CAIE Further Paper 3 2020 November — Question 3 6 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2020
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeElastic string – conical pendulum (string inclined to vertical)
DifficultyChallenging +1.2 This is a standard conical pendulum problem with elastic string requiring resolution of forces in vertical and horizontal directions, application of Hooke's law, and circular motion equations. While it involves multiple concepts (elasticity, circular motion, equilibrium), the setup is conventional and the solution follows a systematic approach with clearly defined steps. The given angular speed and specific parameters guide students toward the solution, making it moderately above average difficulty but not requiring novel insight.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.05c Horizontal circles: conical pendulum, banked tracks

One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(4mg\), is attached to a fixed point \(O\). The other end of the string is attached to a particle of mass \(m\). The particle moves in a horizontal circle with a constant angular speed \(\sqrt{\frac{g}{a}}\) with the string inclined at an angle \(\theta\) to the downward vertical through \(O\). The length of the string during this motion is \((k+1)a\).
  1. Find the value of \(k\). [4]
  2. Find the value of \(\cos\theta\). [2]

Question 3:

AnswerMarks
3(a)ka
T =4mg.
AnswerMarks Guidance
aB1 Use Hooke’s law
Tsinθ=  mrg =m ( k+1 ) asinθ. g
 
AnswerMarks Guidance
 a  aM1 N2L horizontally. Must see T and k .
T =mg ( k+1 )A1
1
Equate: k =
AnswerMarks
3A1
4

AnswerMarks Guidance
3(b)↑Tcosθ=mg M1
4 mg 3
(T = mg) cosθ= =
3 4 4
mg
AnswerMarks
3A1
2
AnswerMarks Guidance
QuestionAnswer Marks
Question 3:
--- 3(a) ---
3(a) | ka
T =4mg.
a | B1 | Use Hooke’s law
Tsinθ=  mrg =m ( k+1 ) asinθ. g
 
 a  a | M1 | N2L horizontally. Must see T and k .
T =mg ( k+1 ) | A1
1
Equate: k =
3 | A1
4
--- 3(b) ---
3(b) | ↑Tcosθ=mg | M1
4 mg 3
(T = mg) cosθ= =
3 4 4
mg
3 | A1
2
Question | Answer | Marks | Guidance
One end of a light elastic string, of natural length $a$ and modulus of elasticity $4mg$, is attached to a fixed point $O$. The other end of the string is attached to a particle of mass $m$. The particle moves in a horizontal circle with a constant angular speed $\sqrt{\frac{g}{a}}$ with the string inclined at an angle $\theta$ to the downward vertical through $O$. The length of the string during this motion is $(k+1)a$.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$. [4]
\item Find the value of $\cos\theta$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q3 [6]}}