CAIE FP2 2015 June — Question 10 EITHER

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2015
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeVertical circle – string/rod (tension and energy)
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and careful geometric analysis across multiple stages of motion. It involves non-trivial setup (string over peg), deriving a given result, then using tension ratios to find initial speed and subsequent motion. The multi-part nature with connected results and the need to track geometry changes makes this substantially harder than typical A-level mechanics questions.
Spec6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

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One end of a light inextensible string of length \(\frac { 3 } { 2 } a\) is attached to a fixed point \(O\) on a horizontal surface. The other end of the string is attached to a particle \(P\) of mass \(m\). The string passes over a small fixed smooth peg \(A\) which is at a distance \(a\) vertically above \(O\). The system is in equilibrium with \(P\) hanging vertically below \(A\) and the string taut. The particle is projected horizontally with speed \(u\) (see diagram). When \(P\) is at the same horizontal level as \(A\), the tension in the string is \(T\). Show that \(T = \frac { 2 m } { a } \left( u ^ { 2 } - a g \right)\). The ratio of the tensions in the string immediately before, and immediately after, the string loses contact with the peg is \(5 : 1\).
  1. Show that \(u ^ { 2 } = 5 a g\).
  2. Find, in terms of \(m\) and \(g\), the tension in the string when \(P\) is next at the same horizontal level as \(A\).

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One end of a light inextensible string of length $\frac { 3 } { 2 } a$ is attached to a fixed point $O$ on a horizontal surface. The other end of the string is attached to a particle $P$ of mass $m$. The string passes over a small fixed smooth peg $A$ which is at a distance $a$ vertically above $O$. The system is in equilibrium with $P$ hanging vertically below $A$ and the string taut. The particle is projected horizontally with speed $u$ (see diagram). When $P$ is at the same horizontal level as $A$, the tension in the string is $T$. Show that $T = \frac { 2 m } { a } \left( u ^ { 2 } - a g \right)$.

The ratio of the tensions in the string immediately before, and immediately after, the string loses contact with the peg is $5 : 1$.\\
(i) Show that $u ^ { 2 } = 5 a g$.\\
(ii) Find, in terms of $m$ and $g$, the tension in the string when $P$ is next at the same horizontal level as $A$.

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