| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation and circular motion equations for part (a), then projectile motion for part (b). While it involves multiple steps and careful application of mechanics principles, the techniques are well-practiced at this level with no novel insights required. The algebra is straightforward and the 'show that' format provides a target to work towards. |
| Spec | 3.02i Projectile motion: constant acceleration model6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| Energy: \(\frac{1}{2}mu^2 + mgl\left(\cos\theta - \frac{1}{4}\right) = \frac{1}{2}mv^2\) | M1A1 |
| Resolving: \(T - mg\cos\theta = \frac{mv^2}{l}\) | M1A1 |
| Eliminate \(v^2\): \(T = mg\cos\theta + \frac{1}{l}(2mgl(\cos\theta - \frac{1}{4}))\) | M1 |
| \(T = 3mg\cos\theta - \frac{mg}{2}\) | A1 |
| Answer | Marks |
|---|---|
| \(\theta = 60° \Rightarrow mv^2 = 2mgl\left(\frac{1}{2} - \frac{1}{4}\right)\) | M1 |
| \(\Rightarrow v^2 = \frac{gl}{2}\) | A1 |
| Vertical motion under gravity: \(0 = (v\cos 30°)^2 - 2gs\) | M1 |
| \(0 = \frac{gl}{2} \times \frac{3}{4} - 2gs \Rightarrow s = \frac{3l}{16}\) | A1 |
| Distance below A \(= \frac{l}{2} - \frac{3l}{16} = \frac{5l}{16}\) | M1A1 |
| Answer | Marks |
|---|---|
| \(\frac{1}{2}mv^2 - mgl\cos 60° = \frac{1}{2}m(v\cos 60°)^2 - mged\) | M1A1 |
| \(\frac{gl}{4} - \frac{gl}{2} = \frac{gl}{4} \times \frac{1}{4} - ed\) | M1 |
| \(d = \frac{1-4+8}{16}l = \frac{5l}{16}\) | A1 |
## Part (a)
Energy: $\frac{1}{2}mu^2 + mgl\left(\cos\theta - \frac{1}{4}\right) = \frac{1}{2}mv^2$ | M1A1 |
Resolving: $T - mg\cos\theta = \frac{mv^2}{l}$ | M1A1 |
Eliminate $v^2$: $T = mg\cos\theta + \frac{1}{l}(2mgl(\cos\theta - \frac{1}{4}))$ | M1 |
$T = 3mg\cos\theta - \frac{mg}{2}$ | A1 |
## Part (b)
$\theta = 60° \Rightarrow mv^2 = 2mgl\left(\frac{1}{2} - \frac{1}{4}\right)$ | M1 |
$\Rightarrow v^2 = \frac{gl}{2}$ | A1 |
Vertical motion under gravity: $0 = (v\cos 30°)^2 - 2gs$ | M1 |
$0 = \frac{gl}{2} \times \frac{3}{4} - 2gs \Rightarrow s = \frac{3l}{16}$ | A1 |
Distance below A $= \frac{l}{2} - \frac{3l}{16} = \frac{5l}{16}$ | M1A1 |
### Alternative for end of (b) using energy
$\frac{1}{2}mv^2 - mgl\cos 60° = \frac{1}{2}m(v\cos 60°)^2 - mged$ | M1A1 |
$\frac{gl}{4} - \frac{gl}{2} = \frac{gl}{4} \times \frac{1}{4} - ed$ | M1 |
$d = \frac{1-4+8}{16}l = \frac{5l}{16}$ | A1 |
---One end of a light inextensible string of length $l$ is attached to a fixed point $A$. The other end is attached to a particle $P$ of mass $m$, which is held at a point $B$ with the string taut and $AP$ making an angle arccos $\frac{1}{4}$ with the downward vertical. The particle is released from rest. When $AP$ makes an angle $\theta$ with the downward vertical, the string is taut and the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$T = 3mg \cos \theta - \frac{mg}{2}.$$ [6]
\end{enumerate}
\includegraphics{figure_3}
At an instant when $AP$ makes an angle of $60°$ to the downward vertical, $P$ is moving upwards, as shown in Figure 3. At this instant the string breaks. At the highest point reached in the subsequent motion, $P$ is at a distance $d$ below the horizontal through $A$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $d$ in terms of $l$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2009 Q5 [11]}}