CAIE FP2 2009 November — Question 1 5 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2009
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeValidity and assumptions questions
DifficultyChallenging +1.2 This is a standard circular motion problem requiring energy conservation and tension analysis. While it involves multiple steps (finding speed when string slackens using T=0 condition, then using energy conservation to find initial speed), the approach is methodical and follows well-established mechanics principles. The 120° angle and the specific answer form suggest this is a textbook-style question, though it requires careful application of Newton's second law in circular motion and energy methods, making it moderately above average difficulty.
Spec6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall

A particle of mass \(m\) is attached to one end \(A\) of a light inextensible string of length \(a\). The other end of the string is attached to a fixed point \(O\) and the particle hangs in equilibrium under gravity. The particle is projected horizontally so that it starts to move in a vertical circle. The string slackens after turning through an angle of \(120°\). Show that the speed of the particle is then \(\sqrt{\left(\frac{4}{3}ga\right)}\) and find the initial speed of projection. [5]

Question 1:
AnswerMarks
1Equate radial forces when string slackens: mv 2 /a = mg cos 60° M1
Rearrange to obtain v: v = √(½ga) A.G. A1
Use conservation of energy: ½ mu 2 = ½ mv 2 + mga(1 + cos 60°) M1 A1
Substitute for v and rearrange to obtain u: u 2 = ½ ga + 3ga, u = √(7ga/2)
AnswerMarks Guidance
or 1⋅87√ (ga) or 5⋅92√a A1(5) [5]
Question 1:
1 | Equate radial forces when string slackens: mv 2 /a = mg cos 60° M1
Rearrange to obtain v: v = √(½ga) A.G. A1
Use conservation of energy: ½ mu 2 = ½ mv 2 + mga(1 + cos 60°) M1 A1
Substitute for v and rearrange to obtain u: u 2 = ½ ga + 3ga, u = √(7ga/2)
or 1⋅87√ (ga) or 5⋅92√a A1 | (5) | [5]
A particle of mass $m$ is attached to one end $A$ of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$ and the particle hangs in equilibrium under gravity. The particle is projected horizontally so that it starts to move in a vertical circle. The string slackens after turning through an angle of $120°$. Show that the speed of the particle is then $\sqrt{\left(\frac{4}{3}ga\right)}$ and find the initial speed of projection. [5]

\hfill \mbox{\textit{CAIE FP2 2009 Q1 [5]}}