| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.8 This is a multi-part vertical circle problem requiring energy conservation, circular motion dynamics, and careful analysis of tension conditions. Part (a) requires showing complete circles by checking the critical top point condition (standard M3 technique), while part (b) adds complexity by requiring verification that maximum tension (which occurs at the lowest point) doesn't exceed the breaking limit. The initial projection angle of 60° and the specific initial speed add computational complexity beyond typical textbook exercises, though the underlying principles are standard M3 content. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}mv^2 + mg(2a) = \frac{1}{2}m(3\sqrt{ag})^2 - mg(2a\cos 60°)\) | M1A1A1 | Energy equation from projection to top. Must have 2 KE terms and difference in GPE |
| \((v^2 = 3ag)\) | ||
| \(T + mg = \frac{mv^2}{2a}\) | M1A1 | Equation of motion towards centre at top. Acceleration must be in form \(\frac{v^2}{r}\). Condone \(2a = r\) if substituted later |
| \(T = \frac{m(3ag)}{2a} - mg = \frac{mg}{2}\) | dM1A1 | Eliminate \(v\) to form equation for \(T\) |
| \(T > 0\), therefore string remains taut and particle performs complete vertical circles | A1 | Correct inequality with concluding statement |
| Answer | Marks |
|---|---|
| - Finds resultant force at top | M1 |
| - \(F = \frac{m(3ag)}{2a} = \frac{3mg}{2}\) | A1 |
| - Compares resultant force to weight | dM1 |
| - \(\frac{3mg}{2} - mg > 0\) or \(\frac{3mg}{2} > mg\) | A1 |
| - Correct concluding statement mentioning tension in string at top | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}mV^2 = \frac{1}{2}m(3\sqrt{ag})^2 + mg(2a - 2a\cos 60°)\) or \(\frac{1}{2}mV^2 = \frac{1}{2}m(3ag) + mg(4a)\) | M1A1 | Energy equation from initial or top to bottom |
| \((V^2 = 11ag)\) | ||
| \(T - mg = \frac{m(11ag)}{2a}\) | M1A1 | Equation of motion at bottom |
| \(T = \frac{13mg}{2} < 7mg\). Tension less than critical value, so particle completes vertical circles | A1 | Correct tension and concluding statement |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}mv^2 + mg(2a\cos 60° - 2a\cos\theta) = \frac{1}{2}m(3\sqrt{ag})^2 \rightarrow v^2 = ag(7 + 4\cos\theta)\) | M1A1A1 | Energy from projection to general point, must have 2 KE terms and GPE difference in \(\theta\) |
| \(T - mg\cos\theta = \frac{mv^2}{2a}\) | M1A1 | Equation of motion towards centre at general point |
| \(T - mg\cos\theta = \frac{mag}{2a}(7 + 4\cos\theta)\) AND \(\theta = \pi\) or \(\cos\theta \geq -1\) | dM1 | Eliminate \(v\), set \(\theta\) or \(\cos\theta\) to evaluate \(T\) at top |
| \(T + mg = \frac{mg}{2}(3) \rightarrow T = \frac{mg}{2}\) | A1 | |
| \(T > 0\), string stays taut and particle completes vertical circles | A1 | |
| \(\theta = 2\pi \rightarrow T - mg\cos 2\pi = \frac{mg}{2}(7 + 4\cos 2\pi)\) | M1, M1 | |
| \(T - mg = \frac{m(11ag)}{2a}\) | A1, A1 | |
| \(T = \frac{13mg}{2} < 7mg\). Tension less than critical value, so particle completes vertical circles | A1 |
# Question 6:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 + mg(2a) = \frac{1}{2}m(3\sqrt{ag})^2 - mg(2a\cos 60°)$ | M1A1A1 | Energy equation from projection to top. Must have 2 KE terms and difference in GPE |
| $(v^2 = 3ag)$ | | |
