| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Maximum/minimum tension or reaction |
| Difficulty | Standard +0.8 This is a multi-part circular motion problem requiring energy conservation and force analysis at multiple points. Part (a) requires resolving forces at release (straightforward). Part (b) requires identifying that maximum tension occurs at the bottom and minimum at the top, using energy conservation to find speeds at both points, then applying F=mv²/r+mg·cos(θ) and showing the ratio equals 4. The algebraic manipulation and conceptual understanding of when max/min tensions occur elevates this above a routine M3 question, but it follows a standard framework for this topic. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Initially \(T_{IN} = mg\cos 60\) | M1 | Equation of motion towards centre at \(A\). Must have \(v = 0\). Weight must be resolved and tension not resolved. Allow with \(\cos\) or \(\sin\) of \(60°\) or \(30°\). |
| \(T_{IN} = \dfrac{mg}{2}\) | A1 | Correct tension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{1}{2}mv^2 - mg(l) = -mg(l\sin 30)\) | M1A1A1 | Energy equation from \(A\) to the lowest point. One KE and a difference in GPE required. Correct equation, \(-1\) for each error. |
| \(T_{FI} - mg = m\dfrac{v^2}{r}\) | M1A1 | Equation of motion at the lowest point, with acceleration in either form. Tension and weight needed. Acceleration must be in correct form. |
| \(T_{FI} = m\left(\dfrac{gl}{l}\right) + mg\) | DM1 | Solve to find tension at lowest point. Must reach \(T = \ldots\) but need not be simplified. Depends on both M marks in (b). |
| \(T_{FI} = 2mg = 4T_{IN}\) ✱ | A1✱ | Achieve the given result, from fully correct working. c.s.o. |
## Question 4:
### Part 4a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Initially $T_{IN} = mg\cos 60$ | M1 | Equation of motion towards centre at $A$. Must have $v = 0$. Weight must be resolved and tension not resolved. Allow with $\cos$ or $\sin$ of $60°$ or $30°$. |
| $T_{IN} = \dfrac{mg}{2}$ | A1 | Correct tension |
**Total: (2)**
### Part 4b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{1}{2}mv^2 - mg(l) = -mg(l\sin 30)$ | M1A1A1 | Energy equation from $A$ to the lowest point. One KE and a difference in GPE required. Correct equation, $-1$ for each error. |
| $T_{FI} - mg = m\dfrac{v^2}{r}$ | M1A1 | Equation of motion at the lowest point, with acceleration in either form. Tension and weight needed. Acceleration must be in correct form. |
| $T_{FI} = m\left(\dfrac{gl}{l}\right) + mg$ | DM1 | Solve to find tension at lowest point. Must reach $T = \ldots$ but need not be simplified. Depends on both M marks in (b). |
| $T_{FI} = 2mg = 4T_{IN}$ ✱ | A1✱ | Achieve the **given** result, from fully correct working. c.s.o. |
**Total: (7)**
**NB:** If equations in (b) found at a general point, mark as above. Final M mark will require some evidence of maximising the tension.
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b99b3eb0-9bca-42e3-bea9-3b0454a872db-12_483_848_306_589}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A circus performer has mass $m$. She is attached to one end of a cable of length $l$. The other end of the cable is attached to a fixed point $O$
Initially she is held at rest at point $A$ with the cable taut and at an angle of $30 ^ { \circ }$ below the horizontal, as shown in Figure 3.
The circus performer is released from $A$ and she moves on a vertical circular path with centre $O$
The circus performer is modelled as a particle and the cable is modelled as light and inextensible.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $g$, the tension in the cable at the instant immediately after the circus performer is released.
\item Show that, during the motion following her release, the greatest tension in the cable is 4 times the least tension in the cable.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q4 [9]}}