CAIE Further Paper 3 2024 November — Question 6 2 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks2
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Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeConical pendulum – horizontal circle in free space (no surface)
DifficultyModerate -1.0 This is a single sub-part worth 2 marks asking to find a specific value (β) in a circular motion context. Without seeing parts (a) and (b), this appears to be a straightforward calculation following from earlier setup work, requiring substitution into a formula or solving a simple equation rather than novel problem-solving.
Spec6.05c Horizontal circles: conical pendulum, banked tracks
  1. Find the value of \(\beta\). [2]
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
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $\beta$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q6 [2]}}
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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