M4 June 2009 Q2
2. At time \(t = 0\), a particle \(P\) of mass \(m\) is projected vertically upwards with speed \(\sqrt{\dfrac{g}{k}}\), where \(k\) is a constant. At time \(t\) the speed of \(P\) is \(v\). The particle \(P\) moves against air resistance whose magnitude is modelled as being \(mkv^2\) when the speed of \(P\) is \(v\). Find, in terms of \(k\), the distance travelled by \(P\) until its speed first becomes half of its initial speed. (9)
| Scheme | Marks |
|---|---|
| \(-mg - mkv^2 = ma\) | M1 A1 |
| \(-(g + kv^2) = v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | M1 |
| \(\displaystyle\pm\int_0^X \mathrm{d}x = \int_{\sqrt{\frac{g}{k}}}^{\frac{1}{2}\sqrt{\frac{g}{k}}} \frac{-v\,\mathrm{d}v}{(g + kv^2)}\) | DM1 A1 (both previous) |
| \(X = \dfrac{1}{2k}\Big[\ln(g + kv^2)\Big]_{\frac{1}{2}\sqrt{\frac{g}{k}}}^{\sqrt{\frac{g}{k}}}\) | M1 A1 |
| \(= \dfrac{1}{2k}\left(\ln 2g - \ln\dfrac{5g}{4}\right)\) | M1 |
| \(= \dfrac{1}{2k}\ln\dfrac{8}{5}\) | A1 |
| (9 marks) |