CAIE M2 2008 November — Question 4 7 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2008
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeString through hole – hanging particle in equilibrium below table
DifficultyStandard +0.3 This is a standard coupled particles problem combining circular motion with equilibrium. Students must equate tension from B's weight to the centripetal force for A's circular motion, then solve for angular speed. It's slightly above average difficulty due to the coupling of two concepts, but follows a well-established problem type with straightforward application of T = mg and T = mrω².
Spec6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

  1. the base of the cylinder,
  2. the curved surface of the cylinder.
    (ii) \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{5109244c-3062-4f5f-9277-fc6b5b28f2d4-3_348_745_1183_740} \captionsetup{labelformat=empty} \caption{Fig. 2}
    \end{figure} Sphere \(A\) is now attached to one end of a light inextensible string. The string passes through a small smooth hole in the middle of the base of the cylinder. Another small sphere \(B\), of mass 0.25 kg , is attached to the other end of the string. \(B\) hangs in equilibrium below the hole while \(A\) is moving in a horizontal circle of radius 0.2 m (see Fig. 2). Find the angular speed of \(A\).

Question 4:
Part (i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
(a) Magnitude is \(1.5\text{ N}\)B1 From \(R = mg\)
(b) \(S = mv^2/r\)M1 For using Newton's second law
Magnitude is \(0.617\text{ N}\)A1 [3]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Tension is \(2.5\text{ N}\)B1
\(T = mr\omega^2\)M1 For using Newton's second law
\(2.5 = 0.15 \times 0.2 \times \omega^2\)A1ft
Angular speed is \(9.13 \text{ rads}^{-1}\)A1 [4]
## Question 4:

### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (a) Magnitude is $1.5\text{ N}$ | B1 | From $R = mg$ |
| (b) $S = mv^2/r$ | M1 | For using Newton's second law |
| Magnitude is $0.617\text{ N}$ | A1 | **[3]** |

### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tension is $2.5\text{ N}$ | B1 | |
| $T = mr\omega^2$ | M1 | For using Newton's second law |
| $2.5 = 0.15 \times 0.2 \times \omega^2$ | A1ft | |
| Angular speed is $9.13 \text{ rads}^{-1}$ | A1 | **[4]** |

---
\begin{enumerate}[label=(\alph*)]
\item the base of the cylinder,
\item the curved surface of the cylinder.\\
(ii)

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{5109244c-3062-4f5f-9277-fc6b5b28f2d4-3_348_745_1183_740}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

Sphere $A$ is now attached to one end of a light inextensible string. The string passes through a small smooth hole in the middle of the base of the cylinder. Another small sphere $B$, of mass 0.25 kg , is attached to the other end of the string. $B$ hangs in equilibrium below the hole while $A$ is moving in a horizontal circle of radius 0.2 m (see Fig. 2). Find the angular speed of $A$.
\end{enumerate}

\hfill \mbox{\textit{CAIE M2 2008 Q4 [7]}}