Question 3 - A-Level Maths

CAIE Further Paper 3 2024 June — Question 3 7 marks

Exam Board	CAIE
Module	Further Paper 3 (Further Paper 3)
Year	2024
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circular Motion 2
Type	Vertical circle: string becomes slack
Difficulty	Standard +0.8 This is a standard circular motion problem requiring energy conservation and Newton's second law in two positions, but the algebra is non-trivial. Part (a) needs careful application of centripetal force at two points plus energy methods to eliminate velocity terms, requiring 3-4 connected steps. Part (b) is straightforward once (a) is established. The question is harder than typical A-level mechanics due to the algebraic manipulation needed, but follows a well-established problem type in Further Maths.
Spec	6.02i Conservation of energy: mechanical energy principle 6.05d Variable speed circles: energy methods

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 \(O\). When the particle is hanging vertically below \(O\), it is projected horizontally with speed \(u\) so that it begins to move along a circular path. When \(P\) is at the lowest point of its motion, the tension in the string is \(T\). When \(OP\) makes an angle \(\theta\) with the upward vertical, the tension in the string is \(S\).

Show that \(S = T - 3mg(1 + \cos\theta)\). [5]
Given that \(u = \sqrt{4ag}\), find the value of \(\cos\theta\) when the string goes slack. [2]

Show mark scheme Show mark scheme source

Question 3:

Answer	Marks
3(a)	mu2

At lowest point, T mg 

Answer	Marks	Guidance
a	B1	Condone r used consistently instead of a

throughout this question.

mv2

When string makes angle  with upward vertical, Smgcos

Answer	Marks
a	B1

1 1

Energy: mu2  mv2 mga1cos

Answer	Marks	Guidance
2 2	M1	Must include m. Allow sin instead of cos for

this mark, allow sign errors.

Answer	Marks	Guidance
Eliminate u2 and v2	M1	Need to see at least one line of working.
S T 3mg1cos	A1	AG

5

Answer	Marks	Guidance
3(b)	When string goes slack, S = 0 so T 3mg1cos	M1

equation.

mu2 2

But T mg mg4mg 5mg, so cos

Answer	Marks
a 3	A1

2

Answer	Marks	Guidance
Question	Answer	Marks

Question 3:
--- 3(a) ---
3(a) | mu2
At lowest point, T mg 
a | B1 | Condone r used consistently instead of a
throughout this question.
mv2
When string makes angle  with upward vertical, Smgcos
a | B1
1 1
Energy: mu2  mv2 mga1cos
2 2 | M1 | Must include m. Allow sin instead of cos for
this mark, allow sign errors.
Eliminate u2 and v2 | M1 | Need to see at least one line of working.
S T 3mg1cos | A1 | AG
5
--- 3(b) ---
3(b) | When string goes slack, S = 0 so T 3mg1cos | M1 | May use v2 agcos substituted into energy
equation.
mu2 2
But T mg mg4mg 5mg, so cos
a 3 | A1
2
Question | Answer | Marks | Guidance

Show LaTeX source

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 $O$. When the particle is hanging vertically below $O$, it is projected horizontally with speed $u$ so that it begins to move along a circular path. When $P$ is at the lowest point of its motion, the tension in the string is $T$. When $OP$ makes an angle $\theta$ with the upward vertical, the tension in the string is $S$.

\begin{enumerate}[label=(\alph*)]
\item Show that $S = T - 3mg(1 + \cos\theta)$. [5]

\item Given that $u = \sqrt{4ag}$, find the value of $\cos\theta$ when the string goes slack. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q3 [7]}}

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