CAIE Further Paper 3 2020 June — Question 1 2 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2020
SessionJune
Marks2
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypePeriod or time for one revolution
DifficultyModerate -0.5 This is a straightforward application of circular motion formulas requiring only two steps: equate tension to centripetal force (4mg = mv²/a) to find speed, then use T = 2πa/v. Standard bookwork with no conceptual challenges, though the Further Mechanics context places it slightly above typical AS-level questions.
Spec6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r

A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\) on a smooth horizontal plane. The particle \(P\) moves in horizontal circles about \(O\). The tension in the string is \(4mg\). Find, in terms of \(a\) and \(g\), the time that \(P\) takes to make one complete revolution. [2]

Question 1:
AnswerMarks
14g
T =4mg =maω2 so ω2 =
AnswerMarks
aB1
2π a
Time per revn = = π
AnswerMarks
ω gB1
2
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | 4g
T =4mg =maω2 so ω2 =
a | B1
2π a
Time per revn = = π
ω g | B1
2
Question | Answer | Marks
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$ on a smooth horizontal plane. The particle $P$ moves in horizontal circles about $O$. The tension in the string is $4mg$.

Find, in terms of $a$ and $g$, the time that $P$ takes to make one complete revolution. [2]

\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q1 [2]}}