CAIE M2 2011 November — Question 5 9 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2011
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeMultiple particles on string
DifficultyStandard +0.8 This is a multi-particle circular motion problem requiring systematic application of Newton's second law to three connected particles, with careful bookkeeping of tensions and radii. While the individual steps are standard (F=mrω²), the three-part structure with interdependent calculations and the need to track multiple tensions and masses makes this more challenging than typical single-particle circular motion questions.
Spec6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

5 \includegraphics[max width=\textwidth, alt={}, center]{a093cbad-3ba0-45ce-a617-d4ecc8cb1ec9-3_927_1022_689_559} One end of a light inextensible string of length 1.2 m is attached to a fixed point \(O\) on a smooth horizontal surface. Particles \(P , Q\) and \(R\) are attached to the string so that \(O P = P Q = Q R = 0.4 \mathrm {~m}\). The particles rotate in horizontal circles about \(O\) with constant angular speed \(\omega \operatorname { rads } ^ { - 1 }\) and with \(O , P\), \(Q\) and \(R\) in a straight line (see diagram). \(R\) has mass 0.2 kg , and the tensions in the parts of the string attached to \(Q\) are 6 N and 10 N .
  1. Show that \(\omega = 5\).
  2. Calculate the mass of \(Q\).
  3. Given that the kinetic energy of \(P\) is equal to the kinetic energy of \(R\), calculate the tension in the part of the string attached to \(O\).

AnswerMarks Guidance
(i) \(0.2\omega^2 \times 1.2 = 6\)M1 Uses radial acceleration on \(R\), 1 force
\(\omega = 5\)A1 [2]
(ii) \(m\omega^2 \times 2 \times 0.4 = 10 - 6\)M1 Uses radial acceleration on \(Q\), 2 forces
\(m = 0.2 \text{ kg}\)A1 A1 [3]
(iii) \(0.2 \times (5 \times 1.2)^2/2 = M(5 \times 0.4)^2/2\)M1
\(M = 1.8 \text{ kg}\)A1
\(1.8 \times 5^2 \times 0.4 = T - 10\)DM1
\(T = 28 \text{ N}\)A1 [4]
**(i)** $0.2\omega^2 \times 1.2 = 6$ | M1 | Uses radial acceleration on $R$, 1 force
$\omega = 5$ | A1 | [2]

**(ii)** $m\omega^2 \times 2 \times 0.4 = 10 - 6$ | M1 | Uses radial acceleration on $Q$, 2 forces
$m = 0.2 \text{ kg}$ | A1 | A1 [3]

**(iii)** $0.2 \times (5 \times 1.2)^2/2 = M(5 \times 0.4)^2/2$ | M1 |
$M = 1.8 \text{ kg}$ | A1 |
$1.8 \times 5^2 \times 0.4 = T - 10$ | DM1 |
$T = 28 \text{ N}$ | A1 | [4]
5\\
\includegraphics[max width=\textwidth, alt={}, center]{a093cbad-3ba0-45ce-a617-d4ecc8cb1ec9-3_927_1022_689_559}

One end of a light inextensible string of length 1.2 m is attached to a fixed point $O$ on a smooth horizontal surface. Particles $P , Q$ and $R$ are attached to the string so that $O P = P Q = Q R = 0.4 \mathrm {~m}$. The particles rotate in horizontal circles about $O$ with constant angular speed $\omega \operatorname { rads } ^ { - 1 }$ and with $O , P$, $Q$ and $R$ in a straight line (see diagram). $R$ has mass 0.2 kg , and the tensions in the parts of the string attached to $Q$ are 6 N and 10 N .\\
(i) Show that $\omega = 5$.\\
(ii) Calculate the mass of $Q$.\\
(iii) Given that the kinetic energy of $P$ is equal to the kinetic energy of $R$, calculate the tension in the part of the string attached to $O$.

\hfill \mbox{\textit{CAIE M2 2011 Q5 [9]}}