| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Parallel or perpendicular vectors condition |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question testing basic vector kinematics. Part (a) is simple position-velocity integration, part (b) requires equating j-components (standard 'due west' condition), and part (c) uses the routine parallel vectors condition (proportional components). All parts are textbook exercises with no problem-solving insight required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{p} = (-5\mathbf{i}+9\mathbf{j}) + t(\mathbf{i}-2\mathbf{j})\) | M1 A1 (2) | Allow slips but must have \(+\) sign |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(2 = 9 - 2t\) | M1 | Equate \(\mathbf{j}\) component to 2 |
| \(t = 3.5\) | A1 | |
| \(\mathbf{p} = (-5\mathbf{i}+9\mathbf{j})+3.5(\mathbf{i}-2\mathbf{j}) = (-1.5\mathbf{i}+2\mathbf{j})\) | M1 A1 (4) | Independent M1 for substituting \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\dfrac{2b-1}{5-2b} = \dfrac{1}{-2}\) | M1 A1 | Must be in \(b\) only; allow slips; or \(\frac{2b-1}{5-2b} = \pm\frac{2}{1}\) |
| \(b = -1.5\) | DM1 A1 (4) | DM1 dependent on first M1 for solving for \(b\) |
## Question 4:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $\mathbf{p} = (-5\mathbf{i}+9\mathbf{j}) + t(\mathbf{i}-2\mathbf{j})$ | M1 A1 (2) | Allow slips but must have $+$ sign |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|---------|
| $2 = 9 - 2t$ | M1 | Equate $\mathbf{j}$ component to 2 |
| $t = 3.5$ | A1 | |
| $\mathbf{p} = (-5\mathbf{i}+9\mathbf{j})+3.5(\mathbf{i}-2\mathbf{j}) = (-1.5\mathbf{i}+2\mathbf{j})$ | M1 A1 (4) | Independent M1 for substituting $t$ |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|---------|
| $\dfrac{2b-1}{5-2b} = \dfrac{1}{-2}$ | M1 A1 | Must be in $b$ only; allow slips; or $\frac{2b-1}{5-2b} = \pm\frac{2}{1}$ |
| $b = -1.5$ | DM1 A1 (4) | DM1 dependent on first M1 for solving for $b$ |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively and position vectors are given relative to a fixed origin $O$ ]
\end{enumerate}
A particle $P$ is moving with velocity $( \mathbf { i } - 2 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$. At time $t = 0$ hours, the position vector of $P$ is $( - 5 \mathbf { i } + 9 \mathbf { j } ) \mathrm { km }$. At time $t$ hours, the position vector of $P$ is $\mathbf { p } \mathrm { km }$.\\
(a) Find an expression for $\mathbf { p }$ in terms of $t$.
The point $A$ has position vector ( $3 \mathbf { i } + 2 \mathbf { j }$ ) km.\\
(b) Find the position vector of $P$ when $P$ is due west of $A$.
Another particle $Q$ is moving with velocity $[ ( 2 b - 1 ) \mathbf { i } + ( 5 - 2 b ) \mathbf { j } ] \mathrm { km } \mathrm { h } ^ { - 1 }$ where $b$ is a constant.
Given that the particles are moving along parallel lines,\\
(c) find the value of $b$.\\
\hfill \mbox{\textit{Edexcel M1 2016 Q4 [10]}}