| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Moderate -0.3 This is a straightforward M2 vector mechanics question requiring differentiation of position to find velocity, then solving simultaneous equations when velocity equals zero. The steps are routine (differentiate each component, set both to zero, solve for k) with no conceptual challenges beyond standard A-level mechanics techniques. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\mathbf{v} = \frac{d\mathbf{r}}{dt} = (3t-3)\mathbf{i} + (t^2-k)\mathbf{j}\) | M2 A1 |
| (b) | At rest when coeffs of \(\mathbf{i}\) and \(\mathbf{j}\) are both zero: \(3t-3=0\) and \(t^2-k=0\); both satisfied when \(k=1\) | M1, M1, A1 |
(a) | $\mathbf{v} = \frac{d\mathbf{r}}{dt} = (3t-3)\mathbf{i} + (t^2-k)\mathbf{j}$ | M2 A1 |
(b) | At rest when coeffs of $\mathbf{i}$ and $\mathbf{j}$ are both zero: $3t-3=0$ and $t^2-k=0$; both satisfied when $k=1$ | M1, M1, A1 | (6 marks total)
---\begin{enumerate}
\item A particle $P$ moves such that at time $t$ seconds its position vector, $\mathbf { r }$ metres, relative to a fixed origin $O$ is given by
\end{enumerate}
$$\mathbf { r } = \left( \frac { 3 } { 2 } t ^ { 2 } - 3 t \right) \mathbf { i } + \left( \frac { 1 } { 3 } t ^ { 3 } - k t \right) \mathbf { j } ,$$
where $k$ is a constant and $\mathbf { i }$ and $\mathbf { j }$ are perpendicular horizontal unit vectors.\\
(a) Find an expression for the velocity of $P$ at time $t$.\\
(b) Given that $P$ comes to rest instantaneously, find the value of $k$.\\
\hfill \mbox{\textit{Edexcel M2 Q1 [6]}}