CAIE Further Paper 3 2023 June — Question 1 5 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest, string starts taut
DifficultyStandard +0.3 This is a straightforward application of Hooke's law and energy conservation in a standard elastic string setup. Part (a) uses energy methods (EPE + GPE = KE) with clear initial and final states, while part (b) requires F=ma with tension from Hooke's law. Both are routine Further Maths mechanics techniques with no novel insight required, making it slightly easier than average.
Spec3.03c Newton's second law: F=ma one dimension6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(3mg\), is attached to a fixed point \(O\). The other end of the string is attached to a particle \(P\) of mass \(m\). The string hangs with \(P\) vertically below \(O\). The particle \(P\) is pulled vertically downwards so that the extension of the string is \(2a\). The particle \(P\) is then released from rest.
  1. Find the speed of \(P\) when it is at a distance \(\frac{3}{4}a\) below \(O\). [3]
  2. Find the initial acceleration of \(P\) when it is released from rest. [2]
Question 1:
AnswerMarks
1(a)3mg
2a2
AnswerMarks Guidance
2aB1 Correct EPE term seen
1 mv2 mg  3a 3 a   3mg 2a2
AnswerMarks Guidance
2  4  2aM1 Dimensionally correct energy equation. Must have
one KE, one EPE term and at least one GPE.
Allow sign errors.
15
v ag 2.74 ag
 
AnswerMarks Guidance
2A1 AEF
3
AnswerMarks
1(b)3mg
T mg mA and T  2a
AnswerMarks Guidance
aM1 N2L and Hooke’s law
Acceleration = 5g [upwards]A1 Allow 50 or 5g
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 1:
--- 1(a) ---
1(a) | 3mg
2a2
2a | B1 | Correct EPE term seen
1 mv2 mg  3a 3 a   3mg 2a2
2  4  2a | M1 | Dimensionally correct energy equation. Must have
one KE, one EPE term and at least one GPE.
Allow sign errors.
15
v ag 2.74 ag
 
2 | A1 | AEF
3
--- 1(b) ---
1(b) | 3mg
T mg mA and T  2a
a | M1 | N2L and Hooke’s law
Acceleration = 5g [upwards] | A1 | Allow 50 or 5g
2
Question | Answer | Marks | Guidance
One end of a light elastic string, of natural length $a$ and modulus of elasticity $3mg$, is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $m$. The string hangs with $P$ vertically below $O$. The particle $P$ is pulled vertically downwards so that the extension of the string is $2a$. The particle $P$ is then released from rest.

\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ when it is at a distance $\frac{3}{4}a$ below $O$. [3]

\item Find the initial acceleration of $P$ when it is released from rest. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q1 [5]}}
This paper (7 questions)
View full paper
Q1 5 Q2 5 Q3 7 Q4 8 Q5 7 Q6 9 Q7 9