| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces (normal reaction and weight) and application of F=mv²/r. The geometry is given explicitly (0.3m below centre), making it straightforward. All three parts follow a predictable sequence with no novel insight required, placing it slightly above average difficulty due to 3D geometry and multi-step calculation. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \([Rx(0.3/0.5) = 0.12g]\) | M1 | For resolving forces vertically |
| Force exerted is \(2\ \text{N}\) | A1 [2] | AG |
| (ii) \([R\cos\alpha = mv^2/r]\) | M1 | For using Newton's second law with \(a = v^2/r\) |
| \(2(0.4/0.5) = 0.12v^2/(0.5 \times 0.4/0.5)\) | A1 | |
| Speed is \(2.31\ \text{ms}^{-1}\) | A1 [3] | |
| (iii) | M1 | For using \(T = 2\pi r/v\); Ft \(T = 0.8\pi/v\) or correct value |
| Time taken is \(1.09\ \text{s}\) | A1\(\sqrt{}\) [2] | From incorrect \(r\) in (ii) and (iii) |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $[Rx(0.3/0.5) = 0.12g]$ | M1 | For resolving forces vertically |
| Force exerted is $2\ \text{N}$ | A1 [2] | AG |
| **(ii)** $[R\cos\alpha = mv^2/r]$ | M1 | For using Newton's second law with $a = v^2/r$ |
| $2(0.4/0.5) = 0.12v^2/(0.5 \times 0.4/0.5)$ | A1 | |
| Speed is $2.31\ \text{ms}^{-1}$ | A1 [3] | |
| **(iii)** | M1 | For using $T = 2\pi r/v$; Ft $T = 0.8\pi/v$ or correct value |
| Time taken is $1.09\ \text{s}$ | A1$\sqrt{}$ [2] | From incorrect $r$ in **(ii)** and **(iii)** |
---4\\
\begin{tikzpicture}[>=Stealth]
% Main circle (sphere cross-section)
\coordinate (C) at (0,0); % centre of sphere
\coordinate (H) at (0,-0.3*4); % centre of horizontal circle, 0.3 m below centre (scaled by 4)
\def\R{2} % radius of sphere scaled (0.5 m * 4)
\def\d{1.2} % 0.3 m * 4 = vertical distance below centre
% Radius of horizontal circle: sqrt(0.5^2 - 0.3^2) = 0.4 m, scaled by 4 = 1.6
\def\r{1.6}
% Draw the outer circle (sphere)
\draw[thick] (C) circle (\R);
% Draw the horizontal dashed circle (as an ellipse seen from an angle)
\draw[dashed] (H) ellipse ({\r} and {0.35});
% Draw radius line from centre to a point on the circle (upper right)
\coordinate (P) at ({\r*cos(30)}, {-\d + \r*sin(30)*0.35/\r});
% Actually, let's draw a line from centre to the sphere surface to show 0.5 m
\coordinate (S) at ({1.2}, {1.6}); % a point on the sphere, upper right area
% Better approach: draw the radius to a point on the sphere
% Point on sphere at some angle
\pgfmathsetmacro{\ang}{50}
\coordinate (S) at ({\R*cos(\ang)}, {\R*sin(\ang)});
\draw[dashed] (C) -- (S);
\node[right] at ({0.5*\R*cos(\ang)+0.15}, {0.5*\R*sin(\ang)+0.1}) {$0.5\,\mathrm{m}$};
% Draw vertical dashed line from centre down to H showing 0.3 m
\draw[dashed] (C) -- (H);
% Mark the 0.3 m distance
\node[left] at (-0.15, {-\d/2}) {$0.3\,\mathrm{m}$};
% Small tick marks or braces for 0.3 m
\draw ({-0.08},0) -- ({0.08},0);
\draw ({-0.08},{-\d}) -- ({0.08},{-\d});
% Draw a line from centre of sphere to the particle on the horizontal circle (on the sphere surface)
% The particle sits where the horizontal circle meets the sphere
\coordinate (Part) at ({\r}, {-\d});
% Draw line from C to particle
\draw[dashed] (C) -- (Part);
% Draw the particle as a filled dot
\fill (Part) circle (0.06);
% Small marks at centre
\fill (C) circle (0.04);
\fill (H) circle (0.04);
\end{tikzpicture}
A particle of mass 0.12 kg is moving on the smooth inside surface of a fixed hollow sphere of radius 0.5 m . The particle moves in a horizontal circle whose centre is 0.3 m below the centre of the sphere (see diagram).\\
(i) Show that the force exerted by the sphere on the particle has magnitude 2 N .\\
(ii) Find the speed of the particle.\\
(iii) Find the time taken for the particle to complete one revolution.
\hfill \mbox{\textit{CAIE M2 2009 Q4 [7]}}