M1 June 2018 Q6
6. [In this question \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal unit vectors due east and due north respectively]
Two forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\) act on a particle \(P\) of mass 0.5 kg.
\[\mathbf{F}_1 = (4\mathbf{i} - 6\mathbf{j})\text{ N and }\mathbf{F}_2 = (p\mathbf{i} + q\mathbf{j})\text{ N}.\]Given that the resultant force of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) is in the same direction as \(-2\mathbf{i} - \mathbf{j}\),
Given that \(q = 3\)
| Scheme | Marks |
|---|---|
| \((4\mathbf{i} - 6\mathbf{j}) + (p\mathbf{i} + q\mathbf{j}) = (4 + p)\mathbf{i} + (q - 6)\mathbf{j}\) | M1 |
| \(\dfrac{(4 + p)}{(q - 6)} = \dfrac{2}{1}\) or \(-\dfrac{2}{1}\) (or \(\dfrac{1}{2}\) or \(-\dfrac{1}{2}\)) | DM1 A1 |
| \(2q - 12 = 4 + p\) | |
| \(p - 2q = -16\) GIVEN ANSWER | DM1 A1 |
| (5) |
Notes
Allow column vectors throughout
First M1 for adding the two forces, with i’s and j’s collected, seen or implied
Second DM1, dependent on first M1, for an equation in \(p\) and \(q\) only. Allow \(\dfrac{1}{2}\) or \(-\dfrac{1}{2}\) or \(-\dfrac{2}{1}\) instead of \(\dfrac{2}{1}\)
First A1 for a correct equation in any form
Third DM1, dependent on the second M1, for (at least)one correct intermediate line of working
Second A1 for correct given answer
| Scheme | Marks |
|---|---|
| \(q = 3 \Rightarrow p = -10\) | B1 |
| EITHER \(\ 0.5\mathbf{a} = -6\mathbf{i} - 3\mathbf{j}\) OR \(\ |\mathbf{R}| = \sqrt{(-6)^2 + (-3)^2}\) | M1 |
| \(\mathbf{a} = -12\mathbf{i} - 6\mathbf{j}\) \(= \sqrt{45}\) oe | A1 |
| \(|\mathbf{a}| = \sqrt{(-12)^2 + (-6)^2}\) \(0.5a = \sqrt{45}\) | M1 |
| \(a = \sqrt{180} = 13.4\) m s\(^{-2}\) \(a = \sqrt{180} = 13.4\) m s\(^{-2}\) | A1 |
| (5) |
Notes
Allow column vectors throughout
B1 for \(p = -10\) seen or implied
EITHER
First M1 for use of \(\mathbf{F} = 0.5\mathbf{a}\) with their resultant force (must be a sum of the two forces)
First A1 for \(\mathbf{a} = -12\mathbf{i} - 6\mathbf{j}\)
Second M1 (independent) for finding magnitude of their \(\mathbf{a}\)
Second A1 for \(\sqrt{180}\) oe or 13.4 or better
OR
First M1 for finding the magnitude of their resultant force \(\mathbf{R}\) (must be a sum of the two forces) \(R = \sqrt{(-6)^2 + (-3)^2}\)
First A1 for \(\sqrt{45}\) oe
Second M1 for using \(R = 0.5a\) to find \(a\)
Second A1 for \(a = 2\sqrt{45}\) oe 13.4 m s\(^{-2}\) or better
| Scheme | Marks |
|---|---|
| e.g. \(\tan\theta = \dfrac{12}{6} \Rightarrow \theta = 63.4^\circ\) | M1A1 |
| Bearing \(= 180^\circ + 63.4^\circ = 243^\circ\) (nearest degree) | A1cao |
| (3) | |
| (13 marks) |
Notes
Allow column vectors throughout
M1 for use of a relevant trig ratio from their \(\mathbf{a}\) or their \(\mathbf{R}\) (may not be the sum of the two forces) or \(-2\mathbf{i} - \mathbf{j}\)
First A1 for any relevant correct angle coming from a correct \(\mathbf{a}\) or \(\mathbf{R}\) or from \(-2\mathbf{i} - \mathbf{j}\)
Second A1 for 243