| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | String through hole – hanging particle in equilibrium below table |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem with coupled particles through a string. It requires setting up force equations for both particles (tension balance for Q, resolving forces and using circular motion equation for P), then using geometry to relate the constraint that the string length is 6l. The problem is methodical rather than insightful—students follow a well-practiced procedure of drawing force diagrams, applying F=ma and F=mrω², and solving simultaneous equations. Slightly easier than average because it's a 'show that' question with given answers and follows a standard template for this topic. |
| Spec | 3.03d Newton's second law: 2D vectors6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
(i)
B1 for using Q to state that $T_Q = 2mg$ or $T_Q = 2mg$ with $T_Q = T_P$ seen or implied later.
M1 for attempting to resolve vertically for P. $T$ must be resolved but sin/cos interchange or omission of $g$ are accuracy errors.
A1cso for combining the two equations to obtain $\theta = 60°$
Note: This is a "show" question, so if no expression is seen for $T$ and just $2mg\cos\theta = mg$ shown, award 0/3 as this equation could have been produced from the required result, so insufficient working.
(ii)
M1 for attempting Newton's second law for P along the radius. The mass used must be $m$ if the particle is not stated to be P; a mass of $2m$ would imply use of Q. $T$ must be resolved. Acceleration can be in either form.
A1 for $T\sin\theta = mr\omega^2$ or $T\sin\theta = mr\omega^2$
M1 dep for eliminating $T$ between the two equations for P and substituting for $r$ in terms of $l$ and $\theta$. Dependent on the second but not the first M mark.
A1 for $2mg\sin\theta = m \times 5l\sin\theta \times \omega^2$ or $\frac{T\sin\theta}{T\cos\theta} = \tan\theta = 5l\sin\theta\left(\frac{\omega^2}{g}\right)$ or $\theta$ or $60°$
A1cso for re-arranging to obtain $\omega = \sqrt{\frac{2g}{5l}}$. Ensure the square root is correctly placed.
**Alternatives:**
M1A1 for $T = m \times 5l\omega^2$
M1A1 for $2mg = 5ml\omega^2$
A1cso as above
**Vector Triangle method:**
Triangle must be seen
B1 $T = 2mg$
M1 $\cos\theta = \frac{mg}{2mg}$
A1 $\theta = 60°$
M1A1 Correct triangle
M1A1 $\sin\theta = \frac{5ml\sin\theta\omega^2}{2mg}$
A1cso $\omega = \ldots$ as above
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f6ab162c-8473-4464-ad62-87a359d85ab3-04_707_1006_258_427}
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\caption{Figure 2}
\end{center}
\end{figure}
Two particles $P$ and $Q$, of mass $m$ and $2 m$ respectively, are attached to the ends of a light inextensible string of length 6l. The string passes through a small smooth fixed ring at the point $A$. The particle $Q$ is hanging freely at a distance $l$ vertically below $A$. The particle $P$ is moving in a horizontal circle with constant angular speed $\omega$. The centre $O$ of the circle is vertically below $A$. The particle $Q$ does not move and $A P$ makes a constant angle $\theta$ with the downward vertical, as shown in Figure 2.
Show that\\
(i) $\theta = 60 ^ { \circ }$\\
(ii) $\omega = \sqrt { } \left( \frac { 2 g } { 5 l } \right)$
\hfill \mbox{\textit{Edexcel M3 2013 Q3 [8]}}