Pre-U Pre-U 9795/2 2014 June — Question 7 8 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2014
SessionJune
Marks8
TopicCircular Motion 1
TypeString through hole – lower particle also moves in horizontal circle (conical pendulum below)
DifficultyChallenging +1.2 This is a standard conical pendulum problem with a twist involving two particles connected by a string through a ring. It requires setting up force equations (tension, weight, centripetal force) and using geometry to find the radius, then solving simultaneous equations. While it involves multiple steps and careful geometric reasoning, the mechanics principles are standard A-level Further Maths content with no novel insight required—just systematic application of Newton's laws in circular motion.
Spec3.03b Newton's first law: equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

\includegraphics{figure_7} A light inextensible string of length 8 m is threaded through a smooth fixed ring, \(R\), and carries a particle at each end. One particle, \(P\), of mass 0.5 kg is at rest at a distance 3 m below \(R\). The other particle, \(Q\), is rotating in a horizontal circle whose centre coincides with the position of \(P\) (see diagram). Find the angular speed and the mass of \(Q\). [8]

Question 7:
AnswerMarks
7Recognising (3, 4, 5) triangle
Resolving vertically for P: T= 0.5g
3
Resolving vertically for Q: T =mg
5
4
Resolving horizontally for Q: T =4mω2
5
⇒ m = 0.3
g
ω = or 1.83
AnswerMarks
3[can award B1 if answer
correctly obtained without
use of T = 5]
Needs resolving with trig (if
AnswerMarks Guidance
resolve QR must have
cos or sin × mrω 2 )B1
B1
M1
A1
M1
A1
A1
A1 [8]
AnswerMarks Guidance
Page 5Mark Scheme Syllabus
Pre-U – May/June 20149795 02
Question 7:
7 | Recognising (3, 4, 5) triangle
Resolving vertically for P: T= 0.5g
3
Resolving vertically for Q: T =mg
5
4
Resolving horizontally for Q: T =4mω2
5
⇒ m = 0.3
g
ω = or 1.83
3 | [can award B1 if answer
correctly obtained without
use of T = 5]
Needs resolving with trig (if
resolve || QR must have
cos or sin × mrω 2 ) | B1
B1
M1
A1
M1
A1
A1
A1 [8]
Page 5 | Mark Scheme | Syllabus | Paper
Pre-U – May/June 2014 | 9795 | 02
\includegraphics{figure_7}

A light inextensible string of length 8 m is threaded through a smooth fixed ring, $R$, and carries a particle at each end. One particle, $P$, of mass 0.5 kg is at rest at a distance 3 m below $R$. The other particle, $Q$, is rotating in a horizontal circle whose centre coincides with the position of $P$ (see diagram). Find the angular speed and the mass of $Q$. [8]

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2014 Q7 [8]}}