| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.8 This is a standard conical pendulum question with routine calculations: converting rpm to rad/s (part a), applying v²/r for acceleration (part b), and resolving forces vertically then using circular motion horizontally (part c). All steps are textbook exercises requiring only direct application of standard formulas with no problem-solving insight needed. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{40 \times 2\pi}{60} = \frac{4\pi}{3} \text{ (rad/sec)}\) | M1, A1 | |
| (b) \(a = \omega^2 r = \left(\frac{4\pi}{3}\right)^2 \times 0.2 = \frac{16\pi^2}{45}\) | M1, A1 | Accept \(0.356\pi^2\) (3sf) |
| (c)(i) Diagram showing forces (mg, T, and direction) | B1 | |
| (ii) Vertically: No acceleration, forces balance \(mg = T\cos\theta\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{16\pi^2}{45g}\) or \(\tan\theta = 0.358(08)\) giving \(\theta = 20°\) | M1, A1F, m1, A1F, A1F | ft acceleration; SC \(\tan\theta = \frac{\omega^2 r}{g}\), 1st 3 marks for quoting and using correctly; ft provided M1 earned in (b) |
**(a)** $\frac{40 \times 2\pi}{60} = \frac{4\pi}{3} \text{ (rad/sec)}$ | M1, A1 | | **Total: 2 marks**
**(b)** $a = \omega^2 r = \left(\frac{4\pi}{3}\right)^2 \times 0.2 = \frac{16\pi^2}{45}$ | M1, A1 | Accept $0.356\pi^2$ (3sf) | **Total: 2 marks**
**(c)(i)** Diagram showing forces (mg, T, and direction) | B1 | | **Total: 1 mark**
**(ii)** Vertically: No acceleration, forces balance $mg = T\cos\theta$ | B1 | | **Total: 1 mark**
**(iii)** Horizontally: $T\sin\theta = m \times \frac{16\pi^2}{45}$
$T\cos\theta = mg$
$\tan\theta = \frac{16\pi^2}{45g}$ or $\tan\theta = 0.358(08)$ giving $\theta = 20°$ | M1, A1F, m1, A1F, A1F | ft acceleration; SC $\tan\theta = \frac{\omega^2 r}{g}$, 1st 3 marks for quoting and using correctly; ft provided M1 earned in (b) | **Total: 5 marks**
**Question 6 Total: 11 marks**
---6 A particle is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point $O$. The particle is set into motion, so that it describes a horizontal circle whose centre is vertically below $O$. The angle between the string and the vertical is $\theta$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{480a817d-074f-440d-829e-c8f8a9746151-6_506_442_534_794}
\begin{enumerate}[label=(\alph*)]
\item The particle completes 40 revolutions every minute.
Show that the angular speed of the particle is $\frac { 4 \pi } { 3 }$ radians per second.
\item The radius of the circle is 0.2 metres.
Find, in terms of $\pi$, the magnitude of the acceleration of the particle.
\item The mass of the particle is $m \mathrm {~kg}$ and the tension in the string is $T$ newtons.
\begin{enumerate}[label=(\roman*)]
\item Draw a diagram showing the forces acting on the particle.
\item Explain why $T \cos \theta = m g$.
\item Find the value of $\theta$, giving your answer to the nearest degree.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q6 [11]}}