CAIE Further Paper 3 2024 November — Question 6 3 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeMultiple particles on string
DifficultyStandard +0.3 This is a standard conical pendulum problem with two particles. Part (a) asks only for tension in OP, requiring resolution of forces on P (horizontal: centripetal force, vertical: weight balance), using given geometry to find angles from radii. The setup is more complex than single-particle conical pendulum, but the actual calculation is straightforward application of F=mrω² and force resolution with clearly specified numerical values.
Spec6.05c Horizontal circles: conical pendulum, banked tracks
\includegraphics{figure_6} A particle \(P\) of mass \(0.05\text{kg}\) is attached to one end of a light inextensible string of length \(1\text{m}\). The other end of the string is attached to a fixed point \(O\). A particle \(Q\) of mass \(0.04\text{kg}\) is attached to one end of a second light inextensible string. The other end of this string is attached to \(P\). The particle \(P\) moves in a horizontal circle of radius \(0.8\text{m}\) with angular speed \(\omega\text{rads}^{-1}\). The particle \(Q\) moves in a horizontal circle of radius \(1.4\text{m}\) also with angular speed \(\omega\text{rads}^{-1}\). The centres of the circles are vertically below \(O\), and \(O\), \(P\) and \(Q\) are always in the same vertical plane. The strings \(OP\) and \(PQ\) remain at constant angles \(\alpha\) and \(\beta\) respectively to the vertical (see diagram).
  1. Find the tension in the string \(OP\). [3]
Question 6:
AnswerMarks
6(a)At P:  T cos=T cos+0.05g
1 2
At Q:  T cos=0.04g
AnswerMarks
2B1
B1OR: whole system: T cos=0.09g
1
T =0.15g =1.5 N
AnswerMarks
1B1
3
AnswerMarks
6(b)T sin−T sin=0.050.82
1 2
T sin=0.041.42
AnswerMarks Guidance
2M1 Allow sin/cos mix
M1
T sin=0.050.82 +0.041.42
1
5
2 =12.5,  = 2
AnswerMarks
2A1
3
AnswerMarks
6(c)T cos=0.04g and T sin=0.041.42
2 2
7
Divide: tan=
AnswerMarks Guidance
4M1 From part (a) and part (b)
=60.3A1
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | At P:  T cos=T cos+0.05g
1 2
At Q:  T cos=0.04g
2 | B1
B1 | OR: whole system: T cos=0.09g
1
T =0.15g =1.5 N
1 | B1
3
--- 6(b) ---
6(b) | T sin−T sin=0.050.82
1 2
T sin=0.041.42
2 | M1 | Allow sin/cos mix
M1
T sin=0.050.82 +0.041.42
1
5
2 =12.5,  = 2
2 | A1
3
--- 6(c) ---
6(c) | T cos=0.04g and T sin=0.041.42
2 2
7
Divide: tan=
4 | M1 | From part (a) and part (b)
=60.3 | A1
2
Question | Answer | Marks | Guidance
\includegraphics{figure_6}

A particle $P$ of mass $0.05\text{kg}$ is attached to one end of a light inextensible string of length $1\text{m}$. The other end of the string is attached to a fixed point $O$. A particle $Q$ of mass $0.04\text{kg}$ is attached to one end of a second light inextensible string. The other end of this string is attached to $P$.

The particle $P$ moves in a horizontal circle of radius $0.8\text{m}$ with angular speed $\omega\text{rads}^{-1}$. The particle $Q$ moves in a horizontal circle of radius $1.4\text{m}$ also with angular speed $\omega\text{rads}^{-1}$. The centres of the circles are vertically below $O$, and $O$, $P$ and $Q$ are always in the same vertical plane. The strings $OP$ and $PQ$ remain at constant angles $\alpha$ and $\beta$ respectively to the vertical (see diagram).

\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string $OP$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q6 [3]}}
This paper (13 questions)
View full paper
Q1 5 Q2 5 Q3 6 Q3 2 Q4 4 Q4 3 Q5 4 Q5 3 Q6 3 Q6 3 Q6 2 Q7 4 Q7 6