CAIE Further Paper 3 2023 November — Question 6 11 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeMaximum/minimum tension or reaction
DifficultyChallenging +1.8 This is a challenging Further Mechanics problem requiring multiple concepts: conservation of angular momentum about the pivot, energy conservation, circular motion with tension analysis, and the key insight that equal tensions at 60° provides a constraint relating x to a. While systematic, it demands careful coordinate geometry, simultaneous equations, and extended multi-step reasoning across three interconnected parts—significantly harder than standard A-level but not requiring exceptional creativity.
Spec6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration
A particle \(P\) of mass \(m\) is attached to one end of a light inextensible rod of length \(3a\). An identical particle \(Q\) is attached to the other end of the rod. The rod is smoothly pivoted at a point \(O\) on the rod, where \(OQ = x\). The system, of rod and particles, rotates about \(O\) in a vertical plane. At an instant when the rod is vertical, with \(P\) above \(Q\), the particle \(P\) is moving horizontally with speed \(u\). When the rod has turned through an angle of \(60°\) from the vertical, the speed of \(P\) is \(2\sqrt{ag}\), and the tensions in the two parts of the rod, \(OP\) and \(OQ\), have equal magnitudes.
  1. Show that the speed of \(Q\) when the rod has turned through an angle of \(60°\) from the vertical is \(\frac{2x}{3a-x}\sqrt{ag}\). [2]
  2. Find \(x\) in terms of \(a\). [5]
  3. Find \(u\) in terms of \(a\) and \(g\). [4]
Question 6:
AnswerMarks
6(a)Angular speeds of P and Q are equal,
v v
so Q = P
AnswerMarks
x 3a−xM1
2x ag
v =
AnswerMarks Guidance
Q 3a−xA1 Shown convincingly: angular speeds equal stated.
AG
2
AnswerMarks
6(b)m4ag
For P: T +mgcos60=
AnswerMarks
3a−xB1
mv2
For Q: T −mgcos60= Q
AnswerMarks
xB1
m.4ag mv2
Eliminate T: −mgcos60+ =mgcos60+ Q
AnswerMarks
3a−x xM1
m4ag mx4ag
=1+
3a−x (3a−x)2
AnswerMarks Guidance
4a(3a−x)=(3a−x)2+4ax, x2 +2ax−3a2 =0M1 Solve to find x.
Obtain 3-term quadratic equation.
AnswerMarks
(x−a)(x+3a)=0, x=aA1
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(c)Energy changes from initial position:
1 ( 4ag−u2)
Gain in KE of P: m
2
1  u 2 
Loss in KE of Q: m    −v Q 2  
2 2
 
Loss in GPE of P = mg(3a−x)(1−cos60) (=mga)
 1 
Gain in GPE of Q = mgx(1−cos60) = mga 
AnswerMarks Guidance
 2 B1 KEs correct.
B1FTGPEs correct.
1 m ( 4ag−u2) − 1 m      u  2 −v Q 2    =−mgx(1−cos60)+mg(3a−x)(1−cos60)
2 2 2
AnswerMarks Guidance
 M1 Energy equation.
5
Simplify: 4ag− u2 +ag =ag
4
16 4
u2 = ag, u= 5ag
AnswerMarks Guidance
5 5A1 AEF
4
Question 6:
--- 6(a) ---
6(a) | Angular speeds of P and Q are equal,
v v
so Q = P
x 3a−x | M1
2x ag
v =
Q 3a−x | A1 | Shown convincingly: angular speeds equal stated.
AG
2
--- 6(b) ---
6(b) | m4ag
For P: T +mgcos60=
3a−x | B1
mv2
For Q: T −mgcos60= Q
x | B1
m.4ag mv2
Eliminate T: −mgcos60+ =mgcos60+ Q
3a−x x | M1
m4ag mx4ag
=1+
3a−x (3a−x)2
4a(3a−x)=(3a−x)2+4ax, x2 +2ax−3a2 =0 | M1 | Solve to find x.
Obtain 3-term quadratic equation.
(x−a)(x+3a)=0, x=a | A1
5
Question | Answer | Marks | Guidance
--- 6(c) ---
6(c) | Energy changes from initial position:
1 ( 4ag−u2)
Gain in KE of P: m
2
1  u 2 
Loss in KE of Q: m    −v Q 2  
2 2
 
Loss in GPE of P = mg(3a−x)(1−cos60) (=mga)
 1 
Gain in GPE of Q = mgx(1−cos60) = mga 
 2  | B1 | KEs correct.
B1FT | GPEs correct.
1 m ( 4ag−u2) − 1 m      u  2 −v Q 2    =−mgx(1−cos60)+mg(3a−x)(1−cos60)
2 2 2
  | M1 | Energy equation.
5
Simplify: 4ag− u2 +ag =ag
4
16 4
u2 = ag, u= 5ag
5 5 | A1 | AEF
4
A particle $P$ of mass $m$ is attached to one end of a light inextensible rod of length $3a$. An identical particle $Q$ is attached to the other end of the rod. The rod is smoothly pivoted at a point $O$ on the rod, where $OQ = x$. The system, of rod and particles, rotates about $O$ in a vertical plane.

At an instant when the rod is vertical, with $P$ above $Q$, the particle $P$ is moving horizontally with speed $u$. When the rod has turned through an angle of $60°$ from the vertical, the speed of $P$ is $2\sqrt{ag}$, and the tensions in the two parts of the rod, $OP$ and $OQ$, have equal magnitudes.

\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $Q$ when the rod has turned through an angle of $60°$ from the vertical is $\frac{2x}{3a-x}\sqrt{ag}$. [2]

\item Find $x$ in terms of $a$. [5]

\item Find $u$ in terms of $a$ and $g$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q6 [11]}}
This paper (6 questions)
View full paper
Q1 7 Q2 7 Q3 8 Q4 8 Q5 9 Q6 11