M4 June 2005 Q7
7. A light elastic string, of natural length \(a\) and modulus of elasticity \(5ma\omega^2\), lies unstretched along a straight line on a smooth horizontal plane. A particle of mass \(m\) is attached to one end of the spring. At time \(t = 0\), the other end of the spring starts to move with constant speed \(U\) along the line of the spring and away from the particle. As the particle moves along the plane it is subject to a resistance of magnitude \(2m\omega v\), where \(v\) is its speed. At time \(t\), the extension of the spring is \(x\) and the displacement of the particle from its initial position is \(y\). Show that
(a) \(x + y = Ut\), (2)
(b) \(\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2\omega\dfrac{\mathrm{d}x}{\mathrm{d}t} + 5\omega^2x = 2\omega U\). (7)
(c) Find \(x\) in terms of \(\omega\), \(U\) and \(t\). (8)
| Scheme | Marks |
|---|---|
| \(a + Ut = y + (a + x)\) | M1 |
| \(Ut = y + x\ \ *\) | A1 |
| (2) |
Notes
The published mark scheme for this paper is handwritten.
| Scheme | Marks |
|---|---|
| For particle, R\((\rightarrow)\), \(\ T - 2m\omega\dot{y} = m\ddot{y}\) | M1 A1 |
| \(T = \dfrac{5ma\omega^2x}{a}\) | M1 |
| \(U = \dot{y} + \dot{x}\ ;\ \ 0 = \ddot{y} + \ddot{x}\) | B1; B1 |
| \(5m\omega^2x - 2m\omega(U - \dot{x}) = m(-\ddot{x})\) | M1 |
| \(\Rightarrow \ddot{x} + 2\omega\dot{x} + 5\omega^2x = 2\omega U\ \ *\) | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| AE: \(\ u^2 + 2\omega u + 5\omega^2 = 0 \Rightarrow (u + \omega)^2 = -4\omega^2\) | B1 |
| \(\Rightarrow u = -\omega \pm 2\mathrm{i}\omega\) | M1 |
| CF: \(\ x = \mathrm{e}^{-\omega t}(A\cos 2\omega t + B\sin 2\omega t)\) | B1 |
| PI: \(\ x = \dfrac{2\omega U}{5\omega^2} = \dfrac{2U}{5\omega}\) | |
| GS: \(\ x = \mathrm{e}^{-\omega t}(A\cos 2\omega t + B\sin 2\omega t) + \dfrac{2U}{5\omega}\) | B1 |
| \(x = 0,\ t = 0\colon\ 0 = A + \dfrac{2U}{5\omega} \Rightarrow A = -\dfrac{2U}{5\omega}\) | M1 A1ft |
| \(\dot{x} = -\omega\mathrm{e}^{-\omega t}(A\cos 2\omega t + B\sin 2\omega t) + \mathrm{e}^{-\omega t}(-2\omega A\sin 2\omega t + 2\omega B\cos 2\omega t)\) | M1 |
| \(t = 0,\ \dot{y} = 0 \Rightarrow \dot{x} = U\) | |
| \(U = -\omega A + 2\omega B \Rightarrow B = \dfrac{3U}{10\omega}\) | |
| \(x = \mathrm{e}^{-\omega t}\left(\dfrac{3U}{10\omega}\sin 2\omega t - \dfrac{2U}{5\omega}\cos 2\omega t\right) + \dfrac{2U}{5\omega}\) | A1 |
| (8) | |
| (17 marks) |