CAIE M2 2012 November — Question 4 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionNovember
Marks6
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TopicCircular Motion 1
TypeParticle on cone surface – with string attached to vertex or fixed point
DifficultyStandard +0.3 This is a standard circular motion problem with a clear setup and straightforward force resolution. Students must resolve forces perpendicular and parallel to the cone surface, apply the circular motion equation (T = mω²r), and use the given condition that tension equals weight. While it requires careful geometry with the 45° angle and systematic application of Newton's laws, it follows a well-practiced method with no novel insights needed. The 6 marks reflect multiple steps rather than conceptual difficulty.
Spec3.03e Resolve forces: two dimensions3.03n Equilibrium in 2D: particle under forces6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks
\includegraphics{figure_4} A particle \(P\) is moving inside a smooth hollow cone which has its vertex downwards and its axis vertical, and whose semi-vertical angle is \(45°\). A light inextensible string parallel to the surface of the cone connects \(P\) to the vertex. \(P\) moves with constant angular speed in a horizontal circle of radius \(0.67\) m (see diagram). The tension in the string is equal to the weight of \(P\). Calculate the angular speed of \(P\). [6]
AnswerMarks Guidance
\(Rcos45 - Tcos45 = mg\)M1 Resolves vertically for \(P\)
\(Rcos45 = mg + mg \cos45\)A1 May be implied for later work
\(Rsin45 + Tsin45 = m\omega^2 \times 0.67\)M1 Uses Newton's Second Law horizontally for \(P\)
M1Obtaining an equation in \(m\) (and \(g\))
\(mg + mg \cos45 + mg \sin45 = m\omega^2 \times 0.67\)A1
\(\omega = 6(00) \, \text{rads}^{-1}\)A1 [6]
OR
AnswerMarks Guidance
\(M1\)Resolves radial acceleration parallel to the slope for \(P\)
Acceleration \(= \omega^2 \times 0.67cos45\)A1 May be implied by later work
\(m\omega^2 \times 0.67\cos45 = T + mg \cos45\)M1 Uses Newton's Second Law parallel to the slope for \(P\)
M1Obtaining an equation in \(m\) (and \(g\))
\(m\omega^2 \times 0.67\cos45 = mg + mg \cos45\)A1
\(\omega = 6(00) \, \text{rads}^{-1}\)A1 
$Rcos45 - Tcos45 = mg$ | M1 | Resolves vertically for $P$
$Rcos45 = mg + mg \cos45$ | A1 | May be implied for later work
$Rsin45 + Tsin45 = m\omega^2 \times 0.67$ | M1 | Uses Newton's Second Law horizontally for $P$
| M1 | Obtaining an equation in $m$ (and $g$)
$mg + mg \cos45 + mg \sin45 = m\omega^2 \times 0.67$ | A1 |
$\omega = 6(00) \, \text{rads}^{-1}$ | A1 [6] |

**OR**

$M1$ | Resolves radial acceleration parallel to the slope for $P$
Acceleration $= \omega^2 \times 0.67cos45$ | A1 | May be implied by later work
$m\omega^2 \times 0.67\cos45 = T + mg \cos45$ | M1 | Uses Newton's Second Law parallel to the slope for $P$
| M1 | Obtaining an equation in $m$ (and $g$)
$m\omega^2 \times 0.67\cos45 = mg + mg \cos45$ | A1 |
$\omega = 6(00) \, \text{rads}^{-1}$ | A1 |
\includegraphics{figure_4}

A particle $P$ is moving inside a smooth hollow cone which has its vertex downwards and its axis vertical, and whose semi-vertical angle is $45°$. A light inextensible string parallel to the surface of the cone connects $P$ to the vertex. $P$ moves with constant angular speed in a horizontal circle of radius $0.67$ m (see diagram). The tension in the string is equal to the weight of $P$. Calculate the angular speed of $P$.
[6]

\hfill \mbox{\textit{CAIE M2 2012 Q4 [6]}}
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