M2 January 2013 Q4
4. At time \(t\) seconds the velocity of a particle \(P\) is \([(4t - 5)\mathbf{i} + 3\mathbf{j}]\) m s\(^{-1}\). When \(t = 0\), the position vector of \(P\) is \((2\mathbf{i} + 5\mathbf{j})\) m, relative to a fixed origin \(O\).
A second particle \(Q\) moves with constant velocity \((-2\mathbf{i} + c\mathbf{j})\) m s\(^{-1}\). When \(t = 0\), the position vector of \(Q\) is \((11\mathbf{i} + 2\mathbf{j})\) m. The particles \(P\) and \(Q\) collide at the point with position vector \((d\mathbf{i} + 14\mathbf{j})\) m.
| Scheme | Marks |
|---|---|
| \(t = \dfrac{5}{4}\) | M1 |
| (1) |
Notes
M1 1.25
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = \left(2t^2 - 5t\right)\mathbf{i} + 3t\mathbf{j}\ (+\mathbf{c})\) | M1 A1 |
| \(t = 0\ \ \ 2\mathbf{i} + 5\mathbf{j} = \mathbf{c}\) | DM1 |
| \(\mathbf{r} = \left(2t^2 - 5t\right)\mathbf{i} + 3t\mathbf{j} + \left(2\mathbf{i} + 5\mathbf{j}\right)\) \((2t^2 - 5t + 2)\mathbf{i} + (3t + 5)\mathbf{j}\) | A1 |
| (4) |
Notes
M1 Integrate the velocity vector
A1 NB Also correct to use suvat with \(\boldsymbol{a} = 4\mathbf{i}\) and \(\boldsymbol{u} = -5\mathbf{i} + 3\mathbf{j}\). Correct
DM1 Use \(\mathbf{r}_0\) to find \(C\)
A1 oe
(The M1 for integrating is not printed in the marks column of the mark scheme; the part is worth 4 marks.)
| Scheme | Marks |
|---|---|
| \(\mathbf{r}_Q = 11\mathbf{i} + 2\mathbf{j} - 2t\mathbf{i} + ct\mathbf{j}\) \((11 - 2t)\mathbf{i} + (2 + ct)\mathbf{j}\) | B1 |
| \(\mathbf{r}_P = \left(2t^2 - 5t + 2\right)\mathbf{i} + \left(3t + 5\right)\mathbf{j}\) | |
| \(\mathbf{r}_Q = \mathbf{r}_P = d\mathbf{i} + 14\mathbf{j}\) | |
| \(3t + 5 = 14\) | M1 |
| \(t = 3\) | A1 |
| \(2 + ct = 14 \Rightarrow c = 4\) | A1 ft |
| \(d = 11 - 2 \times 3 = 5\) or \(d = 2 \times 3^2 - 5 \times 3 + 2 \Rightarrow d = 5\) | A1 ft |
| (5) | |
| (10 marks) |
Notes
B1 Correct \(\mathbf{j}\) component of \(\mathbf{r}_Q\). Do not actually require the whole thing - can answer the Q by considering only the \(\mathbf{j}\) component.
M1 Form an equation in \(t\) only
A1 ft Their \(t\)
A1 ft Their \(t\)
Printed beside \(3t + 5 = 14\) (using the \(\mathbf{i}\) components): \(2t^2 - 3t - 9,\ \ (2t + 3)(t - 3) = 0,\ \ t = 3\) A1 ft
Alt: \(2t^2 - 5t + 2 = 11 - 2t = d \Rightarrow t = \dfrac{11 - d}{2}\)
\(2\left(\dfrac{11 - d}{2}\right)^2 - 5\left(\dfrac{11 - d}{2}\right) + 2 = d,\)
\(d^2 - 19d + 70 = 0 = (d - 5)(d - 14)\)
(The final A1 ft for \(d\) is not printed in the marks column of the mark scheme; the part is worth 5 marks.)