Challenging +1.8 This is a challenging circular motion problem requiring energy conservation and circular motion dynamics at two positions, with the key insight that tension becomes zero at B. Students must set up equations at both A and B, use the perpendicular geometry, and solve simultaneously for the tension at A. The 8-mark allocation and the need to connect multiple concepts (energy, centripetal force, string becoming slack) place this well above average difficulty, though it follows a recognizable framework for Further Maths circular motion problems.
\includegraphics{figure_4}
A particle 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 is initially held with the string taut at the point \(A\), where \(OA\) makes an angle \(\theta\) with the downward vertical through \(O\). The particle is then projected with speed \(u\) perpendicular to \(OA\) and begins to move upwards in part of a vertical circle. The string goes slack when the particle is at the point \(B\) where angle \(AOB\) is a right angle. The speed of the particle when it is at \(B\) is \(\frac{1}{2}u\) (see diagram).
Find the tension in the string at \(A\), giving your answer in terms of \(m\) and \(g\). [8]
Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4
2
1 mu2 − 1 m u =mg ( acosθ+asinθ)
Answer
Marks
Guidance
2 2 2
*M1
Energy equation, with 2 KE terms and a two-part GPE term,
allow cos/sin mix.
3 ( )
u2 =2ag cosθ+sinθ
Answer
Marks
4
A1
2
u
m
2
mgsinθ=
At B, tension in string is zero, so
a
(u2=4agsinθ
Answer
Marks
Guidance
)
B1
N2L
u2
Answer
Marks
Eliminate
DM1
tanθ=2
Answer
Marks
OE
A1
mu2
At A, T −mgcosθ=
Answer
Marks
Guidance
a
B1
N2L
9 5
T = mg ( = 4.02mg)
Answer
Marks
Guidance
5
M1 A1
Substitute to find T.
Alternative method for question 4
2
1 mu2 − 1 m u =mg ( acosθ+asinθ)
Answer
Marks
Guidance
2 2 2
*M1
Energy equation, with 2 KE terms and a two-part GPE term,
allow cos/sin mix.
3 u2 =2ag ( cosθ+sinθ)
Answer
Marks
Guidance
4
A1
Question
Answer
Marks
4
2
u
m
2
mgsinθ=
At B, tension in string is zero, so
a
Answer
Marks
Guidance
(u2=4agsinθ)
B1
N2L
Eliminate θ : u2 = 8ag 5
Answer
Marks
5
DM1 A1
mu2
At A, T −mgcosθ=
Answer
Marks
Guidance
a
B1
N2L
9 5
T = mg ( = 4.02mg)
Answer
Marks
Guidance
5
M1 A1
Substitute to find T.
8
Answer
Marks
Guidance
Question
Answer
Marks
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4 | 2
1 mu2 − 1 m u =mg ( acosθ+asinθ)
2 2 2 | *M1 | Energy equation, with 2 KE terms and a two-part GPE term,
allow cos/sin mix.
3 ( )
u2 =2ag cosθ+sinθ
4 | A1
2
u
m
2
mgsinθ=
At B, tension in string is zero, so
a
(u2=4agsinθ
) | B1 | N2L
u2
Eliminate | DM1
tanθ=2
OE | A1
mu2
At A, T −mgcosθ=
a | B1 | N2L
9 5
T = mg ( = 4.02mg)
5 | M1 A1 | Substitute to find T.
Alternative method for question 4
2
1 mu2 − 1 m u =mg ( acosθ+asinθ)
2 2 2 | *M1 | Energy equation, with 2 KE terms and a two-part GPE term,
allow cos/sin mix.
3 u2 =2ag ( cosθ+sinθ)
4 | A1
Question | Answer | Marks | Guidance
4 | 2
u
m
2
mgsinθ=
At B, tension in string is zero, so
a
(u2=4agsinθ) | B1 | N2L
Eliminate θ : u2 = 8ag 5
5 | DM1 A1
mu2
At A, T −mgcosθ=
a | B1 | N2L
9 5
T = mg ( = 4.02mg)
5 | M1 A1 | Substitute to find T.
8
Question | Answer | Marks | Guidance
\includegraphics{figure_4}
A particle 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 is initially held with the string taut at the point $A$, where $OA$ makes an angle $\theta$ with the downward vertical through $O$. The particle is then projected with speed $u$ perpendicular to $OA$ and begins to move upwards in part of a vertical circle. The string goes slack when the particle is at the point $B$ where angle $AOB$ is a right angle. The speed of the particle when it is at $B$ is $\frac{1}{2}u$ (see diagram).
Find the tension in the string at $A$, giving your answer in terms of $m$ and $g$. [8]
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q4 [8]}}