| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 4 |
| Topic | Moments |
| Type | Particle suspended by strings |
| Difficulty | Challenging +1.2 This is a static equilibrium problem requiring resolution of forces and geometric reasoning. Part (i) involves standard force resolution with symmetry simplifying the setup. Part (ii) requires algebraic manipulation combining force equations with geometric constraints, which is more demanding but still follows a clear methodical approach typical of mechanics problems. |
| Spec | 3.03n Equilibrium in 2D: particle under forces3.04b Equilibrium: zero resultant moment and force |
**(i)** $AC$ and $BD$ are inclined at 60 degrees to the horizontal. B1
Using $T$ and $S$ as the tensions in $AC/BD$ and $CD$, respectively, resolving the forces at $C$, vertically and horizontally: M1
$T\sin 60 - 2g = 0$, (system in equilibrium)
$S - T\cos 60 = 0$ A1
Obtain $T = 23.1$ N and $S = 11.5$ N A1 **[4]**
**(ii)** If $x$ is the inclination of $AC$ to the horizontal, then
$\cos x = \frac{1}{2}(20 - L) \div 10 = 1 - \frac{L}{20}$ B1
Resolving vertically: M1
$50\sin x = 20$
$\Rightarrow \sin x = \frac{2}{5}$ A1
$\left(1 - \frac{L}{20}\right)^2 = 1 - \frac{4}{25}$ A1
$\Rightarrow L = 20 - 4\sqrt{21}$ **AG** **[4]**11 Three light inextensible strings $A C , C D$ and $D B$, each of length 10 cm , are joined as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{85043199-527d-4105-aa0b-c913dec0e35b-5_300_670_475_699}
The ends $A$ and $B$ are fixed to points 20 cm apart on the same horizontal level. Two heavy particles, each of mass 2 kg , are attached at $C$ and $D$. The system remains in a vertical plane.\\
(i) Determine the tension in each string.\\
(ii) The string $C D$ is replaced by one of length $L \mathrm {~cm}$, made of the same material. If the tension in $A C$ is 50 N , show that $L = 20 - 4 \sqrt { 21 }$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q11 [4]}}