June 2022 Paper 2 Q16
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 \((3\mathbf{i} - 4\mathbf{j})\) m s−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.
(a) Show that \(c = -13\) [4 marks]
(b)
(i) Find an expression for the position vector of \(P\) at time \(t\) seconds. [1 mark]
(ii) Hence, prove that the paths of \(P\) and \(Q\) are not collinear. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| States or uses the direction of motion is \(\begin{bmatrix} 3 \\ -4 \end{bmatrix}\) or \(\begin{bmatrix} 9 \\ c + 1 \end{bmatrix}\) Or States or uses the gradient of the direction of motion is \(-\dfrac{4}{3}\) or \(\dfrac{c + 1}{9}\) | M1 | 3.1a |
| Obtains a correct vector equation eg \(\begin{bmatrix} 10 \\ c \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \end{bmatrix} + k\begin{bmatrix} 3 \\ -4 \end{bmatrix}\) OE Or Obtains both gradients or both direction vectors \(\begin{bmatrix} 3 \\ -4 \end{bmatrix}\), \(\begin{bmatrix} 9 \\ c + 1 \end{bmatrix}\) or Obtains a correct cartesian equation for \(Q\). eg \(y + 1 = -\dfrac{4}{3}(x - 1)\) | A1 | 1.1b |
| Obtains or eliminates parameter in their vector equation Or Equates gradients or the reciprocals \(\dfrac{c + 1}{9} = -\dfrac{4}{3}\) Or substitutes \(x\) = 10 into their cartesian equation | M1 | 1.1a |
| Shows that \(c = -13\) AG A correct verification method using the given \(c = -13\) scores a maximum of M1A1M0A0 | A1 | 1.1b |
| (4) |
Typical solution
\[\begin{bmatrix} 10 \\ c \end{bmatrix} - \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 9 \\ c + 1 \end{bmatrix}\]\[\begin{bmatrix} 9 \\ c + 1 \end{bmatrix} = k\begin{bmatrix} 3 \\ -4 \end{bmatrix}\]\[k = 3\]\[c + 1 = -12\]\[\Rightarrow c = -13\]| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \(\begin{bmatrix} -4 \\ 5 \end{bmatrix} + \dfrac{1}{2}\begin{bmatrix} 3 \\ -4 \end{bmatrix}t^2\) metres OE Condone missing units | B1 | 2.2a |
| (1) | ||
| (ii) Equates their position vector from (b)(i) to one of the two known position vectors given for \(Q\). Their position vector must be quadratic in \(t\) for both components Or Substitutes a known point for \(P\) into their Cartesian equation for the path of \(Q\) from part (a) OE Or Substitutes a known point for \(Q\) into their Cartesian equation for the path of \(P\) from part (a) OE Or forms Cartesian equations for the path of \(P\) and the path of \(Q\) Or Calculates the 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 | 3.1b |
| Obtains \(t^2 = \dfrac{10}{3}\) or 3 or \(t = \sqrt{\dfrac{10}{3}} = 1.82\ldots\) or \(\sqrt{3} = 1.73\ldots\) Or Shows that \(y \ne 5\) for \(x = -4\) OE Or Writes the two correct cartesian equations in a comparable form eg \(y = -\dfrac{4}{3}x + \dfrac{1}{3}\) and \(y = -\dfrac{4}{3}x - \dfrac{1}{3}\) Or Compares two appropriate direction vectors | A1 | 1.1b |
| Completes reasoned argument by explaining that there is an inconsistency and deduces that paths are not collinear CSO | R1 | 2.1 |
| (3) | ||
| (8 marks) |
Typical solution
(i)
\[\mathbf{r} = \begin{bmatrix} -4 \\ 5 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 3 \\ -4 \end{bmatrix}t^2\](ii)
\[\begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -4 \\ 5 \end{bmatrix} + \frac{1}{2}t^2\begin{bmatrix} 3 \\ -4 \end{bmatrix}\]\[1 = 1.5t^2 - 4\]\[t^2 = \frac{10}{3}\]\[5 - 2t^2 = -1\]\[t^2 = 3\]Since the \(t^2\) values are not the same no single value of \(t\) exists which satisfies both components.
Therefore, the paths are not collinear.