M1 June 2014 (R) Q2
2. Two forces \((4\mathbf{i} - 2\mathbf{j})\) N and \((2\mathbf{i} + q\mathbf{j})\) N act on a particle \(P\) of mass 1.5 kg. The resultant of these two forces is parallel to the vector \((2\mathbf{i} + \mathbf{j})\).
At time \(t = 0\), \(P\) is moving with velocity \((-2\mathbf{i} + 4\mathbf{j})\) m s\(^{-1}\).
| Scheme | Marks |
|---|---|
| \((4\mathbf{i} - 2\mathbf{j}) + (2\mathbf{i} + q\mathbf{j}) = (6\mathbf{i} + (q - 2)\mathbf{j})\) | M1A1 |
| \(6 = 2(q - 2)\) ratio 2:1 | DM1 |
| \(q = 5\) | A1 |
| (4) |
Notes
First M1 for \((4\mathbf{i} - 2\mathbf{j}) + (2\mathbf{i} + q\mathbf{j})\)
First A1 for \((6\mathbf{i} + (q - 2)\mathbf{j})\) (seen or implied)
Second M1, dependent on first M1, for using ‘parallel to \((2\mathbf{i} + \mathbf{j})\)’ to obtain an equation in \(q\) only.
Second A1 for \(q = 5\)
| Scheme | Marks |
|---|---|
| \(6\mathbf{i} + 3\mathbf{j} = 1.5\mathbf{a}\) | M1 |
| \(\mathbf{a} = (4\mathbf{i} + 2\mathbf{j})\) m s\(^{-2}\) | A1 |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t = (-2\mathbf{i} + 4\mathbf{j}) + 2(4\mathbf{i} + 2\mathbf{j})\) | M1 |
| \(= 6\mathbf{i} + 8\mathbf{j}\) | A1ft |
| speed \(= \sqrt{6^2 + 8^2}\) | M1 |
| \(= 10\) m s\(^{-1}\) | A1 |
| (6) | |
| (10 marks) |
Notes
First M1 for their resultant force \(= 1.5\mathbf{a}\)
First A1 for \(\mathbf{a} = 4\mathbf{i} + 2\mathbf{j}\)
Second M1 for \((-2\mathbf{i} + 4\mathbf{j}) + 2 \times\) (their \(\mathbf{a}\)) (M0 if force is used instead of \(\mathbf{a}\))
Second A1 ft for their velocity at \(t = 2\)
Third M1 for finding the magnitude of their velocity at \(t = 2\)
Third A1 for 10 (m s\(^{-1}\))
N.B. In (b), if they use scalars throughout, M0A0M0A0M0A0