| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.3 This is a straightforward conical pendulum problem requiring resolution of forces in two directions (vertical equilibrium and horizontal centripetal force) with standard circular motion formulas. The setup is clearly defined with given values, making it slightly easier than average but still requiring proper application of mechanics principles. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resolving forces vertically (3 term equation) | M1 | |
| \(T\cos 60° + 0.5 \times 10 = 8\) | A1 | |
| Tension is 6 N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Radius of circle is \(9\sin 60°\) \((= 7.7942)\) | B1 | |
| Using Newton's second law horizontally with \(a = \dfrac{v^2}{r}\) | M1 | |
| \(6\sin 60° = 0.5\dfrac{v^2}{(9\sin 60°)}\) | A1 ft | |
| Alternative: Using Newton's second law perpendicular to string with \(a = \dfrac{v^2}{r}\) | M1 | |
| \((8 - 0.5 \times 10)\sin 60° = 0.5\dfrac{v^2}{(9\sin 60°)}\cos 60°\) | A1 ft | |
| Speed is 9 ms\(^{-1}\) | A1 | NB: Use of \(mr\omega^2\), M1 withheld until \(v = r\omega\) is used |
# Question 5:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Resolving forces vertically (3 term equation) | M1 | |
| $T\cos 60° + 0.5 \times 10 = 8$ | A1 | |
| Tension is 6 N | A1 | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Radius of circle is $9\sin 60°$ $(= 7.7942)$ | B1 | |
| Using Newton's second law horizontally with $a = \dfrac{v^2}{r}$ | M1 | |
| $6\sin 60° = 0.5\dfrac{v^2}{(9\sin 60°)}$ | A1 ft | |
| **Alternative:** Using Newton's second law perpendicular to string with $a = \dfrac{v^2}{r}$ | M1 | |
| $(8 - 0.5 \times 10)\sin 60° = 0.5\dfrac{v^2}{(9\sin 60°)}\cos 60°$ | A1 ft | |
| Speed is 9 ms$^{-1}$ | A1 | NB: Use of $mr\omega^2$, M1 withheld until $v = r\omega$ is used |
---5\\
\begin{tikzpicture}[>=Stealth]
% Define coordinates
\coordinate (O) at (0,0);
\coordinate (top) at (0,5);
\coordinate (aircraft) at (4.5,3.5);
\coordinate (center) at (0,3.5);
% Dashed vertical line upward from O
\draw[dashed] (O) -- (center);
% String from O to aircraft
\draw (O) -- (aircraft);
% Horizontal dashed circle (ellipse for perspective)
% The aircraft moves in a horizontal circle at the height of the aircraft
% Draw the ellipse centered on the vertical line at the aircraft's height
% Draw back part of ellipse (dashed)
\draw[dashed] (center) ellipse (4.5 and .8);
% Draw front part of ellipse (solid) - over the top
% Filled circle for aircraft
\fill (aircraft) circle (3pt);
% Label for string length: 9 m
\node[right] at (2.8,1.8) {$9\,\mathrm{m}$};
% Angle arc for 60 degrees from upward vertical
\draw (0,1.2) arc (90:37:1.2);
\node[above right] at (0.3,0.6) {$60^\circ$};
% Label O
\node[below left] at (O) {$O$};
\end{tikzpicture}
A toy aircraft of mass 0.5 kg is attached to one end of a light inextensible string of length 9 m . The other end of the string is attached to a fixed point $O$. The aircraft moves with constant speed in a horizontal circle. The string is taut, and makes an angle of $60 ^ { \circ }$ with the upward vertical at $O$ (see diagram). In a simplified model of the motion, the aircraft is treated as a particle and the force of the air on the aircraft is taken to act vertically upwards with magnitude 8 N . Find\\
(i) the tension in the string,\\
(ii) the speed of the aircraft.
\hfill \mbox{\textit{CAIE M2 2003 Q5 [7]}}