| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.5 This is a standard conical pendulum problem requiring resolution of forces and application of circular motion equations. While it involves two parts and requires careful angle work, the setup is straightforward with given values, and the solution follows a well-practiced method (resolve vertically for tension, then horizontally for radius/length). Slightly easier than average due to its routine nature. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (a) resolve \(\uparrow\): \(T \cos 60° - mg = 0\) | M1 | |
| \(\therefore \frac{1}{2}T = 0.1 \times 9.8\) | A1 | |
| giving \(T = 1.96 \text{ N}\) | A1 | |
| (b) resolve \(\leftarrow\): \(T \sin 60° = mr\omega^2\) | M1 A1 | |
| let length of string \(= l\) \(\therefore r = l \sin 60°\) so \(1.96 = 0.1 \times l \times 4^2\) | M1 A1 | |
| giving \(l = 1.225 \text{ m}\) | A1 | (8) |
(a) resolve $\uparrow$: $T \cos 60° - mg = 0$ | M1 |
$\therefore \frac{1}{2}T = 0.1 \times 9.8$ | A1 |
giving $T = 1.96 \text{ N}$ | A1 |
(b) resolve $\leftarrow$: $T \sin 60° = mr\omega^2$ | M1 A1 |
let length of string $= l$ $\therefore r = l \sin 60°$ so $1.96 = 0.1 \times l \times 4^2$ | M1 A1 |
giving $l = 1.225 \text{ m}$ | A1 | (8)3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{00776cc0-0214-4029-8ef1-c1cba89f4b87-2_382_796_1640_479}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A popular racket game involves a tennis ball of mass 0.1 kg which is attached to one end of a light inextensible string. The other end of the string is attached to the top of a fixed rigid pole.
A boy strikes the ball such that it moves in a horizontal circle with angular speed $4 \mathrm { rad } \mathrm { s } ^ { - 1 }$ and the string makes an angle of $60 ^ { \circ }$ with the downward vertical as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string.
\item Find the length of the string.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q3 [8]}}