M5 June 2015 Q4
4. A particle \(P\), whose initial mass is \(m_0\), is projected vertically upwards from the ground at time \(t = 0\) with speed \(\dfrac{g}{k}\), where \(k\) is a constant. As the particle moves upwards it gains mass by picking up small droplets of moisture from the atmosphere. The droplets are at rest before they are picked up. At time \(t\) the speed of \(P\) is \(v\) and its mass has increased to \(m_0\mathrm{e}^{kt}\). Assuming that, during the motion, the acceleration due to gravity is constant,
| Scheme | Marks |
|---|---|
| \((m + \delta m)(v + \delta v) - mv = -mg\delta t\) | M1 A1 |
| \(m\delta v + v\delta m = -mg\delta t\) | DM1 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{v}{m}\dfrac{\mathrm{d}m}{\mathrm{d}t} = -g\) | |
| \(m = m_0e^{kt} \Rightarrow \dfrac{\mathrm{d}m}{\mathrm{d}t} = m_0ke^{kt}\ (= km)\) | M1 A1 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{v}{m}km = -g\quad\) i.e. \(\ kv + \dfrac{\mathrm{d}v}{\mathrm{d}t} = -g\quad\) PRINTED | A1 |
| (6) |
Notes
First M1 for momentum equation (correct number of terms, excluding any \(\delta m\delta v\) terms)
First A1 for a correct equation
Second M1, dependent on first M1, for simplifying and dividing by \(\delta t\) and taking limits
Third M1 for differentiating the mass equation
Second A1 for \(\dfrac{\mathrm{d}m}{\mathrm{d}t} = km\)
Third A1 for PRINTED ANSWER
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_{\frac{g}{k}}^{v}\frac{\mathrm{d}v}{kv + g} = -\int_0^T \mathrm{d}t\) | M1 |
| \(\dfrac{1}{k}\left[\ln(kv + g)\right]_v^{\frac{g}{k}} = T\) | M1 A1 |
| \(v = 0 \Rightarrow \dfrac{1}{k}\ln 2 = T\) | DM1 |
| \(m = m_0e^{kT} = 2m_0\) | M1 A1 |
| (6) | |
| (12 marks) |
Notes
First M1 for separating and integrating
First A1 correct equation (without \(C\))
Second M1 for using limits or conditions
Third M1, dependent on first M1, for putting \(v = 0\) to give an equation in \(k\) and \(t\) only
Fourth M1 for solving for \(m\)
Second A1 for correct answer
N.B. If they put \(v = 0\) in DE and use \(\mathrm{d}v/\mathrm{d}t = -g\), NO MARKS
OR
| \(\displaystyle\int\frac{\mathrm{d}v}{g + kv} = -\int \mathrm{d}t\) | M1 |
| \(\dfrac{1}{k}\ln(g + kv) = -t + (C)\) | A1 |
| \(t = 0,\ v = \dfrac{g}{k} \Rightarrow C = \left(\dfrac{1}{k}\ln 2g\right)\) | M1 |
| \(\dfrac{1}{k}\ln(g + kv) = -t + \dfrac{1}{k}\ln 2g\) | |
| Put \(v = 0,\ t = \dfrac{1}{k}\ln 2\) | DM1 |
| \(m = m_0\mathrm{e}^{kt} = 2m_0\) | M1 A1 |
| (6) |
OR
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + kv = -g\) | |
| IF \(= \mathrm{e}^{kt}\) | |
| \(\dfrac{\mathrm{d}(v\mathrm{e}^{kt})}{\mathrm{d}t} = -g\mathrm{e}^{kt}\) | |
| \(v\mathrm{e}^{kt} = \displaystyle\int -g\mathrm{e}^{kt}\,\mathrm{d}t\) | M1 |
| \(v\mathrm{e}^{kt} = -\dfrac{g}{k}\mathrm{e}^{kt} + C\) | A1 |
| \(t = 0,\ v = \dfrac{g}{k} \Rightarrow C = \left(\dfrac{2g}{k}\right)\) | M1 |
| \(v\mathrm{e}^{kt} = -\dfrac{g}{k}\mathrm{e}^{kt} + \dfrac{2g}{k}\) | |
| Put \(v = 0,\ \mathrm{e}^{kt} = 2\) | DM1 |
| \(m = m_0\mathrm{e}^{kt} = 2m_0\) | M1 A1 |
| (6) |