| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem with a particle on a string. Part (a) uses Pythagoras' theorem (2a, a, a√3 forms a right triangle), part (b) is basic trigonometry, parts (c) requires resolving forces and applying F=mrω² which is routine for M3, and part (d) involves finding the condition for the string to remain taut. All steps follow standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (a) string taut \(\therefore PR = a\), \(PR^2 + QR^2 = a^2 + 3a^2 = 4a^2 = PQ^2\) by converse of Pythag. \(\angle PRQ = 90°\) | M1 A1 | |
| (b) \(\sin \angle PQR = \frac{a}{2a} = \frac{1}{2} \therefore \angle PQR = 30°\) | B1 | |
| (c) (i) resolve \(\uparrow\): \(T_1 \sin 60 - mg = 0\) | M1 A1 | |
| \(\therefore T_1 = \frac{2mg}{\sqrt{3}}\) (or \(\frac{2}{3}\sqrt{3}mg\)) | A1 | |
| (ii) resolve \(\leftarrow\): \(T_2 + T_1 \cos 60 = \frac{mv^2}{r}\) | M1 A1 | |
| \(\therefore T_2 = \frac{mv^2}{a} - \frac{1}{2} \times \frac{2mg}{\sqrt{3}} = \frac{mv^2}{a} - \frac{mg}{\sqrt{3}}\) (or \(\frac{mv^2}{a} - \frac{1}{3}\sqrt{3}mg\)) | M1 A1 | |
| (d) PR taut \(\therefore T_2 \geq 0\) | M1 | |
| giving \(\frac{mv^2}{a} \geq \frac{mg}{\sqrt{3}}\) so \(u^2 \geq \frac{a g}{\sqrt{3}}\) | M1 A1 | (13) |
(a) string taut $\therefore PR = a$, $PR^2 + QR^2 = a^2 + 3a^2 = 4a^2 = PQ^2$ by converse of Pythag. $\angle PRQ = 90°$ | M1 A1
(b) $\sin \angle PQR = \frac{a}{2a} = \frac{1}{2} \therefore \angle PQR = 30°$ | B1
(c) (i) resolve $\uparrow$: $T_1 \sin 60 - mg = 0$ | M1 A1
$\therefore T_1 = \frac{2mg}{\sqrt{3}}$ (or $\frac{2}{3}\sqrt{3}mg$) | A1
(ii) resolve $\leftarrow$: $T_2 + T_1 \cos 60 = \frac{mv^2}{r}$ | M1 A1
$\therefore T_2 = \frac{mv^2}{a} - \frac{1}{2} \times \frac{2mg}{\sqrt{3}} = \frac{mv^2}{a} - \frac{mg}{\sqrt{3}}$ (or $\frac{mv^2}{a} - \frac{1}{3}\sqrt{3}mg$) | M1 A1
(d) PR taut $\therefore T_2 \geq 0$ | M1
giving $\frac{mv^2}{a} \geq \frac{mg}{\sqrt{3}}$ so $u^2 \geq \frac{a g}{\sqrt{3}}$ | M1 A1 | (13)
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8b85b908-bb74-4532-a1b4-3826946bd43b-4_437_364_196_717}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
The two ends of a light inextensible string of length $3 a$ are attached to fixed points $Q$ and $R$ which are a distance of $a \sqrt { } 3$ apart with $R$ vertically below $Q$. A particle $P$ of mass $m$ is attached to the string at a distance of $2 a$ from $Q$.\\
$P$ is given a horizontal speed, $u$, such that it moves in a horizontal circle with both sections of the string taut as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that $\angle P R Q$ is a right angle.
\item Find $\angle P Q R$ in degrees.
\item Find, in terms of $a , g , m$ and $u$, the tension in the section of string
\begin{enumerate}[label=(\roman*)]
\item $P Q$,
\item $P R$.
\end{enumerate}\item Show that $u ^ { 2 } \geq \frac { g a } { \sqrt { 3 } }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [13]}}