CAIE Further Paper 3 2024 June — Question 2 6 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
DifficultyStandard +0.8 This is a Further Maths mechanics question combining circular motion with elastic strings and equilibrium. Part (a) requires setting up the geometry correctly (natural length a, HQ = a/4, finding HP) and using Hooke's law with the centripetal force equation. Part (b) involves balancing forces on Q. While the individual concepts are standard, the multi-body setup with the string through a hole and the need to carefully track extensions makes this moderately challenging, though still a fairly routine Further Maths question.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.05c Horizontal circles: conical pendulum, banked tracks
A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(2mg\). A particle \(Q\) of mass \(km\) is attached to the other end of the string. Particle \(P\) lies on a smooth horizontal table. The string passes through a small smooth hole \(H\) in the table and then passes through a small smooth hole \(H\) in the table. Particle \(P\) moves in a horizontal circle on the surface of the table with constant speed \(\sqrt{\frac{1}{3}ga}\). Particle \(Q\) hangs in equilibrium vertically below the hole with \(HQ = \frac{1}{4}a\).
  1. Find, in terms of \(a\), the extension in the string. [4]
  2. Find the value of \(k\). [2]
Question 2:
AnswerMarks
2(a)2mgx
Hooke’s law: T 
AnswerMarks Guidance
aB1 2mgx
seen anywhere.
a
mga
2
 T 
a
a x
AnswerMarks Guidance
4B1 RHS seen anywhere.
May be in terms of radius or extended length, for
mga mga
2 2
example , .
1 r
l a
4
mga
Equate: 2mgx  2 , 4x2 3axa2 0
a 3a
x
AnswerMarks Guidance
4M1 Equate two expressions for T and obtain a
simplified homogeneous quadratic equation
4l2 5al0,
4r2 3ara2 0,
2k2 3k20
a
x
AnswerMarks Guidance
4A1 Single correct answer only.
4
AnswerMarks Guidance
2(b) T kmg B1
seen in part (a). Note that no response in part (b)
can earn B1 if kmg seen in part (a).
2mgx 1
T  , k 
AnswerMarks Guidance
a 2B1 CWO. Part (a) needs to be correct.
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | 2mgx
Hooke’s law: T 
a | B1 | 2mgx
seen anywhere.
a
mga
2
 T 
a
a x
4 | B1 | RHS seen anywhere.
May be in terms of radius or extended length, for
mga mga
2 2
example , .
1 r
l a
4
mga
Equate: 2mgx  2 , 4x2 3axa2 0
a 3a
x
4 | M1 | Equate two expressions for T and obtain a
simplified homogeneous quadratic equation
4l2 5al0,
4r2 3ara2 0,
2k2 3k20
a
x
4 | A1 | Single correct answer only.
4
--- 2(b) ---
2(b) |  T kmg | B1 | kmg seen anywhere in an equation. This may be
seen in part (a). Note that no response in part (b)
can earn B1 if kmg seen in part (a).
2mgx 1
T  , k 
a 2 | B1 | CWO. Part (a) needs to be correct.
2
Question | Answer | Marks | Guidance
A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length $a$ and modulus of elasticity $2mg$. A particle $Q$ of mass $km$ is attached to the other end of the string. Particle $P$ lies on a smooth horizontal table. The string passes through a small smooth hole $H$ in the table and then passes through a small smooth hole $H$ in the table.

Particle $P$ moves in a horizontal circle on the surface of the table with constant speed $\sqrt{\frac{1}{3}ga}$. Particle $Q$ hangs in equilibrium vertically below the hole with $HQ = \frac{1}{4}a$.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the extension in the string. [4]

\item Find the value of $k$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q2 [6]}}
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