CAIE M2 2010 November — Question 3 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2010
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeConical pendulum – particle on horizontal surface
DifficultyStandard +0.3 This is a standard conical pendulum problem with straightforward application of circular motion principles. Part (i) requires resolving forces and using F=mrω² with given values. Part (ii) involves recognizing that maximum speed occurs when normal reaction becomes zero. Both parts use routine mechanics techniques with no novel insight required, making it slightly easier than average.
Spec6.05c Horizontal circles: conical pendulum, banked tracks

\includegraphics{figure_3} One end of a light inextensible string of length \(0.2 \text{ m}\) is attached to a fixed point \(A\) which is above a smooth horizontal surface. A particle \(P\) of mass \(0.6 \text{ kg}\) is attached to the other end of the string. \(P\) moves in a circle on the surface with constant speed \(v \text{ m s}^{-1}\), with the string taut and making an angle of \(30°\) to the horizontal (see diagram).
  1. Given that \(v = 1.5\), calculate the magnitude of the force that the surface exerts on \(P\). [4]
  2. Given instead that \(P\) moves with its greatest possible speed while remaining in contact with the surface, find \(v\). [3]

Question 3:
AnswerMarks Guidance
3(i) 0.6x1.5 2 /(0.2cos30°) = Tcos30° M1
of tension
AnswerMarks Guidance
T = 9 NA1
R = 0.6g – 9sin30°M1 Resolves vertically, 3 terms
R = 1.5 NA1
[4]
AnswerMarks Guidance
(ii) Tsin30° = 0.6gM1 Resolves vertically, 2 terms
2
AnswerMarks
0.6v /(0.2cos30°) = 12cos30°M1
2
AnswerMarks
v = 3, v =1.73A1
[3]
AnswerMarks Guidance
Page 5Mark Scheme: Teachers’ version Syllabus
GCE A LEVEL – October/November 20109709 52
Question 3:
3 | (i) 0.6x1.5 2 /(0.2cos30°) = Tcos30° | M1 | Uses N2L horizontally with component
of tension
T = 9 N | A1
R = 0.6g – 9sin30° | M1 | Resolves vertically, 3 terms
R = 1.5 N | A1
[4]
(ii) Tsin30° = 0.6g | M1 | Resolves vertically, 2 terms
2
0.6v /(0.2cos30°) = 12cos30° | M1
2
v = 3, v =1.73 | A1
[3]
Page 5 | Mark Scheme: Teachers’ version | Syllabus | Paper
GCE A LEVEL – October/November 2010 | 9709 | 52
\includegraphics{figure_3}

One end of a light inextensible string of length $0.2 \text{ m}$ is attached to a fixed point $A$ which is above a smooth horizontal surface. A particle $P$ of mass $0.6 \text{ kg}$ is attached to the other end of the string. $P$ moves in a circle on the surface with constant speed $v \text{ m s}^{-1}$, with the string taut and making an angle of $30°$ to the horizontal (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Given that $v = 1.5$, calculate the magnitude of the force that the surface exerts on $P$. [4]
\item Given instead that $P$ moves with its greatest possible speed while remaining in contact with the surface, find $v$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2010 Q3 [7]}}