M4 June 2013 Q6
6. A particle \(P\) of mass \(m\) kg is attached to the end \(A\) of a light elastic string \(AB\), of natural length \(a\) metres and modulus of elasticity \(9ma\) newtons. Initially the particle and the string lie at rest on a smooth horizontal plane with \(AB = a\) metres. At time \(t = 0\) the end \(B\) of the string is set in motion and moves at a constant speed \(U\) m s\(^{-1}\) in the direction \(AB\). The air resistance acting on \(P\) has magnitude \(6mv\) newtons, where \(v\) m s\(^{-1}\) is the speed of \(P\). At time \(t\) seconds, the extension of the string is \(x\) metres and the displacement of \(P\) from its initial position is \(y\) metres.
Show that, while the string is taut,
You are given that the general solution of the differential equation in (b) is \[x = (A + Bt)U\mathrm{e}^{-3t} + \frac{2U}{3}\] where \(A\) and \(B\) are arbitrary constants.
| Scheme | Marks |
|---|---|
| \(a + Ut = y + (a + x)\) | M1 |
| \(Ut = x + y\) *Answer Given* | A1 |
Notes
M1 Diagram or clear explanation using distances
A1 Watch out for fudges.
| Scheme | Marks |
|---|---|
| \(T = \dfrac{9ma \times x}{a} = 9mx\) | B1 |
| \(T - 6m\dot{y} = m\ddot{y}\) | M1 |
| \(9mx - 6m(U - \dot{x}) = -m\ddot{x}\) | A2 |
| \(\ddot{x} + 6\dot{x} + 9x = 6U\) | A1 |
Notes
M1 Equation of motion of \(P\). Requires all 3 terms in terms of \(x\) and/or \(y\)
A2 Expressed in terms of \(x\). -1 each error
A1 Answer given. Watch out for fudges
| Scheme | Marks |
|---|---|
| \(t = 0,\ x = 0,\ \dot{x} = U \qquad 0 = AU + \dfrac{2U}{3},\ \ A = -\dfrac{2}{3}\) | M1 A1 |
| \(\dot{x} = BUe^{-3t} - 3(A + Bt)Ue^{-3t}\) | M1 A1 |
| \(U = BU - 3AU,\ \ B = 3A + 1 = -1\) | A1 |
Notes
M1 Use initial conditions to find \(A\)
M1 Differentiate
| Scheme | Marks |
|---|---|
| \(\dot{y} = U - \dot{x} = U - \left(-Ue^{-3t} + 2Ue^{-3t} + 3Ute^{-3t}\right)\) | M1 |
| \(= U\left(1 - e^{-3t} - 3te^{-3t}\right)\) | A1 |
| (14 marks) |
Notes
A1 Or equivalent