CAIE Further Paper 3 2023 November — Question 7 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeParticle in circular motion with string/rod
DifficultyChallenging +1.8 This is a multi-part mechanics problem requiring energy conservation with elastic potential energy, force resolution in equilibrium, and tension calculation. While it involves several steps and careful geometry (finding spring extensions using Pythagoras), the techniques are standard for Further Maths mechanics: energy equation setup, resolving forces, and applying Hooke's law. The geometry is straightforward (3-4-5 triangle), and each part follows logically from the previous one without requiring novel insight.
Spec6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_7} A particle \(P\) of mass \(m\) is attached to one end of a light rod of length \(3a\). The other end of the rod is able to pivot smoothly about the fixed point \(A\). The particle is also attached to one end of a light spring of natural length \(a\) and modulus of elasticity \(kmg\). The other end of the spring is attached to a fixed point \(B\). The points \(A\) and \(B\) are in a horizontal line, a distance \(5a\) apart, and these two points and the rod are in a vertical plane. Initially, \(P\) is held in equilibrium by a vertical force \(F\) with the stretched length of the spring equal to \(4a\) (see diagram). The particle is released from rest in this position and has a speed of \(\frac{6}{5}\sqrt{2ag}\) when the rod becomes horizontal.
  1. Find the value of \(k\). [5]
  2. Find \(F\) in terms of \(m\) and \(g\). [2]
  3. Find, in terms of \(m\) and \(g\), the tension in the rod immediately before it is released. [2]
Question 7:
AnswerMarks
7(a)2
1 6 
Gain in KE = m  2ag  and Gain in GPE = mg3asin
AnswerMarks
2 5 B1
1
kmg
Loss in EPE = 2 ( (3a)2 −a2 )
AnswerMarks
aB1
1
kmg
2
Energy equation: 1 m   6 2ag   +mg3asin= 2 ( (3a)2 −a2 )
AnswerMarks Guidance
2 5  aM1 KE, GPE and at least one EPE present, allow sign errors,
dimensionally correct.
AnswerMarks
A1All correct.
36 12
mg+ mg =4kmg
25 5
24
k=
AnswerMarks
25A1
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)3a 72
In lower position, tension T in spring = kmg =3kmg = mg
a 25
AnswerMarks Guidance
Perpendicular to rod, T =(F +mg)cosM1 Hooke’s law.
72
So, (F+mg)cos= mg,
25
19
F = mg
AnswerMarks
5A1
Alternative method for question 7(b)
3a 72
In lower position, tension T in spring = kmg =3kmg = mg
a 25
Tcos=Tsin
AnswerMarks Guidance
Tsin+Tcos =F+mgM1 Hooke’s law.
Eliminate T.
19
F = mg
AnswerMarks
5A1
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(c)Let tension in rod = T'
Parallel to rod, T=(F +mg)sinM1
96
T= mg
AnswerMarks
25A1
Alternative method for question 7(c)
Tcos=Tsin
AnswerMarks Guidance
Tsin+Tcos =F+mgM1 At least one equation seen with their T and/or F.
96
T= mg
AnswerMarks
25A1
2
Question 7:
--- 7(a) ---
7(a) | 2
1 6 
Gain in KE = m  2ag  and Gain in GPE = mg3asin
2 5  | B1
1
kmg
Loss in EPE = 2 ( (3a)2 −a2 )
a | B1
1
kmg
2
Energy equation: 1 m   6 2ag   +mg3asin= 2 ( (3a)2 −a2 )
2 5  a | M1 | KE, GPE and at least one EPE present, allow sign errors,
dimensionally correct.
A1 | All correct.
36 12
mg+ mg =4kmg
25 5
24
k=
25 | A1
5
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | 3a 72
In lower position, tension T in spring = kmg =3kmg = mg
a 25
Perpendicular to rod, T =(F +mg)cos | M1 | Hooke’s law.
72
So, (F+mg)cos= mg,
25
19
F = mg
5 | A1
Alternative method for question 7(b)
3a 72
In lower position, tension T in spring = kmg =3kmg = mg
a 25
Tcos=Tsin
Tsin+Tcos =F+mg | M1 | Hooke’s law.
Eliminate T.
19
F = mg
5 | A1
2
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | Let tension in rod = T'
Parallel to rod, T=(F +mg)sin | M1
96
T= mg
25 | A1
Alternative method for question 7(c)
Tcos=Tsin
Tsin+Tcos =F+mg | M1 | At least one equation seen with their T and/or F.
96
T= mg
25 | A1
2
\includegraphics{figure_7}

A particle $P$ of mass $m$ is attached to one end of a light rod of length $3a$. The other end of the rod is able to pivot smoothly about the fixed point $A$. The particle is also attached to one end of a light spring of natural length $a$ and modulus of elasticity $kmg$. The other end of the spring is attached to a fixed point $B$. The points $A$ and $B$ are in a horizontal line, a distance $5a$ apart, and these two points and the rod are in a vertical plane.

Initially, $P$ is held in equilibrium by a vertical force $F$ with the stretched length of the spring equal to $4a$ (see diagram). The particle is released from rest in this position and has a speed of $\frac{6}{5}\sqrt{2ag}$ when the rod becomes horizontal.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$. [5]
\item Find $F$ in terms of $m$ and $g$. [2]
\item Find, in terms of $m$ and $g$, the tension in the rod immediately before it is released. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q7 [9]}}
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