| $T + mg = \frac{mv^2}{2a}$ | M1A1 | Equation of motion towards centre at top. Acceleration must be in form $\frac{v^2}{r}$. Condone $2a = r$ if substituted later |
| $T = \frac{m(3ag)}{2a} - mg = \frac{mg}{2}$ | dM1A1 | Eliminate $v$ to form equation for $T$ |
| $T > 0$, therefore string remains taut and particle performs complete vertical circles | A1 | Correct inequality with concluding statement |
**SC:** Uses $T > 0$ without $T =$ for last five marks:
- Finds resultant force at top | M1 |
- $F = \frac{m(3ag)}{2a} = \frac{3mg}{2}$ | A1 |
- Compares resultant force to weight | dM1 |
- $\frac{3mg}{2} - mg > 0$ or $\frac{3mg}{2} > mg$ | A1 |
- Correct concluding statement mentioning tension in string at top | A1 |
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mV^2 = \frac{1}{2}m(3\sqrt{ag})^2 + mg(2a - 2a\cos 60°)$ or $\frac{1}{2}mV^2 = \frac{1}{2}m(3ag) + mg(4a)$ | M1A1 | Energy equation from initial or top to bottom |
| $(V^2 = 11ag)$ | | |
| $T - mg = \frac{m(11ag)}{2a}$ | M1A1 | Equation of motion at bottom |
| $T = \frac{13mg}{2} < 7mg$. Tension less than critical value, so particle completes vertical circles | A1 | Correct tension and concluding statement |
### ALT (a and b) — General Point
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 + mg(2a\cos 60° - 2a\cos\theta) = \frac{1}{2}m(3\sqrt{ag})^2 \rightarrow v^2 = ag(7 + 4\cos\theta)$ | M1A1A1 | Energy from projection to general point, must have 2 KE terms and GPE difference in $\theta$ |
| $T - mg\cos\theta = \frac{mv^2}{2a}$ | M1A1 | Equation of motion towards centre at general point |
| $T - mg\cos\theta = \frac{mag}{2a}(7 + 4\cos\theta)$ AND $\theta = \pi$ or $\cos\theta \geq -1$ | dM1 | Eliminate $v$, set $\theta$ or $\cos\theta$ to evaluate $T$ at top |
| $T + mg = \frac{mg}{2}(3) \rightarrow T = \frac{mg}{2}$ | A1 | |
| $T > 0$, string stays taut and particle completes vertical circles | A1 | |
| $\theta = 2\pi \rightarrow T - mg\cos 2\pi = \frac{mg}{2}(7 + 4\cos 2\pi)$ | M1, M1 | |
| $T - mg = \frac{m(11ag)}{2a}$ | A1, A1 | |
| $T = \frac{13mg}{2} < 7mg$. Tension less than critical value, so particle completes vertical circles | A1 | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-20_789_858_121_536}
\captionsetup{labelformat=empty}
\caption{Figure 7}
\end{center}
\end{figure}
A particle of mass $m$ is attached to one end of a light inextensible string of length $2 a$. The other end of the string is attached to a fixed point $O$. The particle is initially held at the point $A$ with the string taut and $O A$ making an angle of $60 ^ { \circ }$ with the downward vertical.
The particle is then projected upwards with a speed of $3 \sqrt { a g }$, perpendicular to $O A$, in the vertical plane containing $O A$, as shown in Figure 7.
In an initial model of the motion of the particle, it is assumed that the string does not break.
Using this model,
\begin{enumerate}[label=(\alph*)]
\item show that the particle performs complete vertical circles.
In a refined model it is assumed that the string will break if the tension in it exceeds 7 mg . Using this refined model,
\item show that the particle still performs complete vertical circles.\\
\includegraphics[max width=\textwidth, alt={}, center]{8a687d17-ec7e-463f-84dd-605f5c230db1-20_2249_50_314_1982}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q6 [13]}}