CAIE FP2 2013 November — Question 3 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeRatio of tensions/forces
DifficultyChallenging +1.8 This is a challenging Further Maths circular motion problem requiring energy conservation and centripetal force equations at multiple points, with geometric constraints linking two positions. Students must set up simultaneous equations involving tensions, speeds, and angles, then eliminate variables systematically—significantly harder than standard single-position circular motion questions but follows established FM2 techniques.
Spec6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods

3 A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\). The path of the particle is a complete vertical circle with centre \(O\). When \(P\) is at its lowest point, its speed is \(u\). When \(P\) is at the point \(A\), the tension in the string is \(T\) and the string makes an angle \(\theta\) with the downward vertical, where \(\cos \theta = \frac { 3 } { 5 }\). When \(P\) is at the point \(B\), above the level of \(O\), the tension in the string is \(\frac { 1 } { 8 } T\) and angle \(B O A = 90 ^ { \circ }\). Find \(u\) in terms of \(a\) and \(g\).

Question 3:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(T - mg\cos\theta = mv_A^2/a\)M1 A1 Equate radial forces at \(A\)
\(T/8 + mg\sin\theta = mv_B^2/a\)A1 Equate radial forces at \(B\)
\(\frac{1}{2}mv_A^2 = \frac{1}{2}mu^2 - mga(1 - \cos\theta)\)M1 A1 Energy equation (M1 for either)
\(\frac{1}{2}mv_B^2 = \frac{1}{2}mu^2 - mga(1 + \sin\theta)\)A1 Energy equation
\(mv_A^2/a + mg\cos\theta = 8mv_B^2/a - 8mg\sin\theta\) Find \(u\) by eliminating \(T\)
\(v_A^2 = 8v_B^2 - 8ga(4/5) - ga(3/5)\)
\(= 8v_B^2 - 7ga\)M1
\(\frac{1}{2}mv_A^2 = \frac{1}{2}mv_B^2 + mga(\cos\theta + \sin\theta)\) Finding one of \(v_A^2\) or \(v_B^2\)
\(v_A^2 = v_B^2 + (14/5)ga\)
\(v_A^2 = (21/5)ga\) or \(v_B^2 = (7/5)ga\)M1
\(u^2 = v_A^2 + (4/5)ga\) or \(v_B^2 + (18/5)ga\)
\(= 5ga,\ u = \sqrt{5ga}\)A1
Part total: 9Total: 9
## Question 3:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $T - mg\cos\theta = mv_A^2/a$ | M1 A1 | Equate radial forces at $A$ |
| $T/8 + mg\sin\theta = mv_B^2/a$ | A1 | Equate radial forces at $B$ |
| $\frac{1}{2}mv_A^2 = \frac{1}{2}mu^2 - mga(1 - \cos\theta)$ | M1 A1 | Energy equation (M1 for either) |
| $\frac{1}{2}mv_B^2 = \frac{1}{2}mu^2 - mga(1 + \sin\theta)$ | A1 | Energy equation |
| $mv_A^2/a + mg\cos\theta = 8mv_B^2/a - 8mg\sin\theta$ | — | Find $u$ by eliminating $T$ |
| $v_A^2 = 8v_B^2 - 8ga(4/5) - ga(3/5)$ | — | |
| $= 8v_B^2 - 7ga$ | M1 | |
| $\frac{1}{2}mv_A^2 = \frac{1}{2}mv_B^2 + mga(\cos\theta + \sin\theta)$ | — | Finding one of $v_A^2$ or $v_B^2$ |
| $v_A^2 = v_B^2 + (14/5)ga$ | — | |
| $v_A^2 = (21/5)ga$ or $v_B^2 = (7/5)ga$ | M1 | |
| $u^2 = v_A^2 + (4/5)ga$ or $v_B^2 + (18/5)ga$ | — | |
| $= 5ga,\ u = \sqrt{5ga}$ | A1 | |

**Part total: 9 | Total: 9**

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3 A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The path of the particle is a complete vertical circle with centre $O$. When $P$ is at its lowest point, its speed is $u$. When $P$ is at the point $A$, the tension in the string is $T$ and the string makes an angle $\theta$ with the downward vertical, where $\cos \theta = \frac { 3 } { 5 }$. When $P$ is at the point $B$, above the level of $O$, the tension in the string is $\frac { 1 } { 8 } T$ and angle $B O A = 90 ^ { \circ }$. Find $u$ in terms of $a$ and $g$.

\hfill \mbox{\textit{CAIE FP2 2013 Q3 [9]}}