Question 6 - A-Level Maths

CAIE Further Paper 3 2024 November — Question 6 10 marks

Exam Board	CAIE
Module	Further Paper 3 (Further Paper 3)
Year	2024
Session	November
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Circular Motion 2
Type	Vertical circle with peg/obstacle
Difficulty	Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and careful analysis of the transition when the string hits the peg. Students must find the speed at two positions, apply the tension equation for circular motion, and use the slack condition (T=0) to find the angle. The multi-stage nature and the need to track energy through the peg collision make this substantially harder than typical A-level questions, though the individual techniques are standard for Further Maths.
Spec	6.05d Variable speed circles: energy methods 6.05e Radial/tangential acceleration

\includegraphics{figure_6} 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 \(O\). The particle \(P\) is held with the string taut and the string makes an angle \(\theta\) with the downward vertical through \(O\). The particle is projected at right angles to the string with speed \(\frac{1}{3}\sqrt{10ag}\) and begins to move downwards along a circular path. When the string is vertical, it strikes a small smooth peg at the point \(A\) which is vertically below \(O\). The circular path and the point \(A\) are in the same vertical plane. After the string strikes the peg, the particle \(P\) begins to move in a vertical circle with centre \(A\). When the string makes an angle \(\theta\) with the upward vertical through \(A\) the string becomes slack (see diagram). The distance of \(A\) below \(O\) is \(\frac{5}{6}a\).

Find the value of \(\cos \theta\). [6]
Find the ratio of the tensions in the string immediately before and immediately after it strikes the peg. [4]

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks
6(a)	For P to lowest point L:

1 1

Energy: mv2 = mu2 +mga(1−cos)

2 2

28 [v2 = ag−2agcos]

Answer	Marks	Guidance
9	M1*	Dimensionally correct, all terms present, allow

sign errors, allow cos/sin error.

From L to string goes slack:

1 1 4a

Energy: mv2 = mw2 + mg(1+cos)

2 2 9

 20 26 

w2 = ag− agcos

 

Answer	Marks	Guidance
 9 9 	M1*	4a

Dimensionally correct, all terms present, with ,

9 allow sign errors, allow cos/sin error.

Answer	Marks
Both equations correct, allow unsimplified.	A1

mw2

4a

When string goes slack: mgcos=

Answer	Marks
9	B1
Equate expressions for w2 to find a value for cos	DM1

2 cos=

Answer	Marks
3	A1

6

Answer	Marks	Guidance
Question	Answer	Marks

Alternative method for question 6(a)

For P from start to string goes slack:

1 1  4 

Energy: mw2 = mu2 +mga  ( 1−cos)− (1+cos) 

Answer	Marks	Guidance
2 2  9 	M2	4

Dimensionally correct, all terms present, with a

9 Allow sign errors, allow cos/sin error

5  4

RHS may appear with a− 1+  acos

9  9

4 Allow M1 if is missing or if an attempt at both

9 heights, but all other conditions are met.

 10 5 13  20 26 

w2 = ag+2ag − cos = ag− agcos

   

Answer	Marks	Guidance
 9 9 9  9 9 	A1	Correct, allow unsimplified.

mw2

4a

When string goes slack: mgcos=

Answer	Marks
9	B1
Equate expressions for w2 to find a value for cos	DM1

2 cos=

Answer	Marks
3	A1

6

Answer	Marks	Guidance
Question	Answer	Marks

Answer	Marks
6(b)	mv2

Tension before: T −mg =

1 a

mv2

Tension after: T −mg =

2 4

a

Answer	Marks	Guidance
9	M1	EITHER equation, dimensionally correct, allow

sign error only

 16  25

v2 = ag, T = mg T =5mg

Answer	Marks	Guidance
  9   1 9 2	A1	EITHER tension correct
Find the other tension (from a valid equation) and find ratio of tensions	M1	mv2

Equation must be of the form T −mg =

r

Answer	Marks	Guidance
Ratio is 5 : 9	A1	5

Any equivalent ratio, allow

9

4

Answer	Marks	Guidance
Question	Answer	Marks

Question 6:
--- 6(a) ---
6(a) | For P to lowest point L:
1 1
Energy: mv2 = mu2 +mga(1−cos)
2 2
28
[v2 = ag−2agcos]
9 | M1* | Dimensionally correct, all terms present, allow
sign errors, allow cos/sin error.
From L to string goes slack:
1 1 4a
Energy: mv2 = mw2 + mg(1+cos)
2 2 9
 20 26 
w2 = ag− agcos
 
 9 9  | M1* | 4a
Dimensionally correct, all terms present, with ,
9
allow sign errors, allow cos/sin error.
Both equations correct, allow unsimplified. | A1
mw2
4a
When string goes slack: mgcos=
9 | B1
Equate expressions for w2 to find a value for cos | DM1
2
cos=
3 | A1
6
Question | Answer | Marks | Guidance
Alternative method for question 6(a)
For P from start to string goes slack:
1 1  4 
Energy: mw2 = mu2 +mga  ( 1−cos)− (1+cos) 
2 2  9  | M2 | 4
Dimensionally correct, all terms present, with a
9
Allow sign errors, allow cos/sin error
5  4
RHS may appear with a− 1+  acos
9  9
4
Allow M1 if is missing or if an attempt at both
9
heights, but all other conditions are met.
 10 5 13  20 26 
w2 = ag+2ag − cos = ag− agcos
   
 9 9 9  9 9  | A1 | Correct, allow unsimplified.
mw2
4a
When string goes slack: mgcos=
9 | B1
Equate expressions for w2 to find a value for cos | DM1
2
cos=
3 | A1
6
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | mv2
Tension before: T −mg =
1 a
mv2
Tension after: T −mg =
2 4
a
9 | M1 | EITHER equation, dimensionally correct, allow
sign error only
 16  25
v2 = ag, T = mg T =5mg
  9   1 9 2 | A1 | EITHER tension correct
Find the other tension (from a valid equation) and find ratio of tensions | M1 | mv2
Equation must be of the form T −mg =
r
Ratio is 5 : 9 | A1 | 5
Any equivalent ratio, allow
9
4
Question | Answer | Marks | Guidance

Show LaTeX source

\includegraphics{figure_6}

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 $O$. The particle $P$ is held with the string taut and the string makes an angle $\theta$ with the downward vertical through $O$. The particle is projected at right angles to the string with speed $\frac{1}{3}\sqrt{10ag}$ and begins to move downwards along a circular path. When the string is vertical, it strikes a small smooth peg at the point $A$ which is vertically below $O$. The circular path and the point $A$ are in the same vertical plane. After the string strikes the peg, the particle $P$ begins to move in a vertical circle with centre $A$. When the string makes an angle $\theta$ with the upward vertical through $A$ the string becomes slack (see diagram). The distance of $A$ below $O$ is $\frac{5}{6}a$.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $\cos \theta$. [6]
\item Find the ratio of the tensions in the string immediately before and immediately after it strikes the peg. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q6 [10]}}

This paper (7 questions)

View full paper

Q1 3 Q2 4 Q3 7 Q4 7 Q5 8 Q6 10 Q7 11