| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: speed at specific point |
| Difficulty | Standard +0.3 This is a standard M2 vertical circle problem requiring energy conservation from lowest point to rest position, then resolving forces at rest. The setup is straightforward with given angle and clear two-part structure. Slightly easier than average due to the particle being at rest (simplifying force equations) and standard application of SUVAT-free energy methods. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using energy conservation from \(P\) to \(Q\) | M1 | |
| Height gained \(= 3 - 3\cos 15° = 3(1-\cos 15°)\) | A1 | |
| \(\frac{1}{2}mv^2 = mg \times 3(1-\cos 15°)\) | M1 | |
| \(v^2 = 2 \times 9.8 \times 3(1-\cos 15°)\) | ||
| \(v = \sqrt{2 \times 9.8 \times 3(1-\cos 15°)} \approx 2.74\ \text{ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At \(Q\), particle at rest, so \(T - mg\cos 15° = 0\) (no centripetal acceleration) | M1 A1 | |
| \(m = \frac{T}{g\cos 15°} = \frac{22}{9.8\cos 15°} = 2.32\) kg | A1 |
# Question 8:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using energy conservation from $P$ to $Q$ | M1 | |
| Height gained $= 3 - 3\cos 15° = 3(1-\cos 15°)$ | A1 | |
| $\frac{1}{2}mv^2 = mg \times 3(1-\cos 15°)$ | M1 | |
| $v^2 = 2 \times 9.8 \times 3(1-\cos 15°)$ | | |
| $v = \sqrt{2 \times 9.8 \times 3(1-\cos 15°)} \approx 2.74\ \text{ms}^{-1}$ | A1 | |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| At $Q$, particle at rest, so $T - mg\cos 15° = 0$ (no centripetal acceleration) | M1 A1 | |
| $m = \frac{T}{g\cos 15°} = \frac{22}{9.8\cos 15°} = 2.32$ kg | A1 | |8 A particle is attached to one end of a light inextensible string of length 3 metres. The other end of the string is attached to a fixed point $O$. The particle is set into motion horizontally at point $P$ with speed $v$, so that it describes part of a vertical circle whose centre is $O$. The point $P$ is vertically below $O$.\\
\includegraphics[max width=\textwidth, alt={}, center]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-16_510_334_493_861}
The particle first comes momentarily to rest at the point $Q$, where $O Q$ makes an angle of $15 ^ { \circ }$ to the vertical.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $v$.
\item When the particle is at rest at the point $Q$, the tension in the string is 22 newtons.
Find the mass of the particle.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{3ffa0a2b-aa7d-46eb-b92b-3e3ee59f235c-17_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2010 Q8 [7]}}