Challenging +1.8 This is a challenging circular motion problem requiring simultaneous analysis of two connected particles with geometric constraints. Students must set up tension equations for both particles, apply circular motion principles (T = mrω²), use the geometric constraint that the string is inextensible (relating the radii via the angles), and solve a system involving trigonometric relationships. The 7-mark allocation and the need to coordinate multiple concepts (forces, circular motion, geometry, and algebra) places this significantly above average difficulty, though it follows a systematic approach once the setup is understood.
\includegraphics{figure_5}
A light inextensible string \(AB\) passes through two small holes \(C\) and \(D\) in a smooth horizontal table where \(AC = 3a\) and \(DB = a\). A particle of mass \(m\) is attached at the end \(A\) and moves in a horizontal circle with angular velocity \(\omega\). A particle of mass \(\frac{3}{4}m\) is attached to the end \(B\) and moves in a horizontal circle with angular velocity \(k\omega\). \(AC\) makes an angle \(\theta\) with the downward vertical and \(DB\) makes an angle \(\theta\) with the horizontal (see diagram).
Find the value of \(k\). [7]
Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
Answer
Marks
Guidance
5
For A: Tsinmr2
M1
r3asin
B1
Correct expression for radius.
T m3a2
A1
3
Similarly, for B: Tcos mrk22
Answer
Marks
Guidance
4
M1
N2L horizontal
3
T mak22
Answer
Marks
4
A1
3
m3a2 mak22
Answer
Marks
Guidance
4
M1
Equate expressions for T.
k2 4, k 2
A1
Question
Answer
Marks
5
Alternative method for question 5
For A: Tcosmg, Tsinmr2
M1
N2L horizontal and vertical.
r3asin
B1
Correct expression for radius.
5 5 g
T m3a2 mg, 2
Answer
Marks
Guidance
4 12 a
A1
Combine to obtain expression for 2.
3
Similarly, for B: Tcos mrk22
Answer
Marks
Guidance
4
M1
N2L horizontal.
3
T mak22
Answer
Marks
4
A1
5 3 5g
mg mak2
Answer
Marks
Guidance
4 4 12a
M1
Substitute for T and .
k2 4, k 2
A1
7
Answer
Marks
Guidance
Question
Answer
Marks
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
5 | For A: Tsinmr2 | M1 | N2L horizontal.
r3asin | B1 | Correct expression for radius.
T m3a2 | A1
3
Similarly, for B: Tcos mrk22
4 | M1 | N2L horizontal
3
T mak22
4 | A1
3
m3a2 mak22
4 | M1 | Equate expressions for T.
k2 4, k 2 | A1
Question | Answer | Marks | Guidance
5 | Alternative method for question 5
For A: Tcosmg, Tsinmr2 | M1 | N2L horizontal and vertical.
r3asin | B1 | Correct expression for radius.
5 5 g
T m3a2 mg, 2
4 12 a | A1 | Combine to obtain expression for 2.
3
Similarly, for B: Tcos mrk22
4 | M1 | N2L horizontal.
3
T mak22
4 | A1
5 3 5g
mg mak2
4 4 12a | M1 | Substitute for T and .
k2 4, k 2 | A1
7
Question | Answer | Marks | Guidance
\includegraphics{figure_5}
A light inextensible string $AB$ passes through two small holes $C$ and $D$ in a smooth horizontal table where $AC = 3a$ and $DB = a$. A particle of mass $m$ is attached at the end $A$ and moves in a horizontal circle with angular velocity $\omega$. A particle of mass $\frac{3}{4}m$ is attached to the end $B$ and moves in a horizontal circle with angular velocity $k\omega$. $AC$ makes an angle $\theta$ with the downward vertical and $DB$ makes an angle $\theta$ with the horizontal (see diagram).
Find the value of $k$. [7]
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q5 [7]}}