| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Two strings/rods system |
| Difficulty | Standard +0.8 This is a standard M3 circular motion problem with two strings requiring resolution of forces in horizontal and vertical directions, followed by a constraint deduction. While it involves multiple steps (geometry, force resolution, two tensions, inequality), the techniques are routine for M3 students and the 'show that' format provides guidance. The geometry is straightforward (3-4-5 triangle), making this above average but not exceptionally challenging. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(\cos\theta = \frac{4}{5}\) or \(\sin\theta = \frac{3}{5}\) | B1 | |
| R(vert): \(T_B\cos 45 + mg = T_A\cos\theta\) leading to \(\frac{1}{\sqrt{2}}T_B + mg = \frac{4}{5}T_A\) | M1 A1 | |
| R(horiz): \(T_A\sin\theta + T_B\cos 45 = m \times 3a\omega^2\) leading to \(\frac{3}{5}T_A + \frac{1}{\sqrt{2}}T_B = 3ma\omega^2\) | M1 A1=A1 | |
| Eliminating \(T_B\): \(\frac{3}{5}T_A - mg = 3ma\omega^2 - \frac{4}{5}T_a\) | M1 | |
| \(\frac{7}{5}T_A = 3ma\omega^2 + mg\) | ||
| \(T_A = \frac{5}{7}m(3a\omega^2 + g)\) | A1 | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(T_b = \sqrt{2}\left(\frac{4}{5}T_a - mg\right)\) | M1 | |
| \(= \sqrt{2}\left(\frac{4}{7}m(3a\omega^2+g)-mg\right)\) | ||
| \(= \frac{3\sqrt{2}}{7}m(4a\omega^2-g)\) | A1 | oe, (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(T_b \geqslant 0 \Rightarrow 4a\omega^2 \geqslant g\) | M1 | |
| \(\omega^2 \geqslant \frac{g}{4a}\) | ||
| \(\omega \geqslant \frac{1}{2}\sqrt{\frac{g}{a}}\) | A1 | Allow strict inequalities. (2) |
# Question 4:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| $\cos\theta = \frac{4}{5}$ or $\sin\theta = \frac{3}{5}$ | B1 | |
| R(vert): $T_B\cos 45 + mg = T_A\cos\theta$ leading to $\frac{1}{\sqrt{2}}T_B + mg = \frac{4}{5}T_A$ | M1 A1 | |
| R(horiz): $T_A\sin\theta + T_B\cos 45 = m \times 3a\omega^2$ leading to $\frac{3}{5}T_A + \frac{1}{\sqrt{2}}T_B = 3ma\omega^2$ | M1 A1=A1 | |
| Eliminating $T_B$: $\frac{3}{5}T_A - mg = 3ma\omega^2 - \frac{4}{5}T_a$ | M1 | |
| $\frac{7}{5}T_A = 3ma\omega^2 + mg$ | | |
| $T_A = \frac{5}{7}m(3a\omega^2 + g)$ | A1 | (8) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| $T_b = \sqrt{2}\left(\frac{4}{5}T_a - mg\right)$ | M1 | |
| $= \sqrt{2}\left(\frac{4}{7}m(3a\omega^2+g)-mg\right)$ | | |
| $= \frac{3\sqrt{2}}{7}m(4a\omega^2-g)$ | A1 | oe, (2) |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| $T_b \geqslant 0 \Rightarrow 4a\omega^2 \geqslant g$ | M1 | |
| $\omega^2 \geqslant \frac{g}{4a}$ | | |
| $\omega \geqslant \frac{1}{2}\sqrt{\frac{g}{a}}$ | A1 | Allow strict inequalities. (2) |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{826ad8ff-6e5c-4224-88ba-e78b79d1bc21-07_805_460_214_740}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A light inextensible string has its ends attached to two fixed points $A$ and $B$. The point $A$ is vertically above $B$ and $A B = 7 a$. A particle $P$ of mass $m$ is fixed to the string and moves in a horizontal circle of radius $3 a$ with angular speed $\omega$. The centre of the circle is $C$ where $C$ lies on $A B$ and $A C = 4 a$, as shown in Figure 4. Both parts of the string are taut.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in $A P$ is $\frac { 5 } { 7 } m \left( 3 a \omega ^ { 2 } + g \right)$.
\item Find the tension in $B P$.
\item Deduce that $\omega \geqslant \frac { 1 } { 2 } \sqrt { } \left( \frac { g } { a } \right)$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2011 Q4 [12]}}