| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Two strings/rods system |
| Difficulty | Standard +0.8 This is a challenging M3 circular motion problem requiring geometric reasoning to find angles, resolution of forces in two directions with two tensions, and inequality derivation from physical constraints (both strings taut). The multi-part structure with 'show that' proofs and the need to consider limiting cases for the inequalities elevates this above standard circular motion exercises, though it follows established M3 techniques. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use cosine rule on triangle \(APB\) OR trig on 'half' of triangle \(APB\) to find one relevant angle | M1 | Either complete method to obtain one relevant angle |
| Given answers correctly obtained | A1* | Correct given angles correctly obtained. Sufficient annotation/justification required. Stating \(\angle OBP=2\times\angle OAP\) alone is not sufficient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_A\cos30°+T_B\cos60°=mg\) | M1 A1 | Resolve vertically, dimensionally correct equation with correct no. of terms, condone sign errors and sin/cos confusion |
| \(T_A\sin30°+T_B\sin60°=mr\omega^2\) | M1 A1 A1 | Equation of motion horizontally, dimensionally correct, condone sign errors and sin/cos confusion. Correct equation with at most one error. If \(r\omega^2\) is never seen, this is an A error |
| \(r=a\sin60°\) (or \(r=a\sqrt{3}\cos30°\) or \(r=a\frac{\sqrt{3}}{2}\)) | B1 | Cao. If seen in (a) must be used in (b) for this mark |
| Solve for \(T_A\) | dM1 | Solve for \(T_A\) in terms of \(m\), \(a\), \(g\) and \(\omega\) |
| \(T_A=\frac{1}{2}m\sqrt{3}(2g-a\omega^2)\) | A1* | Given answer correctly obtained. Must see exactly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt to obtain one inequality on \(\omega^2\) | M1 | Correct use of either \(T_A>0\) or \(T_B>0\) to obtain one inequality on \(\omega^2\) |
| Correct inequality | A1 | Correct inequality |
| Attempt to obtain another inequality on \(\omega^2\) and use both to obtain answer | M1 | Use both \(T_A>0\) and \(T_B>0\) to form inequalities. Note \(T_B=\frac{3}{2}ma\omega^2-mg\) |
| \(\frac{2g}{3a}<\omega^2<\frac{2g}{a}\) | A1* | Given answer correctly obtained |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use cosine rule on triangle $APB$ OR trig on 'half' of triangle $APB$ to find one relevant angle | M1 | Either complete method to obtain one relevant angle |
| Given answers correctly obtained | A1* | Correct given angles correctly obtained. Sufficient annotation/justification required. Stating $\angle OBP=2\times\angle OAP$ alone is not sufficient |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_A\cos30°+T_B\cos60°=mg$ | M1 A1 | Resolve vertically, dimensionally correct equation with correct no. of terms, condone sign errors and sin/cos confusion |
| $T_A\sin30°+T_B\sin60°=mr\omega^2$ | M1 A1 A1 | Equation of motion horizontally, dimensionally correct, condone sign errors and sin/cos confusion. Correct equation with at most one error. If $r\omega^2$ is never seen, this is an A error |
| $r=a\sin60°$ (or $r=a\sqrt{3}\cos30°$ or $r=a\frac{\sqrt{3}}{2}$) | B1 | Cao. If seen in (a) must be used in (b) for this mark |
| Solve for $T_A$ | dM1 | Solve for $T_A$ in terms of $m$, $a$, $g$ and $\omega$ |
| $T_A=\frac{1}{2}m\sqrt{3}(2g-a\omega^2)$ | A1* | Given answer correctly obtained. Must see exactly |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt to obtain one inequality on $\omega^2$ | M1 | Correct use of either $T_A>0$ or $T_B>0$ to obtain one inequality on $\omega^2$ |
| Correct inequality | A1 | Correct inequality |
| Attempt to obtain another inequality on $\omega^2$ and use both to obtain answer | M1 | Use both $T_A>0$ and $T_B>0$ to form inequalities. Note $T_B=\frac{3}{2}ma\omega^2-mg$ |
| $\frac{2g}{3a}<\omega^2<\frac{2g}{a}$ | A1* | Given answer correctly obtained |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-12_535_674_283_699}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $\mathrm { a } \sqrt { 3 }$. The other end of the string is attached to a fixed point A . The particle P is also attached to one end of a second light inextensible string of length a. The other end of this string is attached to a fixed point B , where B is vertically below A , with $\mathrm { AB } = \mathrm { a }$.
The particle $P$ moves in a horizontal circle with centre 0 , where 0 is vertically below $B$.\\
The particle P moves with constant angular speed $\omega$, with both strings taut, as shown in Figure 5.
\begin{enumerate}[label=(\alph*)]
\item Show that the upper string makes an angle of $30 ^ { \circ }$ with the downward vertical and the lower string makes an angle of $60 ^ { \circ }$ with the downward vertical.
\item Show that the tension in the upper string is $\frac { 1 } { 2 } m \sqrt { 3 } \left( 2 g - a \omega ^ { 2 } \right)$.
\item Show that $\frac { 2 g } { 3 a } < \omega ^ { 2 } < \frac { 2 g } { a }$
\begin{center}
\end{center}
$\_\_\_\_$ VIAV SIHI NI JIIHM ION OC\\
VILU SIHIL NI GLIUM ION OC\\
VEYV SIHI NI III HM ION OC
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2023 Q5 [14]}}