| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle inside smooth hollow cylinder |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces in two directions (horizontal for centripetal force, vertical for equilibrium) with the cone angle providing the geometric constraint. The setup is clearly defined, the approach is routine (resolve perpendicular and parallel to surfaces or horizontally/vertically), and the calculations are straightforward. Slightly easier than average due to clear diagram and standard method. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 06.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving forces vertically — equation must contain weight and component of R | |
| \(R\cos 25° = 0.1g = 1\) | A1 | |
| \(R = 1.10\) | A1ft | ft for \(35°\) instead of \(25°\) (1.22) or sin/cos mix (2.37) |
| \(mv^2/r = S + R\sin 25°\) | M1 | For using Newton's second law and \(a = v^2/r\) (3 terms) |
| \(0.1 \times 2.5^2/0.5 = S + 1.10\sin 25°\) \([1.25 = S + 0.466]\) | A1ft | ft from ans (i) with consistency in sin/cos mix case |
| \(S = 0.784\) or \(0.785\) | A1 | 6 |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces vertically — equation must contain weight and component of R |
| $R\cos 25° = 0.1g = 1$ | A1 | |
| $R = 1.10$ | A1ft | ft for $35°$ instead of $25°$ (1.22) or sin/cos mix (2.37) |
| $mv^2/r = S + R\sin 25°$ | M1 | For using Newton's second law and $a = v^2/r$ (3 terms) |
| $0.1 \times 2.5^2/0.5 = S + 1.10\sin 25°$ $[1.25 = S + 0.466]$ | A1ft | ft from ans (i) with consistency in sin/cos mix case |
| $S = 0.784$ or $0.785$ | A1 | 6 | |
**Total: 6 marks**
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\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{57f7ca89-f028-447a-9ac9-55f931201e6b-2_561_597_1585_406}
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\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{57f7ca89-f028-447a-9ac9-55f931201e6b-2_447_387_1726_1354}
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\caption{Fig. 2}
\end{center}
\end{figure}
A hollow container consists of a smooth circular cylinder of radius 0.5 m , and a smooth hollow cone of semi-vertical angle $65 ^ { \circ }$ and radius 0.5 m . The container is fixed with its axis vertical and with the cone below the cylinder. A steel ball of weight 1 N moves with constant speed $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a horizontal circle inside the container. The ball is in contact with both the cylinder and the cone (see Fig. 1). Fig. 2 shows the forces acting on the ball, i.e. its weight and the forces of magnitudes $R \mathrm {~N}$ and $S \mathrm {~N}$ exerted by the container at the points of contact. Given that the radius of the ball is negligible compared with the radius of the cylinder, find $R$ and $S$.
\hfill \mbox{\textit{CAIE M2 2007 Q3 [6]}}