CAIE Further Paper 3 2024 June — Question 5 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
DifficultyChallenging +1.2 This is a standard circular motion problem with friction and connected particles. It requires setting up force equations for two particles in limiting equilibrium, applying F=mrω² and friction laws, then solving simultaneous equations. The multi-step nature and need to consider both particles systematically makes it moderately above average, but the techniques are all standard for Further Maths mechanics with no novel insight required.
Spec3.03v Motion on rough surface: including inclined planes6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks
Two particles \(A\) and \(B\) of masses \(m\) and \(km\) respectively are connected by a light inextensible string of length \(a\). The particles are placed on a rough horizontal circular turntable with the string taut and lying along a radius of the turntable. Particle \(A\) is at a distance \(a\) from the centre of the turntable and particle \(B\) is at a distance \(2a\) from the centre of the turntable. The coefficient of friction between each particle and the turntable is \(\frac{1}{3}\). When the turntable is made to rotate with angular speed \(\frac{2}{5}\sqrt{\frac{g}{a}}\), the system is in limiting equilibrium.
  1. Find the tension in the string, in terms of \(m\) and \(g\). [4]
  2. Find the value of \(k\). [3]
Question 5:
AnswerMarks Guidance
5(a)For A: F T ma2
AM1 Only allow sign errors.
F mg  1mg
AnswerMarks Guidance
A 5B1 Accept with g replaced by 10.
Combine: T  1mg 4 mg
AnswerMarks Guidance
5 25M1 To reach an equation in T and mg only.
Accept with g replaced by 10.
T  1 mg
AnswerMarks Guidance
25A1 CAO
4
AnswerMarks Guidance
5(b)For B: F T km2a2
BM1 Only allow sign errors.
F kmg 1kmg and combine to find k
AnswerMarks Guidance
B 5M1 To reach an equation in k only.
k  1
AnswerMarks
3A1
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | For A: F T ma2
A | M1 | Only allow sign errors.
F mg  1mg
A 5 | B1 | Accept with g replaced by 10.
Combine: T  1mg 4 mg
5 25 | M1 | To reach an equation in T and mg only.
Accept with g replaced by 10.
T  1 mg
25 | A1 | CAO
4
--- 5(b) ---
5(b) | For B: F T km2a2
B | M1 | Only allow sign errors.
F kmg 1kmg and combine to find k
B 5 | M1 | To reach an equation in k only.
k  1
3 | A1
3
Question | Answer | Marks | Guidance
Two particles $A$ and $B$ of masses $m$ and $km$ respectively are connected by a light inextensible string of length $a$. The particles are placed on a rough horizontal circular turntable with the string taut and lying along a radius of the turntable. Particle $A$ is at a distance $a$ from the centre of the turntable and particle $B$ is at a distance $2a$ from the centre of the turntable. The coefficient of friction between each particle and the turntable is $\frac{1}{3}$.

When the turntable is made to rotate with angular speed $\frac{2}{5}\sqrt{\frac{g}{a}}$, the system is in limiting equilibrium.

\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string, in terms of $m$ and $g$. [4]
\item Find the value of $k$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q5 [7]}}
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Q1 6 Q2 7 Q3 5 Q4 7 Q5 7 Q6 9 Q7 9