| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Standard +0.3 This is a standard energy conservation problem with circular motion. Part (a) requires applying conservation of energy between two points (routine for M2), and part (b) needs resolving forces at the highest point using the result from (a). Both are textbook applications 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 methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Part (a): Using conservation of energy (lowest and highest points): \(\frac{1}{2}m(7v)^2 = \frac{1}{2}mv^2 + 2mga\); \(\frac{48}{2}v^2 = 2ga\); \(\therefore v = \sqrt{\frac{ag}{12}}\) | M1, A1A1, M1, A1 | A1 for 7v and v; Needs 48 or 24; AG |
| Part (b): Velocity at \(A\) is \(\sqrt{\frac{ag}{12}}\); Resolving vertically at \(A\): \(m\frac{v^2}{a} + R = mg\); \(R = mg - \frac{m}{a} \times \frac{ag}{12} = \frac{11}{12}mg\) | M1, A1,A1, A1 | 3 terms; A1 correct 3 terms, A1 correct signs; \((1-\frac{1}{12})mg\) M1A2; Condone \(-\frac{11}{12}mg\) |
| Total: 9 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Part (a):** Using conservation of energy (lowest and highest points): $\frac{1}{2}m(7v)^2 = \frac{1}{2}mv^2 + 2mga$; $\frac{48}{2}v^2 = 2ga$; $\therefore v = \sqrt{\frac{ag}{12}}$ | M1, A1A1, M1, A1 | A1 for 7v and v; Needs 48 or 24; AG |
| **Part (b):** Velocity at $A$ is $\sqrt{\frac{ag}{12}}$; Resolving vertically at $A$: $m\frac{v^2}{a} + R = mg$; $R = mg - \frac{m}{a} \times \frac{ag}{12} = \frac{11}{12}mg$ | M1, A1,A1, A1 | 3 terms; A1 correct 3 terms, A1 correct signs; $(1-\frac{1}{12})mg$ M1A2; Condone $-\frac{11}{12}mg$ |
| | | **Total: 9** |5 A bead of mass $m$ moves on a smooth circular ring of radius $a$ which is fixed in a vertical plane, as shown in the diagram. Its speed at $A$, the highest point of its path, is $v$ and its speed at $B$, the lowest point of its path, is $7 v$.\\
\includegraphics[max width=\textwidth, alt={}, center]{676e753d-1b80-413c-a4b9-21861db8dde5-4_419_317_484_842}
\begin{enumerate}[label=(\alph*)]
\item Show that $v = \sqrt { \frac { a g } { 12 } }$.
\item Find the reaction of the ring on the bead, in terms of $m$ and $g$, when the bead is at $A$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q5 [9]}}