M4 June 2011 Q5
5. A particle \(Q\) of mass 6 kg is moving along the \(x\)-axis. At time \(t\) seconds the displacement of \(Q\) from the origin \(O\) is \(x\) metres and the speed of \(Q\) is \(v\) m s\(^{-1}\). The particle moves under the action of a retarding force of magnitude \((a + bv^2)\) N, where \(a\) and \(b\) are positive constants. At time \(t = 0\), \(Q\) is at \(O\) and moving with speed \(U\) m s\(^{-1}\) in the positive \(x\)-direction. The particle \(Q\) comes to instantaneous rest at the point \(X\).
(a) Show that the distance \(OX\) is\[\frac{3}{b}\ln\left(1 + \frac{bU^2}{a}\right)\text{ m}\] (6)
Given that \(a = 12\) and \(b = 3\),
(b) find, in terms of \(U\), the time taken to move from \(O\) to \(X\). (5)
| Scheme | Marks |
|---|---|
| Need an equation linking speed and displacement, so \(mv\dfrac{dv}{dx} = -(a + bv^2)\) | M1 A1 |
| Separating the variables: \(\displaystyle\int \frac{6v}{a + bv^2}\,dv = \int -1\,dx\) | M1 |
| Integrating : \(\ \dfrac{3}{b}\ln(a + bv^2) = -x + (C)\) | A1 |
| \(X = \dfrac{3}{b}\left[\ln(a + bU^2) - \ln(a)\right] = \dfrac{3}{b}\ln\left[1 + \dfrac{bU^2}{a}\right]\quad **\) as required | M1 A1 |
| (6) |
| Scheme | Marks |
|---|---|
| Equation connecting \(v\) and \(t\): \(\ 6\dfrac{dv}{dt} = -(12 + 3v^2)\) | M1 |
| Separate the variables: \(\ \displaystyle\int \frac{-6}{12 + 3v^2}\,dv = \int 1\,dt\) | M1, A1 |
| \(\displaystyle\int_U^0 \frac{-2}{4 + v^2}\,\mathrm{d}v = \int_0^U \frac{2}{4 + v^2}\,\mathrm{d}v = T\) | M1 |
| \(T = \dfrac{2}{2}\tan^{-1}\dfrac{U}{2} = \tan^{-1}\dfrac{U}{2}\) (s) | A1 |
| (5) | |
| (11 marks) |