| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard differentiation of position to find velocity, then differentiation again to find acceleration, followed by applying F=ma. The calculations are routine with no conceptual challenges or problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v = (4t-4)i - 2tj\), \(t = 3\): \(v = (8i - 6j) \text{ m s}^{-1}\) | M1 A1 A1 | |
| (b) \(a = 4i - 2j\), \( | a | = \sqrt{20}\), constant, \(F = 3\sqrt{20} = 13.4 \text{ N}\) |
(a) $v = (4t-4)i - 2tj$, $t = 3$: $v = (8i - 6j) \text{ m s}^{-1}$ | M1 A1 A1 |
(b) $a = 4i - 2j$, $|a| = \sqrt{20}$, constant, $F = 3\sqrt{20} = 13.4 \text{ N}$ | M1 A1 M1 A1 | Total: 7 marksA particle $P$ of mass 3 kg has position vector $\mathbf{r} = (2t^2 - 4t)\mathbf{i} + (1 - t^2)\mathbf{j}$ m at time $t$ seconds.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity vector of $P$ when $t = 3$. [3 marks]
\item Find the magnitude of the force acting on $P$, showing that this force is constant. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [7]}}