CAIE Further Paper 3 2023 June — Question 5 8 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeElastic string – conical pendulum (string inclined to vertical)
DifficultyChallenging +1.2 This is a standard circular motion problem with elastic strings requiring resolution of forces, application of Hooke's law, and centripetal force equations. While it involves multiple steps (geometry, vertical/horizontal force balance, and solving simultaneous equations), the approach is methodical and follows a well-established template for this topic. The algebra is moderate but not particularly challenging. For Further Maths students, this represents a routine application of learned techniques rather than requiring novel insight.
Spec6.02h Elastic PE: 1/2 k x^26.05c Horizontal circles: conical pendulum, banked tracks
One end of a light elastic string, of natural length \(12a\) and modulus of elasticity \(kmg\), is attached to a fixed point \(O\). The other end of the string is attached to a particle of mass \(m\). The particle moves with constant speed \(\frac{2}{3}\sqrt{3ag}\) in a horizontal circle with centre at a distance \(12a\) below \(O\). The string is inclined at an angle \(\theta\) to the downward vertical through \(O\).
  1. Find, in terms of \(a\), the extension of the string. [5]
  2. Find the value of \(k\). [3]
Question 5:
AnswerMarks
5Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
AnswerMarks Guidance
5(a)Tcosmg B1
mv2  27ag
Tsin  m
r 4r
AnswerMarks
 B1
r 12atan usedM1
27 3
Divide: tan , so tan
AnswerMarks Guidance
412tan 4M1 Finds value for tan OE.
Reduces to equation in  or x, no k.
AnswerMarks
r 9a, extension of string = 3aA1
Alternative method for question 5(a)
Let L be stretched length of string.
AnswerMarks Guidance
TcosmgB1 12a
Or T mg
L
mv2  27ag
Tsin  m
r 4r
AnswerMarks
 B1
r  Lsin usedM1
 L2 144a20.5
12a
Use cos and sin and eliminate T.
AnswerMarks
L LM1
[L2 144a2 81a2] L15a, extension of string = 3aA1
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
5(b)kmgL12a
Hooke’s law: T 
AnswerMarks
12aB1
kmgL12a mgL
Eliminate T: 
AnswerMarks
12a 12aM1
L
k  5
AnswerMarks
L12aA1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | Tcosmg | B1
mv2  27ag
Tsin  m
r 4r
  | B1
r 12atan used | M1
27 3
Divide: tan , so tan
412tan 4 | M1 | Finds value for tan OE.
Reduces to equation in  or x, no k.
r 9a, extension of string = 3a | A1
Alternative method for question 5(a)
Let L be stretched length of string.
Tcosmg | B1 | 12a
Or T mg
L
mv2  27ag
Tsin  m
r 4r
  | B1
r  Lsin used | M1
 L2 144a20.5
12a
Use cos and sin and eliminate T.
L L | M1
[L2 144a2 81a2] L15a, extension of string = 3a | A1
5
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | kmgL12a
Hooke’s law: T 
12a | B1
kmgL12a mgL
Eliminate T: 
12a 12a | M1
L
k  5
L12a | A1
3
Question | Answer | Marks | Guidance
One end of a light elastic string, of natural length $12a$ and modulus of elasticity $kmg$, is attached to a fixed point $O$. The other end of the string is attached to a particle of mass $m$. The particle moves with constant speed $\frac{2}{3}\sqrt{3ag}$ in a horizontal circle with centre at a distance $12a$ below $O$. The string is inclined at an angle $\theta$ to the downward vertical through $O$.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the extension of the string. [5]
\item Find the value of $k$. [3]
\end{enumerate}

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