| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Moderate -0.3 This is a straightforward M1 question testing standard vector mechanics and conservation of momentum. Parts (a)-(d) involve routine vector arithmetic (distance formula, velocity = displacement/time, position vector equation), while part (e) applies basic momentum conservation to a collision. All techniques are standard textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors3.02a Kinematics language: position, displacement, velocity, acceleration6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\overrightarrow{XY} = 8\mathbf{i} + 6\mathbf{j}\) → \(XY = \sqrt{8^2 + 6^2} = 10\) m | M1 A1 | |
| (b) \(\mathbf{v}_P = (8\mathbf{i} + 6\mathbf{j}) + 4 = (2\mathbf{i} + 1.5\mathbf{j})\) ms\(^{-1}\) | M1 A1 A1 | |
| (c) \(4\mathbf{i} - 5\mathbf{j} + t(2\mathbf{i} + 1.5\mathbf{j})\) or \((4 + 2t)\mathbf{i} + (1.5t - 5)\mathbf{j}\) | M1 A1 | |
| (d) \(4 + 2t = 16\) and \(1.5t - 5 = 4\) when \(t = 6\) | M1 A1 | |
| (e) Momentum: \(2(2.5) + 4(0) = 6v\) → \(v = \frac{5}{6}\) | M1 A1 | |
| \(\mathbf{v} = \frac{2}{6}(2\mathbf{i} + 1.5\mathbf{j}) + 2.5 = \frac{2}{3}\mathbf{i} + \frac{1}{2}\mathbf{j}\) | M1 A1 A1 | 14 marks |
**(a)** $\overrightarrow{XY} = 8\mathbf{i} + 6\mathbf{j}$ → $XY = \sqrt{8^2 + 6^2} = 10$ m | M1 A1 |
**(b)** $\mathbf{v}_P = (8\mathbf{i} + 6\mathbf{j}) + 4 = (2\mathbf{i} + 1.5\mathbf{j})$ ms$^{-1}$ | M1 A1 A1 |
**(c)** $4\mathbf{i} - 5\mathbf{j} + t(2\mathbf{i} + 1.5\mathbf{j})$ or $(4 + 2t)\mathbf{i} + (1.5t - 5)\mathbf{j}$ | M1 A1 |
**(d)** $4 + 2t = 16$ and $1.5t - 5 = 4$ when $t = 6$ | M1 A1 |
**(e)** Momentum: $2(2.5) + 4(0) = 6v$ → $v = \frac{5}{6}$ | M1 A1 |
$\mathbf{v} = \frac{2}{6}(2\mathbf{i} + 1.5\mathbf{j}) + 2.5 = \frac{2}{3}\mathbf{i} + \frac{1}{2}\mathbf{j}$ | M1 A1 A1 | 14 marks |Relative to a fixed origin $O$, the points $X$ and $Y$ have position vectors $(4\mathbf{i} - 5\mathbf{j})$ m and $(12\mathbf{i} + \mathbf{j})$ m respectively, where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular unit vectors.
\begin{enumerate}[label=(\alph*)]
\item Find the distance $XY$. [2 marks]
\end{enumerate}
A particle $P$ of mass $2$ kg moves from $X$ to $Y$ in $4$ seconds, in a straight line at a constant speed.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the velocity vector of $P$ is $(2\mathbf{i} + 1.5\mathbf{j}) \text{ ms}^{-1}$. [3 marks]
\end{enumerate}
The particle continues beyond $Y$ with the same constant velocity.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Write down an expression for the position vector of $P$ $t$ seconds after leaving $X$. [2 marks]
\item Find the value of $t$ when $P$ is at the point with position vector $(16\mathbf{i} + 4\mathbf{j})$ m. [2 marks]
\end{enumerate}
When it is moving with the same constant speed, $P$ collides directly with another particle $Q$, of mass $4$ kg, which is at rest. $P$ and $Q$ coalesce and move together as a single particle.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item Find the velocity vector of the combined particle after the collision. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [14]}}