CAIE M2 2012 November — Question 4 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2012
SessionNovember
Marks6
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Mark schemeDownload PDF ↗
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 particle on a cone surface. It requires resolving forces in horizontal and vertical directions, applying Newton's second law for circular motion, and using the given constraint that tension equals weight. The geometry is straightforward with a 45° angle, and the problem follows a well-established template for conical pendulum questions. While it involves multiple steps (force resolution, circular motion equation, algebraic manipulation), these are routine techniques for M2 students with no novel insight required.
Spec6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks
4 \includegraphics[max width=\textwidth, alt={}, center]{2c6b2e42-09cb-4653-9378-6c6add7771cc-2_538_885_1809_628} 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 \(45 ^ { \circ }\). 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 \(P\).
AnswerMarks Guidance
\(R\cos45 - T\cos45 = mg\)M1 Resolves vertically for \(P\)
\(R\cos45 = mg + mg\cos45\)A1 May be implied for later work
\(R\sin45 + T\sin45 = m\omega^2 \times 0.67\)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
M1Resolves radial acceleration parallel to the slope for \(P\)
Acceleration \(= \omega^2 \times 0.67\cos45\)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 
$R\cos45 - T\cos45 = mg$ | M1 | Resolves vertically for $P$
$R\cos45 = mg + mg\cos45$ | A1 | May be implied for later work
$R\sin45 + T\sin45 = m\omega^2 \times 0.67$ | 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 $= \omega^2 \times 0.67\cos45$ | 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 |
4\\
\includegraphics[max width=\textwidth, alt={}, center]{2c6b2e42-09cb-4653-9378-6c6add7771cc-2_538_885_1809_628}

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 $45 ^ { \circ }$. 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 $P$.

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