| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Standard +0.3 This is a standard M1 kinematics question with constant velocity in 2D. Parts (a)-(b) are routine magnitude and bearing calculations. Part (c) requires setting up position equations with two velocity phases, which is straightforward bookwork. Part (d) needs identifying when the i-component returns to its initial value, requiring basic algebraic manipulation. All techniques are standard M1 fare with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Speed \(= \sqrt{5^2 + 8^2} \approx 9.43 \text{ ms}^{-1}\) | M1 A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Forming \(\arctan\frac{8}{5}\) or \(\arctan\frac{5}{8}\) or equivalent | M1 | |
| Bearing \(= 360 - \arctan\frac{5}{8}\) or \(270 + \arctan\frac{8}{5} = 328\) | DM1 A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| At \(t=3\), p.v. of \(P = (7-15)\mathbf{i} + (-10+24)\mathbf{j} = -8\mathbf{i} + 14\mathbf{j}\) | M1 A1 | |
| Hence \(-8\mathbf{i} + 14\mathbf{j} + 4(u\mathbf{i} + v\mathbf{j}) = \mathbf{0}\) | M1 | |
| \(\Rightarrow u = 2,\ v = -3.5\) | DM1 A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| p.v. of \(P\), \(t\) secs after changing course \(= (-8\mathbf{i}+14\mathbf{j}) + t(2\mathbf{i} - 3.5\mathbf{j})\) | M1 | |
| \(= 7\mathbf{i} + \ldots\) | DM1 | |
| Hence total time \(= 10.5\text{ s}\) | A1 (3) | Total: 13 |
## Question 6:
### Part (a)
| Working | Marks | Notes |
|---------|-------|-------|
| Speed $= \sqrt{5^2 + 8^2} \approx 9.43 \text{ ms}^{-1}$ | M1 A1 (2) | |
### Part (b)
| Working | Marks | Notes |
|---------|-------|-------|
| Forming $\arctan\frac{8}{5}$ or $\arctan\frac{5}{8}$ or equivalent | M1 | |
| Bearing $= 360 - \arctan\frac{5}{8}$ or $270 + \arctan\frac{8}{5} = 328$ | DM1 A1 (3) | |
### Part (c)
| Working | Marks | Notes |
|---------|-------|-------|
| At $t=3$, p.v. of $P = (7-15)\mathbf{i} + (-10+24)\mathbf{j} = -8\mathbf{i} + 14\mathbf{j}$ | M1 A1 | |
| Hence $-8\mathbf{i} + 14\mathbf{j} + 4(u\mathbf{i} + v\mathbf{j}) = \mathbf{0}$ | M1 | |
| $\Rightarrow u = 2,\ v = -3.5$ | DM1 A1 (5) | |
### Part (d)
| Working | Marks | Notes |
|---------|-------|-------|
| p.v. of $P$, $t$ secs after changing course $= (-8\mathbf{i}+14\mathbf{j}) + t(2\mathbf{i} - 3.5\mathbf{j})$ | M1 | |
| $= 7\mathbf{i} + \ldots$ | DM1 | |
| Hence total time $= 10.5\text{ s}$ | A1 (3) | **Total: 13** |
---6. [In this question, the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are due east and due north respectively.]
A particle $P$ is moving with constant velocity $( - 5 \mathbf { i } + 8 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. Find
\begin{enumerate}[label=(\alph*)]
\item the speed of $P$,
\item the direction of motion of $P$, giving your answer as a bearing.
At time $t = 0 , P$ is at the point $A$ with position vector ( $7 \mathbf { i } - 10 \mathbf { j }$ ) m relative to a fixed origin $O$. When $t = 3 \mathrm {~s}$, the velocity of $P$ changes and it moves with velocity $( u \mathbf { i } + v \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$, where $u$ and $v$ are constants. After a further 4 s , it passes through $O$ and continues to move with velocity ( $u \mathbf { i } + v \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
\item Find the values of $u$ and $v$.
\item Find the total time taken for $P$ to move from $A$ to a position which is due south of A.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2008 Q6 [13]}}