CAIE Further Paper 3 2024 June — Question 2 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeParticle at midpoint of string between two horizontal fixed points: vertical motion
DifficultyChallenging +1.2 This is a standard Further Mechanics elastic string problem requiring equilibrium analysis with geometry (part a) and energy conservation (part b). The geometry is straightforward (Pythagoras on symmetric configuration), and both parts follow well-established methods. While it requires multiple steps and careful algebra, it demands no novel insight beyond applying standard techniques, placing it moderately above average difficulty.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
The points \(A\) and \(B\) are at the same horizontal level a distance \(4a\) apart. The ends of a light elastic string, of natural length \(4a\) and modulus of elasticity \(\lambda\), are attached to \(A\) and \(B\). A particle \(P\) of mass \(m\) is attached to the midpoint of the string. The system is in equilibrium with \(P\) at a distance \(\frac{5}{8}a\) below \(M\), the midpoint of \(AB\).
  1. Find \(\lambda\) in terms of \(m\) and \(g\). [3]
The particle \(P\) is pulled down vertically and released from rest at a distance \(\frac{8}{5}a\) below \(M\).
  1. Find, in terms of \(a\) and \(g\), the speed of \(P\) as it passes through \(M\) in the subsequent motion. [4]
Question 2:
AnswerMarks Guidance
2(a)In equilibrium: 2Tcosmg M1
 5a  
Hooke’s law: T    2a  
AnswerMarks
2a  2  4B1
Equate and use cos 3: 10mg
AnswerMarks
5 3A1
3
AnswerMarks
2(b)2
1 20 
EPE loss =  a4a
AnswerMarks Guidance
2 4a 3 B1 80
mga
27
1 8a 80
Energy equation: mv2 mg ' mga'
AnswerMarks Guidance
2 3 27M1 A1 All 3 terms required, dimensionally correct, their
.
4 ga
v
AnswerMarks Guidance
3 3A1 Any equivalent form.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
--- 2(a) ---
2(a) | In equilibrium: 2Tcosmg | M1
 5a  
Hooke’s law: T    2a  
2a  2  4 | B1
Equate and use cos 3: 10mg
5 3 | A1
3
--- 2(b) ---
2(b) | 2
1 20 
EPE loss =  a4a
2 4a 3  | B1 | 80
mga
27
1 8a 80
Energy equation: mv2 mg ' mga'
2 3 27 | M1 A1 | All 3 terms required, dimensionally correct, their
.
4 ga
v
3 3 | A1 | Any equivalent form.
4
Question | Answer | Marks | Guidance
The points $A$ and $B$ are at the same horizontal level a distance $4a$ apart. The ends of a light elastic string, of natural length $4a$ and modulus of elasticity $\lambda$, are attached to $A$ and $B$. A particle $P$ of mass $m$ is attached to the midpoint of the string. The system is in equilibrium with $P$ at a distance $\frac{5}{8}a$ below $M$, the midpoint of $AB$.

\begin{enumerate}[label=(\alph*)]
\item Find $\lambda$ in terms of $m$ and $g$. [3]
\end{enumerate}

The particle $P$ is pulled down vertically and released from rest at a distance $\frac{8}{5}a$ below $M$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, in terms of $a$ and $g$, the speed of $P$ as it passes through $M$ in the subsequent motion. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q2 [7]}}
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