| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a multi-part vertical circle problem requiring energy conservation, circular motion dynamics, and projectile motion after the string slackens. While the individual techniques are standard M3 content, the question requires careful coordination across multiple parts, finding when tension becomes zero, then analyzing subsequent projectile motion. The algebraic manipulation and conceptual understanding needed place it moderately above average difficulty. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks |
|---|---|
| (a) Energy: \(\frac{1}{2}m \cdot 3ag - \frac{1}{2}mv^2 = mga(1 + \cos \theta)\) | M1 A1 |
| \(v^2 = ag(1 - 2\cos \theta)\) (o.e.) | A1 (3) |
| (b) \(T + mg \cos \theta = m\frac{v^2}{a}\) | M1 A1 |
| Hence \(T = (1 - 3\cos \theta)mg\) (*) | A1 cso (3) |
| (c) Using \(T = 0\) to find \(\cos \theta\) | M1 |
| Hence height above \(A = \frac{4}{3}a\) Accept 1.33a (but must have 3+ s.f.) | A1 (2) |
| (d) \(v^2 = \frac{1}{3}ag\) (o.e.) f.t. using \(\cos \theta = \frac{1}{3}\) in \(v^2\) | B1√ |
| consider vert motion: \((\nu \sin \theta)^2 = 2gh\) (with \(v\) resolved) | M1 A1 ↓ M1 |
| \(\sin^2 \theta = \frac{8}{9}\) (or \(\theta = 70.53, \sin \theta = 0.943\)) and solve for \(h\) (as ka) | |
| \(h = \frac{4}{27}a\) or 0.148a (awrt) | A1 |
| OR consider energy: \(\frac{1}{2}m(v\cos \theta)^2 + mgh = \frac{1}{2}mv^2\) (3 non-zero terms) | M1 A1 ↓ M1 |
| Sub for v, \(\theta\) and solve for \(h\) | |
| \(h = \frac{4}{27}a\) or 0.148a (awrt) | A1 |
(a) Energy: $\frac{1}{2}m \cdot 3ag - \frac{1}{2}mv^2 = mga(1 + \cos \theta)$ | M1 A1 |
$v^2 = ag(1 - 2\cos \theta)$ (o.e.) | A1 (3) |
(b) $T + mg \cos \theta = m\frac{v^2}{a}$ | M1 A1 |
Hence $T = (1 - 3\cos \theta)mg$ (*) | A1 cso (3) |
(c) Using $T = 0$ to find $\cos \theta$ | M1 |
Hence height above $A = \frac{4}{3}a$ Accept 1.33a (but must have 3+ s.f.) | A1 (2) |
(d) $v^2 = \frac{1}{3}ag$ (o.e.) f.t. using $\cos \theta = \frac{1}{3}$ in $v^2$ | B1√ |
consider vert motion: $(\nu \sin \theta)^2 = 2gh$ (with $v$ resolved) | M1 A1 ↓ M1 |
$\sin^2 \theta = \frac{8}{9}$ (or $\theta = 70.53, \sin \theta = 0.943$) and solve for $h$ (as ka) | |
$h = \frac{4}{27}a$ or 0.148a (awrt) | A1 |
OR consider energy: $\frac{1}{2}m(v\cos \theta)^2 + mgh = \frac{1}{2}mv^2$ (3 non-zero terms) | M1 A1 ↓ M1 |
Sub for v, $\theta$ and solve for $h$ | |
$h = \frac{4}{27}a$ or 0.148a (awrt) | A1 |
---4.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{25b3ece7-69ed-4ec4-a6c7-4cd83ec2cc5e-05_574_510_324_726}
\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 attached to a point $O$. The point $A$ is vertically below $O$, and $O A = a$. The particle is projected horizontally from $A$ with speed $\sqrt { } ( 3 a g )$. When $O P$ makes an angle $\theta$ with the upward vertical through $O$ and the string is still taut, the tension in the string is $T$ and the speed of $P$ is $v$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a , g$ and $\theta$, an expression for $v ^ { 2 }$.
\item Show that $T = ( 1 - 3 \cos \theta ) m g$.
The string becomes slack when $P$ is at the point $B$.
\item Find, in terms of $a$, the vertical height of $B$ above $A$.
After the string becomes slack, the highest point reached by $P$ is $C$.
\item Find, in terms of $a$, the vertical height of $C$ above $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2007 Q4 [13]}}