| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 5 |
| 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 application of the SUVAT equation v = u + at in vector form. Students must equate components and solve two simultaneous equations for the initial velocity components, then find the magnitude. It requires standard technique with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(-6\mathbf{i} + \mathbf{j} = \mathbf{u} + 3(2\mathbf{i} - 5\mathbf{j})\) | M1 A1 | |
| \(\Rightarrow \mathbf{u} = -12\mathbf{i} + 16\mathbf{j}\) | A1 cso | |
| \(\Rightarrow u = \sqrt{(-12)^2 + 16^2} = 20\) | M1 A1 | [5] |
## Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $-6\mathbf{i} + \mathbf{j} = \mathbf{u} + 3(2\mathbf{i} - 5\mathbf{j})$ | M1 A1 | |
| $\Rightarrow \mathbf{u} = -12\mathbf{i} + 16\mathbf{j}$ | A1 cso | |
| $\Rightarrow u = \sqrt{(-12)^2 + 16^2} = 20$ | M1 A1 | **[5]** |
---\begin{enumerate}
\item A particle $P$ moves with constant acceleration $( 2 \mathbf { i } - 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$. At time $t = 0 , P$ has speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At time $t = 3 \mathrm {~s} , P$ has velocity $( - 6 \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\end{enumerate}
Find the value of $u$.\\
(5)\\
\hfill \mbox{\textit{Edexcel M1 2009 Q1 [5]}}