| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.8 This is a comprehensive circular motion problem requiring energy conservation, circular motion dynamics, and analysis of complete revolution conditions. While the techniques are standard M3 content (energy methods, resolving forces radially), it requires careful coordinate setup, multi-part reasoning across connected sub-questions, and proving complete circular motion by checking tension remains positive throughout - more demanding than routine exercises but uses established methods without requiring novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Energy: \(mga\sin\theta = \frac{1}{2}m \times 5ag - \frac{1}{2}mv^2\) | M1 A1 | |
| \(v^2 = 5ag - 2ag\sin\theta\) | A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Eqn of motion along radius: \(T + mg\sin\theta = \frac{mv^2}{a}\) | M1 A1 | |
| \(T = \frac{m}{a}(5ag - 2ag\sin\theta) - mg\sin\theta\) | M1 | |
| \(T = mg(5 - 3\sin\theta)\) | A1 | (4 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| At C, \(\theta = 90°\): \(T = mg(5 - 3) = 2mg\) | M1 A1 | |
| \(T > 0 \therefore\) P reaches C | A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Max speed at lowest point (\(\theta = 270°\); \(v^2 = 5ag - 2ag\sin 270°\)) | M1 | |
| \(v^2 = 5ag + 2ag = 7ag\) | ||
| \(v = \sqrt{(7ag)}\) | A1 | (2 marks total) |
## Part (a)
Energy: $mga\sin\theta = \frac{1}{2}m \times 5ag - \frac{1}{2}mv^2$ | M1 A1 |
$v^2 = 5ag - 2ag\sin\theta$ | A1 | (3 marks total)
## Part (b)
Eqn of motion along radius: $T + mg\sin\theta = \frac{mv^2}{a}$ | M1 A1 |
$T = \frac{m}{a}(5ag - 2ag\sin\theta) - mg\sin\theta$ | M1 |
$T = mg(5 - 3\sin\theta)$ | A1 | (4 marks total)
## Part (c)
At C, $\theta = 90°$: $T = mg(5 - 3) = 2mg$ | M1 A1 |
$T > 0 \therefore$ P reaches C | A1 | (3 marks total)
## Part (d)
Max speed at lowest point ($\theta = 270°$; $v^2 = 5ag - 2ag\sin 270°$) | M1 |
$v^2 = 5ag + 2ag = 7ag$ | |
$v = \sqrt{(7ag)}$ | A1 | (2 marks total)
**[12 marks total]**\includegraphics{figure_5}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is fixed at the point $O$. The particle is initially held with $OP$ horizontal and the string taut. It is then projected vertically upwards with speed $u$, where $u^2 = 5ag$. When $OP$ has turned through an angle $\theta$ the speed of $P$ is $v$ and the tension in the string is $T$, as shown in Figure 5.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, $g$ and $\theta$, an expression for $v^2$. [3]
\item Find, in terms of $m$, $g$ and $\theta$, an expression for $T$. [4]
\item Prove that $P$ moves in a complete circle. [3]
\item Find the maximum speed of $P$. [2]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [12]}}