| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle problem with straightforward energy conservation and circular motion equations. Parts (a)-(c) follow routine procedures (energy equation, resolving forces radially, checking tension ≥ 0), while part (d) requires recognizing maximum speed occurs at the bottom. The given initial condition u² = 5ag makes calculations clean, and proving complete revolution is a textbook exercise requiring no novel insight. |
| Spec | 3.02h Motion under gravity: vector form6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| 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) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| 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) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| At \(C\), \(\theta = 90°\): \(T = mg(5-3) = 2mg\) | M1 A1 | |
| \(T > 0\) \(\therefore P\) reaches \(C\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\theta = 270°\); \(v^2 = 5ag - 2ag\sin 270°\) | M1 | Max speed at lowest point |
| \(v^2 = 5ag + 2ag\) | ||
| \(v = \sqrt{7ag}\) | A1 | (2) [12] |
## Question 5:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| 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) |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| 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) |
### Part (c):
| Answer/Working | Marks | Notes |
|---|---|---|
| At $C$, $\theta = 90°$: $T = mg(5-3) = 2mg$ | M1 A1 | |
| $T > 0$ $\therefore P$ reaches $C$ | A1 | (3) |
### Part (d):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\theta = 270°$; $v^2 = 5ag - 2ag\sin 270°$ | M1 | Max speed at lowest point |
| $v^2 = 5ag + 2ag$ | | |
| $v = \sqrt{7ag}$ | A1 | (2) **[12]** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{049ea68c-d15f-41f8-860e-0816d36a2748-10_474_465_269_735}
\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 $a$. The other end of the string is fixed at the point $O$. The particle is initially held with $O P$ horizontal and the string taut. It is then projected vertically upwards with speed $u$, where $u ^ { 2 } = 5 a g$. When $O P$ 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 $v ^ { 2 }$.
\item Find, in terms of $m , g$ and $\theta$, an expression for $T$.
\item Prove that $P$ moves in a complete circle.
\item Find the maximum speed of $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2010 Q5 [12]}}