| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2002 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring energy conservation and circular motion equations. Part (a) uses energy methods with given amplitude, part (b) applies Newton's second law in circular motion (routine derivation), and part (c) finds max/min values using the given cosine range. All techniques are standard M3 material with no novel insights required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mu^2 = mgl(1 - \cos \theta)\) | M1 A1 A1 | |
| \(u = \sqrt{\frac{4}{3}gl}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(T - mg\cos \theta = \frac{mv^2}{l}\) | M1 A1 | |
| \(\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mgl(1 - \cos \theta)\) | M1 A1 | |
| eliminating \(v^2\), \(T = \frac{mg}{3}(9\cos \theta - 4)\) | M1, A1 cso | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| max \(T\), \(\theta = 0\), \(T_{MAX} = \frac{5mg}{3}\) | M1 | |
| min \(T\), \(\cos \theta = \frac{2}{3}\), \(T_{MIN} = \frac{2mg}{3}\) | M1 A1 | |
| \(\frac{2mg}{3} \leq T \leq \frac{5mg}{3}\) | A1 | (4) |
## Part (a)
$\frac{1}{2}mu^2 = mgl(1 - \cos \theta)$ | M1 A1 A1
$u = \sqrt{\frac{4}{3}gl}$ | A1 | (4)
## Part (b)
$T - mg\cos \theta = \frac{mv^2}{l}$ | M1 A1
$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mgl(1 - \cos \theta)$ | M1 A1
eliminating $v^2$, $T = \frac{mg}{3}(9\cos \theta - 4)$ | M1, A1 cso | (6)
## Part (c)
max $T$, $\theta = 0$, $T_{MAX} = \frac{5mg}{3}$ | M1
min $T$, $\cos \theta = \frac{2}{3}$, $T_{MIN} = \frac{2mg}{3}$ | M1 A1
$\frac{2mg}{3} \leq T \leq \frac{5mg}{3}$ | A1 | (4)
**Total: (14 marks)**\includegraphics{figure_3}
A particle of mass $m$ is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point $O$. The particle is hanging at the point $A$, which is vertically below $O$. It is projected horizontally with speed $u$. When the particle is at the point $P$, $\angle AOP = \theta$, as shown in Fig. 3. The string oscillates through an angle $\alpha$ on either side of $OA$ where $\cos \alpha = \frac{2}{3}$.
\begin{enumerate}[label=(\alph*)]
\item Find $u$ in terms of $g$ and $l$.
[4]
\end{enumerate}
When $\angle AOP = \theta$, the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $T = \frac{mg}{3}(9\cos\theta - 4)$.
[6]
\item Find the range of values of $T$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2002 Q7 [14]}}