CAIE M2 2014 November — Question 7 8 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2014
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle: tension at specific point
DifficultyStandard +0.3 This is a standard vertical circular motion problem requiring application of Newton's second law for circular motion and energy conservation. Part (i) is direct substitution into T - mg = mv²/r. Part (ii) uses energy conservation between two positions. Part (iii) applies the tension formula again. All steps are routine textbook exercises with no novel problem-solving required, making it slightly easier than average.
Spec6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall
\includegraphics{figure_7} A particle of mass \(0.4\) kg is attached to one end of a light inextensible string of length \(2\) m. The other end of the string is attached to a fixed point \(O\). The particle moves in a vertical circle and passes through the lowest point of the circle with speed \(6\) m s\(^{-1}\).
  1. Find the tension in the string when the particle is at the lowest point. [2]
  2. Find the speed of the particle when the string makes an angle of \(60°\) with the downward vertical. [4]
  3. Hence find the tension in the string at this position. [2]
Question 7:
(ii) ---
7 (i)
AnswerMarks
(ii) 0.4 
15 
cosθ 
T =
2
3
T = AG
cosθ
Tcosθ = mg
m = 0.3
r = 0.4tanθ
0.3v2
=Tsinθ OR 0.3ω 2 r = Tsinθ
r
3
0.3ω 2 (0.4tanθ) = × sinθ
cosθ
ω = 5
SC
Candidates who choose at least two specific
values of θ:
Calculation of r twice
AnswerMarks
Both calculations give ω = 5M1
A1
M1
A1
[4]
B1
M1
A1
A1
[4]
B1
AnswerMarks
B1λext
Uses T =
2
Resolves vertically for P
nd
Newton’s 2 law with correct
expression for radial accn, ft cv(m(i))
AnswerMarks
(iii) 0.4 2
15 
cosθ 
EPE =
2×2
( )2
0.3 5×0.4tanθ
KE =
2
 0.4 2
15 
cosθ  0.3(2tanθ)2 
= ×2
2×2  2 
2 2 2
cos θ tan θ = 0.5 OR sin θ = 0.5
AnswerMarks
θ = 45B1
B1
M1
A1
AnswerMarks
[4]ft candidate’s value of ω
Award if × 2 is with wrong term
www
Question 7:
--- 7 (i)
(ii) ---
7 (i)
(ii) |  0.4 
15 
cosθ 
T =
2
3
T = AG
cosθ
Tcosθ = mg
m = 0.3
r = 0.4tanθ
0.3v2
=Tsinθ OR 0.3ω 2 r = Tsinθ
r
3
0.3ω 2 (0.4tanθ) = × sinθ
cosθ
ω = 5
SC
Candidates who choose at least two specific
values of θ:
Calculation of r twice
Both calculations give ω = 5 | M1
A1
M1
A1
[4]
B1
M1
A1
A1
[4]
B1
B1 | λext
Uses T =
2
Resolves vertically for P
nd
Newton’s 2 law with correct
expression for radial accn, ft cv(m(i))
(iii) |  0.4 2
15 
cosθ 
EPE =
2×2
( )2
0.3 5×0.4tanθ
KE =
2
 0.4 2
15 
cosθ  0.3(2tanθ)2 
= ×2
2×2  2 
2 2 2
cos θ tan θ = 0.5 OR sin θ = 0.5
θ = 45 | B1
B1
M1
A1
[4] | ft candidate’s value of ω
Award if × 2 is with wrong term
www
\includegraphics{figure_7}

A particle of mass $0.4$ kg is attached to one end of a light inextensible string of length $2$ m. The other end of the string is attached to a fixed point $O$. The particle moves in a vertical circle and passes through the lowest point of the circle with speed $6$ m s$^{-1}$.

\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string when the particle is at the lowest point. [2]
\item Find the speed of the particle when the string makes an angle of $60°$ with the downward vertical. [4]
\item Hence find the tension in the string at this position. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2014 Q7 [8]}}
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Q1 7 Q2 6 Q3 6 Q4 8 Q5 7 Q6 12 Q7 8