| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 9 |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a classic circular motion problem requiring energy conservation and Newton's second law in a non-trivial context. Part (i)(a) involves deriving the tension formula using both energy methods and centripetal force (5 marks suggests multi-step derivation). Part (i)(b) requires understanding that T≥0 for complete circles. Part (ii) involves finding when T=0 for a specific case and describing projectile motion. While the techniques are standard A-level mechanics, the problem requires careful application across multiple parts with some physical insight about slack strings and projectile motion, placing it moderately above average difficulty. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg l(1-\cos\theta) = \frac{1}{2}mu^2 - \frac{1}{2}mv^2 \Rightarrow v^2 = u^2 - 2gl(1-\cos\theta)\) | M1, A1 | |
| \(T - mg\cos\theta = \frac{mv^2}{l}\) | M1, A1 | |
| \(\therefore T = mg\cos\theta + \frac{mu^2}{l} - 2mg + 2mg\cos\theta = mg(3\cos\theta - 2) + \frac{mu^2}{l}\) | M1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore \frac{mu^2}{l} - 5mg \geq 0 \Rightarrow u^2 \geq 5gl\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \cos\theta = -\frac{1}{3} \Rightarrow \theta = 109.5°\) or \(1.91\) rads | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Particle assumes projectile motion (with slack string) | B1 | 1 |
### (i)(a)
$mg l(1-\cos\theta) = \frac{1}{2}mu^2 - \frac{1}{2}mv^2 \Rightarrow v^2 = u^2 - 2gl(1-\cos\theta)$ | M1, A1 |
$T - mg\cos\theta = \frac{mv^2}{l}$ | M1, A1 |
$\therefore T = mg\cos\theta + \frac{mu^2}{l} - 2mg + 2mg\cos\theta = mg(3\cos\theta - 2) + \frac{mu^2}{l}$ | M1 | **5**
### (i)(b)
For complete circles $T \geq 0$ when $\cos\theta = -1$
$\therefore \frac{mu^2}{l} - 5mg \geq 0 \Rightarrow u^2 \geq 5gl$ | B1 | **1**
### (ii)(a)
Putting $u^2 = 3gl$ and $T = 0$ in (*):
$3mg\cos\theta = 2mg - 3mg$
$\Rightarrow \cos\theta = -\frac{1}{3} \Rightarrow \theta = 109.5°$ or $1.91$ rads | M1, A1 | **2**
### (ii)(b)
Particle assumes projectile motion (with slack string) | B1 | **1**One end of a light inextensible string of length $l$ is attached to a fixed point $O$. The other end is attached to a particle $P$ of mass $m$. The particle hangs at rest vertically below $O$. The particle is then given a horizontal speed $u$.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Show that when $OP$ has turned through an angle $\theta$ the tension in the string is given by
$$T = mg(3\cos \theta - 2) + \frac{mu^2}{l}$$
as long as the string remains taut. [5]
\item Deduce that $u^2 \geq 5gl$ in order for the particle to perform complete circles. [1]
\end{enumerate}
\item \begin{enumerate}[label=(\alph*)]
\item In the case $u^2 = 3gl$, find the angle that $OP$ makes with the downward vertical at $O$ at the instant when the string becomes slack. [2]
\item Describe the nature of the motion while the string is slack. [1]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q2 [9]}}