| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Particle motion: 2D direction/bearing |
| Difficulty | Standard +0.3 Part (a) is a straightforward 'show that' requiring conversion of direction to unit vector and applying v = u + at. Part (b) requires differentiating speed or recognizing when velocity is perpendicular to acceleration, which is a standard M1 technique. The multi-step nature and vector manipulation elevate it slightly above average, but it follows predictable M1 patterns without requiring novel insight. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors3.02e Two-dimensional constant acceleration: with vectors3.03r Friction: concept and vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) let acc⁰ be \(k(2\hat{i} + \hat{j})\) so magnitude is \(k(\sqrt{2^2 + 1^2}) = k\sqrt{5}\) | M2 | |
| \(\Rightarrow k = 3\), so \(a = 6\hat{i} + 3\hat{j}\) | A1 | |
| using \(v = u + at, v = (1 - 5\hat{j}) + t(6\hat{i} + 3\hat{j})\) | M1 | |
| so \(v = [(6t + 1)\hat{i} + (3t - 5)\hat{j}]\) m s⁻¹ | M1 A1 | |
| (b) speed² = \((6t + 1)^2 + (3t - 5)^2 = 45t^2 - 18t + 26\) | M1 A1 | |
| by calculus or completing square, \(t = \frac{1}{5}\) | M2 A1 | (11 marks) |
**(a)** let acc⁰ be $k(2\hat{i} + \hat{j})$ so magnitude is $k(\sqrt{2^2 + 1^2}) = k\sqrt{5}$ | M2 |
$\Rightarrow k = 3$, so $a = 6\hat{i} + 3\hat{j}$ | A1 |
using $v = u + at, v = (1 - 5\hat{j}) + t(6\hat{i} + 3\hat{j})$ | M1 |
so $v = [(6t + 1)\hat{i} + (3t - 5)\hat{j}]$ m s⁻¹ | M1 A1 |
**(b)** speed² = $(6t + 1)^2 + (3t - 5)^2 = 45t^2 - 18t + 26$ | M1 A1 |
by calculus or completing square, $t = \frac{1}{5}$ | M2 A1 | (11 marks)7. A particle has an initial velocity of $( \mathbf { i } - 5 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$ and is accelerating uniformly in the direction $( 2 \mathbf { i } + \mathbf { j } )$ where $\mathbf { i }$ and $\mathbf { j }$ are perpendicular unit vectors.
Given that the magnitude of the acceleration is $3 \sqrt { } 5 \mathrm {~ms} ^ { - 2 }$,
\begin{enumerate}[label=(\alph*)]
\item show that, after $t$ seconds, the velocity vector of the particle is
$$[ ( 6 t + 1 ) \mathbf { i } + ( 3 t - 5 ) \mathbf { j } ] \mathrm { ms } ^ { - 1 }$$
\item Using your answer to part (a), or otherwise, find the value of $t$ for which the speed of the particle is at its minimum.\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [11]}}