| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.8 This is a multi-step circular motion problem requiring geometric proof (showing the right angle using Pythagoras converse), then setting up force equations in both vertical and horizontal directions with two tensions, finding the radius of circular motion, and finally deriving an inequality for the period involving angular speed. It goes beyond routine M3 questions by requiring the inequality manipulation and combining geometry with dynamics, but the individual techniques are standard for this module. |
| Spec | 6.04a Centre of mass: gravitational effect6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((6a)^2 + (8a)^2 = (10a)^2\) | M1 | For squaring the sides and showing they fit Pythagoras' theorem or ratio of sides 3:4:5 or use the cosine rule |
| By Pythagoras (converse), \(APB = 90°\) | A1 (2) | For stating that the converse of Pythagoras' theorem shows that \(APB = 90°\) or appropriate conclusion for their method |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_1 \sin\alpha + T_2 \cos\alpha = mr\omega^2\) | M1 A2 | For NL2 horizontally. Two tensions both resolved, acceleration in form \(mr\omega^2\). A1 for any two correct terms; A1 for third correct term with different tensions |
| \(T_1 \cos\alpha - T_2 \sin\alpha = mg\) | M1 A1 | For resolving vertically. Two tensions both resolved; A1 for fully correct equation with different tensions |
| \(r = 8a\sin\alpha\) | M1 A1 | M1 for finding radius; A1 for \(r = 8a\sin\alpha\) (may not be shown explicitly) |
| \(\sin\alpha = \frac{3}{5}\) or \(\cos\alpha = \frac{4}{5}\) | B1 | For a correct value of \(\sin\alpha\) or \(\cos\alpha\) |
| Solving: \(T_2 = \frac{3m}{25}(32a\omega^2 - 5g)\) | M1 | Dependent on all M marks above and two different tensions. Or making \(T_2 = 0\) and solving for \(\omega\) |
| \(T_2 \geq 0 \Rightarrow \omega = \sqrt{\dfrac{5g}{32a}}\) | M1 A1 | M1dep for using \(T_2 \geq 0\) in their expression; A1 for \(\omega_{\min} = \sqrt{\dfrac{5g}{32a}}\) |
| max time \(= \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{32a}{5g}}\) | M1 A1 (13) | M1 for using \(\dfrac{2\pi}{\omega}\) with their \(\omega\); A1cso for max time \(= 2\pi\sqrt{\dfrac{32a}{5g}}\) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(6a)^2 + (8a)^2 = (10a)^2$ | M1 | For squaring the sides and showing they fit Pythagoras' theorem or ratio of sides 3:4:5 or use the cosine rule |
| By Pythagoras (converse), $APB = 90°$ | A1 (2) | For stating that the converse of Pythagoras' theorem shows that $APB = 90°$ or appropriate conclusion for their method |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_1 \sin\alpha + T_2 \cos\alpha = mr\omega^2$ | M1 A2 | For NL2 horizontally. Two tensions both resolved, acceleration in form $mr\omega^2$. A1 for any two correct terms; A1 for third correct term with different tensions |
| $T_1 \cos\alpha - T_2 \sin\alpha = mg$ | M1 A1 | For resolving vertically. Two tensions both resolved; A1 for fully correct equation with different tensions |
| $r = 8a\sin\alpha$ | M1 A1 | M1 for finding radius; A1 for $r = 8a\sin\alpha$ (may not be shown explicitly) |
| $\sin\alpha = \frac{3}{5}$ or $\cos\alpha = \frac{4}{5}$ | B1 | For a correct value of $\sin\alpha$ or $\cos\alpha$ |
| Solving: $T_2 = \frac{3m}{25}(32a\omega^2 - 5g)$ | M1 | Dependent on all M marks above and two different tensions. Or making $T_2 = 0$ and solving for $\omega$ |
| $T_2 \geq 0 \Rightarrow \omega = \sqrt{\dfrac{5g}{32a}}$ | M1 A1 | M1dep for using $T_2 \geq 0$ in their expression; A1 for $\omega_{\min} = \sqrt{\dfrac{5g}{32a}}$ |
| max time $= \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{32a}{5g}}$ | M1 A1 (13) | M1 for using $\dfrac{2\pi}{\omega}$ with their $\omega$; A1cso for max time $= 2\pi\sqrt{\dfrac{32a}{5g}}$ |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2c0bb9ea-31a6-42f1-9e2e-d792eee8fd10-09_1089_1072_278_466}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A light inextensible string of length $14 a$ has its ends attached to two fixed points $A$ and $B$, where $A$ is vertically above $B$ and $A B = 10 a$. A particle of mass $m$ is attached to the string at the point $P$, where $A P = 8 a$. The particle moves in a horizontal circle with constant angular speed $\omega$ and with both parts of the string taut, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that angle $A P B = 90 ^ { \circ }$.
\item Show that the time for the particle to make one complete revolution is less than
$$2 \pi \sqrt { \left( \frac { 32 a } { 5 g } \right) } .$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q6 [15]}}