setting $S \geq 0$; $\omega^2 \geq \frac{4g}{3l}$ (*) | (no wrong working seen) | M1 A1 | (2 marks)
**(11 marks total)**
---\includegraphics{figure_1}
A particle $P$ of mass $m$ is attached to the ends of two light inextensible strings $AP$ and $BP$ each of length $l$. The ends $A$ and $B$ are attached to fixed points, with $A$ vertically above $B$ and $AB = \frac{3}{4}l$, as shown in Fig. 1. The particle $P$ moves in a horizontal circle with constant angular speed $\omega$. The centre of the circle is the mid-point of $AB$ and both strings remain taut.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension $AP$ is $\frac{1}{6}m(3l\omega^2 + 4g)$. [7]
\item Find, in terms of $m$, $l$, $\omega$ and $g$, an expression for the tension in $BP$. [2]
\item Deduce that $\omega^2 \geq \frac{4g}{3l}$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2003 Q4 [11]}}