| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Velocity from two position vectors |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on vector motion requiring routine application of position vectors, velocity calculation, and interception conditions. Parts (a)-(c) involve straightforward algebraic manipulation, while part (d) requires calculating and comparing speeds—all standard textbook techniques with no novel insight needed. Slightly easier than average due to clear structure and step-by-step guidance. |
| Spec | 1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\mathbf{v} = \frac{8\mathbf{i} + 11\mathbf{j} - (3\mathbf{i} - 4\mathbf{j})}{2.5}\) or any equivalent | M1 A1 | (3 marks) |
| \(\mathbf{v} = 2\mathbf{i} + 6\mathbf{j}\) | A1 | |
| (b) \(\mathbf{b} = 3\mathbf{i} - 4\mathbf{j} + v\) ft their \(\mathbf{v}\) | M1 A1 ft | (3 marks) |
| \(= 3\mathbf{i} - 4\mathbf{j} + (2\mathbf{i} + 6\mathbf{j})\) | A1cao | |
| (c) i component: \(-9 + 6t = 3 + 2t\) | M1 | (5 marks) |
| \(t = 3\) | M1 A1 | |
| j component: \(20 + 3\lambda = -4 + 18\) | M1 | |
| \(\lambda = -2\) | A1 | |
| (d) \(v_B = \sqrt{(2^2 + 6^2)}\) or \(v_C = \sqrt{(6^2 + (-2)^2)}\) | M1 | (3 marks) |
| Both correct | A1 | |
| The speeds of \(B\) and \(C\) are the same | cso A1 | |
| [14] |
**(a)** $\mathbf{v} = \frac{8\mathbf{i} + 11\mathbf{j} - (3\mathbf{i} - 4\mathbf{j})}{2.5}$ or any equivalent | M1 A1 | (3 marks)
$\mathbf{v} = 2\mathbf{i} + 6\mathbf{j}$ | A1 |
**(b)** $\mathbf{b} = 3\mathbf{i} - 4\mathbf{j} + v$ ft their $\mathbf{v}$ | M1 A1 ft | (3 marks)
$= 3\mathbf{i} - 4\mathbf{j} + (2\mathbf{i} + 6\mathbf{j})$ | A1cao |
**(c)** i component: $-9 + 6t = 3 + 2t$ | M1 | (5 marks)
$t = 3$ | M1 A1 |
j component: $20 + 3\lambda = -4 + 18$ | M1 |
$\lambda = -2$ | A1 |
**(d)** $v_B = \sqrt{(2^2 + 6^2)}$ or $v_C = \sqrt{(6^2 + (-2)^2)}$ | M1 | (3 marks)
Both correct | A1 |
The speeds of $B$ and $C$ are the same | cso A1 |
| | [14] |\begin{enumerate}
\item A boat $B$ is moving with constant velocity. At noon, $B$ is at the point with position vector $( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { km }$ with respect to a fixed origin $O$. At 1430 on the same day, $B$ is at the point with position vector $( 8 \mathbf { i } + 11 \mathbf { j } ) \mathrm { km }$.\\
(a) Find the velocity of $B$, giving your answer in the form $p \mathbf { i } + q \mathbf { j }$.
\end{enumerate}
At time $t$ hours after noon, the position vector of $B$ is $\mathbf { b } \mathrm { km }$.\\
(b) Find, in terms of $t$, an expression for $\mathbf { b }$.
Another boat $C$ is also moving with constant velocity. The position vector of $C$, $\mathbf { c k m }$, at time $t$ hours after noon, is given by
$$\mathbf { c } = ( - 9 \mathbf { i } + 20 \mathbf { j } ) + t ( 6 \mathbf { i } + \lambda \mathbf { j } ) ,$$
where $\lambda$ is a constant. Given that $C$ intercepts $B$,\\
(c) find the value of $\lambda$,\\
(d) show that, before $C$ intercepts $B$, the boats are moving with the same speed.\\
\hfill \mbox{\textit{Edexcel M1 2007 Q7 [14]}}