CAIE Further Paper 3 2023 November — Question 5 8 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
DifficultyChallenging +1.8 This is a challenging circular motion problem requiring multiple equations (centripetal force at two positions, energy conservation, geometric relationships) and careful algebraic manipulation. The constraint that reactions are in a 1:6 ratio adds complexity, and students must correctly handle the geometry where AOB is perpendicular. However, the solution path is systematic once the setup is recognized, making it hard but not exceptional for Further Maths.
Spec6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration
\includegraphics{figure_5} A bead of mass \(m\) moves on a smooth circular wire, with centre \(O\) and radius \(a\), in a vertical plane. The bead has speed \(v_A\) when it is at the point \(A\) where \(OA\) makes an angle \(\alpha\) with the downward vertical through \(O\), and \(\cos \alpha = \frac{2}{3}\). Subsequently the bead has speed \(v_B\) at the point \(B\), where \(OB\) makes an angle \(\theta\) with the upward vertical through \(O\). Angle \(AOB\) is a right angle (see diagram). The reaction of the wire on the bead at \(B\) is in the direction \(OB\) and has magnitude equal to \(\frac{1}{6}\) of the magnitude of the reaction when the bead is at \(A\).
  1. Find, in terms of \(m\) and \(g\), the magnitude of the reaction at \(B\). [6]
  2. Given that \(v_A = \sqrt{kag}\), find the value of \(k\). [2]
Question 5:
AnswerMarks
5(a)mv2
At A: R −mgcos= A
AnswerMarks
A aB1
mv2
At B: −R +mgcos= B
AnswerMarks
B aB1
1 1
Energy: mv2 − mv2 =mga(cos+cos)
A B
AnswerMarks Guidance
2 2M1 All terms present, allow sign errors, cos/sin mix.
A1AEF
1
Eliminate velocities, use R = R and substitute angle values
B A
6
1 1
a( R −mgcos)− a(−R +mgcos)=mga(cos+cos)
A B
2 2
3 4 7
6R − mg+R − mg=2mg
B B
AnswerMarks Guidance
5 5 5M1 Any equivalent working, leading to a dimensionally
correct equation in R and mg.
B
3
R = mg
B
AnswerMarks
5A1
6
AnswerMarks
5(b)3 3 mv2
From first equation in (a), 6 mg− mg = A
AnswerMarks
5 5 aM1
v2 =3ag, so
A
AnswerMarks Guidance
k=3A1 CAO
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | mv2
At A: R −mgcos= A
A a | B1
mv2
At B: −R +mgcos= B
B a | B1
1 1
Energy: mv2 − mv2 =mga(cos+cos)
A B
2 2 | M1 | All terms present, allow sign errors, cos/sin mix.
A1 | AEF
1
Eliminate velocities, use R = R and substitute angle values
B A
6
1 1
a( R −mgcos)− a(−R +mgcos)=mga(cos+cos)
A B
2 2
3 4 7
6R − mg+R − mg=2mg
B B
5 5 5 | M1 | Any equivalent working, leading to a dimensionally
correct equation in R and mg.
B
3
R = mg
B
5 | A1
6
--- 5(b) ---
5(b) | 3 3 mv2
From first equation in (a), 6 mg− mg = A
5 5 a | M1
v2 =3ag, so
A
k=3 | A1 | CAO
2
Question | Answer | Marks | Guidance
\includegraphics{figure_5}

A bead of mass $m$ moves on a smooth circular wire, with centre $O$ and radius $a$, in a vertical plane. The bead has speed $v_A$ when it is at the point $A$ where $OA$ makes an angle $\alpha$ with the downward vertical through $O$, and $\cos \alpha = \frac{2}{3}$. Subsequently the bead has speed $v_B$ at the point $B$, where $OB$ makes an angle $\theta$ with the upward vertical through $O$. Angle $AOB$ is a right angle (see diagram). The reaction of the wire on the bead at $B$ is in the direction $OB$ and has magnitude equal to $\frac{1}{6}$ of the magnitude of the reaction when the bead is at $A$.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $g$, the magnitude of the reaction at $B$. [6]
\item Given that $v_A = \sqrt{kag}$, find the value of $k$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q5 [8]}}
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