CAIE FP2 2008 November — Question 2 8 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2008
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeVertical circle – string/rod (tension and energy)
DifficultyChallenging +1.2 This is a standard vertical circular motion problem requiring energy conservation and Newton's second law at two symmetric positions. While it involves multiple steps (finding speeds at both positions, applying circular motion equations, and algebraic manipulation), the approach is methodical and follows a well-established template for this topic. The symmetry simplifies the algebra, and the constraint θ < cos⁻¹(2/3) ensures the bead remains in contact throughout. This is moderately above average difficulty due to the multi-step nature and need to handle two positions systematically, but it's a classic Further Maths mechanics question without requiring novel insight.
Spec6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods

2 A small bead \(B\) of mass \(m\) is threaded on a smooth wire fixed in a vertical plane. The wire forms a circle of radius \(a\) and centre \(O\). The highest point of the circle is \(A\). The bead is slightly displaced from rest at \(A\). When angle \(A O B = \theta\), where \(\theta < \cos ^ { - 1 } \left( \frac { 2 } { 3 } \right)\), the force exerted on the bead by the wire has magnitude \(R _ { 1 }\). When angle \(A O B = \pi + \theta\), the force exerted on the bead by the wire has magnitude \(R _ { 2 }\). Show that \(R _ { 2 } - R _ { 1 } = 4 m g\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(\frac{1}{2}mv_1^2 = mga(1 - \cos\theta)\)M1 A1 M1 for either energy equation
\(\frac{1}{2}mv_2^2 = mga(1 + \cos\theta)\)A1
\(R_1 = mg\cos\theta - mv_1^2/a\)M1 A1 M1 for either radial resolution
\(R_2 = mg\cos\theta + mv_2^2/a\)A1
\(R_1 = 3mg\cos\theta - 2mg\) EITHER: substitute in \(R_1\), \(R_2\) and combine
\(R_2 = 3mg\cos\theta + 2mg\)
\(R_2 - R_1 = 4mg\) A.G.M1 A1
OR: \(R_2 - R_1 = m(v_2^2 + v_1^2)/a\) Combine \(R_1\), \(R_2\) and substitute
\(= 4mg\) A.G.(M1 A1) Total: 8
## Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv_1^2 = mga(1 - \cos\theta)$ | M1 A1 | M1 for either energy equation |
| $\frac{1}{2}mv_2^2 = mga(1 + \cos\theta)$ | A1 | |
| $R_1 = mg\cos\theta - mv_1^2/a$ | M1 A1 | M1 for either radial resolution |
| $R_2 = mg\cos\theta + mv_2^2/a$ | A1 | |
| $R_1 = 3mg\cos\theta - 2mg$ | | EITHER: substitute in $R_1$, $R_2$ and combine |
| $R_2 = 3mg\cos\theta + 2mg$ | | |
| $R_2 - R_1 = 4mg$ **A.G.** | M1 A1 | |
| OR: $R_2 - R_1 = m(v_2^2 + v_1^2)/a$ | | Combine $R_1$, $R_2$ and substitute |
| $= 4mg$ **A.G.** | (M1 A1) | **Total: 8** |

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2 A small bead $B$ of mass $m$ is threaded on a smooth wire fixed in a vertical plane. The wire forms a circle of radius $a$ and centre $O$. The highest point of the circle is $A$. The bead is slightly displaced from rest at $A$. When angle $A O B = \theta$, where $\theta < \cos ^ { - 1 } \left( \frac { 2 } { 3 } \right)$, the force exerted on the bead by the wire has magnitude $R _ { 1 }$. When angle $A O B = \pi + \theta$, the force exerted on the bead by the wire has magnitude $R _ { 2 }$. Show that $R _ { 2 } - R _ { 1 } = 4 m g$.

\hfill \mbox{\textit{CAIE FP2 2008 Q2 [8]}}