AQA M2 2008 January — Question 7 8 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2008
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle: speed at specific point
DifficultyStandard +0.3 This is a standard vertical circle problem requiring conservation of energy between two points and Newton's second law for circular motion. The steps are routine: apply energy conservation from B to A, then use T - mg = mv²/r at the lowest point. While it involves two parts and careful algebra, it follows a well-practiced template with no novel insight required, making it slightly easier than average.
Spec6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration

7 A light inextensible string, of length \(a\), has one end attached to a fixed point \(O\). A particle, of mass \(m\), is attached to the other end. The particle is moving in a vertical circle, centre \(O\). When the particle is at \(B\), vertically above \(O\), the string is taut and the particle is moving with speed \(3 \sqrt { a g }\). \includegraphics[max width=\textwidth, alt={}, center]{1bc18163-b20e-4dc6-bd35-496efec8dc73-5_422_399_497_778}
  1. Find, in terms of \(g\) and \(a\), the speed of the particle at the lowest point, \(A\), of its path.
  2. Find, in terms of \(g\) and \(m\), the tension in the string when the particle is at \(A\).

(a) Conservation of energy:
\(\frac{1}{2}m(3\sqrt{ag})^2 + mg \cdot 2a = \frac{1}{2}mv^2\)
\(\frac{9}{2}mga + 2mga = \frac{1}{2}mv^2\)
AnswerMarks Guidance
\(v = \sqrt{13ag}\)M1A1 A1 A1 4 marks
(b) At A, consider vertical forces:
\(T - mg = \frac{mv^2}{a}\)
\(T = mg + 13mg\)
AnswerMarks Guidance
\(T = 14mg\)M1A1 m1 A1ft 4 marks
Total: 8 marks
**(a)** Conservation of energy:
$\frac{1}{2}m(3\sqrt{ag})^2 + mg \cdot 2a = \frac{1}{2}mv^2$

$\frac{9}{2}mga + 2mga = \frac{1}{2}mv^2$

$v = \sqrt{13ag}$ | M1A1 A1 A1 | 4 marks | M1 for 3 terms: 2 KE and PE

**(b)** At A, consider vertical forces:
$T - mg = \frac{mv^2}{a}$

$T = mg + 13mg$

$T = 14mg$ | M1A1 m1 A1ft | 4 marks | M1 for 3 terms, 2 correct; ft from (a)

**Total: 8 marks**

---
7 A light inextensible string, of length $a$, has one end attached to a fixed point $O$. A particle, of mass $m$, is attached to the other end. The particle is moving in a vertical circle, centre $O$. When the particle is at $B$, vertically above $O$, the string is taut and the particle is moving with speed $3 \sqrt { a g }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{1bc18163-b20e-4dc6-bd35-496efec8dc73-5_422_399_497_778}
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $g$ and $a$, the speed of the particle at the lowest point, $A$, of its path.
\item Find, in terms of $g$ and $m$, the tension in the string when the particle is at $A$.
\end{enumerate}

\hfill \mbox{\textit{AQA M2 2008 Q7 [8]}}