| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.8 This M3 circular motion problem requires resolving forces in two directions for a particle on two strings, applying Newton's second law for circular motion, and proving an inequality involving the period. It demands careful geometric reasoning with the 30° angles, systematic force resolution, and algebraic manipulation to reach the given expressions. While methodical, it's more demanding than standard single-string conical pendulum questions due to the two-string configuration and the inequality proof in part (b). |
| Spec | 3.03d Newton's second law: 2D vectors6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_A\cos 30° = mg + T_B\cos 30°\) | M1, A1 | Attempt vertical equation (can have \(\theta\) for angle); completely correct equation with numerical angle |
| \(T_A - T_B = \frac{2mg}{\sqrt{3}}\) | — | |
| Radius \(= \frac{1}{2}l\tan 30° \left(= \frac{\sqrt{3}}{6}l\ \text{oe}\right)\) | B1 | Correct radius seen anywhere |
| \(T_A\cos 60° + T_B\cos 60° = mr\omega^2 = m\left(\frac{1}{2}l\tan 30°\right)\omega^2\) | M1, A1, A1ft | NL2 along radius; correct sum of tensions; correct mass × acceleration ft their radius |
| \(T_A + T_B = \frac{ml\omega^2}{\sqrt{3}}\) | — | |
| (i) \(T_A = \frac{1}{2}\left(\frac{2mg}{\sqrt{3}}+\frac{ml\omega^2}{\sqrt{3}}\right) = \frac{m\sqrt{3}}{6}(2g+l\omega^2)\) | DM1, A1cso | Solve equations to \(T_A=\ldots\); correct given answer (no equivalents) |
| (ii) \(T_B = \frac{1}{2}\left(\frac{ml\omega^2}{\sqrt{3}}-\frac{2mg}{\sqrt{3}}\right)\) oe | A1cso (9) | Correct expression for \(T_B\); any equivalent 2-term expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_B > 0 \Rightarrow 2mg < ml\omega^2\) | M1 | Deducing inequality from expression for \(T_B\); can have \(2(m)g < (m)l\omega^2\) or \(\leq\) or \(=\) |
| \(\omega^2 > \frac{2g}{l}\) | A1 | \(\omega^2 > \frac{2g}{l}\) or \(\geq\), oe inc equivalent in words |
| \(T = \frac{2\pi}{\omega}\); \(T < 2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}\) | DM1, A1cso (4) | Use \(T=\frac{2\pi}{\omega}\) with their \(\omega\) to form inequality for \(T\); correct final statement. Must be \(T<\ldots\) or equivalent in words |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_A\cos 30° = mg + T_B\cos 30°$ | M1, A1 | Attempt vertical equation (can have $\theta$ for angle); completely correct equation with numerical angle |
| $T_A - T_B = \frac{2mg}{\sqrt{3}}$ | — | |
| Radius $= \frac{1}{2}l\tan 30° \left(= \frac{\sqrt{3}}{6}l\ \text{oe}\right)$ | B1 | Correct radius seen anywhere |
| $T_A\cos 60° + T_B\cos 60° = mr\omega^2 = m\left(\frac{1}{2}l\tan 30°\right)\omega^2$ | M1, A1, A1ft | NL2 along radius; correct sum of tensions; correct mass × acceleration ft their radius |
| $T_A + T_B = \frac{ml\omega^2}{\sqrt{3}}$ | — | |
| **(i)** $T_A = \frac{1}{2}\left(\frac{2mg}{\sqrt{3}}+\frac{ml\omega^2}{\sqrt{3}}\right) = \frac{m\sqrt{3}}{6}(2g+l\omega^2)$ | DM1, A1cso | Solve equations to $T_A=\ldots$; correct given answer (no equivalents) |
| **(ii)** $T_B = \frac{1}{2}\left(\frac{ml\omega^2}{\sqrt{3}}-\frac{2mg}{\sqrt{3}}\right)$ oe | A1cso (9) | Correct expression for $T_B$; any equivalent 2-term expression |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_B > 0 \Rightarrow 2mg < ml\omega^2$ | M1 | Deducing inequality from expression for $T_B$; can have $2(m)g < (m)l\omega^2$ or $\leq$ or $=$ |
| $\omega^2 > \frac{2g}{l}$ | A1 | $\omega^2 > \frac{2g}{l}$ or $\geq$, oe inc equivalent in words |
| $T = \frac{2\pi}{\omega}$; $T < 2\pi\sqrt{\frac{l}{2g}} = \pi\sqrt{\frac{2l}{g}}$ | DM1, A1cso (4) | Use $T=\frac{2\pi}{\omega}$ with their $\omega$ to form inequality for $T$; correct final statement. Must be $T<\ldots$ or equivalent in words |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c1c51ff-6ae8-402d-b303-b656d26e4230-07_842_449_248_826}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to the ends of two light inextensible strings. The other ends of the strings are attached to fixed points $A$ and $B$, where $B$ is vertically below $A$ and $A B = l$. The particle is moving with constant angular speed $\omega$ in a horizontal circle. Both strings are taut and inclined at $30 ^ { \circ }$ to $A B$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the tension in $A P$ is $\frac { m \sqrt { 3 } } { 6 } \left( 2 g + l \omega ^ { 2 } \right)$
\item Find the tension in $B P$.
\end{enumerate}\item Show that the time taken by $P$ to complete one revolution is less than $\pi \sqrt { \frac { 2 l } { g } }$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2016 Q5 [13]}}