| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2016 |
| Session | Specimen |
| Marks | 11 |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring resolution of forces in two directions, use of friction laws, and calculus optimization. While it involves several steps and coordinate geometry with the inclined string, the techniques are standard A-level mechanics (resolving perpendicular/parallel to plane, limiting friction μR, differentiation to find maximum). The 'show that' structure guides students through the solution systematically. Slightly above average difficulty due to the geometric setup and algebraic manipulation required, but well within typical Further Maths mechanics scope. |
| Spec | 3.03b Newton's first law: equilibrium3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
**Question 10**
(i) As system is in equilibrium, tension in string is $T = mg$ [B1]
Resolving at right angles to the plane: $R + T\sin\alpha = 2mg\cos\alpha$ [M1]
giving $R = mg(2\cos\alpha - \sin\alpha)$ AG [A1]
(ii) By implication $\alpha \leqslant 45°$ [M1]
$\cos\alpha \geqslant \frac{1}{\sqrt{2}}$; $\sin\alpha \leqslant \frac{1}{\sqrt{2}}$ [A1]
$R \geqslant mg\left(\frac{2}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)$ AG [A1]
(iii) Resolving up the slope $F = 2mg\sin\alpha - T\cos\alpha = mg(2\sin\alpha - \cos\alpha)$ [M1]
For this to be non-negative [A1]
and combined with first line of solution to (ii) $0.5 \leqslant \tan\alpha \leqslant 1$ AG [A1]
(iv) Using $F = \mu R$ [M1]
$\mu = \frac{2\sin\alpha - \cos\alpha}{2\cos\alpha - \sin\alpha} = \frac{2t-1}{2-t}$ [A1]
Max value of $\mu$ is $1$ when $t = 1$ [A1]
**Total: 11 marks**10\\
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Particles $A$ and $B$ of masses $2 m$ 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 $\alpha \leqslant 45 ^ { \circ }$ and $B$ is above the plane. The vertical plane defined by $A P B$ contains a line of greatest slope of the plane, and $P A$ is inclined at angle $2 \alpha$ to the horizontal (see diagram).\\
(i) Show that the normal reaction $R$ between $A$ and the plane is $m g ( 2 \cos \alpha - \sin \alpha )$.\\
(ii) Show that $R \geqslant \frac { 1 } { 2 } m g \sqrt { 2 }$.
The coefficient of friction between $A$ and the plane is $\mu$. The particle is about to slip down the plane.\\
(iii) Show that $0.5 < \tan \alpha \leqslant 1$.\\
(iv) Express $\mu$ as a function of $\tan \alpha$ and deduce its maximum value as $\alpha$ varies.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q10 [11]}}