| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Topic | Momentum and Collisions 1 |
| Type | Direct collision with direction reversal |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring resolution of forces in 3D geometry, but follows standard techniques. Parts (i)-(ii) involve routine force resolution and inequality manipulation. Parts (iii)-(iv) require setting up friction conditions and optimization, which is more challenging but still within standard A-level mechanics scope. The 3D geometry (string at angle 2α to horizontal on a plane at angle α) adds moderate complexity, but the problem guides students through each step with 'show that' prompts. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks |
|---|---|
| As the system is in equilibrium, the tension in the string is \(T = mg\) | B1 |
| Resolving at right angles to the plane | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| giving \(R = mg(2\cos\alpha - \sin\alpha)\) | A1 AG | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| By implication \(\alpha \le 45°\) (condone boundary case only) | M1 | |
| \(\cos\alpha \ge \frac{1}{\sqrt{2}}; \sin\alpha \le \frac{1}{\sqrt{2}}\) | A1 | |
| \(R \ge mg\left(\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\) | A1 AG | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving up the slope | ||
| \(F = 2mg\sin\alpha - T\cos\alpha = mg(2\sin\alpha - \cos\alpha)\) | M1 | |
| For this to be positive | A1 | |
| and combined with first line of solution of (ii) | ||
| \(0.5 < \tan\alpha \le 1\) | A1 AG | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(F = \mu R\) | M1 | |
| \(\mu = \frac{2\sin\alpha - \cos\alpha}{2\cos\alpha - \sin\alpha} = \frac{2\tan\alpha - 1}{2 - \tan\alpha}\) | A1 | |
| Max value of \(\mu\) is 1 when \(\tan\alpha = 1\) | A1 | [3] |
**Part (i)**
As the system is in equilibrium, the tension in the string is $T = mg$ | B1
Resolving at right angles to the plane | M1
$R + T\sin\alpha = 2mg\cos\alpha$
giving $R = mg(2\cos\alpha - \sin\alpha)$ | A1 AG | [3]
**Part (ii)**
By implication $\alpha \le 45°$ (condone boundary case only) | M1
$\cos\alpha \ge \frac{1}{\sqrt{2}}; \sin\alpha \le \frac{1}{\sqrt{2}}$ | A1
$R \ge mg\left(\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)$ | A1 AG | [3]
**Part (iii)**
Resolving up the slope |
$F = 2mg\sin\alpha - T\cos\alpha = mg(2\sin\alpha - \cos\alpha)$ | M1
For this to be positive | A1
and combined with first line of solution of (ii) |
$0.5 < \tan\alpha \le 1$ | A1 AG | [3]
**Part (iv)**
Using $F = \mu R$ | M1
$\mu = \frac{2\sin\alpha - \cos\alpha}{2\cos\alpha - \sin\alpha} = \frac{2\tan\alpha - 1}{2 - \tan\alpha}$ | A1
Max value of $\mu$ is 1 when $\tan\alpha = 1$ | A1 | [3]\includegraphics{figure_13}
Particles $A$ and $B$ of masses $2m$ and $m$, respectively, are attached to the ends of a light inextensible string. The string passes over a smooth fixed pulley $P$. The particle $A$ rests in equilibrium on a rough plane inclined at an angle $\alpha$ to the horizontal, where $\alpha \leqslant 45°$ and $B$ is above the plane. The vertical plane defined by $APB$ contains a line of greatest slope of the plane, and $PA$ is inclined at angle $2\alpha$ to the horizontal (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the normal reaction $R$ between $A$ and the plane is $mg(2 \cos \alpha - \sin \alpha)$. [3]
\item Show that $R \geqslant \frac{1}{2}mg\sqrt{2}$. [3]
\end{enumerate}
The coefficient of friction between $A$ and the plane is $\mu$. The particle $A$ is about to slip down the plane.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{2}
\item Show that $0.5 < \tan \alpha \leqslant 1$. [3]
\item Express $\mu$ as a function of $\tan \alpha$ and deduce its maximum value as $\alpha$ varies. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q13 [12]}}