Challenging +1.2 This is a standard circular motion problem requiring force resolution and geometric constraints. Students must balance tensions (using the smooth ring condition that tensions are equal), apply circular motion formula (T = mrω²), and use string length constraint. While it involves multiple steps and both mechanics and geometry, the approach is methodical and follows standard A-level Further Maths techniques without requiring novel insight.
\includegraphics{figure_2}
A light inextensible string of length \(a\) is threaded through a fixed smooth ring \(R\). One end of the string is attached to a particle \(A\) of mass \(3m\). The other end of the string is attached to a particle \(B\) of mass \(m\). The particle \(A\) hangs in equilibrium at a distance \(x\) vertically below the ring. The angle between \(AR\) and \(BR\) is \(\theta\) (see diagram). The particle \(B\) moves in a horizontal circle with constant angular speed \(2\sqrt{\frac{g}{a}}\).
Show that \(\cos \theta = \frac{1}{3}\) and find \(x\) in terms of \(a\). [5]
\includegraphics{figure_2}
A light inextensible string of length $a$ is threaded through a fixed smooth ring $R$. One end of the string is attached to a particle $A$ of mass $3m$. The other end of the string is attached to a particle $B$ of mass $m$. The particle $A$ hangs in equilibrium at a distance $x$ vertically below the ring. The angle between $AR$ and $BR$ is $\theta$ (see diagram). The particle $B$ moves in a horizontal circle with constant angular speed $2\sqrt{\frac{g}{a}}$.
Show that $\cos \theta = \frac{1}{3}$ and find $x$ in terms of $a$. [5]
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q2 [5]}}