| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.8 This is a non-trivial circular motion problem requiring careful force resolution in both vertical and horizontal directions, geometric reasoning to find the radius, and algebraic manipulation with the given tension relationship. It goes beyond standard single-string conical pendulum questions but uses established M3 techniques without requiring exceptional insight. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2T\cos 30 = T\cos 30 + mg\) | M1A1 | M1 resolve vertically - both tensions resolved; A1 fully correct equation |
| \(\frac{T\sqrt{3}}{2} = mg\) | A1 | Substitute for trig function |
| \(3T\cos 60 = mr\omega^2\) | M1A1 | M1 NL2 horizontally - tensions resolved, acceleration in either form; A1 correct equation, \(r\) still present, acceleration \(r\omega^2\) |
| \(\frac{3}{2} \times \frac{2mg}{\sqrt{3}} = mr\omega^2\) | A1 | Correct equation with no trig function |
| \(AB = 4a\), \(\frac{r}{2a} = \tan 30 = \frac{1}{\sqrt{3}}\), \(r = \frac{2a}{\sqrt{3}}\) | ||
| \(\frac{3g}{\sqrt{3}} = \frac{2a}{\sqrt{3}}\omega^2\) | DM1 | Dep on both prev M marks; eliminate \(r\) and \(T\) to obtain equation with \(\omega\), \(a\), \(g\) only; \(r\) need not be correct but do not allow \(r=a\) |
| \(\omega^2 = \frac{3g}{2a}\), \(\omega = \sqrt{\frac{3g}{2a}}\) | A1 (8) | Correct result |
# Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2T\cos 30 = T\cos 30 + mg$ | M1A1 | M1 resolve vertically - both tensions resolved; A1 fully correct equation |
| $\frac{T\sqrt{3}}{2} = mg$ | A1 | Substitute for trig function |
| $3T\cos 60 = mr\omega^2$ | M1A1 | M1 NL2 horizontally - tensions resolved, acceleration in either form; A1 correct equation, $r$ still present, acceleration $r\omega^2$ |
| $\frac{3}{2} \times \frac{2mg}{\sqrt{3}} = mr\omega^2$ | A1 | Correct equation with no trig function |
| $AB = 4a$, $\frac{r}{2a} = \tan 30 = \frac{1}{\sqrt{3}}$, $r = \frac{2a}{\sqrt{3}}$ | | |
| $\frac{3g}{\sqrt{3}} = \frac{2a}{\sqrt{3}}\omega^2$ | DM1 | Dep on both prev M marks; eliminate $r$ and $T$ to obtain equation with $\omega$, $a$, $g$ only; $r$ need not be correct but do not allow $r=a$ |
| $\omega^2 = \frac{3g}{2a}$, $\omega = \sqrt{\frac{3g}{2a}}$ | A1 (8) | Correct result |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3706a02d-95c6-4e7a-bf38-88b338d77892-05_828_624_264_676}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A light inextensible string has one end attached to a fixed point $A$ and the other end attached to a particle $P$ of mass $m$. An identical string has one end attached to the fixed point $B$, where $B$ is vertically below $A$ and $A B = 4 a$, and the other end attached to $P$, as shown in Figure 2. The particle is moving in a horizontal circle with constant angular speed $\omega$, with both strings taut and inclined at $30 ^ { \circ }$ to the vertical. The tension in the upper string is twice the tension in the lower string.
Find $\omega$ in terms of $a$ and $g$.
\hfill \mbox{\textit{Edexcel M3 2015 Q3 [8]}}