M4 June 2013 Q1
1. A particle \(P\) of mass 0.5 kg falls vertically from rest. After \(t\) seconds it has speed \(v\) m s\(^{-1}\). A resisting force of magnitude \(1.5v\) newtons acts on \(P\) as it falls.
| Scheme | Marks |
|---|---|
| Equation of motion: \(\dfrac{1}{2}g - \dfrac{3}{2}v = \dfrac{1}{2}\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | M1 A1 |
| \(\displaystyle\int 1\,\mathrm{d}t = \int \frac{1}{9.8 - 3v}\,\mathrm{d}v\) | M1 |
| \(t + (C) = -\dfrac{1}{3}\ln(9.8 - 3v)\) | A1 A1 |
| \(t = 0,\ v = 0 \Rightarrow C = -\dfrac{1}{3}\ln 9.8\) | M1 |
| \(t = -\dfrac{1}{3}\ln\left(\dfrac{9.8 - 3v}{9.8}\right)\) | A1 |
| \(3v = 9.8\left(1 - e^{-3t}\right)\) *Given Answer* | A1 |
| (8) |
Notes
M1 A1 Differential equation. All 3 terms required but condone sign errors
NB: these two marks are available in (b) if not scored in (a)
M1 Separate the variables and attempt to integrate
A1 A1 A1 for each side. \(C\) not needed
M1 Use initial conditions to evaluate \(C\) or limits on a definite integral.
A1 Or equivalent
A1 Watch out. cwo
(a) alt
| Equation of motion: \(\dfrac{1}{2}g - \dfrac{3}{2}v = \dfrac{1}{2}\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | M1 A1 |
| \(e^{3t}\dfrac{\mathrm{d}v}{\mathrm{d}t} + 3e^{3t}v = ge^{3t},\ \ \dfrac{\mathrm{d}}{\mathrm{d}t}\left(ve^{3t}\right) = ge^{3t}\) | M1 |
| \(ve^{3t} = \dfrac{1}{3}ge^{3t}\ (+c)\) | A1 A1 |
| \(t = 0,\ v = 0 \Rightarrow 0 = \dfrac{1}{3}g + C\) | M1 |
| \(\Rightarrow ve^{3t} = \dfrac{1}{3}g\left(e^{3t} - 1\right),\ \ 3v = 9.8\left(1 - e^{-3t}\right)\) | A1 A1 |
M1 A1 All 3 terms required but condone sign errors
M1 Use of integrating factor \(e^{3t}\)
A1 A1 A1 for each side. \(+C\) not required.
M1 Use initial conditions to evaluate \(C\)
A1 A1 Correct equation in any equivalent form. Given form cwo
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{9.8}{3}\left(1 - e^{-3t}\right) \Rightarrow x = \dfrac{9.8}{3}\left(t + \dfrac{1}{3}e^{-3t}\right)(+C)\) | M1 A1 |
| \(t = 0,\ x = 0 \Rightarrow C = -\dfrac{9.8}{9}\) | M1 A1 |
| \(t = 2,\ x \approx 5.4\) (m) | A1 |
| (5) | |
| (13 marks) |
Notes
M1 A1 Integrate the given \(v\) to find \(x\). \(C\) not needed
M1 A1 Use the initial conditions to evaluate \(C\) or use limits correctly in a definite integral
A1 5.45, \(\dfrac{g}{9}\left(5 + e^{-6}\right)\) or equivalent
(b) alt
| \(g - 3v = v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | |
| \(\displaystyle\int 1\,\mathrm{d}x = \int \frac{v}{g - 3v}\,\mathrm{d}v = \int -\frac{1}{3} + \frac{g}{3(g - 3v)}\,\mathrm{d}v\) | M1 |
| \(x = -\dfrac{v}{3} - \dfrac{g}{9}\ln(g - 3v) + C\) | A1 |
| \(x = 0,\ v = 0 \Rightarrow C = \dfrac{g}{9}\ln g\) and \(t = 2,\ v = \dfrac{g}{3}\left(1 - e^{-6}\right)\ (= 3.258\ldots)\) | M1 A1 |
| \(x = \dfrac{g}{9}\left(1 - e^{-6}\right) - \dfrac{g}{9}\ln\left(e^{-6}\right) = 5.4\) | A1 |
| (5) |
M1 Separate the variables and rearrange the RHS
A1 \(+C\) not needed
M1 A1 Use the initial conditions to find C & find the value of \(v\) when \(t = 2\)