M5 June 2014 Q1
1. A small bead is threaded on a smooth, straight horizontal wire which passes through the point \(A(-3, 1)\) and the point \(B(2, 5)\) in the \(x\)-\(y\) plane. The bead moves under the action of a horizontal force \(\mathbf{F}\) of magnitude 8.5 N whose line of action is parallel to the line with equation \(15x - 8y + 4 = 0\). The unit on both the \(x\) and \(y\) axes has length one metre. Find the work done by \(\mathbf{F}\) as it moves the bead from \(A\) to \(B\). (8)
| Scheme | Marks |
|---|---|
| \(\mathbf{F} = \lambda(\pm 8\mathbf{i} \pm 15\mathbf{j})\) | M1 A1 |
| \(\lambda^2(8^2 + 15^2) = 8.5^2\) | M1 A1 |
| \(\mathbf{F} = \tfrac{1}{2}(8\mathbf{i} + 15\mathbf{j})\) | A1 |
| \(\mathbf{AB} = (5\mathbf{i} + 4\mathbf{j})\) | B1 |
| Work done \(= \tfrac{1}{2}(8\mathbf{i} + 15\mathbf{j}) \cdot (5\mathbf{i} + 4\mathbf{j})\) | M1 |
| \(= 50\) (J) | A1 |
| (8 marks) |
Notes
First M1 for \(\lambda(\pm 8\mathbf{i} \pm 15\mathbf{j})\)
First A1 for correct expression
Second M1 for \(\lambda^2(8^2 + 15^2) = 8.5^2\) from previous incorrect vector
Second A1 for \(\lambda = \pm\tfrac{1}{2}\)
Third A1 for correct \(\mathbf{F}\)
B1 for correct \(\mathbf{AB}\)
Third M1 for their \(\mathbf{F}\cdot\mathbf{AB}\)
Fourth A1 for 50 (J). (−50 is A0)
(Corrected from the printed mark scheme: the work done line is printed as \(\tfrac{1}{2}(8\mathbf{i} + 15\mathbf{j})\) without the \(\cdot(5\mathbf{i} + 4\mathbf{j})\).)