M5 June 2016 Q1
1. [In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors in a horizontal plane.]
A bead \(P\) of mass 0.4 kg is threaded on a smooth straight horizontal wire. The wire lies along the line with vector equation \(\mathbf{r} = (\mathbf{i} + 2\mathbf{j}) + \lambda(-2\mathbf{i} + 3\mathbf{j})\). The bead is initially at rest at the point \(A\) with position vector \((-\mathbf{i} + 5\mathbf{j})\) m. A constant horizontal force \((0.5\mathbf{i} + \mathbf{j})\) N acts on \(P\) and moves it along the wire to the point \(B\). At \(B\) the speed of \(P\) is 5 m s\(^{-1}\).
Find the position vector of \(B\). (7)
| Scheme | Marks |
|---|---|
| Let pv of \(B = a\mathbf{i} + b\mathbf{j}\) | |
| \((0.5\mathbf{i} + \mathbf{j}) \cdot ((a + 1)\mathbf{i} + (b - 5)\mathbf{j}) = \dfrac{1}{2} \times 0.4 \times 5^2\) | M1 A1 |
| \(a + 2b = 19\) | M1 A1 |
| \(a = 1 - 2\lambda;\ b = 2 + 3\lambda\) | B1 |
| solving for \(a\) and \(b\) | M1 |
| pv of \(B = (-6\mathbf{i} + 12.5\mathbf{j})\) m | A1 |
| (7 marks) |
Notes
First M1 for attempt at using work-energy principle, with usual rules
First A1 for a correct unprocessed equation
Second M1 for producing an equation in \(a\) and \(b\) (seen or implied)
Second A1 for a correct equation
B1 for \(a = 1 - 2\lambda\) and \(b = 2 + 3\lambda\) seen or implied
Third M1 for solving for \(a\) and \(b\)
Third A1 for correct answer (must be a vector)
OR
| \((0.5\mathbf{i} + \mathbf{j}) \cdot (-2\lambda\mathbf{i} + 3\lambda\mathbf{j}) = \dfrac{1}{2} \times 0.4 \times 5^2\) | M1 A1 |
| \(-\lambda + 3\lambda = 5\) | M1 A1 |
| \(\lambda = \dfrac{5}{2}\) | B1 |
| pv of \(B = (-\mathbf{i} + 5\mathbf{j}) + \dfrac{5}{2}(-2\mathbf{i} + 3\mathbf{j})\) | M1 |
| \(= (-6\mathbf{i} + 12.5\mathbf{j})\) m | A1 |
| (7 marks) |
Alternative using forces and acceleration
| Resolving along the wire: \(\ (0.5\mathbf{i} + \mathbf{j}) \cdot \dfrac{1}{\sqrt{13}}(-2\mathbf{i} + 3\mathbf{j}) = 0.4a\) | M1 |
| \(\dfrac{5}{\sqrt{13}} = a\) | A1 |
| \(v^2 = u^2 + 2as:\quad 5^2 = 2 \times \dfrac{5}{\sqrt{13}}s\) | M1 |
| \(\dfrac{5\sqrt{13}}{2} = s\) | A1 |
| \(\mathbf{AB} = \dfrac{5}{2}(-2\mathbf{i} + 3\mathbf{j}) = (-5\mathbf{i} + 7.5\mathbf{j})\) | B1 |
| pv of \(B = (-\mathbf{i} + 5\mathbf{j}) + (-5\mathbf{i} + 7.5\mathbf{j})\) | M1 |
| \(= (-6\mathbf{i} + 12.5\mathbf{j})\) | A1 |
| (7 marks) |
First M1 for resolving along the wire, with usual rules
First A1 for a correct acceleration seen or implied
Second M1 for a complete method, using suvat or calculus, to find the distance along the wire
Second A1 for a correct distance
B1 for \(\mathbf{AB} = (-5\mathbf{i} + 7.5\mathbf{j})\) seen or implied
Third M1 for finding the answer
Third A1 for correct answer (must be a vector)