M4 January 2006 Q1
1. A particle \(P\) of mass 0.5 kg is released from rest at time \(t = 0\) and falls vertically through a liquid. The motion of \(P\) is resisted by a force of magnitude \(2v\) N, where \(v\) m s\(^{-1}\) is the speed of \(v\) at time \(t\) seconds.
(a) Show that \(5\dfrac{\mathrm{d}v}{\mathrm{d}t} = 49 - 20v\). (2)
(b) Find the speed of \(P\) when \(t = 1\). (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2}\dfrac{dv}{dt} = \dfrac{1}{2}g - 2v\) | M1 |
| \(\Rightarrow 5\dfrac{dv}{dt} = 49 - 20v\qquad (*)\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{5\,dv}{49 - 20v} = \int dt\qquad\) (separate variables) | M1 |
| \(\dfrac{-5}{20}\ln(49 - 20v) = t\ \ (+c)\) | A1 |
| \(t = 0,\ v = 0 \Rightarrow c = -\dfrac{1}{4}\ln 49\qquad\) (attempt to get c) | M1 |
| \(t = \dfrac{1}{4}\ln\left(\dfrac{49}{49 - 20v}\right)\) | |
| \(t = 1\colon\ 1 = \dfrac{1}{4}\ln\left(\dfrac{49}{49 - 20v}\right)\qquad\) (correct use of logs/exp) | M1 |
| \(\rightarrow v \approx 2.41\,ms^{-1}\) or \(2.4\,ms^{-1}\) | A1 |
| (5) | |
| (7 marks) |