M4 June 2018 Q4
4. A particle \(P\) of mass 0.5 kg moves in a horizontal straight line. At time \(t\) seconds \((t \geqslant 0)\), the displacement of \(P\) from a fixed point \(O\) of the line is \(x\) metres, the speed of \(P\) is \(v\) m s\(^{-1}\) and \(P\) is moving in the direction of \(x\) increasing. A force of magnitude \(kx\) newtons acts on \(P\) in the direction \(PO\). The motion of \(P\) is also subject to a resistance of magnitude \(\lambda v\) newtons.
Given that \[x = (1.5 + 10t)\mathrm{e}^{-4t}\] find
(a) the value of \(k\) and the value of \(\lambda\), (8)
(b) the distance from \(P\) to \(O\) when \(P\) is instantaneously at rest. (3)
| Scheme | Marks |
|---|---|
| Equation of motion: \(\dfrac{1}{2}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}\left(= -28\mathrm{e}^{-4t} + 80t\mathrm{e}^{-4t}\right) = -kx - \lambda v\) | M1A2 |
| Differentiate: \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = -4(1.5 + 10t)\mathrm{e}^{-4t} + 10\mathrm{e}^{-4t} = 4\mathrm{e}^{-4t} - 40t\mathrm{e}^{-4t}\) | M1 |
| \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -16\mathrm{e}^{-4t} - 40\mathrm{e}^{-4t} + 160t\mathrm{e}^{-4t} = -56\mathrm{e}^{-4t} + 160t\mathrm{e}^{-4t}\) | A1 |
| Substitute and compare coefficients: \(-28\mathrm{e}^{-4t} + 80t\mathrm{e}^{-4t} = \mathrm{e}^{-4t}(-1.5k - 10kt - 4\lambda + 40\lambda t)\) | M1 |
| \(1.5k + 4\lambda = 28\) \(-10k + 40\lambda = 80\) | |
| \(k = 8,\quad \lambda = 4\) | A1 A1 |
| (8) |
Alt
| Equation of motion: \(\dfrac{1}{2}\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2}\left(= -28\mathrm{e}^{-4t} + 80t\mathrm{e}^{-4t}\right) = -kx - \lambda v\) | M1A2 |
| \(\ddot{x} + 2\lambda\dot{x} + 2kx = 0\) | |
| \(m^2 + 2\lambda m + 2k = 0 \Rightarrow m = \dfrac{-2\lambda \pm \sqrt{4\lambda^2 - 8k}}{2} = -\lambda \pm \sqrt{\lambda^2 - 2k}\) | M1A1 |
| \(\Rightarrow \lambda = 4\), and \(\lambda^2 - 2k = 0 \Rightarrow k = 8\) | M1A1 A1 |
| (8) |
Alternative for the last 5 marks in (a):
| AE has a repeated root \(m = -4\) | M1A1 |
| \(\Rightarrow m^2 + 2\lambda m + 2k = m^2 + 8m + 16\) | M1 |
| \(k = 8,\quad \lambda = 4\) | A1A1 |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\) when \(t = \dfrac{1}{10}\) | B1 |
| \(x = 2.5\mathrm{e}^{-0.4} = 1.68\) | M1A1 |
| (3) | |
| (11 marks) |