| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Moderate -0.3 This is a straightforward vector kinematics problem requiring application of v = u + at in component form, followed by basic trigonometry to find bearings. The two parts involve routine calculations with no conceptual challenges beyond standard M1 material, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = (10\mathbf{i} + 4\mathbf{j}) + 6(-2\mathbf{i} + 3\mathbf{j})\) | M1 | Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) with \(t = 6\) |
| \(= -2\mathbf{i} + 22\mathbf{j}\) | A1 | |
| \(\tan\theta = \pm\frac{22}{2}\) or \(\tan\theta = \pm\frac{2}{22}\) | M1 | Correct use of trig to find a relevant angle for their \(\mathbf{v}\) |
| \(\theta = 85°\) or \(5°\) | A1 | Seen or implied |
| bearing is \(355°\) | A1 | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = (10\mathbf{i} + 4\mathbf{j}) + t(-2\mathbf{i} + 3\mathbf{j})\) | M1 | Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) |
| \(= (10 - 2t)\mathbf{i} + (4 + 3t)\mathbf{j}\) | A1 | Correct unsimplified |
| \((10 - 2t) = (4 + 3t)\) | DM1 | Equate coefficients to give equation in \(t\) only |
| \(t = 1.2\) | A1 | |
| (4) | ||
| [9] |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = (10\mathbf{i} + 4\mathbf{j}) + 6(-2\mathbf{i} + 3\mathbf{j})$ | M1 | Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ with $t = 6$ |
| $= -2\mathbf{i} + 22\mathbf{j}$ | A1 | |
| $\tan\theta = \pm\frac{22}{2}$ or $\tan\theta = \pm\frac{2}{22}$ | M1 | Correct use of trig to find a relevant angle for their $\mathbf{v}$ |
| $\theta = 85°$ or $5°$ | A1 | Seen or implied |
| bearing is $355°$ | A1 | |
| | **(5)** | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = (10\mathbf{i} + 4\mathbf{j}) + t(-2\mathbf{i} + 3\mathbf{j})$ | M1 | Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ |
| $= (10 - 2t)\mathbf{i} + (4 + 3t)\mathbf{j}$ | A1 | Correct unsimplified |
| $(10 - 2t) = (4 + 3t)$ | DM1 | Equate coefficients to give equation in $t$ only |
| $t = 1.2$ | A1 | |
| | **(4)** | |
| | **[9]** | |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively.]
\end{enumerate}
A particle $P$ moves with constant acceleration $( - 2 \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$. At time $t$ seconds, the velocity of $P$ is $\mathbf { v m ~ s } ^ { - 1 }$. When $t = 0 , \mathbf { v } = 10 \mathbf { i } + 4 \mathbf { j }$.\\
(a) Find the direction of motion of $P$ when $t = 6$, giving your answer as a bearing to the nearest degree.\\
(b) Find the value of $t$ when $P$ is moving north east.\\
\hfill \mbox{\textit{Edexcel M1 2017 Q6 [9]}}