| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 10 |
| 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 question requiring standard application of v = u + at in component form. Part (a) requires setting the j-component to zero, part (b) needs magnitude calculation, and part (c) involves basic trigonometry. All steps are routine M1 techniques with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(v = u + at\): \(\mathbf{v} = (-2+2t)\mathbf{i} + (7-3t)\mathbf{j}\) | M1 A1 | |
| \(\mathbf{v}\) parallel to \(\mathbf{i} \Rightarrow 7-3t=0 \Rightarrow t=2\frac{1}{3}\) s | M1 A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(t=3\), \(\mathbf{v} = 4\mathbf{i} - 2\mathbf{j}\) | M1 | |
| \( | \mathbf{v} | = \sqrt{20} \approx 4.47\) m s\(^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Angle \(= (\arctan\frac{2}{4}) + 90° = 116.6°\) | M1, M1 A1 | accept 117° |
| \([\text{or } 180° - (\arctan\frac{4}{2})]\) | [M1 M1 A1] | (3) |
# Question 5:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $v = u + at$: $\mathbf{v} = (-2+2t)\mathbf{i} + (7-3t)\mathbf{j}$ | M1 A1 | |
| $\mathbf{v}$ parallel to $\mathbf{i} \Rightarrow 7-3t=0 \Rightarrow t=2\frac{1}{3}$ s | M1 A1 | (4) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $t=3$, $\mathbf{v} = 4\mathbf{i} - 2\mathbf{j}$ | M1 | |
| $|\mathbf{v}| = \sqrt{20} \approx 4.47$ m s$^{-1}$ | M1 A1 | (3) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| Angle $= (\arctan\frac{2}{4}) + 90° = 116.6°$ | M1, M1 A1 | accept 117° |
| $[\text{or } 180° - (\arctan\frac{4}{2})]$ | [M1 M1 A1] | (3) |
---5. A particle $P$ moves with constant acceleration $( 2 \mathbf { i } - 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$. At time $t$ seconds, its velocity is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$. When $t = 0 , \mathbf { v } = - 2 \mathbf { i } + 7 \mathbf { j }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$ when $P$ is moving parallel to the vector $\mathbf { i }$.
\item Find the speed of $P$ when $t = 3$.
\item Find the angle between the vector $\mathbf { j }$ and the direction of motion of $P$ when $t = 3$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q5 [10]}}