| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.8 This is a standard M3 circular motion problem requiring resolution of forces in two directions, use of Pythagoras to find string lengths, and application of the condition for both strings to remain taut. While it involves multiple steps and careful geometry, the techniques are routine for this module with no novel insight required beyond standard circular motion principles. |
| Spec | 3.03d Newton's second law: 2D vectors3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(T_a\sin\theta = mg + T_b\sin\phi\) | M1 | Resolve vertically; both tensions to be resolved. |
| \(\frac{3}{5}T_a = T_b\sin\phi + mg\) | A1A1 | Both tension terms correct; \(\theta, \phi\) allowed. Completely correct equation. |
| \(T_a\cos\theta + T_b\cos\phi = m\times 4a\omega^2\) | M1 | NL2 along radius; both tensions resolved. |
| \(\frac{4}{5}T_a + T_b\cos\phi = 4ma\omega^2\) | A1A1 | Both terms correct; completely correct with \(\cos\theta = \frac{4}{5}\); acceleration \(4a\omega^2\). |
| \(\frac{3}{5}T_a + \frac{4}{5}T_a = mg + 4ma\omega^2\) (using \(\cos\phi = \sin\phi\)) | dM1 | Eliminate \(T_b\); depends on both M marks; use of \(\cos\phi = \sin\phi\). |
| \(T_a = \frac{5}{7}m(4a\omega^2 + g)\) | A1cso (8) | Given result from correct working. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{\sqrt{2}}T_b = 4ma\omega^2 - \frac{4}{7}m(4a\omega^2 + g)\) | M1 | Obtain \(T_b\) by valid means; must use given \(T_a\) expression. |
| \(T_b = \sqrt{2}\left(\frac{12}{7}ma\omega^2 - \frac{4}{7}mg\right)\) or \(\frac{4\sqrt{2}}{7}m(3a\omega^2 - g)\) | A1cao (2) | Correct expression in any equivalent form. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(T_b \geq 0 \Rightarrow 3a\omega^2 \geq g\) | M1 | Set \(T_b \geq 0\); deduce inequality connecting \(\omega^2\) (or \(\omega\)) and \(g\). |
| \(\omega \geq \sqrt{\frac{g}{3a}}, \quad k=3\) | A1 (2) [12] | Complete to required form with correct \(k\); need not be shown explicitly. |
# Question 4:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $T_a\sin\theta = mg + T_b\sin\phi$ | M1 | Resolve vertically; both tensions to be resolved. |
| $\frac{3}{5}T_a = T_b\sin\phi + mg$ | A1A1 | Both tension terms correct; $\theta, \phi$ allowed. Completely correct equation. |
| $T_a\cos\theta + T_b\cos\phi = m\times 4a\omega^2$ | M1 | NL2 along radius; both tensions resolved. |
| $\frac{4}{5}T_a + T_b\cos\phi = 4ma\omega^2$ | A1A1 | Both terms correct; completely correct with $\cos\theta = \frac{4}{5}$; acceleration $4a\omega^2$. |
| $\frac{3}{5}T_a + \frac{4}{5}T_a = mg + 4ma\omega^2$ (using $\cos\phi = \sin\phi$) | dM1 | Eliminate $T_b$; depends on both M marks; use of $\cos\phi = \sin\phi$. |
| $T_a = \frac{5}{7}m(4a\omega^2 + g)$ | A1cso (8) | Given result from correct working. |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{\sqrt{2}}T_b = 4ma\omega^2 - \frac{4}{7}m(4a\omega^2 + g)$ | M1 | Obtain $T_b$ by valid means; must use given $T_a$ expression. |
| $T_b = \sqrt{2}\left(\frac{12}{7}ma\omega^2 - \frac{4}{7}mg\right)$ or $\frac{4\sqrt{2}}{7}m(3a\omega^2 - g)$ | A1cao (2) | Correct expression in any equivalent form. |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $T_b \geq 0 \Rightarrow 3a\omega^2 \geq g$ | M1 | Set $T_b \geq 0$; deduce inequality connecting $\omega^2$ (or $\omega$) and $g$. |
| $\omega \geq \sqrt{\frac{g}{3a}}, \quad k=3$ | A1 (2) [12] | Complete to required form with correct $k$; need not be shown explicitly. |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{85d8fc7d-8863-419e-8eef-8751a6fb6315-05_654_515_267_712}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\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 with constant angular speed $\omega$ in a horizontal circle of radius $4 a$. The centre of the circle is $C$, where $C$ lies on $A B$ and $A C = 3 a$, as shown in Figure 3. Both parts of the string are taut.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in $A P$ is $\frac { 5 } { 7 } m \left( 4 a \omega ^ { 2 } + g \right)$.
\item Find the tension in $B P$.
\item Deduce that $\omega \geqslant \sqrt { \frac { g } { k a } }$, stating the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q4 [12]}}