M4 June 2010 Q4
4. A particle of mass \(m\) is projected vertically upwards, at time \(t = 0\), with speed \(U\). The particle is subject to air resistance of magnitude \(\dfrac{mgv^2}{k^2}\), where \(v\) is the speed of the particle at time \(t\) and \(k\) is a positive constant.
(a) Show that the particle reaches its greatest height above the point of projection at time \(\dfrac{k}{g}\tan^{-1}\left(\dfrac{U}{k}\right)\). (6)
(b) Find the greatest height above the point of projection attained by the particle. (6)
| Scheme | Marks |
|---|---|
| \(-mg\left(1 + \dfrac{v^2}{k^2}\right) = m\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | M1 A1 |
| \(\displaystyle g\int_0^T \mathrm{d}t = \int_U^0 \frac{-k^2\,\mathrm{d}v}{(k^2 + v^2)}\) | DM1 |
| \(T = \dfrac{k}{g}\left[\tan^{-1}\dfrac{v}{k})\right]_0^U\) | A1 |
| \(= \dfrac{k}{g}\tan^{-1}\dfrac{U}{k}\) | DM1 A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(-mg\left(1 + \dfrac{v^2}{k^2}\right) = mv\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | M1 A1 |
| \(\displaystyle g\int_0^H \mathrm{d}x = \int_U^0 \frac{-k^2v\,\mathrm{d}v}{(k^2 + v^2)}\) | DM1 |
| \(H = \dfrac{k^2}{2g}\left[\ln(k^2 + v^2\right]_0^U\) | A1 |
| \(H = \dfrac{k^2}{2g}\ln\dfrac{(k^2 + U^2)}{k^2}\) | DM1 A1 |
| (6) | |
| (12 marks) |