AQA M2 2007 January — Question 3 6 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2007
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeVertical circle – string/rod (tension and energy)
DifficultyModerate -0.3 This is a standard vertical circular motion problem requiring energy conservation (part a) and circular motion force equation (part b). Both are routine M2 techniques with straightforward application—no novel insight needed, though it requires careful execution of two distinct methods across the two parts.
Spec6.02i Conservation of energy: mechanical energy principle

3 A light inextensible string has length \(2 a\). One end of the string is attached to a fixed point \(O\) and a particle of mass \(m\) is attached to the other end. Initially, the particle is held at the point \(A\) with the string taut and horizontal. The particle is then released from rest and moves in a circular path. Subsequently, it passes through the point \(B\), which is directly below \(O\). The points \(O , A\) and \(B\) are as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{480a817d-074f-440d-829e-c8f8a9746151-3_426_437_575_772}
  1. Show that the speed of the particle at \(B\) is \(2 \sqrt { a g }\).
  2. Find the tension in the string as the particle passes through \(B\). Give your answer in terms of \(m\) and \(g\).

AnswerMarks Guidance
(a) \(mg \cdot 2a = \frac{1}{2}mv^2\) giving \(v = 2\sqrt{ga}\)M1, A1, A1 Energy equation
(b) \(T - mg = \frac{mv^2}{2a}\) giving \(T = 3mg\)M1, A1, A1F All terms for M1, no component; if \(T > 0\)
Question 3 Total: 6 marks
**(a)** $mg \cdot 2a = \frac{1}{2}mv^2$ giving $v = 2\sqrt{ga}$ | M1, A1, A1 | Energy equation | **Total: 3 marks**

**(b)** $T - mg = \frac{mv^2}{2a}$ giving $T = 3mg$ | M1, A1, A1F | All terms for M1, no component; if $T > 0$ | **Total: 3 marks**

**Question 3 Total: 6 marks**

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3 A light inextensible string has length $2 a$. One end of the string is attached to a fixed point $O$ and a particle of mass $m$ is attached to the other end. Initially, the particle is held at the point $A$ with the string taut and horizontal. The particle is then released from rest and moves in a circular path. Subsequently, it passes through the point $B$, which is directly below $O$. The points $O , A$ and $B$ are as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{480a817d-074f-440d-829e-c8f8a9746151-3_426_437_575_772}
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of the particle at $B$ is $2 \sqrt { a g }$.
\item Find the tension in the string as the particle passes through $B$. Give your answer in terms of $m$ and $g$.
\end{enumerate}

\hfill \mbox{\textit{AQA M2 2007 Q3 [6]}}