| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a straightforward circular motion problem requiring students to apply Newton's second law with tension and normal reaction forces, then find the limiting case when the particle loses contact with the floor (R=0). The geometry is given (forming a 60° angle), and the method is standard M3 content with no novel insight required—slightly easier than average. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Any correct sin or cos ratio stated | B1 | May be seen explicitly or used in equation. |
| \(T\cos 60° + N = mg\) | M1A1 | M1: Attempt vertical resolution, 3 terms needed. A1: Correct equation. |
| \(T\sin 60° = mr\omega^2 = m\omega^2 \times 2l\sin 60°\) | M1A1 | M1: Attempt NL2 along radius, acceleration in \(r\omega^2\) form (not \('a'\)). May have \(r\) and \(v\). A1: Fully correct with acceleration in \(r\omega^2\) form, radius in terms of \(l\). |
| \(\Rightarrow ml\omega^2 + N = mg\) | DM1 | Eliminate \(T\). Depends on both M marks. Must still involve \(N\). |
| \(N \geq 0 \Rightarrow l\omega^2 \leq g\) | DM1 | Use \(N \geq 0\). Depends on all 3 M marks. Must see correct inequality, not \(N=0\) or \(N>0\). |
| \(\omega \leq \sqrt{\dfrac{g}{l}}\) * | A1* (8) | Reach given result from fully correct working. |
# Question 2:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Any correct sin or cos ratio stated | B1 | May be seen explicitly or used in equation. |
| $T\cos 60° + N = mg$ | M1A1 | M1: Attempt vertical resolution, 3 terms needed. A1: Correct equation. |
| $T\sin 60° = mr\omega^2 = m\omega^2 \times 2l\sin 60°$ | M1A1 | M1: Attempt NL2 along radius, acceleration in $r\omega^2$ form (not $'a'$). May have $r$ and $v$. A1: Fully correct with acceleration in $r\omega^2$ form, radius in terms of $l$. |
| $\Rightarrow ml\omega^2 + N = mg$ | DM1 | Eliminate $T$. Depends on both M marks. Must still involve $N$. |
| $N \geq 0 \Rightarrow l\omega^2 \leq g$ | DM1 | Use $N \geq 0$. Depends on all 3 M marks. Must see correct inequality, not $N=0$ or $N>0$. |
| $\omega \leq \sqrt{\dfrac{g}{l}}$ * | A1* (8) | Reach given result from fully correct working. |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bd1e2b07-4a87-49d6-addd-c9f67467ef2f-04_351_993_246_536}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $2 l$. The other end of the string is attached to a fixed point $A$ above a smooth horizontal floor. The particle moves in a horizontal circle on the floor with the string taut. The centre $O$ of the circle is vertically below $A$ with $O A = l$, as shown in Figure 2 .
The particle moves with constant angular speed $\omega$ and remains in contact with the floor.\\
Show that
$$\omega \leqslant \sqrt { \frac { g } { l } }$$
\hfill \mbox{\textit{Edexcel M3 2022 Q2 [8]}}