| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.5 This is a standard conical pendulum problem requiring routine application of circular motion formulas and resolving forces. Part (a) is trivial unit conversion, parts (b) and (c) involve straightforward resolution of tension and applying F=mrω², with all steps clearly signposted. Slightly easier than average due to the 'show that' scaffolding and standard setup. |
| Spec | 3.03b Newton's first law: equilibrium6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega = 40 \text{ rev/min} = \frac{40}{60} \text{ rev/s} = \frac{40 \times 2\pi}{60}\) | M1 | Convert to rad/s |
| \(= \frac{80\pi}{60} = \frac{4\pi}{3}\) rad/s | A1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving vertically: \(T\cos 30° = mg = 6(9.8)\) | M1 | Correct vertical equation |
| \(T = \frac{6 \times 9.8}{\cos 30°}\) | M1 | Rearrangement |
| \(T = \frac{58.8}{0.8660} = 67.9\) N | A1 | Correct answer shown |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\sin 30° = m\omega^2 r\) | M1 | Correct horizontal equation |
| \(67.9 \times 0.5 = 6 \times \left(\frac{4\pi}{3}\right)^2 \times r\) | M1 | Substitution |
| \(33.95 = 6 \times \frac{16\pi^2}{9} \times r\) | A1 | Correct working |
| \(r = \frac{33.95 \times 9}{6 \times 16\pi^2} = 0.322\) m | A1 | Correct answer |
# Question 5:
## Part (a):
| $\omega = 40 \text{ rev/min} = \frac{40}{60} \text{ rev/s} = \frac{40 \times 2\pi}{60}$ | M1 | Convert to rad/s |
|---|---|---|
| $= \frac{80\pi}{60} = \frac{4\pi}{3}$ rad/s | A1 | Correctly shown |
## Part (b):
| Resolving vertically: $T\cos 30° = mg = 6(9.8)$ | M1 | Correct vertical equation |
|---|---|---|
| $T = \frac{6 \times 9.8}{\cos 30°}$ | M1 | Rearrangement |
| $T = \frac{58.8}{0.8660} = 67.9$ N | A1 | Correct answer shown |
## Part (c):
| $T\sin 30° = m\omega^2 r$ | M1 | Correct horizontal equation |
|---|---|---|
| $67.9 \times 0.5 = 6 \times \left(\frac{4\pi}{3}\right)^2 \times r$ | M1 | Substitution |
| $33.95 = 6 \times \frac{16\pi^2}{9} \times r$ | A1 | Correct working |
| $r = \frac{33.95 \times 9}{6 \times 16\pi^2} = 0.322$ m | A1 | Correct answer |
---5 A particle, of mass 6 kg , is attached to one end of a light inextensible string. The other end of the string is attached to the fixed point $O$. The particle is set in motion, so that it moves in a horizontal circle at constant speed, with the string at an angle of $30 ^ { \circ }$ to the vertical. The centre of this circle is vertically below $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{851cb2a3-5bc8-4af9-b1fc-a143d37beebe-4_586_490_541_767}
The particle moves in a horizontal circle with an angular speed of 40 revolutions per minute.
\begin{enumerate}[label=(\alph*)]
\item Show that the angular speed of the particle is $\frac { 4 \pi } { 3 }$ radians per second.
\item Show that the tension in the string is 67.9 N , correct to three significant figures.
\item Find the radius of the horizontal circle.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q5 [9]}}