| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Session | Specimen |
| Marks | 11 |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: equilibrium (find unknowns) |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring resolution of forces in equilibrium (part i) followed by basic kinematics with constant acceleration (part ii). Both parts use standard A-level techniques with no novel insight required—just systematic application of F=ma and SUVAT equations. The vector notation and multiple forces add minor complexity but this remains easier than average. |
| Spec | 3.02e Two-dimensional constant acceleration: with vectors3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Use of 'vector force = magnitude times unit vector' | M1 | Obtain \(\mathbf{F}_1 = 2(3\mathbf{i} + 4\mathbf{j})\) |
| (ii) The new net force vector is \(6\mathbf{i} + 8\mathbf{j} + 4\mathbf{j} + \sqrt{52}\left(-\frac{3}{\sqrt{13}}\mathbf{i} + \frac{2}{\sqrt{13}}\mathbf{j}\right)\) | M1A1 | \(= 16\mathbf{j}\) |
(i) Use of 'vector force = magnitude times unit vector' | M1 | Obtain $\mathbf{F}_1 = 2(3\mathbf{i} + 4\mathbf{j})$ | A1 | $\mathbf{F}_2 = -12\mathbf{j}$ and $F(-\cos\theta\mathbf{i} + \sin\theta\mathbf{j})$ | A1 | Equilibrium ⇒ $6 - F\cos\theta = 0, 8 - 12 + F\sin\theta = 0$ | M1 | Obtain $\theta = \tan^{-1}\left(\frac{4}{3}\right) = 0.588$ and $F = \sqrt{52} = 7.21$ | A1A1 | 6 marks
(ii) The new net force vector is $6\mathbf{i} + 8\mathbf{j} + 4\mathbf{j} + \sqrt{52}\left(-\frac{3}{\sqrt{13}}\mathbf{i} + \frac{2}{\sqrt{13}}\mathbf{j}\right)$ | M1A1 | $= 16\mathbf{j}$ | A1 | Acceleration is $\frac{16}{3}\mathbf{j}$ | A1 | Position vector is $48\mathbf{j}$ | A1 | 5 marksA particle $P$ of mass $1.5$ kg is placed on a smooth horizontal table. The particle is initially at the origin of a $2$-dimensional vector system defined by perpendicular unit vectors $\mathbf{i}$ and $\mathbf{j}$ in the plane of the table. The particle is subject to three forces of magnitudes $10$ N, $12$ N and $F$ N, acting in the directions of the vectors $3\mathbf{i} + 4\mathbf{j}$, $-\mathbf{j}$ and $-\cos \theta \mathbf{i} + \sin \theta \mathbf{j}$ respectively, and no others.
\begin{enumerate}[label=(\roman*)]
\item Given that the system is in equilibrium, determine $F$ and $\theta$. [6]
\end{enumerate}
The force of magnitude $12$ N is replaced by one of magnitude $4$ N, but in the opposite direction. The particle is initially at rest.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the position vector of the particle $3$ seconds later. [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 Q11 [11]}}