| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Topic | Moments |
| Type | Particle suspended by strings |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem combining basic vector geometry with equilibrium of forces. Part (i) is routine dot product calculation to verify perpendicularity. Parts (ii) and (iii) involve standard resolution of forces in perpendicular directions and converting scalar tensions to vector form using unit vectors—all standard A-level mechanics techniques with no novel insight required. The multi-part structure and 8 total marks indicate moderate length, but each step follows textbook methods. |
| Spec | 1.10g Problem solving with vectors: in geometry3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Any valid method, for example | ||
| \(\vec{AB} \cdot \vec{AC} = (-4i + 3k)(3i + 4k)\) | M1 | |
| \(= -12 + 12 = 0\) | Hence result | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving along \(AB\): \(T_{AB} = 20\cos\left(\tan^{-1}\frac{4}{3}\right)\) | M1 | |
| Obtain 12N | A1 | |
| Resolving along \(AC\): \(T_{AC} = 20\sin\left(\tan^{-1}\frac{4}{3}\right) = 16\text{N}\) | A1 | [3] |
| Answer | Marks |
|---|---|
| The vector tension is 12 x unit vector in \(AB\) direction | M1 |
| \(= -9.6i + 7.2j\) | A1 ∨ A1 ∨ |
| Or \(= -ai + bj\) where \(\frac{a}{b} = \frac{4}{3}\) and \(a^2 + b^2 = (\text{their } T_{AB})^2\) | [3] |
**Part (i)**
Any valid method, for example |
$\vec{AB} \cdot \vec{AC} = (-4i + 3k)(3i + 4k)$ | M1
$= -12 + 12 = 0$ | Hence result | A1 | [2]
**Part (ii)**
Resolving along $AB$: $T_{AB} = 20\cos\left(\tan^{-1}\frac{4}{3}\right)$ | M1
Obtain 12N | A1
Resolving along $AC$: $T_{AC} = 20\sin\left(\tan^{-1}\frac{4}{3}\right) = 16\text{N}$ | A1 | [3]
*SC: Both answers either unassigned or swapped – B1*
**Part (iii)**
The vector tension is 12 x unit vector in $AB$ direction | M1
$= -9.6i + 7.2j$ | A1 ∨ A1 ∨
Or $= -ai + bj$ where $\frac{a}{b} = \frac{4}{3}$ and $a^2 + b^2 = (\text{their } T_{AB})^2$ | [3]The points $A$, $B$ and $C$ lie in a vertical plane and have position vectors $4\mathbf{i}$, $3\mathbf{j}$ and $7\mathbf{i} + 4\mathbf{j}$, respectively. The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertically upwards, respectively. The units of the components are metres.
\begin{enumerate}[label=(\roman*)]
\item Show that angle $BAC$ is a right angle. [2]
\end{enumerate}
\includegraphics{figure_10}
Strings $AB$ and $AC$ are attached to $B$ and $C$, and joined at $A$. A particle of weight 20 N is attached at $A$ (see diagram). The particle is in equilibrium.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{1}
\item By resolving in the directions $AB$ and $AC$, determine the magnitude of the tension in each string. [3]
\item Express the tension in the string $AB$ as a vector, in terms of $\mathbf{i}$ and $\mathbf{j}$. [3]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q10 [8]}}