| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Maximum/minimum tension or reaction |
| Difficulty | Standard +0.3 This is a standard vertical circular motion problem requiring energy conservation between top and bottom positions, then applying Newton's second law at each point to find tensions. The two-part structure and straightforward application of well-practiced techniques (energy equation, centripetal force) make it slightly easier than average, though it requires careful algebraic manipulation and understanding of when tension is maximum/minimum. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using energy conservation between B (bottom) and top: \(\frac{1}{2}m(5v)^2 = \frac{1}{2}m(3v)^2 + mg(2a)\) | M1 A1 | Correct energy equation |
| \(25v^2 - 9v^2 = 4ga\) | A1 | Simplifying |
| \(v^2 = \frac{ga}{4}\), so \(v = \frac{1}{2}\sqrt{ga}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Greatest tension at bottom: \(T_{max} - mg = \frac{m(5v)^2}{a}\) | M1 A1 | Equation at bottom |
| \(T_{max} = mg + \frac{25mv^2}{a} = mg + \frac{25mg}{4a} \cdot a = mg\left(1 + \frac{25}{4}\right) = \frac{29mg}{4}\) | A1 | |
| Least tension at top: \(T_{min} + mg = \frac{m(3v)^2}{a}\) | M1 | |
| \(T_{min} = \frac{9mv^2}{a} - mg = \frac{9mg}{4} - mg = \frac{5mg}{4}\) | A1 | |
| Ratio \(= \frac{29mg/4}{5mg/4} = \frac{29}{5}\) | A1 | cao |
# Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using energy conservation between B (bottom) and top: $\frac{1}{2}m(5v)^2 = \frac{1}{2}m(3v)^2 + mg(2a)$ | M1 A1 | Correct energy equation |
| $25v^2 - 9v^2 = 4ga$ | A1 | Simplifying |
| $v^2 = \frac{ga}{4}$, so $v = \frac{1}{2}\sqrt{ga}$ | A1 | cao |
# Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Greatest tension at bottom: $T_{max} - mg = \frac{m(5v)^2}{a}$ | M1 A1 | Equation at bottom |
| $T_{max} = mg + \frac{25mv^2}{a} = mg + \frac{25mg}{4a} \cdot a = mg\left(1 + \frac{25}{4}\right) = \frac{29mg}{4}$ | A1 | |
| Least tension at top: $T_{min} + mg = \frac{m(3v)^2}{a}$ | M1 | |
| $T_{min} = \frac{9mv^2}{a} - mg = \frac{9mg}{4} - mg = \frac{5mg}{4}$ | A1 | |
| Ratio $= \frac{29mg/4}{5mg/4} = \frac{29}{5}$ | A1 | cao |6 A light inextensible string, of length $a$, has one end attached to a fixed point $O$. A small bead, of mass $m$, is attached to the other end of the string. The bead is moving in a vertical circle, centre $O$. When the bead is at $B$, vertically below $O$, the string is taut and the bead is moving with speed $5 v$.\\
\includegraphics[max width=\textwidth, alt={}, center]{9d039ec3-fd0a-40ae-9afe-7627439081df-14_536_554_502_774}
\begin{enumerate}[label=(\alph*)]
\item The speed of the bead at the highest point of its path is $3 v$.
Find $v$ in terms of $a$ and $g$.
\item Find the ratio of the greatest tension to the least tension in the string, as the bead travels around its circular path.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9d039ec3-fd0a-40ae-9afe-7627439081df-14_1261_1709_1446_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2011 Q6 [9]}}