CAIE M2 2010 June — Question 3 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2010
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeConical pendulum – horizontal circle in free space (no surface)
DifficultyModerate -0.8 This is a straightforward conical pendulum problem requiring standard application of Newton's second law in circular motion. Students resolve forces vertically and horizontally, use the given acceleration to find tan θ (which is shown), then find tension and speed using basic trigonometry and circular motion formulas. All steps are routine with no problem-solving insight required.
Spec3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

3 \includegraphics[max width=\textwidth, alt={}, center]{ae809dfc-c5af-4c0a-9c88-009949d3e9f9-3_456_511_260_817} A particle of mass 0.24 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. The particle moves with constant speed in a horizontal circle. The string makes an angle \(\theta\) with the vertical (see diagram), and the tension in the string is \(T \mathrm {~N}\). The acceleration of the particle has magnitude \(7.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }\).
  1. Show that \(\tan \theta = 0.75\) and find the value of \(T\).
  2. Find the speed of the particle.

(i)
\(mg = T\cos\theta\)
\(ma = T\sin\theta\)
\(\tan\theta = a/g = 0.75\)
AnswerMarks Guidance
\(T = 0.24 \times 10/\cos\theta = 3\)\(B1, B1, B1, B1\) [4] SR \(B1\) not \(B2\) for \(\tan\theta = \sqrt{3}/gr\) or \(a/g\) used. AG. For using \(T\cos\theta = mg\) to find \(T\)
(ii)
\([v^2 = 7.5 \times 2\sin\theta]\)
AnswerMarks Guidance
Speed is 3ms\(^{-1}\)\(M1, A1\) [2] For using \(v^2 = ar\) to find \(v\)
**(i)**

$mg = T\cos\theta$
$ma = T\sin\theta$
$\tan\theta = a/g = 0.75$
$T = 0.24 \times 10/\cos\theta = 3$ | $B1, B1, B1, B1$ [4] | SR $B1$ not $B2$ for $\tan\theta = \sqrt{3}/gr$ or $a/g$ used. AG. For using $T\cos\theta = mg$ to find $T$

**(ii)**

$[v^2 = 7.5 \times 2\sin\theta]$
Speed is 3ms$^{-1}$ | $M1, A1$ [2] | For using $v^2 = ar$ to find $v$

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\includegraphics[max width=\textwidth, alt={}, center]{ae809dfc-c5af-4c0a-9c88-009949d3e9f9-3_456_511_260_817}

A particle of mass 0.24 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. The particle moves with constant speed in a horizontal circle. The string makes an angle $\theta$ with the vertical (see diagram), and the tension in the string is $T \mathrm {~N}$. The acceleration of the particle has magnitude $7.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Show that $\tan \theta = 0.75$ and find the value of $T$.\\
(ii) Find the speed of the particle.

\hfill \mbox{\textit{CAIE M2 2010 Q3 [6]}}