M4 June 2008 Q3
3. At time \(t = 0\), a particle of mass \(m\) is projected vertically downwards with speed \(U\) from a point above the ground. At time \(t\) the speed of the particle is \(v\) and the magnitude of the air resistance is modelled as being \(mkv\), where \(k\) is a constant.
Given that \(U \lt \dfrac{g}{2k}\), find, in terms of \(k\), \(U\) and \(g\), the time taken for the particle to double its speed. (8)
| Scheme | Marks |
|---|---|
| \(mg - mkv = m\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | M1* A1 A1 |
| \(\displaystyle\int \mathrm{d}t = \int \dfrac{\mathrm{d}v}{g - kv}\) | DM1* |
| \(t = -\dfrac{1}{k}\ln(g - kv) + c\) | A1cao |
| \(t = 0,\ v = u \Rightarrow c = \dfrac{1}{k}\ln(g - ku)\) | M1† |
| \(T = \dfrac{1}{k}\ln(g - ku) - \dfrac{1}{k}\ln(g - 2ku)\) | DM1† |
| \(= \dfrac{1}{k}\ln\left(\dfrac{g - ku}{g - 2ku}\right)\) | A1 |
| (8 marks) |