| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – string/rod (tension and energy) |
| Difficulty | Standard +0.3 This is a standard vertical circular motion problem requiring conservation of energy and Newton's second law for circular motion. The steps are routine: apply energy conservation to find speed at B, then use centripetal force equation with normal reaction. While it involves multiple steps, the techniques are well-practiced M2 material with no novel insight required, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Using conservation of energy: \(\frac{1}{2}mv_B^2 = \frac{1}{2}mv_A^2 - mgh\) | M1 | Energy conservation |
| \(\frac{1}{2}(6)v_B^2 = \frac{1}{2}(6)(64) - 6(9.8)(2)\) | M1 | Correct substitution |
| \(3v_B^2 = 192 - 117.6 = 74.4\) | A1 | Correct working |
| \(v_B = \sqrt{24.8} = 4.98 \text{ m s}^{-1}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| At \(B\): \(R - mg\cos\theta = \frac{mv_B^2}{r}\) | M1 | Correct equation for normal reaction |
| \(B\) is at height 2m, so position angle: \(\cos\theta = \frac{2}{4} = 0.5\), \(\theta = 60°\) from bottom | M1 | Correct geometry |
| \(R - 6(9.8)\cos 60° = \frac{6(24.8)}{4}\) | A1 | Correct substitution |
| \(R = \frac{148.8}{4} + 29.4 = 37.2 + 29.4 = 66.6\) N | A1 | Correct answer |
# Question 7:
## Part (a):
| Using conservation of energy: $\frac{1}{2}mv_B^2 = \frac{1}{2}mv_A^2 - mgh$ | M1 | Energy conservation |
|---|---|---|
| $\frac{1}{2}(6)v_B^2 = \frac{1}{2}(6)(64) - 6(9.8)(2)$ | M1 | Correct substitution |
| $3v_B^2 = 192 - 117.6 = 74.4$ | A1 | Correct working |
| $v_B = \sqrt{24.8} = 4.98 \text{ m s}^{-1}$ | A1 | Correct answer |
## Part (b):
| At $B$: $R - mg\cos\theta = \frac{mv_B^2}{r}$ | M1 | Correct equation for normal reaction |
|---|---|---|
| $B$ is at height 2m, so position angle: $\cos\theta = \frac{2}{4} = 0.5$, $\theta = 60°$ from bottom | M1 | Correct geometry |
| $R - 6(9.8)\cos 60° = \frac{6(24.8)}{4}$ | A1 | Correct substitution |
| $R = \frac{148.8}{4} + 29.4 = 37.2 + 29.4 = 66.6$ N | A1 | Correct answer |
---7 A hollow cylinder, of internal radius 4 m , is fixed so that its axis is horizontal. The point $O$ is on this axis. A particle, of mass 6 kg , is set in motion so that it moves on the smooth inner surface of the cylinder in a vertical circle about $O$. Its speed at the point $A$, which is vertically below $O$, is $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{851cb2a3-5bc8-4af9-b1fc-a143d37beebe-5_746_739_504_662}
When the particle is at the point $B$, at a height of 2 m above $A$, find:
\begin{enumerate}[label=(\alph*)]
\item its speed;
\item the normal reaction between the cylinder and the particle.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q7 [7]}}