| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Parallel or perpendicular vectors condition |
| Difficulty | Standard +0.3 This question involves standard vector mechanics (SUVAT with vectors) and parallel vector conditions. Part (a) requires recognizing that parallel paths mean direction vectors are scalar multiples, which is straightforward. Parts (b)(i) and (b)(ii) use routine kinematic equations and checking collinearity by testing if a point lies on a line. All techniques are standard A-level mechanics with no novel insight required, making it slightly easier than average. |
| Spec | 1.10g Problem solving with vectors: in geometry1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| States/uses direction of motion is \(\begin{bmatrix}3\\-4\end{bmatrix}\) or \(\begin{bmatrix}9\\c+1\end{bmatrix}\), or gradient \(-\frac{4}{3}\) or \(\frac{c+1}{9}\) | M1 | AO 3.1a |
| Correct vector equation e.g. \(\begin{bmatrix}10\\c\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix} + k\begin{bmatrix}3\\-4\end{bmatrix}\), or both direction vectors/gradients, or correct Cartesian equation e.g. \(y+1=-\frac{4}{3}(x-1)\) | A1 | AO 1.1b |
| Eliminates parameter in vector equation, or equates gradients/reciprocals \(\frac{c+1}{9}=-\frac{4}{3}\), or substitutes \(x=10\) into Cartesian equation | M1 | AO 1.1a |
| Shows \(c=-13\) AG. Typical solution: \(\begin{bmatrix}10\\c\end{bmatrix}-\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}9\\c+1\end{bmatrix}\), \(\begin{bmatrix}9\\c+1\end{bmatrix}=k\begin{bmatrix}3\\-4\end{bmatrix}\), \(k=3\), \(c+1=-12\), \(\Rightarrow c=-13\) | A1 | AO 1.1b. A correct verification method using given \(c=-13\) scores max M1A1M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r} = \begin{bmatrix}-4\\5\end{bmatrix} + \frac{1}{2}\begin{bmatrix}3\\-4\end{bmatrix}t^2\) metres | B1 | AO 2.2a. Condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equates position vector from (b)(i) to one of the two known position vectors for \(Q\). Position vector must be quadratic in \(t\) for both components. Or substitutes known point for \(P\)/\(Q\) into Cartesian equation for path of \(Q\)/\(P\). Or forms Cartesian equations for both paths. Or calculates difference between \((-4\mathbf{i}+5\mathbf{j})\) and \((\mathbf{i}-\mathbf{j})\) or between \((-4\mathbf{i}+5\mathbf{j})\) and \((10\mathbf{i}-13\mathbf{j})\) | M1 | AO 3.1b |
| Obtains \(t^2=\frac{10}{3}\) or \(3\); i.e. \(t=\sqrt{\frac{10}{3}}=1.82...\) or \(\sqrt{3}=1.73...\). Or shows \(y\neq 5\) for \(x=-4\). Or writes two correct Cartesian equations e.g. \(y=-\frac{4}{3}x+\frac{1}{3}\) and \(y=-\frac{4}{3}x-\frac{1}{3}\). Or compares two appropriate direction vectors. Typical: \(1=1.5t^2-4 \Rightarrow t^2=\frac{10}{3}\); \(5-2t^2=-1 \Rightarrow t^2=3\) | A1 | AO 1.1b |
| Completes reasoned argument explaining inconsistency and deduces paths are not collinear. Since the \(t^2\) values are not the same, no single value of \(t\) exists which satisfies both components. Therefore paths are not collinear. | R1 | AO 2.1. CSO |
# Question 16(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| States/uses direction of motion is $\begin{bmatrix}3\\-4\end{bmatrix}$ or $\begin{bmatrix}9\\c+1\end{bmatrix}$, or gradient $-\frac{4}{3}$ or $\frac{c+1}{9}$ | M1 | AO 3.1a |
| Correct vector equation e.g. $\begin{bmatrix}10\\c\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix} + k\begin{bmatrix}3\\-4\end{bmatrix}$, or both direction vectors/gradients, or correct Cartesian equation e.g. $y+1=-\frac{4}{3}(x-1)$ | A1 | AO 1.1b |
| Eliminates parameter in vector equation, or equates gradients/reciprocals $\frac{c+1}{9}=-\frac{4}{3}$, or substitutes $x=10$ into Cartesian equation | M1 | AO 1.1a |
| Shows $c=-13$ **AG**. Typical solution: $\begin{bmatrix}10\\c\end{bmatrix}-\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}9\\c+1\end{bmatrix}$, $\begin{bmatrix}9\\c+1\end{bmatrix}=k\begin{bmatrix}3\\-4\end{bmatrix}$, $k=3$, $c+1=-12$, $\Rightarrow c=-13$ | A1 | AO 1.1b. A correct verification method using given $c=-13$ scores max M1A1M0A0 |
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# Question 16(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = \begin{bmatrix}-4\\5\end{bmatrix} + \frac{1}{2}\begin{bmatrix}3\\-4\end{bmatrix}t^2$ metres | B1 | AO 2.2a. Condone missing units |
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# Question 16(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equates position vector from (b)(i) to one of the two known position vectors for $Q$. Position vector must be quadratic in $t$ for both components. Or substitutes known point for $P$/$Q$ into Cartesian equation for path of $Q$/$P$. Or forms Cartesian equations for both paths. Or calculates difference between $(-4\mathbf{i}+5\mathbf{j})$ and $(\mathbf{i}-\mathbf{j})$ or between $(-4\mathbf{i}+5\mathbf{j})$ and $(10\mathbf{i}-13\mathbf{j})$ | M1 | AO 3.1b |
| Obtains $t^2=\frac{10}{3}$ or $3$; i.e. $t=\sqrt{\frac{10}{3}}=1.82...$ or $\sqrt{3}=1.73...$. Or shows $y\neq 5$ for $x=-4$. Or writes two correct Cartesian equations e.g. $y=-\frac{4}{3}x+\frac{1}{3}$ and $y=-\frac{4}{3}x-\frac{1}{3}$. Or compares two appropriate direction vectors. Typical: $1=1.5t^2-4 \Rightarrow t^2=\frac{10}{3}$; $5-2t^2=-1 \Rightarrow t^2=3$ | A1 | AO 1.1b |
| Completes reasoned argument explaining inconsistency and deduces paths are not collinear. Since the $t^2$ values are not the same, no single value of $t$ exists which satisfies both components. Therefore paths are not collinear. | R1 | AO 2.1. CSO |
---16 Two particles, $P$ and $Q$, move in the same horizontal plane.
Particle $P$ is initially at rest at the point with position vector $( - 4 \mathbf { i } + 5 \mathbf { j } )$ metres and moves with constant acceleration $( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$
Particle $Q$ moves in a straight line, passing through the points with position vectors $( \mathbf { i } - \mathbf { j } )$ metres and $( 10 \mathbf { i } + c \mathbf { j } )$ metres.\\
$P$ and $Q$ are moving along parallel paths.\\
16
\begin{enumerate}[label=(\alph*)]
\item Show that $c = - 13$\\
16
\item (i) Find an expression for the position vector of $P$ at time $t$ seconds.\\
16 (b) (ii) Hence, prove that the paths of $P$ and $Q$ are not collinear.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q16 [8]}}