CAIE Further Paper 3 2023 June — Question 3 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle: string becomes slack
DifficultyChallenging +1.2 This is a standard circular motion problem requiring energy conservation and centripetal force equations at two positions. Part (a) uses energy conservation between A and B (straightforward application), while part (b) requires resolving forces at A and applying Newton's second law radially. The problem is well-structured with clear conditions and uses familiar Further Maths mechanics techniques, though it requires careful coordinate geometry and multiple equations. The 7 total marks and multi-step nature place it above average difficulty, but it follows standard circular motion methodology without requiring novel insight.
Spec6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration
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 particle \(P\) is held at the point \(A\), where \(OA\) makes an angle \(\theta\) with the downward vertical through \(O\), and with the string taut. The particle \(P\) is projected perpendicular to \(OA\) in an upwards direction with speed \(u\). It then starts to move along a circular path in a vertical plane. The string goes slack when \(P\) is at \(B\), where angle \(AOB\) is \(90°\) and the speed of \(P\) is \(\sqrt{\frac{1}{3}ag}\).
  1. Find the value of \(\sin\theta\). [2]
  2. Find, in terms of \(m\) and \(g\), the tension in the string when \(P\) is at \(A\). [5]
Question 3:
AnswerMarks
3(a)m4ag
At B, mgsin
AnswerMarks Guidance
5aM1 Allow cos instead of sin for M1. Do not award
until tension 0 used. Mass must be seen. No sign
error.
4
sin
AnswerMarks
5A1
2
AnswerMarks
3(b)mu2
At A, T mgcos
AnswerMarks
aB1
1 1 4ag
Energy mu2  m mgacossin
AnswerMarks Guidance
2 2 5M1 A1 Energy equation with 4 terms, dimensionally
correct. Mass must be present, allow sign errors.
1
Must see in the KE terms.
2
AnswerMarks Guidance
Solve to find TM1 Complete method leading to an expression in mg
for T .
21
T  mg
AnswerMarks Guidance
5A1 CWO
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | m4ag
At B, mgsin
5a | M1 | Allow cos instead of sin for M1. Do not award
until tension 0 used. Mass must be seen. No sign
error.
4
sin
5 | A1
2
--- 3(b) ---
3(b) | mu2
At A, T mgcos
a | B1
1 1 4ag
Energy mu2  m mgacossin
2 2 5 | M1 A1 | Energy equation with 4 terms, dimensionally
correct. Mass must be present, allow sign errors.
1
Must see in the KE terms.
2
Solve to find T | M1 | Complete method leading to an expression in mg
for T .
21
T  mg
5 | A1 | CWO
5
Question | Answer | Marks | Guidance
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 particle $P$ is held at the point $A$, where $OA$ makes an angle $\theta$ with the downward vertical through $O$, and with the string taut. The particle $P$ is projected perpendicular to $OA$ in an upwards direction with speed $u$. It then starts to move along a circular path in a vertical plane. The string goes slack when $P$ is at $B$, where angle $AOB$ is $90°$ and the speed of $P$ is $\sqrt{\frac{1}{3}ag}$.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $\sin\theta$. [2]

\item Find, in terms of $m$ and $g$, the tension in the string when $P$ is at $A$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q3 [7]}}
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