| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2004 |
| Session | January |
| Marks | 14 |
| 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 vertical circle problem with multiple parts requiring energy conservation, circular motion equations, and projectile motion. While it involves several steps and concepts (energy, tension formula, slack condition, projectile trajectory), each part follows well-established methods taught in M3. The slack angle calculation and projectile phase are routine applications. More challenging than basic C1/C2 questions but typical for M3 mechanics. |
| Spec | 3.02h Motion under gravity: vector form6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks |
|---|---|
| (a) Energy: \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga\sin\theta\) | M1 |
| \(v^2 = \frac{3}{2}ga + 2ga\sin\theta\) | A1 (2) |
| (b) Radial equation: \(T - mg\sin\theta = m\frac{v^2}{a}\) | M1A1 |
| \(T = \frac{3mg}{2}(1 + 2\sin\theta)\) any form | A1 ⋈ (3) |
| (c) Setting \(T = 0\) and solving trig. equation; \((\sin\theta = -\frac{1}{2}) \Rightarrow \theta = 210°\) * | M1;A1(2) |
| (d) Setting \(v = 0\) in (a) and solving for \(\theta\) | M1 |
| \(\sin\theta = -\frac{3}{4}\) so not complete circle | A1 (2) |
| OR Substituting \(\theta = 270°\) in (a); \(v^2 < 0\) so not possible to complete | |
| (e) No change in PE \(\Rightarrow\) no change in KE (Cof E) so \(v = u\) | B1 (1) |
| (f) When string becomes slack, \(V^2 = \frac{1}{2}ga\) \([\sin\theta = -\frac{1}{2}\) in (a)] | B1 ⋈ |
| Using fact that horizontal component of velocity is unchanged | M1 |
| \(\sqrt{\frac{ga}{2}}\cos 60° = \sqrt{\frac{3ga}{2}}\cos\phi\) | |
| \(\cos\phi = \sqrt{\frac{1}{12}} \Rightarrow \phi = 73.2°\) | M1A1 (4) |
| [14] |
(a) Energy: $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga\sin\theta$ | M1 |
$v^2 = \frac{3}{2}ga + 2ga\sin\theta$ | A1 (2) |
(b) Radial equation: $T - mg\sin\theta = m\frac{v^2}{a}$ | M1A1 |
$T = \frac{3mg}{2}(1 + 2\sin\theta)$ any form | A1 ⋈ (3) |
(c) Setting $T = 0$ and solving trig. equation; $(\sin\theta = -\frac{1}{2}) \Rightarrow \theta = 210°$ * | M1;A1(2) |
(d) Setting $v = 0$ in (a) and solving for $\theta$ | M1 |
$\sin\theta = -\frac{3}{4}$ so not complete circle | A1 (2) |
OR Substituting $\theta = 270°$ in (a); $v^2 < 0$ so not possible to complete | |
(e) No change in PE $\Rightarrow$ no change in KE (Cof E) so $v = u$ | B1 (1) |
(f) When string becomes slack, $V^2 = \frac{1}{2}ga$ $[\sin\theta = -\frac{1}{2}$ in (a)] | B1 ⋈ |
Using fact that horizontal component of velocity is unchanged | M1 |
$\sqrt{\frac{ga}{2}}\cos 60° = \sqrt{\frac{3ga}{2}}\cos\phi$ | |
$\cos\phi = \sqrt{\frac{1}{12}} \Rightarrow \phi = 73.2°$ | M1A1 (4) |
| [14] |7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{c4b453e7-8a32-458b-8041-58c9e4ef9533-6_710_729_172_672}
\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 a point $O$. The particle is held with the string taut and $O P$ horizontal. It is then projected vertically downwards with speed $u$, where $u ^ { 2 } = \frac { 3 } { 2 } g a$. When $O P$ has turned through an angle $\theta$ and the string is still taut, the speed of $P$ is $v$ and the tension in the string is $T$, as shown in Fig. 3.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v ^ { 2 }$ in terms of $a , g$ and $\theta$.
\item Find an expression for $T$ in terms of $m , g$ and $\theta$.
\item Prove that the string becomes slack when $\theta = 210 ^ { \circ }$.
\item State, with a reason, whether $P$ would complete a vertical circle if the string were replaced by a light rod.
After the string becomes slack, $P$ moves freely under gravity and is at the same level as $O$ when it is at the point $A$.
\item Explain briefly why the speed of $P$ at $A$ is $\sqrt { } \left( \frac { 3 } { 2 } g a \right)$.
The direction of motion of $P$ at $A$ makes an angle $\varphi$ with the horizontal.
\item Find $\varphi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2004 Q7 [14]}}